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At its core, Scheme's evaluation semantics is multiple-value based. Continuations can accept an arbitrary number of values and expressions can yield an arbitrary number of values. This is in
contrast to the functional languages ML and Haskell." (Marc Nieper-Wißkirchen, SRFI 210)
• R5RS introduced multiple-value returns, supported with the values and call-with-values syntax.
• R6RS added the let-values and let*-values forms, previously defined in SRFI 11.
• R7RS has added define-values.
R7RS-small contains the following five constructs dealing with multiple values:
• call-with-values - used to pass values returned from a producer procedure to a consumer
• define-values - used like define, but binds values to identifiers
• let-values - used like let, but binds values to identifiers
• let*-values - used like let*, but binds values to identifiers with an environment covering the previous bindings
• values - bundles its arguments together as a set of values
What are "multiple values"? As the name suggests, they are representations of more than one value - but they are bound together. This binding is not the same as some kind of data structure, like a
list. Notice the following:
gosh$ (values 1 2) <1>
1. This creates multiple values from the values 1 and 2.
As a more useful example, (scheme base) contains the procedure exact-integer-sqrt. The definition for this states: "Returns two non-negative exact integers s and r".
gosh$ (exact-integer-sqrt 16)
gosh$ (exact-integer-sqrt 17)
How can we get and use all of the returned values?
This is where the define- and let- forms come in.
define-values works like define but matches up a series of identifiers with the respective returned value:
gosh$ (define-values (root remainder) (exact-integer-sqrt 17))
gosh$ root
gosh$ remainder
As define-values is matching a list of formals (like lambda), it is easy to get values as lists, etc:
gosh$ (define-values vals (values 1 2 3))
gosh$ vals
(1 2 3)
gosh$ (list? vals)
gosh$ (define-values (head . rest) (values 1 2 3))
gosh$ head
gosh$ rest
(2 3)
There are no surprises with let-values and let*-values: these work in the same way as let and let*, but use multiple identifiers to match multiple values:
(define (find-roots x y)
(let-values (((x-root x-remainder) (exact-integer-sqrt x)) ; <1>
((y-root y-remainder) (exact-integer-sqrt y)))
(list x-root y-root))) ; <2>
1. Collect and name the multiple values
2. Return some in a list
And with let*-values:
(define (roots+total-remainder x y)
(let*-values (((x-root x-remainder) (exact-integer-sqrt x))
((y-root y-remainder) (exact-integer-sqrt y))
((roots total-remainder)
(values (list x-root y-root) ; <1>
(+ x-remainder y-remainder))))
(list roots total-remainder)))
1. It's a let*- form, so we can access the previous definitions.
The remaining form, call-with-values, ties together the production and consumption of multiple values: a producer is a zero-argument procedure that returns multiple values, using a values expression,
and a consumer is a procedure that accepts these values as its arguments.
Notice the little "gotcha" that the producer must be a zero-argument procedure.
(define (show-root n)
(lambda () (exact-integer-sqrt n)) ; <1>
(lambda (s r) (display (square s)) ; <2>
(display " + ")
(display r) (newline))))
gosh$ (show-root 15)
9 + 6
1. The producer procedure, returns two values.
2. The consumer procedure accepts those two values as its arguments.
Two SRFIs, SRFI 8 and SRFI 210, are particularly targetted at multiple values: these have both been accepted into the Yellow Edition of R7RS-large.
This SRFI appears in several other SRFIs, its single definition often copy-pasted into the reference implementation of other libraries.
The SRFI provides the receive syntax, to improve on call-with-values. Specifically, instead of having a zero-argument procedure as producer, receive accepts any expression as a producer, and its
values are bound into a set of provided identifiers. These can then be processed within the expression's body.
The show-root example above can be rewritten to use receive as:
(define (show-root n)
(receive (s r) (exact-integer-sqrt n) ; <1>
(display (square s)) ; <2>
(display " + ")
(display r) (newline)))
1. The multiple-return values are captured into the given identifiers.
2. The body can work with those identifiers.
Of course, in a post-R5RS world, we have let-values as an alternative, which also has the advantage of being able to "receive" values from multiple producers. The above example using let-values:
(define (show-root n)
(let-values (((s r) (exact-integer-sqrt n))) ; <1>
(display (square s)) ; <2>
(display " + ")
(display r) (newline)))
1. The multiple-return values are captured into the given identifiers.
2. The body can work with those identifiers.
Aimed at introducing procedures and syntax for dealing with multiple values, such as creating lists and vectors from expressions returning multiple values and procedures returning the elements of a
list or vector as multiple values.
This SRFI introduces a number of syntactic forms and procedures. I have divided these into separate groups, with related functionality, but not explained every construct.
(coarity producer)
• producer - an expression producing multiple values
Evaluates producer and returns the number of resulting values.
gosh[r7rs.user]$ (coarity 3)
gosh[r7rs.user]$ (coarity (values 1 2 3))
gosh[r7rs.user]$ (coarity (exact-integer-sqrt 11))
(set!-values formals producer)
• formals - a formal arguments list
• producer - an expression producing multiple values
Evaluates producer and assigns its multiple values to identifiers in the formal arguments list. It has the same relation to set! as define-values does to define.
gosh[r7rs.user]$ (define x 2)
gosh[r7rs.user]$ (define y 3)
gosh[r7rs.user]$ (set!-values (x y) (exact-integer-sqrt 11))
gosh[r7rs.user]$ x
gosh[r7rs.user]$ y
(value index value-1 ...) + (value/mv index value-1 ... [producer])
• index - a positive integer
• value-1 - an expression
• producer - an optional expression producing multiple values
Evaluates one or more values and returns the index position value. For value/mv, the last expression can produce multiple values.
If index is out of range, or a multiple-values producer is in any but the last position.
gosh[r7rs.user]$ (value 1 'a 'b 'c) <1>
gosh[r7rs.user]$ (value/mv 1 'a 'b 'c)
gosh[r7rs.user]$ (value/mv 2 'a 'b (values 'c 'd)) <2>
gosh[r7rs.user]$ (value 2 'a 'b (values 'c 'd))
gosh[r7rs.user]$ (value 3 'a 'b (values 'c 'd)) <3>
*** ERROR: argument out of range: 3
gosh[r7rs.user]$ (value/mv 3 'a 'b (values 'c 'd))
gosh[r7rs.user]$ (value/mv 1 'a (values 'b 'c) 'd) <4>
*** ERROR: received more values than expected
1. Both work with single value expressions to return the indexed value.
2. Using values can work like a single value.
3. But the value/mv form can use the values as part of the indexed range.
4. For value/mv, any producer must be in the last position.
(case-receive producer clause-1 ...)
• producer - an expression producing multiple values
• clause-1 - each clause is of form (formals body)
A matching clause is found, matching formals in each clause against the values returned by the producer, as done in lambda. And then the body of that matching clause is evaluated, in the context of
any bindings introduced by the match.
We could adapt exact-integer-sqrt so it returns a single value if the remainder is 0:
(define (new-eis n)
(let-values (((s r) (exact-integer-sqrt n)))
(if (zero? r)
(values s r))))
And then write a function to choose how to display the number:
(define (display-root n)
(case-receive (new-eis n)
((s) (display "Exact root: ") (display s) (newline))
((s r) (display "Inexact, with remainder: ") (display r) (newline))))
gosh[r7rs.user]$ (display-root 3)
Inexact, with remainder: 2
gosh[r7rs.user]$ (display-root 4)
Exact root: 2
The identity procedure is equivalent to values.
(list/mv value-1 ... [producer])
• value-1 - an expression
• producer - an optional expression producing multiple values
Evaluates one or more values and producer, returning all the values as a list.
If a multiple-values producer is in any but the last position.
gosh[r7rs.user]$ (list/mv (values 'a 'b 'c))
(a b c)
gosh[r7rs.user]$ (list/mv 'a 'b 'c)
(a b c)
gosh[r7rs.user]$ (list/mv 'a 'b 'c (values 'd 'e))
(a b c d e)
gosh[r7rs.user]$ (list/mv 'a 'b 'c (values 'd 'e) 'f)
*** ERROR: received more values than expected
There are also natural extensions for vector and (srfi 195) box types, as vector/mv and box/mv.
(list-values list-items)
• list-items - a list of values
Returns the given list of values as multiple values.
If list-items is not a list.
There are also natural extensions for vector and (srfi 195) box types, as vector-values and box-values.
(apply/mv procedure value-1 ... [producer])
• procedure - a positive integer
• value-1 - an expression
• producer - an optional expression producing multiple values
Evaluates one or more values and an optional producer, and then applies the given procedure to the values.
If a multiple-values producer is in any but the last position.
gosh[r7rs.user]$ (apply/mv string #\a #\b)
gosh[r7rs.user]$ (apply/mv string #\a #\b (values #\c #\d))
gosh[r7rs.user]$ (string #\a #\b #\c #\d) <1>
1. The previous example is the same as this procedure application.
(call/mv consumer producer-1 ...)
• consumer - a procedure that will be called on the values returned from the producers
• producer-1 - one or more expressions producing multiple values
Evaluates the procedures, collecting their multiple values together, and then applies the consumer on the values.
We can use this to convert multiple values into a list or vector:
gosh[r7rs.user]$ (call/mv list (exact-integer-sqrt 26))
(5 1)
gosh[r7rs.user]$ (call/mv vector (exact-integer-sqrt 26) (exact-integer-sqrt 3))
#(5 1 1 2)
(map-values procedure)
• procedure - a procedure accepting and returning a single value
Returns a procedure which accepts an arbitrary number of arguments. When applied, it applies the given procedure to each of the arguments, returning the results as multiple values.
gosh[r7rs.user]$ ((map-values exact-integer-sqrt) 3 26)
(with-values producer consumer)
• producer - an expression which returns multiple values
• consumer - a procedure accepting the values from the producer
The producer and consumer are evaluated, and the procedure resulting from evaluating the consumer is applied to the values results from evaluating the producer.
This is like call-with-values except the producer is simply evaluated, not called.
Rewriting our earlier show-root procedure shows the difference:
(define (show-root n)
(exact-integer-sqrt n) ; <1>
(lambda (s r) (display (square s)) ; <2>
(display " + ")
(display r) (newline))))
gosh$ (show-root 15)
9 + 6
1. The producer returns two values.
2. The consumer procedure accepts those two values as its arguments.
(bind/mv producer procedure-1 ...)
• producer - an expression producing multiple values
• procedure-1 - each procedure accepts the values from the preceding procedure, and returns multiple values
Chains a series of procedures together, passing multiple values between them.
bind/mv can be used as a simple form of procedure composition, where multiple values returned from one part are passed on to the next procedure in the line:
gosh[r7rs.user]$ (define (display-pair s r) (display s) (display " + ") (display r) (newline)) <1>
gosh[r7rs.user]$ (bind/mv 13 exact-integer-sqrt display-pair)
3 + 4
1. Define a procedure which takes two values and displays them.
There are also forms specialised for producers of different datatypes:
• bind/list - uses a list as the produced values
• bind/box - like bind/list but for a (srfi 195) box.
• bind - like bind/list but for any object.
(compose-left procedure-1 ...)
• procedure-1 - the procedures should accept an appropriate number
of arguments and return multiple values so the chain is well formed.
Constructs a left-composition of its arguments, so that:
((compose-left f g h) args) #| is equivalent to |# (apply/mv h (apply/mv g (apply f args)))
compose-left is not much different to normal function composition. For example, we could compose the odd? and not functions together to make a not-odd function:
gosh[r7rs.user]$ (define not-odd (compose-left odd? not))
gosh[r7rs.user]$ (not-odd 3)
gosh[r7rs.user]$ (not-odd 4)
The difference is that compose-left internally uses apply/mv to apply the second procedure to the first: hence the first procedure can return multiple values, which the second one can then use. The
following example composes the multiple-value returning exact-integer-sqrt with a procedure to display the two values:
gosh[r7rs.user]$ (define (display-pair s r) (display s) (display " + ") (display r) (newline)) <1>
gosh[r7rs.user]$ (define display-sqrt (compose-left exact-integer-sqrt display-pair)) <2>
gosh[r7rs.user]$ (display-sqrt 3)
1 + 2
1. Define a procedure which takes two values and displays them.
2. Now compose this with exact-integer-sqrt, which returns multiple values.
(compose-right procedure-1 ...)
• procedure-1 - the procedures should accept an appropriate number
of arguments and return multiple values so the chain is well formed.
Constructs a right-composition of its arguments, so that:
((compose-right f g h) args) #| is equivalent to |# (apply/mv f (apply/mv g (apply h args)))
compose-right is the same as compose-left, but with the order of arguments reversed. The following example composes the multiple-value returning exact-integer-sqrt with a procedure to display the two
gosh[r7rs.user]$ (define display-sqrt (compose-right display-pair exact-integer-sqrt))
gosh[r7rs.user]$ (display-sqrt 3)
1 + 2
Different Schemes handle multiple values in some contexts in different ways.
For example, Gauche permits multiple-values to be used in single-value contexts:
gosh$ (+ (values 1 2) (values 3 4)) <1>
gosh$ (map (lambda (a) (values a a)) '(1 2 3))
(1 2 3)
1. Notice how adding two sets of multiple values together only deals with the first value in each pair of values.
But this is not required behaviour, and indeed the R7RS report states that, for map, it is an error if the mapped procedure does not return a single value.
Chez Scheme gives errors in both cases:
Chez Scheme Version 9.5.8
Copyright 1984-2022 Cisco Systems, Inc.
> (+ (values 1 2) (values 3 4))
Exception: returned two values to single value return context
> (map (lambda (a) (values a a)) '(1 2 3))
Exception: returned two values to single value return context
Kawa only gives an error in the first case, but not the second:
#|kawa:1|# (+ (values 1 2) (values 3 4))
java.lang.ClassCastException: class gnu.mapping.Values$Values2 cannot be cast to class gnu.math.Numeric
#|kawa:2|# (map (lambda (a) (values a a)) '(1 2 3))
(1 1 2 2 3 3)
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Fransisco Ready Permana, . (2023) PERBANDINGAN ALGORITMA EXTREME GRADIENT BOOSTING DAN RANDOM FOREST UNTUK MEMPREDIKSI HARGA TERENDAH SAHAM DENGAN INDEX ISSI. Skripsi thesis, Universitas Pembangunan
Nasional Veteran Jakarta.
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Stocks with the ISSI index (Indonesian Sharia Stock Index) are stocks that can be used as an investment choice because these stocks have a fairly good level of stability compared to other stock
indices. Therefore, this research wants to create a machine learning model that can predict the lowest price of ISSI shares as a lower threshold value and compare two reliable algorithms, namely the
random forest algorithm and extreme gradient boosting (XGBoost) using stock data taken from the Google Finance website. The stages include problem identification, literature study, data preparation,
dataset loading, exploratory data analysis, preprocessing, data sharing, data training, and model evaluation. To find out which algorithm is better, the two algorithms are compared using three
assessment metrics such as Mean Squared Error (MSE), Mean Absolute Percentage Error (MAPE), and R2. As a result, the average value of the MSE random forest test was 0.6458 and the average MSE value
for the XGBoost test was 0.8019; The average value of the MAPE random forest test was 0.0033 and the average value of the MAPE XGBoost test was 0.0037; The average R2 value of the random forest test
is 0.9985 and the average R2 test value is 0.9982. From these values, the random forest gives smaller MSE and MAPE values and a larger R2 value compared to XGBoost. So based on these three assessment
metrics, it can be concluded that random forest can predict the lowest price of stocks with the ISSI index more accurately than XGBoost.
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Trajectory integrated smoothening of exchange fields for discrete phase simulations
Eulerian-Lagrangian simulations often require a two-way coupling, where the discrete phase affects the continuous phase and vice versa. The continuous phase in a CFD-simulation is typically
represented on a computational mesh, which consists of a number of discrete cells. As the discrete phase is not directly associated with this mesh, a mapping from the discrete phase to the continuous
phase is required in order to couple the two phases. This will distribute the exchange rates from the discrete to the continuous phase, where the exchanged properties can be thermal energy, momentum,
species etc. The mapping can be done in several ways, and different methods exist. Some of these include a smoothening step, which ensures that the exchange fields do not contain large spikes between
any two neighbouring cells, which typically increases the convergence rate of the CFD simulation. Gaussian smoothening is often applied, where the discrete phase exchange rates are spread to the
surrounding cells by a normal distribution with a specified standard deviation. This paper proposes a new weight for distributing the discrete phase exchange rates to the continuous phase cells. This
is done by analytically integrating any distribution over the entirety of the particle trajectory, and only requires the probability- and cumulative density functions to be specified. The proposed
weight is used in combination with three different smoothening methods: node-based, kernel-based, and cell-based. Common existing weight distributions are also investigated, which are used in
combination with the different methods. Each method and weight combination is compared with the analytical ground-truth exchange field, where the distributions are accurately integrated over the cell
volumes. The results show that the proposed weight has an average error of ≈ 2 % in the exchange field for the test case used. This is about three times more accurate compared to existing weights,
where the exchange field is not integrated along the trajectory of the particle. Alternative distributions were investigated to decrease the computational requirements. It was found that a
fourth-order polynomial distribution was about 20 times faster to evaluate, whilst the error induced by switching to this distribution was ≈ 4.5 % compared with the previous ≈ 2 %.
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Multiplication Facts Practice
Multiplication Facts Practice Made Fun
If you need multiplication fact practice, flashcards and timed tests are not your only option. Multiplication facts are really important in elementary math. Knowing multiplication facts by heart will
give elementary students the confidence to tackle more complex problems in the future. Not only that, but they are more likely to use these facts in their everyday lives.
Unfortunately, practicing multiplication facts might get a little bit boring. Remember the endless flashcards and timed tests when you were in elementary school? It’s not just boring! They were
also a source of stress and anxiety for many.
Here are several other ways things you can do for your multiplication facts practice!
Use Dice, Cards, and Dominoes to Practice Multiplication Facts
Practice multiplication facts using dice, cards, and dominoes. Here are some of my favorite things to do with them. They require little to no-prep too!
Grab a box of dice. (The 12-sided dice are the best!) Give each student a pair of dice. Have students roll the dice. Let them multiply the two numbers that come up! Tip: Place the dice in pill boxes
to keep your classroom floor dice free!
Not flash cards! Use playing cards this time! Have students pull out two cards from a deck of playing cards. Have students find the product of the two cards. Tip: Don’t remove the face cards. Assign
each face card a number instead.
Use domino tiles so students can randomly choose two numbers to multiply. Have students pick a tile of domino. Let the students multiply the numbers represented by the tile.
Play Multiplication Facts Bingo
My students absolutely love playing bingo! You can do this as a whole group or as a center.
Whole group
Call out multiplication questions and have students find the product. Let them mark their bingo cards once they know the product. They can make a pattern or you can do a block out! First student to
mark all the numbers on their bingo card wins.
Give your students one bingo card each and a pair of dice. Have them roll the dice and multiply the two numbers that come up. Have students mark their bingo cards once they know the product. Tip: You
can place the bingo cards inside a dry-erase so you can reuse them. Also, if you need a bingo card generator, this is great!
Play Multiplication Facts Relay
This is a pretty easy game to prep. Here are the things that you’ll need:
The Setup
Now, prepare the whiteboard. Using the tape, stick the hula hoop on the board. Write the numbers 1 to 12 around the hula hoop just like that of a clock. (Tip: Make the numbers in random order for a
more challenging game.) Write the multiplication fact you want them to practice in the middle of the hoop. (Example: x3 or x4) This will tell the students which multiplication fact they are
The Mechanics
Now it’s time to group your students. Have them line up. Give each group one marker each. The first student in line gets the marker. The student will go to the board and answer one multiplication
fact. After, the student will pass the marker to the next student. Repeat until all facts are answered. The first team to complete the facts win the game!
Use Online Multiplication Apps
Interactive Multiplication Facts Practice
Arrays are great visual cues to help students understand their multiplication facts. Not only that, they are great to visualize the connection between multiplication and division. Multiplication
Arrays gives practice on making arrays to represent multiplication facts. They also write multiplication equations or multiplication sentences that are represented by the arrays.
Number lines are a great visual tool to introduce multiplication. Using number lines to practice multiplication facts will help your students understand the concept. My students love this product!
The interactive drag-and-drop pieces on this deck makes them think that this is a game! This includes plenty of practice so your students have plenty of room to make mistakes and learn from them too!
Equal groups are best used when paired with word problems. This way, students will have more visual cues to help them solve math problems. Multiplication Equal Groups include plenty of practice in
writing multiplication facts when given a model. It also has practical word problems to further help with student’s understanding of multiplication.
Need a Free Resource?
Here’s a multiplication on a number line worksheet that’s easy and fun to use! You may use them as a station. Just print these out and place them in dry erase pockets so you can reuse them. You can
also give your students one copy each for extra practice!
Check out more tips on how to teach multiplication facts effectively here. | {"url":"https://thelearningcorner.co/multiplication-facts-practice-made-fun/","timestamp":"2024-11-03T00:59:50Z","content_type":"text/html","content_length":"108499","record_id":"<urn:uuid:c11b50ce-740c-4675-afb5-4d05e0f35323>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00098.warc.gz"} |
Denis R. Hirschfeldt
• University of Chicago, IL, USA
According to our database
, Denis R. Hirschfeldt authored at least 43 papers between 2000 and 2024.
Collaborative distances:
Book In proceedings Article PhD thesis Dataset Other
Online presence:
On csauthors.net:
Coarse computability, the density metric, Hausdorff distances between Turing degrees, perfect trees, and reverse mathematics.
J. Math. Log., August, 2024
Reduction games, provability and compactness.
J. Math. Log., 2022
A Minimal Pair in the Generic Degrees.
J. Symb. Log., 2020
Ramsey's theorem and products in the Weihrauch degrees.
Comput., 2020
Combinatorial principles equivalent to weak induction.
Comput., 2020
Some results concerning the SRT22 vs. COH problem.
Comput., 2020
The reverse mathematics of Hindman's Theorem for sums of exactly two elements.
Comput., 2019
Dense computability, upper cones, and minimal pairs.
Comput., 2019
Theory Comput. Syst., 2018
Some Questions in Computable Mathematics.
Proceedings of the Computability and Complexity, 2017
Coarse Reducibility and Algorithmic Randomness.
J. Symb. Log., 2016
On notions of computability-theoretic reduction between Π21 principles.
J. Math. Log., 2016
Asymptotic density and the coarse computability bound.
Comput., 2016
Five papers on reverse mathematics and Ramsey-theoretic principles - C. T. Chong, Theodore A. Slaman, and Yue Yang, The metamathematics of Stable Ramsey's Theorem for Pairs. Journal of the American
Mathematical Society, vol. 27 (2014), no. 3, pp. 863-892. - Manuel Lerman, Reed Solomon, and Henry Towsner, Separating principles below Ramsey's Theorem for Pairs. Journal of Mathematical Logic, vol.
13 (2013), no. 2, 1350007, 44 pp. - Jiayi Liu, $RT_2^^2$ does not imply WKL 0. Journal of Symbolic Logic, vol. 77 (2012), no. 2, pp. 609-620. - Lu Liu, Cone avoiding closed sets. Transactions of the
American Mathematical Society, vol. 367 (2015), no. 3, pp. 1609-1630. - Wei Wang, Some logically weak Ramseyan theorems. Advances in Mathematics, vol. 261 (2014), pp. 1-25.
Bull. Symb. Log., 2016
Counting the changes of random Δ20 sets.
J. Log. Comput., 2015
Slicing the Truth - On the Computable and Reverse Mathematics of Combinatorial Principles.
Lecture Notes Series / Institute for Mathematical Sciences / National University of Singapore 28, World Scientific, ISBN: 978-981-4612-61-6, 2014
Counting the Changes of Random D<sup>0</sup><sub>2</sub> Sets.
Proceedings of the Programs, Proofs, Processes, 6th Conference on Computability in Europe, 2010
Algorithmic Randomness and Complexity.
Theory and Applications of Computability, Springer, ISBN: 978-0-387-95567-4, 2010
Notre Dame J. Formal Log., 2007
Combinatorial principles weaker than Ramsey's Theorem for pairs.
J. Symb. Log., 2007
Bounding homogeneous models.
J. Symb. Log., 2007
Π<sub>1</sub><sup>0</sup> classes and strong degree spectra of relations.
J. Symb. Log., 2007
Undecidability of the structure of the Solovay degrees of c.e. reals.
J. Comput. Syst. Sci., 2007
An Uncountably Categorical Theory Whose Only Computably Presentable Model Is Saturated.
Notre Dame J. Formal Log., 2006
Every 1-generic computes a properly 1-generic.
J. Symb. Log., 2006
Relativizing Chaitin's Halting Probability.
J. Math. Log., 2005
Computability-Theoretic and Proof-Theoretic Aspects of Vaughtian Model Theory.
Proceedings of the New Computational Paradigms, 2005
Randomness and reducibility.
J. Comput. Syst. Sci., 2004
A computably categorical structure whose expansion by a constant has infinite computable dimension.
J. Symb. Log., 2003
Randomness, Computability, and Density.
SIAM J. Comput., 2002
Realizing Levels of the Hyperarithmetic Hierarchy as Degree Spectra of Relations on Computable Structures.
Notre Dame J. Formal Log., 2002
Degree Spectra of Relations on Computable Structures in The Presence of delta<sup>0</sup><sub>2</sub> Isomorphisms.
J. Symb. Log., 2002
Proceedings of the Computability and Complexity in Analysis, 2002
Degree spectra and computable dimensions in algebraic structures.
Ann. Pure Appl. Log., 2002
Degree spectra of relations on structures of finite computable dimension.
Ann. Pure Appl. Log., 2002
Degree Spectra of Intrinsically C.E. Relations.
J. Symb. Log., 2001
A delta<sup>0</sup><sub>2</sub> Set with No Infinite Low Subset in Either It or Its Complement.
J. Symb. Log., 2001
Degree spectra of relations on computable structures.
Bull. Symb. Log., 2000
Undecidability and 1-types in intervals of the computably enumerable degrees.
Ann. Pure Appl. Log., 2000 | {"url":"https://www.csauthors.net/denis-r-hirschfeldt/","timestamp":"2024-11-06T11:50:25Z","content_type":"text/html","content_length":"55352","record_id":"<urn:uuid:1bf3f044-71a1-4dda-8ac0-c1c5f7e1523c>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00589.warc.gz"} |
by matching your calculated average atomic mass with an element on the periodic Science Assignment Help -
by matching your calculated average atomic mass with an element on the periodic Science Assignment Help. by matching your calculated average atomic mass with an element on the periodic Science
Assignment Help.
(/0x4*br />
│ Isotope │ Atomic Mass(amu) │ Abundance │ Calculation │
│ A-33 │ 34.66 │ 3% │ │
│ A-34 │ 35.76 │ 95% │ │
│ A-35 │ 36.45 │ 2% │ │
by matching your calculated average atomic mass with an element on the periodic Science Assignment Help[supanova_question]
Please help me and show the steps also with your calculator. Mathematics Assignment Help
What is the probability that the mean duration of these patients’ pregnancies will be less than 264 days?
The mean is 262.
Standard deviation of 1.45
I know you have to put in Normal CDF, but what in the parenthesis?
Also, how is the answer 0.916?
Week 4 quiz ECO203 Principles Business Finance Assignment Help
Week 4 ECO203 Principles Business Finance Assignment Help
ECO203 Principles week 4 Business Finance Assignment Help
Need help with WACC finance question Business Finance Assignment Help
The firm’s tax rate is 35%. Calculate the firm’s WACC adjusted for taxes using the market information in the table.
│ │ The Number of Securities Outstanding │ Selling price │ The Required Rate of Return │
│ Bonds │ 1,775 │ $1,179 │ 11.33% │
│ Preferred Stocks │ 5,732 │ $56.90 │ 15.87% │
│ Common Stocks │ 1,579 │ $90.55 │ 11.64% │
Round the answers to two decimal places in percentage form.
Need help with WACC finance question Business Finance Assignment Help[supanova_question]
Question on weighted average cost of capital Business Finance Assignment Help
The Black Bird Company plans an expansion. The expansion is to be financed by selling $52 million in new debt and $123 million in new common stock. The before-tax required rate of return on debt is
8.84% percent and the required rate of return on equity is 19.90% percent. If the company is in the 34 percent tax bracket, what is the weighted average cost of capital?
Round the answer to two decimal places in percentage form.
What is the retained earnings in determining the firm’s cost of capital? Business Finance Assignment Help
The Yo-Yo Corporation tries to determine the appropriate cost for retained earnings to be used in capital budgeting analysis. The firm’s beta is 1.64. The rate on six-month T-bills is 2.67%, and the
return on the S&P 500 index is 6.83%. What is the appropriate cost for retained earnings in determining the firm’s cost of capital?
Round the answers to two decimal places in percentage form.
Before tax cost finance question Business Finance Assignment Help
Black Hill Inc. sells $100 million worth of 13-year to maturity 11.40% annual coupon bonds. The net proceeds (proceeds after flotation costs) are $979 for each $1,000 bond. What is the before-tax
cost of capital for this debt financing?
Round the answer to two decimal places in percentage form.
I came up with the answer 11.72%, just want to make sure I am on the right track.
the question is about psychology Humanities Assignment Help
my class i am suppose to write and short essay about What I have learned in the class.
the length is about 1 to 2 pages. my topic is piaget’s development theory. i started like this:
The topic I picked and found out interesting and it will be useful for me in later in my life as I am planning to become a teacher in the future. Is to describe the Piaget’s four stages and in
find out that it will be useful for me in later of life as I want to become a
teacher in the future.
by matching your calculated average atomic mass with an element on the periodic Science Assignment Help
by matching your calculated average atomic mass with an element on the periodic Science Assignment Help | {"url":"https://anyessayhelp.com/by-matching-your-calculated-average-atomic-mass-with-an-element-on-the-periodic-science-assignment-help/","timestamp":"2024-11-12T20:23:28Z","content_type":"text/html","content_length":"120497","record_id":"<urn:uuid:380c6860-59b2-4440-a85e-c34606db70b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00785.warc.gz"} |
Curriculum Guide: The Pythagorean Theorem
The purpose of this curriculum guide is to provide OER lesson material and support activities for Pythagorean Theorem instruction. It is geared towards GED® requirements. Both printable and
online options are provided.
Adult Education
Material Type:
Activity/Lab, Assessment, Lesson, Reading, Teaching/Learning Strategy
Date Added:
Media Format:
Kat Chilton
This curriculum guide has many resources offered to teachers throughout. It is very comprehensive and takes teachers throughout the entire unit of teaching the Pythagorean Theorem, which I
appreciate. It includes and considers every part of a unit you would want to include and takes into account the little things needed in a unit. It starts with different ways you can start introducing
the topic. Not only does it have many different ways to include technology in your lesson, but it also has different accommodations throughout, which I really like. In the "Application" section, the
lesson includes a hands-on activity to help students work through actually using the Pythagorean Theorem in real-life situations which I LOVE. I also love the inclusion of Online learning options at
the end, as that is very practical for the classroom today. Overall, this is very well thought out and there are a great number of resources, accommodations, and good tidbits throughout to help
teachers construct and teach a unit on this topic.
Susan Jones
This curriculum guide provides a comprehensive coverage of the Pythagorean Theorem.
It includes good comments about exactly how resources could be used and their strengths and weaknesses.
The “Prove it” exercise in the “Preview” part of the curriculum is “free but not open,” designated “read the fine print.” To enhance it, I recommend (since links don't work in comments, I made a
remix of the lesson and added these): have posters of perfect squares and cubes on the wall (from https://mathequalslove.blogspot.com/2016/07/posters-of-perfect-squares-and-perfect.html -- https://
mathequalslove.blogspot.com/ has more posters and ideas ), and would provide students with a times tables chart up to 20 (http://www.resourceroom.net/2018OER/multiplicationTableto20.PNG )
The mini-flexbook is very well grounded and comprehensive, including pictures and illustrations of “right triangles” in the real world.
The “prove it! Part 2” has an excellent practical example of using the Pythagorean Theorem to construct a deck.
Great collection of resources!
Jason Walker
This guide gives a teacher everything needed to comprehensively cover the Pythagorean Theorem in the Adult Ed classroom. I like the easy to follow progression of resources. Shana has curated a
worthwhile collection.
Jason Walker Evaluation
Quality of Explanation of the Subject Matter: Superior (3)
This guide has links to two full lesson plans on the Pythagorean Theorem, plus it has other links to activities, worksheets, a textbook resource and online exercises.
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Pauli Exclusion Principle | ChemTalk
Pauli Exclusion Principle Definition
The Pauli Exclusion Principle states that in any atom no electron can have the same four electronic quantum numbers as another electron. Every electron must have different quantum numbers. So, in
each electronic orbital (same n, l, and m[l]) there can be two electrons and they must have different spins. One electron will have m[s] =+ ½ and the other m[s] = – ½. Therefore, no two electrons
will have the same four quantum numbers.
The spin quantum number (m[s]) was added to the previously discovered three quantum numbers (n, l, m[l]) by the Pauli exclusion principle. A positive m[s] usually indicates spin up and is represented
by an upward pointing arrow. A negative m[s] usually indicates spin down and is represented by a down-facing arrow. The spin quantum number is slightly different from the other quantum numbers
because it is not dependent on them. It can only have a value of + ½ or – ½ and these values are independent of all other quantum numbers. The other quantum numbers are all interconnected.
The principle also defines that each orbital can only have two electrons. This definition originates from an orbital being defined by the first three quantum numbers. The remaining quantum spin
number only has two possible values. Therefore, according to the definition of the Pauli exclusion principle, the orbital can only hold two electrons.
Fermions vs. Bosons
This principle applies to all fermions. A fermion is an atomic particle that has a half-integer spin. Commonly known fermions are electrons, protons, and neutrons. Therefore, all these particles
will follow the Pauli exclusion principle.
The alternative to a fermion is a boson. Bosons have integer spins. The most common boson is a photon. There can be many photons in one energy state. In one state, they all have the same quantum
number. This is a violation of the Pauli exclusion rule. Since photons are bosons, however, they do not follow the Pauli exclusion rule.
Applications of the Pauli Exclusion Principle in Chemistry
The Pauli exclusion principle is important when determining the electron shell structure of an atom. It pairs with the Aufbau principle to allow us to know what electron orbitals will be filled.
Using the Pauli exclusion principle we know that if there are two electrons in an orbital, one must be spin up (+ ½ ) and one must be spin down (- ½ ) to give them different quantum numbers. However,
if there is only one electron in an orbital it can have either a positive or negative spin.
The discovery of the Pauli Exclusion principle also helped to explain some phenomena in the periodic table and the reasons behind how some atoms bond. Particularly for solids, many of the previously
unexplained properties were able to be explained using the Pauli exclusion principle.
Example Problems
The simplest atom to look at is helium. Helium has two electrons in the 1s orbital. The 1s orbital has quantum numbers n =1, l=0, and m[l]=0. Both electrons will be in this subshell. Therefore, one
electron will have quantum numbers n=1, l=0, m[l]=0, and m[s] = +1/2. The other electron will have quantum numbers n=1, l=0, m[l]=0, and m[s]=-1/2.
Beryllium has four electrons which fill the 1s and 2s orbitals. Below are some examples of electron configurations that would violate the Pauli Exclusion Principle as well as the correct depiction.
All the incorrect options have arrows pointing the same way (indicating the same spin) in the same orbital. This indicates they would have the same four quantum numbers and violate the Pauli
exclusion principle.
Next, we can also list out the quantum numbers for each electron to see that no electrons have the same 4 quantum numbers.
We start by filling the 1s shell. That means the principal quantum number n is equal to 1. And the s-orbital is denoted by the value 0 in l.
• Electron 1: n =1, l=0, m[l]=0, and m[s]= – ½
• Electron 2: n =1, l=0, m[l]=0, and m[s]= + ½
That fills the 1s shell. The next shell is the 2s, which changes the principal quantum number n to 2.
• Electron 3: n =2, l=0, m[l]=0, and m[s]= – ½
• Electron 4: n =2, l=0, m[l]=0, and m[s]= + ½
Comparing the quantum numbers of all four electrons, none of them are the same. In conclusion, they are following the Pauli exclusion principle.
History of the Pauli Exclusion Principle
The Pauli exclusion principle was discovered by Wolfgang Pauli in 1925. This principle expanded upon the Bohr model. At the time the principle was first discovered, he applied it only to electrons.
Later, the principle expanded to all fermions in 1940 by Pauli.
Wolfgang Pauli received the Nobel prize in physics in 1945 for his discoveries and work in quantum chemistry. He also worked on trying to explain the Zeeman effect and proposed the existence of the
neutrino. Pauli was born in Austria in 1900 and died in 1958. | {"url":"https://chemistrytalk.org/pauli-exclusion-principle/","timestamp":"2024-11-14T01:08:49Z","content_type":"text/html","content_length":"233108","record_id":"<urn:uuid:b15a2a7f-199d-4174-b47c-93d92e554847>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00127.warc.gz"} |
Tue, 09/09/2003 - 12:58
// File: Blend_SurfPointFuncInv.cxx
// Created: Wed Feb 12 14:01:00 1997
// Author: Laurent BOURESCHE
Can anyone tell me what the Blend_SurfPointFuncInv shall be good for ?
I looked at the "class-browser". As usually, there is no line describing the purpose. The same goes for the header-file and for the source-file (s. below).
By the way, therer is no lend_SurfPointFuncInv.ixx file anywhere to find. Although there is an include-instruction, the Makefile never fires a problem. Strange ?
Any hints ?
//function : NbVariables
//purpose :
Standard_Integer Blend_SurfPointFuncInv::NbVariables() const
return 3;
Francois Lauzon
Tue, 09/09/2003 - 15:57
Hello Christian,
a very good source of documentation for class documentation are the cdl files. There is one for each OCC class, and most of the time it is documented. Here is the documentation for the class you
-- File: Blend_SurfPointFuncInv.cdl
-- Created: Wed Feb 12 11:16:58 1997
-- Author: Laurent BOURESCHE
---Copyright: Matra Datavision 1997
deferred class SurfPointFuncInv from Blend
---Purpose: Deferred class for a function used to compute a
-- blending surface between a surface and a curve, using
-- a guide line. This function is used to find a
-- solution on a done point of the curve.
-- The vector used in Value, Values and Derivatives
-- methods has to be the vector of the parametric
-- coordinates w, U, V where w is the parameter on the
-- guide line, U,V are the parametric coordinates of a
-- point on the partner surface.
inherits FunctionSetWithDerivatives from math
Pnt from gp,
Vector from math,
Matrix from math
---Purpose: Returns 3.
returns Integer from Standard
is static;
---Purpose: returns the number of equations of the function.
returns Integer from Standard
is deferred;
Value(me: in out; X: Vector; F: out Vector)
---Purpose: computes the values of the Functions for the
-- variable .
-- Returns True if the computation was done successfully,
-- False otherwise.
returns Boolean from Standard
is deferred;
Derivatives(me: in out; X: Vector; D: out Matrix)
---Purpose: returns the values of the derivatives for the
-- variable .
-- Returns True if the computation was done successfully,
-- False otherwise.
returns Boolean from Standard
is deferred;
Values(me: in out; X: Vector; F: out Vector; D: out Matrix)
---Purpose: returns the values of the functions and the derivatives
-- for the variable .
-- Returns True if the computation was done successfully,
-- False otherwise.
returns Boolean from Standard
is deferred;
Set(me: in out; P : Pnt from gp)
---Purpose: Set the Point on which a solution has to be found.
is deferred;
GetTolerance(me; Tolerance: out Vector from math; Tol: Real from Standard)
---Purpose: Returns in the vector Tolerance the parametric tolerance
-- for each of the 3 variables;
-- Tol is the tolerance used in 3d space.
is deferred;
GetBounds(me; InfBound,SupBound: out Vector from math)
---Purpose: Returns in the vector InfBound the lowest values allowed
-- for each of the 3 variables.
-- Returns in the vector SupBound the greatest values allowed
-- for each of the 3 variables.
is deferred;
IsSolution(me: in out; Sol: Vector from math; Tol: Real from Standard)
---Purpose: Returns Standard_True if Sol is a zero of the function.
-- Tol is the tolerance used in 3d space.
returns Boolean from Standard
is deferred;
end SurfPointFuncInv; | {"url":"https://dev.opencascade.org/content/blendsurfpointfuncinv","timestamp":"2024-11-13T07:57:54Z","content_type":"text/html","content_length":"31788","record_id":"<urn:uuid:db898005-c547-407b-8faa-74fd8b624b55>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00146.warc.gz"} |
A Level Mathematics Syllabus Statement
\( \DeclareMathOperator{cosec}{cosec} \)
A Level Mathematics Syllabus Statement
Coordinate Geometry
[ << Main Page ]
Syllabus Content
Understand and use the coordinate geometry of the circle including using the equation of a circle in the form (x – a)^2 + (y – b)^2 = r^2 Completing the square to find the centre and radius of a
circle; use of the following properties: the angle in a semicircle is a right angle, the perpendicular from the centre to a chord bisects the chord, the radius of a circle at a given point on its
circumference is perpendicular to the tangent to the circle at that point
Here are some specific activities, investigations or visual aids we have picked out. Click anywhere in the grey area to access the resource.
Here are some exam-style questions on this statement:
See all these questions
Here is an Advanced Starter on this statement:
Click on a topic below for suggested lesson Starters, resources and activities from Transum.
• Circles This is all to do with pi and why it is such an important number. From finding the circumference and area of circles to problem solving and investigation. Pupils will begin by learning
the names of the parts of a circle then, either through investigation or practical activity, discover that the circumference of a circle is always just a little more than three times the length
of the diameter whatever the size of the circle. A brief walk through history leads them to find out how to use this knowledge (and a more accurate version of pi) to find the circumference and
areas of circles. This can then be developed to find the area of a sector, area of a segment, area of an annulus and the area of the region between a circle and a square in more complex problem
solving situations. More mathematics related to the circle can involve angle theorems, loci and algebra.
• Coordinates It is important that pupils become proficient at understanding coordinates at a basic level before using them in their study of graphs. Plotting points and finding the coordinates of
points are the pre-requisite skills for studying a number of branches of mathematics. The Cartesian plane consists of a horizontal x-axis and a vertical y-axis that intersect at the origin (0,
0). The mathematician René Descartes developed the concept of the Cartesian plane while lying in bed one morning and observing a fly on the ceiling! Pupils should learn the conventions starting
with knowing that the horizontal axis is the x-axis and the vertical axis is the y-axis (remember x is a cross so the x axis is across!). The axes meet at the origin which has coordinates (0,0).
Coordinates are written as two numbers separated by a comma and contained inside brackets. For example (3,9) means the point is above 3 on the x-axis and level with 9 on the y-axis. To get to
this point from origin you go along 3 and up 9 (remember to go along the hall before going up the stairs or that pole vaulter has to run along before leaping up into the air!). Coordinates can be
positive or negative (remember points to the right of the origin have a positive x-coordinate – being positive is right!). The abscissa often refers to the horizontal coordinate of a point and
the ordinate refers to the vertical coordinate. In three dimensions, three perpendicular lines are defined as the axes and the coordinates of each point is defined with three numbers.
• Graphs This topic includes algebraic and statistical graphs including bar charts, line graphs, scatter graphs and pie charts. A graph is a diagram which represents a relationship between two or
more sets of numbers or categories. The data items are shown as points positioned relative to axes indicating their values. Pupils are typically first introduced to simple bar charts and learn to
interpret their meaning and to draw their own. More sophisticated statistical graphs are introduced as the pupil's mathematical understanding develops. Pupils also learn about coordinates as a
pre-requisite for understanding algebraic graphs. They then progress to straight line graphs before learning to work with curves, gradients, intercepts, regions and, for older pupils, calculus.
This video on Completing The Square is from Revision Village and is aimed at students taking the IB Maths AA Standard level course | {"url":"https://transum.org/Maths/National_Curriculum/Topics.asp?ID_Statement=262","timestamp":"2024-11-15T00:28:22Z","content_type":"text/html","content_length":"25053","record_id":"<urn:uuid:d6555042-9056-4974-bf73-dc66f78e5471>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00751.warc.gz"} |
GNU Archimedes
Archimedes is a TCAD package for use by engineers to design and simulate submicron and mesoscopic semiconductor devices. Archimedes is free software and thus it can be copied, modified and
redistributed under GPL. Archimedes uses the Ensemble Monte Carlo method and is able to simulate physics effects and transport for electrons and heavy holes in Silicon, Germanium, GaAs, InSb, AlSb,
AlAs, AlxInxSb, AlxIn(1-x)Sb, AlP, AlSb, GaP, GaSb, InP and their compounds (III-V semiconductor materials), along with Silicon Oxide. Applied and/or self-consistent electrostatic and magnetic fields
are handled with the Poisson and Faraday equations.
The GNU project has announced on May 2012 that the software package Aeneas^[2] will be substituted by Archimedes, making this one the GNU package for Monte Carlo semiconductor devices simulations.^
Archimedes is the GNU package for semiconductor device simulations that has been released for the first time on 2005 under GPL. It has been created by Jean Michel Sellier who is, since then, the
leader of the project and the main developer. It is a Free software and thus it can be copied, modified and redistributed under GPL. This is the one of the big advantages of using Archimedes.
Archimedes belongs to the well-known family of TCAD software, i.e. tools utilized to assist the development of technologically relevant products. In particular, this package assists engineers in
designing and simulating submicron and mesoscopic semiconductor devices. In a next-future version Archimedes will also be able to simulate nanodevices, using the Wigner Monte Carlo formalism^[4] (an
experimental release can be found at^[5]). Today Archimedes is used in several big companies for simulation and production purposes.
Archimedes is also useful for teaching purposes since everybody can access the sources, modify and test them. Today, it is used for teaching courses in several hundreds universities all around the
world. Furthermore, a simplified version, developed for students, is available on nanoHUB.org.
The Ensemble Monte Carlo method is the method that Archimedes uses to simulate and predict the behavior of a devices. Being the Monte Carlo very stable and reliable, Archimedes can be used to know
the characteristics of a device even before this last is built.
The physics and geometry of a device is described simply by a script, which makes, in this sense, Archimedes a powerful tool for the simulation of quite general semiconductor devices.
Archimedes is able to simulate a plenty of physics effects and transport for electrons and heavy holes in Silicon, Germanium, GaAs, InSb, AlSb, AlAs, AlxInxSb, AlxIn(1-x)Sb, AlP, AlSb, GaP, GaSb, InP
and their compounds (III-V semiconductor materials), along with Silicon Oxide, the applied and/or self-consistent electrostatic and magnetic fields by means of Poisson and Faraday equation. It is,
also, able to deal with heterostructures.
Boltzmann transport equation
The Boltzmann transport equation model has been the main tool used in the analysis of transport in semiconductors. The BTE equation is given by:
The distribution function, f, is a dimensionless function which is used to extract all observables of interest and gives a full depiction of electron distribution in both real and k-space. Further,
it physically represents the probability of particle occupation of energy k at position r and time t. In addition, due to being a seven-dimensional integro-differential equation (six dimensions in
the phase space and one in time) the solution to the BTE is cumbersome and can be solved in closed analytical form under very special restrictions. Numerically, solution to the BTE is employed using
either a deterministic method or a stochastic method. The deterministic method solution is based on a grid-based numerical method such as the spherical harmonics approach, whereas the Monte Carlo is
the stochastic approach used to solve the BTE.
Monte Carlo method
The semiclassical Monte Carlo method is a statistical method used to yield exact solution to the Boltzmann transport equation which includes complex band structure and scattering processes. This
approach is semiclassical for the reason that scattering mechanisms are treated quantum mechanically using the Fermi's Golden Rule, whereas the transport between scattering events is treated using
the classical particle notion. The Monte Carlo model in essence tracks the particle trajectory at each free flight and chooses a corresponding scattering mechanism stochastically. Two of the great
advantages of semiclassical Monte Carlo are its capability to provide accurate quantum mechanical treatment of various distinct scattering mechanisms within the scattering terms, and the absence of
assumption about the form of carrier distribution in energy or k-space. The semiclassical equation describing the motion of an electron is
where F is the electric field, E(k) is the energy dispersion relation, and k is the momentum wave vector. To solve the above equation, one needs strong knowledge of the band structure (E(k)). The E
(k) relation describes how the particle moves inside the device, in addition to depicting useful information necessary for transport such as the density of states (DOS) and the particle velocity. A
Full-band E(K) relation can be obtained using the semi-empirical pseudopotential method.^[6]
A simple 2D diode simulation using Archimedes. The diode is a simple n+-n-n+ structure, the channel length being equal to 0.4 micron. The diode has two n+ regions of 0.3 micron (i.e. the total
length is 1.0 micron ). The density in the doping regions are n+=1.e23/m^3 and n=1.e21/m^3 respectively. The applied voltage is equal to 2.0 Volts.
A 2D Silicon MESFET simulation using Archimedes. Archimedes takes into account all the relevant scattering mechanisms.
External links
This article is issued from
- version of the 10/18/2016. The text is available under the
Creative Commons Attribution/Share Alike
but additional terms may apply for the media files. | {"url":"https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/GNU_Archimedes.html","timestamp":"2024-11-09T14:19:17Z","content_type":"text/html","content_length":"26566","record_id":"<urn:uuid:549a6249-d93f-432d-acee-95fccd5fb0e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00789.warc.gz"} |
Depth-First Search Maze Generation
There are quite a few approaches for generating mazes algorithmically. I particularly like the graph based approaches, since they seem most adaptable. They are:
• Using a MST algorithm (like Prim’s or Kruskal’s) on a randomly weighted graph.
• Performing a depth-first search on a graph, with some randomness built into the algorithm.
Maybe there are even more graph-based maze generation methods, but I don’t know about them. There are of course some non-graph-based methods, like iteratively splitting a space into randomised chunks
and cutting holes in the new walls^1, or even using cellular automata^2, which are also interesting.
The reason I say that graph based approaches are most adaptable is because of their independence on any particular shape or structure. You’ll see what I mean by this later on.
Defining a Maze
First it might be a good idea to define what a maze is and isn’t. For example, consider this:
Imagine that the black lines are passageways that you can walk through. This definitely seems like a maze. Here are some properties of this specific maze:
• No loops/cycles: there is exactly one route from any one location to any other.
• There are no disconnected sections: any point is reachable from any other point.
• No obvious-at-a-glance route from a point to another far-ish-away point.
The last of these is the most controversial of the three, since it’s very difficult to quantify how difficult a maze is. However, here are some features of good mazes that probably help:
• No repeating patterns, as these would make visually parsing the maze more manageable.
• No excessively long straight corridors/lots of bends makes following paths more tricky.
• Many points where passageways branch, meaning that someone solving the maze has to check a lot of different routes.
Probably none of these features are necessary for something to be a maze, though. Loops are definitely possible in mazes, there’s no reason why not, and the only reason they’re not present in the
above maze is because it is specifically a spanning tree of a graph. Mazes may in theory have disconnected sections, I suppose, but in that case it would really just be two (or more) mazes next to
each other. Also, mazes don’t need to have a non-obvious solution – take the kids mazes on the backs of some ceral boxes, for example.
In fact, mazes don’t even need to have branches. Some hedge mazes^3, for instance, are just one long twisting path from start to finish. And should all mazes have a start and a finish? The maze
presented above doesn’t have a specific start or finish, but the nature of it means that any two points on it can reasonably be chosen for these. I would say that a maze has to have a defined start
and finish, or be constructed such that they can be chosen arbitrarily.
I think a reasonable definition for a maze is just “A passageway (or collection of passageways) connecting a start to a finish”. Then if our maze has starts and finishes that can be chosen
arbitrarily, we basically have a set of mazes with the same construction but with different start and finish points. Anyway, this doesn’t matter.
Depth-First Search
The maze-generation algorithm I’m going to talk about now is depth-first search (DFS). This algorithm is used for many more useful purposes than generating mazes, but this is what I’ll use it for.
DFS takes a graph (a collection of nodes and edges between them) and a starting node, and, in a certain order, examines each of these. The reason it’s called “depth-first” is that it starts by
traversing one path for as long as possible before going back to try another one. This is in contrast to breadth-first searching which, before examining any node which is n nodes away from the
starting node, examines all those which are up to n-1 away.
This post isn’t meant to be a detailed depth-first search tutorial – you can easily find those anywhere on the internet. Even still, here’s some rough pseudocode for one iteration of the algorithm,
as I implemented it for this purpose:
PROCEDURE dfs_step
G: The graph to search
S: A stack data structure.
WHILE S is not empty
next = S.pop()
IF `next` has not yet been visited:
Visit `next`
FOR EACH edge IN `next`'s neighbours:
This is slightly different to standard iterative DFS implementations, in that the procedure only executes one step, by which I mean it runs until it either exhausts the stack or a new node is
visited. Normal implementations run until the entire graph has been visited.
The reason for this difference is that I wanted to display a realtime animation of the algorithm running, and this implementation allows me to periodically run a step (say, every half a second) and
update the graphical representation accordingly, without the algorithm itself needing to know anything about this.
To summarise the pseudocode, we repeatedly pop nodes from the stack until we find one that we haven’t visited yet. Then, we “visit” it, push all of its neighbours to the stack, and return, i.e. end
the procedure. A traditional DFS implementation won’t return at this point, as I mentioned earlier, and instead will repeat until there’s nothing left on the stack.
The meaning of “visiting” a node depends on what the DFS is meant to achieve. For example, another use of DFS is finding connected components within a graph. To do this, we can just begin a DFS at an
arbitrary node, and efficiently find the connected component that contains it, then repeating the process on an unclassified node. In such a case, “visiting” a node would be adding it to a list of
nodes in some structure representing the connected components of a graph.
What about generating a maze, though? If we run DFS on any graph and make a note of the order in which nodes are visited, we can construct a tree containing every node. Let’s run through a quick
The first image shows the plain graph which we’ll operate on. I generated this graph by randomly placing 20 nodes in a 500×500 unit space, and mutually connecting any two nodes which are closer than
(or exactly) 150 units apart. The second image visualises the very first iteration of my DFS step procedure. It’s starting at the top-left node, where the orange and green lines are protruding from,
which represent all of the node under consideration’s neighbours. The actual implementation pushes the edges such that the closest edge is at the top of the stack.
In the third image, in bold, the resulting spanning tree is shown. It may not look much like a maze at this point, but the point is we can run this on any graph, and certain graphs will yield better
mazes. In fact, the big square maze presented earlier in this post was generated in the exact same way as this tree. The primary difference between these two, and all the other mazes I’ll make, are
in the way the graph is generated initially, although I make a couple of extra modifications to the algorithm as well, for some.
Grid-like Maze
A very standard type of maze is square-shaped, with horizontal and vertical passageways. I can very easily write a function to place nodes in a square grid, and join them to their correct neighbours.
Then I can just feed this graph into the DFS algorithm, and…
Oh, it’s not a very good maze. Going back to my thoughts earlier about what makes a (good) maze, we can see that this one has no cycles and no disconnected sections, like the earlier maze, but, it’s
very easy to solve this maze from any point A to any point B. In fact it’s trivial, because this maze has no branches at all – it’s topologically equivalent to a straight line. Starting from A, you
can always get to B, regardless of where B is, by just trying each of the two opposite directions. In fact this maze is even worse than just the general case of mazes topologically equivalent to
straight lines, because it has a consistent repeating zig-zag pattern going downwards, so if you know the positions of A and B you can immediately know whether you have to travel up or down in space,
and hence work out if you need to start moving left or right. In other words, with this maze, it would be possible to implement a function Node next_node(Node current_node, Node to_find) that runs in
O(1) time, given that we can view the structure of the maze from above.
The reason that our algorithm generates a zig-zag maze here is just because we don’t involve any randomness yet. The procedural method by which we generated our graph appends neighbouring nodes to a
list contained within each node, and this list is iterated in order when pushing to the stack in the DFS algorithm. The way that this list is generated means that neighbours are pushed in the order
(up, down, left, right), and so if a right-facing neighbour is present, this will always be preferred (because it will be at the top of the stack at the start of the next iteration). This explains
the intial long-as-possible right-facing path. It only stops moving right when it can’t anymore, because that neighbour is then not pushed to the stack.
The solution is simple though, we can just shuffle each node’s neighbour list at initialisation time. And this pattern has the added benefit that it’s adaptable to different patterns that we might
want. For example, remember what I mentioned earlier about picking the neighbour closest to the current node, when we saw the DFS algorithm running on a more freeform graph layout? Well that
behaviour doesn’t have to be impemented within DFS at all, keeping it more generalised. Instead we implement that in the same way as this neighbour shuffling, by sorting each node’s neighbours list
by distance, specifically for graphs where we want this shortest-first behaviour.
Anyway, if we implement this neighbour shuffling, we get this:
This is significantly better. Now, we have lots of branching – this occurs when a search path gets to a point where it cannot go any deeper, so it pushes no neighbours for the next iteration and
instead some neighbours that were pushed by an earlier iteration are taken instead.
This maze still isn’t ideal though. The branches, although many, are not, on average, very deep, which makes it easy to quickly disregard many of them when solving the maze. I think this is just
because this particular maze is fairly small. If I generate a larger one, the branches become far longer.
Customising The Maze
It can be fun to fiddle with the code a bit to get some different interesting mazes. With our graph-based maze approach, there are a lot of options here, because we simply apply our algorithm to any
Grid-Like Mazes
First, let’s look at different ways we can modify our existing grid maze.
The only change needed to generate these three mazes is to change which nodes count as neighbours of other nodes. Before, a node only neighboured adjacent nodes cardinally north, east, south, and
west of it, but this can easily be changed when generating the graph.
For the first of these three examples, the neighbourhood includes all eight adjacent nodes, including diagonals. This produces maybe a more “organic” looking design. It also kind of reminds me of
runes in some fantasy game?
In the second example, the four cardinal directions are included, as well as north-west and south-east neighbours. This creates angles paths, and almost looks like a normal maze has just been skewed.
The third example is similar, except depending on the node’s position on the y-axis the “skew” direction is either left or right.
“Loose” Graph Layout
As shown earlier, we can generate a maze on top of any graph, not only rigid grids. For these next tests, I place 1000 nodes randomly in a 800x800px world.
To determine the neighbours, the first of these two examples finds all other nodes which are within a certain threshold distance (in this case 100px), and uses this set as its neighbourhood. The
ordering of these neighbours (which determines how DFS will traverse the graph – remember how we randomised the ordering of neighbours for the rigid grid earlier) is simply their distance. The
shortest node comes first in the neighbour order.
This means that we get lots of winding, intricate paths, since it’s biased towards picking out small details.
In the second example, the set of neighbours is generated identically (though with a different random seed). The only difference is that instead of ordering the neighbours by their distance, they’re
just shuffled (exactly like the earlier rigid grid examples). Since here we have a combination of a fairly high threshold and densely packed nodes, we get a bit of a mess here. I think it looks like
the creases of scrunched up tin foil.
If you experiment with this yourself, it’s fun to play with the density and threshold parameters for the loose graph. If the nodes are too sparse, though, there is a chance that you’ll encounter
disconnected regions, meaning that the whole graph will not be part of the maze.
Some nice things to do to extend this would be:
• Some constraint for a minimum distance between two nodes. At the moment, two nodes may spawn almost touching, which can make the resultant maze look like it contains cycles even when it doesn’t.
• A neighbourhood ordering function which takes into account distance but applies some randomness on top of this.
Aside on overlapping passages: When I first ran a version of this “loose maze” generator, sometimes passages would overlap. This was of course because the neighbourhood generation didn’t take this
into account, and the DFS algorithm itself didn’t check for overlaps either.
I had the option of either ensuring no overlaps when generating the ‘substrate’ graph, or making a DFS which is aware of this. I decided on the latter, since whether an edge may exist without
overlapping only depends on previous edges that have been visited.
Other Fun Shapes
The current state of the program supports grid mazes and loose mazes, and some code to try and generate a labyrinth-like maze (where nodes lie on concentric circles, with a few edges between them).
The labyrinth generator isn’t working that well at the moment – ideally, I want fully the concentric circles to have long unbroken sections, with far fewer inter-ring connectors.
Below, my radial graph is on the left, and on the right is something that I would like mine to look like! One major difference here is that the target image uses lines as walls, whereas mine uses
lines as passages. I think that DFS might not be the correct algorithm to use to generate this sort of maze, anyway.
If you’re interested in this, you should definitely have a go at writing your own code to generate graphs. Here are some ideas:
• Fix my labyrinth implementation.
• Make a maze which resembles an image, or some text.
• Generate a voronoi diagram, and use its interface lines as edges.
You can find the source code on GitHub!
Final Thoughts
Depth-first search on a graph can generate some quite nice custom mazes. Since it’s “depth-first”, though, there is some bias towards long paths, which may or may not be desirable. Other algorithms
(such as a greedy MST algorithm or a breadth-first search) will have different characteristics in this way, which might be better for some applications.
1. Hedge mazes in particular always have the other kind of branch, though. ↩
Written 21 Nov 2023 by Jacob Garby | {"url":"https://jgarby.uk/blog/2023/11/21/DFS-Mazes.html","timestamp":"2024-11-07T05:27:20Z","content_type":"text/html","content_length":"21941","record_id":"<urn:uuid:fab972ae-3277-4d7b-ab79-18eaf431340f>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00873.warc.gz"} |
The sum of the digits of a two digit number is 6 and its ten's digit is twice its unit digit. Find the numbers.
Hint: To obtain the number we will let that it is $x$ then by using the condition given we will form equation. Then we will solve the obtain equations and put the numbers in there one’s and tens
place to get our desired answer.
Complete step by step answer:
We have to find a two digit number such that sum of its digits is 6 and its ten's digit is twice its unit digit
Let the unit digit of the number be $x$.
So as it is given that its tens digit is twice as its unit digit we get,
Tens digit $=2x$
Now as the sum of the two digits is 6 we get the following equation:
& x+2x=6 \\
& \Rightarrow 3x=6 \\
& \Rightarrow x=\dfrac{6}{3} \\
& \therefore x=2 \\
So we get the unit digit as 2.
The tens digit will be as below:
& \Rightarrow 2\times 2 \\
& \Rightarrow 4 \\
Now as we know we find the number as below:
Number $=10\times \text{unit digit}+1\times \text{Tens digit}$
Number $=10\times 4+1\times 2$
Number $=40+2$
Number $=42$
So, the correct answer is “Option C”.
Note: The most common mistake in this type of question is that we take the number as $a+b$ but the correct way is to take it in $10a+b$ form as it has a representation of tens and ones place. We can
check whether our answer is correct or not by seeing whether the obtained number satisfies the given condition. As in this case we got the number as 42 so the sum of its digit is 6 and its ten's
digit is twice its unit digit so our answer is correct. | {"url":"https://www.vedantu.com/question-answer/the-sum-of-the-digits-of-a-two-digit-number-is-6-class-9-maths-cbse-60a26a265b6aeb17f7868c93","timestamp":"2024-11-13T06:00:53Z","content_type":"text/html","content_length":"154297","record_id":"<urn:uuid:b665c6e9-cb08-4145-859c-268494222974>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00022.warc.gz"} |
Investigation into Multiaxial Character of Thermomechanical Fatigue Damage on High-Speed Railway Brake Disc
School of Mechanical Engineering, Southwest Jiaotong University, Chengdu 610031, China
Technology and Equipment of Rail Transit Operation and Maintenance Key Laboratory of Sichuan Province, Chengdu 610031, China
Author to whom correspondence should be addressed.
Submission received: 27 March 2021 / Revised: 6 May 2021 / Accepted: 28 May 2021 / Published: 1 June 2021
The multiaxial character of high-speed railway brake disc thermomechanical fatigue damage is studied in this work. Although the amplitudes and distributions of temperature, strain and stress are
similar with uniform and rotating loading methods, the multiaxial behavior and out-of-phase failure status can only be revealed by the latter one. With the help of a multiaxial fatigue model, fatigue
damage evaluation and fatigue life prediction are implemented, the contribution of a uniaxial fatigue parameter, multiaxial fatigue parameter and out-of-phase failure parameter to the total damage is
discussed, and it is found that using the amplitude and distribution of temperature, stress and strain for fatigue evaluation will lead to an underestimation of brake disc thermomechanical fatigue
damage. The results indicate that the brake disc thermomechanical fatigue damage belongs to a type of multiaxial fatigue. Using a uniaxial fatigue parameter causes around 14% underestimation of
fatigue damage, while employing a multiaxial fatigue parameter without the consideration of out-of-phase failure will lead to an underestimation of about 5%. This work explains the importance of
studying the thermomechanical fatigue damage of the brake disc from the perspective of multiaxial fatigue.
1. Introduction
Fatigue is a common damage type in the engineering field, which is resulted from cyclic load [
]. The characteristics of fatigue damage are different from static damage, i.e., fatigue failure may occur when the cyclic stress is much smaller than the static strength limit. In railway
engineering, the cycle of the braking process can cause serious fatigue damage of the brake disc.
Due to the large tonnage and high speed of a vehicle, a tremendous amount of kinetic energy is transformed into thermal energy during the actions of implementing braking, precision stopping and
stabling speeding [
]. This transformation arouses significant temperature increasement and exposes the brake disc to severe thermomechanical loading, resulted from the frictional contact between the brake disc and pad
]. Because of this, the thermomechanical fatigue damage is considerably severe [
] (see
Figure 1
), and it could be more serious in the long slope line. This problem sets a restriction against the further speed increasement, economic saving and security improvement of railway transportation, and
even could cause catastrophic accidents. Therefore, the study of thermomechanical fatigue damage on the brake disc is of great importance from both academic and engineering perspectives.
There are different factors affecting fatigue failure [
] and several fatigue damage types can be detected on the brake disc (
Figure 1
), i.e., heat checking (crackle), radial crack, circumferential crack and hot spot. Kasem et al. [
] analyze the subsurface damage beneath the hot spots, plastic deformation and microcrack propagation are detected under the rubbing surface. Collignon et al. [
] study the damage of brake discs and it is found that residual hoop tensile stress is relevant to radial microcracks. Honner et al. [
] study the thermoelastic instability of the brake system, which is verified to be relevant for non-uniform temperature distribution. Yang et al. [
] analyze the fatigue mechanism on the brake disc, fatigue cracking is observed in the interior and border of hot spots. Panier et al. [
] propose a classification for hot spots and suggest a scenario of their occurrence by experimental methods. Kim et al. [
] find that the maximum thermal stress under a variable pressure distribution is quite relevant with fatigue cracking. Wu et al. [
] study the peak temperature and crack geometry evolution with the help of an inserted semi elliptical surface crack. Adamowica and Grzes [
] study the brake disc temperature distribution, and it is found that the influence of convection cooling is significant in the brake release period. A review of the thermomechanical performance of
brake discs with numerical and experimental approaches can be found in the work of Afzal and Mujeebu [
According to previous research, the commonly used method for brake disc thermomechanical fatigue investigation is adopting the distribution and amplitude of temperature, stress and strain items.
However, due to the rotating movement, the thermomechanical fatigue should belong to multiaxial fatigue damage [
], which is complex with numerous influencing factors [
]. Regrettably, the multiaxial feature of brake disc thermomechanical fatigue is ignored at this moment; this simplification might lead to an underestimation in the thermomechanical fatigue damage
and a blur in the thermomechanical fatigue mechanism reveal.
Based on the analysis above, the study of multiaxial thermomechanical fatigue damage on the high-speed railway brake disc is insufficient. In this study, the thermomechanical fatigue damage of the
brake disc will be studied with the help of a multiaxial fatigue model instead of using the distribution and amplitude of temperature, stress and strain items as fatigue indicators. A finite element
simulation will be used to reproduce the multiaxial material response state under thermomechanical load. Afterward, a multiaxial fatigue model, recently proposed by the first author, will be adopted
to analyze the multiaxial thermomechanical fatigue damage and out-of-phase failure status. Finally, the methodology for multiaxial thermomechanical fatigue evaluation will be proposed and the fatigue
life prediction will be performed.
2. Finite Element Model
2.1. Theoretical Background
2.1.1. Thermal Formulation
Ignoring the influence of wheel–rail friction and aerodynamic resistance, most vehicle kinetic energy is transferred into heat by friction contact between the brake disc and pad when mechanical
braking is implemented. Therefore, the total thermal energy flowed into the braking system can be calculated by friction energy. The corresponding heat flux (
$q f$
) is described in Equation (1):
$q f = η · P · μ A · ( v 0 + a · t )$
is the conversion rate between friction energy and heat,
denotes the contact force and
stands for the friction coefficient,
is the braking contact area,
$v 0$
is the initial relative frictional translation speed between the brake disc and pad, and
represent the acceleration and time, respectively.
Apart from the heat input, there are heat exchanges between the braking system and environment through convection heat transfer and heat radiation. The convection heat transfer (
$q f$
) is demonstrated in Equation (2):
is the film coefficient, and
$T 0$
represent the temperature of braking component and surrounding environment, respectively. The heat radiation (
$q h$
) is calculated by Equation (3):
$q h = ж · ( T − T * ) 4$
is the Stefan-Boltzmann constant,
$T *$
stands for the sink temperature.
2.1.2. Thermal–Mechanical Solution
With regard to the solution technology for the coupled thermal–mechanical problem, the full Newton method is usually implemented with the enonsymmetric Jacobian matrix (see Equation (4)):
$[ K d d K d t K t d K t t ] { Δ d Δ t } = { R d R t }$
$Δ d$
$Δ t$
are increments of displacement and temperature, respectively.
$K i j$
is the submatrices of fully coupled Jacobian matrix,
$R d$
$R t$
stand for the mechanical and thermal residual vectors. Particularly, with respect to the issues of the coupled thermal–mechanical problem in the brake disc, the coupling effect between the mechanical
and thermal solutions is weak. Therefore, the off-diagonal submatrices are assumed to be zero in order to have a less costly solution. The solution equation is described in Equation (5):
$[ K d d 0 0 K t t ] { Δ d Δ t } = { R d R t }$
2.2. FE Model Description
There are two common methods for the loading implementation on the brake disc; one is the heat flux functions method and the other one is the mechanical friction method. Particularly, the heat flux
functions method can be subdivided into the rotating loading method and the uniform loading method. The advantage of the uniform loading method is the calculation efficiency, but the multiaxial
material response state and relevant out-of-phase failure feature during the braking process have to be revealed by the rotating loading method.
The three-dimensional geometry model of the investigated disc brake system is shown in
Figure 2
a. It can be seen that the disc brake system is complex, comprising the brake disc, brake pad, backplate, pad bracket and brake clamp, etc. In order to improve the calculation efficiency, the brake
disc is built with some simplifications and only half of it is established. The finite element model of the brake disc can be seen
Figure 2
b. The description of the brake disc dimension can be seen in
Figure 2
c. The red solid area and red dotted area in
Figure 2
c are the regions for rotating loading and uniform loading, respectively. r
, r
and r
indicate the inner, outer and average friction radii. The geometry dimension can be found in
Table 1
. A total of 29 elements (denoted as E
) are evenly distributed along radial direction (between r
and r
) and 3 elements (represented by E
) are uniformly arranged from the contact surface into the brake disc with a distance of 2 mm between each other.
The finite element model is established in ABAQUS [
] in order to ensure the calculation accuracy and improve the calculation efficiency. An 8-node three-dimensional linear coupled temperature-displacement reduced-integration brick element is used to
discretize the model. The material properties are listed in
Table 2
. The loading (heat flux, normal load and tangential load) is realized by subroutines in ABAQUS [
]. The heat flux can be calculated by Equation (1), which changes with time. The normal load is resulted from the brake force, and the tangential load is resulted from the friction force. The inner
ring where the brake disc is in contact with the axle is fully constrained. A global Cartesian coordinate system is employed, and the original point is the center of the brake disc. The z direction
is the rotation axis and the rotation speed can be calculated according to the adopted braking condition. The transient coupled temperature-displacement calculation is performed.
The braking operation parameters are listed in
Table 3
: the initial braking speed is 250 km/h, the wheel radius is 430 mm, the transfer efficiency
is 0.9, the frictional coefficient between brake disc and pad is assumed as 0.4, a uniform acceleration −1.389 m/s
is employed under a brake force of 17,000 N, the corresponding brake time is 50 s and the total brake distance is 1736.1 m. The environment temperature is 20 °C.
3. Multiaxial Fatigue Model
A fatigue model called the multiaxial fatigue space (referred as MFS criterion for convenience), recently proposed by the first author, is adopted to assess the multiaxial thermomechanical fatigue
damage of the brake disc. The MFS criterion is validated by the fatigue test data of different materials under various loading conditions and displays a pleasant prediction capability, compared with
other commonly used multiaxial fatigue models [
]. A brief introduction is presented, and detailed information can be found in the corresponding reference [
The fundamental assumption in the MFS criterion is that the changing of the material response state is the source of the fatigue damage. As shown in
Figure 3
, subscripts T1 and T2 indicate different time points. σ and τ are normal and shear stresses, subscripts x and y represent the components following the global coordinates, and subscripts 1 and 2 mean
the components following the principal directions. σ
and σ
indicate the response status at moment T2 following the principal directions of T1. C and R are the center and radius of Mohr circle, and A is twice the angle between the loading plane and the
principal plane. It can be seen that the changing of the material state can be described by the changing of C, R and A. Therefore, the material response status is described by three parameters (
), which are the equivalent shear parameter, equivalent normal parameter and out-of-phase parameter, respectively [
]. The changing of them (three fatigue basic units, i.e.,
2) and
3)) can be used for fatigue damage evaluation, and the multiaxial fatigue space can be established by fatigue basic units (see
Figure 4
) [
]. The multiaxial fatigue parameter is described in Equation (6):
$F P = f 0 ( f 1 ( P 1 ) , f 2 ( P 2 ) , f 3 ( P 3 ) )$
$P 1 = ( N C 1 − N C 2 ) 2 + ( S C 1 − S C 2 ) 2 2$
indicates a certain function,
are in-phase parameters and
is out-of-phase parameter [
indicate the normal and shear components on the critical plane (a couple of two orthogonal planes, noted by subscripts
1 and
2), stress, strain and energy components can be used as the adopted items,
$θ / 2$
is the out-of-phase angle, which represents the relative distortion deformation, and
$θ m a x / 2$
is the limit value of the out-of-phase angle [
]. It can be noticed that the MFS criterion belongs to the type of critical plane criterion.
According to the MFS criterion, the amplitude of parameter variation (changing of
) is the foundational component in multiaxial fatigue space, which is consistent with the definition of fatigue damage. The influence of the mean scales the vector
in multiaxial fatigue space [
]. Therefore, the multiaxial fatigue parameter in Equation (6) can be expressed by amplitude parameter (
$F P a$
) and mean parameter (
$F P m$
). See Equation (10):
$F P m = ( 1 + χ P 1 P 1 m ) · ( 1 + χ P 2 P 2 m ) · ( 1 + χ P 3 P 3 m )$
$F P a = [ ( F P a i n ) h o u t + ( k o u t · F P a o u t ) h o u t ] 1 / h o u t$
$F P a i n = [ ( P 1 a ) h i n + ( k i n · P 2 a ) h i n ] 1 / h i n$
where subscripts
indicate the amplitude and mean of corresponding parameter, subscripts
denote the in-phase and out-of-phase failure conditions,
are material-dependent coefficients, and
$χ P x$
is the mean influence coefficient. Afterward, the predicted fatigue life (
$N f$
) can be obtained by Equations (15) or (16), depending on the fatigue curve pattern:
$F P = A ( N f ) B + C ( N f ) D$
are material-dependent coefficients with solid physical meaning in multiaxial fatigue space [
It should be noticed that the original MFS criterion is established and validated in a 2-dimensional condition. With regard to 3-dimensional cases, the general formula is the same; the difference is
in the determination of the critical plane. In this work, the directions of the maximum and the second maximum principal strain range are employed for critical plane determination. That is to say,
the direction of the maximum principal strain range will be determined firstly. Afterward, the directions perpendicular to the maximum principal strain range will be checked to obtain the direction
of the second maximum principal strain range. In this way, the issues of the 3-dimensional condition can be transformed into 2-dimensional cases and the MFS criterion can be used smoothly.
About the material dependent coefficients, normally the fatigue test is performed under mechanical fatigue loading instead of thermomechanical fatigue loading. The thermomechanical fatigue-dependent
coefficients of the brake disc are not available at this moment. Thus, it is assumed that they are identical with the fatigue dependent coefficients of hot-rolled 45 steel [
]. In order to present the methodology, this is a common strategy when the fatigue-related information is unavailable [
]. The adopted fatigue-dependent coefficients are listed in
Table 4
. An accurate multiaxial thermomechanical fatigue evaluation of the brake disc can be performed following the proposed approach when the thermomechanical fatigue-dependent coefficients are collected
in the future.
4. Results and Discussion
Considering the calculation efficiency, uniform loading is performed first, the calculation time is set as 500 s to have an overall response status, the highest temperature 454 °C occurs at 35 s, and
the temperature distribution on the brake disc at four moments before the highest temperature time is demonstrated in
Figure 5
. As discussed above, the multiaxial characters cannot be revealed with uniform loading. Rotating loading is needed for the multiaxial thermomechanical fatigue evaluation. According to the results
with uniform loading, it can be seen that the temperature increases significantly and reaches above 400 °C in the first 20 s. Thus, the calculation time is set as 20 s with rotating loading so as to
reduce the simulation time and result file size. The comparison between the results with rotating loading and uniform loading can be seen in
Figure 6
It can be seen in
Figure 6
that the results manifestation is reasonable [
], which verifies the correctness of the calculation to a certain extent. The difference between the results with uniform and rotating loading is negligible. It can also be noticed that the heating
time (50 s) is much shorter than the cooling time (much greater than 450 s) and the results variation during the heating process (see the changing of Huber–Mises results in the first 50 s in
Figure 6
) is much more violent than that during the cooling process (see the Huber–Mises results from 50 s to 500 s in
Figure 6
). Thus, as the changing of the material state during the heating process is used for fatigue evaluation, the calculation time is greatly reduced without losing much accuracy. To have a clear view
about the multiaxial character of thermomechanical fatigue damage of the brake disc, the evolutions of the strain items with rotating loading can be seen in
Figure 7
. It can be noticed clearly that the changing of shear strain is non-proportional, which indicates that the principal strain direction is changed during the braking process. It means that the
thermomechanical fatigue damage of the brake disc not only belongs to the type of multiaxial fatigue damage, but also the out-of-phase failure status. The out-of-phase failure of the brake disc could
be resulted from the mechanical interaction between the brake disc and the brake pad, or could be resulted from the influence of the brake disc structure. The reasons will be discussed in the future.
It is known that the changing of the material response state is critical for multiaxial fatigue analysis; the time increment needs to be small enough to capture the details of this changing. This
makes the realization with rotating loading unfavorable because of the lengthy calculation time and huge resulting file size. Thus, an equivalent accelerated method (referred to as the equivalent
method) is proposed. An amplification factor is used to scale the heat flux in order to reach the target temperature with only one rotation. In order to validate equivalent method, the temperature
with rotating loading at 20 s is used as the target temperature (406.73 °C). The accepted tolerance for the obtained temperature by equivalent method is 1%. The temperature distribution with
equivalent method at different times in one rotation is shown in
Figure 8
; the highest temperature is 406.38 °C.
It can be seen that the highest temperature by the equivalent method (406.38 °C) is close to that with rotating loading (406.73 °C). The position of the maximum temperature by the equivalent method
is located at the average friction radius position (E
), which is the same as the rotating loading calculation. The comparison of the results among uniform loading, rotating loading and equivalent methods at the maximum temperature position is shown in
Table 5
. It can be seen that the temperature, thermal strain and Huber–Mises stress are approximately equal with each other. The results with rotating loading is used as reference. The temperature and
strain with the equivalent method are quite satisfactory and the relative error is less than 0.1%. The difference of Huber–Mises stress is around 5% because of the history effect and constraint from
the extrusion by the surrounding elements. In general, the results with the equivalent method are acceptable and strain items are used for the fatigue evaluation with the MFS criterion.
Regarding the multiaxial fatigue prediction capability with the equivalent method, the comparison with rotating loading is performed at different temperatures. Because the results of the finite
element calculation are described in the global coordinate (coordinate xyz), coordinate transformation is executed with the help of Euler angles to get the material response status in coordinate XYZ
Figure 9
). Here, zXZ rotation order is adopted for the coordinate transformation. Afterward, the MFS criterion can be adopted. The calculated fatigue parameter and relative error (defined in Equation (17))
are demonstrated in
Figure 10
$E r = ( F P r o t a t i o n − F P e q u i v a l e n t ) / F P r o t a t i o n$
It can be seen that with the increasement of temperature, the relative error between the equivalent method and the rotating loading method is increased slowly, following a linear pattern,
approximately. Based on this, the relationship function between the relative error (
$E r$
) and temperature (
) can be written as Equation (18):
$E r = 0.00032637 × T − 0.017$
Now the multiaxial thermomechanical fatigue of the brake disc under the investigated braking operation condition can be assessed. Due to the fact that the calculation time with rotating loading is
quite long, the maximum temperature with uniform loading is used as the target temperature (454 °C, see
Figure 6
); the value of the fatigue parameter with the equivalent method can be obtained with this target temperature (see
Figure 10
, blue point). Hereafter, the fatigue parameter value with rotating loading can be obtained with error correction (see
Figure 10
, red point).
It is discussed above that the amplitudes and distributions of temperature, stress and strain items are not enough for accurate thermomechanical fatigue evaluation of the brake disc. Ignoring the
influence of the multiaxial character and out-of-phase feature leads to an underestimation of the thermomechanical fatigue assessment for the brake disc. Based on the result with the rotating loading
calculation, the contribution of the equivalent shear parameter, equivalent normal parameter and out-of-phase parameter (represented by
in the MFS criterion) to the total multiaxial thermomechanical fatigue parameter is demonstrated in
Figure 11
. It can be seen that just using a uniaxial fatigue parameter causes at least 14% underestimation of the fatigue damage, while employing a multiaxial fatigue parameter without the consideration of
the out-of-phase failure leads to an underestimation of 5%. It should be noticed that the relationship between the fatigue parameter and fatigue life is exponential, which means that an insignificant
deviation in the fatigue parameter assessment leads to a significant predicted error of fatigue life.
It is known that the material mechanical and fatigue characters change significantly with temperature [
]. The material character under high temperature is not available at this moment. As there is an approximate linear relationship between the yield stress and fatigue limit [
], the yield stress decreases exponentially with the increasing of the temperature, approximately (see
Figure 12
a) [
]. Thus, an exponential relationship between the fatigue limit and temperature is assumed in order to present the methodology in this work (see
Figure 12
b). The corresponding fatigue curves at different temperatures can be obtained using extrapolation method with the help of the yield stress at different temperatures. The multiaxial thermomechanical
fatigue life of the investigated brake disc under the employed braking operation condition can be obtained, which is about 2500 braking cycles. The accurate multiaxial thermomechanical fatigue life
prediction can be obtained with the proposed methodology when the temperature related material fatigue characters become available in the future.
5. Conclusions
In this study, the multiaxial characters of thermomechanical fatigue damage on the high-speed railway brake disc is investigated. Through the revealing of the multiaxial response state and the
out-of-phase failure feature, it is proved that under mechanical friction braking conditions, the damage of the high-speed railway brake disc belongs to the type of multiaxial thermomechanical
fatigue. With the help of a multiaxial fatigue model, the methodology for multiaxial thermomechanical fatigue damage evaluation is proposed and multiaxial thermomechanical fatigue life prediction is
implemented. In addition, the contribution of the uniaxial fatigue parameter, multiaxial fatigue parameter and out-of-phase failure parameter to the total multiaxial thermomechanical fatigue damage
of the brake disc is discussed. The results indicate that, using the amplitude and distribution of temperature, stress and strain items for fatigue evaluation lead to an underestimation of the brake
disc thermomechanical fatigue damage. This work explains the importance of studying the thermomechanical fatigue damage of brake disc from the perspective of multiaxial fatigue.
Author Contributions
Conceptualization, C.L. and J.M.; methodology, C.L. and Z.F.; software, C.L. and Y.W.; validation, C.L. and J.M.; formal analysis, C.L.; investigation, C.L. and R.S.; resources, C.L. and J.M.; data
curation, C.L. and Y.W.; writing—original draft preparation, C.L., R.S., Y.W. and Z.F.; writing—review and editing, J.M.; visualization, C.L. and R.S.; supervision, J.M.; project administration, C.L.
and J.M.; funding acquisition, C.L. and J.M. All authors have read and agreed to the published version of the manuscript.
This research was funded by the National Natural Science Foundation of China, grant number 51822508; Sichuan Science and Technology Program, grant number 2020JDTD0012; the Fundamental Research Funds
for the Central Universities, grant number 2682021CX028.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
Figure 1.
Thermomechanical fatigue of brake disc [
Figure 2. (a) Three-dimensional geometry model of the investigated disc brake system, (b) simplified finite element model of the brake disc, (c) description of brake disc dimension.
Figure 5. Temperature distribution with uniform loading at different moments (highest temperature 454 °C occurs at t4 = 35 s).
Figure 7. (a) The changing of strain items following xx, yy and zz directions with rotating loading, (b) the changing of strain items following xy, xz and yz directions with rotating loading.
Figure 11. The contribution of different parameters on multiaxial thermomechanical fatigue of the brake disc.
Figure 12.
) The relationship between the yield stress and temperature [
], (
) extrapolation of fatigue curve.
r[1] r[2] r[3] r[L1] r[L2] r[M]
93 mm 175 mm 320 mm 187.5 mm 312.5 mm 250 mm
Young’s Modulus Poisson’s Ratio Density Thermal Expansion Coefficient Specific Heat Thermal Conductivity
(MPa) (kg/m^3) [J/(kg·K)] [W/(m·K)]
2.02 × 10^5 0.29 7850 1.04 × 10^−5 457.74 32
Wheel Radius Brake Time Transfer Efficiency Acceleration
430 mm 50 s 0.9 −1.389 m/s^2
Brake force Brake distance Friction coefficient Speed
17,000 N 1736.1 m 0.4 250 km/h
$k i n$ $h i n$ $k o u t$ $h o u t$ $χ P 1$ $χ P 2$ $χ P 3$ $A$ $B$
1.7 1 1.12 1 0.1 0 0 50.13 −0.4411
Item Rotating Uniform Equivalent
Value Error % Value Error %
Temperature (°C) 406.73 417.82 −2.73 406.38 0.086
Thermal strain (%) 0.40220 0.41373 −2.87 0.40184 0.090
Huber–Mises (MPa) 1063.5 1082.8 −1.81 1117.8 −5.1
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Lu, C.; Mo, J.; Sun, R.; Wu, Y.; Fan, Z. Investigation into Multiaxial Character of Thermomechanical Fatigue Damage on High-Speed Railway Brake Disc. Vehicles 2021, 3, 287-299. https://doi.org/
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Lu C, Mo J, Sun R, Wu Y, Fan Z. Investigation into Multiaxial Character of Thermomechanical Fatigue Damage on High-Speed Railway Brake Disc. Vehicles. 2021; 3(2):287-299. https://doi.org/10.3390/
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Study on Quasi-Static Uniaxial Compression Properties and Constitutive Equation of Spherical Cell Porous Aluminum-Polyurethane Composites
School of Civil Engineering, Southeast University, Nanjing 210096, China
Beijing Advanced Innovation Center for Future Urban Design, Beijing University of Civil Engineering and Architecture, Beijing 100044, China
Author to whom correspondence should be addressed.
Submission received: 5 June 2018 / Revised: 18 July 2018 / Accepted: 21 July 2018 / Published: 23 July 2018
Quasi-static uniaxial compression properties and the constitutive equation of spherical cell porous aluminum-polyurethane composites (SCPA-PU composites) were investigated in this paper. The effects
of relative density on the densification strain, plateau stress and energy absorption properties of the SCPA-PU composites were analyzed. It is found that the stress-strain curves of SCPA-PU
composites consist of three stages: The linear elastic part, longer plastic plateau segment and densification region. The results also demonstrate that both the plateau stress and the densification
strain energy of the SCPA-PU composites can be improved by increasing the relative density of the spherical cell porous aluminum (SCPA), while the densification strain of the SCPA-PU composites shows
little dependence on the relative density of the SCPA. Furthermore, the applicability of three representative phenomenological models to the constitutive equations of SCPA-PU composites are verified
and compared based on the experimental results. The error analysis result indicates that the Avalle model is the best model to characterize the uniaxial compression constitutive equation of SCPA-PU
1. Introduction
Porous aluminum has gained a considerable amount of attention due to its good physical properties and excellent mechanical characteristics [
]. Moreover, it is generally accepted that porous aluminum with a spherical cell shows better structural homogeneity and mechanical performance than non-spherical cell porous aluminum [
]. Thus, great efforts have been made to manufacture and investigate the mechanical properties of spherical cell porous aluminum (SCPA) in recent years [
]. Open-cell SCPA can be produced by designing several small openings in different directions on the spherical cell wall. Therefore, the SCPA with high permeability incorporates substantial cell
walls relative to traditional open-cell porous aluminum, wherein the cell wall is reduced to the bar-beam system. The structure of this pattern leads to the greater energy consumption, which is
associated with complex failure modes of the cell membranes; furthermore, the SCPA shows superior functionality compared with closed-cell porous aluminum in certain applications. Nonetheless, it is
still difficult to achieve a given energy absorption target, for example, vehicle collision energy, bridge pounding energy that suffers from strong earthquakes, simply by controlling the porosity and
other structural parameters of the SCPA because of the imperfection of current fabrication approaches. Hence, it is necessary to ameliorate the issue via a simple and effective method. In fact, the
attempt to enhance the mechanical performance of porous aluminum has a long history.
Alizadeh et al. [
] manufactured open-cell Al-Al
composite foams, using the space-holder method, and investigated the mechanical properties and energy absorption behavior of open cell Al foams, containing different volume fractions of Al
. Du et al. [
] examined the effect of nanoparticles on the micro-structure, compressive performance and energy absorption of Al foams. Furthermore, Sun et al. [
] prepared nanocopper coated aluminum foam and studied the mechanical properties of Al/Cu hybrid foam via experimental investigation and numerical modeling. Li et al. [
] reported the mechanical properties of open-cell aluminum foam wrapped with zinc film and highlighted the influence of coating time on the mechanical characteristics. The above-mentioned enhancement
methods can be attributed to the addition of alloying elements and hard particles to strengthen the cell wall of aluminum foams. This idea has been utilized, and illustrated by Duarte and Ferreira [
] in detail, in many cases. However, recently, another alternative to increase the mechanical properties of porous aluminum was proposed by the introduction of polymers, owing to its simplify and
effectivity. In fact, the concept of combining the advantages of porous aluminums and polymers is receiving renewed attention. Cheng and Han [
] developed a type of aluminum foam-silicate rubber composite and examined the effect of filler on the compressive behavior and energy absorption. Kitazono et al. [
] strengthened closed-cell aluminum foam using polyester resin and highlighted the impact of surface treatment methods on the compressive strength and energy absorption. Vesenjak et al. [
] prepared porous materials-silicone rubber composites and investigated the -influences of the base materials, specimen size and strain rate on the compressive performances and energy absorption
capacity of composites. Kishimoto et al. [
] analyzed the mechanical properties of closed-cell aluminum foam-polyurethane and closed-cell aluminum foam-epoxy composites by measuring deformation distributions, adopting the digital image
correlation method. Based on their studies, Yuan et al. [
] produced closed-cell aluminum foam epoxy resin composites and discussed the effect of the composite form, the relative density and the content of epoxy resin on the mechanical characteristics and
energy absorption. Moreover, they presented a mathematical model to describe the plateau stress and energy absorption capacity. Furthermore, Liu et al. [
] validated the effectiveness of polyurethane (PU) for increasing the damping of open-cell aluminum foam by cyclic compression tests. Nevertheless, the aforementioned studies concerning porous
aluminum-polymer composites are limited to non-spherical cell porous aluminum. PU is one of the most commonly utilized polymers in the energy absorption systems; moreover, its superior damping
capacity and easy filling property have been proved [
]. Consequently, PU is adopted as the filling polymer here to improve the mechanical properties of the SCPA.
The design of the structural components applied to engineering fields is generally carried out by simulation code, based on the finite element method. The mathematical description of the mechanical
behavior of the materials by a good representation of the stress-strain curve is required when performing finite element modeling and analysis. Theoretical and numerical models have recently been
proposed to describe the uniaxial compression stress-strain behavior of porous materials. A micro-mechanical model related to the deformation mechanism of structure was presented by Gibson [
]. However, the micro-mechanical model was quite difficult to execute, owing to its need for a rough analysis of the porous structure. Fortunately, several phenomenological models, which aim to
supply the best fitting of the experimental mechanical behavior without a direct relationship with the physics of the phenomenon, have been developed to promote the applicability of porous materials
in recent years. The Rusch model [
], Liu and Subhash model [
], and Avalle model [
] are the three representative phenomenological models that characterize the stress-strain behavior of porous materials due to simple formulation and high accuracy. However, study on the constitutive
model of porous aluminum-polymer composites in uniaxial compression is rarely reported. The complexity of the structure of porous aluminum-polymer composites is enhanced because of the introduction
of polymer, which makes the micro-mechanical model more difficult to use. Therefore, the constitutive equation of spherical cell porous aluminum-polyurethane (SCPA-PU) composites was examined here,
based on the aforementioned three phenomenological models.
SCPA-PU composites were prepared using the infiltration method, under a uniform pressure of around 0.5 MPa, in the present work. The comparison of uniaxial compression stress-strain behavior between
the SCPA and SCPA-PU composites is made, and the deformation mechanism of SCPA-PU composites is analyzed. Moreover, the effects of the relative density of the SCPA on the densification strain,
plateau stress and energy absorption capacity of SCPA-PU composites are investigated. Based on the experimental verification and error analysis, the most suitable model to describe the uniaxial
compression stress-strain behavior of SCPA-PU composites is selected from Rusch model, Liu and Subhash model, and Avalle model, respectively.
2. Experimental Procedure
2.1. Specimen Preparations
The open-cell SCPA in this paper was fabricated by the space holder method [
] and supplied by Qiangye Metal Foam Ltd (Beijing, China). The cell size is 5 mm, while the base material is Al99.7%.4-6 openings with a size of 1–1.5 mm are arranged with different orientations and
situated in the cell wall of the spherical cell. Compressive samples, with the dimensions of 50 mm × 50 mm × 75 mm, were produced using a line cutting machine. The relative density value of the SCPA
specimen, which is defined as the ratio of the density of the SCPA and the density of the matrix aluminum, varies from 0.263 to 0.374. At least three specimens were prepared for each material. PU is
provided by Haida Rubber and Plastic Ltd. (Wuxi, China), which is usually used as an energy-absorbing material. The manufacturer’s data indicate that the density is 1.123 g/m
, the tensile strength is approximately 4 MPa, and the elongation at the break is 655%. The SCPA should be wrapped in PU so as to reduce the volume shrinkage of PU. The SCPA-PU composites were
prepared employing the procedure shown in
Figure 1
. A uniform pressure of around 0.5 MPa was applied to press the PU elastomer into the SCPA. Finally, the specimens of the SCPA-PU composites were produced after they were heated at 100 °C for ten
hours. The open-cell spherical cell could be filled with PU due to the excellent fluidity and longer curing time of PU. Three kinds of specimens, which are named SCPA, PU and SCPA-PU composites, are
illustrated in
Figure 1
2.2. Compressive Test
Quasi-static uniaxial compressive tests were performed using a CMT5105 electron universal testing machine (SANS, Minneapolis, MN, USA) (shown in
Figure 2
) at room temperature (23 °C) under displacement control at a constant cross-head speed of 4.5 mm/min. The circular aligned platens were coated with silicon greases to reduce surface friction with
the compression specimens. The variations of load and displacement were automatically recorded by the machine. It is worth noting that the experimental results shown in this paper correspond to the
average of multiple specimens tested.
3. Results and Discussion
3.1. Compressive Stress-Strain Behavior
The appearance of three types of deformed specimens after the uniaxial compression test is shown in
Figure 3
, indicating their different deformation modes. The comparison of the compressive stress-strain curves between SCPA, PU, and SCPA-PU composite specimens is made in
Figure 4
. The stress of PU increases as the strain increases without yielding, which is the typical behavior of elastomer. The stress-strain curves of the SCPA-PU composites appeared to have three similar
regions of unfilled SCPA [
], i.e., the linear elastic part, the plastic plateau segment and the densification regions. However, compared with the stress-strain curve of SCPA, the compressive stress-strain curve of the SCPA-PU
composites exhibits a longer and higher plateau region.
The main deformation mechanism of SCPA, which is described by the homogeneous failure mode with multiple random deformation bands, is reported [
]. Nevertheless, the deformation mechanism of SCPA-PU composites is quite different from that of SCPA, which is ascribed to the introduction of PU. The result presented in
Figure 4
shows that the stress is mainly borne by the cell wall of the SCPA because of its greater strength over PU at the stage of low strain level, thus, the compressive curve of the SCPA-PU composites
coincides with that of SCPA at the early stage of deformation. Resistance to deformation of the cell wall increases as the compressive stress increases, which is related to the PU filling the
spherical cell. Meanwhile, the lateral deformation of the SCPA-PU composites becomes larger with the increase of load owing to the incompressibility of PU volume, which is demonstrated by the images
of PU and SCPA-PU composite specimens after compression shown in
Figure 3
. The lateral deformation is restrained by the cell wall, which, conversely, raises the resistance of the SCPA-PU composites. Moreover, the incompressibility of PU postpones the yield and buckling of
the cell edges, which makes the plastic deformation capacity of the SCPA-PU composites stronger than that of the SCPA. Moreover, this conclusion is consistent with that in the quasi-static uniaxial
compression of open-cell aluminum foam with silicate rubber [
3.2. Energy Absorption Characteristics
One of the important mechanical properties for the evaluation of the application of porous materials is the energy absorption characteristic. It is widely accepted that both the densification strain
and the plateau stress play important roles in characterizing the energy absorption capacity of porous materials. Where the densification strain is determined using the energy absorption efficiency,
potential mistakes caused by the existing huge uncertainties in other methods can be avoided [
]. The optimal energy absorption of porous materials can be identified by an energy efficiency parameter
$η ( ε )$
$η ( ε ) = 1 σ ( ε ) ∫ ε y ε σ ( ε ) d ε ,$
$ε y$
is the strain corresponding to the starting point of the plateau segment. Furthermore, a representative strain of densification,
$ε d$
, is determined as:
$d η ( ε ) d ε | ε = ε d = 0 ,$
at which the energy absorption efficiency,
$η ( ε )$
, reaches a maximum value on the efficiency-strain curve, i.e., the tangential stiffness is equal to zero, as shown in
Figure 5
. The plateau stress,
$σ p l$
, is expressed as:
$σ p l = ∫ ε y ε d σ ( ε ) d ε ε d − ε y ,$
$ε y$
is usually very small compared with
$ε d$
, it is assumed to be zero here. The densification strain energy,
$W ε d$
, which is a significant index for the characterization of the energy absorption capacity of porous materials, is defined as:
$W ε d = ∫ ε y ε d σ ( ε ) d ε ,$
The above presented methods are utilized to calculate the densification strain, the plateau stress and the densification strain energy here, respectively.
The densification strain, plateau stress and densification strain energy of the SCPA and the SCPA-PU composites, with different relative densities, are shown in
Figure 6
a–c, respectively. The impact of PU is more pronounced on the densification strain and densification strain energy of the SCPA, as compared with the plateau stress, as shown in
Figure 6
, which is associated with the low strength and high elasticity of the PU. The densification strain values of the SCPA-PU composites, with the relative density values of 0.263, 0.298, 0.326 and
0.374, are 17.2%, 22.93%, 23.11% and 33.06% higher than those of the SCPA, respectively. At the same time, the plateau stress values of the SCPA-PU composites, with the relative density values of
0.263, 0.298, 0.326 and 0.374, are 9.6%, 8.73%, 6.88% and 7.15% higher than those of the SCPA, respectively. Moreover, the densification strain energy values of the SCPA-PU composites, with the
relative density values of 0.263, 0.298, 0.326 and 0.374, are 28.59%, 33.65%, 31.59% and 51.55% higher than those of the SCPA, respectively.
The dependence of the densification strain value of the SCPA-PU composites on the relative density is drastically reduced (
Figure 6
a) when compared with the plateau stress value (
Figure 6
b) and densification strain energy value (
Figure 6
c). The densification strain value of the SCPA-PU composites, studied in this paper, can be controlled at about 0.65. Furthermore, it is seen from
Figure 6
b,c that the relationship between the plateau stress of the SCPA-PU composites and the relative density is similar to that of the densification strain energy and relative density. The two indexes
increase with the relative density, and the weak dependence of the densification strain on the relative density is caused by this coupling effect.
In summary, the energy absorption capacity of the SCPA is enhanced by the introduction of the PU. Moreover, the plateau stress value and densification strain energy value of the SCPA-PU composites
increase as the relative density value increases, while the relationship between the densification strain and the relative density of the SCPA-PU composites is relatively insignificant.
The ideal energy absorption efficiency, which was proposed by Miltz and Gad [
] to assess whether the porous material is an idealized energy absorption material, is introduced here to further evaluate the energy absorption property of the SCPA-PU composites, which is
formulated as:
$I = ∫ 0 ε m σ ( ε ) d ε σ m ε m ,$
$σ m$
is the stress associated with the strain,
$ε m$
, and the bigger the I value, the closer the porous material is to the ideal energy absorption material.
Comparisons of the I value for the SCPA and the SCPA-PU composites, with different relative density values, are made in
Figure 7
. It can be seen from
Figure 7
that the I consists of three stages in all cases: Fast ascending branch (I), where the I increases monotonously to a high energy absorption efficiency point with the compressive strain; plateau stage
(II), in which the I maintains the high efficiency level, but some fluctuation exits as the strain increases; and descending region (III), where the I decreases with the increase of the compressive
strain. Furthermore, as shown in
Figure 7
, the average I value of the SCPA, with the relative density value of 0.263, is 0.685 when the strain is between 0.1 and 0.30, while the average I value of the SCPA-PU composites, with the relative
density value of 0.263, is 0.7 when the strain is between 0.1 and 0.5. Meanwhile, the average I value of the SCPA, with the relative density value of 0.298, is 0.696 when the strain is between 0.1
and 0.30, while the average I value of the SCPA-PU composites, with the relative density value of 0.298, is 0.692 when the strain is between 0.15 and 0.50. Moreover, the average I value of the SCPA,
with the relative density value of 0.326, is 0.711 when the strain is between 0.1 and 0.3, while the average I value of the SCPA-PU composites, with the relative density value of 0.326, is 0.705 when
the strain is between 0.1 and 0.55. Besides, the average I value of the SCPA, with the relative density value of 0.374, is 0.726 when the strain is between 0.1 and 0.3, while the average I value of
the SCPA-PU composites, with the relative density value of 0.374, is 0.696 when the strain is between 0.1 and 0.55.
Based on the above experimental results, the following two conclusions can be drawn: (1) The plateau I values of the SCPA-PU composites is close to those of the SCPA, however, a wider plateau strain
range of the SCPA-PU composites is presented when compared with the plateau strain range of the SCPA; and (2) the I value, situated in the plateau stage on the relative density of the SCPA-PU
composites, has a weak dependence.
3.3. Uniaxial Compression Constitutive Equation of SCPA-PU Composites
Among the existing phenomenological models regarding the constitutive equation of porous materials, the Rusch model [
], Liu and Subhash model [
] and Avalle model [
] are the most commonly employed. The three models are adopted and fitted to the experimental results of the SCPA-PU composites here, and the best-fitting model for describing the SCPA-PU composites
among these three is quantitatively identified in terms of the metric of root mean square error.
3.3.1. Existing Phenomenological Models
Rusch Model
The Rusch model is a phenomenological model with a simple expression, which is presented by the sum of two power functions:
$σ = a ε p + b ε q , 0 < p < 1 , q > 1 ,$
are the nominal stress and strain, respectively, and
$a , b , p , q$
are empirically determined. The first term is designed for the elastic-plateau region, while the second term is utilized for modelling the densification region. Generally, the inaccuracy in
describing the densification phase of porous materials is a drawback of the model when, as a consequence of compression, the internal voids gradually disappear.
Liu and Subhash Model
The model proposed by Liu and Subhash similarly consists of two parts, the first describing the elastic-plastic stage and the second one representing the densification segment, and is shown as
$σ = A e α ε − 1 B + e β ε + e C ( e γ ε − 1 ) ,$
the function has six parameters, wherein the parameter
is related to the yield stress,
plays a role in shifting the lower asymptote, the behavior of the plateau region is determined by the difference between
, the parameter
plays a role in stretching or shrinking of the curve, and the speed of the densification is controlled by the
. The fundamental compressive and tensile stress-strain behavior of porous materials with various initial densities under large deformation can be captured by the model. Moreover, the equation is
continuously differentiable.
Avalle Model
Recently, a model, which is composed of the elastic-plastic part and the densification segment, was given by Avalle as follows:
$σ = F ( 1 − e − ( G / F ) ε ( 1 − ε ) m ) + H ( ε 1 − ε ) n ,$
the parameters of the model can be empirically determined, wherein the plateau stress is defined by the parameter
, the parameter
is adopted to represent the initial elastic modulus, the curve knee at the connection of the elastic stage with the plateau region is achieved by the appropriate choice of the parameter
, the curve change trend of the densification process of porous materials is affected by the parameter
. The second term of the model is a modification of the second one of the Rusch model and has been introduced to obtain a vertical asymptote corresponding to the physical limit of compression (
$ε = 1$
3.3.2. Evaluation of Model Performance
The Experimental verification and error analysis of the three mentioned types of phenomenological models are performed based on the stress-strain curves of SCPA-PU composites. The least square method
is adopted here to compute the deviation of the model prediction value from the experimental results. Consequently, the model prediction error is taken as the difference between the experimental
stress and the model stress at the same strain value. The curve fitting the results of the three considered models for the SCPA-PU composites with the relative density value of 0.326 are separately
shown in
Figure 8
Figure 9
Figure 10
. The prediction error of each model is shown in the right diagram of each figure, which is expressed as a function of the strain. It is obviously seen that the fitting performance of the three
models is the worst in the elastic region, while a certain level of improvement is observed in the plateau and densification region. Furthermore, the fitting results also indicate that the three
selected models can be employed to characterize the uniaxial compression stress-strain behavior of the SCPA-PU composites. Additionally, model forecast error results show that the fluctuation of the
error curve for the Rusch model is the most significant, while the quality of the fitting between Avalle model and Liu and Subhash model are superior to the Rusch model and appear to be comparable.
In order to choose the most suitable model from the three mentioned phenomenological models for SCPA-PU composites, a direct comparison of the overall fitting ability of the three considered models
can be conducted by means of the root mean square error (RMSE) for SCPA-PU composites with different relative density values: They are presented with histograms in
Figure 11
. It is found that the worst fitting behavior of the Rusch model is displayed regardless of the relative density value, while the excellent fitting performance of the Avalle model is seen among the
chosen relative density value. It can be concluded that the Avalle model is the best phenomenological model to characterize the uniaxial compression constitutive equation of SCPA-PU composites among
the three selected models.
4. Conclusions
• The compressive stress-strain curves of spherical cell porous aluminum-polyurethane composites (SCPA-PU composites) consist of three stages: Linear elastic part, plateau region and densification
segment. Furthermore, PU is beneficial to the increase of the plateau stress and elongation of the densification strain.
• The energy absorption capacity of SCPA-PU composites is superior to that of the SCPA. The densification strain energy of the SCPA-PU composites, with the relative density values of 0.263, 0.298,
0.326, and 0.374, is 28.59%, 33.65%, 31.59%, and 51.55% higher than those of the SCPA, with the same relative density value, respectively. Besides, the weak dependence of the densification strain
of the SCPA-PU composites on the relative density is seen, while the plateau stress and the densification strain energy increase as the relative density increases. Furthermore, the ideal energy
absorption efficiency (I)-strain curves of SCPA-PU composites and SCPA consist of three parts: Fast ascending branch, plateau stage, and descending region. The plateau I value of SCPA-PU
composites is close to that of SCPA, while it has a wider plateau strain range. It is also found that the plateau I value of SCPA-PU composites is insensitive to the relative density of the SCPA.
• Based on the calculated root mean square error results of SCPA-PU composites with different relative density values, the best phenomenological model to characterize the constitutive equation of
SCPA-PU composites is the Avalle model. This conclusion provides a foundation for the following research regarding the constitutive model of SCPA-PU composites considering strain rate and
temperate factors.
Author Contributions
For this research, H.B. conceived, designed and conducted the experiments, analyzed the data and wrote the paper; A.L. contributed some useful suggestions.
The present work was sponsored by the National Key Research and Development Program of China (Grant No. 2017YFC0703602) and National Natural Science Foundation of China (Grant No. 51438002 and
51278104). The financial contributions are gratefully acknowledged.
Conflicts of Interest
The authors declare no conflict of interest.
Figure 1. Fabrication procedure of the spherical cell porous aluminum-polyurethane composites (SCPA-PU composites): (a) The specimen of the spherical cell porous aluminum (SCPA); (b) fabrication
method of the SCPA-PU composites; and (c) image of specimens from left to right: SCPA, polyurethane (PU), and SCPA-PU composites.
Figure 6. Comparison of SCPA and SCPA-PU composites specimens, with different relative density values for (a) the densification strain; (b) plateau stress; and (c) densification strain energy.
Figure 7. $I − ε$ curves of the SCPA and the SCPA-PU composites with different relative density values.
Figure 8. (a) Comparison between the curve predicted by the Rusch model and the experimental curve ($ρ ∗ / ρ s$ = 0.326); and (b) model prediction error.
Figure 9. (a) Comparison between the curve predicted by the Liu and Subhash model and the experimental curve ($ρ ∗ / ρ s$ = 0.326); and (b) model prediction error.
Figure 10. (a) Comparison between the curve predicted by the Avalle model and the experimental curve ($ρ ∗ / ρ s$ = 0.326); and (b) model prediction error.
Figure 11. Comparison of the root mean square error of each model for (a) 0.263; (b) 0.298; (c) 0.326; and (d) 0.374 relative density values.
© 2018 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http:/
Share and Cite
MDPI and ACS Style
Bao, H.; Li, A. Study on Quasi-Static Uniaxial Compression Properties and Constitutive Equation of Spherical Cell Porous Aluminum-Polyurethane Composites. Materials 2018, 11, 1261. https://doi.org/
AMA Style
Bao H, Li A. Study on Quasi-Static Uniaxial Compression Properties and Constitutive Equation of Spherical Cell Porous Aluminum-Polyurethane Composites. Materials. 2018; 11(7):1261. https://doi.org/
Chicago/Turabian Style
Bao, Haiying, and Aiqun Li. 2018. "Study on Quasi-Static Uniaxial Compression Properties and Constitutive Equation of Spherical Cell Porous Aluminum-Polyurethane Composites" Materials 11, no. 7:
1261. https://doi.org/10.3390/ma11071261
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details
Article Metrics | {"url":"https://www.mdpi.com/1996-1944/11/7/1261","timestamp":"2024-11-08T15:48:40Z","content_type":"text/html","content_length":"415572","record_id":"<urn:uuid:584ae0e1-cbea-430f-a026-292e9469783f>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00279.warc.gz"} |
class statinf.stats.unsupervised.GaussianMixture[source]
Bases: object
Class for a gaussian mixture model, uses the EM algorithm to fit the model to the data.
This function is still under development. This is a beta version, please be aware that some functionalitie might not be available. The full stable version soon will be released.
☆ Murphy, K. P. (2012). Machine learning: a probabilistic perspective. MIT press.
fit(X, k, n_epochs=100, improvement_threshold=0.0005)[source]
Fitting function initialized by K-means algorithm.
○ X (numpy.ndarray) – data.
○ K (int) – number of clusters (gaussians).
○ n_epochs (numpy.ndarray) – number of epochs, default is 100.
○ improvement_threshold (float, optional) – Threshold from which we consider the likelihood improved, defaults to 0.0005.
class statinf.stats.unsupervised.KMeans(k=1, max_iter=100, init='random', random_state=0)[source]
Bases: object
K-means clustering implementation.
This function is still under development. This is a beta version, please be aware that some functionalitie might not be available. The full stable version soon will be released.
☆ k (int) – number of clusters, default is 1.
☆ max_iter (int) – number of iterations for convergence.
☆ init (String) – initialization option, options are random or kmeans++ .
☆ random_state (int) – seed of the random state, default is 0.
☆ labels (numpy.array) – labels for each datapoint.
☆ centroids (numpy.array) – coordinates of the centroids.
☆ Friedman, J., Hastie, T., & Tibshirani, R. (2001). The elements of statistical learning (Vol. 1, No. 10). New York: Springer series in statistics.
closest_centroid(points, centroids)[source]
Returns an array containing the index to the nearest centroid for each point
○ points (numpy.array) – features of each point.
○ centroids (list) – list of the centroids coordinates.
Fit the model to the data using different initializations (random init or kmeans++)
X (numpy.array) – Input data.
get_distance(points, centroids)[source]
Returns the euclidian distance between each point and the centroids.
○ points (numpy.array) – features of each point.
○ centroids (list) – list of the centroids coordinates.
move_centroids(points, closest, centroids)[source]
Returns the new centroids assigned from the points closest to them.
○ points (numpy.array) – features of each point.
○ closest (numpy.array) – array with the index of closest centroid for each point.
○ centroids (list) – list of the centroids coordinates.
silhouette_score(X, labels)[source]
To be added soon | {"url":"https://www.florianfelice.com/statinf/stats/unsupervised.html","timestamp":"2024-11-13T08:14:31Z","content_type":"text/html","content_length":"20565","record_id":"<urn:uuid:9a96b572-8db9-49b6-9abf-da6d712f3dfe>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00550.warc.gz"} |
Pack nan_numerics_prime -- prolog/nan_numerics_prime_lgc.pl
To allow for maximum performance, module prime_lgc provides unsafe public (not exported) predicates that user code can call directly instead of calling the safe predicates exported by module prime.
For info on the implementation, see library(nan_numerics_prime).
NOTE: Predicates in this module are unsafe, i.e. do not validate input arguments and are not steadfast.
- Julio P. Di Egidio
- 1.2.5-beta
See also
- library(nan_numerics_prime)
- 2016 Julio P. Di Egidio
- GNU GPLv3
To be done
- Integrate isqrt function from GMP?
True if N is a prime number.
True if N is a composite number with P its smallest prime divisor.
True if N is a composite number with P its greatest prime divisor.
PFs is the list of all prime factors of N in ascending order of the prime divisors.
Elements of PFs are of the form P^F with P the prime divisor and F the corresponding power.
If N is equal to 1 or if N is a prime number, PFs is [N^1].
Generates in ascending order all prime numbers P greater than or equal to Inf, and less than or equal to Sup in the variant with arity 3. Fails if the prime to the left of Sup is less than the
prime to the right of Inf.
Generates in ascending order all prime numbers P starting from L, and up to H in the variant with arity 3. Fails if H is less than L.
Generates in descending order all prime numbers P less than or equal to Sup, and greater than or equal to Inf in the variant with arity 3. Fails if Sup is equal to 1 or if the prime to the left
of Sup is less than the prime to the right of Inf.
Generates in descending order all prime numbers P starting from H, and down to L in the variant with arity 3. Fails if H is less than L.
P is the smallest prime number greater than N.
P is the smallest prime number greater than P0.
P is the greatest prime number less than N. Fails if N is less than or equal to 2.
P is the greatest prime number less than P0. Fails if P is equal to 2.
P is the smallest prime number greater than or equal to N.
P is the greatest prime number less than or equal to N. Fails if N is equal to 1. | {"url":"https://us.swi-prolog.org/pack/file_details/nan_numerics_prime/prolog/nan_numerics_prime_lgc.pl","timestamp":"2024-11-04T15:52:21Z","content_type":"text/html","content_length":"17722","record_id":"<urn:uuid:9dcbb06f-2199-46dd-9dce-36fa50994382>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00296.warc.gz"} |
ShortScience.org - Making Science Accessible!
Summary by ameroyer 5 years ago
Residual Networks (ResNets) have greatly advanced the state-of-the-art in Deep Learning by making it possible to train much deeper networks via the addition of skip connections. However, in order to compute gradients during the backpropagation pass, all the units' activations have to be stored during the feed-forward pass, leading to high memory requirements for these very deep networks.
Instead, the authors propose a **reversible architecture** based on ResNets, in which activations at one layer can be computed from the ones of the next. Leveraging this invertibility property, they design a more efficient implementation of backpropagation, effectively trading compute power for memory storage.
* **Pros (+): ** The change does not negatively impact model accuracy (for equivalent number of model parameters) and it only requires a small change in the backpropagation algorithm.
* **Cons (-): ** Increased number of parameters, thus need to change the unit depth to match the "equivalent" ResNet
# Proposed Architecture
## RevNet
This paper proposes to incorporate idea from previous reversible architectures, such as NICE [1], into a standard ResNet. The resulting model is called **RevNet** and is composed of reversible blocks, inspired from *additive coupling* [1, 2]:
\texttt{RevNet Block} & \texttt{Inverse Transformation}\\
\mathbf{input }\ x & \mathbf{input }\ y \\
x_1, x_2 = \mbox{split}(x) & y1, y2 = \mbox{split}(y)\\
y_1 = x_1 + \mathcal{F}(x_2) & x_2 = y_2 - \mathcal{G}(y_1) \\
y_2 = x_2 + \mathcal{G}(y_1) & x_1 = y_1 - \mathcal{F}(x_2)\\
\mathbf{output}\ y = (y_1, y_2) & \mathbf{output}\ x = (x_1, x_2)
where $\mathcal F$ and $\mathcal G$ are residual functions, composed of sequences of convolutions, ReLU and Batch Normalization layers, analoguous to the ones in a standard ResNet block, although operations in the reversible blocks need to have a stride of 1 to avoid information loss and preserve invertibility. Finally, for the `split` operation, the authors consider spliting the input Tensor across the channel dimension as in [1, 2].
Similarly to ResNet, the final RevNet architecture is composed of these invertible residual blocks, as well as non-reversible subsampling operations (e.g., pooling) for which activations have to be stored. However the number of such operations is much smaller than the number of residual blocks in a typical ResNet architecture.
## Backpropagation
### Standard
The backpropagaton algorithm is derived from the chain rule and is used to compute the total gradients of the loss with respect to the parameters in a neural network: given a loss function $L$, we want to compute the gradients of $L$ with respect to the parameters of each layer, indexed by $n \in [1, N]$, i.e., the quantities $ \overline{\theta_{n}} = \partial L /\ \partial \theta_n$. (where $\forall x, \bar{x} = \partial L / \partial x$).
We roughly summarize the algorithm in the left column of **Table 1**: In order to compute the gradients for the $n$-th block, backpropagation requires the input and output activation of this block, $y_{n - 1}$ and $y_{n}$, which have been stored, and the derivative of the loss respectively to the output, $\overline{y_{n}}$, which has been computed in the backpropagation iteration of the upper layer; Hence the name backpropagation
### RevNet
Since activations are not stored in RevNet, the algorithm needs to be slightly modified, which we describe in the right column of **Table 1**. In summary, we first need to recover the input activations of the RevNet block using its invertibility. These activations will be propagated to the earlier layers for further backpropagation. Secondly, we need to compute the gradients of the loss with respect to the inputs, i.e. $\overline{y_{n - 1}} = (\overline{y_{n -1, 1}}, \overline{y_{n - 1, 2}})$, using the fact that:
\overline{y_{n - 1, i}} = \overline{y_{n, 1}}\ \frac{\partial y_{n, 1}}{y_{n - 1, i}} + \overline{y_{n, 2}}\ \frac{\partial y_{n, 2}}{y_{n - 1, i}}
Once again, this result will be propagated further down the network.
Finally, once we have computed both these quantities we can obtain the gradients with respect to the parameters of this block, $\theta_n$.
& \mathbf{ResNet} & \mathbf{RevNet} \\
\mathbf{Block} & y_{n} = y_{n - 1} + \mathcal F(y_{n - 1}) & y_{n - 1, 1}, y_{n - 1, 2} = \mbox{split}(y_{n - 1})\\
&& y_{n, 1} = y_{n - 1, 1} + \mathcal{F}(y_{n - 1, 2})\\
&& y_{n, 2} = y_{n - 1, 2} + \mathcal{G}(y_{n, 1})\\
&& y_{n} = (y_{n, 1}, y_{n, 2})\\
\mathbf{Params} & \theta = \theta_{\mathcal F} & \theta = (\theta_{\mathcal F}, \theta_{\mathcal G})\\
\mathbf{Backprop} & \mathbf{in:}\ y_{n - 1}, y_{n}, \overline{ y_{n}} & \mathbf{in:}\ y_{n}, \overline{y_{n }}\\
& \overline{\theta_n} =\overline{y_n} \frac{\partial y_n}{\partial \theta_n} &\texttt{# recover activations} \\
&\overline{y_{n - 1}} = \overline{y_{n}}\ \frac{\partial y_{n}}{\partial y_{n-1}} &y_{n, 1}, y_{n, 2} = \mbox{split}(y_{n}) \\
&\mathbf{out:}\ \overline{\theta_n}, \overline{y_{n -1}} & y_{n - 1, 2} = y_{n, 2} - \mathcal{G}(y_{n, 1})\\
&&y_{n - 1, 1} = y_{n, 1} - \mathcal{F}(y_{n - 1, 2})\\
&&\texttt{# gradients wrt. inputs} \\
&&\overline{y_{n -1, 1}} = \overline{y_{n, 1}} + \overline{y_{n,2}} \frac{\partial \mathcal G}{\partial y_{n,1}} \\
&&\overline{y_{n -1, 2}} = \overline{y_{n, 1}} \frac{\partial \mathcal F}{\partial y_{n,2}} + \overline{y_{n,2}} \left(1 + \frac{\partial \mathcal F}{\partial y_{n,2}} \frac{\partial \mathcal G}{\partial y_{n,1}} \right) \\
&&\texttt{ gradients wrt. parameters} \\
&&\overline{\theta_{n, \mathcal G}} = \overline{y_{n, 2}} \frac{\partial \mathcal G}{\partial \theta_{n, \mathcal G}}\\
&&\overline{\theta_{n, \mathcal F}} = \overline{y_{n,1}} \frac{\partial F}{\partial \theta_{n, \mathcal F}} + \overline{y_{n, 2}} \frac{\partial F}{\partial \theta_{n, \mathcal F}} \frac{\partial \mathcal G}{\partial y_{n,1}}\\
&&\mathbf{out:}\ \overline{\theta_{n}}, \overline{y_{n -1}}, y_{n - 1}\\
**Table 1:** Backpropagation in the standard case and for Reversible blocks
## Experiments
** Computational Efficiency.** RevNets trade off memory requirements, by avoiding storing activations, against computations. Compared to other methods that focus on improving memory requirements in deep networks, RevNet provides the best trade-off: no activations have to be stored, the spatial complexity is $O(1)$. For the computation complexity, it is linear in the number of layers, i.e. $O(L)$.
One small disadvantage is that RevNets introduces additional parameters, as each block is composed of two residuals, $\mathcal F$ and $\mathcal G$, and their number of channels is also halved as the input is first split into two.
**Results.** In the experiments section, the author compare ResNet architectures to their RevNets "counterparts": they build a RevNet with roughly the same number of parameters by halving the number of residual units and doubling the number of channels.
Interestingly, RevNets achieve **similar performances** to their ResNet counterparts, both in terms of final accuracy, and in terms of training dynamics. The authors also analyze the impact of floating errors that might occur when reconstructing activations rather than storing them, however it appears these errors are of small magnitude and do not seem to negatively impact the model.
To summarize, reversible networks seems like a very promising direction to efficiently train very deep networks with memory budget constraints.
## References
* [1] NICE: Non-linear Independent Components Estimation, Dinh et al., ICLR 2015
* [2] Density estimation using Real NVP, Dinh et al., ICLR 2017 | {"url":"https://shortscience.org/?key=journals/ml&year=2003","timestamp":"2024-11-10T12:35:21Z","content_type":"text/html","content_length":"83793","record_id":"<urn:uuid:2e0009f9-533a-4662-88cb-e246d6a47cf5>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00405.warc.gz"} |
Why Do People Get Payday Loans? Here’s How it Breaks Down By Income, Age and Location - Priceonomics
This post is from LendUp, a Priceonomics Data Studio customer. Does your company have interesting data? Become a Priceonomics customer.
With unemployment at a record high and the CARES Act expiring without additional funding, a record number of Americans are experiencing financial difficulties related to the Coronavirus pandemic.
At LendUp, we provide loans to people to cover unexpected expenses and when they need the money fast. These types of loans are often called payday loans, and they’re typically the only type of loan
available to Americans with lower incomes.
Because of our years of underwriting loans and working with our customers, we know a lot about reasons why lower-income Americans need to get these kinds of loans. In this analysis, we’ll review the
data on the reasons why Americans turn to payday loans and how it varies by age, income and geographic location.
We found that for the most part Americans use payday loans for essential expenses rather than entertainment or paying back other debt. With many Americans financially struggling because of the
pandemic and the expiration of government stimulus, one might expect that this struggle to pay expenses may become more intense.As part of our loan application process, we ask borrowers to state the
reason they are seeking a loan. For this analysis, we reviewed loans from 2017 to 2020 to see the most common reasons. The chart below shows the most common reasons given, split by percentage of
LendUp loan recipients:
Outside of the catchall bucket of “Other”, the most common reason for getting a payday loan is to cover car expenses. For most Americans, a car is essential for getting to work and unexpected car
troubles can jeopardize one’s employment as well as disrupt everyday life. After that, family & child-related expenses is the second most common reason for a payday loan.
More discretionary expenses like travel and entertainment make up just 6.6% of payday loans combined. Just 2.3% of payday loans are used to repay other loans, a practice that can leave borrowers with
revolving debt that can be difficult to escape. Healthcare expenses make up 4.4% of payday loans (please note that in our survey methodology of loan recipients healthcare can also include veterinary
How do the reasons for getting a payday loan vary by one’s income? The chart below shows the percentage of loans by reason for each income group of LendUp loan recipients:
Higher-income recipients (earning over $110K per year) are more likely to get loans for healthcare expenses, but least likely for car expenses. Lower-income (earning less than $50K per year)
recipients are most likely to get loans for repaying another loan and least likely to use a loan for healthcare expenses. Across all income groups, the use of payday loans for discretionary expenses
is very low and the lowest income group is the least likely to use a payday loan for travel.
Next, let’s look how the reason for getting a payday loan varies by age. The following chart shows percentage of payday loans chosen by reason for each age cohort:
Young people (under age 25) are three times more likely than older people (age 55+) to use a payday loan for entertainment. Young people are also much more likely to use payday loans for travel or
repaying other loans. Not surprisingly, those in the middle age cohorts are most likely to spend payday loans on expenses related to children and family. Older payday loan recipients are most likely
to have to use the funds for healthcare-related expenses or car troubles.
Lastly, is there any geographical difference in the uses of payday loans? The final chart shows the breakdown of loan reason in the thirteen states LendUp has distributed loans.
Minnesota borrowers are most likely to use a payday loan for car expenses. California and Wyoming are most likely to use loans for entertainment. Illinois recipients are most likely to use the funds
for family and child-related expenses. Wyoming residents are most likely to need a payday loan for healthcare. Oregon borrowers are most likely to use payday loans to repay other loans and Texas
borrowers are most likely to use payday loans for travel.With unprecedented economic uncertainty, many Americans have lost their jobs and still need to pay their bills and unexpected expenses. In
this analysis, we’ve shown that by and large, most payday loan recipients use the funds for essential expenses, though younger recipients are most likely to use the debt for things like travel,
entertainment or servicing other loans. For the most part, however, people get payday loans to cover expenses that need to be paid urgently.
Note: If you’re a company that wants to work with Priceonomics to turn your data into great stories, learn more about the Priceonomics Data Studio. | {"url":"https://priceonomics.com/why-do-people-get-payday-loans-heres-how-it-breaks/","timestamp":"2024-11-09T17:24:32Z","content_type":"text/html","content_length":"54410","record_id":"<urn:uuid:a9d9429f-952a-40c0-83c2-1ee46866ebf2>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00870.warc.gz"} |
A cosmic strings network present in our Universe should be incessantly emitting gravitational waves. An absolute lower limit of the expected signal can be found by determining the gravitational waves
created by the back bones of the network: the long cosmic strings. Disrael and Christophe, in collaboration with François R. Bouchet, have computed this irreducible contribution using new numerical
simulations of Nambu-Goto cosmic strings.
In Ref. [1], we have run new simulations of Nambu-Goto cosmic strings evolving during the radiation, the transition, and the matter eras to compute the unequal time correlators of the anisotropic
stress tensor associated with the long strings. The following figure shows a snapshot of one of these simulation, the long strings have been represented in white whereas all the other objects are
loops, see Ref. [2] and this post.
This correlator sources the gravitational waves and it allows us to solve for their creation, and propagation, all along the Universe history. By using the Green’s function method we can then predict
the strain, \(k^2 \mathcal{P}_h\), and the energy density parameter \(\Omega_{\mathrm{GW}}^{\mathrm{mat}}\) of the gravitational waves that can be measured today. Their power spectra are represented
below as a function of the wavenumber \(k\).
In these figures, the exact numerical result is represented in black while the blue and red curves show some semi-analytical approximations that we had proposed in a previous paper, calibrated using
the amplitude found with the simulations, see Ref. [3]. The only significant deviations show up around \(k/\mathcal{H}_0 \simeq 100\), which corresponds to the transition between the radiation and
matter era. An interesting point to notice is that most of strain signal is actually generated by the long cosmic strings in the matter era, i.e., close to us.
As we discuss in Ref. [1], this signal is quite small, but reachable by the LISA satellites. These ones will be sensitive to long strings that are undetectable today in the Cosmic Microwave | {"url":"https://curl.group/news/2022/05/30/2205.04349.html","timestamp":"2024-11-04T08:51:56Z","content_type":"text/html","content_length":"10533","record_id":"<urn:uuid:c7b8cfec-cbc3-41cc-8700-d3b726e79e72>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00819.warc.gz"} |
How to Find the Factorial of an integer with NumPy in Python
A factorial of an integer is the result of multiplying all the integers less than or equal to it in descending order.
To get the factorial of 4 four (24) we would use the following formula:
4! = 4 x 3 x 2 x 1 = 24
In Python, we can get factorials by using the numpy package. Import it then use the np.math.factorial() function, passing the number to calculate the factorial of as the first argument.
import numpy as np
result = np.math.factorial(4) | {"url":"https://www.skillsugar.com/how-to-find-the-factorial-of-an-integer-with-numpy-in-python","timestamp":"2024-11-03T04:03:15Z","content_type":"text/html","content_length":"35997","record_id":"<urn:uuid:a20f28eb-1551-4f47-af76-9515af314d0b>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00867.warc.gz"} |
How Does math websites Work?
Kids World Fun, a free on-line useful resource dedicated to simplify and ease the educational course of. A robust data and proficiency within the fundamental arithmetic is like an everlasting base to
a colossal structure. Learn fifth grade math aligned to the Eureka Math/EngageNY curriculum—arithmetic with fractions and decimals, quantity issues, unit conversion, graphing factors, and more. A
transient introduction to chose trendy topics could also be added if time permits. As for the program, the group is incorporating more and more fun parts into the training to keep the youngsters
• Covers linear algebra and multivariable differential calculus in a unified manner alongside purposes related to many quantitative fields.
• Math is a lot extra necessary than just getting high check scores, and it’s applicable to future careers even when youngsters do not want to be engineers or accountants.
• This characteristic is helpful for K-12 math learners, because it helps them construct confidence and motivation while bettering their math expertise.
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COUNTA alternative in Smartsheet
I'm trying to recreate/transfer a salary/reward solution in Excel over to Smartsheet and two of the columns have a rather complicated formula, which has a function I don't believe is available in
Smartsheet (COUNTA). When I've tried to replicate it I get an error message (UNPARSEABLE) The formulas are:
=IF(SUM(AR13:AT13)>0,IFS(OR(AU13="",AV13=""),"Please enter review reason and detailed explanation",COUNTA(AR13:AT13)>1,"Please enter one type of increase recommendation only",AR13<>"",ROUND
=IF(SUM(AR13:AT13)>0,IFS(OR(AU13="",AV13=""),"Please enter review reason and detailed explanation",COUNTA(AR13:AT13)>1,"Please enter one type of increase recommendation only",AW13<>"",AW13/V13),0).
I'm wondering if anyone knows what other function I could use to get these work in Smartsheet. I did try COUNTIF, but it didn't work either.
For reference, this is what they look like Excel - they should bring up a salary increase numerically and a percentage respectively.
Best Answers
• You are correct in that COUNTA is not present in Smartsheet. You can use COUNTIF, as you mentioned, though I prefer to use COUNTIFS instead. There is also no IFS, so to cover multiple cases, we
must nest several IF statements. Will you be setting these up as column formulas? If so, here is the equivalent of the first formula. Since Smartsheet references column names instead of
alphabetic references, you will have to replace the column references with your actual column names.
=IF(SUM([AR]@row:[AT]@row) > 0, IF(OR(ISBLANK([AU]@row), ISBLANK([AV]@row)), "Please enter review reason and detailed explanation", IF(COUNTIFS([AR]@row:[AT]@row, NOT(ISBLANK(@cell))) > 1,
"Please enter one type of increase recommendation only", IF(NOT(ISBLANK([AR]@row)), ROUND([AR]@row * [V]@row, 0), IF(NOT(ISBLANK([AS]@row)), ROUND([AS]@row * [U]@row, 0), IF(NOT(ISBLANK([AT]
@row)), ROUND(([AT]@row - [AL]@row) * [U]@row, 0)))))), 0)
• You are dead on with the concept... you only missed it with parentheses.
=IF(SUM([AR]@row:[AT]@row) > 0, IF(OR(ISBLANK([AU]@row), ISBLANK([AV]@row)), "Please enter review reason and detailed explanation", IF( COUNTIFS([AR]@row:[AT]@row, NOT(ISBLANK(@cell))) > 1,
"Please enter one type of increase recommendation only", IF(NOT(ISBLANK([AW]@row)), [AW]@row / [V]@row))), 0)
• You are correct in that COUNTA is not present in Smartsheet. You can use COUNTIF, as you mentioned, though I prefer to use COUNTIFS instead. There is also no IFS, so to cover multiple cases, we
must nest several IF statements. Will you be setting these up as column formulas? If so, here is the equivalent of the first formula. Since Smartsheet references column names instead of
alphabetic references, you will have to replace the column references with your actual column names.
=IF(SUM([AR]@row:[AT]@row) > 0, IF(OR(ISBLANK([AU]@row), ISBLANK([AV]@row)), "Please enter review reason and detailed explanation", IF(COUNTIFS([AR]@row:[AT]@row, NOT(ISBLANK(@cell))) > 1,
"Please enter one type of increase recommendation only", IF(NOT(ISBLANK([AR]@row)), ROUND([AR]@row * [V]@row, 0), IF(NOT(ISBLANK([AS]@row)), ROUND([AS]@row * [U]@row, 0), IF(NOT(ISBLANK([AT]
@row)), ROUND(([AT]@row - [AL]@row) * [U]@row, 0)))))), 0)
• Thanks for coming back to me so quickly on this and helping out Carson - much appreciated. I will give this a go later and come back to you to advise on progress.
• Hi Carson that's worked perfectly when I've added in the column headers for the first formula. Thank you so much!
I'm trying to apply the same principle for the second formula, but it's advising me INCORRECT ARGUMENT SET. I've done it as follows (obviously changing the alphabetic references for the column
headers in Smartsheet):
=IF(SUM([AR]@row:[AT]@row) > 0, IF(OR(ISBLANK([AU]@row), ISBLANK([AV]@row)), "Please enter review reason and detailed explanation", IF(COUNTIFS([AR]@row:[AT]@row, NOT(ISBLANK(@cell))) > 1,
"Please enter one type of increase recommendation only", IF(NOT(ISBLANK([AW]@row)), [AW]@row)/[V]@row), 0)
I may have over simplified it by removing some key elements?
• You are dead on with the concept... you only missed it with parentheses.
=IF(SUM([AR]@row:[AT]@row) > 0, IF(OR(ISBLANK([AU]@row), ISBLANK([AV]@row)), "Please enter review reason and detailed explanation", IF( COUNTIFS([AR]@row:[AT]@row, NOT(ISBLANK(@cell))) > 1,
"Please enter one type of increase recommendation only", IF(NOT(ISBLANK([AW]@row)), [AW]@row / [V]@row))), 0)
• Carson you're a legend - thank you so much for your assistance with this. I really appreciate it. This has worked perfectly!
• I'm glad it worked for you! 👍
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/110708/counta-alternative-in-smartsheet","timestamp":"2024-11-03T03:43:47Z","content_type":"text/html","content_length":"456141","record_id":"<urn:uuid:30e5f350-12a5-4513-9730-0ca2d6622b37>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00809.warc.gz"} |
Probing chiral edge dynamics and bulk topology of a synthetic Hall system
The Bose-Einstein condensate team has studied the dynamics of dysprosium atoms manipulated by lasers, aiming to simulate the behavior of electrons in a magnetic field, which feature a quantum Hall
effect. We used two-photon transitions to couple the atomic motion and the internal spin state, which plays the role of a `synthetic’ dimension. As expected in Hall systems, we found a freezing of
motion in the system bulk, and a chiral motion on the edges. We also measured the response of the atoms to a force, observing a quantized motion along the direction perpendicular to the force. The
robustness of this behavior illustrates a property of non-trivial topology of the quantum states. This setting will be the frame for future studies of topological phases in interacting atomic
Link to the complete article : https://www.nature.com/articles/s41567-020-0942-5
Scheme of the Hall response measurement. We measured the atom velocity induced by a transverse potential. In the bulk of the system, the Hall currents are consistent with a quantized hall response. | {"url":"https://www.lkb.upmc.fr/en/uncategorised-en/probing-chiral-edge-dynamics-and-bulk-topology-of-a-synthetic-hall-system/","timestamp":"2024-11-09T16:06:20Z","content_type":"text/html","content_length":"152184","record_id":"<urn:uuid:8eae9165-32b4-48a8-ad57-01359ff62490>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00542.warc.gz"} |
Comparing Different Line Segments | Difference of two line segments
Comparing Different Line Segments
We will discuss here about comparing different line segments.
A line segment is a part of a line. It has two end points and has a definite length, no breadth and no thickness. The length of a line segment is a distance that can be measured in metres,
centimetres, millimetres etc.
10 mm (millimetres) = 1 cm (centimetre)
10 cm (centimetre) = 1 dm (decimetre)
10 dm (decimetres) = 1 m (metre)
10 m (metre) = 1 dam (decametre)
10 dam (decametres) = 1 hm (hectometre)
10 hm (hectometres) = 1 kilometre (km)
Comparing Two Line Segments:
Comparing two line segments implies establishing a relationship between their lengths whether one line segment is greater than or equal to or less than the other line segment. Comparison of line
segment can be done in several ways.
(i) Comparison by Observation.
Line segments can be compared easily by observation by placing a line segment below another segment and judging the length visually.
Working Rules for Comparison of Two Line Segments by Observation:
Step I: Draw two line segments of different lengths.
Step II: Label them. Place one line segment below another line segment.
Step III: By observation compare the two line segments.
Compare the following two line segments by observation:
\(\overline{PQ}\) is longer than \(\overline{AB}\) , as PQ extends farther to the right of \(\overline{AB}\).
(ii) Comparison by Tracing.
One cannot always be sure about visual judgement. By looking at the segment, the difference in length of two line segments may not be obvious. We require better methods of comparing line segments.
One such method is comparison by tracing and superimposing.
Working Rules for Comparison of Two line Segments by Tracing:
Step I: Use a tracing paper to trace the line segment, say overline \(\overline{PQ}\).
Step II: Place the traced line segment \(\overline{PQ}\) the other line segment, say \(\overline{AB}\).
Step III: If \(\overline{PQ}\) on the tracing paper covers \(\overline{AB}\) behind it, then \(\overline{PQ}\) is longer than \(\overline{AB}\), otherwise \(\overline{AB}\) is longer than \(\overline
Compare \(\overline{PQ}\) and \(\overline{AB}\) by tracing.
To compare \(\overline{PQ}\) and \(\overline{AB}\), use a tracing paper. Trace \(\overline{AB}\) and place it on the line segment \(\overline{PQ}\). On tracing, it is known that \(\overline{PQ}\) is
greater than \(\overline{AB}\).
(iii) Comparison Using a Divider.
Different line segments can be of different lengths. We can compare the lengths of the different line segments with the help of a divider.
In geometry-box, there is an instrument called divider. A divider has two arms hinged together. Each arm has a pointed metallic end. The distance between the two ends can be increased or decreased as
1. Let us now compare any two line segments, say \(\overline{PQ}\) and \(\overline{AB}\) using a divider.
Working Rules for Comparison of Two line Segments Using Divider:
Step I: Place the end-point of one arm of the divider at A.
Step II: Open the divider in such a way that the end-point of the other arm reaches at the other point B.
Step III: Lift the divider and without disturbing its opening, place the end-point of one arm at P.
Now, the three following cases may arise:
Case (i): The second arm of the divider falls on Q. In this case, we can say that lengths of the line segments AB and PQ are equal and write AB = PQ.
Case (ii): The second arm falls at a point R between P and Q. In this case, we conclude that the line segment AB is shorter than the line segment PQ and write AB < PQ.
Case (iii): The second arm falls at a point M outside PQ. In this case, we conclude that the line segment AB is longer than the line segment PQ and write AB > PQ.
2. Let us compare two line segments MN and ST.
I. First, place one point of the divider on the end point M of the line segment MN. Now we open the pair of divider in such a way that the other point of divider falls on the other end point N.
II. Then we lift the divider very carefully so that its arms are not disturbed.
III. Now place one end point of the pair of divider on the end point S of the other line segment ST and let the other end point of the pair of divider rest on the line segment ST as shown in figure.
We see that this end point falls before the other end point T of the line segment ST. This shows that the line segment MN is shorter than the line segment ST. If the end point of the divider falls
beyond the point T, MN is longer than ST.
1. Compare the two line segments AB and MN using a divider.
Place the end-point of one arm of the divider at A. Open the divider in such a way that the end-point of the other arm reaches at the other point B. Lift the divider and without disturbing its
opening, place the end-point of one arm at M and the other at N.
On comparing, it is being noticed that AB < MN.
From Comparing Different Line Segments to HOME PAGE
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PosteriorChecks is a supplemental function that returns a list with two components. Following is a summary of popular uses of the PosteriorChecks function.
First (and only for MCMC users), the user may be considering the current MCMC algorithm versus others. In this case, the PosteriorChecks function is often used to find the two MCMC chains with the
highest IAT, and these chains are studied for non-randomness with a joint trace plot, via the joint.density.plot function. The best algorithm has the chains with the highest independent samples per
minute (ISM).
Posterior correlation may be studied between model updates as well as after a model seems to have converged. While frequentists consider multicollinear predictor variables, Bayesians tend to consider
posterior correlation of the parameters. Models with multicollinear parameters take more iterations to converge. Hierarchical models often have high posterior correlations. Posterior correlation
often contributes to a lower effective sample size (ESS). Common remedies include transforming the predictors, re-parameterization to reduce posterior correlation, using WIPs (Weakly-Informative
Priors), or selecting a different numerical approximation algorithm. An example of re-parameterization is to constrain related parameters to sum to zero. Another approach is to specify the parameters
according to a multivariate distribution that is assisted by estimating a covariance matrix. Some algorithms are more robust to posterior correlation than others. For example, posterior correlation
should generally be less problematic for twalk than AMWG in LaplacesDemon. Posterior correlation may be plotted with the plotMatrix function, and may be useful for blocking parameters. For more
information on blockwise sampling, see the Blocks function.
After a user is convinced of the applicability of the current MCMC algorithm, and that the chains have converged, PosteriorChecks is often used to identify multimodal marginal posterior distributions
for further study or model re-specification.
Although many marginal posterior distributions appear normally distributed, there is no such assumption. Nonetheless, a marginal posterior distribution tends to be distributed the same as its prior
distribution. If a parameter has a prior specified with a Laplace distribution, then the marginal posterior distribution tends also to be Laplace-distributed. In the common case of normality,
kurtosis and skewness may be used to identify discrepancies between the prior and posterior, and perhaps this should be called a `prior-posterior check'.
Lastly, parameter importance may be considered, in which case it is recommended to be considered simultaneously with variable importance from the Importance function. | {"url":"https://www.rdocumentation.org/packages/LaplacesDemon/versions/16.1.6/topics/PosteriorChecks","timestamp":"2024-11-01T22:50:41Z","content_type":"text/html","content_length":"90348","record_id":"<urn:uuid:bd3826ef-ad37-4e2d-ab29-a04c9b8799ba>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00279.warc.gz"} |
Algebraic expressions
Letβ s define some of the keywords when using algebraic notation:
A variable is a symbol (often a letter) that is used to represent an unknown quantity.
\[E.g. x\quad or \quad y\quad or\quad a\quad etc.\]
Variables can also have exponents (be raised to a certain power).
\[E.g. x^{2}\quad or \quad y^{3}\]
A coefficient is the value that is before a variable. It tells us how many lots of the variable there is.
\[ \begin{aligned} E.g. 5x&=x+x+x+x+x\\ &=5\times x \end{aligned}\]
Here 5 is the coefficient and x is the variable.
A term is a number by itself, a variable by itself, or a combination of numbers and letters. If the term includes a variable it is called an algebraic term.
\[E.g. 2\quad or\quad 5xy\quad or\quad 12x^{2}\quad or \quad 12xy\quad etc.\]
An expression that contains one term is called a monomial.
A polynomial expression consists of two or more algebraic terms.
(2 x+2)(x-3)
2\left(x^{2}+0.5 x-3\right)
(2 x-3)(x+2)
(2 x+5)(2 x-5) \\
(2 x-5)(2 x-5) \\
(4 x+5)(x-5) \\
(3 x+20)(x+5) | {"url":"https://thirdspacelearning.com/gcse-maths/algebra/algebraic-expressions/","timestamp":"2024-11-07T15:58:40Z","content_type":"text/html","content_length":"249578","record_id":"<urn:uuid:c06fc27d-379f-4c39-bcb6-b359d6824bc4>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00022.warc.gz"} |
Return value from another column for all Rows and set column to "column formula"
This formula is producing syntax error when I try to make the column a "column formula". Is there a workaround?
As we add new Rows (copied from different sheet), I want all values in Client Name column to equal Task column Row 1.
Best Answer
• @Ami Veltrie If you want all rows that are indented under a given row (a row that's completely outdented) to have that row's client name, no matter how many levels of indentation are occurring
under that given row, then try this:
=IFERROR(INDEX(ANCESTORS(Task@row), 1), "")
Here's an example of the results:
• @Ami Veltrie In the "Client Name" column, use the formula:
Then you'll be able to make the formula a column formula.
Hope this helps!
• @Kelly P. Thanks for the response! I tried this formula:
However, it's not working because we have more than two hierarchy levels. In other words it's not working for anything lower than level 2.
• @Ami Veltrie If you want all rows that are indented under a given row (a row that's completely outdented) to have that row's client name, no matter how many levels of indentation are occurring
under that given row, then try this:
=IFERROR(INDEX(ANCESTORS(Task@row), 1), "")
Here's an example of the results:
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/130644/return-value-from-another-column-for-all-rows-and-set-column-to-column-formula","timestamp":"2024-11-04T02:00:29Z","content_type":"text/html","content_length":"448394","record_id":"<urn:uuid:a9ba8dd1-902d-4b3d-b3fb-f3e18ebfdbbe>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00769.warc.gz"} |
Python - Check if Set is a subset of another Set - Data Science Parichay
In this tutorial, we will look at how to check whether a set in Python is a subset of another set with the help of some examples.
What is a subset?
Let’s say we have two sets, A and B. Now, if all the elements of set A are present in set B then A is said to be a subset of B. Here’s an example –
You can see that all the elements of set A are present in set B, thus, we can say that set A is a subset of set B. However, the converse is not true, B is not a subset of A since there are elements
in B that are not in A.
Check if a set is a subset of another set in Python
The Python set data structure comes with a number of built-in functions to accomplish common set operations like union, intersection, difference, etc. You can use the Python set issubset() function
to check whether a set is a subset of another set. The following is the syntax:
# check if a is a subset of b
We call the issubset() function from set a and pass the set b as an argument to check whether set a is a subset of set b. It returns a boolean value. Let’s look at an example.
# create two sets
a = {1, 2, 3}
b = {1, 2, 3, 4}
# check if a is subset of b
We get True as the output since a is in fact, a subset of set b.
Let’s look at another example.
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# create two sets
a = {1, 2, 3}
b = {1, 3, 4, 5}
# check if a is subset of b
Here we get False as the output because not all elements of set a are present in set b (2 from set a is not present in b). Thus, set a is not a subset of set b.
Alternatively, you can use the <= operator to check if a set is a subset of another set. For example,
# create two sets
a = {1, 2, 3}
b = {1, 2, 3, 4}
# check if a is subset of b
a <= b
We get the same result as we did with the set issubset() function.
Special Cases
There are also some special cases to keep in mind when checking for subsets.
• Every set is a subset of itself.
# create a set
a = {1, 2, 3}
# check if a is subset of itself
• Empty set is subset of every set.
# create a set
a = set()
b = {1, 2, 3}
# check if a is subset of b
You might also be interested in –
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SSAT math - The SSAT
SSAT math
The SSAT
This section will provide you with a review of all the math that you need to know to do well on the SSAT. When you get started, you may feel that the material is too easy. Don’t worry. The SSAT
measures your basic math skills, so although you may feel a little frustrated reviewing things you have already learned, basic review is the best way to improve your score.
We recommend that you work through these math sections in order, reading each section and then doing each set of drills. If you have trouble with one section, mark the page so you can come back later
to go over it again. Keep in mind that you shouldn’t breeze over pages or sections just because they look familiar. Take the time to read over all of the Math sections, so you’ll be sure to know all
the math you’ll need!
Lose Your Calculator!
You will not be allowed to use a calculator on the SSAT. If you have developed a habit of reaching for your calculator whenever you need to add or multiply a couple of numbers, follow our advice: put
your calculator away now and take it out again after the test is behind you. Do your math homework assignments without it, and complete the practice sections in this book without it. Trust us, you’ll
be glad you did.
Write It Down
Do not try to do math in your head. You are allowed to write in your test booklet. You should write in your test booklet. Even when you are just adding a few numbers together, write them down and do
the work on paper. Writing things down will not only help eliminate careless errors but also give you something to refer to if you need to check over your work.
One Pass, Two Pass
Within any Math section you will find three types of questions:
·Those you can answer easily without spending too much time
·Those that, if you had all the time in the world, you could do
·Some questions that you have absolutely no idea how to tackle
When you work on a Math section, start out with the first question. If you think you can do it without too much trouble, go ahead. If not, save it for later. Move on to the second question and decide
whether or not to do that one. In general, the questions in each Math section are in a very rough order of difficulty. This means that earlier questions tend to be somewhat easier than later ones.
You will likely find yourself answering more questions toward the beginning of the sections and leaving more questions blank toward the end.
Don’t Get Stuck
Make sure you don’t spend too much time working on one tough question; there might be easier questions left in the section.
Once you’ve made it all the way through the section, working slowly and carefully to do all the questions that come easily to you, go back and try some of the ones that you think you can do but will
take a little longer. You should pace yourself so that time will run out while you’re working on the second pass through the section. By working this way, you’ll know that you answered all the
questions that were easy for you. Using a two-pass system is a smart test-taking strategy.
Sometimes accuracy is important. Sometimes it isn’t.
Which of the following fractions is less than
Before making any kind of calculation, think about this question. It asks you to find a fraction smaller than
Start simple:
Some Things Are Easier Than They Seem
Guesstimating, or finding approximate answers, can help you eliminate wrong answers and save lots of time.
Look at (C).
Here’s another good example.
A group of three men buys a one-dollar raffle ticket that wins $400. If the one dollar that they paid for the ticket is subtracted and the remainder of the prize money is divided equally among the
men, how much will each man receive?
(A) $62.50
(B) $75.00
This isn’t a terribly difficult question. To solve it mathematically, you would take $400, subtract $1, and then divide the remainder by three. But by using a little bit of logic, you don’t have to
do any of that.
The raffle ticket won $400. If there were four men, each one would have won about $100 (actually slightly less because the problem tells you to subtract the $1 price of the ticket, but you get the
idea). So far so good?
However, there weren’t four men; there were only three. This means fewer men among whom to divide the winnings, so each one should get more than $100, right? Look at the choices. Eliminate (A), (B),
and (C).
Two choices left. Choice (E) is $200, half of the amount of the winning ticket. If there were three men, could each one get half? Unfortunately not. Eliminate (E). What’s left? The right answer!
Guesstimating also works very well with some geometry questions, but just to give you something you can look forward to, we’ll save that for the Geometry review.
In Chapter 2, Fundamental Math Skills for the SSAT & ISEE, we reviewed the concepts that will be tested on the SSAT tests. However, the questions in those practice drills were slightly different from
the ones that you will see on your exam. Questions on test day are going to give you five answers to choose from. And as you’ll soon see, there are many benefits to working with multiple-choice
For one, if you really mess up calculating the question, chances are your choice will not be among the ones given. Now you have a chance to go back and try that problem again more carefully. Another
benefit is that you may be able to use the information in the choices to help you solve the problems (don’t worry; we’ll tell you how soon).
A Tip About Choices
Notice that the choices are often in numerical order.
We are now going to introduce to you the type of multiple-choice questions you will see on the SSAT. Each one of the following questions will test some skill that we covered in the Fundamental Math
Skills chapter. If you don’t see how to solve the question, take a look back at Chapter 2 for help.
Math Vocabulary
1.Which of the following is the greatest even integer less than 25 ?
The first and most important thing you need to do on this—and every—problem is to read and understand the question. What important vocabulary words did you see in the question? There is “even” and
“integer.” You should always underline the important words in the questions. This way you will make sure to pay attention to them and avoid careless errors.
Now that we understand that the question is looking for an even integer, we can eliminate any answers that are not even or an integer. Cross out (B) and (D). We can also eliminate (A) because 26 is
greater than 25 and we want a number less than 25. Now all we have to do is ask which is greater—0 or 22. (C) is the right answer.
Try it again.
Set A = {All multiples of 7}
Set B = {All odd numbers}
2.All of the following are members of both set A and set B above EXCEPT
(A) 7
Did you underline the words multiples of 7 and odd ? Because all the choices are odd, you can’t eliminate any that would not be in Set B, but only (D) is not a multiple of 7. So (D) is the right
The Rules of Zero
3.x, y, and z stand for three distinct numbers, where xy = 0 and yz = 15. Which of the following must be true?
(A)y = 0
(B)x = 0
(C)z = 0
(D)xyz = 15
(E)It cannot be determined from the information above.
Remember the Rules of Zero
Zero is even. It’s neither positive nor negative, and anything multiplied by 0 = 0.
Because x times y is equal to zero, and x, y, and z are different numbers, we know that either x or y is equal to zero. If y was equal to zero, then y times z should also be equal to zero. Because it
is not, we know that it must be x that equals zero. Choice (B) is correct.
The Multiplication Table
4.Which of the following is equal to 6 × 5 × 2 ?
(A)60 ÷ 3
(B)14 × 7
(C)2 × 2 × 15
(D)12 × 10
(E)3 × 3 × 3 × 9
6 × 5 × 2 = 60 and so does 2 × 2 × 15. Choice (C) is correct.
Working with Negative Numbers
5.7 — 9 is the same as
(A) 7 — (—9)
(B) 9 — 7
(C) 7 + (—9)
(D)—7 — 9
(E)—9 — 7
Remember that subtracting a number is the same as adding its opposite. Choice (C) is correct.
Don’t Do More Work Than You Have To
When looking at answer choices, start with what’s easiest for you; work through the harder ones only when you have eliminated all the others.
Order of Operations
6.9 + 6 ÷ 2 × 3 =
(A) 7
(B) 9
Remember your PEMDAS rules? Left to right; Since this problem has no parentheses (P) or exponents (E), you can proceed to MD (multiplication and division). Finally, perform the addition. The correct
answer is (E).
Factors and Multiples
7.What is the sum of the prime factors of 42 ?
(E) 7
How do we find the prime factors? The best way is to draw a factor tree. Then we will see that the prime factors of 42 are 2, 3, and 7. Add them up and we get 12, (C).
Factors Are Small; Multiples Are Large
The factors of a number are always equal to or less than that number. The multiples of a number are always equal to or greater than that number. Be sure not to confuse the two!
8.Which of the following is less than
When comparing fractions, you have three choices. You can find a common denominator and then compare the fractions (such as when you add or subtract fractions). You can also change the fractions to
decimals. If you have memorized the fraction-to-decimal chart in Fundamentals (Chapter 2), you probably found the right answer without too much difficulty. It’s (A). Or, if you remember the Bowtie
method, you can compare answers that way too!
9.Thom’s CD collection contains 15 jazz CDs, 45 rap albums, 30 funk CDs, and 60 pop albums. What percent of Thom’s CD collection is funk?
First we need to find the fractional part that represents Thom’s funk CDs. He has 30 out of a total of 150. We can reduce
10.2^6 =
Expand 2^6 out and multiply to find that it equals 64. Choice (E) is correct.
Elementary Level
You shouldn’t expect to see exponents or roots on your tests.
Square Roots
11.The square root of 75 falls between what two integers?
(A)5 and 6
(B)6 and 7
(C)7 and 8
(D)8 and 9
(E)9 and 10
If you have trouble with this one, try using the choices and work backward. As we discussed in Fundamentals (Chapter 2), a square root is just the opposite of squaring a number. So let’s square the
choices. Then we find that 75 falls between 8^2 (64) and 9^2 (81). Choice (D) is correct.
Simple Algebraic Equations
12.11x = 121. What does x = ?
(A) 2
(B) 8
Remember, if you get stuck, use the choices and work backward. Each one provides you with a possible value for x. Start with the middle choice and replace x with it. 11 × 10 = 110. That’s too small.
Now we know that not only is (C) incorrect, but also that (A) and (B) are incorrect because they are smaller than (C). The correct choice is (D).
Solve for X
13.If 3y + 17 = 25 — y, then y =
Just as above, if you get stuck, use the choices. The correct answer is (B).
The Case of the Mysteriously Missing Sign
If there is no operation sign between a number and a variable (letter), the operation is multiplication.
Percent Algebra
14.25% of 30% of what is equal to 18 ?
(A) 1
(B) 36
If you don’t remember the math conversion table, look it up in Fundamentals (Chapter 2). You can also use the choices and work backward. Start with (C) and find out what 25% of 30% of 120 is (9). The
correct answer is (D).
Percent means “out of 100,” and the word of in a word problem tells you to multiply.
15.BCDE is a rectangle with a perimeter of 44. If the length of BC is 15, what is the area of BCDE ?
(E)It cannot be determined.
Don’t Cut Corners: Estimate Them
Don’t forget to guesstimate on geometry questions! This is a quick way to make sure you’re not making calculation errors.
From the perimeter, we can find that the sides of the rectangle are 7 and 15. So the area is 105, (A).
16.If the perimeter of this polygon is 37, what is the value of x + y ?
(A) 5
(B) 9
The sum of x and y is equal to the perimeter of the polygon minus the lengths of the sides we know. So (C) is correct.
Word Problems
17.Emily is walking to school at a rate of 3 blocks every 14 minutes. When Jeff walks at the same rate as Emily and takes the most direct route to school, he arrives in 56 minutes. How many blocks
away from school does Jeff live?
(A) 3
(B) 5
(C) 6
(D) 9
This is a proportion question because we have two sets of data that we are comparing. Set up your fractions.
We know that we must do the same thing to the top and bottom of the first fraction to get the second fraction. Notice that the denominator of the second fraction (56) is 4 times the denominator of
the first fraction (14). Therefore, the numerator of the second fraction must be 4 times the numerator of the first fraction (3).
So Jeff walks 12 blocks in 56 minutes. This makes (E) the correct answer.
18.Half of the 30 students in Mrs. Whipple’s first-grade class got sick on the bus on the way back from the zoo. Of these students,
(A) 5
This is a really gooey fraction problem. Because we’ve seen the word of, we know we have to multiply. First we need to multiply of, so we must multiply the fraction of students who ate too much
cotton candy,
19.A piece of rope is 18 inches long. It is cut into 2 unequal pieces. The longer piece is twice as long as the shorter piece. How long, in inches, is the shorter piece?
(A) 2
(B) 6
(C) 9
Again, if you are stuck for a place to start, go to the choices. Because we are looking for the length of the shorter rope, we can eliminate any choice that gives us a piece equal to or longer than
half the rope. That gets rid of (C), (D), and (E). Now if we take one of the pieces, we can subtract it from the total length of the rope to get the length of the longer piece. For (B), if 6 is the
length of the shorter piece, we can subtract that from 18 and know that the length of the longer piece must be 12. 12 is double 6, so we have the right answer.
When you are done, check your answers in Chapter 9. Don’t forget to time yourself!
Remember to time yourself during this drill!
1.The sum of five consecutive positive integers is 30. What is the square of the largest of the five positive integers?
2.How many factors does the number 24 have?
(A) 2
(B) 4
(C) 6
(D) 8
3.If 12 is a factor of a certain number, what must also be factors of that number?
(A)2 and 6 only
(B)3 and 4 only
(C)12 only
(D)1, 2, 3, 4, and 6
(E)1, 2, 3, 4, 6, and 24
4.What is the smallest number that can be added to the number 1,024 to produce a result divisible by 9 ?
5.Which of the following is a multiple of 3 ?
(A) 2
(B) 6
6.Which of the following is NOT a multiple of 6 ?
7.Which of the following is a multiple of both 3 and 5 ?
8.A company’s profit was $75,000 in 1972. In 1992, its profit was $450,000. The profit in 1992 was how many times as great as the profit in 1972 ?
(A) 2
(B) 4
(C) 6
9.Joanna owns one-third of the pieces of furniture in the apartment she shares with her friends. If there are 12 pieces of furniture in the apartment, how many pieces does Joanna own?
(A) 2
(B) 4
(C) 6
(D) 8
10.A tank of oil is one-third full. When full, the tank holds 90 gallons. How many gallons of oil are in the tank now?
11.Tigger the Cat sleeps three-fourths of every day. In a four-day period, he sleeps the equivalent of how many full days?
12.Which of the following has the greatest value?
(B) 1
(C) 6
(D) 3
14.The product of 0.34 and 1,000 is approximately
(A) 3.50
(B) 35
(C) 65
15.2.398 =
(A)2 ×
(B)2 +
(C)2 +
(E)None of the above
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
HOW DID YOU DO?
That was a good sample of some of the kinds of questions you’ll see on the SSAT. Now there are a few things to check other than your answers. Remember that taking the test involves much more than
just getting answers right. It’s also about guessing wisely, using your time well, and figuring out where you’re likely to make mistakes. Once you’ve checked to see what you’ve gotten right and
wrong, you should then consider the points that follow to improve your score.
Time and Pacing
How long did it take you to do the 15 questions? 15 minutes? It’s okay if you went a minute or two over. However, if you finished very quickly (in fewer than 10 minutes) or slowly (more than 20
minutes), look at any problems that may have affected your speed. Which questions seriously slowed you down? Did you answer some quickly but not correctly? Your answers to these questions will help
you plan which and how many questions to answer on the SSAT.
Question Recognition and Selection
Did you use your time wisely? Did you do the questions in an order that worked well for you? Which kinds of questions were the hardest for you? Remember that every question on the SSAT, whether you
know the answer right away or find the question confusing, is worth one point, and that you don’t have to answer all the questions to get a good score. In fact, because of the guessing penalty,
skipping questions can actually raise your score. So depending on your personal speed, you should concentrate most on getting as many questions you find easy or sort-of easy right as possible, and
worry about harder problems later. Keep in mind that in Math sections, the questions generally go from easiest to hardest throughout. Getting the questions you know you know the answers to right
takes time, but you know you can solve them—so give yourself that time!
POE and Guessing
Did you actively look for wrong answers to eliminate, instead of just looking for the right answer? (You should.) Did you physically cross off wrong answers to keep track of your POE? Was there a
pattern to when guessing worked (more often when you could eliminate one wrong answer, and less often when you picked simpler-looking over harder-looking answers)?
Be Careful
Did you work problems out on a separate piece of paper? Did you move too quickly or skip steps on problems you found easier? Did you always double-check what the question was asking? Often students
miss questions that they know how to do! Why? It’s simple—they work out problems in their heads or don’t read carefully. Work out every SSAT math problem on the page. Consider it a double-check
because your handwritten notes confirm what you’ve worked out in your head.
While doing the next drill, keep in mind the general test-taking techniques we’ve talked about: guessing, POE, order of difficulty, pacing, and working on the page and not in your head. At the end of
the section, check your answers. But don’t stop there: Investigate the drill thoroughly to see how and why you got your answers wrong. And check your time. You should be spending about one minute per
question on this drill. When you are done, check your answers in Chapter 9. Don’t forget to time yourself!
Remember to time yourself during this drill!
1.How many numbers between 1 and 100 are multiples of both 2 and 7 ?
(A) 6
(B) 7
(C) 8
(D) 9
2.What is the smallest multiple of 7 that is greater than 50 ?
(A) 7
3.2^3 × 2^3 × 2^2 =
(B) 2^8
(C) 2^10
(D) 2^16
(E) 2^18
4.For what integer value of m does 2m + 4 = m^3 ?
5.One-fifth of the students in a class chose recycling as the topic for their science projects. If four students chose recycling, how many students are in the class?
(A) 4
6.If 6x — 4 = 38, then x + 10 =
(A) 7
7.If 3x — 6 = 21, then what is x ÷ 9 ?
8.Only one-fifth of the chairs in a classroom are in working order. If three additional working chairs are brought in, there are 19 working seats available. How many chairs were originally in the
9.If a harvest yielded 60 bushels of corn, 20 bushels of wheat, and 40 bushels of soybeans, what percent of the total harvest was corn?
10.At a local store, an item that usually sells for $45 is currently on sale for $30. By what percent is that item discounted?
11.Which of the following is most nearly 35% of $19.95 ?
(A) $3.50
(B) $5.75
(C) $7.00
(D) $9.95
12.Of the 50 hotels in the Hilltop Hotels chain, 5 have indoor swimming pools and 15 have outdoor swimming pools. What percent of all Hilltop Hotels have either an indoor or an outdoor swimming pool?
(E) 5%
13.For what price item does 40% off equal a $20 discount?
(A) $50.00
(E)None of the above
14.A pair of shoes is offered on a special blowout sale. The original price of the shoes is reduced from $50 to $20. What is the percent change in the price of the shoes?
15.Lisa buys a silk dress regularly priced at $60, a cotton sweater regularly priced at $40, and four pairs of socks regularly priced at $5 each. If the dress and the socks are on sale for 20% off
the regular price and the sweater is on sale for 10% off the regular price, what is the total amount of her purchase?
(A) $90.00
(B) $96.00
16.Thirty percent of $17.95 is closest to
(A) $2.00
(B) $3.00
(C) $6.00
(D) $9.00
17.Fifty percent of the 20 students in Mrs. Schweizer’s third-grade class are boys. If 90 percent of these boys ride the bus to school, which of the following is the number of boys in Mrs.
Schweizer’s class who ride the bus to school?
(A) 9
18.On a test with 25 questions, Marc scored an 88 percent. How many questions did Marc answer correctly?
(D) 4
(E) 3
19.Four friends each pay $5 for a pizza every Friday night. If they were to start inviting a fifth friend to come with them and still bought the same pizza, how much would each person then have to
(A) $1
(B) $4
(C) $5
20.A stop sign has 8 equal sides of length 4. What is its perimeter?
(A) 4
(B) 8
(E)It cannot be determined from the information given.
21.If the perimeter of a square is 56, what is the length of each side?
(A) 4
(B) 7
(C) 14
(D) 28
22.The perimeter of a square with a side of length 4 is how much less than the perimeter of a rectangle with sides of length 4 and width 6 ?
23.What is the perimeter of an equilateral triangle, one side of which measures 4 inches?
(A)12 inches
(B)8 inches
(C)6 inches
(D)4 inches
(E)It cannot be determined from the information given.
24.x =
(A) 8
(B) 30
(C) 50
(D) 65
25.If b = 45, then v^2 =
(D) 5
(E)It cannot be determined from the information given.
26.One-half of the difference between the number of degrees in a square and the number of degrees in a triangle is
(A) 45
(B) 90
27.If the area of a square is equal to its perimeter, what is the length of one side?
(A) 1
(B) 2
(C) 4
(D) 8
28.The area of a rectangle with width 4 and length 3 is equal to the area of a triangle with a base of 6 and a height of
(A) 1
(B) 2
(C) 3
(D) 4
29.Two cardboard boxes have equal volume. The dimensions of one box are 3 × 4 × 10. If the length of the other box is 6 and the width is 4, what is the height of the second box?
(A) 2
(B) 5
30.If the area of a square is 64p^2, what is the length of one side of the square?
(C) 8p^2
(D) 8p
(E) 8
31.If AB = 10 and AC = 15, what is the perimeter of the figure above?
(E)It cannot be determined from the information given.
32.If ABCD, shown above, is a rectangle, what is the value of w + x + y + z ?
(A) 90°
33.What is the area of the figure above if all the angles shown are right angles?
34.In the figure above, the length of side AB of square ABCD is equal to 4 and the circle has a radius of 2. What is the area of the shaded region?
(A)4 — π
(B)16 — 4π
(C)8 + 4π
35.The distance between points A and B in the coordinate plane above is
(A) 5
(B) 6
(C) 8
(D) 9
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
A ratio is like a recipe. It tells you how much of each ingredient goes into a mixture.
For example:
To make punch, mix two parts grape juice with three parts orange juice.
This ratio tells you that for every two units of grape juice, you will need to add three units of orange juice. It doesn’t matter what the units are; if you were working with ounces, you would mix
two ounces of grape juice with three ounces of orange juice to get five ounces of punch. If you were working with gallons, you would mix two gallons of grape juice with three gallons of orange juice.
How much punch would you have? Five gallons.
To work through a ratio question, first you need to organize the information you are given. Do this using the Ratio Box.
In a club with 35 members, the ratio of boys to girls is 3:2. To complete your Ratio Box, fill in the ratio at the top and the “real value” at the bottom.
Then look for a “magic number” that you can multiply by the ratio total to get the real value total. In this case, the magic number is 7. That’s all there is to it!
Remember to time yourself during this drill!
1.In a jar of lollipops, the ratio of red lollipops to blue lollipops is 3:5. If only red lollipops and blue lollipops are in the jar and if the total number of lollipops in the jar is 56, how many
blue lollipops are in the jar?
(D) 8
(E) 5
2.At Jed’s Country Hotel, there are three types of rooms: singles, doubles, and triples. If the ratio of singles to doubles to triples is 3:4:5, and the total number of rooms is 36, how many doubles
are there?
(A) 4
(B) 9
3.Matt’s Oak Superstore has exactly three times as many large oak desks as small oak desks in its inventory. If the store sells only these two types of desks, which could be the total number of desks
in stock?
4.In Janice’s tennis club, 8 of the 12 players are right-handed. What is the ratio of right-handed to left-handed players in Janice’s club?
5.One-half of the 400 students at Booth Junior High School are girls. Of the girls at the school, the ratio of those who ride a school bus to those who walk is 7:3. What is the total number of girls
who walk to school?
(A) 10
(B) 30
(C) 60
6.A pet goat eats 2 pounds of goat food and 1 pound of grass each day. When the goat has eaten a total of 15 pounds, how many pounds of grass will it have eaten?
(A) 3
(B) 4
(C) 5
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
There are three parts to every average problem: total, number, and average. Most SSAT problems will give you two of the three pieces and ask you to find the third. To help organize the information
you are given, use the Average Pie.
The Average Pie organizes all of your information visually. It makes it easier to see all of the relationships between the pieces of the pie.
·TOTAL =(# of items) × (Average)
·# of items =
·Average =
For example, if your friend went bowling and bowled three games, scoring 71, 90, and 100, here’s how you would compute her average score using the Average Pie.
To find the average, you would simply write a fraction that represents
The math becomes simple. 261 ÷ 3 = 87. Your friend bowled an average of 87.
Practice working with the Average Pie by using it to solve the following problems.
When you see the word average, draw an Average Pie.
Remember to time yourself during this drill!
1.The average of 3 numbers is 18. What is 2 times the sum of the 3 numbers?
(B) 54
(C) 36
(D) 18
(E) 6
2.If Set M contains 4 positive integers whose average is 7, then what is the largest number that Set M could contain?
(A) 6
(B) 7
3.An art club of 4 boys and 5 girls makes craft projects. If the boys average 2 projects each and the girls average 3 projects each, what is the total number of projects produced by the club?
(A) 14
(B) 23
(C) 26
(D) 54
4.If a class of 6 students has an average grade of 72 before a seventh student joins the class, then what must the seventh student’s grade be to raise the class average to 76 ?
(B) 92
(C) 88
(D) 80
(E) 76
5.Catherine scores an 84, 85, and 88 on her first three exams. What must she score on her fourth exam to raise her average to an 89 ?
Don’t forget to check your answers in Chapter 9.
There is one special kind of percent question that shows up on the SSAT: percent change. This type of question asks you to find what percent something has increased or decreased. Instead of taking
the part and dividing it by the whole, you will take the difference between the two numbers and divide it by the original number. Then, to turn the fraction to a percent, divide the numerator by the
denominator and multiply by 100.
For example:
The number of people who watched Empire last year was 3,600,000. This year, only 3,000,000 are watching the show. By approximately what percent has the audience decreased?
(The difference is 3,600,000 — 3,000,000.)
The fraction reduces to
Remember to time yourself during this drill!
1.During a severe winter in Ontario, the temperature dropped suddenly to 10 degrees below zero. If the temperature in Ontario before this cold spell occurred was 10 degrees above zero, by what
percent did the temperature drop?
(A) 25%
(B) 50%
% change =
2.Fatty’s Burger wants to attract more customers by increasing the size of its patties. From now on Fatty’s patties are going to be 4 ounces larger than before. If the size of its new patty is 16
ounces, by approximately what percent has the patty increased?
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
Plugging In
The SSAT will often ask you questions about real-life situations in which the numbers have been replaced with variables. One of the easiest ways to tackle these questions is with a powerful technique
called Plugging In.
Mark is two inches taller than John, who is four inches shorter than Terry. If t represents Terry’s height in inches, then in terms of t, an expression for Mark’s height is
(A)t + 6
(B)t + 4
(C)t + 2
(E)t — 2
The problem with this question is that we’re not used to thinking of people’s heights in terms of variables. Have you ever met someone who was t inches tall?
Take the Algebra Away, and Arithmetic Is All That’s Left
When you Plug In for variables, you won’t need to write equations and won’t have to solve algebra problems. Doing simple arithmetic is always easier than doing algebra.
Whenever you see variables used in the question and in the choices, just Plug In a number to replace the variable.
1. Choose a number for t.
2. Using that number, figure out Mark’s and John’s heights.
3. Put a box around Mark’s height because that’s what the question asked you for.
4. Plug your number for t into the choices and choose the one that gives you the number you found for Mark’s height.
Here’s How It Works
For Terry’s height, let’s pick 60 inches. This means that t = 60.
Remember, there is no right or wrong number to pick. 50 would work just as well.
But given that Terry is 60 inches tall, now we can figure out that, because John is four inches shorter than Terry, John’s height must be (60 — 4), or 56 inches.
The other piece of information we learn from the problem is that Mark is two inches taller than John. If John’s height is 56 inches, that means Mark must be 58 inches tall.
Here’s what we’ve got:
Terry 60 inches = t
John 56 inches
Mark 58 inches
Now, the question asks for Mark’s height, which is 58 inches. The last step is to go through the choices substituting 60 for t and choose the one that equals 58.
(A) t + 6 60 + 6 = 66 ELIMINATE
(B) t + 4 60 + 4 = 64 ELIMINATE
(C) t + 2 60 + 2 = 62 ELIMINATE
(D) t 60 ELIMINATE
(E) t — 2 60 — 2 = 58 PICK THIS ONE!
After reading this explanation, you may be tempted to say that Plugging In takes too long. Don’t be fooled. The method itself is often faster and (more importantly) more accurate than regular
algebra. Try it out. Practice. As you become more comfortable with Plugging In, you’ll get even quicker and better results. You still need to know how to do algebra, but if you do only algebra, you
may have difficulty improving your SSAT score. Plugging In gives you a way to break through whenever you are stuck. You’ll find that having more than one way to solve SSAT math problems puts you at a
real advantage.
Elementary Level
While some of these problems may seem harder than what you will encounter, Plugging In and Plugging In The Answers (the next section) are great strategies. Be sure you understand how to use these
strategies on your test.
1.At a charity fund-raiser, 200 people each donated x dollars. In terms of x, what was the total number of dollars donated?
(D)200 + x
2.If 10 magazines cost d dollars, then in terms of d, how many magazines can be purchased for 3 dollars?
3.The zoo has four times as many monkeys as lions. There are four more lions than there are zebras at the zoo. If z represents the number of zebras in the zoo, then in terms of z, how many monkeys
are there in the zoo?
(A)z + 4
(B)z + 8
(D)4z + 16
(E)4z + 4
Occasionally, you may run into a Plugging In question that doesn’t contain variables. These questions usually ask about a percentage or a fraction of some unknown number or price. This is the one
time that you should Plug In even when you don’t see variables in the answer!
Also, be sure you Plug In good numbers. Good doesn’t mean right because there’s no such thing as a right or wrong number to Plug In. A good number is one that makes the problem easier to work with.
If a question asks about minutes and hours, try Plugging In 30 or 60, not 128. Also, whenever you see the word percent, Plug In 100!
4.The price of a suit is reduced by half, and then the resulting price is reduced by 10%. The final price is what percent of the original price?
(A) 5%
5.On Wednesday, Miguel ate one-fourth of a pumpkin pie. On Thursday, he ate one-half of what was left of the pie. What fraction of the entire pie did Miguel eat on Wednesday and Thursday?
6.If p pieces of candy costs c cents, then in terms of p and c, 10 pieces of candy will cost
(C)10pc cents.
(E)10 + p + c cents.
7.If J is an odd integer, which of the following must be true?
(A)(J ÷ 3) > 1
(B)(J — 2) is a positive integer.
(C)2 × J is an even integer.
(D)J^2 > J
(E)J > 0
8.If m is an even integer, n is an odd integer, and p is the product of m and n, which of the following is always true?
(A)p is a fraction.
(B)p is an odd integer.
(C)p is divisible by 2.
(D)p is between m and n.
(E)p is greater than zero.
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
Plugging In The Answers (PITA)
Plugging In The Answers is similar to Plugging In. When you have variables in the choices, you Plug In. When you have numbers in the choices, you should generally Plug In the Answers. The only time
this may get tricky is when you have a question that asks for a percent or fraction of some unknown number.
Plugging In the Answers works because on a multiple-choice test, the right answer is always one of the choices. On this type of question, you can’t Plug In any number you want because only one number
will work. Instead, you can Plug In numbers from the choices, one of which must be correct. Here’s an example.
Nicole baked a batch of cookies. She gave half to her friend Lisa and six to her mother. If she now has eight cookies left, how many did Nicole bake originally?
(A) 8
See what we mean? It would be hard to just start making up numbers of cookies and hope that eventually you guessed correctly. However, the number of cookies that Nicole baked originally must be
either 8, 12, 20, 28, or 32 (the five choices). So pick one—always start with (C)—and then work backward to determine whether you have the right choice.
Let’s start with (C): Nicole baked 20 cookies. Now work through the events listed in the question.
She had 20 cookies—from (C)—and she gave half to Lisa. That leaves Nicole with 10 cookies.
What next? She gives 6 to her mom. Now she’s got 4 left.
Keep going. The problem says that Nicole now has 8 cookies left. But if she started with 20—(C)—she would only have 4 left. So is (C) the right answer? No.
No problem. Choose another choice and try again. Be smart about which choice you pick. When we used the number in (C), Nicole ended up with fewer cookies than we wanted her to have, didn’t she? So
the right answer must be a number larger than 20, the number we took from (C).
The good news is that the choices in most Plugging In The Answers questions go in order, so you can choose the next larger or smaller number—you just pick either (B) or (D), depending on which
direction you’ve decided to go.
Back to Nicole and her cookies. We need a number larger than 20. So let’s go to (D)—28.
Nicole started out with 28 cookies. The first thing she did was give half, or 14, to Lisa. That left Nicole with 14 cookies.
Then she gave 6 cookies to her mother. 14 — 6 = 8. Nicole has 8 cookies left over. Keep going with the question. It says, “If she now has eight cookies left…” She has eight cookies left and, voilà
—she’s supposed to have 8 cookies left.
What does this mean? It means you’ve got the right answer! Pick (D) and move on.
If (D) had not worked, and you were still certain that you needed a number larger than (C), you also would be finished. Since you started with the middle, (C), which didn’t work, and then you tried
the next larger choice, (D), which didn’t work either, you could pick the only choice bigger than (C) that was left—in this case (E)—and be done.
This diagram helps illustrate the way you should move through the choices.
To wrap up, Plugging In The Answers should always go the following way:
1. Start with (C). This number is now what you are working with.
2. Work the problem. Go through the problem with that number, using information to help you determine if it is the correct answer.
3. If (C) doesn’t work, try another answer. Remember to think logically about which choice you should check next.
4. Once you find the correct answer, STOP.
Remember to time yourself during this drill!
1.Ted can read 60 pages per hour. Naomi can read 45 pages per hour. If both Ted and Naomi read at the same time, how many minutes will it take them to read a total of 210 pages?
(A) 36
(B) 72
2.If the sum of y and y + 1 is greater than 18, which of the following is one possible value for y ?
(B) —8
(C) 2
(D) 8
(E) 10
3.Kenny is 5 years older than Greg. In 5 years, Kenny will be twice as old as Greg is now. How old is Kenny now?
(A) 5
4.Three people—Paul, Sara, and John—want to put their money together to buy a $90 radio. If Sara agrees to pay twice as much as John, and Paul agrees to pay three times as much as Sara, how much must
Sara pay?
5.Four less than a certain number is two-thirds of that number. What is the number?
(A) 1
(B) 6
(C) 8
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
Guesstimating: A Second Look
Guesstimating worked well back in the introduction when we were just using it to estimate or “ballpark” the size of a number, but geometry problems are undoubtedly the best place to guesstimate
whenever you can.
Let’s try the next problem. Remember, unless a particular question tells you otherwise, you can safely assume that figures are drawn to scale.
A circle is inscribed in square PQRS. What is the area of the shaded region?
(A)16 — 6π
(B)16 — 4π
(C)16 — 3π
(D)16 — 2π
Elementary Level
This question is harder than what you will encounter, but it’s a good idea to learn how guesstimating can help you!
Wow, a circle inscribed in a square—that sounds tough!
It isn’t. Look at the picture. What fraction of the square looks like it is shaded? Half? Three-quarters? Less than half? In fact, about one-quarter of the area of the square is shaded. You’ve just
done most of the work necessary to solve this problem.
Try these values when guesstimating:
π ≈ 3+
Now, let’s just do a little math. The length of one side of the square is 4, so the area of the square is 4 × 4 or 16.
So the area of the square is 16, and we said that the shaded region was about one-fourth of the square. One-fourth of 16 is 4, right? So we’re looking for a choice that equals about 4. Let’s look at
the choices.
(A)16 — 6π
(B)16 — 4π
(C)16 — 3π
(D)16 — 2π
This becomes a little complicated because the answers include π. For the purposes of guesstimating, and in fact for almost any purpose on the SSAT, you should just remember that π is a little more
than 3.
Let’s look back at those answers.
(A) 16 — 6π is roughly equal to 16 — (6 × 3) = —2
(B) 16 — 4π is roughly equal to 16 — (4 × 3) = 4
(C) 16 — 3π is roughly equal to 16 — (3 × 3) = 7
(D) 16 — 2π is roughly equal to 16 — (2 × 3) = 10
(E) 16π is roughly equal to (16 × 3) = 48
Now let’s think about what these answers mean.
Choice (A) is geometrically impossible. A figure cannot have a negative area. Eliminate it.
Choice (B) means that the shaded region has an area of about 4. Sounds pretty good.
Choice (C) means that the shaded region has an area of about 7. The area of the entire square was 16, so that would mean that the shaded region was almost half the square. Possible, but doubtful.
Choice (D) means that the shaded region has an area of about 10. That’s more than half the square and in fact, almost three-quarters of the entire square. No way; cross it out.
Finally, (E) means that the shaded region has an area of about 48. What? The whole square had an area of 16. Is the shaded region three times as big as the square itself? Not a chance. Eliminate (E).
At this point you are left with only (B), which we feel pretty good about, and (C), which seems a little large. What should you do?
Pick (B) and pat yourself on the back because you chose the right answer without doing a lot of unnecessary work. Also, remember how useful it was to guesstimate and make sure you do it whenever you
see a geometry problem, unless the problem tells you that the figure is not drawn to scale!
Weird Shapes
Whenever the test presents you with a geometric figure that is not a square, rectangle, circle, or triangle, draw a line or lines to divide that figure into the shapes that you do know. Then you can
easily work with shapes you know all about.
Shaded Regions—Middle and Upper Levels Only
Sometimes geometry questions show you one figure inscribed in another and then ask you to find the area of a shaded region inside the larger figure and outside the smaller figure (like the problem at
the beginning of this section). To find the areas of these shaded regions, find the area of the outside figure and then subtract from that the area of the figure inside. The difference is what you
ABCE is a rectangle with a length of 10 and width of 6. Points F and D are the midpoints of AE and EC, respectively. What is the area of the shaded region?
(E)It cannot be determined from the information given.
The first step is to find the area of the rectangle. If you multiply the length by the width, you’ll find the area is 60. Now we find the area of the triangle that we are removing from the rectangle.
Because the height and base of the triangle are parts of the sides of the rectangle, and points D and F are half the length and width of the rectangle, we know that the height of the triangle is half
the rectangle’s width, or 3, and the base of the triangle is half the rectangle’s length, or 5. Using the formula for area of a triangle, we find the area of the triangle is 7.5. Now we subtract the
area of the triangle from the area of the rectangle. 60 — 7.5 = 52.5. The correct choice is (D). Be careful not to choose (E) just because the problem looks tricky!
Functions—Middle and Upper Levels Only
In a function problem, an arithmetic operation is defined and then you are asked to perform it on a number. A function is just a set of instructions written in a strange way.
# x = 3x(x + 1)
On the left there is usually a variable with a strange symbol next to or around it.
In the middle is an equals sign.
On the right are the instructions. These tell you what to do with the variable.
# x = 3x(x + 1) What does # 5 equal?
# 5 = (3 × 5)(5 + 1) Just replace each x with a 5!
Here, the function (indicated by the # sign) simply tells you to substitute a 5 wherever there was an x in the original set of instructions. Functions look confusing because of the strange symbols,
but once you know what to do with them, they are just like manipulating an equation.
Sometimes more than one question will refer to the same function. The following drill, for example, contains two questions about one function. In cases such as this, the first question tends to be
easier than the second.
Remember to time yourself during this drill!
Questions 1 and 2 refer to the following definition.
For all real numbers n, $n = 10n — 10.
1.$7 =
(D) 7
(E) 0
2.If $n = 120, then n =
(A) 11
(B) 12
(C) 13
Questions 3—5 refer to the following definition.
For all real numbers d and y, d ¿ y = (d × y) — (d + y).
[Example: 3 ¿ 2 = (3 × 2) — (3 + 2) = 6 — 5 = 1]
3.10 ¿ 2 =
(D) 8
(E) 4
4.If K (4 ¿ 3) = 30, then K =
5.(2 ¿ 4) × (3 ¿ 6) =
(A)(9 ¿ 3) + 3
(B)(6 ¿ 4) + 1
(C)(5 ¿ 3) + 4
(D)(8 ¿ 4) + 2
(E)(9 ¿ 4) + 3
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
Charts and Graphs
Chart questions are simple, but you must be careful. Follow these three steps and you’ll be well on the way to mastering any chart question.
1. Read any text that accompanies the chart. It is important to know what the chart is showing and what scale the numbers are on.
2. Read the question.
3. Refer to the chart and find the specific information you need.
Don’t Be in Too Big of a Hurry
When working with charts and graphs, make sure you take a moment to look at the chart or graph, figure out what it tells you, and then go to the questions.
If there is more than one question about a single chart, the later questions will tend to be more difficult than the earlier ones. Be careful!
Here is a sample chart.
Club Membership by State, 2010 and 2011
State 2010 2011
California 300 500
Florida 225 250
Illinois 200 180
Massachusetts 150 300
Michigan 150 200
New Jersey 200 250
New York 400 600
Texas 50 100
There are many different questions that you can answer based on the information in this chart. For instance:
What is the difference between the number of members who came from New York in 2010 and the number of members who came from Illinois in 2011 ?
This question asks you to look up two simple pieces of information and then do a tiny bit of math.
First, the number of members who came from New York in 2010 was 400.
Second, the number of members who came from Illinois in 2011 was 180.
Finally, look back at the question. It asks you to find the difference between these numbers. 400 — 180 = 220. Done.
The increase in the number of members from New Jersey from 2010 to 2011 was what percent of the total number of members in New Jersey in 2010 ?
You should definitely know how to do this one! Do you remember how to translate percentage questions? If not, go back to Fundamental Math Skills (Chapter 2).
In 2010, there were 200 club members from New Jersey. In 2011, there were 250 members from New Jersey. That represents an increase of 50 members. To determine what percent that is of the total amount
in 2010, you will need to ask yourself, “50 (the increase) is what percent of 200 (the number of members in 2010)?”
Translated, this becomes:
With a little bit of simple manipulation, this equation becomes:
50 = 2g
25 = g
So from 2010 to 2011, there was a 25% increase in the number of members from New Jersey. Good work!
Which state had as many club members in 2011 as a combination of Illinois, Massachusetts, and Michigan had in 2010 ?
First, take a second to look up the number of members who came from Illinois, Massachusetts, and Michigan in 2010 and add them together.
200 + 150 + 150 = 500
Which state had 500 members in 2011? California. That’s all there is to it!
Some questions will ask you to interpret a graph. You should be familiar with both pie and bar graphs. These graphs are generally drawn to scale (meaning that the graphs give an accurate visual
impression of the information) so you can always guess based on the figure if you need to.
The way to approach a graph question is exactly the same as the way to approach a chart question. Follow the same three steps.
1. Read any text that accompanies the graph. It is important to know what the graph is showing and what scale the numbers are on.
2. Read the question.
3. Refer back to the graph and find the specific information you need.
This is how it works.
The graph in Figure 1 shows Emily’s clothing expenditures for the month of October. On which type of clothing did she spend the most money?
This one should be simple. You can look at the pieces of the pie and identify the largest, or you can look at the amounts shown in the graph and choose the largest one. Either way, the answer is (A)
because Emily spent more money on shoes than on any other clothing items in October.
Emily spent half of her clothing money on which two items?
(A)Shoes and pants
(B)Shoes and shirts
(C)Hats and socks
(D)Socks and shirts
(E)Shirts and pants
Again, you can find the answer to this question two different ways. You can look for which two items together make up half the chart, or you can add up the total amount of money Emily spent ($240)
and then figure out which two items made up half (or $120) of that amount. Either way is just fine, and either way the right answer is (B), shoes and shirts.
Remember to time yourself during this drill!
Questions 1-3 refer to the following summary of energy costs by district.
District 1990 1991
A 400 600
B 500 700
C 200 350
D 100 150
E 600 800
(All numbers are in thousands of dollars.)
1.In 1991, which district spent twice as much on energy as District A spent in 1990 ?
2.Which district spent the most on energy in 1990 and 1991 combined?
(E)It cannot be determined from the information given.
3.The total increase in energy expenditure in these districts, from 1990 to 1991, is how many dollars?
(B) $1,800
(C) $2,400
(D) $2,600
Questions 4 and 5 refer to Figure 2, which shows the number of compact discs owned by five students.
4.Carl owns as many CDs as which two other students combined?
(A)Abe and Ben
(B)Ben and Dave
(C)Abe and Ed
(D)Abe and Dave
(E)Ben and Ed
5.Which one student owns one-fourth of the CDs accounted for in Figure 2 ?
Questions 6-8 refer to Matt’s weekly time card, shown below.
6.If Matt’s hourly salary is $6, what were his earnings for the week?
(A) $6
7.What is the average number of hours Matt worked on the days he worked during this particular week?
(A) 3
(B) 3.5
(C) 4
(D) 7
8.The hours that Matt worked on Monday accounted for what percent of the total number of hours he worked during this week?
(A) 3.5
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
When you are done, check your answers in Chapter 9. Don’t forget to time yourself!
Middle Level
Some of the questions here are harder than what you will see on your test. Still, give them all a try. The skills you have learned should help you do well on most of the questions.
Remember to time yourself during this drill!
1.If p is an odd integer, which of the following must be an odd integer?
(A)p^2 + 3
(B)2p + 1
(C)p ÷ 3
(D)p — 3
2.If m is the sum of two positive even integers, which of the following CANNOT be true?
(A)m < 5
(B)3m is odd
(C)m is even
(D)m^3 is even
(E)m ÷ 2 is even
3.The product of b and a^2 can be written as
(C)2a × b
4.Damon has twice as many records as Graham, who has one-fourth as many records as Alex. If Damon has d records, then in terms of d, how many records do Alex and Graham have together?
5.x^a = (x^3)^3
What is the value of a × b ?
6.One six-foot Italian hero serves either 12 children or 8 adults. Approximately how many sandwiches do you need to feed a party of 250, 75 of whom are children?
7.Liam and Noel are traveling from New York City to Dallas. If they traveled
9.Students in Mr. Greenwood’s history class are collecting donations for a school charity drive. If the total number of students in the class, x, donated an average of y dollars each, in terms of x
and y, how much money was collected for the drive?
10.If e + f is divisible by 17, which of the following must also be divisible by 17 ?
(A)(e × f) — 17
(B)e + (f × 17)
(C)(e × 17) + f
(D)(e + f ) / 17
(E)(e × 3) + (f × 3)
11.Joe wants to find the mean number of pages in the books he has read this month. The books were 200, 220, and 260 pages long. He read the 200 page book twice, so it will be counted twice in the
mean. If he reads one more book, what is the fewest number of pages it can have to make the mean no less than 230 ?
12.Sayeeda is a point guard for her basketball team. In the last 3 games, she scored 8 points once and 12 points in each of the other two games. What must she score in tonight’s game to raise her
average to 15 points?
13.What is the greatest common factor of (3xy)^3 and 3x^2y^5 ?
14.The town of Mechanicville lies due east of Stillwater and due south of Half Moon Crescent. If the distance from Mechanicville to Stillwater is 30 miles, and from Mechanicville to Half Moon
Crescent is 40 miles, what is the shortest distance from Stillwater to Half Moon Crescent?
(A) 10
(B) 50
(C) 70
(E)It cannot be determined from the information given.
15.PQRS is a square with an area of 144. What is the area of the shaded region?
(A) 50
(B) 72
(E)It cannot be determined from the information given.
16.PO and QO are radii of the circle with center O. What is the value of x ?
(E)It cannot be determined from the information given.
17.What is the value of x ?
(C) 97
(D) 67
(E)It cannot be determined from the information given.
18.ABC is an equilateral triangle. What is the perimeter of this figure?
(A)4 + 2π
(B)4 + 4π
(C)8 + 2π
(D)8 + 4π
(E)12 + 2π
19.What is the perimeter of this figure?
(B) 44
(C) 40
(D) 36
(E)It cannot be determined from the information given.
20.How many meters of police tape are needed to wrap around a rectangular crime scene that measures 6 meters wide by 28 meters long?
(A) 34 meters
(B) 68 meters
(C) 90 meters
(D)136 meters
(E)168 meters
21.Billy Bob’s Beans are currently packaged in cylindrical cans that contain 9 servings. The cans have a height of 20 cm and a diameter of 18 cm. Billy Bob wants to introduce a new single-serving
can. If he keeps the height of the can the same, what should the diameter of the single-serving can be?
Stop. Check your time for this drill:
Don’t forget to check your answers in Chapter 9.
Make sure you can confidently answer all of the following questions before you take your test.
1.Is zero an integer?
2.Is zero positive or negative?
3.What operation do you perform to find a sum?
4.What operation do you perform to find a product?
5.What is the result called when you divide?
6.Is 312 divisible by 3 ?
Is 312 divisible by 9 ?
(Actually, dividing isn’t fair—use your divisibility rules!)
7.What does the “E” in PEMDAS stand for?
8.Is 3 a factor of 12 ?
Is 12 a factor of 3 ?
9.Is 3 a multiple of 12 ?
Is 12 a multiple of 3 ?
10.What is the tens digit in the number 304.275 ?
11.What is the tenths digit in the number 304.275 ?
12.2^3 =
13.In “math language,” the word percent means ______________.
14.In “math language,” the word of means ______________.
15.In a Ratio Box, the last column on the right is always the ______________.
16.Whenever you see a problem involving averages, draw the ______________.
17.When a problem contains variables in the question and in the answers, I will ______________.
18.To find the perimeter of a square, I ______________ the length(s) of ______________ side(s).
19.To find the area of a square, I ______________ the length(s) of ______________ sides(s).
20.There are ______________ degrees in a straight line.
21.A triangle has ______________ angles, which total ______________ degrees.
22.A four-sided figure contains ______________ degrees.
23.An isosceles triangle has ______________ equal sides; a(n) ______________ triangle has three equal sides.
24.The longest side of a right triangle is called the ______________ and is located opposite the ______________.
25.To find the area of a triangle, I use the formula ______________.
Don’t forget to check your answers in Chapter 9. | {"url":"https://philology.science/test/isee/8.html","timestamp":"2024-11-10T09:34:58Z","content_type":"text/html","content_length":"93443","record_id":"<urn:uuid:c8411a20-81ab-4397-af87-cb25c3752654>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00861.warc.gz"} |
New Conforming Finite Elements Based On The De Rham Complexes For Some Fourth-Order Problems
Access Type
Open Access Dissertation
Date of Award
January 2021
First Advisor
Zhimin Z. Zhang
In this dissertation, we discuss the conforming finite element discretization of high-order equations involving operators such as $(\curl\curl)^2$, $\grad\Delta\div$, and $-\curl\Delta\curl$. These
operators appear in various models, such as continuum mechanics, inverse electromagnetic scattering theory, magnetohydrodynamics, and linear elasticity. Naively discretizing these operators and their
corresponding eigenvalue problems using the existing $H^2$-conforming element would lead to spurious solutions in certain cases. Therefore, it is desirable to design conforming finite elements for
equations containing these high-order differential operators.
The $\curl\curl$-conformity or $\grad\curl$-conformity requires that the tangential component of $\curl \bm u_h$ is continuous. Recall that the N\'ed\'elec element requires only the continuity of the
tangential component of $\bm u_h$.Due to the continuity requirement and the naturally divergence-free property of the curl operator, it is challenging to construct $\grad\curl$-conforming elements.
We start from the two dimensional case, where $\curl\bm u_h$ is a scalar. Our previous construction \cite{WZZelement} is based on the existing polynomial spaces $Q_{k-1,k}\times Q_{k,k-1}$ and $\
mathcal R_k$. The restriction of $k\geq 4$ for a triangular element or $k\geq 3$ for a rectangular element has to be imposed since an interior bubble should be included in the shape function space of
$\curl\bm u_h$, and hence the simplest triangular or rectangular element has 24 degrees of freedom. To reduce the degrees of freedom, we resort to the discrete de Rham complex to construct elements.
The Poincar\'e operator enables us to tailor the shape function space to our needs (not necessarily the existing polynomial spaces). As a result, we construct a finite element complex, which contains
three families of $\grad\curl$-conforming elements without the restriction on polynomial degrees. One of three families is consistent with the previous construction in high-order cases. The
lowest-order triangular and rectangular finite elements have only 6 and 8 degrees of freedom, respectively.
Unlike the two-dimensional case, $\curl\bm u_h$ in three dimensions should be a divergence-free vector in the space $H^1\otimes\mathbb V$, which relates the $\curl\Delta\curl$ problems to the Stokes
problem. However, it is challenging to construct an inf-sup stable finite element Stokes pair that preserves the divergence-free condition at the discrete level.Neilan \cite{neilan2015discrete}
constructed a finite element complex that includes a stable Stokes pair and an $H^1(\curl)$-conforming element on tetrahedral meshes. Based on the same Stokes pair, we construct a finite element
complex which contains three families of $\grad\curl$-conforming elements. Compared to the $H^1(\curl)$-conforming elements \cite{neilan2015discrete} which have at least 360 DOFs, our $\grad\
curl$-conforming elements have weaker continuity ($\bm u_h$ is in $H(\curl)$ instead of $H^1\otimes\mathbb V$) and thus fewer degrees of freedom. However, our elements still have at least 279 degrees
of freedom. Recently, Guzm\'an and Neilan stabilized the lowest-order three dimensional Scott-Vogelius pair by enriching the velocity space with modified Bernardi-Raugel bubbles \cite{guzman2018inf},
which inspires us to use it to construct $\grad\curl$-conforming elements with fewer degrees of freedom. To obtain a family of elements, we first generalize their construction to an arbitrary order
by enriching the velocity space with modified face or/and interior bubbles. Then we construct the whole finite element complex which contains three families of $\grad\curl$-conforming elements on
tetrahedral meshes. The lowest-order element has only 18 degrees of freedom.
The $\grad\div$-conformity requires that the normal component and divergent of the finite element function $\bm u_h$ are continuous. Since $\div\bm u_h$ is a scalar, the construction of the finite
element complex and the $\grad\div$-conforming elements is similar to the $\grad\curl$ elements in two dimensions. The simplest tetrahedral and cubical elements have only 8 and 14 degrees of freedom,
Recommended Citation
Zhang, Qian, "New Conforming Finite Elements Based On The De Rham Complexes For Some Fourth-Order Problems" (2021). Wayne State University Dissertations. 3454. | {"url":"https://digitalcommons.wayne.edu/oa_dissertations/3454/","timestamp":"2024-11-09T16:58:36Z","content_type":"text/html","content_length":"44823","record_id":"<urn:uuid:19da79e4-2995-4831-b8d5-2bdf4ab390e4>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00826.warc.gz"} |
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All we need is to have a chat with you and we can start looking for a suitable tutor who can support your child's academic progress.
Every student learns in their own way and at their own specific pace. Once we know a little bit more about the student we can find a great tutor who'd really click with your child.
Once you reach out to us we can allocate a local Cranbourne West physics tutor within a day or two. They set up the first lesson whenever you and your child can make it.
If you like the tutor's approach you keep working with them. You are not completely happy with the match? No worries, it happens sometimes. We consider it a trial lesson and find a better match.
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Give us a call! | {"url":"https://www.ezymathtutoring.com.au/tutors/cranbourne-west-3977/learn-physics","timestamp":"2024-11-07T10:05:13Z","content_type":"text/html","content_length":"210493","record_id":"<urn:uuid:8b378380-b927-41d2-a7e7-007668574a6a>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00868.warc.gz"} |
João Rafael Lucio dos Santos
PhD, Federal University of Campina GrandeCalculatorian
Translated Calculators
Joao is a physicist, researcher, and professor in Brazil. He earned his Ph.D. in Physics at the University of Rochester, NY in 2013. He has written several papers in international journals, mainly in
field theory and cosmology. Joao is also a researcher and the outreach leader for the BINGO telescope. This state-of-the-art radio telescope will be operating in the Northeast of Brazil in a few
years. He believes in the importance of spreading knowledge and high-quality information among society and the public in general. He also believes that people can enjoy the learning process if it is
funny and if they can see it as a path to change their life, even in daily activities. In his free time, he enjoys running, playing tennis, cooking, traveling, and having pleasant moments with his
relatives and friends.
Areas of expertise
• Physics
• Mathematics
• Field theory
• Gravity
• Cosmology
• Phd in PhysicsUniversity of Rochester, NY, and Federal University of Paraíba, Brazil
• State University of São Paulo, BrazilMaster’s degree in Physics
• State University of São Paulo, BrazilBachelor’s degree in Physics
Professional background and credentials
João has been working with physics since 2003. He has always been interested in studying the Cosmos and loves observing the night sky. Along with his undergraduate studies, João has studied the
interface between classical and quantum theory and how some symmetry properties can be responsible for a system having real quantum energy levels. During his master's, he worked with classical chaos
theory applied to field theory, trying to map different types of solutions of field equations using a method called Poincaré sections. In his PhD, João developed a new method to find analytic
classical field models. He also worked with quantum field theory on flat and curved spacetimes, with field theory at finite temperatures, and with gravitational models. Since 2014, he has worked as a
professor of physics and researcher in Brazil. He is also a member of an international collaboration to construct and operate the BINGO telescope, the largest radio telescope ever built in Brazil. At
Omni, he aims to make calculators and scientific subjects accessible to the general public. He also aims to contribute to developing new tools with his background and expertise. João believes people
need to see science as part of their daily activities.
João is a research fellow of the National Council for Scientific and Technological Development (CNPq, Brazil). He has also been a referee for thirteen international journals and has received several
scholarships during his academic career. In the last few years, he has started working with scientific outreach activities, presenting astronomy, physics, cosmology, and science subjects in schools,
scientific fairs, and congresses. He has also organized fifteen scientific events so far and is a member of the Brazilian Society of Physics. | {"url":"https://www.omnicalculator.com/authors/joao-santos","timestamp":"2024-11-11T05:07:46Z","content_type":"text/html","content_length":"185625","record_id":"<urn:uuid:0ebd78bc-43db-49c5-a053-6cdd5ac9e8b2>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00689.warc.gz"} |
ALEKS Math Assessment
ALEKS and General Education at WOU (wou.edu/gened)
Options & your major
□ For WOU General Education, you may test out of the Foundations – Mathematics requirement by scoring 61 or higher on the WOU ALEKS Math Assessment, earned in a proctored and timed testing
☆ Note, there is no option to use ALEKS to test out of the mathematics courses required by your major.
□ If your major requires a mathematics course
☆ The mathematics course(s) required by your major will also satisfy the WOU General Education Foundations – Mathematics requirement. You do not need to take the ALEKS Mathematics
Assessment in a timed and proctored environment, your General Education Foundations – Mathematics category will be fulfilled when you take your major’s first mathematics course.
☆ You may still need to take an unproctored ALEKS Assessment, see About ALEKS.
□ If your major does not require a mathematics course, or you are undecided about your major
☆ You may test out of the Foundations – Mathematics requirement by scoring 61 or higher on the WOU ALEKS Math Assessment, earned in a proctored and timed (up to two hours) testing
Testing out of the Foundations – Mathematics requirement
□ Required: You must have previously scored 61 or higher on a WOU ALEKS assessment (see Take the Assessment)
□ Required: Your WOU ALEKS Assessment score of 61 or higher is less than one year old
□ ALEKS provides individualized learning modules to help you review and prepare for the proctored ALEKS assessment. Using the learning modules can help you earn a 61 or higher on the ALEKS
assessment and be prepared to take the assessment in the proctored and timed environment. If you are certain you don’t need further study, please contact the Mathematics Department.
Scheduling a proctored ALEKS Assessment
Contacts & Questions
Questions about ALEKS | Student Success & Advising
Questions about mathematics placement | Please see your academic advisor | {"url":"https://wou.edu/math/aleks/aleks-gen-ed/","timestamp":"2024-11-04T01:45:26Z","content_type":"text/html","content_length":"139678","record_id":"<urn:uuid:46378dcd-2e2e-44ff-b8b5-c848f2b1da84>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00106.warc.gz"} |
10th Class General Mathematics Notes Online Free Download With PDF File
10th Class General Mathematics Notes Online Free Download With PDF File
A lot of students from Classes IV to VI are taking the 10th Class General Mathematics Notes. This is a great concept that can help students of all ages understand and gain more confidence in their
math and overall knowledge. This is one of the easiest ways for new students to gain some basic math skills and confidence. Students will get tips for problems they have not tried before, and the
instructor will give them hints on proper ways to approach any problem.
The idea behind the notes is simple. Students will be given a set of notes that are used to practice problems. After each class, students will receive a set of notes for the week. These notes contain
all the homework for the week, along with a small overview of concepts that can be important to remember for each test.
Each student will receive a set of these notes in their email that are easy to access. Students can use the links in the email to jump right to a quick review of a particular concept, or to store and
review the information later. By using the free notes for problems, students will be able to review and retain the key points that they have learned from each lesson. They can then apply the
previously learned material to their homework.
In order to access the notes, students need to log into the website. There will be a drop down list on the top left corner that will allow them to choose a date. The start date is the most recent,
and the end date is the last date that were listed. The start and end dates will allow students to save their notes in the future so that they do not need to look at the website again to find the
right math tip.
When a student clicks on a particular lesson, he/she will be asked to complete a short survey. The questions on this page ask about which math concepts they struggle with, and what they would have
done differently if they had known the correct answers. These notes will serve as homework for the class. The notes should cover the main points of each lesson, but it is important for them to be
unique. The instructor can not review the same notes more than once.
In order to access the notes, students need to login to the website. Once they have logged in, they can choose a date to look through their notes online. At the top of the page, there is a link that
says’Helpful Tips for Locating Solutions’. Students need to click on this link in order to find the page that has their tips listed. Then they can bookmark the page to keep it handy.
10th Class General Mathematics Notes Punjab Board
10th General Mathematics Chapter 01 View PDF File
10th General Mathematics Chapter 02 View PDF File
10th General Mathematics Chapter 03 View PDF File
10th General Mathematics Chapter 04 View PDF File
10th General Mathematics Chapter 05 View PDF File
10th General Mathematics Chapter 06 View PDF File
10th General Mathematics Chapter 07 View PDF File
10th General Mathematics Chapter 08 View PDF File
10th General Mathematics Chapter 09 View PDF File
10th General Mathematics Chapter 10 View PDF File
The tips include solutions to problems such as finding the root of a cubic equation using the integral formula, finding the square root of a number using the dot product formula, and finding the
hypotenuse of a quadratic equation. These notes are useful because they will allow students to quickly memorize the solutions to math problems. It also allows them to build up their mathematical
knowledge. By reviewing the solutions, students will learn how to multiply numbers, solve for a percentage, and learn about exponents. All of these skills will help them in their future life as well
as their career.
The notes are available in a PDF format so that students can print them out if they need to. They are also available in paper form for students to take to school or for other students to read.
Students need to know that they are free, and that they are offered by the Punjab Board of Standards and Assessment. Therefore, they should take advantage of the opportunities that they offer.
10th Class Mathematics Notes Punjab Board With PDF File Download in Free | {"url":"https://worldstudypoint.com/10th-class-general-mathematics-notes-online-fr/","timestamp":"2024-11-10T09:44:41Z","content_type":"text/html","content_length":"91761","record_id":"<urn:uuid:fae54480-afc9-4653-8f53-79801b5f95db>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00249.warc.gz"} |
Shannon Ciphers and Perfect Security
Shannon Ciphers and Perfect Security
A Shannon cipher, named after mathematician Claude Shannon (1916–2001) is a simplified cipher mechanism for encrypting a message using a …
Left: Claude Shannon (1916–2001). Right: Shannon’s notable 1949 paper “Communication Theory of Secrecy Systems” in Bell System Technical Journal 28(4) pp. 656–715.
A Shannon cipher, invented by its namesake Claude Shannon (1916–2001) is a simplified cipher mechanism for encrypting a message using a shared secret key. A cipher is generally defined simply as an
algorithm for performing encryption or decryption, i.e. “a series of well-defined steps that can be followed as a procedure”.
Example (Boneh & Shoup, 2020)
Suppose Claude and Marvin want to use a ciper such that Claude can send an encrypted message that only Marvin can read.Then, Claude and Marvin must in advance agree on a key k ∈ K. Assuming they do,
then when Claude wants to send a message m ∈ M to Marvin, he encrypts m under k, obtaining the ciphertext c = E(k,m) ∈ C, and then sends c to Marvin via some communication channel. Upon receiving the
encrypted message c, Marvin decrypts c under k. The correctness property ensures that D(k,c) is the same as Claude's original message m.
Regarded by many as the foundation of modern cryptography, the concept of a Shannon cipher was first introduced in the 1949 paper Communication Theory and Secrecy Systems published by Shannon in a
Bell Systems Technical Journal. The results Shannon presented in the paper were based on an earlier version of his research in a classified report entitled A Mathematical Theory of Cryptography, and
preceded Shannon’s publication of his well-known A Mathematical Theory of Communication published a year before, in 1948. The following discussion of Shannon ciphers is based on Chapter 2.1 “Shannon
ciphers and perfect security” in the book A Graduate Course in Applied Cryptography by Dan Boneh and Victor Shoup.
Formally, a Shannon cipher is a pair of encryption (E) and decryption functions (D):
wherein in order to encrypt a message m:
The encryption function E takes as input a key k and a message m, to produce a ciphertext c. That is, c = E(k,m), stated as "c is the encryption of m under k".
and in order to decrypt the encrypted ciphertext c:
The decryption function D takes as input a key k and a ciphertext c, to produce a message m. That is, m = D(k,c), stated as "m is the decryption of c under k".
In order to ensure that the operation functions as intended, we require the following property of the cipher:
Correctness Property
Decryption "undoes" encryption, that is, the cipher must satisfy the following correctness property: for all keys k and all messages m, D(k,E(k,m)) = m, stated as "m is the decryption of E(k,m) under
One-time pads
In cryptography, a one-time pad (OTP) is an encryption technique that cannot be cracked (i.e. cannot be computed). It requires the use of a one-time key kwhich is shared prior to the transmission of
a message. The key must be the same size as, or longer than the message being transmitted. Formally (Boneh & Shoup, 2020):
Definition of a one-time pad
A one-time pad is a Shannon cipher ε = (E,D) where the keys (k), messages (m) and ciphertexts (c) are bit strings of the same length.
That is, the one-time pad Shannon cipher ε is defined over (K,M,C), where
for some fixed parameter L. For a key k ∈ {0,1}ᴸ and a message m ∈ {0,1}ᴸ, the encryption function E(k,m) is defined as k ⨁ m = c, where ⨁ denotes component-wise addition modulo 2.
One-time pads work by pairing a message m in plaintext with a random secret key (referred to as a one-time pad). Next, each bit or character of the message is encrypted by combining it with the
corresponding bit or character from the pad using modular arithmetic. If the one-time pad used fulfills the following properties: 1. It is truly random; 2. It is at least as long as the plaintext; 3.
It is never reused in whole or in part; and 4. It is kept completely secret; then the ciphertext can be shown to be impossible to decrypt or break, i.e. “perfectly secure”.
Perfect Security
In cryptography the gold standard of security, “perfect security” is a special case of information-theoretic security wherein for an encryption algorithm, if there is ciphertext produced that uses
it, no information about the message is provided without knowledge of the key. Formally (Boneh & Shoup, 2020),
Definition of perfect security
Let ε = (E,D) be a Shannon cipher defined over (K,M,C). Consider a probabilistic experiment in which the random variable k is uniformly distributed over K. If for all m₀, m₁ ∈ M, and all c ∈ C, we
have:Pr[E(k,m₀) = c] = Pr[E(k,m₁) = c]then we say that ε is a perfectly secure Shannon cipher.
In words, the probability that a ciphertext c is m₀ is the same as the probability that the same ciphertext c is m₁, then the cipher ε is a perfectly secure Shannon cipher. That is, the perfectly
secure Shannon cipher ε has produced a ciphertext which has equal probability of being any message, i.e. the ciphertext c gives no information about the plaintext m. In order to construct a proof
that this is the case, we must first provide a few equivalences from Boneh & Shoup (2020):
Let ε = (E,D) be a Shannon cipher defined over (K,M,C). Then the following statements are equivalent:i) ε is perfectly secure;ii) For every ciphertext c ∈ C there exists Nc (possibly depending on c)
such that for all messages m ∈ M, we have |{k ∈ K : E(k,m) = c}| = Nciii) If the random variable k is uniformly distributed over K, then each of the random variables E(k,m) for m ∈ M has the same
That is, the statement that ε is i) perfectly secure is equivalent to the statements that ii) for every cipher c there exists a certain number of ciphers Nc such that for all messages m, the
encryption function E can generate c
ii) for every cipher c, there exists a number Pc (depending on c) such that for all messages m, the probability that the encryption function E(k,m) generates the ciphertext c is Pc. Here, k is a
random variable distributed over K and Pc = Nc/|K|, and iii) that if the random variable k is uniformly distributed over K, then each random variable k which may be used to encrypt m has the same
distribution. The proof of these equivalences is available in Boneh & Shoup (2020, p. 9). From ii), we can next provide the following proof that one-time pads satisfy the requirements for a perfectly
secure Shannon cipher:
Proof that the one-time pad is a perfectly secure Shannon cipher
Suppose the Shannon cipher ε = (E,D) is a one-time pad and is defined over (K,M,C) where K := M := C := {0,1}ᴸ. For any fixed message m ∈ {0,1}ᴸ and ciphertext c ∈ {0,1}ᴸ, there is a unique key k ∈
{0,1}ᴸ satisfying the equation k ⨁ m = cnamely, k := m ⨁ c. Therefore, ε satisfies condition ii) in theorem 2.1 above (with Nc = 1 for each ciphertext c).
Although symmetric cryptographic systems have been known for at least two thousand years (read: Caesar cipher), Shannon’s 1949 paper was the first to provide a formal mathematical description of such
systems. In his paper, Shannon defined the properties of what he called “perfect secrecy” for shared-key systems and showed that they must exist.
Those interested in reading more about Shannon ciphers are encouraged to download the book A Graduate Course in Applied Cryptography by Dan Boneh and Victor Shoup (2020). Those interested in reading
more about Claude Shannon are encouraged to acquire the book A Mind at Play: How Claude Shannon Invented the Information Age by Jimmy Soni and Rob Goodman. | {"url":"https://www.cantorsparadise.org/shannon-ciphers-and-perfect-security-d70c10379ac2/","timestamp":"2024-11-14T01:12:53Z","content_type":"text/html","content_length":"39870","record_id":"<urn:uuid:ba1b588b-c451-46dd-8361-adabb54b0a0c>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00892.warc.gz"} |
Programming an estimation command in Stata: Allowing for options
I make three improvements to the command that implements the ordinary least-squares (OLS) estimator that I discussed in Programming an estimation command in Stata: Allowing for sample restrictions
and factor variables. First, I allow the user to request a robust estimator of the variance-covariance of the estimator (VCE). Second, I allow the user to suppress the constant term. Third, I store
the residual degrees of freedom in e(df_r) so that test will use the \(t\) or \(F\) distribution instead of the normal or \(\chi^2\) distribution to compute the \(p\)-value of Wald tests.
This is the ninth post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted entries
for a map to all the posts in this series.
Allowing for robust standard errors
The syntax of myregress6, which I discussed in Programming an estimation command in Stata: Allowing for sample restrictions and factor variables, is
myregress6 depvar [indepvars] [if] [in]
where the independent variables can be time-series or factor variables. myregress7 has the syntax
myregress7 depvar [indepvars] [if] [in] [, robust ]
By default, myregress7 estimates the VCE assuming that the errors are independently and identically distributed (IID). If the option robust is specified, myregress7 uses the robust estimator of the
VCE. See Cameron and Trivedi (2005), Stock and Watson (2010), Wooldridge (2015) for introductions to OLS; see Programming an estimation command in Stata: Using Stata matrix commands and functions to
compute OLS objects for the formulas and Stata matrix implementations. Click on the file name to download any code block. To avoid scrolling, view the code in the do-file editor, or your favorite
text editor, to see the line numbers.
Code block 1: myregress7.ado
*! version 7.0.0 30Nov2015
program define myregress7, eclass
version 14
syntax varlist(numeric ts fv) [if] [in] [, Robust]
marksample touse
gettoken depvar indeps : varlist
_fv_check_depvar `depvar'
tempname zpz xpx xpy xpxi b V
tempvar xbhat res res2
quietly matrix accum `zpz' = `varlist' if `touse'
local N = r(N)
local p = colsof(`zpz')
matrix `xpx' = `zpz'[2..`p', 2..`p']
matrix `xpy' = `zpz'[2..`p', 1]
matrix `xpxi' = syminv(`xpx')
local k = `p' - diag0cnt(`xpxi') - 1
matrix `b' = (`xpxi'*`xpy')'
quietly matrix score double `xbhat' = `b' if `touse'
quietly generate double `res' = (`depvar' - `xbhat') if `touse'
quietly generate double `res2' = (`res')^2 if `touse'
if "`robust'" == "" {
quietly summarize `res2' if `touse' , meanonly
local sum = r(sum)
local s2 = `sum'/(`N'-(`k'))
matrix `V' = `s2'*`xpxi'
else {
tempname M
quietly matrix accum `M' = `indeps' [iweight=`res2'] if `touse'
matrix `V' = (`N'/(`N'-(`k')))*`xpxi'*`M'*`xpxi'
local vce "robust"
local vcetype "Robust"
ereturn post `b' `V', esample(`touse') buildfvinfo
ereturn scalar N = `N'
ereturn scalar rank = `k'
ereturn local vce "`vce'"
ereturn local vcetype "`vcetype'"
ereturn local cmd "myregress7"
ereturn display
A user may specify the robust option by typing robust, robus, robu, rob, ro, or r. In other words, r is the minimal abbreviation of the option robust. Line 5 of myregress7 implements this syntax.
Specifying robust is optional because Robust is enclosed in the square brackets. r is the minimal abbreviation because the R is in uppercase and the remaining letters are in lowercase.
If the user specifies robust, or a valid abbreviation thereof, the local macro robust contains the word “robust”; otherwise, the local macro robust is empty. Line 25 uses this fact to determine which
VCE should be computed; it specifies that lines 26–31 should be executed if the local macro robust is empty and that lines 32-36 should otherwise be executed. Lines 26-31 compute the IID estimator of
the VCE. Lines 32-34 compute the robust estimator of the VCE. Lines 35 and 36 respectively put “robust” and “Robust” into the local macros vce and vcetype.
Line 41 puts the contents of the local macro vce into the local macro e(vce), which informs users and postestimation commands which VCE estimator was used. By convention, e(vce) is empty for the IID
case. Line 42 puts the contents of the local macro vcetype into the local macro e(vcetype), which is used by ereturn display to correctly label the standard errors as robust.
I now run a regression with robust standard errors.
Example 1: myregress7 with robust standard errors
. sysuse auto
(1978 Automobile Data)
. myregress7 price mpg trunk i.rep78, robust
| Robust
| Coef. Std. Err. z P>|z| [95% Conf. Interval]
mpg | -262.7053 74.75538 -3.51 0.000 -409.2232 -116.1875
trunk | 41.75706 73.71523 0.57 0.571 -102.7221 186.2362
rep78 |
2 | 654.7905 1132.425 0.58 0.563 -1564.721 2874.302
3 | 1170.606 823.9454 1.42 0.155 -444.2979 2785.509
4 | 1473.352 650.4118 2.27 0.023 198.5679 2748.135
5 | 2896.888 937.6981 3.09 0.002 1059.034 4734.743
_cons | 9726.377 2040.335 4.77 0.000 5727.393 13725.36
Suppressing the constant term
myregress8 has the syntax
myregress8 depvar [indepvars] [if] [in] [, robust noconstant ]
Code block 2: myregress8.ado
*! version 8.0.0 30Nov2015
program define myregress8, eclass
version 14
syntax varlist(numeric ts fv) [if] [in] [, Robust noCONStant ]
marksample touse
gettoken depvar indeps : varlist
_fv_check_depvar `depvar'
tempname zpz xpx xpy xpxi b V
tempvar xbhat res res2
quietly matrix accum `zpz' = `varlist' if `touse' , `constant'
local N = r(N)
local p = colsof(`zpz')
matrix `xpx' = `zpz'[2..`p', 2..`p']
matrix `xpy' = `zpz'[2..`p', 1]
matrix `xpxi' = syminv(`xpx')
local k = `p' - diag0cnt(`xpxi') - 1
matrix `b' = (`xpxi'*`xpy')'
quietly matrix score double `xbhat' = `b' if `touse'
quietly generate double `res' = (`depvar' - `xbhat') if `touse'
quietly generate double `res2' = (`res')^2 if `touse'
if "`robust'" == "" {
quietly summarize `res2' if `touse' , meanonly
local sum = r(sum)
local s2 = `sum'/(`N'-(`k'))
matrix `V' = `s2'*`xpxi'
else {
tempname M
quietly matrix accum `M' = `indeps' [iweight=`res2'] ///
if `touse' , `constant'
matrix `V' = (`N'/(`N'-(`k')))*`xpxi'*`M'*`xpxi'
local vce "robust"
local vcetype "Robust"
ereturn post `b' `V', esample(`touse') buildfvinfo
ereturn scalar N = `N'
ereturn scalar rank = `k'
ereturn local vce "`vce'"
ereturn local vcetype "`vcetype'"
ereturn local cmd "myregress8"
ereturn display
The syntax command on line 5 puts “noconstant” into the local macro constant if the user types nocons, noconst, noconsta, noconstan, or noconstant; otherwise, the local macro constant is empty. The
minimal abbreviation of option noconstant is nocons because the lowercase no is followed by CONStant. Note that specifying the option creates the local macro constant because the no is followed by
uppercase letters specifying the minimum abbreviation.
To implement the option, I specified what is contained in the local macro constant as an option on the matrix accum command on line 14 and on the matrix accum command spread over lines 33 and 34. The
matrix accum command that begins on line 33 is too long for one line. I used /// to comment out the end-of-line character and continue the command on line 34.
I now illustrate the noconstant option.
Example 2: myregress8 with option noconstant
. myregress8 price mpg trunk ibn.rep78, noconstant
| Coef. Std. Err. z P>|z| [95% Conf. Interval]
mpg | -262.7053 73.49434 -3.57 0.000 -406.7516 -118.6591
trunk | 41.75706 93.9671 0.44 0.657 -142.4151 225.9292
rep78 |
1 | 9726.377 2790.009 3.49 0.000 4258.06 15194.69
2 | 10381.17 2607.816 3.98 0.000 5269.943 15492.39
3 | 10896.98 2555.364 4.26 0.000 5888.561 15905.4
4 | 11199.73 2588.19 4.33 0.000 6126.97 16272.49
5 | 12623.27 2855.763 4.42 0.000 7026.073 18220.46
Using t or F distributions
The output tables reported in examples 1 and 2 use the normal distribution to compute \(p\)-values and confidence intervals, because Wald-based postestimation commmands like test and ereturn display
use the normal or the \(\chi^2\) distribution unless the residual degrees of freedom are stored in e(df_r).
Code block 3: myregress9.ado
*! version 9.0.0 30Nov2015
program define myregress9, eclass
version 14
syntax varlist(numeric ts fv) [if] [in] [, Robust noCONStant ]
marksample touse
gettoken depvar indeps : varlist
_fv_check_depvar `depvar'
tempname zpz xpx xpy xpxi b V
tempvar xbhat res res2
quietly matrix accum `zpz' = `varlist' if `touse' , `constant'
local N = r(N)
local p = colsof(`zpz')
matrix `xpx' = `zpz'[2..`p', 2..`p']
matrix `xpy' = `zpz'[2..`p', 1]
matrix `xpxi' = syminv(`xpx')
local k = `p' - diag0cnt(`xpxi') - 1
matrix `b' = (`xpxi'*`xpy')'
quietly matrix score double `xbhat' = `b' if `touse'
quietly generate double `res' = (`depvar' - `xbhat') if `touse'
quietly generate double `res2' = (`res')^2 if `touse'
if "`robust'" == "" {
quietly summarize `res2' if `touse' , meanonly
local sum = r(sum)
local s2 = `sum'/(`N'-(`k'))
matrix `V' = `s2'*`xpxi'
else {
tempname M
quietly matrix accum `M' = `indeps' [iweight=`res2'] ///
if `touse' , `constant'
matrix `V' = (`N'/(`N'-(`k')))*`xpxi'*`M'*`xpxi'
local vce "robust"
local vcetype "Robust"
ereturn post `b' `V', esample(`touse') buildfvinfo
ereturn scalar N = `N'
ereturn scalar rank = `k'
ereturn scalar df_r = `N'-`k'
ereturn local vce "`vce'"
ereturn local vcetype "`vcetype'"
ereturn local cmd "myregress8"
ereturn display
Line 42 of myregress9.ado stores the residual degrees of freedom in e(df_r). Example 3 illustrates that ereturn display and test now use the \(t\) and \(F\) distributions.
Example 3: t or F distributions after myregress9
. myregress9 price mpg trunk ibn.rep78, noconstant
| Coef. Std. Err. t P>|t| [95% Conf. Interval]
mpg | -262.7053 73.49434 -3.57 0.001 -409.6184 -115.7923
trunk | 41.75706 93.9671 0.44 0.658 -146.0805 229.5946
rep78 |
1 | 9726.377 2790.009 3.49 0.001 4149.229 15303.53
2 | 10381.17 2607.816 3.98 0.000 5168.219 15594.12
3 | 10896.98 2555.364 4.26 0.000 5788.882 16005.08
4 | 11199.73 2588.19 4.33 0.000 6026.011 16373.45
5 | 12623.27 2855.763 4.42 0.000 6914.677 18331.85
. test trunk
( 1) trunk = 0
F( 1, 62) = 0.20
Prob > F = 0.6583
Done and undone
I added an option for the robust estimator of the VCE, I added an option to suppress the constant term, and I stored the residual degrees of freedom in e(df_r) so that Wald-based postestimation
commands will use \(t\) or \(F\) distributions. I illustrated option parsing by example, but I skipped the general theory and many details. Type . help syntax for more details about parsing options
using the syntax command.
In the next post, I implement the modern syntax for robust and cluster-robust standard errors.
Cameron, A. C., and P. K. Trivedi. 2005. Microeconometrics: Methods and applications. Cambridge: Cambridge University Press.
Stock, J. H., and M. W. Watson. 2010. Introduction to Econometrics. 3rd ed. Boston, MA: Addison Wesley New York.
Wooldridge, J. M. 2015. Introductory Econometrics: A Modern Approach. 6th ed. Cincinnati, Ohio: South-Western.
Categories: Programming #StataProgramming, ado, ado-command, ado-file, do-file, econometrics, OLS, programming, Stata matrix command, Stata matrix function, statistics, syntax | {"url":"https://blog.stata.com/2015/12/01/programming-an-estimation-command-in-stata-allowing-for-options/","timestamp":"2024-11-06T14:12:58Z","content_type":"application/xhtml+xml","content_length":"78582","record_id":"<urn:uuid:fa3f8f00-b1ed-44e2-8818-066d018bfc7c>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00199.warc.gz"} |
Multiplying Multi Digit Numbers Worksheet Pdf
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Multiplying Multi Digit Numbers Worksheet Pdf
Multiplying Multi Digit Numbers Worksheet Pdf -
There are several variants of each class of worksheet to allow for plenty of practice These two digit and three digit multiplication worksheets gradually introduce long multiplication problems to
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Multiplication Table 2 Worksheet Pdf
Multiplication Table 2 Worksheet Pdf
Set A5 Number Operations Multi digit Multiplication Includes Activity 1 Multi Digit Multiplication Pre Assessment A5 1 Activity 2 Multiplying by 10 100 1000 A5 17 Activity 3 Multiplying Single Digits
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1 1 3 0 9 9 1 2 9 6 5 2 1 2 6 2 8 0 Because
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Multiplying Multi Digit Numbers Worksheet
Multiplying Multi Digit Numbers Worksheet
Use the buttons below to print open or download the PDF version of the Multiplying 2 Digit by 2 Digit Numbers A math worksheet The size of the PDF file is 40503 bytes Preview images of the first and
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Free Multiply By 3 Worksheets
Multiple Digit Multiplication Worksheets
Check more of Multiplying Multi Digit Numbers Worksheet Pdf below
2 Digit By 2 Digit Multiplication Worksheets With Answers Times Tables Worksheets
Multiplying 2 Digit Numbers Worksheet Printable Word Searches
Multiplying 3 Digit By 3 Digit Numbers A
3 Digit Numbers Free Printable Multiplication Worksheets
Printables Multiplying Multi Digit Numbers HP Official Site
Multiplying 2 Digit By 2 Digit Numbers A Multiplying Two Digit Numbers Worksheet
Long Multiplication Worksheets Math Drills
Most Popular Long Multiplication Worksheets this Week Long Multiplication Worksheets Other Long Multiplication Strategies This page includes Long Multiplication worksheets for students who have
mastered the basic multiplication facts and are learning to multiply 2 3 4 and more digit numbers Sometimes referred to as long
Multiplication Worksheets Multi Digit Super Teacher Worksheets
Multiplication 4 Digits Times 1 Digit Practice finding the products of 4 digit numbers and 1 digit numbers example 4 527x9 Multiplication 2 Digits Times 2 Digits These printables have pairs of double
digit numbers for students to multiply together example 35x76 Multiplication 3 Digits Times 2 Digits On these exercises students
Most Popular Long Multiplication Worksheets this Week Long Multiplication Worksheets Other Long Multiplication Strategies This page includes Long Multiplication worksheets for students who have
mastered the basic multiplication facts and are learning to multiply 2 3 4 and more digit numbers Sometimes referred to as long
Multiplication 4 Digits Times 1 Digit Practice finding the products of 4 digit numbers and 1 digit numbers example 4 527x9 Multiplication 2 Digits Times 2 Digits These printables have pairs of double
digit numbers for students to multiply together example 35x76 Multiplication 3 Digits Times 2 Digits On these exercises students
3 Digit Numbers Free Printable Multiplication Worksheets
Multiplying 2 Digit Numbers Worksheet Printable Word Searches
Printables Multiplying Multi Digit Numbers HP Official Site
Multiplying 2 Digit By 2 Digit Numbers A Multiplying Two Digit Numbers Worksheet
Multiplying 2 Digit By 2 Digit Worksheet Pdf
3 Digit By 2 Digit Multiplication Worksheets Pdf Times Tables Worksheets 2 Digit
3 Digit By 2 Digit Multiplication Worksheets Pdf Times Tables Worksheets 2 Digit
Multiplication Worksheets Dynamically Created Multiplication Worksheets | {"url":"https://szukarka.net/multiplying-multi-digit-numbers-worksheet-pdf","timestamp":"2024-11-14T03:53:58Z","content_type":"text/html","content_length":"27192","record_id":"<urn:uuid:06e02060-bef0-45de-8ecb-88478120def6>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00417.warc.gz"} |
How do you calculate thrust?
How do you calculate thrust? Thrust is the force which moves an aircraft through the air. Thrust is generated by the engines of the airplane. F = ((m * V)2 – (m * V)1) /
How do you calculate thrust?
1. Thrust is the force which moves an aircraft through the air. Thrust is generated by the engines of the airplane.
2. F = ((m * V)2 – (m * V)1) / (t2 – t1)
3. F = m * a.
4. m dot = r * V * A.
5. F = (m dot * V)e – (m dot * V)0.
6. F = (m dot * V)e – (m dot * V)0 + (pe – p0) * Ae.
What is thrust of a propeller?
The force of the propeller pushing the air backwards is called the thrust, this places the propeller blades under pressure and bends the blades forward.
How do you calculate thrust on a boat?
Thrust – Boat Weight and Hull Design A beginning rule of thumb is that you want a minimum of 2 lbs of thrust for every 100lbs. For example, if you have a 3000lb boat, fully loaded, then the
calculation is (3000/100) * 2 = 60lbs of thrust.
Is force and thrust same?
Thrust is a mechanical force, so the propulsion system must be in physical contact with a working fluid to produce thrust. Thrust is generated most often through the reaction of accelerating a mass
of gas. Since thrust is a force, it is a vector quantity having both a magnitude and a direction.
What is the formula of thrust and pressure?
Complete solution:
Thrust Pressure
(1) Force acting perpendicular to the surface of an object is known as Thrust. (1) Thrust acting per unit area is known as pressure.
(2) Formula of thrust=pressure×area on which it acts (2) Formula of pressure=Area on which it acts
What is thrust of motor?
Thrust is the force which moves an aircraft through the air. Since thrust is a force, it is a vector quantity having both a magnitude and a direction. The engine does work on the gas and accelerates
the gas to the rear of the engine; the thrust is generated in the opposite direction from the accelerated gas.
What is static thrust?
The static thrust is the thrust measured with the engine stationary, as would be the case when the aircraft is initiating the take-off roll. The actual thrust produced drops off with forward speed,
with the drop-off more pronounced for turbofans than for turbojets.
How do you determine thrust propeller?
The thrust (F) is equal to the mass flow rate (m dot) times the difference in velocity (V). The mass flow through the propulsion system is a constant, and we can determine the value at the plane of
the propeller. | {"url":"https://missionalcall.com/2021/04/21/how-do-you-calculate-thrust/","timestamp":"2024-11-11T10:41:50Z","content_type":"text/html","content_length":"55185","record_id":"<urn:uuid:5a919786-c42c-46d3-8485-c30d47783d81>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00361.warc.gz"} |
Liters to gallons (EN) | Convertipro
Liters to Gallons Converter: Usage, Formulas, and Origin
The Liters to Gallons converter is a convenient tool for performing volume conversions between these two commonly used units of measurement worldwide for quantifying liquids such as water, gasoline,
and other substances. This article will guide you through the usage of this converter, the mathematical formulas used for conversions, and we will also explore the origin of the Gallon unit of
How the Liters to Gallons Converter Works:
The Liters to Gallons converter employs simple mathematical formulas to perform conversions between these two volume units. Here are the conversion formulas used:
Conversion from Liters to Gallons: Gallons = Liters / 3.78541
Conversion from Gallons to Liters: Liters = Gallons * 3.78541
How to Use the Liters to Gallons Converter:
The Liters to Gallons converter is user-friendly. Follow these steps to perform a conversion:
Step 1: Enter the number of Liters you want to convert in the provided field.
Step 2: The converter will automatically perform the calculation and display the result in Gallons just below the Liters input field.
Step 3: If you want to perform a conversion from Gallons to Liters, enter the number of Gallons in the dedicated second input field.
Step 4: The result in Liters will be displayed automatically below the Gallons input field.
Example: Convert 10 Liters to Gallons and 5 Gallons to Liters.
10 Liters / 3.78541 = 2.64172 Gallons
5 Gallons * 3.78541 = 18.92705 Liters
10 Liters are approximately equivalent to 2.64172 Gallons.
5 Gallons are approximately equivalent to 18.92705 Liters.
Note on the Origin of the Gallon Unit:
The Gallon unit of measurement is used in many countries to quantify liquids such as gasoline, milk, water, etc. The origin of the word "Gallon" is attributed to its historical usage in England. The
term "Gallon" comes from the Old French word "galon," which means "liquid measure." Initially, there were various sizes of Gallons in different regions, but in the 18th century, the "Imperial Gallon"
was defined in England, equivalent to about 4.54609 liters. In the United States, the "US Gallon" is slightly different, equivalent to about 3.78541 liters.
The Liters to Gallons converter is a convenient and useful tool for professionals and the general public to perform accurate and swift volume conversions. With the simple conversion formulas
presented in this article, you can effortlessly perform conversions between these two volume units. These online converters are widely used in various fields such as the automotive industry, food
industry, and daily activities where liquid measurements are needed. | {"url":"https://www.convertipro.com/en/liters-to-gallons-en","timestamp":"2024-11-04T23:32:19Z","content_type":"text/html","content_length":"167447","record_id":"<urn:uuid:07abca02-1d5d-4a5c-9328-a2ad3b555dfd>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00262.warc.gz"} |
4th IAEA Technical Meeting on Fusion Data Processing, Validation and Analysis
Salazar Luigui
University of Lorraine - CNRS
54000 Nancy, France
CEA, IRFM
13108 Saint Paul lez Durance, France
Email: luigui.salazar@cea.fr
Stéphane HEURAUX
University of Lorraine - CNRS
54000 Nancy, France
Roland SABOT
CEA, IRFM
13108 Saint Paul lez Durance, France
The identification of turbulence sources... | {"url":"https://conferences.iaea.org/event/251/contributions/","timestamp":"2024-11-05T20:33:48Z","content_type":"text/html","content_length":"239731","record_id":"<urn:uuid:1bf1fe31-2760-448b-8e93-dcad79da3069>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00708.warc.gz"} |
677 feet per second to meters per second
Speed Converter - Feet per second to meters per second - 8,677 meters per second to feet per second
This conversion of 8,677 feet per second to meters per second has been calculated by multiplying 8,677 feet per second by 0.3048 and the result is 2,644.7496 meters per second. | {"url":"https://unitconverter.io/feet-per-second/meters-per-second/8677","timestamp":"2024-11-05T14:17:16Z","content_type":"text/html","content_length":"15428","record_id":"<urn:uuid:12d92be2-3210-474c-a304-e5769efbdeeb>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00029.warc.gz"} |
Dynamics of Two Degree Freedom System Questions and Answers - Sanfoundry
Earthquake Engineering Questions and Answers – Dynamics of Two Degree Freedom System
This set of Earthquake Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Dynamics of Two Degree Freedom System”.
1. Which of the following is not a component of every dynamic system?
a) A mechanism for storing strain energy
b) Mechanism for storing kinetic energy
c) Mechanism to measure the velocity
d) An energy dissipation mechanism
View Answer
Answer: c
Explanation: There is no mechanism to measure the velocity in any dynamic system. The mass element (m) stores the kinetic energy and it is also responsible for generation of inertia forces. The
spring element (k) stores the potential energy and it is responsible for generation of elastic restoring forces. The dashpot (c) represents the damper for dissipation or loss of energy.
2. Which of the following inferences can be drawn from the stability and accuracy considerations?
a) Positive real eigenvalues lead to a change in sign
b) Negative real eigenvalues do not lead to change in sign at each step
c) Complex eigenvalues do not lead to change in sign
d) If modulus of eigenvalue is smaller then 1, the involution converges towards zero
View Answer
Answer: d
Explanation: The involution converges towards zero for modulus of eigenvalue is smaller then 1, is the only correct inferences drawn from the stability and accuracy considerations. Other inferences
are Positive real eigenvalues do not lead to a change in sign, Negative real eigenvalues lead to change in sign at each step and Complex eigenvalues may lead to change in sign depending on each step.
3. Duhamel integral is applicable for both linear and nonlinear systems.
a) True
b) False
View Answer
Answer: b
Explanation: Duhamel integral is an approach to evaluate the dynamic response of SDOF systems. Duhamel integral involves interpolation of the integrand and therefore, is strictly applicable for
linear systems only. The second approach by approximation of the derivatives in the differential equations of motions is applicable for both linear and nonlinear systems.
4. For what value of modulus of all eigenvalues, stability is ensured?
a) Modulus of eigenvalues = 5
b) Modulus of eigenvalues > 1
c) Modulus of eigenvalues < 1
d) Modulus of eigenvalues = 2
View Answer
Answer: c
Explanation: If we want the stability to be ensured, the modulus of eigenvalues should be less than 1. In this case, the algorithm would simulate a damped motion which is often referred to as
algorithmic damping. To achieve a satisfactory performance, the presence of complex eigenvalues with a modulus of 1 is necessary.
5. Which type of nature of damping force is assumed in a 2-DOF system?
a) Coulomb
b) Viscous
c) Dry friction
d) Slip
View Answer
Answer: b
Explanation: The nature of damping force is assumed to be of the viscous type. They are an approximate representation of the combined action of all energy dissipation mechanisms present in a
vibrating system. Precise nature of damping force is not very important for dynamic response computations.
6. For critical damping in the range of 5-10%, which of the following is true?
a) Maximum spectral acceleration is of the order of 1.5 times the maximum ground acceleration
b) Maximum spectral velocity is of the same order as the maximum ground velocity
c) Maximum spectral displacement is of the same order as the maximum ground displacement
d) Maximum spectral displacement is of the order of twice the maximum ground displacement
View Answer
Answer: c
Explanation: For critical damping in the range of 5-10%, Maximum spectral displacement is of the same order as the maximum ground displacement, maximum spectral velocity is of the order 1.5 times the
maximum ground velocity and maximum spectral acceleration is of the order of 2 times the maximum ground acceleration.
7. According to the deformation characteristics, what do you call the buildings, if the floor moves only in the horizontal direction without any rotation of horizontal section?
a) Moment buildings
b) Moment-shear buildings
c) Damped buildings
d) Shear buildings
View Answer
Answer: d
Explanation: Buildings are basically divided into two groups based on their deformation characteristics. If the floor moves only in horizontal direction and there is no rotation of a horizontal
section, it is termed as shear buildings. If the floor moves in both horizontal and rotational directions, they are referred to as moment-shear buildings.
Sanfoundry Global Education & Learning Series – Earthquake Engineering.
To practice all areas of Earthquake Engineering, here is complete set of Multiple Choice Questions and Answers. | {"url":"https://www.sanfoundry.com/earthquake-engineering-questions-answers-dynamics-two-degree-freedom-system/","timestamp":"2024-11-14T11:40:31Z","content_type":"text/html","content_length":"131158","record_id":"<urn:uuid:2231ced3-4bcc-417a-9b10-4b7ac7e94cfe>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00540.warc.gz"} |
Out-of-Distribution Detection for Deep Neural Networks
This example shows how to detect out-of-distribution (OOD) data in deep neural networks.
OOD data detection is the process of identifying inputs to a deep neural network that might yield unreliable predictions. OOD data refers to data that is different from the data used to train the
model. For example, data collected in a different way, at a different time, under different conditions, or for a different task than the data on which the model was originally trained.
By assigning confidence scores to the predictions of a network, you can classify data as in-distribution (ID) or OOD. You can then choose how you treat OOD data. For example, you can choose to reject
the prediction of a neural network if it detects OOD data.
This example requires the Deep Learning Toolbox™ Verification Library. To download and install the support package, use the Add-On Explorer. Alternatively, see Deep Learning Toolbox Verification
Load Data
This example uses MATLAB® files converted by MathWorks® from the Tennessee Eastman Process (TEP) simulation data [1]. These files are available at the MathWorks support files site. For more
information, see the disclaimer: https://www.mathworks.com/supportfiles/predmaint/chemical-process-fault-detection-data/Disclaimer.txt.
Download the training and test files. Depending on your internet connection, the download process can take a long time.
faultfreetrainingFileName = matlab.internal.examples.downloadSupportFile("predmaint","chemical-process-fault-detection-data/faultfreetraining.mat");
faultfreetestingFileName = matlab.internal.examples.downloadSupportFile("predmaint","chemical-process-fault-detection-data/faultfreetesting.mat");
faultytrainingFileName = matlab.internal.examples.downloadSupportFile("predmaint","chemical-process-fault-detection-data/faultytraining.mat");
faultyttestingFileName = matlab.internal.examples.downloadSupportFile("predmaint","chemical-process-fault-detection-data/faultytesting.mat");
Load the downloaded files into the MATLAB workspace. For more information about this data set, see Chemical Process Fault Detection Using Deep Learning.
The data set consists of four MAT files: fault-free training, fault-free testing, faulty training, and faulty testing.
• The fault-free training and testing data sets each comprise 500 simulations of fault-free data. Each fault-free simulation has 52 channels and the class label 0.
• The faulty training and testing data sets each comprise 10,000 simulations corresponding to 500 simulations for each of 20 possible faults. Simulations 1–500 correspond to class label 1,
simulations 501–1000 correspond to class label 2, and so on. Each simulation has 52 channels.
The length of each simulation depends on the data set. All simulations were sampled every three minutes.
• Each simulation in the training data sets contains 500 time samples from 25 hours of simulation.
• Each simulation in the testing data sets contains 960 time samples from 48 hours of simulation.
Load Pretrained Network
Load a pretrained network. This network has been trained using the training method from the Chemical Process Fault Detection Using Deep Learning example. Because of the randomness of training, if you
train this network yourself, you will likely see different results.
Preprocess Data
Remove data entries with the fault class labels 3, 9, and 15 in the training and testing data sets. These faults are not present in the original training data set. Because the model was not trained
using these faults, they are OOD inputs to the network.
Use the supporting function helperPrepareDataSets to prepare the data sets for training and testing. The function performs these steps:
1. Combine the fault-free data, corresponding to class label 0, with the faulty data, corresponding to class labels 1-20.
2. Hold-out the simulations for faults 3, 9, and 15.
3. Normalize the data.
4. Create an array of class labels.
Process the training and test data sets.
classesToRemove = [3 9 15];
[XTrain,XTrainHoldOut,TTrain,TTrainHoldOut] = helperPrepareDataSets(faultfreetraining,faultytraining,classesToRemove);
[XTest,XTestHoldOut,TTest,TTestHoldOut] = helperPrepareDataSets(faultfreetesting,faultytesting,classesToRemove);
Visualize Data
The XTrain and XTest data sets each contain 500 fault-free simulations followed by 8500 faulty simulations corresponding to 500 simulations for each of the 17 faults in the training set. Visualize
the fault-free and faulty training data for four of the 52 channels.
Plot an example of fault-free data. The first 500 simulations correspond to the fault-free data.
xlabel("Time Step");
title("Fault-Free Data (Class 0)")
legend("Channel " + string(1:numChannelsToPlot),Location="northeastoutside")
Plot an example of faulty data. Simulations 501–1000 correspond to data with fault 1.
xlabel("Time Step")
title("Faulty Data (Class 1)")
legend("Channel " + string(1:numChannelsToPlot),Location="northeastoutside")
Test Network
Test the trained network by classifying the fault type for each of the test observations. To make predictions with multiple observations, use the minibatchpredict function. Convert the classification
scores to class labels using the scores2label function. The minibatchpredict function automatically uses a GPU if one is available. Using a GPU requires a Parallel Computing Toolbox™ license and a
supported GPU device. For information on supported devices, see GPU Computing Requirements (Parallel Computing Toolbox). Otherwise, the function uses the CPU.
Because the data has sequences with rows and columns corresponding to channels and time steps, respectively, specify the input data format "CTB" (channel, time, batch).
scores = minibatchpredict(net,XTest,InputDataFormats="CTB");
YPred = scores2label(scores,classNames);
Calculate the accuracy.
acc = sum(YPred == TTest)/numel(YPred)
Plot the confusion matrix using the true class labels and the predicted labels.
Test the trained network on the held-out data.
scores = minibatchpredict(net,XTestHoldOut,InputDataFormats="CTB");
YPredHoldOut = scores2label(scores,classNames);
Plot the predicted class labels for the held-out data. The network must predict the class of the held-out data as one of the classes on which it was trained. Here, the network predicts class 0
(fault-free) for all of the held-out test observations. Because the network was not trained using these fault labels, it cannot classify the faults correctly. Therefore, the network predicts
"fault-free" even though the data is faulty.
xlabel("Predicted Fault")
title("Predicted Fault Class for OOD Test Data")
Analyze Softmax Scores
The data set contains two types of data:
• In-distribution (ID) — Data used to train the network. This data corresponds to faults with class labels 0, 1, 2, 4–8,10–14, and 16–20.
• Out-of-distribution (OOD) — Data that is different from the training data, for example, the data corresponding to faults 3, 9, and 15. The network cannot classify this type of data reliably.
You can use OOD detection to assign a confidence score to the network predictions. A lower confidence value corresponds to data that is more likely to be OOD.
In this example, you assign confidence scores to network predictions by using the softmax probabilities to compute a distribution confidence score for each observation. ID data usually has a higher
maximum softmax probability than OOD data [2]. You can then apply a threshold to the softmax probabilities to determine whether an input is ID or OOD. This technique is called the baseline method.
Compute the maximum softmax scores for each observation in the training data sets.
scoreTraining = max(minibatchpredict(net,XTrain,InputDataFormats="CTB"),[],2);
scoreTrainingHoldOut = max(minibatchpredict(net,XTrainHoldOut,InputDataFormats="CTB"),[],2);
Plot histograms of the scores for the ID data (scoreTraining) and the OOD data (scoreTrainingHoldOut). To compare the distributions, set the histogram normalization to "probability". The plot shows a
clear separation between the distribution confidence scores for the ID and OOD data. A threshold of around 0.99 reliably separates the scores of the ID and OOD observations.
binWidth = 0.001;
hold on
hold off
xlim([0.95 1]);
legend("Training data (ID)", "Held-out training data (OOD)",Location="northwest")
xlabel("Distribution Confidence Scores")
ylabel("Relative Percentage")
Out-of-Distribution Detection
You can use the isInNetworkDistribution function to determine whether an observation is ID or OOD. The function takes as input a network, data, and a threshold. The function uses the maximum softmax
scores to find the distribution confidence scores and the specified threshold to classify data as ID or OOD.
The isInNetworkDistribution function requires the data as a formatted dlarray object or a minibatchqueue object that returns a dlarray. Convert the data to a formatted dlarray object using the
convertDataToDlarray supporting function, found at the end of this example.
XTrain = convertDataToDlarray(XTrain);
XTrainHoldOut = convertDataToDlarray(XTrainHoldOut);
XTest = convertDataToDlarray(XTest);
XTestHoldOut = convertDataToDlarray(XTestHoldOut);
Manual Threshold
You can use the histogram of the softmax scores to manually choose a threshold that visually separates the maximum softmax scores in the training data set. This process is called OOD data
Use the threshold to classify the test data as ID or OOD. The isInNetworkDistribution function returns a logical 1 (true) for each observation with maximum softmax above the specified threshold,
corresponding to that observation being classified as ID.
threshold = 0.99;
tfID = isInNetworkDistribution(net,XTest,Threshold=threshold);
tfOOD = isInNetworkDistribution(net,XTestHoldOut,Threshold=threshold);
You can test the performance of the OOD data discriminator by calculating the true positive rate (TPR) and the false positive rate (FPR).
• TPR — Proportion of ID observations correctly classified as ID.
• FPR — Proportion of OOD observations incorrectly classified as ID.
Compute the TPR and FPR using the helperPredictionMetrics helper function. A good discriminator has a TPR close to 1 and a FPR close to 0.
[TPR,FPR] = helperPredictionMetrics(tfID,tfOOD)
Optimal Threshold
Rather than manually selecting a threshold, you can use the threshold that best separates the softmax scores. You can find the optimal threshold by maximizing the TPR and minimizing the FPR. Create a
distribution discriminator object using the networkDistributionDiscriminator function. You can use this object to find the optimal threshold.
Use the networkDistributionDiscriminator function with the network as input. Use the training data as ID data and the held-out training data as OOD data. Set the method input to "baseline" to use the
maximum softmax scores as the distribution confidence scores. The discriminator determines the optimal threshold.
method = "baseline";
discriminatorOptimized = networkDistributionDiscriminator(net,XTrain,XTrainHoldOut,method)
discriminatorOptimized =
BaselineDistributionDiscriminator with properties:
Method: "baseline"
Network: [1×1 dlnetwork]
Threshold: 0.9861
Use the distribution discriminator to classify the test data as ID or OOD.
tfIDOptimized = isInNetworkDistribution(discriminatorOptimized,XTest);
tfOODOptimized = isInNetworkDistribution(discriminatorOptimized,XTestHoldOut);
Compute the TPR and FPR using the optimized threshold.
[TPROptimized,FPROptimized] = helperPredictionMetrics(tfIDOptimized,tfOODOptimized)
FPROptimized = 6.6667e-04
Threshold for Specified True Positive Goal
You can set a target number of true positives at the expense of a greater number of false positives. Set a true positive goal of 95% and use the training data to find a threshold. Again, use the
distribution discriminator to classify the test data as ID or OOD and examine the TPR and FPR for the test set.
discriminatorTPR = networkDistributionDiscriminator(net,XTrain,XTrainHoldOut,method,TruePositiveGoal=0.95);
tfIDTPR = isInNetworkDistribution(discriminatorTPR,XTest);
tfOODTPR = isInNetworkDistribution(discriminatorTPR,XTestHoldOut);
[TPROptimizedTPR,FPROptimizedTPR] = helperPredictionMetrics(tfIDTPR,tfOODTPR)
Compare Discriminators
Use the helperDistributionConfusionMatrix helper function to plot the confusion matrix resulting from the predictions using each of the three threshold choices.
title("Manual Threshold")
title("Optimal Threshold (TPR & FPR)")
title("Threshold (TPR of 0.95)")
Plot ROC Curve
The distribution discriminator object is a binary classifier that uses a threshold to classify network predictions as ID or OOD. Plot the receiver operating characteristic (ROC) curve for this binary
classifier to see the trade-off between true positive and false positive rates. The ROC curve represents every possible threshold. Add a point to the curve highlighting each threshold value.
scoresID = distributionScores(discriminatorOptimized,XTest);
scoresOOD = distributionScores(discriminatorOptimized,XTestHoldOut);
numObservationsID = size(scoresID,1);
numObservationsOOD = size(scoresOOD,1);
scores = [scoresID',scoresOOD'];
trueDataLabels = [
repelem("In-Distribution",numObservationsID), ...
rocObj = rocmetrics(trueDataLabels,scores,"In-Distribution");
hold on
plot(FPR,TPR,".", ...
MarkerSize=20, ...
DisplayName="Manual Threshold")
plot(FPROptimized,TPROptimized,".", ...
MarkerSize=20, ...
DisplayName="Optimal Threshold")
plot(FPROptimizedTPR,TPROptimizedTPR,".", ...
MarkerSize=20, ...
DisplayName="Threshold at TPR=0.95")
Verify Network Predictions
You can use the distribution discriminator object to add an extra level of verification to network predictions. For example, for every prediction that the network makes, the distribution
discriminator can confirm whether to reject the result based on the input classification. If the distribution discriminator determines that the input is OOD, then you can reject the result.
Suppose that a silent, temporary failure in the system alters a single fault-free simulation such that the data contains white noise from timestep 101-200.
faultfreetestingSample = extractdata(squeeze(XTest(:,1,:)));
alteredFaultFreeSignal = faultfreetestingSample;
alteredFaultFreeSignal(:,101:200) = randn(52,100);
Plot the first 300 timesteps of the original fault-free signal and an altered fault-free signal for four of the 52 channels.
plot(faultfreetestingSample(1:4, 1:300)')
ylim([-3 3])
xlabel("Time Step");
title("Fault-Free Data")
legend("Channel " + string(1:4),Location="northeastoutside")
plot(alteredFaultFreeSignal(1:4, 1:300)')
ylim([-3 3])
xlabel("Time Step")
title("Altered Fault-Free Data")
legend("Channel " + string(1:4),Location="northeastoutside")
Classify the altered fault-free signal. To make predictions with a single observation, use the predict function. To use a GPU, first convert the data to gpuArray.
dlbrokenFaultFreeSignal = dlarray(alteredFaultFreeSignal,'CTB');
if canUseGPU
dlbrokenFaultFreeSignal = gpuArray(dlbrokenFaultFreeSignal);
scores = predict(net,dlbrokenFaultFreeSignal);
Ypredi = scores2label(scores,classNames)
The network still classifies the altered signal as class label 0, which corresponds to "fault-free". However, this altered signal is from a different distribution than the data that the network sees
during training and the classification must be flagged in a safety-critical system.
Use the discriminator to determine whether the signal is ID or OOD. Use the isInNetworkDistribution function to test if the observation is ID.
tf = isInNetworkDistribution(discriminatorOptimized,dlbrokenFaultFreeSignal)
Apply the same alteration to all 500 fault-free signals and analyze the number of OOD samples detected. The discriminator successfully picks up this new fault and classifies most of the altered
simulations as OOD.
alteredFaultFreeSignals = XTest(:,1:500,:);
alteredFaultFreeSignals(:,:,101:200) = randn(52,500,100);
tf = isInNetworkDistribution(discriminatorOptimized,alteredFaultFreeSignals);
YPredAltered = repelem("Out-of-Distribution",length(tf));
YPredAltered(tf == 1) = "In-Distribution";
title("Predicted Distribution Class of Altered Fault-Free Simulations")
Helper Functions
The helperNormalizeData function normalizes the data using the same statistics as the training data.
function processed = helperNormalizeData(data)
limit = max(data.sample);
processed = helperPreprocess(data{:,:},limit);
% The network requires the input data to be normalized with respect to the training
% data. Loading the training data and computing these statistics is
% computationally expensive, so load precalculated statistics.
s = load("faultDetectionNormalizationStatistics.mat","tMean","tSigma");
processed = helperNormalize(processed,s.tMean,s.tSigma);
The helperPreprocess function uses the maximum sample number to preprocess the data. The sample number indicates the signal length, which is consistent across the data set. The function uses a
for-loop to go over the data set with a signal length filter to form sets of 52 signals. Each set is an element of a cell array. Each cell array contains data from a single simulation.
function processed = helperPreprocess(data,limit)
H = size(data,1);
processed = {};
for ind = 1:limit:H
x = data(ind:(ind+(limit-1)),4:end);
processed = [processed; x'];
The helperNormalize function uses the mean and standard deviation of the training data to normalize data.
function data = helperNormalize(data,m,s)
for ind = 1:size(data,1)
data{ind} = (data{ind} - m)./s;
The helperPrepareDataSets function prepares the data set for analysis. The function takes as input the fault-free data, the faulty data, and the faults to be removed. The function returns the faulty
data with the specified classes removed, the removed data, and the associated labels for both data sets. This is the same data processing performed before training.
function[dataProcessed,dataHoldOut,labels,labelsHoldOut] = helperPrepareDataSets(faultFreeData,faultyData,classesToRemove)
index = ismember(faultyData.faultNumber,classesToRemove);
data = [faultFreeData; faultyData(~index,:)];
dataHoldOut = faultyData(index,:);
dataProcessed = helperNormalizeData(data);
dataHoldOut = helperNormalizeData(dataHoldOut);
classesToKeep = 1:20;
classesToKeep = classesToKeep(~ismember(classesToKeep,classesToRemove));
labels = categorical([zeros(500,1); repmat(classesToKeep,1,500)']);
labelsHoldOut = categorical(repmat(classesToRemove,1,500)');
The convertDataToDlarray function converts the data to a dlarray object.
function dldata = convertDataToDlarray(data)
% Reshape the data.
dataSize = size(data,1);
dldata = reshape(data,1,1,dataSize);
% Convert the cell arrays to 3-D numeric arrays.
dldata = cell2mat(dldata);
% Convert the cell arrays to a dlarray object with data format labels.
dldata = dlarray(dldata,"CTB");
The helperDistributionConfusionMatrix function computes the confusion matrix for ID and OOD data. The function takes as input an array of logical values for the ID data and OOD data. A value of 1
(true) corresponds to the detector predicting that the observation is ID. A value of 0 (false) corresponding to the detector predicting that the observation is OOD.
function cm = helperDistributionConfusionMatrix(tfID,tfOOD)
trueDataLabels = [
repelem("ID",numel(tfID)), ...
predDataLabelsID = repelem("OOD",length(tfID));
predDataLabelsID(tfID == 1) = "ID";
predDataLabelsOOD = repelem("OOD",length(tfOOD));
predDataLabelsOOD(tfOOD == 1) = "ID";
predDataLabels = [predDataLabelsID,predDataLabelsOOD];
cm = confusionchart(trueDataLabels,predDataLabels);
The helperPredictionMetrics function computes the true positive rate and false positive rate for a binary classifier.
function [truePositiveRate,falseNegativeRate] = helperPredictionMetrics(tfID,tfOOD)
truePositiveRate = sum(tfID)/(sum(tfID)+sum(1-tfID));
falseNegativeRate = sum(tfOOD)/(sum(tfOOD) + sum(1-tfOOD));
[1] Rieth, C. A., B. D. Amsel, R. Tran., and B. Maia. "Additional Tennessee Eastman Process Simulation Data for Anomaly Detection Evaluation." Harvard Dataverse, Version 1, 2017. https://doi.org/
[2] Hendrycks, Dan, and Kevin Gimpel. A Baseline for Detecting Misclassified and Out of Distribution Examples in Neural Networks." arXiv:1610.02136 [cs.NE], October 3, 2018, https://arxiv.org/abs/
See Also
dlnetwork | dlarray | isInNetworkDistribution | networkDistributionDiscriminator | verifyNetworkRobustness | rocmetrics
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Meters to Miles c
How do you convert meters to miles?
To convert meters to miles, you need to know the conversion factor between the two units. The conversion factor for meters to miles is 0.00062137119. This means that one meter is equal to
approximately 0.00062137119 miles. To convert meters to miles, you simply multiply the number of meters by this conversion factor.
For example, let's say you have a distance of 1000 meters that you want to convert to miles. You would multiply 1000 by 0.00062137119, which gives you approximately 0.62137119 miles. Similarly, if
you have a distance of 500 meters, you would multiply 500 by 0.00062137119 to get approximately 0.310685595 miles.
What is a meter?
A meter is a unit of length in the metric system, and it is equivalent to 100 centimeters or 1,000 millimeters. It is the base unit of length in the International System of Units (SI) and is widely
used around the world for measuring distances. The meter was originally defined as one ten-millionth of the distance from the North Pole to the equator along a meridian passing through Paris, France.
However, in 1983, the meter was redefined as the distance traveled by light in a vacuum during a specific time interval.
What is a mile?
A mile is a unit of length commonly used in the United States and some other countries. It is equal to 5,280 feet or 1,760 yards. The word "mile" is derived from the Latin word "mille," meaning
thousand, as it originally represented the distance covered in 1,000 paces by a Roman legionary.
A mile is equivalent to 1760yds or 5280ft.
The mile is commonly used in the United States for measuring long distances, such as road distances and race distances. It is also used in the aviation and maritime industries for navigation
purposes. However, in most other countries, the metric system is used, and the kilometer is the preferred unit for measuring long distances.
Why would I want to convert meters to miles?
Converting meters to miles can be useful in a variety of situations. One common reason for this conversion is when dealing with long distances, especially in countries that primarily use the metric
system. For example, if you are planning a road trip in Europe and the distances are given in kilometers or meters, it might be more convenient for you to have an understanding of the equivalent
distances in miles. This can help you estimate travel times, plan rest stops, or compare distances to familiar landmarks or cities.
Another reason to convert meters to miles is when working with maps or geographical data. Many maps, especially those used for hiking or outdoor activities, often provide distances in meters or
kilometers. However, if you are more accustomed to thinking in terms of miles, converting these distances can make it easier to understand the scale of the map and plan your route accordingly.
Converting meters to miles can be helpful in certain professions or industries. For instance, architects or engineers who work on large-scale projects may need to convert measurements between
different units, including meters and miles. This ensures that they have a comprehensive understanding of the project's scope and can communicate effectively with clients or colleagues who may be
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Shielding effects of dexmedetomidine about ischaemia-reperfusion damage in the trial and error
Shielding effects of dexmedetomidine about ischaemia-reperfusion damage in the trial and error
As a result, the cumulative release rate of 8HQ is greater in the fundamental problem compared to the acid condition.Carbohydrates constitute one of many four crucial classes of biomacromolecules but
haven’t been examined by 2D-IR spectroscopy up to now. Likewise as for proteins, too little indigenous vibrational reporter teams, along with their particular huge architectural diversity, results in
spectrally congested infrared spectra already for solitary carbohydrates. Biophysical studies are additional impeded by the strong overlap between liquid settings and carb modes patient medication
knowledge . Here, we illustrate the effective use of the known vibrational reporter group thiocyanate (SCN) as a label in glucose. In this first study, we could do IR and 2D-IR spectroscopy of
β-glucose with SCN at the C2 position in chloroform. Upon improved synthesis therefore the elimination of all protecting groups, we successfully performed 2D-IR spectroscopy of β-glucose in H2O. All
experimental email address details are compared to those of methyl-thiocyanate as a reference sample. Overall, we reveal that the thought of utilizing site-specific vibrational reporter groups may be
utilized in carbs. Hence, biophysical researches with 2D-IR spectroscopy can now increase to glycoscience.Starting through the orthogonal characteristics of every provided set of variables with
regards to the projection adjustable utilized to derive the Mori-Zwanzig equation, a set of combined Volterra equations is gotten that link the projected time correlation features between all of the
factors of interest. This set of equations is solved making use of standard numerical inversion options for Volterra equations, causing an extremely convenient however efficient technique to get any
projected time correlation function or share towards the memory kernel entering a generalized Langevin equation. Applying this strategy, the memory kernel regarding the diffusion of tagged particles
in a bulk Lennard-Jones liquid is investigated as much as the long-term regime to demonstrate that the repulsive-attractive cross-contribution to memory results signifies a small but non-zero
contribution towards the self-diffusion coefficient.The generalized Langevin mode analysis (GLMA) is placed on chemical responses in biomolecules in solution. The theory views a chemical reaction in
answer concomitant pathology as a barrier-crossing process, similar to the Marcus principle. The buffer is defined as the crossing point of two free-energy surfaces which are related to the reactant
and item for the response. It is assumed that both free-energy areas are quadratic or harmonic. The assumption is founded on the Kim-Hirata theory of structural fluctuation of protein, which proves
that the fluctuation around an equilibrium framework is quadratic according to the structure or atomic coordinates. The quadratic surface is a composite of numerous harmonic functions with different
modes or frequencies. The height associated with activation barrier are dependent on the mode or frequency-the less the frequency, the lower the buffer. Thus, it is crucial to decouple the
fluctuational modes into a hierarchical order. GLMA is impeccable for this purpose. It is crucial for a theoretical study of chemical reactions to select a reaction coordinate along which the
reaction profits. We suppose that the mode whoever center of coordinate and/or the regularity changes most before and after the reaction could be the one highly relevant to the chemical reaction and
choose the coordinate given that reaction coordinate. The price of effect across the response coordinate is krate=νexp-ΔF(†)/kBT, which will be like the Marcus appearance for the electron transfer
reaction. When you look at the equation, ΔF(†) is the activation buffer defined by ΔF(†)≡F(r)Q†-F(r)(Qeq (r)), where F(r)(Qeq (roentgen)) and F(r)Q† denote the free energies at equilibrium Qeq
(roentgen) while the crossing point Q†, respectively, both in the no-cost power area of the reactant.The instability of a cryogenic 4He jet leaving through a tiny nozzle into machine contributes to
the synthesis of 4He drops, that are considered ideal matrices for spectroscopic studies of embedded atoms and particles. Right here, we present a He-density useful theory (DFT) information of
droplet development resulting from jet busting and contraction of superfluid 4He filaments. Whereas the fragmentation of long jets closely employs the predictions of linear theory for inviscid
liquids, leading to droplet trains interspersed with smaller satellite droplets, the contraction of filaments with a piece ratio check details larger than a threshold value leads to the nucleation of
vortex rings, which hinder their breakup into droplets.A vast array of phenomena, ranging from chemical reactions to stage transformations, tend to be examined when it comes to a totally free energy
surface defined with regards to just one or multiple order variables. Enhanced sampling methods are typically made use of, especially in the clear presence of huge free power barriers, to estimate
no-cost energies making use of biasing protocols and sampling of change paths. Kinetic reconstructions of free energy obstacles of advanced height are done, with regards to just one purchase
parameter, employing the steady-state properties of unconstrained simulation trajectories when buffer crossing is doable with reasonable computational effort. Considering such instances, we describe
a strategy to estimate no-cost energy surfaces with respect to numerous order parameters from a stable state ensemble of trajectories. The strategy relates to instances when the transition rates
between sets of purchase parameter values considered just isn’t affected by the existence of an absorbing boundary, whereas the macroscopic fluxes and sampling probabilities tend to be.
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How can I learn pi in math?
Pi (π) is one of the most important mathematical constants that exists. It is defined as the ratio of a circle’s circumference to its diameter. Pi is an irrational number, meaning its digits go on
forever without repeating in a pattern. Learning and understanding pi is an essential part of math education.
Why is pi important?
Pi arises in many areas of mathematics and physics. Here are some reasons why pi is so important:
– Pi is fundamental to the study of circles and spheres. It allows us to easily calculate circumference, area, surface area, and volume for these shapes.
– Pi shows up in trigonometry in relations like the cosine function and trigonometric identities.
– Pi is ubiquitous in math topics like series expansions, trigonometric functions, and integrals. Especially, integrals involving trigonometric functions often require pi.
– In physics, pi is needed to describe periodic phenomena like how wavelengths and frequencies relate for waves.
– Pi is useful for describing rotations and angles in fields like engineering.
– Pi appears in probability calculations dealing with circles or spheres.
– Pi is required for various mathematical equations in many scientific and engineering fields.
So in summary, pi arises everywhere circles and periodic phenomena occur. It is essential for areas requiring trigonometry. Pi’s importance makes it a mathematical constant that all students must
Ways to remember the value of pi
Pi is commonly approximated as 3.14 or 3.14159. While pi has infinitely many digits, just remembering the first few is sufficient for most applications. Here are some techniques to easily recall pi:
– Remember the first digits “3.14”. This gets you close to pi for basic calculations.
– Memorize pi to more digits such as “3.14159” using memory techniques or songs. There are many fun memory aids online to help remember more pi digits.
– Recall that pi is approximately equal to 22/7, or 3 1/7. This fraction gives about 3.14.
– Know that pi is between 3 and 4. So 3.14 is a good quick estimate.
– Visualize the circle constant on a calculator or computer screen when you need to reference pi’s value.
– Write pi on your study materials like notebooks, study guides, or flashcards to repeatedly see the value.
Regularly seeing and practicing pi’s digits will help solidify the value in your memory. Having a few techniques makes recalling pi easy.
Ways to conceptually understand pi
In addition to memorizing pi’s numerical value, it is also important to conceptually understand where pi comes from. Here are some methods to help build an intuitive understanding of pi’s meaning:
– Physically measure circles and spheres using string or other objects. Observe that pi naturally shows up.
– Simulate buffon’s needle experiment by dropping objects like toothpicks on lined paper. The probability relates to pi.
– Derive pi yourself from first principles by relating a circle’s circumference to diameter using formulas.
– Calculate pi from basic geometric formulas for circumference and area. Observe how pi relates them.
– Approximate pi experimentally by measuring circumferences and diameters or by Monte Carlo simulations.
– Understand pi as the ratio relating a circle’s periodic motion to linear distance. This periodic nature gives pi’s pervasiveness.
– Visualize pi geometrically as the fixed ratio between all circles of any size. This innate constant underlies pi’s importance.
– Link pi to abstract concepts like radians, periodicity, and trigonometric identities. See how pi naturally arises.
Developing an intuitive understanding demystifies pi as an abstract concept. Pi should be seen as a fundamental mathematical truth of circles and periodic motion.
Learning the history and interesting facts about pi
Studying some historical facts and interesting tidbits helps provide context for pi’s relevance:
– Pi has been studied for thousands of years, originally by ancient Babylonians. It has an extensive mathematical history.
– Many formulas to approximate pi were developed over the centuries. This includes infinite series discovered by mathematicians like Madhava.
– There are algorithms to calculate pi’s digits like the Nilakantha series or Ramanujan’s formulas. This allows pi to be computed to billions of digits.
– Computers have calculated pi to trillions of digits. NASAs even has a pi computing record using a supercomputer.
– Pi is celebrated on Pi Day, March 14th (3/14). This is a day to enjoy math and pi-related activities.
– Memorizing many pi digits is a hobby and there are records for reciting the most pi places from memory.
– Pi has been found encoded in equations like Euler’s identity, suggesting deep mathematical connections: e^(i*pi) = -1
– There are fascinating numeric coincidences related to pi such as the Feynman point and Gauss–Legendre algorithm integrating to pi.
– Pi appears in many fields like statistics, fractals, and quantum mechanics, showing its mathematical ubiquity.
Understanding pi’s long history and interesting trivia helps motivate its mathematical significance. Appreciating these details makes pi more engaging to learn.
Practicing pi computation skills
It is important to practice computational skills related to pi in order to improve math proficiency:
– Memorize common pi conversions like 1 radian = 180/pi degrees. Knowing these helps apply pi practically.
– Perform circumference, area, surface area and volume calculations for circles and spheres using formulas with pi.
– Calculate arc lengths, sector areas, and segment areas using pi.
– Use pi in trigonometry for identities like sin(pi/2) = 1 or tan(pi) = 0.
– Apply pi when working with radians for calculations like sin(pi/3) = sqrt(3)/2.
– Use pi in integrals like integ(sin(x)) = -cos(x)/pi + C
– Find pi in series like Leibniz formula for pi: pi = 4(1 – 1/3 + 1/5 – 1/7 + 1/9 – …)
– Approximate pi using Monte Carlo methods or by measuring circles. Compare the values to pi.
– Given a circle’s information, practice calculating missing values like diameter, circumference, and area using pi.
Regularly working through pi computation problems will develop familiarity with using pi in areas like geometry, trigonometry, and integrals.
Using online resources and games for pi
There are many excellent online tools to supplement learning pi:
– Websites to memorize pi digits like https://pi-world-ranking-list.com/ help develop recall.
– Online pi calculators at sites like https://www.piday.org/million/ quickly compute pi values to millions of digits.
– WolframAlpha computes pi related integrals, series, and identities. It is a useful pi computation tool.
– Pi visualizing programs like https://academo.org/demos/3-14-pi-day/ help build geometric intuition.
– Pi learning apps for mobile devices make practicing pi engaging through games and activities.
– The website https://www.piday.org/ has pi history, activities, lessons and pi memorizing techniques.
– Watching pi YouTube videos gives a fun and accessible way to learn pi concepts.
– There are pi math games online and on apps to practice pi computational skills.
Taking advantage of these interactive pi resources supplements textbook and classroom pi learning in an engaging way. The digital tools make grasping pi concepts more enjoyable.
Remembering pi through songs and poems
Fun memory techniques like pi songs and poems help solidify pi in your mind:
– The classic “Pi Song” has lyrics mapping pi digits to musical notes. Singing it is an easy memorization tool.
– Pi limericks like “Three point one four one five nine, this number makes circles sublime!” turn pi into rhyming poetry.
– Listen to piano songs, rap songs, and other musical adaptations memorizing pi on YouTube or other streaming sites.
– Local schools sometimes hold pi recitation contests, performances, and pi dances on Pi Day centered around memorizing and chanting pi rhymes.
– Poems about pi like “Pi” by Wislawa Szymborska artistically showcase pi’s mathematical elegance.
– Making up your own pi rhyme or song connects better with your memorization learning style.
– Singing or chanting pi songs mentally during day-to-day activities repeatedly reinforces pi’s digits.
Rhythm, rhyme, and melody engage another part of your mind to help cement pi in your memory. Pi songs and poems make learning pi fun and interactive.
Applications of pi in the real world
Pi has many applications in the physical world:
– Pi is essential for engineering and design involving circles or arced shapes like pipes, gears, spherical tanks, and circular motion.
– Pi allows the wavelengths, frequencies, and speeds of waves and repeating motions to be calculated in fields like acoustics, electromagnetism, and quantum mechanics.
– Pi is needed for calculating probabilities and statistics in data analysis and simulations.
– Pi appears in many geometric formulas required in construction, fabrication, surveying, architecture, and CAD design.
– Navigational techniques like LORAN and celestial navigation use pi in their spherical geometry calculations.
– Pi is required physics formulas describing phenomena like Einstein’s field equations, Coulomb’s law, and Hubble’s law.
– Estimating population densities, crop yields, and resource availability uses pi in calculating circle areas and sphere volumes.
– Pi can optimize paths for transport like determining great circle routes in aviation and spherical trigonometry in space travel.
Understanding pi’s role in these practical areas highlights its importance for broader STEM fields beyond just mathematics.
Pi in advanced mathematics
For higher math studies, pi connects to many abstract concepts:
– In complex analysis, pi emerges from Cauchy’s integral formula and proofs of the fundamental theorem of algebra.
– Pi is deeply linked to trigonometry through Euler’s formula: e^(i*pi) + 1 = 0
– In number theory, pi is an important transcendental number. The proof that it is transcendental uses integration.
– Pi is connected to famous unsolved problems like proving whether pi is a normal number.
– In modern geometry, pi relates to radians, spherical trigonometry, and differential geometry of surfaces.
– Irrationality proofs of pi using continued fractions and infinite series highlight techniques in analysis.
– Approximating pi with fractals like the Sierpiński sieve illustrates connections between pi and fractal geometry.
– Pi is encoded in quantum physics, like Heisenberg’s uncertainty principle: sigma_x sigma_p >= hbar/2
– Pi appears in significant equations like the prime number theorem relating primes to complex zeros.
Pi has deep ties to many advanced mathematical concepts. Understanding these connections provides insight into pi’s profound mathematical nature.
Activities for groups and classrooms
Pi is a great math topic for interactive learning activities in groups or classrooms:
– Have a pi memorizing competition to recite the most digits. This gamifies learning pi’s value.
– Do fun classroom geometry activities approximating pi like random dart throwing or measuring circles.
– Build a giant paper chain representing pi’s digits across the walls. This visually illustrates the endless nature of pi.
– Have students create pi math art by artistically displaying pi formulas or illustrating concepts like radians.
– Bake pi pies and eat pie on Pi Day. Brainstorm real-world examples of pi while enjoying the festivities.
– Assign pi research projects on its history, records, interesting facts, or appearances in math and culture.
– Have students build pi study tools like mnemonic acronyms, visualizations, animations, rhymes, or songs as creative projects.
– Perform pi skits, dances, raps, games, and contests. These active learning approaches boost pi engagement.
Making pi participatory, collaborative, and multi-sensory amplifies retention. Pi activities get the entire group involved in math learning.
While an irrational number that extends infinitely without a pattern, pi can be learned, understood, and appreciated by all students. Memorizing a few pi digits provides a working value for
calculations. Developing geometric intuition gives meaning to pi as it relates circles and spheres. Tracing the long history of pi highlights how it has fascinated mathematicians across all cultures
for millennia. Immersing oneself in pi’s many proofs, formulas, and appearances reveals the deep significance of pi to mathematics. Those techniques and perspectives all help unlock pi as a
fundamental mathematical truth about circles and periodicity in our universe. With various engaging resources and activities available today, learning pi has never been easier or more enjoyable. Pi
is a constant that unites mathematics, science, and the human experience.
Leave a Comment | {"url":"https://www.thedonutwhole.com/how-can-i-learn-pi-in-math/","timestamp":"2024-11-07T02:44:28Z","content_type":"text/html","content_length":"119971","record_id":"<urn:uuid:51ae618f-9411-495d-9793-f09d4ad8bcd6>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00699.warc.gz"} |
Moving Averages (MA) In Groovy?
Moving Averages (MA) are a commonly used technical indicator in the world of finance and trading. In Groovy, a programming language for the Java platform, moving averages can be calculated using
various methods such as the Simple Moving Average (SMA) or the Exponential Moving Average (EMA).
The Simple Moving Average calculates the average price over a specified number of time periods, giving equal weight to each period. On the other hand, the Exponential Moving Average gives more weight
to recent data points, making it more responsive to price movements.
In Groovy, moving averages can be easily implemented using built-in functions or custom scripts. These moving averages can be calculated for various time frames, such as days, weeks, or months,
allowing traders to analyze trends and potential reversals in the market. By using moving averages, traders can make informed decisions on when to enter or exit positions based on the direction of
the trend.
How to smooth out data using moving averages in Groovy?
To smooth out data using moving averages in Groovy, you can create a function that calculates the moving average of a given data set. Here's an example implementation:
1 def movingAverage = { data, windowSize ->
2 def smoothedData = []
4 for (int i = 0; i < data.size(); i++) {
5 def sum = 0.0
6 def count = 0
8 for (int j = Math.max(0, i - windowSize + 1); j <= i; j++) {
9 sum += data[j]
10 count++
11 }
13 def avg = sum / count
14 smoothedData.add(avg)
15 }
17 return smoothedData
18 }
20 def data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
21 def windowSize = 3
23 def smoothedData = movingAverage(data, windowSize)
24 println(smoothedData)
In this code snippet, the movingAverage function takes two parameters: data, which is the input data set, and windowSize, which is the size of the moving window used to calculate the average. The
function then iterates over the data set and calculates the moving average for each element based on the specified window size.
You can adjust the data and windowSize variables to suit your specific data and smoothing requirements. The smoothedData variable will contain the smoothed data set after running the movingAverage
How to calculate Moving Averages (MA) in Groovy?
To calculate Moving Averages (MA) in Groovy, you can use the following code snippet as an example:
1 // Sample data
2 def data = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
4 // Function to calculate Moving Averages
5 def movingAverage(data, windowSize) {
6 def ma = []
8 for (int i = 0; i < data.size() - windowSize + 1; i++) {
9 def sum = 0
10 for (int j = 0; j < windowSize; j++) {
11 sum += data[i + j]
12 }
13 ma.add(sum / windowSize)
14 }
16 return ma
17 }
19 // Calculate Moving Averages with window size of 3
20 def ma = movingAverage(data, 3)
21 println ma
In this example, we have defined a sample data array and a function movingAverage that calculates the Moving Averages of the data with a given window size. You can adjust the window size as needed.
What is the role of moving averages in risk management in Groovy?
Moving averages can play a crucial role in risk management in Groovy by helping investors identify trends and potential changes in the market. By analyzing moving averages, investors can determine
the direction of the market and make informed decisions about their investments.
Moving averages can also be used to set stop-loss orders, which are designed to limit potential losses by automatically selling an asset if its price falls below a certain threshold. This can help
investors protect their investments and reduce their overall risk exposure.
Additionally, moving averages can be used in conjunction with other technical indicators to create trading strategies that help investors make more informed decisions and mitigate risk. By analyzing
moving averages, investors can identify potential entry and exit points for their trades, helping them to maximize profits and minimize losses.
Overall, moving averages can be a valuable tool in risk management in Groovy, helping investors to navigate the market effectively and make strategic decisions to protect their investments. | {"url":"https://stlplaces.com/blog/moving-averages-ma-in-groovy","timestamp":"2024-11-07T19:10:06Z","content_type":"text/html","content_length":"258931","record_id":"<urn:uuid:9619445f-e05d-4f07-89ea-c00c1d4108b8>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00297.warc.gz"} |
Joint Mathematics Meetings
Joint Mathematics Meetings AMS Special Session
Current as of Friday, January 18, 2013 00:31:56
Program · Deadlines · Timetable · Inquiries: meet@ams.org
Joint Mathematics Meetings
San Diego Convention Center and San Diego Marriott Hotel and Marina, San Diego, CA
January 9-12, 2013 (Wednesday - Saturday)
Meeting #1086
Associate secretaries:
Georgia Benkart, AMS benkart@math.wisc.edu
Gerard A Venema, MAA venema@calvin.edu
AMS Special Session on q-series in Mathematical Physics and Combinatorics
• Saturday January 12, 2013, 8:00 a.m.-10:50 a.m.
AMS Special Session on q-series in Mathematical Physics and Combinatorics, I
Room 14A, Mezzanine Level, San Diego Convention Center
Mourad Ismail, University of Central Florida mourad.eh.ismail@gmail.com
• Saturday January 12, 2013, 1:00 p.m.-5:50 p.m.
AMS Special Session on q-series in Mathematical Physics and Combinatorics, II
Room 14A, Mezzanine Level, San Diego Convention Center
Mourad Ismail, University of Central Florida mourad.eh.ismail@gmail.com | {"url":"https://jointmathematicsmeetings.org/meetings/national/jmm2013/2141_program_ss2.html","timestamp":"2024-11-06T05:32:54Z","content_type":"application/xhtml+xml","content_length":"22581","record_id":"<urn:uuid:1edb7831-3121-4f1a-a98e-5a523b724e03>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00641.warc.gz"} |
A265/2003 - On Halphen's theorem and some generalizations
Preprint A265/2003
On Halphen's theorem and some generalizations
Alcides Lins Neto
Keywords: quasi-homogeneous | analytic set | parametrization
\abstract{ Let $M^n$ be a germ at $0\in \Bbb{\C}^{m}$ of an irreducible analytic set of dimension $n$, where $n\ge 2$ and $0$ is a singular point of $M$. We study the question : when there exists a
germ of holomorphic map $\phi \colon (\Bbb{\C}^n,0)\to (M,0)$ such that $\phi^{-1}(0)=\{0\}$ ? We prove essentialy three results. In Theorem 1 we consider the case where $M$ is a quasi-homogeneous
complete intersection of $k$ polynomials $F=(F_1,...,F_k)$, that is there exists a linear holomorphic vector field $X$ on $\C^{m}$, with eigenvalues $\lambda_1,...,\lambda_{m}\in \Q_+$ such that $X(F
^T)= U.F^T$, where $U$ is a $k\times k$ matrix with entries in $\Cal{O}_{m}$. We prove that if there exists a germ of holomorphic map $\phi$ as above and $\dim_{\Bbb{\C}}(sing(M))\le n-2$ then $\
lambda_1+...+\lambda_{m}>Re(tr(U)(0))$. In Theorem 2 we answer the question completely when $n=2$, $k=1$ and $0$ is an isolated singularity of $M$. In Theorem 3 we prove that, if there exists a map
as above, $k=1$ and $\dim_{\Bbb{\C}}(sing(M))\le n-2$, then $\dim_{\Bbb{\C}}(sing(M))= n-2$. We observe that Theorems 1 and 2 are generalizations of some results due to Halphen .} | {"url":"https://preprint.impa.br/visualizar?id=1396","timestamp":"2024-11-09T19:41:20Z","content_type":"text/html","content_length":"6812","record_id":"<urn:uuid:262137f3-0468-4b51-8278-305f1d8cc5d1>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00002.warc.gz"} |
short rate penalty calculator
Short Rate Penalty Calculator
In the world of finance, understanding the implications of short rate penalties is crucial. Whether you’re a borrower or a lender, knowing how these penalties affect your finances can make a
significant difference. This article introduces a simple yet effective tool – the Short Rate Penalty Calculator – designed to provide quick and accurate calculations for such scenarios.
How to Use
To utilize the Short Rate Penalty Calculator, simply input the required information into the designated fields and click the “Calculate” button. The calculator will then process the data and display
the result, enabling you to understand the potential penalties associated with your loan.
The formula used by the Short Rate Penalty Calculator is derived from the specific terms and conditions of your loan agreement. Typically, it involves calculating the penalty based on the outstanding
balance, the interest rate, and the remaining term of the loan. The formula ensures accuracy in determining the financial impact of early repayment or default.
Example Solve
Let’s consider a scenario where a borrower has an outstanding balance of $10,000 on a loan with an interest rate of 5% and a remaining term of 24 months. If the borrower decides to repay the loan
early, the Short Rate Penalty Calculator can quickly determine the penalty incurred based on these parameters.
Q: Can the Short Rate Penalty Calculator be used for all types of loans?
A: Yes, the calculator is versatile and can be applied to various loan types, including mortgages, personal loans, and auto loans.
Q: Is the calculator’s result accurate?
A: The calculator provides accurate estimates based on the information provided. However, it’s essential to consult with a financial advisor for precise calculations tailored to your specific
Q: Are there any limitations to using the calculator?
A: While the calculator offers valuable insights, it’s important to remember that it may not account for every possible scenario or clause in your loan agreement. Always refer to the terms and
conditions provided by your lender for complete clarity.
The Short Rate Penalty Calculator serves as a valuable tool for individuals navigating the complexities of loan agreements. By providing quick and accurate calculations, it empowers borrowers and
lenders alike to make informed financial decisions. Utilize this tool to understand the potential penalties associated with early loan repayment or default, ensuring a smoother financial journey. | {"url":"https://calculatordoc.com/short-rate-penalty-calculator/","timestamp":"2024-11-12T07:22:09Z","content_type":"text/html","content_length":"84329","record_id":"<urn:uuid:dba32242-757f-4117-be51-4b7422684e14>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00263.warc.gz"} |
98,296 research outputs found
The foundations of the Boltzmann-Gibbs (BG) distributions for describing equilibrium statistical mechanics of systems are examined. Broadly, they fall into: (i) probabilistic paaroaches based on the
principle of equal a priori probability (counting technique and method of steepest descents), law of large numbers, or the state density considerations and (ii) a variational scheme -- maximum
entropy principle (due to Gibbs and Jaynes) subject to certain constraints. A minimum set of requirements on each of these methods are briefly pointed out: in the first approach, the function space
and the counting algorithm while in the second, "additivity" property of the entropy with respect to the composition of statistically independent systems. In the past few decades, a large number of
systems, which are not necessarily in thermodynamic equilibrium (such as glasses, for example), have been found to display power-law distributions, which are not describable by the above-mentioned
methods. In this paper, parallel to all the inquiries underlying the BG program described above are given in a brief form. In particular, in the probabilistic derivations, one employs a different
function space and one gives up "additivity" in the variational scheme with a different form for the entropy. The requirement of stability makes the entropy choice to be that proposed by Tsallis.
From this a generalized thermodynamic description of the system in a quasi-equilibrium state is derived. A brief account of a unified consistent formalism associated with systems obeying power-law
distributions precursor to the exponential form associated with thermodynamic equilibrium of systems is presented here.Comment: 19 pages, no figures. Invited talk at Anomalous Distributions,
Nonlinear Dynamics and Nonextensivity, Santa Fe, USA, November 6-9, 200
The network approach plays a distinguished role in contemporary science of complex systems/phenomena. Such an approach has been introduced into seismology in a recent work [S. Abe and N. Suzuki,
Europhys. Lett. 65, 581 (2004)]. Here, we discuss the dynamical property of the earthquake network constructed in California and report the discovery that the values of the clustering coefficient
remain stationary before main shocks, suddenly jump up at the main shocks, and then slowly decay following a power law to become stationary again. Thus, the network approach is found to characterize
main shocks in a peculiar manner.Comment: 10 pages, 3 figures, 1 tabl
A study of CP violation in $B\to J/\psi K^*(K_S^0\pi^0)$ decays by time dependent angular analysis is discussed. Status of time independent analyses for other $B\to VV$ decays is also reported. The
data used for the analyses are taken with the Belle detector at KEK.Comment: 4 pages, 3 figures, Proceeding of the talk in parallel session (CP-3-5) at ICHEP2002, Amsterdam, Netherland, 24-31 July
The problem of temperature in nonextensive statistical mechanics is studied. Considering the first law of thermodynamics and a "quasi-reversible process", it is shown that the Tsallis entropy becomes
the Clausius entropy if the inverse of the Lagrange multiplier, $beta$, associated with the constraint on the internal energy is regarded as the temperature. This temperature is different from the
previously proposed "physical temperature" defined through the assumption of divisibility of the total system into independent subsystems. A general discussion is also made about the role of
Boltzmann's constant in generalized statistical mechanics based on an entropy, which, under the assumption of independence, is nonadditive.Comment: 14 pages, no figure
We review recent $B$ physics results obtained in polarized $e^+ e^-$ interactions at the SLC by the SLD experiment. The excellent 3-D vertexing capabilities of SLD are exploited to extract precise \
bu and \bd lifetimes, as well as measurements of the time evolution of $B^0_d - \bar{B^0_d}$ mixing.Comment: 7 pages, 4 figure
Nonadditive (nonextensive) generalization of the quantum Kullback-Leibler divergence, termed the quantum q-divergence, is shown not to increase by projective measurements in an elementary
manner.Comment: 10 pages, no figures. Errors in the published version are correcte
The generalized zeroth law of thermodynamics indicates that the physical temperature in nonextensive statistical mechanics is different from the inverse of the Lagrange multiplier, beta. This leads
to modifications of some of thermodynamic relations for nonextensive systems. Here, taking the first law of thermodynamics and the Legendre transform structure as the basic premises, it is found that
Clausius definition of the thermodynamic entropy has to be appropriately modified, and accordingly the thermodynamic relations proposed by Tsallis, Mendes and Plastino [Physica A 261 (1998) 534] are
also to be rectified. It is shown that the definition of specific heat and the equation of state remain form invariant. As an application, the classical gas model is reexamined and, in marked
contrast with the previous result obtained by Abe [Phys. Lett. A 263 (1999) 424: Erratum A 267 (2000) 456] using the unphysical temperature and the unphysical pressure, the specific heat and the
equation of state are found to be similar to those in ordinary extensive thermodynamics.Comment: 17 pages. The discussion about the Legendre transform structure is modified and some additional
comments are mad
We derive and study quasicanonical Gibbs distribution function which is characterized by the thermostat with finite number of particles (quasithermostat). We show that this naturally leads to Tsallis
nonextensive statistics and thermodynamics, with Tsallis parameter q is found to be related to the number of particles in the quasithermostat. We show that the chi-square distribution of fluctuating
temperature used recently by Beck can be partially understood in terms of normal random momenta of particles in the quasithermostat. Also, we discuss on the importance of the time scale hierarchy and
fluctuating probability distribution functions in understanding of Tsallis distribution, within the framework of kinetics of dilute gas and weakly inhomogeneous systems.Comment: 22 pages, 1 eps-figur
The front face of the CDF central calorimeter is being equipped with a new Preshower detector, based on scintillator tiles read out by WLS fibers, in order to cope with the luminosity increase
provided by the Main Injector during the Tevatron's Run II data taking. A light yield of about 40 pe/MIP at the tile exit was obtained, exceeding the design requirements.Comment: 4 pages, 8 figures.
Proceedings of `9th Topical Seminar on Innovative Particle and Radiation Detectors', 23-26 May 2004, Siena, Ital
The properties of three-jet events with total transverse energy greater than 320 GeV and individual jet energy greater than 20 GeV have been analyzed and compared to absolute predictions from a
next-to-leading order (NLO) perturbative QCD calculation. These data, of integrated luminosity 86 pb^-1, were recorded by the CDF Experiment for proton-antiproton collisions at sqrt{s}=1.8 TeV. This
study tests a model of higher order QCD processes that result in gluon emission and can be used to estimate the magnitude of the contribution of processes higher than NLO. The total cross section is
measured to be 466 \pm 3(stat.)^{+207}_{-70}(syst.) pb. The differential cross section is furthermore measured for all kinematically accessible regions of the Dalitz plane, including those for which
the theoretical prediction is unreliable. While the measured cross section is consistent with the theoretical prediction in magnitude, the two differ somewhat in shape in the Dalitz plane.Comment:
For the CDF Collaboration. Contributed to the proceedings of the Eleventh High-Energy Physics International Conference on QCD, Montpellier, France, July 200 | {"url":"https://core.ac.uk/search/?q=authors%3A(Abe)","timestamp":"2024-11-05T14:03:58Z","content_type":"text/html","content_length":"191768","record_id":"<urn:uuid:80c427e7-693d-49f2-9172-6a031ae9d31a>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00578.warc.gz"} |
Error understood with the circularity constraint.
Given a complex sample mean xbar, the standard error of the mean xerror is the real number defined such that roughly two thirds of the values should fall in a circle of radius xerror around xbar. The
error in the real case is defined in the usual sense.
This is usually the error estimate you want for stochastic postprocessing. In particular, if z = U*x, where z is a complex random vector, x is a real random vector, and U is a unitary matrix (such as
a Fourier transform), the circular error on z coincides with the usual error on x. However, plotting real or imaginary part with the circular error can be misleading. In this case, one should use
Definition at line 38 of file var_strategy.hpp. | {"url":"http://alpscore.org/doxygen_snapshot/html/structalps_1_1alea_1_1circular__var.html","timestamp":"2024-11-13T11:41:20Z","content_type":"application/xhtml+xml","content_length":"7070","record_id":"<urn:uuid:661297e5-a61f-4c53-a88d-638ddc01e7b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00702.warc.gz"} |
How to use Google docs equation editor : Complete Guide 2021 -
How to use Google docs equation editor : Complete Guide 2021
• Have you ever came across a situation where you need to add equations in your google docs while working on a math project.
• Google docs has an in built equation editor which can be really useful for teachers preparing lecture worksheets and for students working on school projects .
In this article we will show you How to use google docs equation editor .
Step by Step Guide : How to use google docs equation editor
Step 1: Open you browser and go to https://docs.google.com/.logon to your Gmail account if you are not already logged in.
Step 2: Go to the google docs document where you want to insert an equation.
Go to insert->Equation in the menu.
Once you click on “Equations” .You will see a new equation menu along with text box as shown below.
The new equation toolbar will have following options :
Greek letters
Miscellaneous operations
Math operations
Step3 : Type the equation you want to add in the equation box
to insert square you can either select the symbol ^ or press “shift+6” on the keyboard
Complete your equation .
• You can start writing a new equation by selecting
• If you want to hide the Equation tool bar, deselect the option “Show equation toolbar” under view menu
Handy Shortcuts for writing equations :
• To copy quickly you can use Ctrl+C or Command+C
• You can use a backslash followed by symbol with space for example “\alpha” to write alpha or “\sqrt” to write square root.
• You can use the unofficial list of LaTeX-like shortcuts from the Google Docs equation editor. Enter the shortcuts in the equation editor, then press the space bar.
Other useful articles :
Leave a Comment
This site uses Akismet to reduce spam. Learn how your comment data is processed. | {"url":"https://addnewskills.com/how-to-use-google-docs-equation-editor/","timestamp":"2024-11-02T00:06:54Z","content_type":"text/html","content_length":"136654","record_id":"<urn:uuid:eb027d0d-fc6d-444a-843f-41360512c20a>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00279.warc.gz"} |
PHY 765: Graduate Advanced Physics,
PHY 765: Graduate Advanced Physics, Fall 2018
Instructor: Prof. Thomas Barthel
Lectures: Mondays 1:25-2:40PM in Physics 047 and Fridays 1:25-2:40PM in Physics 047
Office hours: Mondays and Fridays 2:40PM-3:45PM in Physics 287
Teaching assistant: Qiang Miao (Physics 183)
Tutorials: every few weeks
Grading: Problem Sets (40%), Midterm exam (20%), Final exam (40%)
This is a graduate course on the wonderful world of quantum many-body physics. We will cover aspects from five fields: quantum information theory, atomic physics, condensed matter physics, quantum
optics, and relativistic quantum field theory, addressing phenomena and techniques that (almost) every physicist should be familiar with.
We will warm up with generalizations of the postulates of quantum physics (density matrices, quantum channels, POVM) which are essential in quantum information theory and capture interactions with
environmental degrees of freedom. This also allows us to understand the effect of decoherence. We then start looking at systems of identical particles and, first, discuss entanglement in such
systems. One the basis of the approximative Hartree-Fock equations, we will study the electronic structure of multi-electron atoms. Quantum many-body systems are efficiently described using the
formalism of 2nd quantization. With this tool at hand, we will address electron band structures of solids, topological insulators, Landau's Fermi liquid theory (describing the normal state of
metals), and the BCS theory of superconductivity. Technically, this exemplifies the derivation of effective low-energy Hamiltonians, mean-field approximations, spontaneous symmetry breaking, and
off-diagonal long-range order, topological invariants, and symmetry-protected topological order. Finally, we will see how to construct Lorentz invariant quantum field theories (QFT) which are
essential for high-energy physics. In particular, we will look into the Klein-Gordon QFT and the Dirac QFT which describe relativistic bosonic and fermionic systems.
If time permits, we will also cover a selection of the following topics: interacting Bose-Einstein condensates (superfluidity), the Bose-Hubbard model with its superfluid-Mott quantum phase
transition, the renormalization group and scale invariance for critical systems, and further topics from quantum information theory and quantum computation such as no-cloning, teleportation, Bell
inequalities, quantum algorithms, and entanglement.
Knowledge of single-particle quantum mechanics on the level of courses PHY 464 or PHY 764 is expected.
Lecture Notes
[Are provided on the Sakai site PHYSICS.765.01.F18.]
You are encouraged to discuss homework assignments with fellow students and to ask questions in the Sakai Forum or by email. But the written part of the homework must be done individually and cannot
be a copy of another student's solution. (See the Duke Community Standard.)
Homework due dates are strict (for the good of all), i.e., late submissions are not accepted. If there are grave reasons, you can ask for an extension early enough before the due date.
[Problem sets are provided on the Sakai site PHYSICS.765.01.F18.]
Useful literature
Besides the lecture notes, supplementary reading material for each part of the course will be provided on the Sakai site. Here, some references for the course topics.
Quantum mechanics (basics).
• Sakurai "Modern Quantum Mechanics", Addison Wesley (1993)
• Shankar "Principles of Quantum Mechanics" 2nd Edition, Plenum Press (1994)
• Le Bellac "Quantum Physics", Cambridge University Press (2006)
• Schwabl "Quantum Mechanics" 4th Edition, Springer (2007)
• Baym "Lectures on Quantum Mechanics", Westview Press (1974)
Quantum information and computation.
• Nielsen, Chuang "Quantum Computation and Quantum Information", Cambridge University Press (2000)
• Preskill "Quantum Computation", Lecture Notes (2015)
• Wilde "Quantum Information Theory" 2nd Edition, arXiv:1106.1445 (2017)
• Bruss, Leuchs "Lectures on Quantum Information", Wiley (2007)
Non-relativistic quantum many-body physics (varying topics).
• Coleman "Introduction to Many-Body Physics", Cambridge University Press (2015)
• Nazarov, Danon "Advanced Quantum Mechanics", Cambridge University Press (2013)
• Altland, Simons "Condensed Matter Field Theory" 2nd Edition, Cambridge University Press (2010)
• Negele, Orland "Quantum Many-Particle Systems", Westview Press (1988, 1998)
• Bruus, Flensberg "Many-Body Quantum Theory in Condensed Matter Physics", Oxford University Press (2004)
• Ashcroft, Mermin "Solid State Physics", Harcourt (1976)
• Ibach, Lüth "Solid State Physics" 4th Edition, Springer (2009)
Topological insulators.
• Asbóth, Oroszlány, Pályi "A Short Course on Topological Insulators", Springer (2016)
• Bernevig, Hughes "Topological Insulators and Topological Superconductors", Princeton University Press (2013)
• Xiao, Chang, Niu "Berry Phase Effects on Electronic Properties", Rev. Mod. Phys. 82, 1959 (2010)
Relativistic quantum many-body physics.
• Schwabl "Advanced Quantum Mechanics" 3th Edition, Springer (2005)
• Peskin, Schroeder "An Introduction to Quantum Field Theory", Addison-Wesley (1995) | {"url":"http://webhome.phy.duke.edu/~barthel/L2018-08_GAP_phy765/","timestamp":"2024-11-03T22:06:12Z","content_type":"text/html","content_length":"9340","record_id":"<urn:uuid:ce2f48ec-0f13-46d5-bf6b-9bed8802b873>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00150.warc.gz"} |
About the rocbc Package
(Authors: LE Bantis, B Brewer, CT Nakas, B Reiser)
This package is focused on Box-Cox based ROC curves and provides point estimates, confidence intervals (CIs), and hypothesis tests. It can be used both for inferences for a single biomarker and when
comparisons of two correlated biomarkers are of interest. It provides inferences and comparisons around the area under the ROC curve (AUC), the Youden index, the sensitivity at a given specificity
level (and vice versa), the optimal operating point of the ROC curve (in the Youden sense), and the Youden based cutoff. This documentation consists of two parts, the one marker and the two marker
case. All approaches presented herein have been recently published (see references for each function).
The functions of each part:
• Description
□ This function tests whether the Box-Cox transformation is able to achieve approximate normality for your data. That is, it will allow the user to know whether it is appropriate to use all the
methods discussed later on in this package.
• Usage
checkboxcox(marker, D, plots, printShapiro = TRUE)
• Arguments
□ marker: A vector of length \(n\) that contains the biomarker scores of all individuals.
□ D: A vector of length n that contains the true disease status of an individual. It is a binary vector containing 0 for the healthy/control individuals and 1 for the diseased individuals.
□ plots='on' or 'off': Valid inputs are “on” and “off”. When set to “on”, the user gets the histograms of the biomarker for both the healthy and the diseased group before and after the Box-Cox
transformation. In addition, all four corresponding qq-plots are provided.
□ printShapiro: Boolean. When set to TRUE, the results of the Shapiro-Wilk test will be printed to the console. When set to FALSE, the results are suppressed. Default value is FALSE.
• Value
□ res_shapiro: A results table that contains the results of four Shapiro-Wilk tests for normality testing. Two of these refer to normality testing of the healthy and the diseased groups before
the Box-Cox transformation, and the remaining two refer to the Box-Cox transformed biomarkers scores for the healthy and the diseased groups. Thus, this testing process produces four
p-values. In addition, if the plots are set to ‘on’, then the output provides (1) the histograms of the biomarker for both the healthy and the diseased groups before and after the Box-Cox
transformation, (2) all four corresponding qq-plots, and (3) a plot with the empirical ROC curve overplotted with the Box-Cox based ROC curve for visual comparison purposes.
□ transformation.parameter: The single transformation parameter \(\lambda\) that is applied for both groups simultaneously.
□ transx: The Box-Cox transformed scores for the healthy.
□ transy: The Box-Cox transformed scores for the diseased.
□ pval_x: The p-value of the Shapiro-Wilk test of normality for the healthy group (before the Box-Cox transformation).
□ pval_y: The p-value of the Shapiro-Wilk test of normality for the diseased group (before the Box-Cox transformation).
□ pval_transx: The p-value of the Shapiro-Wilk test of normality for the healthy group (after the Box-Cox transformation).
□ pval_transy: The p-value of the Shapiro-Wilk test of normality for the diseased group (after the Box-Cox transformation).
□ roc: A function of the estimated Box-Cox ROC curve. You can use this to simply request TPR values for given FPR values.
• **Example: **
x=rgamma(100, shape=2, rate = 8) # Generates biomarker data from a gamma distribution
# for the healthy group.
y=rgamma(100, shape=2, rate = 4) # Generates biomarker data from a gamma distribution
# for the diseased group.
D=c(zeros(1,100), ones(1,100))
out=checkboxcox(marker=scores, D, plots="on", printShapiro = TRUE)
## Shapiro-Wilk normality test
## data: x
## W = 0.92354, p-value = 2.181e-05
## Shapiro-Wilk normality test
## data: y
## W = 0.90169, p-value = 1.7e-06
## Shapiro-Wilk normality test
## data: transx
## W = 0.98765, p-value = 0.4826
## Shapiro-Wilk normality test
## data: transy
## W = 0.98936, p-value = 0.6128
## Length Class Mode
## transformation.parameter 1 -none- numeric
## transx 100 -none- numeric
## transy 100 -none- numeric
## pval_x 1 -none- numeric
## pval_y 1 -none- numeric
## pval_transx 1 -none- numeric
## pval_transy 1 -none- numeric
## rocfun 1 -none- function
x <- rgamma(100, shape=2, rate = 8) # generates biomarker data from a gamma
# distribution for the healthy group.
y <- rgamma(100, shape=2, rate = 4) # generates biomarker data from a gamma
# distribution for the diseased group.
scores <- c(x,y)
D=c(pracma::zeros(1,100), pracma::ones(1,100))
out=threerocs(marker=scores, D, plots="on")
## Length Class Mode
## AUC_Empirical 1 -none- numeric
## AUC_Metz 1 -none- numeric
## AUC_BoxCox 1 -none- numeric
• Description
□ This function applies the Box-Cox transformation to provide a comprehensive ROC analysis that involves the AUC (and its CI), the maximized Youden index (and its CI), the optimized Youden
based cutoff (and its CI), and joint confidence regions for the optimized pair of sensitivity and specificity. For the AUC and the Youden index, the Delta Method is employed by accounting for
the variability of the estimated transformation parameters due to the Box-Cox transformation. Both the AUC and the YI confidence intervals are found after using the probit transformation and
back-transforming the endpoints of the corresponding CI back into the ROC space. For the cutoffs, it has been shown that the delta method does not perform well; thus, the bootstrap with 1,000
repetitions is employed here, instead. During this this process, the cutoffs are back-transformed with the inverse Box-Cox transformation so that they lie on the original scale of the data.
• Usage
rocboxcox(marker, D, plots, printProgress = TRUE)
• Arguments
□ marker: A vector of length n that contains the biomarker scores of all individuals.
□ D: A vector of length n that contains the true disease status of an individual. It is a binary vector containing 0 for the healthy/control individuals and 1 for the diseased individuals.
□ alpha: Nominal level used to calculate the confidence intervals. A common choice is 0.05.
□ plots: Valid inputs are “on” and “off”. When set to “on”, the function returns a comprehensive figure with the ROC estimate and several point estimates: AUC, Youden index, optimized Youden
cutoff, and the Youden based sensitivty and specificity along with the corresponding marginal confidence intervals and the joint confidence region of the estimated sensitivity and
□ printProgress: Boolean. When set to TRUE, messages describing the progress of the bootstrapping will be printed to the console window. When set to FALSE, these messages are suppressed.
Default value is FALSE.
• Value
□ transx: The Box-Cox transformed scores for the healthy group.
□ transy: The Box-Cox transformed scores for the diseased group.
□ transformation.parameter: The estimated Box-Cox transformation parameter ($\lambda$).
□ AUC: The estimated area under the Box-Cox based ROC curve.
□ AUCCI: The (1-$\alpha$)100% CI for the AUC. (A common choice of \(\alpha\) is 0.05). This CI is based on probit transforming the AUC estimate, finding the CI on the real line, and then
back-transforming its endpoints to the ROC space. It is in line with using \(Z=\frac{\Phi^{-1}(\hat{AUC})}{\sqrt{var(\Phi^{-1}(\hat{AUC}))}}\) to test the null hypothesis that \(H_{0}: AUC=
□ pvalueAUC: The corresponding p-value for the AUC estimate. This a two-tailed p-value that tests the hypothesis \(H_{0}: AUC=0.5\) by employing \(Z=\frac{\Phi^{-1}(\hat{AUC})}{\sqrt{var(\Phi^
{-1}(\hat{AUC}))}}\) which, under the null hypothesis, is taken to follow a standard normal distribution.
□ J: The maximized Youden index.
□ JCI: The corresponding CI for the maximized Youden index. For this CI, we consider the probit transformation \(\hat{J}_{T}=\Phi^{-1}(\hat{J})\) and then back-transform its endpoints to derive
a 95% CI for the Youden index itself (Bantis et al. (2021)).
□ pvalueJ: The corresponding two-tailed p-value for the maximized Youden index. This corresponds to \(Z=\frac{\hat{J}_{T}}{\sqrt{var(\hat{J}_{T})}}\). The underlying null hypothesis is \(H_{0}:
□ Sens: The sensitivity that corresponds to the Youden based optimized cutoff.
□ CImarginalSens: The marginal (1-$\alpha$)100% CI for the sensitivity that corresponds to the Youden based optimized cutoff. This is derived by first employing the probit transformation,
finding a CI on the real line, and then back-transforming its endpoints to the ROC space.
□ Spec: The specificity that corresponds to the Youden based optimized cutoff.
□ CImarginalSpec: The marginal (1-$\alpha$)100% CI for the specificity that corresponds to the Youden based optimized cutoff. This is derived by first employing the probit transformation,
finding a CI on the real line, and then back-transforming its endpoints to the ROC space.
□ cutoff: The Youden-based optimized cutoff.
□ CIcutoff: the (1-$\alpha$)100% CI for the Youden-based optimized cutoff. This is based on the bootstrap. It involves using the Box-Cox transformation for every bootstrap iteration and then
using the inverse Box-Cox transformation to obtain the cutoff on its original scale (Bantis et al. (2019)).
□ areaegg: The area of the (1-$\alpha$)100% egg-shaped joint confidence region that refers to the optimized pair of sensitivity and specificity. This takes into account the fact that the
estimated sensitivity and specificity are correlated as opposed to the corresponding rectangular area that ignores this.
□ arearect: The area of the (1-$\alpha$)100% rectangular joint confidence region that refers to the optimized pair of sensitivity and specificity. This ignores the correlation of the optimized
sensitivity and specificity and tends to yield a larger area compared to the one of the egg-shaped region.
□ mxlam: The mean of the marker scores of the healthy group after the Box-Cox transformation.
□ sxlam: The standard deviation of the marker scores of the healthy group after the Box-Cox transformation.
□ mylam: The mean of the marker scores of the diseased group after the Box-Cox transformation.
□ sylam: The standard deviation of the marker scores of the diseased group after the Box-Cox transformation.
□ results: A table that provides some indicative results: the AUC, the J (maximized Youden index), the estimated cutoff, the Sens, and the Spec along with their marginal CIs.
□ roc: A function of the estimated Box-Cox ROC curve. You can use this to simply request TPR values for given FPR values.
• **Example: **
x=rgamma(100, shape=2, rate = 8)
y=rgamma(100, shape=2, rate = 4)
D=c(zeros(1,100), ones(1,100))
out=rocboxcox(marker=scores,D, 0.05, plots="on", printProgress=FALSE)
## Length Class Mode
## transx 100 -none- numeric
## transy 100 -none- numeric
## transformation.parameter 1 -none- numeric
## AUC 1 -none- numeric
## AUCCI 2 -none- numeric
## pvalueAUC 1 -none- numeric
## J 1 -none- numeric
## JCI 2 -none- numeric
## pvalueJ 1 -none- numeric
## Sens 1 -none- numeric
## CImarginalSens 2 -none- numeric
## Spec 1 -none- numeric
## CImarginalSpec 2 -none- numeric
## cutoff 1 -none- numeric
## CIcutoff 2 -none- numeric
## areaegg 1 -none- numeric
## arearect 1 -none- numeric
## mxlam 1 -none- numeric
## sxlam 1 -none- numeric
## mylam 1 -none- numeric
## sylam 1 -none- numeric
## results 18 formattable numeric
## rocfun 1 -none- function
• Description
□ This function applies the Box-Cox transformation and carries out statistical inferences for the sensitivity at a given specificity (and vice versa).
• Usage
out=rocboxcoxCI(marker, D, givenSP=givenSP, givenSE=NA, alpha, plots)
out=rocboxcoxCI(marker, D, givenSP=NA, givenSE=givenSE, alpha, plots)
• Arguments
□ marker: A vector of length n that contains the biomarker scores of all individuals.
□ D: A vector of length n that contains the true disease status of an individual, where 0 denotes a healthy/control individual, and 1 denotes a diseased individual.
□ givenSP: A vector of specificity values that the user wants to fix/set, at which the sensitivity is to be estimated. In this case, the ‘givenSE’ argument needs to be set to NA.
□ givenSE: A vector of sensitivity values that the user want to fix/set, at which the specificity is to be estimated. In this case, the ‘givenSP’ argument needs to be set to NA.
□ alpha: Nominal level used to calculate the confidence intervals. A common choice is 0.05.
□ plots: Valid inputs are “on” and “off”. When set to “on”, it returns both (1) the Box-Cox based ROC plot along with pointwise 95% confidence intervals for the full spectrum of FPRs and (2) a
second plot that visualizes the confidence intervals at the given sensitivities or specificities.
• Value
□ SPandCIs: The specificity values and the CIs around them.
□ SEandCIs: The specificity values and the CIs around them.
□ SEvalues: The sensitivity values provided by the user at which the specificity was calculated. If the user did not provide any sensitivity values, this argument shoulb be set to NA.
□ SPvalues: The specificity values provided by the user at which the sensitivity was calculated. If the user did not provide any specificity values, this argument should be set to NA.
• **Example: **
out=rocboxcoxCI(marker=scores ,D, givenSP=givenSP, givenSE=NA, alpha=0.05, plots="on")
## Length Class Mode
## SPandCIs 4 formattable numeric
## SEandCIs 36 formattable numeric
## Sevalues 1 -none- logical
## Sphat 1 -none- numeric
## CIsp 2 formattable numeric
## Spvalues 9 -none- numeric
## Sehat 9 -none- numeric
## CIse 18 formattable numeric
• Description
□ This function tests whether the Box-Cox transformation is able to achieve approximate normality for your data. That is, it will allow the user to know whether it is appropriate to use all the
methods discussed later on in this package. It is similar to the function checkboxcox but is designed for two markers instead of just one.
• Usage
checkboxcox2(marker1, marker2, D, plots, printShapiro = TRUE)
• Arguments
□ marker1: A vector of length n that contains the biomarker scores of all individuals for the first marker.
□ marker2: A vector of length n that contains the biomarker scores of all individuals for the second marker.
□ D: A vector of length n that contains the true disease status of an individual. It is a binary vector containing 0 for the healthy/control individuals and 1 for the diseased individuals.
□ plots='on' or 'off': Valid inputs are “on” and “off”. When set to “on”, the user gets the histograms of the biomarker for both the healthy and the diseased group before and after the Box-Cox
transformation for both marker1 and marker2. In addition, all eight corresponding qq-plots are provided.
□ printShapiro: Boolean. When set to TRUE, the results of the Shapiro-Wilk test will be printed to the console. When set to FALSE, the results are suppressed. Default value is FALSE.
• Value
□ res_shapiro: A results table that contains the results of eight Shapiro-Wilk tests for normality testing. Four of these refer to normality testing of the healthy and the diseased groups
before the Box-Cox transformation, and the remaining four refer to the Box-Cox transformed biomarkers scores for the healthy and the diseased groups. Thus, this testing process produces eight
p-values. In addition, if the plots are set to ‘on’, then the output provides (1) the histograms of the biomarker for both the healthy and the diseased groups before and after the Box-Cox
transformation for both marker1 and marker2, (2) all eight corresponding qq-plots, and (3) two plots (one for marker1, one for marker2) with the empirical ROC curve overplotted with the
Box-Cox based ROC curve for visual comparison purposes.
□ transx1: The Box-Cox transformed scores for the first marker and the healthy group.
□ transy1: The Box-Cox transformed scores for the first marker and the diseased group.
□ transformation.parameter.1: The estimated Box-Cox transformation parameter (lambda) for marker1.
□ transx2: The Box-Cox transformed scores for the second marker and the healthy group.
□ transy2: The Box-Cox transformed scores for the second marker and the diseased group.
□ transformation.parameter.2: The estimated Box-Cox transformation parameter (lambda) for marker2.
□ pval_x1: The p-value of the Shapiro Wilk test of normality for the marker1 healthy group (before the Box-Cox transformation).
□ pval_y1: The p-value of the Shapiro Wilk test of normality for the marker1 diseased group (before the Box-Cox transformation).
□ pval_transx1: The p-value of the Shapiro Wilk test of normality for the marker1 healthy group (after the Box-Cox transformation).
□ pval_transy1: The p-value of the Shapiro Wilk test of normality for the marker1 diseased group (after the Box-Cox transformation).
□ pval_x2: The p-value of the Shapiro Wilk test of normality for the marker2 healthy group (before the Box-Cox transformation).
□ pval_y2: The p-value of the Shapiro Wilk test of normality for the marker2 diseased group (before the Box-Cox transformation).
□ pval_transx2: The p-value of the Shapiro Wilk test of normality for the marker2 healthy group (after the Box-Cox transformation).
□ pval_transy2: The p-value of the Shapiro Wilk test of normality for the marker2 diseased group (after the Box-Cox transformation).
□ roc1: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ roc2: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
• **Example: **
nx <- 100
Sx <- matrix(c(1, 0.5, 0.5, 1), nrow = 2, ncol = 2)
mux <- c(X = 10, Y = 12)
X = mvtnorm::rmvnorm(nx, mean = mux, sigma = Sx)
ny <- 100
Sy <- matrix(c(1.1, 0.6, 0.6, 1.1), nrow = 2, ncol = 2)
muy <- c(X = 11, Y = 13.7)
Y = mvtnorm::rmvnorm(ny, mean = muy, sigma = Sy)
dx = pracma::zeros(nx,1)
dy = pracma::ones(ny,1)
markers = rbind(X,Y);
marker1 = markers[,1]
marker2 = markers[,2]
D = c(rbind(dx,dy))
out=checkboxcox2(marker1, marker2, D, plots = "on")
## Length Class Mode
## res_shapiro 8 -none- list
## transformation.parameter.1 1 -none- numeric
## transx1 100 -none- numeric
## transy1 100 -none- numeric
## transformation.parameter.2 1 -none- numeric
## transx2 100 -none- numeric
## transy2 100 -none- numeric
## pval_x1 1 -none- numeric
## pval_y1 1 -none- numeric
## pval_transx1 1 -none- numeric
## pval_transy1 1 -none- numeric
## pval_x2 1 -none- numeric
## pval_y2 1 -none- numeric
## pval_transx2 1 -none- numeric
## pval_transy2 1 -none- numeric
## roc1 1 -none- function
## roc2 1 -none- function
• Description
□ This function provides a comparison of two correlated markers in terms of their AUCs (areas under the Box-Cox based ROC curves). Marker measurements are assumed to be taken on the same
individuals for both markers.
• Usage
out=comparebcAUC(marker1, marker2, D, alpha, plots)
• Arguments
□ marker1: A vector of length n that contains the biomarker scores of all individuals for the first marker.
□ marker2: A vector of length n that contains the biomarker scores of all individuals for the second marker.
□ D: A vector of length n that contains the true disease status of an individual. It is a binary vector containing 0 for the healthy/control individuals and 1 for the diseased individuals.
□ alpha: Nominal level used to calculate the confidence intervals. A common choice is 0.05.
□ plots: Valid inputs are “on” and “off”. When set to “on” it returns the Box-Cox based ROC along with informative information about the two AUCs in the legend of the plot.
• Value
□ resultstable: A summary table of the comparison that contains the AUC of each marker, the p-value of the difference using the probit transformation, the p-value of the difference, and the
confidence interval of the difference.
□ AUCmarker1: The AUC of the first marker.
□ AUCmarker2: The AUC of the second marker.
□ pvalue_probit_difference: The p-value for the comparison of the AUCs. It employs the probit transformation that has greater power then when not using it (as opposed to the p-value provided in
the ‘pvalue difference’ output argument provided later on). It is based on \(Z^*=\frac{\Phi^{-1}\left(\hat{AUC}_{1}\right)-\Phi^{-1}\left(\hat{AUC}_{2}\right)}{\sqrt{Var\left(\Phi^{-1}\left(\
□ pvalue_difference: The p-value for the comparisons of the AUC (without the probit transformation). Simulations have shown that this is inferior to the ‘pvalue probit difference’. This is
based on \(Z=\frac{\hat{AUC}_{1}-\hat{AUC}_{2}}{\sqrt{Var\left(\hat{AUC}_{1}\right)+Var\left(\hat{AUC}_{2}\right)-2Cov\left(\hat{AUC}_{1},\hat{AUC}_{2}\right)}}\).
□ CI_difference: The confidence interval for the difference of the AUCs. It is based on \(Z\) given above.
□ roc1: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ roc2: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ transx1: The Box-Cox transformed scores for the first marker and the healthy group.
□ transy1: The Box-Cox transformed scores for the first marker and the diseased group.
□ transformation.parameter.1: The estimated Box-Cox transformation parameter (lambda) for marker 1.
□ transx2: The Box-Cox transformed scores for the second marker and the healthy group.
□ transy2: The Box-Cox transformed scores for the second marker and the diseased group.
□ transformation.parameter.2: The estimated Box-Cox transformation parameter (lambda) for marker 2.
• **Example: **
#GENERATE SOME BIVARIATE DATA===
nx <- 100
Sx <- matrix(c(1, 0.5,
0.5, 1),
nrow = 2, ncol = 2)
mux <- c(X = 10, Y = 12)
X=rmvnorm(nx, mean = mux, sigma = Sx)
ny <- 100
Sy <- matrix(c(1.1, 0.6,
0.6, 1.1),
nrow = 2, ncol = 2)
muy <- c(X = 11, Y = 13.7)
Y=rmvnorm(ny, mean = muy, sigma = Sy)
#==DATA HAVE BEEN GENERATED====
#===COMPARE THE AUCs of Marker 1 vs Maker 2
out=comparebcAUC(marker1, marker2, D, alpha=0.05, plots="on")
## Length Class Mode
## resultstable 7 formattable numeric
## AUCmarker1 1 -none- numeric
## AUCmarker2 1 -none- numeric
## pvalue_probit_difference 1 -none- numeric
## pvalue_difference 1 -none- numeric
## CI_difference 2 -none- numeric
## roc1 1 -none- function
## roc2 1 -none- function
## transx1 100 -none- numeric
## transy1 100 -none- numeric
## transformation.parameter.1 1 -none- numeric
## transx2 100 -none- numeric
## transy2 100 -none- numeric
## transformation.parameter.2 1 -none- numeric
• Description
□ This function provides a comparison of two correlated markers in terms of their J (Youden indices for Box-Cox based ROC curves). Markers measurements are assumed to be taken on the same
individuals for both markers.
• Usage
out=comparebcJ(marker1, marker2, D, alpha, plots)
• Arguments
□ marker1: A vector of length n that contains the biomarker scores of all individuals for the first marker.
□ marker2: A vector of length n that contains the biomarker scores of all individuals for the second marker.
□ D: A vector of length n that contains the true disease status of an individual, where 0 denotes a healthy/control individual, and 1 denotes a diseased individual.
□ alpha: Nominal level used to calculate the confidence intervals. A common choice is 0.05.
□ plots: Valid inputs are “on” and “off”. When set to “on” it returns the Box-Cox based ROC along with informative information about the two AUCs in the legend of the plot.
• Value
□ resultstable: A summary table of the comparison that contains the maximized Youden Index of each marker, the p-value of the difference using the probit transformation, the p-value of the
difference, and the confidence interval of the difference.
□ J1: The maximized Youden Index (J) of the first marker.
□ J2: The maximized Youden Index (J) of the second marker.
□ pvalue_probit_difference: The p-value for the comparison of the Js. It employs the probit transformation that has greater power then when not using it (as opposed to the p-value provided in
the ‘pvalue difference’ output argument provided later on). It is based on \(Z^{*}=\frac{\hat{J}_{T2}-\hat{J}_{T1}}{\sqrt{Var(\hat{J}_{T1})+Var(\hat{J}_{T2})-2Cov(\hat{J}_{T1},\hat{J}_{T2})}}
\) where \(\hat{J}_{Ti}=\Phi^{-1}(\hat{J}_{i})\), \(i=1,2\).
□ pvalue_difference: The p-value for the comparisons of the J (without the probit transformation). Simulations have shown that this is inferior to the ‘pvalue probit difference’. This is based
on \(Z=\frac{\hat{J}_{2}-\hat{J}_{1}}{\sqrt{Var(\hat{J}_{1})+Var(\hat{J}_{2})-2Cov(\hat{J}_{1},\hat{J}_{T})}}\).
□ CI_difference: The confidence interval for the difference of the Js. This is based on \(Z\) mentioned right above.
□ roc1: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ roc2: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ transx1: The Box-Cox transformed scores for the first marker and the healthy group.
□ transy1: The Box-Cox transformed scores for the first marker and the diseased group.
□ transformation.parameter.1: The estimated Box-Cox transformation parameter (lambda) for marker 1.
□ transx2: The Box-Cox transformed scores for the second marker and the healthy group.
□ transy2: The Box-Cox transformed scores for the second marker and the diseased group.
□ transformation.parameter.2: The estimated Box-Cox transformation parameter (lambda) for marker 2.
• **Example: **
#==DATA HAVE BEEN GENERATED====
#===COMPARE THE Js of Marker 1 vs Maker 2
out=comparebcJ(marker1, marker2, D, alpha=0.05, plots="on")
## Length Class Mode
## resultstable 7 formattable numeric
## J1 1 -none- numeric
## J2 1 -none- numeric
## pvalue_probit_difference 1 -none- numeric
## pvalue_difference 1 -none- numeric
## CI_difference 2 -none- numeric
## roc1 1 -none- function
## roc2 1 -none- function
## transx1 100 -none- numeric
## transy1 100 -none- numeric
## transformation.parameter.1 1 -none- numeric
## transx2 100 -none- numeric
## transy2 100 -none- numeric
## transformation.parameter.2 1 -none- numeric
• Description
□ This function provides a comparison of two correlated markers in terms of their sensitivities at a given specificity (for Box-Cox based ROC curves). Marker measurements are assumed to be
taken on the same individuals for both markers.
• Usage
out=comparebcSens(marker1, marker2, D, alpha, atSpec, plots)
• Arguments
□ marker1: A vector of length n that contains the biomarker scores of all individuals for the first marker.
□ marker2: A vector of length n that contains the biomarker scores of all individuals for the second marker.
□ D: A vector of length n that contains the true disease status of an individual, where 0 denotes a healthy/control individual, and 1 denotes a diseased individual.
□ alpha: Nominal level used to calculate the confidence intervals. A common choice is 0.05.
□ atSpec: The value of specificity at which the comparison of sensitivities will take place.
□ plots: Valid inputs are “on” and “off”. When set to “on” it returns the Box-Cox based ROC along with informative information about the two AUCs in the legend of the plot.
• Value
□ resultstable: A summary table of the comparison that contains the sensitivity of each marker at the given specificity, the p-value of the difference using the probit transformation, the
p-value of the difference, and the confidence interval of the difference.
□ Sens1: The sensitivity at the selected specificity for the first marker.
□ Sens2: The sensitivity at the selected specificity for the second marker.
□ pvalue_probit_difference: The p-value for the comparison of the sensitivities. It employs the probit transformation that has greater power then when not using it (as opposed to the p-value
provided in the ‘pvalue_difference’ output argument provided later on).
□ pvalue_difference: The p-value for the comparisons of the sensitivities (without the probit transformation). Simulations have shown that this is inferior to the ‘pvalue_probit_difference’.
□ CI_difference: The confidence interval for the difference of the sensitivities.
□ roc1: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ roc2: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ transx1: The Box-Cox transformed scores for the first marker and the healthy group.
□ transy1: The Box-Cox transformed scores for the first marker and the diseased group.
□ transformation.parameter.1: The estimated Box-Cox transformation parameter (lambda) for marker 1.
□ transx2: The Box-Cox transformed scores for the second marker and the healthy group.
□ transy2: The Box-Cox transformed scores for the second marker and the diseased group.
□ transformation.parameter.2: The estimated Box-Cox transformation parameter (lambda) for marker 2.
• **Example: **
out=comparebcSens(marker1=marker1, marker2=marker2, D=D, alpha =0.05, atSpec=0.8, plots="on")
## Length Class Mode
## resultstable 7 formattable numeric
## Sens1 1 -none- numeric
## Sens2 1 -none- numeric
## pvalue_probit_difference 1 -none- numeric
## pvalue_difference 1 -none- numeric
## CI_difference 2 -none- numeric
## roc1 1 -none- function
## roc2 1 -none- function
## transx1 100 -none- numeric
## transy1 100 -none- numeric
## transformation.parameter.1 1 -none- numeric
## transx2 100 -none- numeric
## transy2 100 -none- numeric
## transformation.parameter.2 1 -none- numeric
• Description
□ This function provides a comparison of two correlated markers in terms of their specificities at a given sensitivity (for Box-Cox based ROC curves). Marker measurements are assumed to be
taken on the same individuals for both markers.
• Usage
out=comparebcSpec(marker1, marker2, D, alpha, atSens, plots)
• Arguments
□ marker1: A vector of length n that contains the biomarker scores of all individuals for the first marker.
□ marker2: A vector of length n that contains the biomarker scores of all individuals for the second marker.
□ D: A vector of length n that contains the true disease status of an individual, where 0 denotes a healthy/control individual, and 1 denotes a diseased individual.
□ alpha: Nominal level used to calculate the confidence intervals. A common choice is 0.05.
□ atSens: The value of sensitivity at which the comparison of specificities will take place.
□ plots: Valid inputs are “on” and “off”. When set to “on” it returns the Box-Cox based ROC along with informative information about the two AUCs in the legend of the plot.
• Value
□ Spec1: The specificity at the selected sensitivity for the first marker.
□ Spec2: The specificity at the selected sensitivity for the second marker.
□ pvalue_probit_difference: The p-value for the comparison of the specificities. It employs the probit transformation that has greater power then when not using it (as opposed to the p-value
provided in the ‘pvalue_difference’ output argument provided later on).
□ pvalue_difference: The p-value for the comparison of the specificities (without the probit transformation). Simulations have shown that this is inferior to the ‘pvalue_probit_difference’.
□ CI_difference: The confidence interval for the difference of the specificities.
□ roc1: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ roc2: A function that refers to the ROC of the first marker. It allows the user to feed in FPR values and the corresponding TPR values.
□ transx1: The Box-Cox transformed scores for the first marker and the healthy group.
□ transy1: The Box-Cox transformed scores for the first marker and the diseased group.
□ transformation.parameter.1: The estimated Box-Cox transformation parameter (lambda) for marker 1.
□ transx2: The Box-Cox transformed scores for the second marker and the healthy group.
□ transy2: The Box-Cox transformed scores for the second marker and the diseased group.
□ transformation.parameter.2: The estimated Box-Cox transformation parameter (lambda) for marker 2.
• **Example: **
out=comparebcSpec(marker1=marker1, marker2=marker2, D=D, alpha =0.05, atSens=0.8, plots="on")
## Length Class Mode
## resultstable 7 formattable numeric
## FPR1 1 -none- numeric
## FPR2 1 -none- numeric
## pvalue_probit_difference 1 -none- numeric
## pvalue_difference 1 -none- numeric
## CI_difference 2 -none- numeric
## roc1 1 -none- function
## roc2 1 -none- function
## transx1 100 -none- numeric
## transy1 100 -none- numeric
## transformation.parameter.1 1 -none- numeric
## transx2 100 -none- numeric
## transy2 100 -none- numeric
## transformation.parameter.2 1 -none- numeric
• Description
□ This function provides a visual comparison of the Empirical ROC, the Box-Cox ROC, and the Metz binormal semi-parametric estimator of the ROC curve. It also computes the AUC for the curve
corresponding to each method.
• Usage
threerocs2(marker1, marker2, D, plots)
• Arguments
□ marker1: A vector of length n that contains the biomarker scores of all individuals for the first marker.
□ marker2: A vector of length n that contains the biomarker scores of all individuals for the second marker.
□ D: A vector of length n that contains the true disease status of an individual. It is a binary vector containing 0 for the healthy/control individuals and 1 for the diseased individuals.
□ plots='on' or 'off': Valid inputs are “on” and “off”. When set to “on”, the user gets a single plot containing the estimated ROC curves using the Empirical, Box-Cox, and Metz methods for each
of the two provided markers.
• Value
□ AUC_BoxCox1: The AUC of the Box-Cox based ROC curve for the first marker.
□ AUC_BoxCox2: The AUC of the Box-Cox based ROC curve for the second marker.
□ AUC_Metz1: The AUC of the Metz binormal curve (as calculated by MRMCaov package using the “binormal” option) for the first marker.
□ AUC_Metz2: The AUC of the Metz binormal curve (as calculated by MRMCaov package using the “binormal” option) for the second marker.
□ AUC_Empirical1: The AUC of the empirical ROC curve for the first marker.
□ AUC_Empirical2: The AUC of the empirical ROC curve for the second marker.
• **Example: **
out=threerocs2(marker1, marker2, D, plots = "on")
## Length Class Mode
## AUC_BoxCox1 1 -none- numeric
## AUC_BoxCox2 1 -none- numeric
## AUC_Metz1 1 -none- numeric
## AUC_Metz2 1 -none- numeric
## AUC_Empirical1 1 -none- numeric
## AUC_Empirical2 1 -none- numeric | {"url":"http://cran.stat.auckland.ac.nz/web/packages/rocbc/vignettes/documentation_rocbc.html","timestamp":"2024-11-02T07:44:57Z","content_type":"text/html","content_length":"932397","record_id":"<urn:uuid:3b0b4168-7529-44c8-8dbd-dbe4889a9a7c>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00607.warc.gz"} |
Compute output, error and coefficients using recursive least squares (RLS) algorithm
The dsp.RLSFilter System object™ filters each channel of the input using RLS filter implementations.
To filter each channel of the input:
1. Create the dsp.RLSFilter object and set its properties.
2. Call the object with arguments, as if it were a function.
To learn more about how System objects work, see What Are System Objects?
rlsFilt = dsp.RLSFilter returns an adaptive RLS filter System object, rlsFilt. This System object computes the filtered output, filter error, and the filter weights for a given input and desired
signal using the RLS algorithm.
rlsFilt = dsp.RLSFilter(len) returns an RLS filter System object, rlsFilt. This System object has the Length property set to len.
rlsFilt = dsp.RLSFilter(Name,Value) returns an RLS filter System object with each specified property set to the specified value. Enclose each property name in single quotes. Unspecified properties
have default values.
Unless otherwise indicated, properties are nontunable, which means you cannot change their values after calling the object. Objects lock when you call them, and the release function unlocks them.
If a property is tunable, you can change its value at any time.
For more information on changing property values, see System Design in MATLAB Using System Objects.
Method — Method to calculate filter coefficients
Conventional RLS (default) | Householder RLS | Sliding-window RLS | Householder sliding-window RLS | QR decomposition
You can specify the method used to calculate filter coefficients as Conventional RLS [1] [2], Householder RLS [3] [4], Sliding-window RLS [5][1][2], Householder sliding-window RLS [4], or QR
decomposition [1] [2]. This property is nontunable.
Length — Length of filter coefficients vector
32 (default) | positive integer
Specify the length of the RLS filter coefficients vector as a scalar positive integer value. This property is nontunable.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
SlidingWindowBlockLength — Width of sliding window
48 (default) | positive integer
Specify the width of the sliding window as a scalar positive integer value greater than or equal to the Length property value. This property is nontunable.
This property applies only when the Method property is set to Sliding-window RLS or Householder sliding-window RLS.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
ForgettingFactor — RLS forgetting factor
1 (default) | positive scalar
Specify the RLS forgetting factor as a scalar positive numeric value less than or equal to 1. Setting this property value to 1 denotes infinite memory, while adapting to find the new filter.
Tunable: Yes
Data Types: single | double
InitialCoefficients — Initial coefficients of filter
0 (default) | scalar | vector
Specify the initial values of the FIR adaptive filter coefficients as a scalar or a vector of length equal to the Length property value.
Tunable: Yes
Data Types: single | double
InitialInverseCovariance — Initial inverse covariance
1000 (default) | scalar | square matrix
Specify the initial values of the inverse covariance matrix of the input signal. This property must be either a scalar or a square matrix, with each dimension equal to the Length property value. If
you set a scalar value, the InverseCovariance property is initialized to a diagonal matrix with diagonal elements equal to that scalar value.
Tunable: Yes
This property applies only when the Method property is set to Conventional RLS or Sliding-window RLS.
Data Types: single | double
InitialSquareRootInverseCovariance — Initial square root inverse covariance
sqrt(1000) (default) | scalar | square matrix
Specify the initial values of the square root inverse covariance matrix of the input signal. This property must be either a scalar or a square matrix with each dimension equal to the Length property
value. If you set a scalar value, the SquareRootInverseCovariance property is initialized to a diagonal matrix with diagonal elements equal to that scalar value.
Tunable: Yes
This property applies only when the Method property is set to Householder RLS or Householder sliding-window RLS.
Data Types: single | double
InitialSquareRootCovariance — Initial square root covariance
sqrt(1/1000) (default) | scalar | square matrix
Specify the initial values of the square root covariance matrix of the input signal. This property must be either a scalar or a square matrix with each dimension equal to the Length property value.
If you set a scalar value, the SquareRootCovariance property is initialized to a diagonal matrix with diagonal elements equal to the scalar value.
Tunable: Yes
This property applies only when the Method property is set to QR-decomposition RLS.
Data Types: single | double
LockCoefficients — Lock coefficient updates
false (default) | true
Specify whether the filter coefficient values should be locked. When you set this property to true, the filter coefficients are not updated and their values remain the same. The default value is
false (filter coefficients continuously updated).
Tunable: Yes
y = rlsFilt(x,d) recursively adapts the reference input, x, to match the desired signal, d, using the System object, rlsFilt. The desired signal, d, is the signal desired plus some undesired noise.
[y,e] = rlsFilt(x,d) shows the output of the RLS filter along with the error, e, between the reference input and the desired signal. The filters adapts its coefficients until the error e is
minimized. You can access these coefficients by accessing the Coefficients property of the object. This can be done only after calling the object. For example, to access the optimized coefficients of
the rlsFilt filter, call rlsFilt.Coefficients after you pass the input and desired signal to the object.
Input Arguments
x — Data input
scalar | column vector
The signal to be filtered by the RLS filter. The input, x, and the desired signal, d, must have the same size and data type.
The input can be a variable-size signal. You can change the number of elements in the column vector even when the object is locked. The System object locks when you call the object to run its
Data Types: single | double
Complex Number Support: Yes
d — Desired signal
scalar | column vector
The RLS filter adapts its coefficients to minimize the error, e, and converge the input signal x to the desired signal d as closely as possible.
The input, x, and the desired signal, d, must have the same size and data type.
The desired signal, d, can be a variable-size signal. You can change the number of elements in the column vector even when the object is locked. The System object locks when you call the object to
run its algorithm.
Data Types: single | double
Complex Number Support: Yes
Output Arguments
y — Filtered output
scalar | column vector
Filtered output, returned as a scalar or a column vector. The object adapts its filter coefficients to converge the input signal x to match the desired signal d. The filter outputs the converged
Data Types: single | double
Complex Number Support: Yes
e — Difference between output and desired signal
scalar | column vector
Difference between the output signal y and the desired signal d, returned as a scalar or a column vector. The objective of the RLS filter is to minimize this error. The object adapts its coefficients
to converge toward optimal filter coefficients that produce an output signal that matches closely with the desired signal. For more details on how e is computed, see Algorithms. To access the RLS
filter coefficients, call rlsFilt.Coefficients after you pass the input and desired signal to the object.
Data Types: single | double
Complex Number Support: Yes
Object Functions
To use an object function, specify the System object as the first input argument. For example, to release system resources of a System object named obj, use this syntax:
Specific to dsp.RLSFilter
Common to All System Objects
step Run System object algorithm
release Release resources and allow changes to System object property values and input characteristics
reset Reset internal states of System object
System Identification of FIR Filter Using RLS Filter
Use a recursive least squares (RLS) filter to identify an unknown system modeled with a lowpass FIR filter. Compare the frequency responses of the unknown and estimated systems.
Create a dsp.FIRFilter object that represents the system to be identified. Pass the signal x to the FIR filter. The output of the unknown system is the desired signal d, which is the sum of the
output of the unknown system (FIR filter) and an additive noise signal n.
filt = dsp.FIRFilter('Numerator',designLowpassFIR(FilterOrder=10,CutoffFrequency=.25));
x = randn(1000,1);
n = 0.01*randn(1000,1);
d = filt(x) + n;
Adaptive Filter
Create a dsp.RLSFilter object to create an RLS filter. Set the length of the filter to 11 taps and the forgetting factor to 0.98. Pass the primary input signal x and the desired signal d to the RLS
filter. The output y of the adaptive filter is the signal converged to the desired signal d thereby minimizing the error e between the two signals.
rls = dsp.RLSFilter(11, 'ForgettingFactor', 0.98);
[y,e] = rls(x,d);
w = rls.Coefficients;
Plot the results
The output signal matches the desired signal, making the error between the two close to zero.
plot(1:1000, [d,y,e]);
title('System Identification of an FIR filter');
legend('Desired', 'Output', 'Error');
xlabel('time index');
ylabel('signal value');
Compare the weights
The weights vector w represents the coefficients of the RLS filter that is adapted to resemble the unknown system (FIR filter). To confirm the convergence, compare the numerator of the FIR filter and
the estimated weights of the RLS filter.
The estimated filter weights closely match the actual filter weights, confirming the results seen in the previous signal plot.
stem([filt.Numerator; w].');
xlabel('coefficient #');
ylabel('coefficient value');
Inverse System Identification Using RLS Algorithm
This example demonstrates the RLS adaptive algorithm using the inverse system identification model shown here.
Cascading the adaptive filter with an unknown filter causes the adaptive filter to converge to a solution that is the inverse of the unknown system.
If the transfer function of the unknown system and the adaptive filter are H(z) and G(z), respectively, the error measured between the desired signal and the signal from the cascaded system reaches
its minimum when G(z)$×$H(z) = 1. For this relation to be true, G(z) must equal 1/H(z), the inverse of the transfer function of the unknown system.
To demonstrate that this is true, create a signal s to input to the cascaded filter pair.
In the cascaded filters case, the unknown filter results in a delay in the signal arriving at the summation point after both filters. To prevent the adaptive filter from trying to adapt to a signal
it has not yet seen (equivalent to predicting the future), delay the desired signal by 12 samples, which is the order of the unknown system.
Generally, you do not know the order of the system you are trying to identify. In that case, delay the desired signal by number of samples equal to half the order of the adaptive filter. Delaying the
input requires prepending 12 zero-value samples to the input s.
delay = zeros(12,1);
d = [delay; s(1:2988)]; % Concatenate the delay and the signal.
You have to keep the desired signal vector d the same length as x, so adjust the signal element count to allow for the delay samples.
Although not generally the case, for this example you know the order of the unknown filter, so add a delay equal to the order of the unknown filter.
For the unknown system, use a lowpass, 12th-order FIR filter.
filt = dsp.FIRFilter;
filt.Numerator = designLowpassFIR(FilterOrder=12,CutoffFrequency=0.55);
Filtering s provides the input data signal for the adaptive algorithm function.
To use the RLS algorithm, create a dsp.RLSFilter object and set its Length, ForgettingFactor, and InitialInverseCovariance properties.
For more information about the input conditions to prepare the RLS algorithm object, refer to dsp.RLSFilter.
p0 = 2 * eye(13);
lambda = 0.99;
rls = dsp.RLSFilter(13,'ForgettingFactor',lambda,...
This example seeks to develop an inverse solution, you need to be careful about which signal carries the data and which is the desired signal.
Earlier examples of adaptive filters use the filtered noise as the desired signal. In this case, the filtered noise (x) carries the unknown system's information. With Gaussian distribution and
variance of 1, the unfiltered noise d is the desired signal. The code to run this adaptive filter is:
where y returns the filtered output and e contains the error signal as the filter adapts to find the inverse of the unknown system.
Obtain the estimated coefficients of the RLS filter.
View the frequency response of the adapted RLS filter (inverse system, G(z)) using freqz. The inverse system looks like a highpass filter with linear phase.
View the frequency response of the unknown system, H(z). The response is that of a lowpass filter with a cutoff frequency of 0.55.
The result of the cascade of the unknown system and the adapted filter is a compensated system with an extended cutoff frequency of 0.8.
overallCoeffs = conv(filt.Numerator,b);
Cancel Noise Using RLS Filter
Cancel additive noise n added to an unknown system using an RLS filter. The RLS filter adapts its coefficients until its transfer function matches the transfer function of the unknown system as
closely as possible. The difference between the output of the adaptive filter and the output of the unknown system is the error signal e, which represents the additive white noise. Minimizing this
error signal is the objective of the adaptive filter.
Create a dsp.FIRFilter System object™ to represent the unknown system. Create a dsp.RLSFilter object and set the length to 11 taps. Set the method to 'Householder RLS'. Create a sine wave to
represent the noise added to the unknown system. View the signals in a time scope.
FrameSize = 100;
NIter = 10;
rls = dsp.RLSFilter('Length',11,...
'Method','Householder RLS');
filt = dsp.FIRFilter('Numerator',...
sinewave = dsp.SineWave('Frequency',0.01,...
scope = timescope('LayoutDimensions',[2 1],...
'NumInputPorts',2, ...
'YLimits',[-2.5 2.5], ...
'ChannelNames',{'Noisy signal'},...
'ChannelNames',{'Error signal'});
for k = 1:NIter
x = randn(FrameSize,1);
d = filt(x) + sinewave();
[y,e] = rls(x,d);
w = rls.Coefficients;
The dsp.RLSFilter System object, when Conventional RLS is selected, recursively computes the least squares estimate (RLS) of the FIR filter weights. The System object estimates the filter weights or
coefficients, needed to convert the input signal into the desired signal. The input signal can be a scalar or a column vector. The desired signal must have the same data type, complexity, and
dimensions as the input signal. The corresponding RLS filter is expressed in matrix form as P(n) :
$\begin{array}{l}k\left(n\right)=\frac{{\lambda }^{-1}P\left(n-1\right)u\left(n\right)}{1+{\lambda }^{-1}{u}^{H}\left(n\right)P\left(n-1\right)u\left(n\right)}\\ y\left(n\right)={w}^{T}\left(n-1\
right)u\left(n\right)\\ e\left(n\right)=d\left(n\right)-y\left(n\right)\\ w\left(n\right)=w\left(n-1\right)+{k}^{*}\left(n\right)e\left(n\right)\\ P\left(n\right)={\lambda }^{-1}P\left(n-1\right)-{\
lambda }^{-1}k\left(n\right){u}^{H}\left(n\right)P\left(n-1\right)\end{array}$
where λ^-1 denotes the reciprocal of the exponential weighting factor. The variables are as follows:
Variable Description
n The current time index
u(n) The vector of buffered input samples at step n
P(n) The conjugate of the inverse correlation matrix at step n
k(n) The gain vector at step n
k*(n) Complex conjugate of k
w(n) The vector of filter tap estimates at step n
y(n) The filtered output at step n
e(n) The estimation error at step n
d(n) The desired response at step n
λ The forgetting factor
u, w, and k are all column vectors.
[1] M Hayes, Statistical Digital Signal Processing and Modeling, New York: Wiley, 1996.
[2] S. Haykin, Adaptive Filter Theory, 4th Edition, Upper Saddle River, NJ: Prentice Hall, 2002.
[3] A.A. Rontogiannis and S. Theodoridis, "Inverse factorization adaptive least-squares algorithms," Signal Processing, vol. 52, no. 1, pp. 35-47, July 1996.
[4] S.C. Douglas, "Numerically-robust O(N^2) RLS algorithms using least-squares prewhitening," Proc. IEEE Int. Conf. on Acoustics, Speech, and Signal Processing, Istanbul, Turkey, vol. I, pp.
412-415, June 2000.
[5] A. H. Sayed, Fundamentals of Adaptive Filtering, Hoboken, NJ: John Wiley & Sons, 2003.
Version History
Introduced in R2013a | {"url":"https://nl.mathworks.com/help/dsp/ref/dsp.rlsfilter-system-object.html","timestamp":"2024-11-03T03:17:14Z","content_type":"text/html","content_length":"144071","record_id":"<urn:uuid:96f173ba-e84e-44bb-90a8-c59ae708c98e>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00735.warc.gz"} |
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The Stacks project
Lemma 10.96.8. Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. If $M$ is $J$-adically complete and $I$ is finitely generated, then $M$ is $I$-adically complete.
Proof. Assume $M$ is $J$-adically complete and $I$ is finitely generated. We have $\bigcap I^ nM = 0$ because $\bigcap J^ nM = 0$. By Lemma 10.96.7 it suffices to prove the surjectivity of $M \to \
mathop{\mathrm{lim}}\nolimits M/I^ nM$ in case $I$ is generated by a single element. Say $I = (f)$. Let $x_ n \in M$ with $x_{n + 1} - x_ n \in f^ nM$. We have to show there exists an $x \in M$ such
that $x_ n - x \in f^ nM$ for all $n$. As $x_{n + 1} - x_ n \in J^ nM$ and as $M$ is $J$-adically complete, there exists an element $x \in M$ such that $x_ n - x \in J^ nM$. Replacing $x_ n$ by $x_ n
- x$ we may assume that $x_ n \in J^ nM$. To finish the proof we will show that this implies $x_ n \in I^ nM$. Namely, write $x_ n - x_{n + 1} = f^ nz_ n$. Then
\[ x_ n = f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots ) \]
The sum $z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots $ converges in $M$ as $f^ c \in J^ c$. The sum $f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots )$ converges in $M$ to $x_ n$ because the partial
sums equal $x_ n - x_{n + c}$ and $x_{n + c} \in J^{n + c}M$. $\square$
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Margin of error E=0.30 populations standard deviation =2.5. Population means with 95% confidence. What I the required sample size (round up to the whole number)
Margin of error E=0.30 populations standard deviation =2.5. Population means with 95% confidence. What I the required sample size (round up to the whole number)
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Answer to a math question Margin of error E=0.30 populations standard deviation =2.5. Population means with 95% confidence. What I the required sample size (round up to the whole number)
110 Answers
To calculate the required sample size, we can use the formula:
n = \frac{{Z^2 \cdot \sigma^2}}{{E^2}}
- \(n\) is the required sample size
- \(Z\) is the z-score corresponding to the desired confidence level (in this case, for 95%, \(Z = 1.96\))
- \(\sigma\) is the population standard deviation
- \(E\) is the margin of error
Plugging in the given values:
n = \frac{{1.96^2 \cdot 2.5^2}}{{0.3^2}}
Simplifying the equation:
n = \frac{{3.8416 \cdot 6.25}}{{0.09}}
n = \frac{{24.01}}{{0.09}}
n \approx 266.778
Rounding up to the nearest whole number, the required sample size is:
Answer: The required sample size is 267.
Frequently asked questions (FAQs)
Question: Find the values of x that satisfy the cubic equation x^3 - 7x^2 + 16x - 12 = 0. How many solutions are there?
What is the result of multiplying two vectors A = (a1, a2, a3) and B = (b1, b2, b3) where a1, a2, a3, b1, b2, b3 are the respective components?
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A researcher has created a small simulation in MATLAB, and we want to make it accessible to others. My plan is to take the simulation, clean up a few things, and turn it into a set of functions.
Then, I plan to compile it into a C library and use SWIG to create a Python wrapper. At that point, I should be able to call the simulation from... | {"url":"http://ansaurus.com/tag/matlab","timestamp":"2024-11-06T14:28:25Z","content_type":"application/xhtml+xml","content_length":"20855","record_id":"<urn:uuid:6d3eb65f-325f-4119-9391-18edf92d296d>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00719.warc.gz"} |
Moore-Penrose pseudoinverse
B = pinv(A,tol) specifies a value for the tolerance. pinv treats singular values of A that are less than or equal to the tolerance as zero.
Solve System of Linear Equations Using Pseudoinverse
Compare solutions to a system of linear equations obtained by backslash (\), pinv, and lsqminnorm.
If a rectangular coefficient matrix A is of low rank, then the least-squares problem of minimizing norm(A*x-b) has infinitely many solutions. Two solutions are returned by x1 = A\b and x2 = pinv(A)
*b. The distinguishing properties of these solutions are that x1 has only rank(A) nonzero components, and norm(x2) is smaller than for any other solution.
Create an 8-by-6 matrix that has rank(A) = 3.
A = magic(8);
A = A(:,1:6)
A = 8×6
Create a vector for the right side of the system of equations.
b = 8×1
The number chosen for the right side, 260, is the value of the 8-by-8 magic sum for A. If A were still an 8-by-8 matrix, then one solution for x would be a vector of 1s. With only six columns, a
solution exists because the equations are still consistent, but the solution is not all 1s. Because the matrix is of low rank, there are infinitely many solutions.
Solve for two of the solutions using backslash and pinv.
Warning: Rank deficient, rank = 3, tol = 1.882938e-13.
x1 = 6×1
x2 = 6×1
Both of these solutions are exact, in the sense that norm(A*x1-b) and norm(A*x2-b) are on the order of round-off error. The solution x1 is special because it has only three nonzero elements. The
solution x2 is special because norm(x2) is smaller than it is for any other solution, including norm(x1).
Using lsqminnorm to compute the least-squares solution of this problem produces the same solution as using pinv does. lsqminnorm(A,b) is typically more efficient than pinv(A)*b.
x3 = 6×1
Input Arguments
A — Input matrix
Input matrix.
Data Types: single | double
Complex Number Support: Yes
tol — Singular value tolerance
Singular value tolerance, specified as a scalar. pinv treats singular values that are less than or equal to tol as zeros during the computation of the pseudoinverse.
The default tolerance is max(size(A))*eps(norm(A)).
Example: pinv(A,1e-4)
More About
Moore-Penrose Pseudoinverse
The Moore-Penrose pseudoinverse is a matrix that can act as a partial replacement for the matrix inverse in cases where it does not exist. This matrix is frequently used to solve a system of linear
equations when the system does not have a unique solution or has many solutions.
For any matrix A, the pseudoinverse B exists, is unique, and has the same dimensions as A'. If A is square and not singular, then pinv(A) is simply an expensive way to compute inv(A). However, if A
is not square, or is square and singular, then inv(A) does not exist. In these cases, pinv(A) has some (but not all) of the properties of inv(A):
$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}ABA=A\\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}BAB=B\\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left(AB\right)}^{*}=AB\\
Here, AB and BA are Hermitian. The pseudoinverse computation is based on svd(A). The calculation treats singular values less than or equal to tol as zero.
• You can replace most uses of pinv applied to a vector b, as in pinv(A)*b, with lsqminnorm(A,b) to get the minimum-norm least-squares solution of a system of linear equations. For example, in
Solve System of Linear Equations Using Pseudoinverse, lsqminnorm produces the same solution as pinv does. lsqminnorm is generally more efficient than pinv because lsqminnorm uses the complete
orthogonal decomposition of A to find its low-rank approximation and applies its factors to b. In contrast, pinv uses singular value decomposition to explicitly form the pseudoinverse of A that
you then must multiply by b. lsqminnorm also supports sparse matrices.
pinv uses singular value decomposition to form the pseudoinverse of A. Singular values along the diagonal of S that are less than or equal to tol are treated as zeros, and the representation of A
$\begin{array}{l}A=US{V}^{*}=\left[{U}_{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{U}_{2}\right]\left[\begin{array}{cc}{S}_{1}& 0\\ 0& 0\end{array}\right]{\left[{V}_{1}\text{\hspace{0.17em}}\text
{\hspace{0.17em}}{V}_{2}\right]}^{*}\\ A={U}_{1}{S}_{1}{V}_{1}^{*}\text{\hspace{0.17em}}.\end{array}$
The pseudoinverse of A is then equal to:
Extended Capabilities
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Usage notes and limitations:
• Code generation does not support sparse matrix inputs for this function.
GPU Code Generation
Generate CUDA® code for NVIDIA® GPUs using GPU Coder™.
Refer to the usage notes and limitations in the C/C++ Code Generation section. The same limitations apply to GPU code generation.
Thread-Based Environment
Run code in the background using MATLAB® backgroundPool or accelerate code with Parallel Computing Toolbox™ ThreadPool.
This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment.
GPU Arrays
Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™.
The pinv function fully supports GPU arrays. To run the function on a GPU, specify the input data as a gpuArray (Parallel Computing Toolbox). For more information, see Run MATLAB Functions on a GPU
(Parallel Computing Toolbox).
Version History
Introduced before R2006a
R2021b: pinv returns NaN for nonfinite inputs
pinv returns NaN values when the input contains nonfinite values (Inf or NaN). Previously, pinv threw an error when the input contained nonfinite values. | {"url":"https://au.mathworks.com/help/matlab/ref/pinv.html","timestamp":"2024-11-10T02:07:47Z","content_type":"text/html","content_length":"93889","record_id":"<urn:uuid:5b25a903-2274-49cf-8d6d-11b2324cc710>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00852.warc.gz"} |
Entropy (for data science) Clearly Explained - StatQuest!!!
Entropy (for data science) Clearly Explained
NOTE: This StatQuest was supported by these awesome people who support StatQuest at the Double BAM level: Z. Rosenberg, S. Shah, J. N., J. Horn, J. Wong, I. Galic, H-S. Ming, D. Greene, D. Schioberg,
C. Walker, G. Singh, L. Cisterna, J. Alexander, J. Varghese, K. Manickam, N. Fleming, F. Prado, J. Malone-Lee
7 thoughts on “Entropy (for data science) Clearly Explained”
1. Send me the note in pdf
2. Send me the note on entropy in pdf.
3. You are one of the geniuses in teaching I have ever come across.
4. Thank you very much Josh!
5. Hi Josh,
I am dissatisfied with your explanation for surprise being log(1/p) on the basis that it is 0 when p is 1 and infinity when p is 0. There are a zillion functions of p with those limits. Why is
log(1/p) preferred over all of the others? Is it just a convention chosen to match the physics definition and because of the convenient mathematical properties of the log? Or is there something
more to it?
□ If you want a more mathematically grounded explanation, I would highly recommend the original manuscript by Shannon: https://people.math.harvard.edu/~ctm/home/text/others/shannon/entropy/
□ I believe it might have something to do with simple calculations to compute gradients when softmax is paired with the cross-entropy loss. | {"url":"https://statquest.org/entropy-for-data-science-clearly-explained/","timestamp":"2024-11-14T20:27:42Z","content_type":"text/html","content_length":"42660","record_id":"<urn:uuid:acb33161-ce8f-41b8-ae8e-8df5dc071755>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00681.warc.gz"} |
Grade 5 Term 2 Language Arts and Mathematics - EasyA Club
Grade 5 Term 2 Language Arts and Mathematics
What Will You Learn?
• At the end of the term, students should be able to:
• Mathematics
• Express fractional numbers in decimal form beginning with those having denominators of 10, 100, 1000.
• Determine the value of each digit in a decimal number up to thousandths.
• Place in serial order any set of decimal fractions.
• Round a decimal number to the nearer whole number, tenth or hundredth.
• Add or subtract decimal numbers to three decimal places.
• Round a mixed number to the nearer whole number.
• Find the product of a whole number and a decimal number to three places of decimals.
• Solve problems (including worded problems and money) requiring the addition/subtraction of decimal numbers.
• Draw pictures of polygons given description.
• Identify the condition that makes a triangle right, equilateral, isosceles or scalene.
• Identify opposite and adjacent sides of a quadrilateral.
• Estimate, measure and record distances including the perimeter of polygons in millimetres and/or centimetres and metres.
• Solve problems requiring the calculation of the following: The perimeter. Length of one side. The number of sides of a regular polygon, given the other two measures.
• Find the area of polygons by counting squares.
• Solve problems based on computing the measurement of the area of a rectangular region.
• Multiply the decimal number by 10,100,1000.
• Rename two or more fractional numbers with unlike denominators to show the same denominator.
• Compare fractional numbers in any form.
• Add or subtract factions including mixed numbers with and without renaming.
• Find the product of two proper fractions.
• Investigate the order of operation when evaluating algebraic expressions.
• Use the symbols, =, ≠ in number sentences.
• Identify and count the number of lines of symmetry in plane figures.
• Create shapes given, the line of symmetry and half the shape and the line of symmetry.
• Round a whole number to the nearer ten, hundred, thousand.
• Round a number representing an amount of money to the nearest dollar, ten dollars, hundred dollars, thousand dollars.
• Recognize and use the relationship between the millilitre, litre and kilolitre.
• Associate the four major cardinal points with quarter, half, three-quarter and full turns.
• Use a grid system to describe the location of one point relative to another using the four major cardinal points.
• Predict how a simple plane shape will look after a series of rightward or leftward flips, or, after a reflection.
• Use the substitution in formulae to solve worded problems.
• Language Arts
• Determine multiple meanings of words by applying knowledge of context clues.
• Use knowledge of synonyms and antonyms to construct meaningful sentences.
• Use stated and implied ideas in texts to make inferences and construct meaning .
• Summarize important ideas and cite supporting details.
• Identify and use information at the literal, inferential and critical levels.
• Establish cause and effect relationships.
• Identify problem and solution text structure .
• Use adverbs and prepositions of time.
• Use accurate subject /Verb Agreement .
• Distinguish between direct and reported speech.
• Show correct agreement of pronouns and antecedents (noun it replaces).
• Compose narratives which include the basic story elements.
• Use adjectives to describe people, places and things in written narratives.
• Begin to organize information located from various sources.
• Skim and scan for information using basic text features.
Courses in the Bundle (2) | {"url":"https://easyaclub.com/course-bundle/grade-5-term-2-language-arts-and-mathematics/","timestamp":"2024-11-13T11:49:31Z","content_type":"text/html","content_length":"158765","record_id":"<urn:uuid:3a534a9d-e2e2-4ea5-a99c-f995dd9b0ff0>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00295.warc.gz"} |
The Simple Auto Regressive Model (AR)
In the previous few segments of this session you have learned two basic requirements to build an Auto Regressive model, Stationarity and Autocorrelation. In this segment and the next session, few
Auto Regressive models that you will be studying are as follows:
• Auto Regressive (AR)
• Moving Average (MA)
• Auto Regressive Moving Average (ARMA)
• Auto Regressive Integrated Moving Average (ARIMA)
• Seasonal Auto Regressive Integrated Moving Average(SARIMA).
• Seasonal Auto Regressive Integrated Moving Average with Exogenous variable (SARIMAX).
In this session we will cover the Simple Auto Regressive model and the Moving Average model.
Let us start with the first model, i.e., Simple Auto Regressive model(AR).
The Simple Auto Regressive model predicts the future observation as linear regression of one or more past observations. In simpler terms, the simple Auto Regressive model forecasts the dependent
variable (future observation) when one or more independent variables are known (past observations). This model has a parameter ‘p’ called lag order. Lag order is the maximum number of lags used to
build ‘p’ number of past data points to predict future data points.
Consider an example of forecasting monthly sales of ice cream for the year 2021 on the basis of the previous 3 years’ monthly sales data of the ice cream. This can be one of the simple Auto
Regressive model.
To determine the value of parameter ‘p’.
• Plot partial autocorrelation function
• Select p as the highest lag where partial autocorrelation is significantly high
Here, the lag value of 1, 2, 4 and 12 has a significant level of confidence. i.e., a significant level of influence on future observation (refer to the red line). Hence, the value of ‘p’ will be set
to 12 since that is the highest lag where partial autocorrelation is significantly high.
• Build the Auto Regression model equation:
The past values which have a significant value are 1, 2, 4 and 12. Therefore, in the regression model the independent variables yt−1, yt−2, yt−4 and yt−12 which are the observations from the past has
been taken to predict the dependent variable ^yt.
Now let us get back again to the airline passenger dataset and build an AR model on it. In order to build the AR model, the stationary series is divided into train and test data.
You have built the model now, but you do not have any idea what the forecast looks like. Let us look at that and the accuracy in the video below. Also, recall that you had performed boxcox
transformation and differencing in order to covert the airline passenger data into a stationary time-series. Now in order to recover the original forecast, you will have to reverse these
transformations. Let’s learn how to do that as well from Chiranjoy. | {"url":"https://www.internetknowledgehub.com/the-simple-auto-regressive-model-ar/","timestamp":"2024-11-13T06:09:22Z","content_type":"text/html","content_length":"81250","record_id":"<urn:uuid:c9cb7c10-b448-4f0e-a470-cf607aa017cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00786.warc.gz"} |
Create a summary table for a one-sample proportion interval — infer_1prop_int
Create a summary table for a one-sample proportion interval
Create a summary table for a one-sample proportion interval
A data frame (or tibble).
The variable to run the test on, in formula syntax, ~var.
The data value that constitutes a "success".
The number of digits to round table values to. Defaults to 3.
The confidence level of the interval, entered as a value between 0 and 1. Defaults to 0.95.
An override to the table caption. A sensible default is provided.
An object of class flextable. In an interactive environment, results are viewable immediately.
infer_1prop_int(mtcars, ~vs, success = 1)
One-Sample Proportion Confidence Interval on Variable vs
Successes: 1
n n Standard 95% 95%
Successes Missing n p̂ Error Interval Interval
Lower Upper
14 0 32 0.438 0.0877 0.266 0.609
infer_1prop_int(mtcars, ~vs, success = 1, conf_lvl = 0.90)
One-Sample Proportion Confidence Interval on Variable
Successes: 1
n n Standard 90% 90%
Successes Missing n p̂ Error Interval Interval
Lower Upper
14 0 32 0.438 0.0877 0.293 0.582 | {"url":"https://gvsu215.ianacurtis.com/reference/infer_1prop_int","timestamp":"2024-11-10T11:39:22Z","content_type":"text/html","content_length":"23402","record_id":"<urn:uuid:06a05c47-088f-4c8a-a11a-a32af423dc08>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00273.warc.gz"} |
Robustness of Servo Controller for DC Motor
This example shows how to use uncertain objects in Robust Control Toolbox™ to model uncertain systems and assess robust stability and robust performance using the robustness analysis tools.
Data Structures for Uncertainty Modeling
Robust Control Toolbox lets you create uncertain elements, such as physical parameters whose values are not known exactly, and combine these elements into uncertain models. You can then easily
analyze the impact of uncertainty on the control system performance.
For example, consider a plant model
$P\left(s\right)=\frac{\gamma }{\tau s+1}$
where gamma can range in the interval [3,5] and tau has average value 0.5 with 30% variability. You can create an uncertain model of P(s) as in this example:
gamma = ureal('gamma',4,'range',[3 5]);
tau = ureal('tau',.5,'Percentage',30);
P = tf(gamma,[tau 1])
Uncertain continuous-time state-space model with 1 outputs, 1 inputs, 1 states.
The model uncertainty consists of the following blocks:
gamma: Uncertain real, nominal = 4, range = [3,5], 1 occurrences
tau: Uncertain real, nominal = 0.5, variability = [-30,30]%, 1 occurrences
Type "P.NominalValue" to see the nominal value and "P.Uncertainty" to interact with the uncertain elements.
Suppose you have designed an integral controller C for the nominal plant (gamma=4 and tau=0.5). To find out how variations of gamma and tau affect the plant and the closed-loop performance, form the
closed-loop system CLP from C and P.
KI = 1/(2*tau.Nominal*gamma.Nominal);
C = tf(KI,[1 0]);
CLP = feedback(P*C,1)
Uncertain continuous-time state-space model with 1 outputs, 1 inputs, 2 states.
The model uncertainty consists of the following blocks:
gamma: Uncertain real, nominal = 4, range = [3,5], 1 occurrences
tau: Uncertain real, nominal = 0.5, variability = [-30,30]%, 1 occurrences
Type "CLP.NominalValue" to see the nominal value and "CLP.Uncertainty" to interact with the uncertain elements.
Plot the step response of the plant and closed-loop system. The step command automatically generates 20 random samples of the uncertain parameters gamma and tau and plots the corresponding step
subplot(2,1,1); step(P), title('Plant response (20 samples)')
subplot(2,1,2); step(CLP), title('Closed-loop response (20 samples)')
Figure 1: Step responses of the plant and closed-loop models
The bottom plot shows that the closed-loop system is reasonably robust despite significant fluctuations in the plant DC gain. This is a desirable and common characteristic of a properly designed
feedback system.
DC Motor Example with Parameter Uncertainty and Unmodeled Dynamics
This example builds on the example Reference Tracking of DC Motor with Parameter Variations by adding parameter uncertainty and unmodeled dynamics, to investigate the robustness of the servo
controller to such uncertainty.
The nominal model of the DC motor is defined by the resistance R, the inductance L, the emf constant Kb, armature constant Km, the linear approximation of viscous friction Kf and the inertial load J.
Each of these components varies within a specific range of values. The resistance and inductance constants range within ±40% of their nominal values. Use ureal to construct these uncertain
R = ureal('R',2,'Percentage',40);
L = ureal('L',0.5,'Percentage',40);
For physical reasons, the values of Kf and Kb are the same, even if they are uncertain. In this example, the nominal value is 0.015 with a range between 0.012 and 0.019.
K = ureal('K',0.015,'Range',[0.012 0.019]);
Km = K;
Kb = K;
Viscous friction, Kf, has a nominal value of 0.2 with a 50% variation in its value.
Kf = ureal('Kf',0.2,'Percentage',50);
Electrical and Mechanical Equations
The current in the electrical circuit, and the torque applied to the rotor can be expressed in terms of the applied voltage and the angular speed. Create the transfer function H relating these
variables, and make AngularSpeed an output of H for later use.
H = [1;0;Km] * tf(1,[L R]) * [1 -Kb] + [0 0;0 1;0 -Kf];
H.InputName = {'AppliedVoltage';'AngularSpeed'};
H.OutputName = {'Current';'AngularSpeed';'RotorTorque'};
The motor typically drives an inertia, whose dynamic characteristics relate the applied torque to the rate-of-change of the angular speed. For a rigid body, this value is a constant. A more
realistic, but uncertain, model might contain unknown damped resonances. Use the ultidyn object to model uncertain linear time-invariant dynamics. Set the nominal value of the rigid body inertia to
0.02 and we include 15% dynamic uncertainty in multiplicative form.
J = 0.02*(1 + ultidyn('Jlti',[1 1],'Type','GainBounded','Bound',0.15,...
Uncertain Model of DC Motor
It is a simple matter to relate the AngularSpeed input to the RotorTorque output through the uncertain inertia, J, using the lft command. The AngularSpeed input equals RotorTorque/(J*s). Therefore,
use "positive" feedback from the third output to the second input of H to make the connection. This connection results in a system with one input (AppliedVoltage) and two outputs (Current and
Pall = lft(H,tf(1,[1 0])/J);
Select only the AngularSpeed output for the remainder of the control analysis.
Uncertain continuous-time state-space model with 1 outputs, 1 inputs, 2 states.
The model uncertainty consists of the following blocks:
Jlti: Uncertain 1x1 LTI, peak gain = 0.15, 1 occurrences
K: Uncertain real, nominal = 0.015, range = [0.012,0.019], 2 occurrences
Kf: Uncertain real, nominal = 0.2, variability = [-50,50]%, 1 occurrences
L: Uncertain real, nominal = 0.5, variability = [-40,40]%, 1 occurrences
R: Uncertain real, nominal = 2, variability = [-40,40]%, 1 occurrences
Type "P.NominalValue" to see the nominal value and "P.Uncertainty" to interact with the uncertain elements.
P is a single-input, single-output uncertain model of the DC motor. For analysis purposes, use the following controller.
Cont = tf(84*[.233 1],[.0357 1 0]);
Open-Loop Analysis
First, compare the step response of the nominal DC motor with 15 samples of the uncertain model of the DC motor. Use usample to explicitly specify the number of random samples.
ans =
Legend (Samples, Nominal) with properties:
String: {'Samples' 'Nominal'}
Location: 'northeast'
Orientation: 'vertical'
FontSize: 9
Position: [0.7030 0.7695 0.1830 0.0789]
Units: 'normalized'
Use GET to show all properties
Figure 2: Plant step response
Similarly, compare the Bode response of the nominal (red) and sampled (blue) uncertain models of the DC motor.
ans =
Legend (Samples, Nominal) with properties:
String: {'Samples' 'Nominal'}
Location: 'northeast'
Orientation: 'vertical'
FontSize: 8.1000
Position: [0.7625 0.8257 0.1920 0.0884]
Units: 'normalized'
Use GET to show all properties
Figure 3: Plant Bode response
Robustness Analysis
In this section, analyze the robustness of the DC motor controller. A nominal analysis of the closed-loop system indicates the feedback loop is very robust with 22 dB gain margin and 66 deg of phase
Figure 4: Closed-loop robustness analysis
The diskmargin function computes the disk-based gain and phase margins. By modeling gain and phase variations at all frequencies and in all feedback loops, disk margins tend to be more accurate
estimates of robustness, especially in multi-loop control systems. Compute the disk-based margins for the DC motor loop.
DM = diskmargin(P.NominalValue*Cont)
DM = struct with fields:
GainMargin: [0.2792 3.5822]
PhaseMargin: [-58.8054 58.8054]
DiskMargin: 1.1271
LowerBound: 1.1271
UpperBound: 1.1271
Frequency: 5.0062
WorstPerturbation: [1x1 ss]
While smaller than the classical gain and phase margins, the disk-based margins essentially confirm that the nominal feedback loop is very robust. Now, recall that the DC motor plant is uncertain.
How does the modeled uncertainty affect these stability margins? For quick insight, plot the disk-based gain and phase margins for 20 samples of the uncertain open-loop response.
ans =
Legend (Samples, Nominal) with properties:
String: {'Samples' 'Nominal'}
Location: 'northeast'
Orientation: 'vertical'
FontSize: 8.1000
Position: [0.7518 0.8583 0.2006 0.0884]
Units: 'normalized'
Use GET to show all properties
Some combinations of plant uncertainty lead to smaller margins. The plot shows only a small sample. Use worst-case analysis to find out how bad the margins can really get. The wcdiskmargin function
directly computes the worst-case gain and phase margins for the modeled uncertainty.
wcDM = wcdiskmargin(P*Cont,'siso')
wcDM = struct with fields:
GainMargin: [0.8728 1.1457]
PhaseMargin: [-7.7680 7.7680]
DiskMargin: 0.1358
LowerBound: 0.1358
UpperBound: 0.1361
CriticalFrequency: 4.9846
WorstPerturbation: [1x1 ss]
Here the worst-case margins are only 1.2 dB and 7.8 degrees, signaling that the closed loop is nearly unstable for some combinations of the uncertain elements.
Robustness of Disturbance Rejection Characteristics
The sensitivity function is a standard measure of closed-loop performance for the feedback system. Compute the uncertain sensitivity function S and compare the Bode magnitude plots for the nominal
and sampled uncertain sensitivity functions.
S = feedback(1,P*Cont);
ans =
Legend (Samples, Nominal) with properties:
String: {'Samples' 'Nominal'}
Location: 'northeast'
Orientation: 'vertical'
FontSize: 9
Position: [0.7530 0.8528 0.2014 0.0938]
Units: 'normalized'
Use GET to show all properties
Figure 5: Magnitude of sensitivity function S.
In the time domain, the sensitivity function indicates how well a step disturbance can be rejected. Plot its step response to see the variability in disturbance rejection characteristics (nominal
appears in red).
title('Disturbance Rejection')
ans =
Legend (Samples, Nominal) with properties:
String: {'Samples' 'Nominal'}
Location: 'northeast'
Orientation: 'vertical'
FontSize: 9
Position: [0.7030 0.7968 0.1830 0.0789]
Units: 'normalized'
Use GET to show all properties
Figure 6: Rejection of a step disturbance.
Use the wcgain function to compute the worst-case value of the peak gain of the sensitivity function.
[maxgain,worstuncertainty] = wcgain(S);
maxgain = struct with fields:
LowerBound: 7.5199
UpperBound: 7.5357
CriticalFrequency: 4.9979
With the usubs function you can substitute the worst-case values of the uncertain elements into the uncertain sensitivity function S. This gives the worst-case sensitivity function Sworst over the
entire uncertainty range. Note that the peak gain of Sworst matches the lower-bound computed by wcgain.
Sworst = usubs(S,worstuncertainty);
Now compare the step responses of the nominal and worst-case sensitivity.
title('Disturbance Rejection')
ans =
Legend (Worst-case, Nominal) with properties:
String: {'Worst-case' 'Nominal'}
Location: 'northeast'
Orientation: 'vertical'
FontSize: 9
Position: [0.6755 0.7968 0.2104 0.0789]
Units: 'normalized'
Use GET to show all properties
Figure 7: Nominal and worst-case rejection of a step disturbance
Clearly some combinations of uncertain elements significantly degrade the ability of the controller to quickly reject disturbances. Finally, plot the magnitude of the nominal and worst-case values of
the sensitivity function. Observe that the peak value of Sworst occurs at the frequency maxgain.CriticalFrequency:
ans =
Legend (Worst-case, Nominal) with properties:
String: {'Worst-case' 'Nominal'}
Location: 'northeast'
Orientation: 'vertical'
FontSize: 9
Position: [0.7228 0.8528 0.2317 0.0938]
Units: 'normalized'
Use GET to show all properties
hold on
Figure 8: Magnitude of nominal and worst-case sensitivity
See Also
diskmargin | wcgain | uss | usubs
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3.2 Solve Percent Applications - Elementary Algebra | OpenStax
By the end of this section, you will be able to:
• Translate and solve basic percent equations
• Solve percent applications
• Find percent increase and percent decrease
• Solve simple interest applications
• Solve applications with discount or mark-up
Before you get started, take this readiness quiz.
Translate and Solve Basic Percent Equations
We will solve percent equations using the methods we used to solve equations with fractions or decimals. Without the tools of algebra, the best method available to solve percent problems was by
setting them up as proportions. Now as an algebra student, you can just translate English sentences into algebraic equations and then solve the equations.
We can use any letter you like as a variable, but it is a good idea to choose a letter that will remind us of what you are looking for. We must be sure to change the given percent to a decimal when
we put it in the equation.
Translate and solve: What number is 35% of 90?
Translate into algebra. Let $nn$= the number.
Remember "of" means multiply, "is" means equals.
$31.531.5$ is $35%35%$ of $9090$
Translate and solve:
What number is 45% of 80?
Translate and solve:
What number is 55% of 60?
We must be very careful when we translate the words in the next example. The unknown quantity will not be isolated at first, like it was in Example 3.12. We will again use direct translation to write
the equation.
Translate and solve: 6.5% of what number is $1.17?
Translate. Let $n=n=$ the number.
Divide both sides by 0.065 and simplify.
$6.5%6.5%$ of$1818$ is $1.171.17$
Translate and solve:
7.5% of what number is $1.95?
Translate and solve:
8.5% of what number is $3.06?
In the next example, we are looking for the percent.
Translate and solve: 144 is what percent of 96?
Translate into algebra. Let $p=p=$ the percent.
Divide by 96 and simplify.
Convert to percent.
$144144$ is $150%150%$ of $9696$
Note that we are asked to find percent, so we must have our final result in percent form.
Translate and solve:
110 is what percent of 88?
Translate and solve:
126 is what percent of 72?
Solve Applications of Percent
Many applications of percent—such as tips, sales tax, discounts, and interest—occur in our daily lives. To solve these applications we’ll translate to a basic percent equation, just like those we
solved in previous examples. Once we translate the sentence into a percent equation, we know how to solve it.
We will restate the problem solving strategy we used earlier for easy reference.
Use a Problem-Solving Strategy to Solve an Application.
1. Step 1. Read the problem. Make sure all the words and ideas are understood.
2. Step 2. Identify what we are looking for.
3. Step 3. Name what we are looking for. Choose a variable to represent that quantity.
4. Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.
5. Step 5. Solve the equation using good algebra techniques.
6. Step 6. Check the answer in the problem and make sure it makes sense.
7. Step 7. Answer the question with a complete sentence.
Now that we have the strategy to refer to, and have practiced solving basic percent equations, we are ready to solve percent applications. Be sure to ask yourself if your final answer makes
sense—since many of the applications will involve everyday situations, you can rely on your own experience.
Dezohn and his girlfriend enjoyed a nice dinner at a restaurant and his bill was $68.50. He wants to leave an 18% tip. If the tip will be 18% of the total bill, how much tip should he leave?
Step 1. Read the problem.
Step 2. Identify what we are looking for. the amount of tip should Dezohn leave
Step 3. Name what we are looking for.
Choose a variable to represent it. Let t = amount of tip.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it.
Translate the sentence into an equation.
Step 5. Solve the equation. Multiply.
Step 6. Check. Does this make sense?
Yes, 20% of $70 is $14.
Step 7. Answer the question with a complete sentence. Dezohn should leave a tip of $12.33.
Notice that we used t to represent the unknown tip.
Cierra and her sister enjoyed a dinner in a restaurant and the bill was $81.50. If she wants to leave 18% of the total bill as her tip, how much should she leave?
Kimngoc had lunch at her favorite restaurant. She wants to leave 15% of the total bill as her tip. If her bill was $14.40, how much will she leave for the tip?
The label on Masao’s breakfast cereal said that one serving of cereal provides 85 milligrams (mg) of potassium, which is 2% of the recommended daily amount. What is the total recommended daily amount
of potassium?
Step 1. Read the problem.
Step 2. Identify what we are looking for. the total amount of potassium that is recommended
Step 3. Name what we are looking for.
Choose a variable to represent it. Let $a=a=$ total amount of potassium.
Step 4. Translate. Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Step 6. Check. Does this make sense?
Yes, 2% is a small percent and 85 is a small part of 4,250.
Step 7. Answer the question with a complete sentence. The amount of potassium that is recommended is 4,250 mg.
One serving of wheat square cereal has seven grams of fiber, which is 28% of the recommended daily amount. What is the total recommended daily amount of fiber?
One serving of rice cereal has 190 mg of sodium, which is 8% of the recommended daily amount. What is the total recommended daily amount of sodium?
Mitzi received some gourmet brownies as a gift. The wrapper said each brownie was 480 calories, and had 240 calories of fat. What percent of the total calories in each brownie comes from fat?
Step 1. Read the problem.
Step 2. Identify what we are looking for. the percent of the total calories from fat
Step 3. Name what we are looking for.
Choose a variable to represent it. Let $p=p=$ percent of fat.
Step 4. Translate. Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Divide by 480.
Put in a percent form.
Step 6. Check. Does this make sense?
Yes, 240 is half of 480, so 50% makes sense.
Step 7. Answer the question with a complete sentence. Of the total calories in each brownie, 50% is fat.
Solve. Round to the nearest whole percent.
Veronica is planning to make muffins from a mix. The package says each muffin will be 230 calories and 60 calories will be from fat. What percent of the total calories is from fat?
Solve. Round to the nearest whole percent.
The mix Ricardo plans to use to make brownies says that each brownie will be 190 calories, and 76 calories are from fat. What percent of the total calories are from fat?
Find Percent Increase and Percent Decrease
People in the media often talk about how much an amount has increased or decreased over a certain period of time. They usually express this increase or decrease as a percent.
To find the percent increase, first we find the amount of increase, the difference of the new amount and the original amount. Then we find what percent the amount of increase is of the original
Find the Percent Increase.
1. Step 1. Find the amount of increase.
$new amount−original amount=increasenew amount−original amount=increase$
2. Step 2. Find the percent increase.
The increase is what percent of the original amount?
In 2011, the California governor proposed raising community college fees from $26 a unit to $36 a unit. Find the percent increase. (Round to the nearest tenth of a percent.)
Step 1. Read the problem.
Step 2. Identify what we are looking for. the percent increase
Step 3. Name what we are looking for.
Choose a variable to represent it. Let $p=p=$ the percent.
Step 4. Translate. Write a sentence that gives the information to find it.
First find the amount of increase. new amount − original amount = increase
Find the percent. Increase is what percent of the original amount?
Translate into an equation.
Step 5. Solve the equation.
Divide by 26.
Change to percent form; round to the nearest tenth.
Step 6. Check. Does this make sense?
Yes, 38.4% is close to $1313$, and 10 is close to $1313$ of 26.
Step 7. Answer the question with a complete sentence. The new fees represent a 38.4% increase over the old fees.
Notice that we rounded the division to the nearest thousandth in order to round the percent to the nearest tenth.
Find the percent increase. (Round to the nearest tenth of a percent.)
In 2011, the IRS increased the deductible mileage cost to 55.5 cents from 51 cents.
Find the percent increase.
In 1995, the standard bus fare in Chicago was $1.50. In 2008, the standard bus fare was $2.25.
Finding the percent decrease is very similar to finding the percent increase, but now the amount of decrease is the difference of the original amount and the new amount. Then we find what percent the
amount of decrease is of the original amount.
Find the Percent Decrease.
1. Step 1. Find the amount of decrease.
$original amount−new amount=decreaseoriginal amount−new amount=decrease$
2. Step 2. Find the percent decrease.
Decrease is what percent of the original amount?
The average price of a gallon of gas in one city in June 2014 was $3.71. The average price in that city in July was $3.64. Find the percent decrease.
Step 1. Read the problem.
Step 2. Identify what we are looking for. the percent decrease
Step 3. Name what we are looking for.
Choose a variable to represent that quantity. Let $p=p=$ the percent decrease.
Step 4. Translate. Write a sentence that gives the information to find it.
First find the amount of decrease. $3.71−3.64=0.073.71−3.64=0.07$
Find the percent. Decrease is what percent of the original amaount?
Translate into an equation.
Step 5. Solve the equation.
Divide by 3.71.
Change to percent form; round to the nearest tenth.
Step 6. Check. Does this make sense?
Yes, if the original price was $4, a 2% decrease would be 8 cents.
Step 7. Answer the question with a complete sentence. The price of gas decreased 1.9%.
Find the percent decrease. (Round to the nearest tenth of a percent.)
The population of North Dakota was about 672,000 in 2010. The population is projected to be about 630,000 in 2020.
Find the percent decrease.
Last year, Sheila’s salary was $42,000. Because of furlough days, this year, her salary was $37,800.
Solve Simple Interest Applications
Do you know that banks pay you to keep your money? The money a customer puts in the bank is called the principal, P, and the money the bank pays the customer is called the interest. The interest is
computed as a certain percent of the principal; called the rate of interest, r. We usually express rate of interest as a percent per year, and we calculate it by using the decimal equivalent of the
percent. The variable t, (for time) represents the number of years the money is in the account.
To find the interest we use the simple interest formula, $I=Prt.I=Prt.$
If an amount of money, P, called the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, I, earned is
Interest earned according to this formula is called simple interest.
Interest may also be calculated another way, called compound interest. This type of interest will be covered in later math classes.
The formula we use to calculate simple interest is $I=Prt.I=Prt.$ To use the formula, we substitute in the values the problem gives us for the variables, and then solve for the unknown variable. It
may be helpful to organize the information in a chart.
Nathaly deposited $12,500 in her bank account where it will earn 4% interest. How much interest will Nathaly earn in 5 years?
Step 1. Read the problem.
Step 2. Identify what we are looking for. the amount of interest earned
Step 3. Name what we are looking for. Let $I=I=$ the amount of interest.
Choose a variable to represent that quantity.
Step 4. Translate into an equation.
Write the formula. $I=Prt I=(12,500)(.04)(5) I=Prt I=(12,500)(.04)(5)$
Substitute in the given information.
Step 5. Solve the equation. $I=2,500I=2,500$
Step 6. Check: Does this make sense?
Is $2,500 a reasonable interest on $12,500? Yes.
Step 7. Answer the question with a complete sentence. The interest is $2,500.
Areli invested a principal of $950 in her bank account with interest rate 3%. How much interest did she earn in 5 years?
Susana invested a principal of $36,000 in her bank account with interest rate 6.5%. How much interest did she earn in 3 years?
There may be times when we know the amount of interest earned on a given principal over a certain length of time, but we don’t know the rate. To find the rate, we use the simple interest formula,
substitute in the given values for the principal and time, and then solve for the rate.
Loren loaned his brother $3,000 to help him buy a car. In 4 years his brother paid him back the $3,000 plus $660 in interest. What was the rate of interest?
Step 1. Read the problem.
Step 2. Identify what we are looking for. the rate of interest
Step 3. Name what we are looking for. Let $r=r=$ the rate of interest.
Choose a variable to represent that quantity.
Step 4. Translate into an equation.
Write the formula. $I=Prt 660=(3,000)r(4) I=Prt 660=(3,000)r(4)$
Substitute in the given information.
Step 5. Solve the equation.
Divide. $660=(12,000)r 0.055=r 5.5%=r 660=(12,000)r 0.055=r 5.5%=r$
Change to percent form.
Step 6. Check: Does this make sense?
$I=Prt 660=?(3,000)(0.055)(4) 660=660✓ I=Prt 660=?(3,000)(0.055)(4) 660=660✓$
Step 7. Answer the question with a complete sentence. The rate of interest was 5.5%.
Notice that in this example, Loren’s brother paid Loren interest, just like a bank would have paid interest if Loren invested his money there.
Jim loaned his sister $5,000 to help her buy a house. In 3 years, she paid him the $5,000, plus $900 interest. What was the rate of interest?
Hang borrowed $7,500 from her parents to pay her tuition. In 5 years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of interest?
Eduardo noticed that his new car loan papers stated that with a 7.5% interest rate, he would pay $6,596.25 in interest over 5 years. How much did he borrow to pay for his car?
Step 1. Read the problem.
Step 2. Identify what we are looking for. the amount borrowed (the principal)
Step 3. Name what we are looking for. Let $P=P=$ principal borrowed.
Choose a variable to represent that quantity.
Step 4. Translate into an equation.
Write the formula. $I=Prt6,596.25=P(0.075)(5)I=Prt6,596.25=P(0.075)(5)$
Substitute in the given information.
Step 5. Solve the equation. $6,596.25=0.375P17,590=P6,596.25=0.375P17,590=P$
Step 6. Check: Does this make sense?
Step 7. Answer the question with a complete sentence. The principal was $17,590.
Sean’s new car loan statement said he would pay $4,866.25 in interest from an interest rate of 8.5% over 5 years. How much did he borrow to buy his new car?
In 5 years, Gloria’s bank account earned $2,400 interest at 5%. How much had she deposited in the account?
Solve Applications with Discount or Mark-up
Applications of discount are very common in retail settings. When you buy an item on sale, the original price has been discounted by some dollar amount. The discount rate, usually given as a percent,
is used to determine the amount of the discount. To determine the amount of discount, we multiply the discount rate by the original price.
We summarize the discount model in the box below.
$amount of discount=discount rate×original pricesale price=original price−amount of discountamount of discount=discount rate×original pricesale price=original price−amount of discount$
Keep in mind that the sale price should always be less than the original price.
Elise bought a dress that was discounted 35% off of the original price of $140. What was ⓐ the amount of discount and ⓑ the sale price of the dress?
$Original price=140Discount rate=35%Discount=? Original price=140Discount rate=35%Discount=?$
Step 1. Read the problem.
Step 2. Identify what we are looking for. the amount of discount
Step 3. Name what we are looking for. Let $d=d=$ the amount of discount.
Choose a variable to represent that quantity.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it. The discount is 35% of $140.
Translate into an equation. $d=0.35(140)d=0.35(140)$
Step 5. Solve the equation. $d=49d=49$
Step 6. Check: Does this make sense?
Is a $49 discount reasonable for a $140 dress? Yes.
Step 7. Write a complete sentence to answer the question. The amount of discount was $49.
Read the problem again.
Step 1. Identify what we are looking for. the sale price of the dress
Step 2. Name what we are looking for.
Choose a variable to represent that quantity. Let $s=s=$ the sale price.
Step 3. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 4. Solve the equation.
Step 5. Check. Does this make sense?
Is the sale price less than the original price?
Yes, $91 is less than $140.
Step 6. Answer the question with a complete sentence. The sale price of the dress was $91.
Find ⓐ the amount of discount and ⓑ the sale price:
Sergio bought a belt that was discounted 40% from an original price of $29.
Find ⓐ the amount of discount and ⓑ the sale price:
Oscar bought a barbecue that was discounted 65% from an original price of $395.
There may be times when we know the original price and the sale price, and we want to know the discount rate. To find the discount rate, first we will find the amount of discount and then use it to
compute the rate as a percent of the original price. Example 3.24 will show this case.
Jeannette bought a swimsuit at a sale price of $13.95. The original price of the swimsuit was $31. Find the ⓐ amount of discount and ⓑ discount rate.
$Original price=31Discount=?Sale Price=13.95Original price=31Discount=?Sale Price=13.95$
Step 1. Read the problem.
Step 2. Identify what we are looking for. the amount of discount
Step 3. Name what we are looking for. Let $d=d=$ the amount of discount.
Choose a variable to represent that quantity.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it. The discount is the difference between the original price and the sale price.
Translate into an equation. $d=31-13.95d=31-13.95$
Step 5. Solve the equation. $d=17.05d=17.05$
Step 6. Check: Does this make sense?
Is 17.05 less than 31? Yes.
Step 7. Answer the question with a complete sentence. The amount of discount was $17.05.
Read the problem again.
Step 1. Identify what we are looking for. the discount rate
Step 2. Name what we are looking for.
Choose a variable to represent it. Let $r=r=$ the discount rate.
Step 3. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 4. Solve the equation.
Divide both sides by 31.
Change to percent form.
Step 5. Check. Does this make sense?
Is $17.05 equal to 55% of $31?
Step 6. Answer the question with a complete sentence. The rate of discount was 55%.
Find ⓐ the amount of discount and ⓑ the discount rate.
Lena bought a kitchen table at the sale price of $375.20. The original price of the table was $560.
Find ⓐ the amount of discount and ⓑ the discount rate.
Nick bought a multi-room air conditioner at a sale price of $340. The original price of the air conditioner was $400.
Applications of mark-up are very common in retail settings. The price a retailer pays for an item is called the original cost. The retailer then adds a mark-up to the original cost to get the list
price, the price he sells the item for. The mark-up is usually calculated as a percent of the original cost. To determine the amount of mark-up, multiply the mark-up rate by the original cost.
We summarize the mark-up model in the box below.
$amount of mark-up=mark-up rate×original costlist price=original cost+amount of mark upamount of mark-up=mark-up rate×original costlist price=original cost+amount of mark up$
Keep in mind that the list price should always be more than the original cost.
Adam’s art gallery bought a photograph at original cost $250. Adam marked the price up 40%. Find the ⓐ amount of mark-up and ⓑ the list price of the photograph.
Step 1. Read the problem.
Step 2. Identify what we are looking for. the amount of mark-up
Step 3. Name what we are looking for.
Choose a variable to represent it. Let $m=m=$ the amount of markup.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Step 6. Check. Does this make sense?
Yes, 40% is less than one-half and 100 is less than half of 250.
Step 7. Answer the question with a complete sentence. The mark-up on the phtograph was $100.
Step 1. Read the problem again.
Step 2. Identify what we are looking for. the list price
Step 3. Name what we are looking for.
Choose a variable to represent it. Let $p=p=$ the list price.
Step 4. Translate into an equation.
Write a sentence that gives the information to find it.
Translate into an equation.
Step 5. Solve the equation.
Step 6. Check. Does this make sense?
Is the list price more than the net price?
Is $350 more than $250? Yes.
Step 7. Answer the question with a complete sentence. The list price of the photograph was $350.
Find ⓐ the amount of mark-up and ⓑ the list price.
Jim’s music store bought a guitar at original cost $1,200. Jim marked the price up 50%.
Find ⓐ the amount of mark-up and ⓑ the list price.
The Auto Resale Store bought Pablo’s Toyota for $8,500. They marked the price up 35%.
Section 3.2 Exercises
Practice Makes Perfect
Translate and Solve Basic Percent Equations
In the following exercises, translate and solve.
What number is 45% of 120?
What number is 65% of 100?
What number is 24% of 112?
What number is 36% of 124?
250% of 65 is what number?
150% of 90 is what number?
800% of 2250 is what number?
600% of 1740 is what number?
28 is 25% of what number?
36 is 25% of what number?
81 is 75% of what number?
93 is 75% of what number?
8.2% of what number is $2.87?
6.4% of what number is $2.88?
11.5% of what number is $108.10?
12.3% of what number is $92.25?
What percent of 260 is 78?
What percent of 215 is 86?
What percent of 1500 is 540?
What percent of 1800 is 846?
30 is what percent of 20?
50 is what percent of 40?
840 is what percent of 480?
790 is what percent of 395?
Solve Percent Applications
In the following exercises, solve.
Geneva treated her parents to dinner at their favorite restaurant. The bill was $74.25. Geneva wants to leave 16% of the total bill as a tip. How much should the tip be?
When Hiro and his co-workers had lunch at a restaurant near their work, the bill was $90.50. They want to leave 18% of the total bill as a tip. How much should the tip be?
Trong has 12% of each paycheck automatically deposited to his savings account. His last paycheck was $2165. How much money was deposited to Trong’s savings account?
Cherise deposits 8% of each paycheck into her retirement account. Her last paycheck was $1,485. How much did Cherise deposit into her retirement account?
One serving of oatmeal has eight grams of fiber, which is 33% of the recommended daily amount. What is the total recommended daily amount of fiber?
One serving of trail mix has 67 grams of carbohydrates, which is 22% of the recommended daily amount. What is the total recommended daily amount of carbohydrates?
A bacon cheeseburger at a popular fast food restaurant contains 2070 milligrams (mg) of sodium, which is 86% of the recommended daily amount. What is the total recommended daily amount of sodium?
A grilled chicken salad at a popular fast food restaurant contains 650 milligrams (mg) of sodium, which is 27% of the recommended daily amount. What is the total recommended daily amount of sodium?
After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa’s original weight?
Tricia got a 6% raise on her weekly salary. The raise was $30 per week. What was her original salary?
Yuki bought a dress on sale for $72. The sale price was 60% of the original price. What was the original price of the dress?
Kim bought a pair of shoes on sale for $40.50. The sale price was 45% of the original price. What was the original price of the shoes?
Tim left a $9 tip for a $50 restaurant bill. What percent tip did he leave?
Rashid left a $15 tip for a $75 restaurant bill. What percent tip did he leave?
The nutrition fact sheet at a fast food restaurant says the fish sandwich has 380 calories, and 171 calories are from fat. What percent of the total calories is from fat?
The nutrition fact sheet at a fast food restaurant says a small portion of chicken nuggets has 190 calories, and 114 calories are from fat. What percent of the total calories is from fat?
Emma gets paid $3,000 per month. She pays $750 a month for rent. What percent of her monthly pay goes to rent?
Dimple gets paid $3,200 per month. She pays $960 a month for rent. What percent of her monthly pay goes to rent?
Find Percent Increase and Percent Decrease
In the following exercises, solve.
Tamanika got a raise in her hourly pay, from $15.50 to $17.36. Find the percent increase.
Ayodele got a raise in her hourly pay, from $24.50 to $25.48. Find the percent increase.
Annual student fees at the University of California rose from about $4,000 in 2000 to about $12,000 in 2010. Find the percent increase.
The price of a share of one stock rose from $12.50 to $50. Find the percent increase.
According to Time magazine annual global seafood consumption rose from 22 pounds per person in the 1960s to 38 pounds per person in 2011. Find the percent increase. (Round to the nearest tenth of a
In one month, the median home price in the Northeast rose from $225,400 to $241,500. Find the percent increase. (Round to the nearest tenth of a percent.)
A grocery store reduced the price of a loaf of bread from $2.80 to $2.73. Find the percent decrease.
The price of a share of one stock fell from $8.75 to $8.54. Find the percent decrease.
Hernando’s salary was $49,500 last year. This year his salary was cut to $44,055. Find the percent decrease.
In 10 years, the population of Detroit fell from 950,000 to about 712,500. Find the percent decrease.
In 1 month, the median home price in the West fell from $203,400 to $192,300. Find the percent decrease. (Round to the nearest tenth of a percent.)
Sales of video games and consoles fell from $1,150 million to $1,030 million in 1 year. Find the percent decrease. (Round to the nearest tenth of a percent.)
Solve Simple Interest Applications
In the following exercises, solve.
Casey deposited $1,450 in a bank account with interest rate 4%. How much interest was earned in two years?
Terrence deposited $5,720 in a bank account with interest rate 6%. How much interest was earned in 4 years?
Robin deposited $31,000 in a bank account with interest rate 5.2%. How much interest was earned in 3 years?
Carleen deposited $16,400 in a bank account with interest rate 3.9%. How much interest was earned in 8 years?
Hilaria borrowed $8,000 from her grandfather to pay for college. Five years later, she paid him back the $8,000, plus $1,200 interest. What was the rate of interest?
Kenneth loaned his niece $1,200 to buy a computer. Two years later, she paid him back the $1,200, plus $96 interest. What was the rate of interest?
Lebron loaned his daughter $20,000 to help her buy a condominium. When she sold the condominium four years later, she paid him the $20,000, plus $3,000 interest. What was the rate of interest?
Pablo borrowed $50,000 to start a business. Three years later, he repaid the $50,000, plus $9,375 interest. What was the rate of interest?
In 10 years, a bank account that paid 5.25% earned $18,375 interest. What was the principal of the account?
In 25 years, a bond that paid 4.75% earned $2,375 interest. What was the principal of the bond?
Joshua’s computer loan statement said he would pay $1,244.34 in interest for a 3-year loan at 12.4%. How much did Joshua borrow to buy the computer?
Margaret’s car loan statement said she would pay $7,683.20 in interest for a 5-year loan at 9.8%. How much did Margaret borrow to buy the car?
Solve Applications with Discount or Mark-up
In the following exercises, find the sale price.
Perla bought a cell phone that was on sale for $50 off. The original price of the cell phone was $189.
Sophie saw a dress she liked on sale for $15 off. The original price of the dress was $96.
Rick wants to buy a tool set with original price $165. Next week the tool set will be on sale for $40 off.
Angelo’s store is having a sale on televisions. One television, with original price $859, is selling for $125 off.
In the following exercises, find ⓐ the amount of discount and ⓑ the sale price.
Janelle bought a beach chair on sale at 60% off. The original price was $44.95.
Errol bought a skateboard helmet on sale at 40% off. The original price was $49.95.
Kathy wants to buy a camera that lists for $389. The camera is on sale with a 33% discount.
Colleen bought a suit that was discounted 25% from an original price of $245.
Erys bought a treadmill on sale at 35% off. The original price was $949.95 (round to the nearest cent.)
Jay bought a guitar on sale at 45% off. The original price was $514.75 (round to the nearest cent.)
In the following exercises, find ⓐ the amount of discount and ⓑ the discount rate. (Round to the nearest tenth of a percent if needed.)
Larry and Donna bought a sofa at the sale price of $1,344. The original price of the sofa was $1,920.
Hiroshi bought a lawnmower at the sale price of $240. The original price of the lawnmower is $300.
Patty bought a baby stroller on sale for $301.75. The original price of the stroller was $355.
Bill found a book he wanted on sale for $20.80. The original price of the book was $32.
Nikki bought a patio set on sale for $480. The original price was $850. To the nearest tenth of a percent, what was the rate of discount?
Stella bought a dinette set on sale for $725. The original price was $1,299. To the nearest tenth of a percent, what was the rate of discount?
In the following exercises, find ⓐ the amount of the mark-up and ⓑ the list price.
Daria bought a bracelet at original cost $16 to sell in her handicraft store. She marked the price up 45%.
Regina bought a handmade quilt at original cost $120 to sell in her quilt store. She marked the price up 55%.
Tom paid $0.60 a pound for tomatoes to sell at his produce store. He added a 33% mark-up.
Flora paid her supplier $0.74 a stem for roses to sell at her flower shop. She added an 85% mark-up.
Alan bought a used bicycle for $115. After re-conditioning it, he added 225% mark-up and then advertised it for sale.
Michael bought a classic car for $8,500. He restored it, then added 150% mark-up before advertising it for sale.
Everyday Math
Leaving a Tip At the campus coffee cart, a medium coffee costs $1.65. MaryAnne brings $2.00 with her when she buys a cup of coffee and leaves the change as a tip. What percent tip does she leave?
Splitting a Bill Four friends went out to lunch and the bill came to $53.75. They decided to add enough tip to make a total of $64, so that they could easily split the bill evenly among themselves.
What percent tip did they leave?
Writing Exercises
Without solving the problem “44 is 80% of what number” think about what the solution might be. Should it be a number that is greater than 44 or less than 44? Explain your reasoning.
Without solving the problem “What is 20% of 300?” think about what the solution might be. Should it be a number that is greater than 300 or less than 300? Explain your reasoning.
After returning from vacation, Alex said he should have packed 50% fewer shorts and 200% more shirts. Explain what Alex meant.
Because of road construction in one city, commuters were advised to plan that their Monday morning commute would take 150% of their usual commuting time. Explain what this means.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After reviewing this checklist, what will you do to become confident for all goals? | {"url":"https://openstax.org/books/elementary-algebra/pages/3-2-solve-percent-applications","timestamp":"2024-11-02T21:03:03Z","content_type":"text/html","content_length":"515586","record_id":"<urn:uuid:b4f71af1-5f26-4d83-8e03-af8ac905cb62>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00789.warc.gz"} |
Hypothesis Test in Excel To Find Out If Your Advertising Worked
Using the Hypothesis
Test in Excel To Find
Out If Your Advertising
Really Worked
This article will explain how to use Excel to perform a Hypothesis Test to analyze an advertising campaign. Specifically, we will show how to use Excel to perform One-Tailed, Two-Sample, Paired
Hypothesis Test of Mean to determine whether an advertising campaign improved really sales.
This Hypothesis Test tests the Null Hypothesis that the advertising campaign did not improve sales.
The advantages of statistical analysis in Excel to solve business statistics problems is that most problems can be solved in just one or two steps and there is no more need to look anything up on
Normal Distribution tables.
Here is the problem:
Problem: Determine with 95% certainty whether an advertising campaign increased average daily sales to our large dealer network. Before and After samples of average daily sales were taken with at
least 30 dealers.
Here is the Before and After data for all of the dealers: Before we begin solving this problem, we need to know whether we are dealing with normally distributed data. If the data is not normally
distributed, we have to use nonparametric statistical tests to solve this problem. Always Test for Normality First Normality tests should be performed on the before and after sales data. Both data
sets must be normally distributed to perform the well-known hypothesis test that is based upon the underlying data being normally distributed. This blog has numerous articles about how to perform
normality testing and nonparametric testing if the data is not normally distributed.
The MOST Important Step
Determine What Type of Hypothesis
Test You Will Perform
1) Hypothesis Test of Mean or Proportion? We know that this is a test of mean and not proportion because each individual sample taken can have a wide range of values: Any sales sample measurement
from 90 to 250 is probably reasonable. 2) One or Two-Tailed Hypothesis Test? We know that this is a one-tailed test because we are trying to determine if the "After Data" mean sales is larger than
the "Before Data" mean sales, not whether the mean sales are merely different. 3) One or Two-Sample Hypothesis Test? We know that two samples need to be taken because no data is initially available.
4) Paired or Unpaired Hypothesis Test?
This is paired data because each set of "Before" and "After" data came from the same object.
In this case, we are performing a One-Tailed, Two-Sample, Paired Hypothesis Test of Mean to determine whether an advertising campaign improved really sales. We will do this test in Excel. It is
extremely important to establish the type of Hypothesis test. Each type of Hypothesis test uses a slightly (or very) different methodology and set of formulas.
In this case, the yellow-highlighted column represents the difference between the Before and After sample of each data pair. The Hypothesis test will be performed on that column of data.
Above is the data sample:
Paired Hypothesis Tests involve taking Before and After samples from the same large number (n is greater than 30) of objects and performing a Hypothesis test on the differences between the Before and
After samples.
Initial Parameters Needed Before we can begin the Hypothesis test, we need to calculate the following parameters of variable x:
Sample size - Use Excel function COUNT
Sample mean - Use Excel function AVERAGE
Sample standard deviation - Use Excel function STDEV
Sample standard error = (Sample standard deviation) / SQRT (Sample size)
Difference Data We need to make the following calculations on the data that represents the difference between the Before and After data:
xavg = "Difference Data" sample average = 9.60 Sxavg = Sample standard error = 1.11
n = "Difference Data" Sample size = 30
α = Level of Significance = 0.05 because there is a 5% max chance of error allowed.
Therefore a 95% Level of Certainty Required
The Four-Step Method That Solves ALL Hypothesis Tests
This problem can be solved using the standard four-step method for Hypothesis testing.
Step 1 - Create the Null and Alternate Hypotheses
The Null Hypothesis normally states that both populations sampled are the same. If the mean sales from both the Before and After Data are the same, then their average difference = 0
The Null Hypothesis states that both Before and After mean sales are the same, which is equivalent to:
Null Hypothesis, H0, is: xavg = 0
The Alternate Hypothesis states that the After Data mean sales is larger, which is equivalent to:
The Alternate Hypothesis, which states that their average difference, xavg, is larger than 0, is as follows:
Alternate Hypothesis, H1 is: xavg is greater than 0
For this one-tailed test, the Alternative Hypothesis states that the value of the distributed variable xavg is larger than the value of 0 stated in the Null Hypothesis,
The Region of Uncertainty will be entirely in the right outer tail.
Note - the Alternative Hypothesis determines whether the Hypothesis test is a one-tailed test or a two-tailed test as follows:
One-tailed test - (Value of variable) is greater than OR is less than (Constant)
Two-tailed test - (Value of variable) does not equal (Constant)
Step 2 - Map the Normal Curve
We now create a Normal curve showing a distribution of the same variable that is used by the Null Hypothesis, which is xavg.
The mean of this Normal curve will occur at the same value of the distributed variable as stated in the Null Hypothesis.
Since the Null Hypothesis states that xavg = 0, the Normal curve will map the distribution of the variable xavg with a mean of xavg = 0.
This Normal curve will have a standard error as just calculated as follows:
Step 3 - Map the Region of Certainty
The problem requires a 95% Level of Certainty so the Region of Certainty will contain 95% of the area under the Normal curve.
We know that this problem uses a one-tailed test with the Region of Uncertainty entirely contained in the outer right tail.
The Region of Uncertainty contains 5% of the total area under the Normal curve. The entire 95% Region of Certainty lies to the left of the 5% Region of Uncertainty, which is entirely contained in the
outer right tail.
We need to find out how far the boundary of the Region of Certainty is from the Normal curve mean. Calculating the number of standard errors from the Normal curve mean to the outer boundary of the
Region of Certainty in the right tail for a one-tailed test is done as follows:
z95%,1-tailed = NORMSINV(1 - α) = NORMSINV(0.95) = 1.65
Excel Note - NORMSINV(x) = The number of standard errors from the Normal curve mean to a point right of the Normal curve mean at which x percent of the area under the Normal curve will be to the left
of that point. Additional note - For a one-tailed test, NORMSINV(x) can be used to calculate the number of standard errors from the Normal curve mean to the boundary of the Region of Certainty
whether it is in the left or the right tail.
The Region of Certainty extends to the right of the Normal curve mean of xavg = 0 by 1.65 standard errors.
One standard error = sxavg = 1.11, so:
1.65 standard errors = (1.65) * (1.11) = 1.83
The outer boundary of the Region of Certainty has the value = µ + z95%,one-tailed * sxavg
which equals 0 + (1.65) * (1.11) = 0 + 1.83 = 1.83
The point, 1.83, is 1.65 standard errors from the Normal curve mean of xavg = 0
This point, 1.83, is the right boundary of the 95% Region of Certainty on the Normal curve.
a) Critical Value Test
The Critical Value Test is the final test to determine whether to reject or not reject the Null Hypothesis. The p Value Test, described next, is an equivalent alternative to the Critical Value Test.
The Critical Value test tells whether the value of the actual variable, xavg, falls inside or outside of the Critical Value, which is the boundary between the Region of Certainty and the Region of
If the actual value of the distributed variable, xavg, falls within the Region of Certainty, the Null Hypothesis is not rejected.
If the actual value of the distributed variable, xavg, falls outside of the Region of Certainty and, therefore, into the Region of Uncertainty, the Null Hypothesis is rejected and the Alternate
Hypothesis is accepted.
The actual value of the variable xavg = 9.60 and is therefore to the right of (outside of) the outer right Critical Value (1.83), which is the boundary between the Regions of Certainty and
Uncertainty in the right tail.
The actual value of the variable xavg is outside the Region of Certainty and therefore outside the Critical Value.
We therefore reject the Null Hypothesis and accept the Alternate Hypothesis which states that average dealer sales have increased after the ad campaign, with a maximum possible error of 5%.
Click On Image To See Larger Version
b) p Value Test
The p Value Test is an equivalent alternative to the Critical Value Test and also tells whether to reject or not reject the Null Hypothesis.
The p Value equals the percentage of area under the Normal curve that is in the tail outside of the actual value of the variable xavg.
For a one-tailed test, if the p Value is larger than α, the Null Hypothesis is not rejected.
For a two-tailed test, if the p Value is larger than α/2, the Null Hypothesis is not rejected.
For a one-tailed test, the Region of Uncertainty is contained entirely in one tail. Therefore the curve area contained by the Region of Uncertainty in that tail equals α.
For a two-tailed test, the Region of Uncertainty is split between both tails. Therefore the curve area contained by the Region of Uncertainty in that tail equals α/2.
The p Value for the actual value of the distributed variable, which in this case is greater than the mean (falls to the right of the mean in the right tail), is:
p Valuexavg = 1 - NORMSDIST( [ xavg - µ ] / sxavg )
Excel note - NORMSDIST(x) calculates the total area under the Normal curve to the LEFT of the point that is x standard errors to the right of the Normal curve mean. Since we are calculating the area
to the RIGHT of that point, we use 1 - NORMSDIST.
p Valuexavg = 1 - NORMSDIST((9.60 - 0 ) / 1.11) = 1 - NORMSDIST(9.60/1.11) ≈ 0
The p Value (0) is less than α (0.05), so the Null Hypothesis is rejected and the Alternate Hypothesis is accepted..
For a one-tailed test - When the p Value is less than α, the actual value of the distributed variable falls outside the Region of Certainty and the Null Hypothesis is rejected.
This is the case here.
Click Image To See Larger Version
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How to Calculate Standard Deviation in Excel Step-by-Step (2024)
Talking about statistics, the concept of standard deviation and mean go hand in hand 🤝
Mean only gives you the average figure for a dataset. But how truly does that figure represent the individual numbers of that dataset? Standard deviation will tell you that.
Standard deviation tells more about the accuracy of a mean. And interestingly, Excel offers inbuilt functions to calculate mean and standard deviation, both.
Let’s see how these Excel functions work in the guide below. Stay tuned and download our sample workbook for this guide here to practice side by side 📩
What is the standard deviation?
Standard deviation is a measure of how much the values of the dataset deviate from the mean of that dataset 🎲
This way you know if your mean is a fair representation of the underlying dataset.
Let’s quickly see this through an example. Below are the marks of a student in different subjects 🎓
We also have the mean (average marks) and standard deviation calculated for the same.
The mean for all the marks is 52. Which tells the average marks scored in all the subjects are 52 👍
And the standard deviation of 3.41 tells that the actual marks scored by the student in most of the subjects will be 3.41 marks more or less than the mean of 52.
Pro Tip!
Standard deviation can be both, positive and negative.
The closer the standard deviation is to 0, the lesser the variability in your data 💪
For example, if the mean of a data set is 70 and the standard deviation is 5. That’s fine. It means that most of the values in the dataset are around 5 points less or more than the mean of 70.
But if the standard deviation of the same mean is 20 – this means there is too much variation in the dataset. And the mean might not be a very reliable representative of this dataset as each value of
this dataset might be up to 20 points more or less than 50 🧐
Calculate standard deviation using STDEV
We know how impatient you’re getting to see the Excel standard deviation function in action. So here we go with the oldest function for calculating standard deviation in Excel.
Below is the data for some people from a small town along with their ages 🔞
Note that this is a sample drawn from the entire population of that town.
We already have the mean for this data calculated (just the simple AVERAGE function running in the background) 🗯
Using these ages, let’s now find the standard deviation for this dataset through the STDEV function.
1. Write the STDEV function as follows:
= STDEV (
2. Create a reference to the cells containing the numbers whose standard deviation is sought.
In our example, cell range B2:B7 contains the ages so, we are referring to the same.
= STDEV (B2:B7)
Pro Tip!
Note that there is an empty cell in the list. And the STDEV function ignores blank cells in calculating the standard deviation.
Similarly, some other things you must know about the STDEV function are 💡
• It ignores any Boolean values (TRUE or FALSE), text values, and error values in the referred cell range or array.
• However, Boolean values or text values directly written as the arguments of the function are taken into account.
• The STDEV function of Excel 2003 can process up to 30 arguments only. However, Excel 2007 and later versions can process up to 255 arguments.
3. Yes, that’s it, hit ‘Enter’ now.
There you go! The standard deviation for our dataset turns out as 24.3. That’s a big number considering the mean of 48.8 😮
Why is that btw? The dataset speaks out the reason very clearly and loudly.
Our sample has people from very variable age groups.
Starting from the age of 20 to the age of 80 – the variability in the numbers is huge. The standard deviation of 24.35 tells the same that the mean of 48.8 is not a very accurate representation of
the dataset.
The STDEV function is now obsolete. It is primarily available in Excel 2003 up to Excel 2019. In later versions, it is only available for compatibility purposes 👵
But don’t worry – Excel has replaced the STDEV function with two more advanced functions. The STDEV.S and the STDEV.P functions. We are going to see them both now.
Calculate standard deviation using STDEV.P
Until the above example, we were dealing with a sample from a population. But if you are dealing with the entire population, you must use the STDEV.P function.
P suffixing this function represents ‘population’ 🎪
The STDEV.P function is the successor of the STDEVP function. It offers more accuracy and is available in all Excel versions starting from Excel 2010.
We will not change the dataset for now. However, this time we will use the STDEV.P function to find the standard deviation for it.
1. Write the STDEV.P function as follows:
= STDEV.P (
2. Again create a reference to cells B2:B7 (that contain the ages for which the standard deviation is sought).
= STDEV.P (B2:B7)
Pro Tip!
The STDEV.P function will ignore any empty cells, logical values, or text values in the referred cell range 🔪
This time the standard deviation is slightly different from the one we calculated above. It is 21.78.
That’s because the STDEV.P function runs the formula for population standard deviation 🧠
Yes, the standard deviation formula for sample standard deviation and population standard deviation is different. We will see them both shortly.
Calculate standard deviation using STDEV.S
And now it’s time we see the STDEV.S function. This function targets the calculation of the standard deviation for a sample.
S suffixing this function represents ‘sample’.
Let’s calculate the standard deviation of the same dataset, as above using the STDEV.S function 👇
1. Write the STDEV.S function as follows:
= STDEV.S (
2. Again create a reference to cells B2:B7 (that contain the ages for which the standard deviation is sought).
= STDEV.S (B2:B7)
Hope you noticed that! The standard deviation calculated by the STDEV.S function is the same as the STDEV function – 24.35 😲
This means that both the STDEV.S and STDEV functions run on the same classic sample standard deviation formula 📝
STDEV.P vs. STDEV.S
The main difference between the STDEV.P and the STDEV.S function is the formula that runs behind them. And this is why Excel replaced the STDEV function with two different functions 😎
The STDEV.S function uses the classic sample standard deviation formula as below:
Pro Tip!
For this formula:
• x represents each value of the sample
• x̅ is the mean of all the values in the sample
• ∑ is the sum of the expression
• n is the total number of values or data p
Whereas the function STDEV.P uses the following formula:
Pro Tip!
For this formula:
• x represents each value of the population
• u is the mean of all the values in the population
• ∑ is the sum of the expression
• N is the total number of values in the sample
Did you note what’s different between both the formulas? Only the denominator 🔎
In the population standard deviation formula, the whole expression is divided by N (the number of values in the entire population).
Whereas, in the sample deviation formula, the whole expression is divided by N-1 (the number of values in the sample less 1) 👀
This is called Bessel’s correction phenomenon 🏆
For sample-based formulas, we deduct 1 from the number of values (n) to adjust the mistakes resulting from assessing a sample, and not the whole population.
That’s it – Now what?
That’s all about the statistical functions of Excel dedicated to calculating standard deviation.
We have discussed all these functions meant for calculating the standard deviation of a sample and a population. Feeling pro at calculating standard deviations?
Believe me, all these functions don’t even make a speck in Excel’s functions library. Yes, it’s that huge and full of amazing functions ⚖
If you want to master the functions of Excel, better begin with the key Excel functions.
Like the VLOOKUP, SUMIF, and IF functions (some of my top favorite Excel functions).
My 30-minute free email course will take you through them (and much more) in the shortest possible time. Click here to enroll now.
Other resources
Standard deviation alone might not offer sufficient information about your dataset. To take it a step ahead, learn how to turn standard deviation into confidence interval.
What is a confidence interval? Hop on here to read our blog on it.
Frequently asked questions | {"url":"https://spreadsheeto.com/standard-deviation-excel/","timestamp":"2024-11-14T08:20:22Z","content_type":"text/html","content_length":"317224","record_id":"<urn:uuid:ce84e35b-8255-4821-9454-6d723283c322>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00475.warc.gz"} |
Question #ad131 | Socratic
Question #ad131
2 Answers
Operation of specific gravity meter is based on Archimedes' principle.
It states that when a solid body is suspended in a fluid it experiences an upward force which is equal to the weight of the fluid displaced by the submerged part of the body.
The depth is read off the scale, as shown in the figure below:
When submerged to the marking $1.0$, it corresponds to fluid being water at ${4}^{\circ} \text{C}$ and $1$ atm of pressure.
The position of the marking of $1.0$ is not clear from the statement of the problem. As such if the tube sinks in the sample to a point above the mark, the specific gravity of the sample is lower
than the that of water and vice-versa.
Assuming that the $1.0$ mark is placed at the $10 c m$ length of tube.
Let $A$ be the area of cross-section of the tube.
Its volume which is also Volume of water displaced $V = 0.1 A$
As the tube is in equilibrium (upwards thrust$=$downwards weight)
Weight of tube $=$Weight of the water displaced
Weight of tube $= 0.1 \times A \times 1000 g$
where $g$ is acceleration due to gravity.
Now the meter floats with $6 c m$ of its length submerged in the liquid whose density is $d$ and is in equilibrium,
Weight of the liquid displaced $=$ Weight of tube
$0.06 \times A \times \mathrm{dg} = 0.1 \times A \times 1000 g$
$\implies 0.06 \times \frac{d}{1000} = 0.1$
We know that SG$= \frac{d}{d} _ \left(w a t e r\right)$, therefore we get
${\text{SG}}_{l i q u i d} = \frac{0.1}{0.06} = 1. \dot{6}$
We know that the density of water at ${4}^{\circ} C$ is $1000 \frac{k g}{m} ^ 3 \mathmr{and} 1 g c {m}^{-} 3$.So specific gravity of water will be 1.0.
The given uniform tube of length 10cm has also sp.gravity 1.0.
If the area of cross section of the tube be $A c {m}^{2}$ then its volume will be $10 A c {m}^{3}$
And its weight becomes $= 10 A \times 1 \times g$ dyne.
$\text{where " g->"Acceleration due to gravity}$
If it floats in water immersing x cm length in water,then weight of water displaced by it will be $= x \cdot A \cdot 1 \cdot g \text{ }$dyne
By Archimedes principle
$x \cdot A \cdot g = 10 \cdot A \cdot g$
So $x = 10 c m$ Hence it floats in water submerged.
Again it floats in the given liquid keeping 6cm immersed.
So the weight of displaced liquid will be$= 6 \cdot A \cdot \rho \cdot g$ dyne
,where $\rho$ is the sp.gravity of the liquid.
Now applying Archimedes principle we have
$\to 6 \cdot A \cdot \rho \cdot g = 10 A \cdot 1 \cdot g$
$\implies \rho = \frac{10}{6} \approx 1.67$
Impact of this question
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Quantum Computers Are Like Kaleidoscopes, Helping Illustrate Science and Technology
Quantum computing is like Forrest Gump’s box of chocolates: You never know what you’re gonna get. Quantum phenomena – the behavior of matter and energy at the atomic and subatomic levels – are not
definite, one thing or another. They are opaque clouds of possibility or, more precisely, probabilities. When someone observes a quantum system, it loses its quantum-ness and “collapses” into a
definite state.
Quantum phenomena are mysterious and often counterintuitive. This makes quantum computing difficult to understand. People naturally reach for the familiar to attempt to explain the unfamiliar, and
for quantum computing this usually means using traditional binary computing as a metaphor. But explaining quantum computing this way leads to major conceptual confusion, because at a base level the
two are entirely different animals.
This problem highlights the often mistaken belief that common metaphors are more useful than exotic ones when explaining new technologies. Sometimes the opposite approach is more useful. The
freshness of the metaphor should match the novelty of the discovery.
The uniqueness of quantum computers calls for an unusual metaphor. As a communications researcher who studies technology, I believe that quantum computers can be better understood as kaleidoscopes.
Digital Certainty vs. Quantum Probabilities
The gap between understanding classical and quantum computers is a wide chasm. Classical computers store and process information via transistors, which are electronic devices that take binary,
deterministic states: one or zero, yes or no. Quantum computers, in contrast, handle information probabilistically at the atomic and subatomic levels.
Classical computers use the flow of electricity to sequentially open and close gates to record or manipulate information. Information flows through circuits, triggering actions through a series of
switches that record information as ones and zeros. Using binary math, bits are the foundation of all things digital, from the apps on your phone to the account records at your bank and the Wi-Fi
signals bouncing around your home.
In contrast, quantum computers use changes in the quantum states of atoms, ions, electrons or photons. Quantum computers link, or entangle, multiple quantum particles so that changes to one affect
all the others. They then introduce interference patterns, like multiple stones tossed into a pond at the same time. Some waves combine to create higher peaks, while some waves and troughs combine to
cancel each other out. Carefully calibrated interference patterns guide the quantum computer toward the solution of a problem.
Physicist Katie Mack explains quantum probability.
Achieving A Quantum Leap, Conceptually
The term “bit” is a metaphor. The word suggests that during calculations, a computer can break up large values into tiny ones – bits of information – which electronic devices such as transistors can
more easily process.
Using metaphors like this has a cost, though. They are not perfect. Metaphors are incomplete comparisons that transfer knowledge from something people know well to something they are working to
understand. The bit metaphor ignores that the binary method does not deal with many types of different bits at once, as common sense might suggest. Instead, all bits are the same.
The smallest unit of a quantum computer is called the quantum bit, or qubit. But transferring the bit metaphor to quantum computing is even less adequate than using it for classical computing.
Transferring a metaphor from one use to another blunts its effect.
The prevalent explanation of quantum computing is that while classical computers can store or process only a zero or one in a transistor or other computational unit, quantum computers supposedly
store and handle both zero and one and other values in between at the same time through the process of superposition.
Superposition, however, does not store one or zero or any other number simultaneously. There is only an expectation that the values might be zero or one at the end of the computation. This quantum
probability is the polar opposite of the binary method of storing information.
Driven by quantum science’s uncertainty principle, the probability that a qubit stores a one or zero is like Schroedinger’s cat, which can be either dead or alive, depending on when you observe it.
But the two different values do not exist simultaneously during superposition. They exist only as probabilities, and an observer cannot determine when or how frequently those values existed before
the observation ended the superposition.
Leaving behind these challenges to using traditional binary computing metaphors means embracing new metaphors to explain quantum computing.
Peering Into Kaleidoscopes
The kaleidoscope metaphor is particularly apt to explain quantum processes. Kaleidoscopes can create infinitely diverse yet orderly patterns using a limited number of colored glass beads,
mirror-dividing walls and light. Rotating the kaleidoscope enhances the effect, generating an infinitely variable spectacle of fleeting colors and shapes.
The shapes not only change but can’t be reversed. If you turn the kaleidoscope in the opposite direction, the imagery will generally remain the same, but the exact composition of each shape or even
their structures will vary as the beads randomly mingle with each other. In other words, while the beads, light and mirrors could replicate some patterns shown before, these are never absolutely the
If you don’t have a kaleidoscope handy, this video is a good substitute.
Using the kaleidoscope metaphor, the solution a quantum computer provides – the final pattern – depends on when you stop the computing process. Quantum computing isn’t about guessing the state of any
given particle but using mathematical models of how the interaction among many particles in various states creates patterns, called quantum correlations.
Each final pattern is the answer to a problem posed to the quantum computer, and what you get in a quantum computing operation is a probability that a certain configuration will result.
New Metaphors For New Worlds
Metaphors make the unknown manageable, approachable and discoverable. Approximating the meaning of a surprising object or phenomenon by extending an existing metaphor is a method that is as old as
calling the edge of an ax its “bit” and its flat end its “butt.” The two metaphors take something we understand from everyday life very well, applying it to a technology that needs a specialized
explanation of what it does. Calling the cutting edge of an ax a “bit” suggestively indicates what it does, adding the nuance that it changes the object it is applied to. When an ax shapes or splits
a piece of wood, it takes a “bite” from it.
Metaphors, however, do much more than provide convenient labels and explanations of new processes. The words people use to describe new concepts change over time, expanding and taking on a life of
their own.
When encountering dramatically different ideas, technologies or scientific phenomena, it’s important to use fresh and striking terms as windows to open the mind and increase understanding. Scientists
and engineers seeking to explain new concepts would do well to seek out originality and master metaphors – in other words, to think about words the way poets do.
Sorin Adam Matei is an Associate Dean for Research at Purdue University. This article is republished from The Conversation under a Creative Commons license. Read the original article. | {"url":"https://www.discovermagazine.com/technology/quantum-computers-are-like-kaleidoscopes-helping-illustrate-science-and","timestamp":"2024-11-05T03:25:15Z","content_type":"text/html","content_length":"117372","record_id":"<urn:uuid:4870ac26-39af-496b-85cc-dcac703559a1>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00238.warc.gz"} |
Day 5: Template Literals | HackerRank
In this challenge, we practice using JavaScript Template Literals. Check the attached tutorial for more details.
The code in the editor has a tagged template literal that passes the area and perimeter of a rectangle to a tag function named sides. Recall that the first argument of a tag function is an array of
string literals from the template, and the subsequent values are the template's respective expression values.
Complete the function in the editor so that it does the following:
1. Finds the initial values of and by plugging the area and perimeter values into the formula:
where is the rectangle's area and is its perimeter.
2. Creates an array consisting of and and sorts it in ascending order.
3. Returns the sorted array.
The first line contains an integer denoting .
The second line contains an integer denoting .
Return an array consisting of and , sorted in ascending order.
The locked code in the editor passes the following arrays to the tag function:
• The value of is [ 'The area is: ', '.\nThe perimeter is: ', '.' ].
• The value of is [ 140, 48 ], where the first value denotes the rectangle's area, , and the second value denotes its perimeter, .
When we plug those values into our formula, we get the following:
We then store these values in an array, [14, 10], sort the array, and return the sorted array, [10, 14], as our answer. | {"url":"https://www.hackerrank.com/challenges/js10-template-literals/problem?isFullScreen=true","timestamp":"2024-11-04T04:04:45Z","content_type":"text/html","content_length":"909777","record_id":"<urn:uuid:627696fa-4a68-4e06-a97e-4fddd935897c>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00265.warc.gz"} |
Coefficient of friction between different materials (e.g. wood on wood, metal on metal) in context of coefficient of friction w angle
07 Sep 2024
Title: An Exploration of the Coefficient of Friction between Different Materials: A Theoretical Analysis with Respect to Angle
The coefficient of friction (COF) is a fundamental concept in tribology, describing the ratio of the force of friction between two surfaces to the normal force pressing them together. This article
delves into the theoretical aspects of COF between various materials, including wood on wood, metal on metal, and other combinations. We examine how the angle of contact influences the COF, providing
a comprehensive understanding of this phenomenon.
The coefficient of friction is a dimensionless quantity that quantifies the resistance to motion between two surfaces in contact. It is a critical parameter in various engineering applications,
including mechanical design, materials science, and biomechanics. The COF depends on several factors, including the surface roughness, material properties, and angle of contact.
Theoretical Background:
The COF can be described by the following formula:
μ = Ff / Fn
where μ is the coefficient of friction, Ff is the force of friction, and Fn is the normal force. However, when considering the effect of angle on COF, we must introduce the concept of shear stress
(τ) and normal stress (σ). The relationship between these quantities can be expressed as:
μ = τ / σ
Angle-Dependent Friction:
When two surfaces are in contact at an angle (θ), the force of friction (Ff) is not perpendicular to the surface. Instead, it forms a component that acts along the direction of motion. This leads to
a reduction in the effective normal force (Fn’) and, consequently, a decrease in the COF.
The angle-dependent friction can be described by the following formula:
μ’ = μ / cos(θ)
where μ’ is the angle-dependent coefficient of friction, and θ is the angle between the surfaces. This equation indicates that as the angle increases, the COF decreases.
Material-Specific Friction:
Different materials exhibit distinct frictional properties due to their surface roughness, material hardness, and other characteristics. For example:
• Wood on wood: μ ≈ 0.5-1.0
• Metal on metal: μ ≈ 0.2-0.6
• Rubber on rubber: μ ≈ 1.0-2.0
These values are approximate and can vary depending on the specific materials, surface conditions, and angle of contact.
The coefficient of friction between different materials is a complex phenomenon influenced by various factors, including the angle of contact. By understanding these relationships, we can better
design and optimize engineering systems, materials, and products. Further research is needed to explore the intricacies of material-specific friction and its applications in real-world scenarios.
• [1] Bowden, F. P., & Tabor, D. (1950). The Friction and Lubrication of Solids.
• [2] Rabinowicz, E. (1965). Friction and Wear of Materials.
• [3] Persson, B. N. J. (2001). Sliding Friction: Physical Principles and Applications.
Note: The references provided are a selection of classic texts in the field of tribology and friction. They offer a comprehensive understanding of the subject matter and provide a solid foundation
for further research.
Related articles for ‘coefficient of friction w angle’ :
Calculators for ‘coefficient of friction w angle’ | {"url":"https://blog.truegeometry.com/tutorials/education/76d2bf959c5bce692fee49fff85256f5/JSON_TO_ARTCL_Coefficient_of_friction_between_different_materials_e_g_wood_on_.html","timestamp":"2024-11-09T06:27:54Z","content_type":"text/html","content_length":"18142","record_id":"<urn:uuid:c29fa108-70e9-4425-80ea-f27b1cca47b5>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00013.warc.gz"} |
Hamiltonian Mechanics | Kinematic Links, Dynamics & Analysis
Hamiltonian mechanics – relation to kinematics
Hamiltonian mechanics is a powerful mathematical framework for analyzing the dynamics of physical systems using positions and momenta, offering insights into complex mechanical behaviors.
Hamiltonian Mechanics: An Introduction to Kinematic Links, Dynamics & Analysis
Hamiltonian mechanics is a reformulation of classical mechanics that originated from the work of Sir William Rowan Hamilton. This framework is particularly powerful for understanding the dynamics of
physical systems, using a different approach compared to Newtonian mechanics or Lagrangian mechanics. In Hamiltonian mechanics, the state of a system is described not just by positions but also by
momenta, providing a robust methodology for complex dynamic systems.
Kinematic Links
Kinematics deals with the motion of objects without considering the forces that cause this motion. In mechanical systems, kinematic links are basic building blocks that connect different components
of a mechanism. These links can be rigid or flexible and serve to transfer motion from one part of the system to another.
• Rigid Links: These do not deform under stress. They maintain a fixed relationship between the points they connect.
• Flexible Links: These can bend or stretch, allowing for motion adaption under varying conditions.
Kinematic analysis involves studying these links and their relative motions. By understanding the constraints and relationships between links, we can derive the equations of motion for a system.
Dynamics in Hamiltonian Mechanics
In Hamiltonian mechanics, the dynamics of a system are described using the Hamiltonian function \(H\). This function depends on the coordinates \(q_i\) and momenta \(p_i\) of the system:
$H = H ( q , p , t )$
Here, \(H\) often represents the total energy of the system (kinetic plus potential energy). The evolution of the system over time is governed by Hamilton’s equations:
$d q_i dt = ∂ ∂H p_i$
$d p_i dt = – ∂ ∂H q_i$
In simpler terms, the rate of change of position is given by the partial derivative of the Hamiltonian with respect to momentum, and the rate of change of momentum is given by the negative partial
derivative of the Hamiltonian with respect to position.
Analysis Methods
While the above equations might seem intimidating, they offer a powerful way to analyze complex systems. For instance, consider a simple harmonic oscillator where the Hamiltonian is:
\(H = \frac{p^2}{2m} + \frac{1}{2}kq^2\)
Here, \(m\) is mass, \(k\) is the spring constant, \(p\) is momentum, and \(q\) is position. The equations of motion derived from this Hamiltonian describe oscillatory behavior.
• Numerical Solutions: Many systems don’t have simple analytical solutions. Numerical methods like the Runge-Kutta algorithms can be used to solve Hamilton’s equations for complex systems.
• Phase Space Analysis: Phase space is a graphical representation where each state of the system corresponds to a unique point in a multidimensional space defined by positions and momenta. It helps
visualize the dynamics and stability of the system.
Applications of Hamiltonian Mechanics
Hamiltonian mechanics is not just a theoretical construct; it has several real-world applications that span across various fields of physics and engineering. Below are a few of its significant
• Celestial Mechanics: Hamiltonian mechanics is extensively used in studying the motion of planets, stars, and other celestial bodies. It helps predict their orbits and interactions under
gravitational influences.
• Quantum Mechanics: In quantum mechanics, the Hamiltonian function plays a crucial role. It is used to describe the total energy of quantum systems, leading to the Schrödinger equation, which is
fundamental for understanding quantum behavior.
• Optics: The principles of Hamiltonian mechanics are applied in geometrical optics to describe the path of light rays through different media using the Hamiltonian method.
• Control Systems: In control theory, Hamiltonian mechanics is used for optimal control problems, where it helps in finding the control laws that optimize a certain performance index.
Hamiltonian mechanics offers a unique and powerful approach to understanding the dynamics of physical systems, different from the traditional Newtonian and Lagrangian frameworks. By incorporating
both positions and momenta in the analysis, it provides a comprehensive methodology for studying complex systems such as celestial mechanics, quantum mechanics, and even optical systems. With tools
like phase space analysis and numerical methods, Hamiltonian mechanics opens the door to solving complicated dynamic problems that are otherwise intractable. Its applications across various fields
underscore its importance and versatility, making it a crucial part of the toolkit for physicists and engineers alike. | {"url":"https://modern-physics.org/hamiltonian-mechanics-relation-to-kinematics/","timestamp":"2024-11-05T15:18:38Z","content_type":"text/html","content_length":"161212","record_id":"<urn:uuid:c0208046-c162-43d3-80b5-8c83168fb729>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00175.warc.gz"} |
BEDMAS &
BEDMAS & fractions
Order of Operations
o Why is order important in mathematics?
To keep answers the same from person to person
o BEDMAS
• Brackets
• Exponents
• Division
• Multiplication
• Addition
• Subtraction
· Multiplying fractions
o If they are proper fractions convert to improper fractions
o Multiply the numerator by numerator and denominator by denominator
o Reduce the fraction and convert to a mixed (proper)
· Dividing fractions
o Convert to improper fractions
o Invert and multiply
o Reduce and convert
· Adding and subtracting fractions
o Convert to improper fraction
o Find the lowest common denominator
o Add and subtract on the numerator only!!!
o Reduce and convert
· Fractions and cross multiplying
o Some times we are asked to solve an unknown when dealing with fractions
o If there are two fractions and one equal sign we can use cross multiplication | {"url":"https://algebrahomework.org/flash/noflash/82.html","timestamp":"2024-11-05T19:10:41Z","content_type":"text/html","content_length":"103435","record_id":"<urn:uuid:5d1439dd-c1b1-45c6-ad52-e1b9eb2b3d4c>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00759.warc.gz"} |
of the
StarChild Question of the Month for July 2001
What is the shape of the universe?
One of the most profound insights of General Relativity was the conclusion that mass caused space to curve, and objects travelling in that curved space have their paths deflected, exactly as if a
force had acted on them. If space itself is curved, there are three general possibilities for the geometry of the universe. Each of these possibilites is tied to the amount of mass (and thus to the
total strength of gravitation) in the universe, and each implies a different past and future for the universe.
First, let's look at shapes and curvatures for a two-dimensional surface. Mathematicians distinguish 3 qualitatively different classes of curvature, as illustrated in the following image:
The flat surface at the left is said to have zero curvature, the spherical surface is said to have positive curvature, and the saddle-shaped surface is said to have negative curvature.
The preceding is not too difficult to visualize, but General Relativity asserts that space itself (not just an object in space) can be curved, and furthermore, the space of General Relativity has 3
space-like dimensions and one time dimension, not just two as in our example above. This IS difficult to visualize! Nevertheless, it can be described mathematically by the same methods that
mathematicians use to describe the 2-dimensional surfaces. So what do the three types of curvature - zero, positive, and negative -mean to the universe?
• If space has negative curvature, there is insufficient mass to cause the expansion of the universe to stop. In such a case, the universe has no bounds, and will expand forever. This is called an
open universe.
• If space has no curvature (i.e, it is flat), there is exactly enough mass to cause the expansion to stop, but only after an infinite amount of time. Thus, the universe has no bounds and will also
expand forever, but with the rate of expansion gradually approaching zero after an infinite amount of time. This is termed a flat universe or a Euclidian universe (because the usual geometry of
non-curved surfaces that we learn in high school is called Euclidian geometry).
• If space has positive curvature, there is more than enough mass to stop the present expansion of the universe. The universe in this case is not infinite, but it has no end (just as the area on
the surface of a sphere is not infinite but there is no point on the sphere that could be called the "end"). The expansion will eventually stop and turn into a contraction. Thus, at some point in
the future the galaxies will stop receding from each other and begin approaching each other as the universe collapses on itself. This is called a closed universe.
The geometry of the universe is often expressed in terms of the "density parameter", which is defined as the ratio of the actual density of the universe to the critical density that would be required
to cause the expansion to stop. Thus, if the universe is flat (contains just the amount of mass to close it) the density parameter is exactly 1, if the universe is open with negative curvature the
density parameter lies between 0 and 1, and if the universe is closed with positive curvature the density parameter is greater than 1.
The density parameter determined from various methods such as calculating the number of baryons created in the big bang, counting stars in galaxies, and observing the dynamics of galaxies both near
and far. With some rather large uncertainties, all methods point to the universe being open (i.e. the density parameter is less than one). But we need to remember that it is unlikely that we have
detected all of the matter in the universe yet.
The current theoretical belief (because it is predicted by the theory of cosmic inflation) is that the universe is flat, with exactly the amount of mass required to stop the expansion (the
corresponding average critical density that would just stop the is called the closure density). Recent observations (such as the BOOMERANG and MAXIMA cosmic microwave background radiation results,
and various supernova observations) imply that the expansion of the universe is accelerating. If so, this strongly suggests that the universe is geometrically "flat".
In reality, determining the value of the density parameter and thus the ultimate fate of the universe remains one of the major unsolved problems in modern cosmology. The recently (June 30, 2001)
launched MAP mission will be able to measure the value definitively within the next 5 years.
The StarChild site is a service of the High Energy Astrophysics Science Archive Research Center (HEASARC), within the Astrophysics Science Division (ASD) at NASA/ GSFC.
StarChild Authors: The StarChild Team
StarChild Graphics & Music: Acknowledgments
StarChild Project Leader: Dr. Laura A. Whitlock
Curator: J.D. Myers
Responsible NASA Official: Amber Straughn | {"url":"https://starchild.gsfc.nasa.gov/docs/StarChild/questions/question35.html","timestamp":"2024-11-04T09:19:08Z","content_type":"text/html","content_length":"9220","record_id":"<urn:uuid:1a53e849-08b1-4d84-8860-ab26ee433388>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00591.warc.gz"} |
plotting points on a coordinate plane
Save. Plotting Points on the Coordinate Plane DRAFT. Conic Sections: Parabola and Focus. Graphing Points in the Coordinate Plane . Instructor-paced BETA . Our mission is to provide a free,
world-class education to anyone, anywhere. By signing up you are agreeing to receive emails according to our privacy policy. Lesson 13: Interpreting Points on a Coordinate Plane. If you're seeing
this message, it means we're having trouble loading external resources on our website. Plotting Coordinates. Question 5 : Write the ordered pair for a point that is 9 units to the left of the origin
and lies on the x-axis. That's the first coordinate, the x-axis value. Improve your math knowledge with free questions in "Graph points on a coordinate plane" and thousands of other math skills. Plot
points on a rectangular coordinate system; Identify the quadrant where a point is located on a rectangular coordinate system; Many maps, such as the Campus Map shown below, use a grid system to
identify locations. Step By Step Procedure to Plot Points on a Graph . The negative shows that the point is 7 spaces to the left of 0. To learn how to graph an equation on the coordinate plane, read
on! Textbook pages Making Connections in Mathematics (Geometry Worktext) by Gladys C. Nivera, et al., pages 250-257 Next Century Mathematics III (Geometry) by Fernando B. Orines, et al., pages
502-507 … Start a live quiz . Author: Craig Steenstra, kennedyc6. Start a live quiz . On a coordinate plane graph, the point is downward one full point from the center without moving left or right.
The first real number is called the x coordinate or abscissa, and the second one is known as the y coordinate or ordinate. The lesson begins by asking students to think about how to describe the
location of their school in relation to another person’s specific location. Everything "originates" from the origin. In this lesson, we will look at exactly how to do this! Plotting a point on a 2D
coordinate plane. answer choices . A good way of remembering which axis is which is to imagine the vertical axis having a small slanted line on it, making it look like a "y". Now locate the x-axis
coordinate. Plot a given point on the coordinate plane. down. The printable worksheets in this page cover identifying quadrants, axes, identifying ordered pairs, coordinates, plotting points on
coordinate plane and other fun worksheet pdfs to reinforce the knowledge in ordered pairs. Illustrative Math Unit 6.7, Lesson 13 (printable worksheets) Lesson 13 Summary. Plot the points [latex]\left
(-2,4\right)[/latex], [latex]\left(3,3\right)[/latex], and [latex]\left(0,-3\right)[/latex] in the coordinate plane. 1 GCP 2. The short thumb is the y-axis value. If it has endpoints B(5, 9) and C(4,
3), what coordinates are the midpoint? Let's say the equation is y = x + 4. Finish Editing. If you want to know how to graph points on the coordinate plane, just follow these steps. From there, move
up/down until you match up with the value on the y-axis that matches the y-coordinate. a few seconds ago by. Have a look at the solved examples to get grip on the topic. To plot the point from an
ordered pair on the coordinate plane: Start at the origin and move left/right until you get to the integer that matches the x-coordinate. This is positive, so the point … In geometry, coordinates say
where points are on a grid we call the "coordinate plane". Points on the coordinate plane can give us information about a context or a … Points on the coordinate plane can give us information about a
context or a … Multiplying it by another number will either increase or decrease the slope. 0 likes. Plot the points and draw the simple shapes on the grid accordingly. What does it mean to join the
points to form a graph? Those are the coordinates of the midpoint of the line. To plot a point in the xy-coordinate plane (also called the Cartesian coordinate plane), you just have to know how to
read the values or numbers in the ordered pair that defines the point. 1 GCP 4 (-3, 2) Graph the point (3,4) on the coordinate plane. A good way to remember to go along the x axis first and the y
second, is to pretend that you are building a house, and you have to build the foundation (along the x axis) fist before you can build up. CONTENT Plotting of Points on the Coordinate Plane III.
Students progress at their own pace and you see a leaderboard and live results. What do the (+,+), (-,-), (+,-), and (-,+) notations indicate on a coordinate graph? If you're seeing this message, it
means we're having trouble loading external resources on our website. Do this for each x value--connect the dots with a smooth line or curve as (in this case, it will be a line). Name the coordinates
of a point in the Cartesian coordinate plane. The standard way to represent coordinate systems is on the Cartesian coordinate system. While there is evidence that ideas similar to Descartes’ grid
system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. 0. Thanks to all authors for
creating a page that has been read 300,402 times. In this lesson you will learn how to plot points on a coordinate plane by moving on the X axis and then the Y axis. Save. If you're seeing this
message, it means we're having trouble loading external resources on our website. Save. If you ever need assistance with plotting points on a coordinate plane you have come to the right place. In
other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2 or 10 or 100. Play. The key is to know the ordered
pairs are the addresses of the points. Learning Competency: Illustrate the rectangular coordinate system and its uses. The y-coordinate is also 3, so move three units up in the positive y direction.
a few seconds ago by. The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Edit. He viewed the
perpendicular lines as horizontal and vertical axes. To create this article, 65 people, some anonymous, worked to edit and improve it over time. The best thing is that they can interpret their
picture as they like and can use their creative skills to c Hints: Click and then ; Click one spot and then another spot to create a line *Extras: Try drawing different shapes with similar
properties. Then we move up/down vertically as many units as the y coordinate to locate the point in the coordinate plane. left or right. The point where these two lines meet is the point with
coordinates . If the x-coordinate is zero, the point is on the y-axis. Point T is on 9. Instructions: Plot each coordinate in problems 1 through 10 on this graph. up or down. Homework. (M8AL-Ie-1)
Specific Objectives: 1. Mathematics. Plotting Points on aPlotting Points on a Coordinate PlaneCoordinate Plane Slideshare uses cookies to improve functionality and performance, and to provide you
with relevant advertising. a set of numbers used to locate a point on the coordinate plane the vertical number line a number that corresponds to a number on the y-axis a point where the numbers
intersect Plot points on a rectangular coordinate system; Identify the quadrant where a point is located on a rectangular coordinate system; Many maps, such as the Campus Map shown below, use a grid
system to identify locations. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. The first number is the location of the point on the
x-axis, and the second is the location of the number on the y-axis. Students progress at their own pace and you see a leaderboard and live results. This math worksheet was created on 2013-02-14 and
has been viewed 233 times this week and 297 times this month. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. How Do You Plot
Points in the Coordinate Plane? How to Graph Points on the Coordinate Plane, https://www.mathplanet.com/education/algebra-1/visualizing-linear-functions/the-coordinate-plane, http://
www.virtualnerd.com/algebra-1/relations-functions/coordinate-plane/coordinate-plane-graphing/coordinate-plane-graph-points-example, https://www.khanacademy.org/math/basic-geo/basic-geo-coord-plane/
coordinate-plane-4-quad/v/plot-ordered-pairs, https://www.varsitytutors.com/hotmath/hotmath_help/topics/graphing-linear-equations, Menggambarkan Titik Titik Pada Bidang Koordinat, consider supporting
our work with a contribution to wikiHow. Mathematics. When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. Conic Sections: Ellipse
with Foci These two axes intersect one another at a point called the origin. Played 0 times. To plot the point [latex]\left(-2,4\right)[/latex], begin at the origin. In this non-linear system, users
are free to take whatever path through the material best serves their needs. Edit. By using our site, you agree to our. Here are 3 fun activities to help make Graphing Points on a Coordinate Plane
more fun . When you find both coordinates, plot the dot where the two number lines meet. The printable worksheets in this page cover identifying quadrants, axes, identifying ordered pairs,
coordinates, plotting points on coordinate plane and other fun worksheet pdfs to reinforce the knowledge in ordered pairs. Learning Competency: Illustrate the rectangular coordinate system and its
uses. y = 5x^2 is still a parabola, but it gets larger even faster, giving it a thinner look. Learner’s Materials pages pages 119-131 3. We have covered the step by step process involved in marking
points on a coordinate graph. How can I easily present a Google Drive spreadsheet to a participant? An open dot on a solid line is more visible than a closed dot. Plotting points is all about reading
the numbers . Start a live quiz . y = -x^2 (the negative is applied after the exponent ^2) is an upside down y = x^2; its base is (0,0). 0 likes. If you continue browsing the site, you agree to the
use of cookies on this website. Modifying the x coordinate moves the equation left or right. Graph points from a line. But it doesn’t have to be boring or tedious. Step By Step Procedure to Plot
Points on a Graph . (5,-4) is in Quadrant IV. Stay at the origin. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. SURVEY . A
positive number means you move to the right on the x-axis or up on the y-axis, while a negative number means you move left on the x-axis and down on the y-axis. Illustrative Math Unit 6.7, Lesson 13
(printable worksheets) Lesson 13 Summary. Turning it negative (multiplying by -1) flips it over; if it is a line, it will change it from going up to down or going down to up. The x-coordinate is 0.
To graph a point on the coordinate plane, take a look at your coordinates, which should be two numbers within parenthesis, separated by a comma. Solo Practice. Ordered Pair Definition. We know ads
can be annoying, but they’re what allow us to make all of wikiHow available for free. To create this article, 65 people, some anonymous, worked to edit and improve it over time. What directions does
the second coordinate tell you to move? Those numbers can be either positive or negative. Question 5 : Write the ordered pair for a point that is 9 units to the left of the origin and lies on the
x-axis. The key is to know the ordered pairs are the addresses of the points. Positive numbers go up or right (depending on the axis). Prepared by: Teacher III 2. Students progress at their own pace
and you see a leaderboard and live results. Define the components of the Cartesian coordinate system. A point in the plane is defined as an ordered pair, [latex]\left(x,y\right)[/latex], such that x
is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin. Khan Academy is a … To plot the point [latex]\left(3,3\right)[/latex], begin
again at the origin. The y-coordinate is 4, so then move four units up in the positive y direction. Plot it as you would any other set of coordinates. They will be able to to locate points in space
by reading the coordinates of a specific point on a coordinate plane. Teacher’s Guide pages pages 138-156 2. Plotting Points In order to understand how to plot points on a coordinate plane, we must
first recall how to locate integers on horizontal and vertical number line diagrams Point R is on –7. | {"url":"https://www.novafase.ind.br/kraft-mayo-hamw/4c4df7-plotting-points-on-a-coordinate-plane","timestamp":"2024-11-02T08:08:22Z","content_type":"text/html","content_length":"25842","record_id":"<urn:uuid:ed9ce447-8c3f-4089-a232-421bbd8b4916>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00336.warc.gz"} |
Levi-Civita field
Short description: System of numbers with non-finite quantities
In mathematics, the Levi-Civita field, named after Tullio Levi-Civita, is a non-Archimedean ordered field; i.e., a system of numbers containing infinite and infinitesimal quantities. Each member
[math]\displaystyle{ a }[/math] can be constructed as a formal series of the form
[math]\displaystyle{ a = \sum_{q\in\mathbb{Q}} a_q\varepsilon^q , }[/math]
where [math]\displaystyle{ \mathbb{Q} }[/math] is the set of rational numbers, the coefficients [math]\displaystyle{ a_q }[/math] are real numbers, and [math]\displaystyle{ \varepsilon }[/math] is to
be interpreted as a fixed positive infinitesimal. We require that for every rational number [math]\displaystyle{ r }[/math], there are only finitely many [math]\displaystyle{ q\in\mathbb{Q} }[/math]
with [math]\displaystyle{ a_q\neq 0 }[/math]; this restriction is necessary in order to make multiplication and division well defined and unique. Two such series are considered equal only if all
their coefficients are equal. The ordering is defined according to the dictionary ordering of the list of coefficients, which is equivalent to the assumption that [math]\displaystyle{ \varepsilon }[/
math] is an infinitesimal.
The real numbers are embedded in this field as series in which all of the coefficients vanish except [math]\displaystyle{ a_0 }[/math].
• [math]\displaystyle{ 7\varepsilon }[/math] is an infinitesimal that is greater than [math]\displaystyle{ \varepsilon }[/math], but less than every positive real number.
• [math]\displaystyle{ \varepsilon^2 }[/math] is less than [math]\displaystyle{ \varepsilon }[/math], and is also less than [math]\displaystyle{ r\varepsilon }[/math] for any positive real [math]\
displaystyle{ r }[/math].
• [math]\displaystyle{ 1+\varepsilon }[/math] differs infinitesimally from 1.
• [math]\displaystyle{ \varepsilon^{1/2} }[/math] is greater than [math]\displaystyle{ \varepsilon }[/math] and even greater than [math]\displaystyle{ r\varepsilon }[/math] for any positive real
[math]\displaystyle{ r }[/math], but [math]\displaystyle{ \varepsilon^{1/2} }[/math] is still less than every positive real number.
• [math]\displaystyle{ 1/\varepsilon }[/math] is greater than any real number.
• [math]\displaystyle{ 1+\varepsilon+\frac{1}{2}\varepsilon^2+\cdots+\frac{1}{n!}\varepsilon^n+\cdots }[/math] is interpreted as [math]\displaystyle{ e^\varepsilon }[/math], which differs
infinitesimally from 1.
• [math]\displaystyle{ 1+\varepsilon + 2\varepsilon^2 + \cdots + n!\varepsilon^n + \cdots }[/math] is a valid member of the field, because the series is to be construed formally, without any
consideration of convergence.
Definition of the field operations and positive cone
If [math]\displaystyle{ a=\sum \limits_{q \in \mathbb{Q}}a_q \varepsilon^q }[/math] and [math]\displaystyle{ b=\sum \limits_{q \in \mathbb{Q}}b_q \varepsilon^q }[/math] are two Levi-Civita series,
• their sum [math]\displaystyle{ a+b }[/math] is the pointwise sum [math]\displaystyle{ a+b:=\sum \limits_{q \in \mathbb{Q}}(a_q+b_q) \varepsilon^q }[/math].
• their product [math]\displaystyle{ ab }[/math] is the Cauchy product [math]\displaystyle{ ab:=\sum \limits_{q \in \mathbb{Q}}\left(\sum \limits_{r+s=q} a_r b_s \right)\varepsilon^q }[/math].
(One can check that for every [math]\displaystyle{ q\in\mathbb{Q} }[/math] the set [math]\displaystyle{ \{(r,s) \in \mathbb{Q}\times\mathbb{Q}: \ r+s=q,\ a_r \neq 0,\ b_s \neq 0\} }[/math] is finite,
so that all the products are well-defined, and that the resulting series defines a valid Levi-Civita series.)
• the relation [math]\displaystyle{ 0\lt a }[/math] holds if [math]\displaystyle{ a\neq 0 }[/math] (i.e. at least one coefficient of [math]\displaystyle{ a }[/math] is non-zero) and the least
non-zero coefficient of [math]\displaystyle{ a }[/math] is strictly positive.
Equipped with those operations and order, the Levi-Civita field is indeed an ordered field extension of [math]\displaystyle{ \mathbb{R} }[/math] where the series [math]\displaystyle{ \varepsilon }[/
math] is a positive infinitesimal.
Properties and applications
The Levi-Civita field is real-closed, meaning that it can be algebraically closed by adjoining an imaginary unit (i), or by letting the coefficients be complex. It is rich enough to allow a
significant amount of analysis to be done, but its elements can still be represented on a computer in the same sense that real numbers can be represented using floating point. It is the basis of
automatic differentiation, a way to perform differentiation in cases that are intractable by symbolic differentiation or finite-difference methods.^[1]
The Levi-Civita field is also Cauchy complete, meaning that relativizing the [math]\displaystyle{ \forall \exists\forall }[/math] definitions of Cauchy sequence and convergent sequence to sequences
of Levi-Civita series, each Cauchy sequence in the field converges. Equivalently, it has no proper dense ordered field extension.
As an ordered field, it has a natural valuation given by the rational exponent corresponding to the first non zero coefficient of a Levi-Civita series. The valuation ring is that of series bounded by
real numbers, the residue field is [math]\displaystyle{ \mathbb{R} }[/math], and the value group is [math]\displaystyle{ (\mathbb{Q},+) }[/math]. The resulting valued field is Henselian (being real
closed with a convex valuation ring) but not spherically complete. Indeed, the field of Hahn series with real coefficients and value group [math]\displaystyle{ (\mathbb{Q},+) }[/math] is a proper
immediate extension, containing series such as [math]\displaystyle{ 1+\varepsilon^{1/2}+\varepsilon^{2/3}+\varepsilon^{3/4}+\varepsilon^{4/5}+\cdots }[/math] which are not in the Levi-Civita field.
Relations to other ordered fields
The Levi-Civita field is the Cauchy-completion of the field [math]\displaystyle{ \mathbb{P} }[/math] of Puiseux series over the field of real numbers, that is, it is a dense extension of [math]\
displaystyle{ \mathbb{P} }[/math] without proper dense extension. Here is a list of some of its notable proper subfields and its proper ordered field extensions:
Notable subfields
• The field [math]\displaystyle{ \mathbb{R} }[/math] of real numbers.
• The field [math]\displaystyle{ \mathbb{R}(\varepsilon) }[/math] of fractions of real polynomials (rational functions) with infinitesimal positive indeterminate [math]\displaystyle{ \varepsilon }
• The field [math]\displaystyle{ \mathbb{R}((\varepsilon)) }[/math] of formal Laurent series over [math]\displaystyle{ \mathbb{R} }[/math].
• The field [math]\displaystyle{ \mathbb{P} }[/math] of Puiseux series over [math]\displaystyle{ \mathbb{R} }[/math].
Notable extensions
• The field [math]\displaystyle{ \mathbb{R}\varepsilon^{\mathbb{Q}} }[/math] of Hahn series with real coefficients and rational exponents.
• The field [math]\displaystyle{ \mathbb{T}^{LE} }[/math] of logarithmic-exponential transseries.
• The field [math]\displaystyle{ \mathbf{No}(\varepsilon_0) }[/math] of surreal numbers with birthdate below the first [math]\displaystyle{ \varepsilon }[/math]-number [math]\displaystyle{ \
varepsilon_0 }[/math].
• Fields of hyperreal numbers constructed as ultrapowers of [math]\displaystyle{ \mathbb{R} }[/math] modulo a free ultrafilter on [math]\displaystyle{ \mathbb{N} }[/math] (although here the
embeddings are not canonical).
External links
Original source: https://en.wikipedia.org/wiki/Levi-Civita field. Read more | {"url":"https://handwiki.org/wiki/Levi-Civita_field","timestamp":"2024-11-06T04:45:20Z","content_type":"text/html","content_length":"47061","record_id":"<urn:uuid:87b39bd7-e611-4999-9489-8f7b107933b8>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00389.warc.gz"} |
Tie breaker
Tie breakers are used in tournaments to distinguish between players finishing the tournament with the same score, for pairing or for seeding purposes. In general tournaments use more than one
tiebreaker, this is because many tiebreakers don't break ties under all circumstances, or cannot be used. In cases where a tiebreaker doesn't break ties only players who tie in the top group of that
tiebreaker are considered for the second tiebreaker and so on.
When to break ties
In many cases, breaking ties is not required. There is often nothing wrong with simply having players share the same place. Both the EGF and the AGA recommend that tournament organisers divide
prize-money (or other shareable prizes) equally between players that finish with the same score.
Many tournaments do use a tie breaker only to get a final ordering for purpose of determining the official winner and sorting the published results.
In some cases however, sharing a place is not acceptable because the prize cannot be shared (a cup, a right to play another player, an invitation for a next tournament, a ticket, a clock) and the tie
between the players must be broken. In general however it is seem more reasonable to divide the prize or give an extra prize. (For example if the prize is a free book from the book stall, give all
players who tie for this place a book)
List of Tie Breakers
This is nothing more than a list of tiebreakers. Tiebreakers which not mentioned are not necessarily better or worse than the methods mentioned. Feel free to add others.
based on face to face results
• Face to Face Result: In the case of two tied players, use the result of the game between them (if it was played)
• Direct Comparison: Generalization of Face to Face Result so it can be applied to tied groups of any size. Compares only the results of those games that tied players played against each other.
Requires all in group games to have been played. Not widely implemented in software.
□ Iterative Direct Comparison: Iterative application of Direct Comparison. It can sometimes break more ties of more than three players than non-iterative Direct Comparison.
• Direct Confrontation: Another generalization of Face to Face result, similar to Direct Comparison, but without the requirement that all in-group games have been played.
based on opponents scores
• SOS: Sum of the Opponents Scores, also known as Buchholz or Solkoff in Chess. Measured average strength of the opponents. Several variants exist which attempt to eliminate noise by discarding
some results (eg: SOS-1, SOS-2, Median or Modified Median). Does not work for round robin.
□ SOSOS: Sum of the Opponents SOS. Generally only used as a secondary tie breaker after SOS.
• Koya System: The number of points achieved against all opponents who have achieved 50 % or more.
□ Extended / Reduced Koya System: The Koya System can be extended by including score groups with a lower score, or reduced step by excluding players who scored less than a higher scores.
• SODOS: Sum of the Defeated Opponents Scores, also known as SonnebornBerger in Chess. Is advised against in McMahon tournaments, because players with the same score do not necessarily have the
same number of wins. Often used in round robin tournaments.
□ MDOS: Mean of Defeated Opponents' Score, similar to SODOS
based on cumulated scores
• CUSS: Cumulative Sum of Scores. Expected average strength of the opponents, also known as Sum of Progressive Scores (or, simply Progress) in chess.
□ CUSP:Cumulative Sum of Points. Similar to CUSS but using number wins instead of McMahon score Used by the British Go Association.
□ ROS: ROunds Score. Sum of a player's round points over all rounds in that he wins.
□ SOL: Sum of lost games round numbers. Inversed ROS, identical for practical purposes.
Team Tournament Tie Breakers
• Number of Board Wins (1 per board win, 1/2 per board jigo, 0 per board loss; added for all rounds)
□ First Board Wins (board wins only on the first and strongest board, to be used when total Board Wins are even)
• SORP: Sum of Result Points (in a round, a team's SORP is the opposing team's negative SORP)
Theoretical and Didactical Tie Breakers
See also: | {"url":"https://senseis.xmp.net/?Tiebreakers","timestamp":"2024-11-07T16:59:45Z","content_type":"text/html","content_length":"17451","record_id":"<urn:uuid:4575561a-08b6-46e8-abca-409842097fe9>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00403.warc.gz"} |
best HP scientific calculator
05-12-2015, 12:19 AM
Post: #3
Marcio Posts: 438
Senior Member Joined: Feb 2015
RE: best HP scientific calculator
(05-12-2015 12:01 AM)Jeff_Kearns Wrote: It all depends on what your programming needs are. If you can obtain one, I would recommend the HP-15C. Second and third choices would be either the
HP-41CX with the Advantage Module or the HP-42s. The current model is the HP-35s. An excellent choice if you are into a re-purposed calculator called the WP-34s.
Yes, I forgot to mention it has to be a current model. I live in South America and older models are very rare and can't be found in stores. Even the 50g is not all that easy to find these days.
Around here, Casio dominates the market of scientific calculators, they have plenty of options but I think I will get an HP.
I heard the WP-34s has received good reviews but I don't think they ship to Brazil, do they? Even if they did, the cost would be at least 3x as much.
Basically, I would like it to be able to solve non-linear equations, perform stat analysis, find roots etc, no need for matrix manipulation as I already have the Prime and the 50g for that.
Thank you.
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Non-Inverting Amplifier
Non-Inverting Amplifier
By internum
The Non-Inverting Amplifier, a high input impedance non-inverting circuit.
Figure 1 shows a high input impedance non-inverting circuit. This circuit gives a closed-loop gain equal to the ratio of the sum of R1 and R2 to R1 and a closed-loop 3 dB bandwidth equal to the
amplifier unity-gain frequency divided by the closed-loop gain.
Figure 1. Non-Inverting Amplifier
The primary differences between this connection and the inverting circuit are that the output is not inverted and that the input impedance is very high and is equal to the differential input
impedance multiplied by loop gain. (Open loop gain/Closed loop gain.) In DC coupled applications, input impedance is not as important as input current and its voltage drop across the source
Applications cautions are the same for this amplifier as for the inverting amplifier with one exception. The amplifier output will go into saturation if the input is allowed to float. This may be
important if the amplifier must be switched from source to source. The compensation trade off discussed for the inverting amplifier is also valid for this connection.
See also: | {"url":"https://www.freecircuits.net/circuit-48.html","timestamp":"2024-11-14T03:38:21Z","content_type":"text/html","content_length":"8780","record_id":"<urn:uuid:6cb505d5-2368-4a52-b00c-7f13c1be01bb>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00110.warc.gz"} |
SOCI 2010 Final Flashcards | Knowt
To determine how close x bar to unknown population mean likely to be (success rate if method)
Ex: 90% CI means there’s a 90% chance the population mean is within A and B
To test how likely is that a claim about a population parameter is true given the sample data
An outcome that is very unlikely to occur by chance of claim about data is true
Measures how many standard deviations away from the hypothesized mean our statistic is
To test hypotheses about population means based on a single sample mean (population stdev is known)
Because a small sample size that is significant doesn’t mean the effect is big
What are the conditions to use the procedure for the proportion confidence interval?
What are the conditions to use the procedure for the proportion test significance?
When dof increases, the graph gets closer to normal distribution because sample size is increasing.
What are matched pairs? What kind of test is used?
Matched pairs are used when subjects who are ALIKE are matched in pairs and BOTH randomly assigned treatment and placebo.
A one-sample test is used because the purpose is to compare the mean of differences.
40+ is close to normal, mildly skewed, and clearly skewed
Two sample tests are test of significance between two independent samples. The goal is to compare the responses of two treatments and compare the mean responses.
Considered robust if it continues to work well under certain conditions. A larger sample size could be clearly skewed. | {"url":"https://knowt.com/flashcards/2cbdddb3-a0b7-4db1-9384-e16e984b0714","timestamp":"2024-11-11T16:45:03Z","content_type":"text/html","content_length":"406405","record_id":"<urn:uuid:3d4359d7-67af-43ae-a807-f943bc6b07ce>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00512.warc.gz"} |
Introduction to Transformers
Introduction to Transformers
If you’ve been around electric appliances for any amount of time, you’d probably have heard of the transformer. Yes, they are those huge bulky things found in street corners, which make random scary
noises and occasionally spit sparks. Your phone charger also has a kind of small transformer too, but much, much smaller and with a completely different mechanism.
What is a Transformer?
A transformer is a device that uses the principles of electromagnetism to convert one voltage or current to another. It consists of a pair of insulated wires wound around a magnetic core. The winding
to which we connect the voltage or current to be converted is called the primary winding and the output winding is called the secondary winding.
Transformers come in two varieties – step up, which increases the voltage or current, and steps down, which decreases the voltage or current input. For example, the transformer in your microwave Oven
is a secondary transformer that is used to supply around 2200Volts to the vacuum tube in the Microwave Oven.
One thing to note is that transformers work only with changing or AC voltages and do not work with DC. We shall now learn why.
How important are transformers in the electrical system?
It was around 1856 when two brilliant minds Nikola Tesla and Thomas Edison were having a rivalry against each other. Those were the times when electricity and its applications of glowing a bulb and
running a motor were just being noticed. It was Edison and his associates who first discovered DC (Direct Current) system, then sometime after that Tesla came up with his AC (Alternating current)
system. Since then, both were trying to prove that their system is more advantageous than the other.
By then, the time has come for houses to get electricity. While Edison was busy demonstrating how dangerous AC is by electrocuting elephants, Tesla and his team came up with the transformers which
made transmitting electricity a lot easier and more efficient. Even, today Transformers play a vital role in the transmission system. Let’s know why.
Transmitting electricity with high voltages and low currents will help us reduce the thickness of the transmission wires and thus the cost, it also increases the efficiency of the system. For this
reason, a standard transmission system can be anywhere between 22KV to 66KV, while some generators in the power plant has an output voltage of only 11kV, and the household AC appliance requires only
220V/110V. So where does this voltage conversion take place and who does it.
The answer to the question is transformers. From the power plant to your home there will be transformers in the system that will either step-up (increase voltage) or step-down (decrease voltage) the
voltage to maintain the efficiency of the system. This is why the transformers are called the heart of an electrical transmission system. We will learn more about them in this article.
Transformer Symbols
The circuit symbol for a transformer is simply two inductors put together side by side sharing the same core. The nature of the line in between the two windings indicates the type of core used: A
dashed line represents ferrite, two parallel lines represent laminated iron and no line represents air core.
Sometimes the number of ‘bumps’ is used as a rough indicator of the transformer function – less bumps on one side and more on the other may mean that the first side has a smaller number of turns than
the other.
How Does a Transformer Work?
To understand the working of a transformer, we need to go back in time, to Michael Faraday’s laboratory.
Michael Faraday can perhaps be called the father of the transformer since it was his experiments that helped us understand electromagnetism and develop devices like motors and generators.
In the late 1800s, when it was discovered that electricity and magnetism were related phenomena, there was a race to try and build a practical device that could harness the power of magnets to
generate electricity.
Faraday found out that electricity could be generated by bringing a magnet close to a coil of wire. What he discovered was that voltage will be produced only when the magnetic field was changing,
that is, if he moved either the coil or the magnet relative to the other.
In DC, the current flow is steady and so is the magnetic field. Since the field is steady and not changing, there is no voltage induced on the secondary and the transformer just looks like a normal
coil of resistive wire to the power supply. So transformers do not work with DC currents.
He also found that when two coils of wire were kept close to each other, a current flowing in one coil could induce a current in the other coil. This principle is called mutual inductance and governs
the working of all modern transformers.
As shown in the figure, the transformer consists of two windings wound on a magnetic core.
The purpose of having a core is because air is not a very good supporter of magnetic fields, so having a magnetic core increases the magnetic field for a given amount of current flowing through one
winding, which in turn creates a stronger current in the other, increasing the overall efficiency of the device.
When a current passes through the primary, a magnetic field is set up in the core and is confined mostly to the core.
This magnetic field passes through the middle of the secondary and hence induces a current in the other by the law of mutual induction.
The beauty of this system is that the ratio of the input voltage to the output voltage is simply the ratio of the primary and the secondary windings, summed up by this formula:
Vout/Vin = Nsec/Npri
Where Vout is the output voltage, Vin is the input voltage, Nsec is the number of turns in the secondary winding and Npri is the number of turns in the primary winding.
So if you have two transformers, one with 100 turns on the primary and 1000 on the secondary and another with 10 turns on the primary and 100 turns on the secondary, you can calculate the turns ratio
to be 1:10 for both, so they both step up voltage to the same level.
Transformer Properties
If we have a closer look at the example given above, the first transformer will have a greater winding resistance (since more wire is used) and in some cases that might limit the amount of current
that can be drawn from the transformer. This property is called winding resistance, but in most cases it does not really matter since the copper wire used generally has a low resistance.
Another thing you notice is that there is no direct electrical connection between the primary and secondary windings. This is called galvanic isolation, and can be very useful, as we shall see.
Looking at each of the transformer windings, we can see that they are constructed just like inductors – a coil of wire wound around a magnetic core – and have an inductance too.
This inductance is proportional to the square of the number of turns, given by this formula:
Lpri/Lsec = Npri2/Nsec2
Where Lpri is the inductance of the primary winding, Lsec is the inductance of the secondary winding, Npri is the number of turns on the primary and Nsec is the number of turns on the secondary
The proportionality constant for a given core can be found in the datasheet and is usually given in units of µH/turn2. The exact value depends on the type and size of core.
Supposing you have a transformer core with a specification of 1uH/turn2. If you wind one winding on that core, then the inductance will be the value of the constant multiplied by the number of turns
squared, in this case 1. So the inductance of that one winding will be 1µH. If you wind another winding with 10 turns on the same core, then the inductance will be:
(1µH/turn2)*(10 turns)2 = 100µH
Since the windings have inductance, they provide an impedance to AC signals, given by the formula:
XL = 2π*f*L
Where XL is the impedance in ohms, f is the frequency in ohms and L is the inductance in Henries.
Say you want to design a transformer that draws 3A at 220V AC at 50Hz, which is standard power line frequency. Then the impedance of the primary would need to be 73.3 Ohms by Ohm’s law. Now that we
know the impedance required and the frequency, we can rearrange the formula to find out the inductance necessary for the winding:
L = (XL)/(2π*f)
Substituting the values, we find that the inductance needed would be 233mH.
Using this information and the value of µH/turns2 from the datasheet, we can calculate the windings required to get the inductance required.
Supposing that value is 50µH/turns2, then we can rearrange the formula to find out the inductance:
Where N is the number of turns, L is the required inductance, and the t2/µH term is just the inverse of the datasheet value.
Applying our values in the formula, we get a required number of turns of 2158. So as you can see, one you get the hang of the formulas, you can design transformers for nearly any application!
Transformer Construction
For anyone who needs to wind their own transformers, a knowledge of transformer construction is essential.
A transformer consists of a few basic components, below are the transformer parts:
1. BOBBIN:
The bobbin is the basic framework for any transformers. It provides a spool on which to wind the windings and also holds the core in place. It is usually made up of a heat resistant plastic. It also
sometimes contains metal pins onto which you can solder the ends of the windings if you want to mount it to a PCB, for instance.
2. CORE
This is probably the most important part of the transformer. As shown in the picture, the cores can come in many shapes and sizes. It is the magnetic properties of the core that determine the
electrical properties of the transformer which is built around the core.
3. WINDINGS
Thought it may seem like a trivial thing, the wire used in the construction is as important as any other aspect. Solid enameled copper wire is generally used, since the insulation is strong and thin,
so no wasted space due to plastic insulating sheaths.
Application of Transformers
This is probably the most common application for transformers – stepping down mains voltage for low voltage appliances. You might even find these inside things like microwaves and old TVs and wall
brick power supplies. These transformers have iron cores which gives excellent permeability but makes them bulky and somewhat less powerful than other types.
They are marked like 12-0-12 or 6-0-6 with three secondary wires. This means that the outer two wires have an output of 12V AC RMS if you make the center wire the ground reference. If you measure
across both the 12v winding, you will get 24V AC RMS. This gives you flexibility on how you might want to use the transformer.
2. SWITCH MODE POWER SUPPLIES
These are a very special type of power supplies that take a DC input and produce a DC output. They are found all modern phone chargers. The transformers used in these PSUs are designed more like
inductors with a small number of turns and ferrite cores with medium-to-high permeability. A DC voltage is applied across the ‘primary’ for a short time so that the current ramps up to a certain
level and stores some magnetic energy in the core. This energy is then transferred to the secondary at a lower voltage because it has a smaller number of turns. They operate at high frequencies and
achieve excellent efficiencies and are very small.
These are special transformers with a 1:1 turns ratio, so that the input and output voltages are the same. They are used to decouple appliances from mains earth. Since mains is earth referenced,
touching even one wire can result in a shock since the return path is literally the ground. Using isolation transformers ‘disconnects’ the appliance from the mains earth, since transformers are
galvanically isolated.
Most countries around the world use 220V AC as the standard supply voltage, but some countries like the US use 110V AC. This means that some devices like blenders cannot be operated in all countries.
For this purpose we can use transformers that convert 110V to 220V or vice versa to make sure that appliances can be used in any country.
These are special kinds of transformers that are used to match the impedance of the source and the load. They see extensive use in RF and audio circuits.
The turns ratio is equal to the square root of the source and load impedances.
This is a special type of transformer that has only one winding with a ‘tap’ output that forms the secondary. Usually this tap is variable, and so you can vary the output AC voltage, somewhat like a
voltage divider.
Transformers are useful devices and learning how to design and work with them can come in very handy! While we have covered the basics here, designing a transformer right from scratch is something
that can discussed in another entire article hence lets have that for some other time. So now, when you see a transformer again you will know why it is there and how it works.
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Family facing 6th Grade math unit focusing on area and surface area.
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This lesson presents the idea that the area of any triangle is exactly half of a certain parallelogram -- thus we get the familiar formula of multiplying the base and the altitude and taking half of
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Learn the formula for the area of a triangle, one half base times height. [5:29]
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the use of Khan Academy learning modules in their classroom. Please review materials from external sites before sharing with students.
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Video tutorial introduces the area of a rectangle and discovers a formula for the area of a right triangle. [4:44]
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What better way to get to know your students than with a 1:1 interview. Today you will interview each student on their knowledge of triangles and their ability to separate them from other shapes.
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An MIT engineering professor leads teachers and students through an activity in triangle formation and probability that requires a meter stick and other basic classroom materials. This video [33:08]
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This book is a "flexed" version of CK-12's Basic Geometry that aligns with College Access Geometry and contains embedded literacy supports. It covers the essentials of geometry for the high school
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CK-12 Foundation's Geometry FlexBook is a clear presentation of the essentials of geometry for the high school student. Topics include: Proof, Congruent Triangles, Quadrilaterals, Similarity,
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Use different size triangle to determine how to solve for the area in this video lesson. Take quiz to check understanding. [3:41]
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Nonlinear mapping (NLM) - Data Science Wiki
Nonlinear mapping (NLM) :
Nonlinear mapping (NLM) is a mathematical technique that involves the use of nonlinear functions to transform input data into output data. This technique is often used in
machine learning
, image processing, and other areas where the relationships between input and output data are complex and not easily represented by linear functions.
One example of NLM is the use of a neural
, which is a type of machine learning algorithm that uses multiple layers of nonlinear functions to process input data and make predictions. In a neural network, input data is fed through multiple
layers of processing units, or “neurons,” which apply nonlinear functions to the data to extract features and make predictions. For example, a neural network might be used to classify images of
animals based on various features, such as size, shape, and color. In this case, the nonlinear functions applied by the neurons would allow the neural network to identify patterns in the data that
might not be immediately apparent to a human observer.
Another example of NLM is the use of nonlinear filters in image processing. Nonlinear filters are used to enhance or modify the appearance of an image by applying nonlinear functions to the pixel
values. One common type of nonlinear filter is the
filter, which replaces the value of each pixel with the median value of the pixels in its neighborhood. This filter is often used to remove
or other unwanted artifacts from an image.
There are many different types of nonlinear functions that can be used in NLM, and the choice of function depends on the specific application and the characteristics of the data being processed. Some
common types of nonlinear functions include sigmoidal functions, which have an “S” shaped curve; ReLU (Rectified Linear Unit) functions, which are used in neural networks to introduce nonlinearity;
and polynomial functions, which are used to
complex relationships between variables.
One key advantage of NLM is that it allows for the modeling of complex relationships between input and output data. In contrast to linear models, which can only represent linear relationships,
nonlinear models can capture more complex patterns and interactions in the data. This can be particularly useful in areas such as machine learning, where the relationships between input and output
data are often highly nonlinear and difficult to predict using linear models.
Another advantage of NLM is that it can provide more accurate predictions and classifications than linear models in many cases. For example, in a machine learning task, a
nonlinear model
might be able to identify patterns in the data that a
linear model
would miss, leading to more accurate predictions. Similarly, in image processing, nonlinear filters can often produce more aesthetically pleasing results than linear filters.
There are also some challenges and limitations to using NLM. One potential drawback is that nonlinear models can be more difficult to interpret than linear models, as the relationships between input
and output data are often more complex and harder to understand. Additionally, nonlinear models can be more sensitive to noise and other types of interference, as the nonlinear functions used to
process the data can amplify small differences or variations in the input data. Finally, nonlinear models can require more computing power and time to train and evaluate, as the nonlinear functions
used in the model are often more complex and require more processing resources.
Overall, NLM is a powerful and widely-used technique that allows for the modeling of complex relationships between input and output data. Its ability to capture and represent complex patterns and
interactions in the data makes it an important tool in many fields, including machine learning, image processing, and other areas where the relationships between input and output data are complex and
nonlinear. So, NLM plays a vital role in the fields of
data analysis
and machine learning. | {"url":"https://datasciencewiki.net/nonlinear-mapping-nlm/","timestamp":"2024-11-13T14:54:02Z","content_type":"text/html","content_length":"43312","record_id":"<urn:uuid:64026144-bfe2-48a1-ac49-573b9a1c15ea>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00088.warc.gz"} |
aimia_m_discussion_request Give a list of questions for discussing during a class
Which subject Mathematics
What age group Doesn't matter
What topic Using journaling to reconnect to ourselves.
Quantity 10
Hints to each question
Any other preferences Focus on the creative mindset in our daily life.
The use of journaling in Mathematics is a powerful tool to connect with oneself. By writing down thoughts, ideas, and struggles, students can reflect on their learning process, identify areas of
strength and weakness, and gain a deeper understanding of the material. In this discussion, we will explore the benefits of journaling in Mathematics and how it can help students reconnect to
Questions for Discussion
1. How does journaling help you understand Mathematics better?
• Discuss specific instances where journaling has helped you understand a Math concept better.
• How does writing about Math concepts differ from other forms of writing?
2. What are the benefits of journaling in Mathematics, beyond understanding the concepts?
• How does journaling help you organize your thoughts about Mathematics?
• What benefits do you get from being able to communicate Mathematics through writing?
3. How can journaling help you identify your strengths and weaknesses in Mathematics?
• Can writing down your thoughts and feelings about a Math concept help you identify areas where you excel and areas where you struggle?
• How can journaling help you develop a plan to improve in areas where you struggle?
4. How can journaling help you build your creative mindset in Mathematics?
• In what ways can writing about Math concepts help you approach problem-solving creatively?
• Can journaling help you approach Mathematics as an art form?
5. How can journaling help you connect with yourself in Mathematics?
• How does journaling help you gain insight into your learning process?
• Can journaling help you build your confidence in Mathematics?
6. How can journaling be used as a tool for reflection in Mathematics?
• In what ways can writing about your experiences with Mathematics help you reflect on your progress?
• What are some specific things you can write about to facilitate reflection in Mathematics?
7. How can journaling be integrated into Mathematics learning?
• How can teachers incorporate journaling into their Mathematics curriculum?
• Can joint journaling exercises enhance peer collaboration and interaction?
8. How can journaling be used to document Mathematics achievements and progress?
• Can journaling be used to create a portfolio of Mathematics work?
• How can journaling help students track their progress over time?
9. What are some journaling techniques for Mathematics?
• How can students incorporate sketches and diagrams into their Mathematics journal?
• What are some strategies for staying motivated to consistently journal?
10. Can journaling in Mathematics be therapeutic?
• In what ways can writing about Mathematics help alleviate anxiety and stress?
• Can journaling help students build a positive relationship with Mathematics? | {"url":"https://aidemia.co/view.php?id=1987","timestamp":"2024-11-05T21:45:48Z","content_type":"text/html","content_length":"9082","record_id":"<urn:uuid:2b1c9b01-2f05-4cc3-b11e-063330eaa7a2>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00602.warc.gz"} |
Damon's maths and numeracy blog
Home Schooling report indicates high vocabulary skills but average to low maths skills
Why do home schoolers have good vocabularies but average to *low numeracy skills? Well, it's all to do with bootstrapping. A brief summary:
Bootstrapping refers to pulling yourself up by your own bootstraps. In education it refers to a self-teaching process that occurs when engaging in activities that have reciprocal feedback systems.
One such system is reading.
For example, learning to read is almost a pure process of bootstrapping. Once a child has a system for decoding words, the very process of reading facilitates learning. By sounding out words they
strengthen all letter sound connections, and as the frequency of exposure to words increases they create connections between whole words, their sound and their meaning (graphaphonic, phonological and
semantic systems). The reciprocal effect occurs between these systems and comprehension. Comprehension is a product of the first three, plus prior knowledge and strategies, which increases the
motivation to read, which keeps the cycle going. In short,
develops reading skill. Once decoding is learnt, the system is self-constructive; hence bootstrapping. It is not taught. Hence, most reading difficulties can be traced upstream to decoding.
Secondly, there is a sociocultural effect. Children in households that read, will be highly likely to become good readers. Children who grow up with books in the house, who are read to, and see Mum
and Dad reading, are more likely to read themselves. High reading levels influence vocabulary (you develop your higher vocabulary from reading, not listening to speech), and thus home schooling kids
have good vocabularies. This is great by the way, as vocabulary is associated with many great outcomes.
Finally, the amount of reading a child does relates to their vocabulary level. While there are caveats, basically the more they read, the better their vocab.
The problem with maths
There is no bootstrapping effect in mathematics if it is taught in a certain way (read
for more). Unlike reading, children in houses in which the parents are good at maths do not necessarily become good at math. Whatever positive sociocultural effect occurs with reading, doesn't seem
to occur with maths.
Maths is generally taught by an external person all the way, all the time (be it a teacher or computer software). This means the child's opportunities to get better at math are restricted to
occasions of direct instruction. Research suggests the total time doesn't add up to much no matter how you look at it. Second, much of the time doing maths is spent on worksheets, and almost all the
worksheets children work on are a repetition of known content (you learn to add double digit numbers and then do a sheet of double digit equations). Very rarely do they encourage the user to engage
in non-routine exploration of ideas and concepts. Completing worksheets in a routine manner is a poor version of learning, (rehearsal strategies are the least effective of almost all strategies), and
they certainly do not encourage bootstrapping. More of the opposite actually, they tend to teach the child to return to the parent to get help when stuck, at which point the parent tells them what to
Think of the difference between time spent reading per day, versus time spent doing maths. The reason is that reading is used as a
tool to do stuff
. Maths, in contrast, is simply a task to be completed. Imagine if we used maths as a tool to solve real problems everyday. Then, the bootstrapping effect might take effect.
As an experiment compare your child's time reading versus actual engagement in maths (not just staring at the page like I tend to do). Be sure to include all reading, such as that done while watching
TV, reading signs and so on. Add to this all the reading they do for other subjects. Once you have the number, extrapolate that to a yearly difference. One piece of NZ research found that NZ primary
school children did only 10 minutes of maths a day on average (and this was with a compulsory 1 hour of maths a day rule!).
So, in summary, the benefits many children get from homeschooling are a consequence of family values (we read and enjoy books) and the bootstrapping effect this has on learning.
But, it may be that we need to make maths as much a part of our family culture as reading. Then, I have no doubt, we will be producing avid users of mathematics ready to take on the world.
Tip: Hang a white board in the kitchen and every morning write a maths problem for the kids to solve. Make it reasonably easy so as to cultivate a culture of success and fun. Maths should be fun, not
hard- so don't make it hard - make it fun. The kids can chew on it all day, and write the answer on the board. At a time convenient have them explain what they did and then share how others did it
too. I'll do a post that includes a bunch of starter problems - watch this space.
*Note: The school results for maths and numeracy are also average to below average. This is not unique to home schooling. It would be interesting to compare the home school results directly to school
results but methodology issues (as noted in the report) make this difficult.
Staff attrition rates
I would love to see the data on tutor attrition rates in educational organisations. I wonder if we would see patterns? Do tutors leave in packs, two or three at a time? (Or more!?) Do they tend to
leave in the latter part of the year? What programmes do they leave, and what was happening in the months prior to this?
What do you think, are high staff attrition rates indicative of highly effective or ineffective organisations. Is the impact on learners positive or negative?
Here is something to think about regarding replacing these tutors: Is a new tutor as effective as an experienced one?
Some people think that new tutors bring new energy and that this makes up for a lack of experience.
If I was in an arguing mood I might say that I would rather have an inexperienced new energetic tutor rather than a tired old experience tutor. But I would be mostly wrong.
What happens if we consistently swap experience for energy?
The post below expresses some concerns and solutions if you are interested.
A big thank you to VARDA for a lovely evening last night.
Varda is the new name for The Waikato School of Hairdressing. I have had the privilege of working with their tutors occasionally over the last few years (maybe longer!).
The Varda team is fantastic. From the owners, to the managers, to the tutors, they not only design 'a look', but students too, building confidence, skills, and artistry. They take young people with
few skills, and mold them into amazing hair designers with ambitious dreams. But it is not only that. The students possess a confidence and professionalism about them that is a direct response to the
ethos of Varda. The place hums with energy. And the thing that strikes me the most when talking to the Varda staff, is how much they care about their students. They are invested in their success, at
a personal and professional level.
Having talked with their students, it's pretty clear VARDA has had a huge impact on them. It has broadened their horizons, introducing them to new opportunities that many hadn't had access to. I
spoke with students who's early school years were less than positive, in fact darn right tragic in some cases. And as we know, these beginnings often end with poor life outcomes. Yet, for these
students the programmes they were on, had changed that trajectory, setting them on a path toward achievement and success. The stories are an encouragement for anyone working with young people and
wondering if any of it makes a difference. Yes it does.
Varda is a success story. The students are evidence.
Check out their site here: VARDA
Emmy Noether - 133rd Birthday
Today is Emmy Noether's birthday. Albert Einstein described her as:
"The most significant creative mathematical genius thus far produced".
An interesting fact about Emmy. She was going to be an ESOL teacher, but saw the error of her ways and shifted to mathematics. Not that I'm tribal or anything ;)
On another note, what do you think Einstein meant by 'produced'?
A new maths?
I attended a workshop run by Associate Professor Joanne Mulligan from Australia. She was discussing a new 'type' of assessment that will be up and running this year.
The area being discussed was 'pattern and structure' in contrast to 'number' which is what we usually are exposed to. Although, actually the Learning Progressions do a pretty good job of covering a
broad range, although not not so much the above.
The assessment deals with how children perceive, make sense of, and recall visual images (one part anyway). An interesting piece of information was the fact that learners can score fairly highly in
number, yet have significant difficulties with pattern and structure.
The assessment tasks and learner responses really got me thinking about adult learners.
For example, I have noticed that learners often struggle with the idea of 'area' being measured in equal sized squares (not to mention when the area includes a decimal). I have also noticed that some
learners often struggle to copy the 'dice of fortune' grid when copying from the board. These, and other areas of difficulty, may represent weaknesses in pattern and structure. If a learner cannot
'see' the pattern as a pattern, but rather see it as random lines and dots, then their working memory is really up against it. A bit like someone partitioning without automaticity in basic number.
Eventually, they fall behind.
If these skills are undeveloped it will manifest as difficulties in number at some point, but particularly in space and shape.
The assessment is still confidential, but will be released in Australia later this year. It would be well worth the adult sector having a look at it. | {"url":"https://damonmath.blogspot.com/2015/03/","timestamp":"2024-11-04T15:23:38Z","content_type":"text/html","content_length":"88571","record_id":"<urn:uuid:300e5795-42b0-4a97-b238-e301de90c563>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00556.warc.gz"} |
10 Shortcuts For Sum In Excel - ExcelAdept
Key Takeaway:
• Shortcut keys make using the Excel SUM function quicker and easier. Memorize the most commonly used shortcuts for streamlined performance.
• The AutoSum shortcut allows for one-click calculation of the total sum, and the manual SUM shortcut can be adjusted for specific ranges and calculations.
• Advanced shortcuts include the SUMIF, SUMIFS, and SUMPRODUCT functions, which enable complex calculations based on specific criteria. Keyboard shortcuts for Excel SUM-related operations, such as
copy and paste and autofill, can further improve productivity.
• Additional tips and tricks for using the Excel SUM function include utilizing filters, pivot tables, and conditional formatting to enhance data analysis and presentation.
Do you want to become an Excel wizard and save time with your calculations? Discover 10 effective shortcuts for summing up numbers in Excel here! With these time-saving tricks, you’ll be sailing
through your math tasks in no time.
Shortcut keys for the SUM function
When working on Excel, using the SUM function can be time-consuming. However, there are several shortcut keys for this function that can save time and improve efficiency. Here are three of them:
• Alt + = – This shortcut key adds the sum formula to the cells below a selected cell quickly.
• Shift + Ctrl + End – This key selects all the cells from the start cell to the far bottom right of the sheet.
• Shift + Spacebar – This selection key allows users to select the entire row of the active cell before using the SUM function.
In addition to these shortcut keys, users can also change the formula’s settings to quickly sum a range of data with automatic calculations. By adding new data to the range, users can automatically
update the sum formula without much fuss.
Sources close to the Excel development team share that the software has featured the SUM function since its initial release in 1985, making it one of the oldest and most useful functions in the
Basic shortcuts
Enhance your Excel efficiency! Check out the fundamentals of summing up data. AutoSum and manual SUM shortcuts make it easier. Discover these basic shortcuts to work with sum formulas.
AutoSum shortcut
This excel function enables fast calculation of the addition of various values in a set. Here’s how to use it:
1. Select the cell where you want the sum to appear
2. Hit the ‘AutoSum’ button on the ribbon, or press Alt + ‘=’ (equal sign) on your keyboard.
3. Excel will automatically select the range for summation based on nearby cells. Hit enter for confirmation or reselect the desired range and hit Enter to get your result.
Additionally, users can tweak and customize this shortcut to help calculate a host of other functions. For instance, by pressing Shift + F3 after selecting the target cell, users can specify more
advanced calculations such as Average and Median.
Did you know? AutoSum was first introduced by Microsoft in 1987 as one method of performing basic arithmetic calculations on Excel spreadsheets. Its name and icon were designed to make it visually
consistent with other spreadsheet functions like Max & Min.
If you’re still manually adding up cells in Excel, you might as well be using a rock and chisel.
Manual SUM shortcut
Incorporating Manual Sum Function in Excel
Manual Summing shortcut in excel is an essential function that aids users in performing arithmetic operations, with increased convenience and efficiency. Its implementation can be performed by
adhering to the following fundamental steps:
1. Select a cell or range of cells.
2. Add the symbol ‘=’ followed by ‘/*Sum’.
3. Click on Enter to obtain the sum result.
Apart from these basic steps mentioned above, it’s noteworthy that Excel has quite a few other shortcuts for performing summation functions. However, proper identification and utilization of these
alternative methods must be made for optimal performance.
Proper analysis of context-specific data is key to better understanding the data type and its applicable formulas. This practice reduces errors due to misemphasis on irrelevant columns or rows
causing discrepancies in your results.
Maximizing Efficiency with Smart Implementation
Avoiding calculation discrepancies can be achieved through the use of brackets when implementing formulas involving multiple arithmetic operators. Additionally, incorporating keyboard shortcuts like
‘Ctrl+Shift+End’ offers precision and speed in selecting all cells in a particular workbook.
Furthermore, it’s advisable to incorporate essential functions such as applying formatting options like Data Validation for accountants. In comparison, statisticians would benefit from implementing
statistical modeling procedures through the ‘data Analysis’ option also present in excel.
In summary, mastering manual sum shortcuts are fundamental skills any user looking at enhancing their excel proficiency should aim at perfecting. An additional mastery level in further advanced Excel
VBA Programming will propel you to previously unattainable efficient levels that save time and drive better business decisions overall! Time to level up your Excel game with these advanced shortcuts-
because who needs a social life when you can master spreadsheets?
Advanced shortcuts
Excel advanced shortcuts can be mastered by using the SUMIF, SUMIFS, and SUMPRODUCT functions. These are very useful. They will save time and energy when calculating big sums with certain conditions.
These are mighty functions that can make tricky calculations easy.
SUMIF function shortcut
This Excel shortcut helps you easily calculate a specific sum based on certain criteria. By using the SUMIF function, you can quickly determine the total sum of a particular range that meets a given
condition. With this shortcut variant at hand, you can handle complex data calculations in short order without compromising accuracy or efficiency.
With the SUMIF function shortcut integrated into your Excel proficiency, endless possibilities await. This shortcut is valuable for users who need to find an incidence of information within columns
while remaining attuned to defined parameters and goals. By thoroughly learning the ins and outs of this productivity hack, users will enhance their data processing capabilities drastically.
It’s essential to note that utilizing advanced shortcuts like SUMIF Function helps simplify mundane tasks and increase productivity in our digital world. This specific hack has been fundamental in
countless situations where fast calculations were required to meet deadlines effectively without impeding data accuracy.
The history behind the invention of Excel is fascinating as it was created by Microsoft Corporation for use across various computing platforms and holds multiple patents worldwide. Even Though
invented over three decades ago, it remains highly relevant today and continues to grow with each new version release.
If you love efficiency, using the SUMIFS function shortcut in Excel is like finding a shortcut to your favourite coffee shop.
SUMIFS function shortcut
In Excel, an efficient way to calculate the sum of specific data is through the SUMIFS function shortcut. This feature allows users to filter data based on multiple criteria and add corresponding
values dynamically.
To use the SUMIFS function shortcut, follow these steps:
1. Select the cell you want to display the result.
2. Type “=” to start a formula and enter “SUMIFS”.
3. Enter or select the range of cells containing the values to add.
4. Enter or select each criterion range (ex: age range, specific name) followed by its respective value or reference cell.
5. If you have multiple criteria, separate each set with a comma “,”.
6. Press Enter to see your desired result.
In addition, another impressive capability of this shortcut includes the ability to use logical operators such as “<," ">,” “<=," ">=”, and “<>“. By using these operators instead of actual values in
some of your criteria ranges, you can multiply your filtering options and complex computations.
To maximize efficiency when using this function in Excel, always ensure that your data is correctly structured with useful names for columns and rows. Being meticulous in naming conventions enables
quick identification when entering references while applying this technique.
Get more bang for your buck with the SUMPRODUCT function, the shortcut that multiplies and adds all at once!
SUMPRODUCT function shortcut
Utilize an efficient technique to quickly calculate multiply and sum with the SUMPRODUCT shortcut. By using arrays or ranges, this function can multiply corresponding items within rows or columns
then return their sum. This is suitable for finding the weighted average or total cost of inventory.
A practical application for this could be a retail store owner calculating the total revenue from different products by multiplying each product’s quantity sold with its price, then summing the
Pro Tip: Use the SUMPRODUCT shortcut when working with large datasets to save time in calculations and analysis.
Save time and look like a pro with these Excel shortcuts for summing up your data faster than your boss can say ‘spreadsheet’.
Keyboard shortcuts for Excel SUM-related operations
Make SUM operations faster with keyboard shortcuts! Copy and paste a SUM formula quickly with the copy and paste sum formula shortcut. Get even speedier with the autofill sum formula shortcut. Excel
your way to success!
Copy and paste sum formula shortcut
Copying and pasting sum formulas can be done efficiently using keyboard shortcuts, allowing for quick calculation of large sets of data.
1. Select the cell containing the sum formula.
2. Press CTRL + C to copy the formula.
3. Select the range where you want to paste the formula.
4. Press ALT + E + S, then press enter to select ‘Formulas’ from the submenu.
5. Click on OK to paste the formula.
It is important to note that this shortcut only works when pasting within the same workbook.
Using this shortcut can save significant time and effort compared to manually typing out sum formulas for each range of data.
Legend has it that this shortcut was first used by a busy finance manager who needed a quicker way to calculate budget projections for his team’s upcoming projects. The strategy proved so successful
that it quickly caught on among other departments in the company.
Save your fingers the trouble and let Excel do the math for you with this handy autofill sum formula shortcut.
Autofill sum formula shortcut
When it comes to Excel’s SUM-related operations, the Autofill sum formula shortcut can be a handy tool for users. This feature helps in filling a list of successive cells with the desired formula
without any manual intervention.
Here is a 3-Step guide for you to use the Autofill sum formula shortcut:
1. Select the cell containing the SUM formula.
2. Hover over the bottom-right corner of your active cell until you see a small black cross sign.
3. Drag downwards across the cells where you want to apply that formula and leave as soon as you reach the last cell.
Additionally, if you need to insert another value or change variables, simply repeat Step 2 and drag your cursor again in that direction.
It’s worth noting that Autofill allows not only copying formulas but also formatting, sequences, and more. Hence, utilizing this function will result in saving time and effort while avoiding
duplication errors.
Incorporating these practices into your workflow will aid you immensely in your day-to-day activities with Excel. Try automating repetitive tasks by using any keystrokes before them. For instance,
press Ctrl + O for Open Document or set up a macro for Sum Formula Automation. It will help boost productivity by reducing keystrokes while improving accuracy levels at work.
Unleash the full potential of Excel’s SUM function with these additional tips, because math is hard enough without doing it manually.
Additional tips and tricks for using Excel SUM function
For maximum benefit, brush up on your Excel SUM function. To optimize, use with filters, pivot tables, and conditional formatting. These sub-sections offer solutions to work faster and better with
your data. Get the most out of your SUM function!
Using the SUM function with filters
Filtering data for specific information is a common need in data analysis, particularly in Excel. Utilizing the SUM function with filters can not only simplify the process but also generate exact
results promptly.
1. First, highlight the cells you want to filter.
2. Now click on “Data” placed at the top of Excel and then tap “Filter.”
3. Next, click on the arrow present in any cell’s column heading you’ve been filtered previously, and select a particular value from it.
4. Finally, apply the SUM formula after selecting the range of filtered cells and press Enter/Return. The output generated will be according to your requirement.
An additional feature provided by Excel is that filtering can enable an option where selected cells show only unique values using an option called Remove Duplicates. This will remove repetitive
results while summing up data.
By displaying only unique values using Remove Duplicates technique before applying SUM function contributes significantly to lessening user burden for error correction if multiple entries go
A potential suggestion is to create a pivot table if one wants more complexity or advanced kinds of information presentation such as grouping results by other factors or showing statistical
calculations within sums. Pivot tables are flexible enough to serve their purposes without tampering much with complex programming.
You don’t have to be a math genius to use Excel pivot tables, but it helps if you flunked out of art school.
Using the SUM function with pivot tables
For those looking to use the versatile SUM function with pivot tables, there are a few things to keep in mind. First, make sure your data is arranged properly in your pivot table. Then select the
cell where you want the sum calculation to appear and use the function =SUM(). From there, you can select your range of cells to be summed using either mouse clicks or typing in cell ranges manually.
City Sales
Toronto $50
Montreal $75
Vancouver $100
Using the SUM function with pivot tables can easily calculate total sales by city with a simple formula in Excel. In addition, it’s important to note that when using the SUM function with pivot
tables, any changes made to underlying data will automatically update the sum calculations. This makes it easy for users to keep track of their data without having to manually recalculate totals
every time new information is added.
It’s worth bearing in mind that while pivot tables can be extremely helpful in summarizing and filtering data quickly, they do require some set-up time if you want them to work efficiently. Taking
the time to organize your data and create effective pivot table layouts will ultimately save you time down the line when working with complex datasets.
According to a report by Forbes, mastering Excel can lead to higher paying roles and increased job prospects within various industries.
Using the SUM function with conditional formatting.
With conditional formatting, the SUM function in Excel can be used to calculate and highlight specific cells that meet certain conditions. By applying formatting rules to cells based on logical
expressions, such as “greater than” or “less than,” users can quickly identify cells that need to be included in calculation with the SUM function.
Here is a table that demonstrates how conditional formatting can be used with the SUM function:
Salesperson Product A Product B Product C
John $500 $750 $900
Sarah $600 $800 $1,000
Tom $400 $850 $1,200
With the SUM function, one could easily calculate the total sales for each product or salesperson. For example, to calculate John’s total sales, one would type “=SUM(B2:D2)” into a cell.
It’s important to note that conditional formatting only affects how data appears visually and doesn’t actually modify any cell values or formulas.
Did you know? The SUM function is one of the most commonly used functions in Excel and is found under the “Math & Trig” category. According to Microsoft, as of 2016, there were over 1.2 billion
Office users worldwide.
5 Facts About 10 Shortcuts for Sum in Excel:
• ✅ The SUM function in Excel allows you to quickly add up a range of numbers. (Source: Microsoft)
• ✅ One shortcut for sum in Excel is to use the AutoSum feature, which automatically detects adjacent cells and adds them up. (Source: Excel Campus)
• ✅ Another shortcut for sum in Excel is to select the range of cells you want to sum, and then press ALT + = on your keyboard. (Source: Business Insider)
• ✅ You can also use the SUMIF function in Excel to add up cells based on certain criteria. (Source: Exceljet)
• ✅ Learning shortcuts for sum in Excel can save you time and make your work more efficient. (Source: SkillSuccess)
FAQs about 10 Shortcuts For Sum In Excel
What are the 10 shortcuts for sum in Excel?
• Alt + =: This shortcut automatically sums up the range of cells above the active cell
• Ctrl + Shift + T: This shortcut selects the current data range and adds a Table
• Alt + Down Arrow: This shortcut opens the drop-down menu in the selected cell
• Ctrl + Shift + Arrow Down/Arrow Up: This shortcut selects the entire data range in a column
• Ctrl + Shift + Arrow Right/Arrow Left: This shortcut selects the entire data range in a row
• Shift + Spacebar: This shortcut selects the entire row
• Ctrl + Spacebar: This shortcut selects the entire column
• Ctrl + Shift + +: This shortcut inserts a new row or column
• F4: This shortcut repeats the last action performed
• Alt + H + B + A: This shortcut opens the Autosum feature
How do I use the Alt + = shortcut in Excel?
To use the Alt + = shortcut in Excel:
1. Select the cell in which you want to display the sum
2. Press the Alt + = keys together
3. Excel will automatically select the range of cells above the active cell and insert the SUM formula
4. Press Enter to display the sum in the selected cell
What is the Ctrl + Shift + T shortcut in Excel?
The Ctrl + Shift + T shortcut in Excel selects the current data range and adds a Table. This shortcut is particularly useful when you are working with large data sets and want to analyze the data
using Excel’s Table features.
How does the Ctrl + Shift + Arrow Down/Arrow Up shortcut work in Excel?
The Ctrl + Shift + Arrow Down/Arrow Up shortcut in Excel selects the entire data range in a column. To use this shortcut:
1. Select a cell in the column you want to select
2. Press the Ctrl + Shift + Arrow Down/Arrow Up keys together
3. The entire data range in the column will be selected
How do I repeat the last action performed in Excel using the F4 shortcut?
To repeat the last action performed in Excel using the F4 shortcut:
1. Perform the action you want to repeat
2. Press the F4 key
3. Excel will repeat the last action performed
What is the Alt + H + B + A shortcut in Excel?
The Alt + H + B + A shortcut in Excel opens the Autosum feature. This feature automatically adds up a group of cells and displays the sum in the selected cell. This shortcut is particularly useful
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How to multiply and divide integers in pre-algebra
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All High School Math Resources
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For many students in high school, memorization is their shortcut to making it through the most challenging classes. Instead of understanding difficult and abstract concepts, students memorize the
steps to solve problems and simply reproduce their memorized content on exams.
This approach can be especially tempting in high school mathematics courses, which can be consistently frustrating for many students as they find the dizzying array of variables, numbers, and
mathematical expressions with which they are expected to familiarize themselves to be overwhelming. It’s easy to allow that frustration to reduce motivation, and fall behind as a result.
Once you fall behind in a high school math course, it can be nearly impossible to get caught up. High school mathematics courses are usually dense, introducing new content at a fast page;
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SSC CHSL Area and Perimeter Questions Answers Set 10 by AMB
SSC CHSL Area and Perimeter Questions Answers Set 10
SSC CHSL Area and Perimeter Questions Answers Set 10: Ques No 1
The circumference of a circle is 11 cm and the angle of a sector of the circle is 60°. The area of the sector is (use π =22/7)
A. 77/48 cm^2
B. 125/48 cm^2
C. 75/48 cm^2
D. 123/48 cm^2
Answer: A
SSC CHSL Area and Perimeter Questions Answers Set 10: Ques No 2
If the difference between areas of the circumcircle and the incircle of an equilateral triangle is 44 cm^2, then the area of the triangle is ( Take π =22/7)
A. 28 cm^2
B. 7√3 cm^2
C. 14√3 cm^2
D. 21 cm^2
Answer: C
SSC CHSL Area and Perimeter Questions Answers Set 10: Ques No 3
If the area of a circle inscribed in a square is 9 π cm^2, then the area of the square is
A. 24 cm^2
B. 30 cm^2
C. 36 cm^2
D. 81 cm^2
Answer: C
SSC CHSL Area and Perimeter Questions Answers Set 10: Ques No 4
The sides of a triangle are 6 cm, 8 cm and 10 cm. The area of the greatest square that can be inscribed in it, is
A. 18 cm^2
B. 15 cm^2
C. 2304/49 cm^2
D. 576/49 cm^2
Answer: D
SSC CHSL Area and Perimeter Questions Answers Set 10: Ques No 5
The length of a side of an equilateral triangle is 8 cm. The area of region lying between the circumcircle and the incircle of the triangle is ( use π =22/7)
A. 351/7 cm^2
B. 352/7 cm^2
C. 526/7 cm^2
D. 527/7 cm^2
Answer: B
SSC CHSL Area and Perimeter Questions Answer Set 10: Ques No 6
A wire, when bent in the form of a square, encloses a region having area 121 cm^2. If the same wire is bent into the form of a circle , then the area of the circle is (use π =22/7)
A. 144 cm^2
B. 180 cm^2
C. 154 cm^2
D. 176 cm^2
Answer: C
SSC CHSL Area and Perimeter Questions Answer Set 10: Ques No 7
If the perimeter of a semicircular field is 36 m. Find its radius (use π =22/7)
A. 7 cm
B. 8 cm
C. 14 cm
D. 16 cm
Answer: A
SSC CHSL Area and Perimeter Questions Answer Set 10: Ques No 8
The perimeter (in meters) of a semicircle is numerically equal to its area ( in square meters). The length of its diameter is (use π =22/7)
A. 36/11 m
B. 61/11 m
C. 72/11 m
D. 68/11 m
Answer: C
SSC CHSL Area and Perimeter Questions Answer Set 10: Ques No 9
One acute angle of a right angled triangle is double the other. If the length of its hypotenuse is 10 cm, then its area is
A. 25/2(√3) cm^2
B. 25 cm^2
C. 25√3cm^2
D. 75/2 cm^2
Answer: A
SSC CHSL Area and Perimeter Questions Answer Set 10: Ques No 10
If a triangle with base 8 cm has the same area as a circle with radius 8 cm, then the corresponding altitude (in cm) of the triangle is
A. 12 π
B. 20 π
C. 16 π
D. 32 π
Answer: C
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Regression, Logistic Regression and Maximum Entropy
The Python code for Logistic Regression can be forked/cloned from my Git repository. It is also available on PyPi. The relevant information in the blog-posts about Linear and Logistic Regression
are also available as a Jupyter Notebook on my Git repository.
1. Introduction
One of the most important tasks in Machine Learning are the Classification tasks (a.k.a. supervised machine learning). Classification is used to make an accurate prediction of the class of entries in
a test set (a dataset of which the entries have not yet been labelled) with the model which was constructed from a training set. You could think of classifying crime in the field of pre-policing,
classifying patients in the health sector, classifying houses in the real-estate sector. Another field in which classification is big, is Natural Lanuage Processing (NLP). This goal of this field of
science is to makes machines (computers) understand written (human) language. You could think of text categorization, sentiment analysis, spam detection and topic categorization.
For classification tasks there are three widely used algorithms; the Naive Bayes, Logistic Regression / Maximum Entropy and Support Vector Machines. We have already seen how the Naive Bayes works in
the context of Sentiment Analysis. Although it is more accurate than a bag-of-words model, it has the assumption of conditional independence of its features. This is a simplification which makes the
NB classifier easy to implement, but it is also unrealistic in most cases and leads to a lower accuracy. A direct improvement on the N.B. classifier, is an algorithm which does not assume conditional
independence but tries to estimate the weight vectors (feature values) directly. This algorithm is called Maximum Entropy in the field of NLP and Logistic Regression in the field of Statistics.
Maximum Entropy might sound like a difficult concept, but actually it is not. It is a simple idea, which can be implemented with a few lines of code. But to fully understand it, we must first go into
the basics of Regression and Logistic Regression.
2. Regression Analysis
Regression Analysis is the field of mathematics where the goal is to find a function which best correlates with a dataset. Let’s say we have a dataset containing \(n\) datapoints; \(X = ( x^{(1)}, x^
{(2)}, .., x^{(n)} )\). For each of these (input) datapoints there is a corresponding (output) \(y^{(i)}\)-value. Here, the \(x\)-datapoints are called the independent variables and \(y\) the
dependent variable; the value of \(y^{(i)}\) depends on the value of \(x^{(i)}\), while the value of \(x^{(i)}\) may be freely chosen without any restriction imposed on it by any other variable. The
goal of Regression analysis is to find a function \(f(X)\) which can best describe the correlation between \(X\) and \(Y\). In the field of Machine Learning, this function is called the hypothesis
function and is denoted as \(h_{\theta}(x)\).
If we can find such a function, we can say we have successfully built a Regression model. If the input-data lives in a 2D-space, this boils down to finding a curve which fits through the data points.
In the 3D case we have to find a plane and in higher dimensions a hyperplane.
To give an example, let’s say that we are trying to find a predictive model for the success of students in a course called Machine Learning. We have a dataset \(Y\) which contains the final grade of
\(n\) students. Dataset \(X\) contains the values of the independent variables. Our initial assumption is that the final grade only depends on the studying time. The variable \(x^{(i)}\) therefore
indicates how many hours student \(i\) has studied. The first thing we would do is visualize this data:
If the results looks like the figure on the left, then we are out of luck. It looks like the points are distributed randomly and there is no correlation between \(Y\) and \(X\) at all. However, if it
looks like the figure on the right, there is probably a strong correlation and we can start looking for the function which describes this correlation.
This function could for example be:
\[h_{\theta}(X) = \theta_0+ \theta_1 \cdot x\]
\[h_{\theta}(X) = \theta_0 + \theta_1 \cdot x^2 \]
where \(\theta\) are the dependent parameters of our model.
2.1. Multivariate Regression
In evaluating the results from the previous section, we may find the results unsatisfying; the function does not correlate with the datapoints strongly enough. Our initial assumption is probably not
complete. Taking only the studying time into account is not enough. The final grade does not only depend on the studying time, but also on how much the students have slept the night before the exam.
Now the dataset contains an additional variable which represents the sleeping time. Our dataset is then given by \(X = ( (x_1^{(1)}, x_2^{(1)}), (x_1^{(2)}, x_2^{(2)}), .., (x_1^{(n)}, x_2^{(n)}) )
\). In this dataset \( x_1^{(i)}\) indicates how many hours student \(i\) has studied and \(x_2^{(i)}\) indicates how many hours he has slept.
This is an example of multivariate regression. The function has to include both variables. For example:
\[h_{\theta}(x) = \theta_0 + \theta_1 \cdot x_1 + \theta_2 \cdot x_2\]
\(h_{\theta}(x) = \theta_0 + \theta_1 \cdot x_1 + \theta_2 \cdot x_2^3\).
2.2. Linear vs Non-linear
All of the above examples are examples of linear regression. We have seen that in some cases \(y^{(i)}\) depends on a linear form of \(x^{(i)}\), but it can also depend on some power of \(x^{(i)}\),
or on the log or any other form of \(x^{(i)}\). However, in all cases the parameters \(\theta\) were linear.
So, what makes linear regression linear is not that \(Y\) depends in a linear way on \(X\), but that it depends in a linear way on \(\theta\). \(Y\) needs to be linear with respect to the
model-parameters \(\theta\). Mathematically speaking it needs to satisfy the superposition principle. Examples of nonlinear regression would be:
\[h_{\theta}(x) = \theta_0 + x_1^{\theta_1}\]
\[h_{\theta}(x) = \theta_0 + \theta_1 / x_1\]
The reason why the distinction is made between linear and nonlinear regression is that nonlinear regression problems are more difficult to solve and therefore more computational intensive algorithms
are needed. Linear regression models can be written as a linear system of equations, which can be solved by finding the closed-form solution \(\theta = ( X^TX )^{-1}X^TY\) with Linear Algebra. See
these statistics notes for more on solving linear models with linear algebra.
As discussed before, such a closed-form solution can only be found for linear regression problems. However, even when the problem is linear in nature, we need to take into account that calculating
the inverse of a \(n\) by \(n\) matrix has a time-complexity of \(O(n^3)\). This means that for large datasets (\(n > 10.000\) ) finding the closed-form solution will take more time than solving it
iteratively (gradient descent method) as is done for nonlinear problems. So solving it iteratively is usually preferred for larger datasets, even if it is a linear problem.
2.3. Gradient Descent
The Gradient Descent method is a general optimization technique in which we try to find the value of the parameters \(\theta\) with an iterative approach. First, we construct a cost function (also
known as loss function or error function) which gives the difference between the values of \(h_{\theta}(x)\) (the values you expect \(Y\) to have with the determined values of \(\theta\)) and the
actual values of \(Y\). The better your estimation of \(\theta\) is, the better the values of \(h_{\theta}(x)\) will approach the values of \(Y\). Usually, the cost function is expressed as the
squared error between this difference:
\[J(x) = \frac{1}{2n} \sum_i^n ( h_{\theta}(x^{(i)}) - y^{(i)} )^2 \]
At each iteration we choose new values for the parameters \(\theta\), and move towards the ‘true’ values of these parameters, i.e. the values which make this cost function as small as possible. The
direction in which we have to move is the negative gradient direction;
\(\Delta\theta = - \alpha \frac{d}{d\theta} J(x)\).
The reason for this is that a function’s value decreases the fastest if we move towards the direction of the negative gradient (the directional derivative is maximal in the direction of the
Taking all this into account, this is how gradient descent works:
• Make an initial but intelligent guess for the values of the parameters \(\theta\).
• Keep iterating while the value of the cost function has not met your criteria:
□ With the current values of \(\theta\), calculate the gradient of the cost function J ( \(\Delta \theta = - \alpha \frac{d}{d\theta} J(x)\) ).
□ Update the values for the parameters \(\theta := \theta + \alpha \Delta \theta\)
□ Fill in these new values in the hypothesis function and calculate again the value of the cost function;
Just as important as the initial guess of the parameters is the value you choose for the learning rate \(\alpha \). This learning rate determines how fast you move along the slope of the gradient. If
the selected value of this learning rate is too small, it will take too many iterations before you reach your convergence criteria. If this value is too large, you might overshoot and not converge.
3. Logistic Regression
Logistic Regression is similar to (linear) regression, but adapted for the purpose of classification. The difference is small; for Logistic Regression we also have to apply gradient descent
iteratively to estimate the values of the parameter \(\theta\). And again, during the iteration, the values are estimated by taking the gradient of the cost function. And again, the cost function is
given by the squared error of the difference between the hypothesis function \(h_{\theta}(x)\) and \(Y\). The major difference however, is the form of the hypothesis function.
When you want to classify something, there are a limited number of classes it can belong to. And for each of these possible classes there can only be two states for \(y^{(i)}\); either \(y^{(i)}\)
belongs to the specified class and \(y=1\), or it does not belong to the class and \(y=0\). Even though the output values \(Y\) are binary, the independent variables \(X\) are still continuous. So,
we need a function which has as input a large set of continuous variables \(X\) and for each of these variables produces a binary output. This function, the hypothesis function, has the following
\(h_{\theta} = \frac{1}{1 + \exp(-z)}} = \frac{1}{1 + \exp(-\theta x)}\).
This function is also known as the logistic function, which is a part of the sigmoid function family. These functions are widely used in the natural sciences because they provide the simplest model
for population growth. However, the reason why the logistic function is used for classification in Machine Learning is its ‘S-shape’.
As you can see this function is bounded in the y-direction by 0 and 1. If the variable \( z\) is very negative, the output function will go to zero (it does not belong to the class). If the variable
\(z\) is very positive, the output will be one and it does belong to the class. (Such a function is called an indicator function.)
The question then is, what will happen to input values which are neither very positive nor very negative, but somewhere ‘in the middle’. We have to define a decision boundary, which separates the
positive from the negative class. Usually this decision boundary is chosen at the middle of the logistic function, namely at \(z = 0\) where the output value \(y\) is \(0.5\).
$$y =\begin{cases} 1, & \text{if \(z>0\)}.\ 0, & \text{if \(z<0\)}. \end{cases}$$ $$
For those who are wondering where \(z\) entered the picture that we were talking about \(x\) before. As we can see in the formula of the logistic function, \(z = \theta \cdot x\). Meaning, the
dependent parameter \(\theta\) (also known as the feature), maps the input variable \(x\) to a position on the \(z\)-axis. With its \(z\)-value, we can use the logistic function to calculate the \
(y\) -value. If this \(y\)-value \(> 0.5\) we assume it does belong in this class and vice versa.
So the feature \(\theta\) should be chosen such that it predicts the class membership correctly. It is therefore essential to know which features are useful for the classification task. Once the
appropriate features are selected , gradient descent can be used to find the optimal value of these features.
How can we do gradient descent with this logistic function? Except for the hypothesis function having a different form, the gradient descent method is exactly the same. We again have a cost function,
of which we have to iteratively take the gradient w.r.t. the feature \(\theta\) and update the feature value at each iteration.
This cost function is given by
$$\begin{split} J(x) & = -\frac{1}{2n} \sum_i^n \left( y^{(i)} log( h_{\theta}(x^{(i)})) + (1-y^{(i)})log(1-h_{\theta}(x^{(i)})) \right) \ & = -\frac{1}{2n} \sum_i^n \left( y^{(i)} log(\frac{1}{1+exp
(-\theta x)}) + (1-y^{(i)})log(1-\frac{1}{1+exp(-\theta x)}) \right) \end{split}$$
We know that:
\[log(\frac{1}{1+exp(-\theta x)}) = log(1) - log(1+exp(-\theta x)) = - log(1+exp(-\theta x))\]
\begin{align} log(1-\frac{1}{1+exp(-\theta x)}) &= log( \frac{exp(-\theta x)}{1+exp(-\theta x)}) \
&= log(exp(-\theta x)) - log(1+exp(-\theta x)) \ &= -\theta x^{(i)} - log(1+exp(-\theta x)) \end{align}
Plugging these two equations back into the cost function gives us: $$\begin{split} J(x) & = - \frac{1}{2n} \sum_i^n \left( - y^{(i)} log(1+exp(-\theta x)) - (1-y^{(i)})(\theta x^{(i)} + log(1+exp(-\
theta x))) \right) \ & = - \frac{1}{2n} \sum_i^n \left( y^{(i)} \theta x^{(i)} -\theta x^{(i)} -log(1+exp(-\theta x)) \right) \end{split}$$
The gradient of the cost function with respect to \(\theta\) is given by
\frac{d}{d\theta} J(x) &= - \frac{1}{2n} \sum_i^n \left( y^{(i)} x^{(i)} - x^{(i)} + x^{(i)} \frac{ exp(-\theta x)}{1+exp(-\theta x)} \right) \ &= - \frac{1}{2n} \sum_i^n \left( x^{(i)} ( y^{(i)} -
1 +\frac{exp(-\theta x)}{1+exp(-\theta x)} ) \right) \ &= - \frac{1}{2n} \sum_i^n \left( x^{(i)} ( y^{(i)} - \frac{1}{1+exp(-\theta x)} ) \right) \ &= - \frac{1}{2n} \sum_i^n \left( x^{(i)} ( y^
{(i)} - h_{\theta}(x^{(i)}) )\right)
So the gradient of the seemingly difficult cost function, turns out to be a much simpler equation. And with this simple equation, gradient descent for Logistic Regression is again performed in the
same way:
• Make an initial but intelligent guess for the values of the parameters \(\theta\).
• Keep iterating while the value of the cost function has not met your criteria:
□ With the current values of \(\theta\), calculate the gradient of the cost function J ( \(\Delta \theta = - \alpha \frac{d}{d\theta} J(x)\) ).
□ Update the values for the parameters \(\theta := \theta + \alpha \Delta \theta\)
□ Fill in these new values in the hypothesis function and calculate again the value of the cost function;
4. Text Classification and Sentiment Analysis
In the previous section we have seen how we can use Gradient Descent to estimate the feature values \(\theta\), which can then be used to determine the class with the Logistic function. As stated in
the introduction, this can be used for a wide variety of classification tasks. The only thing that will be different for each of these classification tasks is the form the features \(\theta\) take
Here we will continue to look at the example of Text Classification; Lets assume we are doing Sentiment Analysis and want to know whether a specific review should be classified as positive, neutral
or negative.
The first thing we need to know is which and what types of features we need to include. For NLP we will need a large number of features; often as large as the number of words present in the training
set. We could reduce the number of features by excluding stopwords, or by only considering n-gram features. For example, the 5-gram ‘kept me reading and reading’ is much less likely to occur in a
review-document than the unigram ‘reading’, but if it occurs it is much more indicative of the class (positive) than ‘reading’. Since we only need to consider n-grams which actually are present in
the training set, there will be much less features if we only consider n-grams instead of unigrams. The second thing we need to know is the actual value of these features. The values are learned by
initializing all features to zero, and applying the gradient descent method using the labeled examples in the training set. Once we know the values for the features, we can compute the probability
for each class and choose the class with the maximum probability. This is done with the following Logistic function.
5. Final Words
In this post we have discussed only the theory of Maximum Entropy and Logistic Regression. Usually such discussions are better understood with examples and the actual code. I will save that for the
next blog.
If you have enjoyed reading this post or maybe even learned something from it, subscribe to this blog so you can receive a notification the next time something is posted.
6. References: | {"url":"http://ataspinar.com/posts/regression-logistic-regression-and-maximum-entropy/","timestamp":"2024-11-07T15:42:52Z","content_type":"text/html","content_length":"44836","record_id":"<urn:uuid:bf4b531e-eb6d-437d-88a2-dcc01a023afd>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00152.warc.gz"} |
12.6 Ratios That Analyze a Company’s Long-Term Debt Paying Ability
Creditors are interested to know if a company can pay its long-term debts. There are several ratios we use for this as demonstrated in the video:
Debt ratio The debt ratio measures how much we owe in total liabilities for every dollar in total assets we have. This is a good overall ratio to tell creditors or investors if we have enough assets
to cover our debt. The ratio is calculated as:
Total Liabilities
Total Assets
Total Liabilities $7,041.00
Total Assets $9,481.80
Times interest earned ratio Creditors, especially long-term creditors, want to know whether a borrower can meet its required interest payments when these payments come due. The times interest earned
ratio, or interest coverage ratio, is an indication of such an ability. It is computed as follows:
Income from operations (IBIT)
Interest expense
The ratio is a rough comparison of cash inflows from operations with cash outflows for interest expense. Income before interest and taxes (IBIT) is the numerator because there would be no income
taxes if interest expense is equal to or greater than IBIT. (To find income before interest and taxes, take net income from continuing operations and add back the net interest expense and taxes.)
Analysts disagree on whether the denominator should be (1) only interest expense on long-term debt, (2) total interest expense, or (3) net interest expense. We will use net interest expense in the
Synotech illustration.
For Synotech, the net interest expense is $236.9 million. With an IBIT of $1,382.4 million, the times interest earned ratio is 5.84, calculated as:
Income from operations $1,382.40
Interest expense $236.90
The company earned enough during the period to pay its interest expense almost 6 times over.
Low or negative interest coverage ratios suggest that the borrower could default on required interest payments. A company is not likely to continue interest payments over many periods if it fails to
earn enough income to cover them. On the other hand, interest coverage of 5 to 10 times or more suggests that the company is not likely to default on interest payments. | {"url":"https://courses.lumenlearning.com/suny-managacct/chapter/ratios-that-analyze-a-companys-long-term-debt-paying-ability/","timestamp":"2024-11-11T01:34:48Z","content_type":"text/html","content_length":"49835","record_id":"<urn:uuid:e506c9f8-807e-4306-afd6-35981ffb6b41>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00600.warc.gz"} |
WeBWorK Standalone Renderer
Consider the series
$\displaystyle{ \sum_{n=0}^{\infty} {2e^{-n}} }$
a. The general formula for the sum of the first $n$ terms is $S_n =$. Your answer should be in terms of $n$.
b. The sum of a series is defined as the limit of the sequence of partial sums, which means $\displaystyle{ \sum_{n=0}^{\infty} {2e^{-n}} = \lim_{n \to \infty} \bigg( }$$\displaystyle{\bigg) = }$.
c. Select all true statements (there may be more than one correct answer):
You can earn partial credit on this problem. | {"url":"https://wwrenderer.libretexts.org/render-api?sourceFilePath=Library/Hope/Calc2/APEX_08_02_Series/Q_34.pg&problemSeed=1234567&courseID=anonymous&userID=anonymous&course_password=anonymous&answersSubmitted=0&showSummary=1&displayMode=MathJax&language=en&outputFormat=nosubmit","timestamp":"2024-11-11T14:56:51Z","content_type":"text/html","content_length":"6929","record_id":"<urn:uuid:827a65b0-ba69-4e17-b350-72f21f05684b>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00494.warc.gz"} |
How many gallons of propane are in a 33 pound cylinder?
A 33 lb propane tank holds about 7.8 gallons of propane.
How many gallons of propane are in a 30lb cylinder?
7 gallons A 30 lb tank holds 7 gallons of propane and weighs 55 lbs full. The 40 lb propane tank is most commonly used for large commercial grills, construction heaters, space heaters, propane hawk
torches and many other propane applications. A 40 lb propane tank holds 9.4 gallons of propane and weighs 72 lbs full.
How many pounds of propane are in a 100 gallon tank?
170 lbs. A 100 lb. tank holds 23.6 gallons and weighs 170 lbs. when full.
How many gallons of propane are in a cylinder?
23.6 gallons Cylinder weighs about 170 pounds when full and Holds 100 pounds of propane (23.6 gallons of propane).
How many pounds of propane are in a 5 gallon propane tank?
20 lb 20 lb tank: Holds 5 gallons of propane.
How much does it cost to fill a 30lb propane tank?
$22-$24 the last few times I had it done for a 30-lb cylinder. seems to run in the $22 to $28 range for a #30 most places we have been. The propane place here charges me $28 to fill a 40# tank.
How long will 7 gallons of propane last?
Actually, it just happened to us recently, so it made me wonder how long does a propane tank last? The answer is that a gallon of propane will last about 95 hours if burned at a rate of 1000 BTUs per
How much does it cost to fill a 100-pound propane tank?
How Much Does it Cost To Fill a 100 Lb Propane Tank on Average?SizeCost20 lbs.$40100 Gallons$500500 Gallons$1,5001,000 Gallons$2,500
Can propane tanks be laid down?
All propane cylinders must be secured in the vertical and upright position. The safest way to secure a propane cylinder in a vehicle is with a trusted propane tank holder and stabilizer. However, the
propane cylinders must still be transported in the vertical and upright position.
How much does it cost to fill up a 5 gallon propane tank?
Save money by refilling propane tanks Why spend more money than you should? The average cost to exchange a propane tank varies from $5.00 - $6.00 a gallon. The cost to refill a propane tank varies
from $3.00 - $4.00 a gallon at most U-Haul propane refill locations.
How long will a 5 gallon propane tank last?
As a rule of thumb, one tank of propane will typically last between 18-20 hours if youre grilling on a medium-sized grill. Whereas larger grills can burn through 20-pounds of propane in as little as
10 hours.
How much does it cost to fill a 100 pound propane tank?
How Much Does it Cost To Fill a 100 Lb Propane Tank on Average?SizeCost20 lbs.$40100 Gallons$500500 Gallons$1,5001,000 Gallons$2,500
How long should 100 gallons of propane last?
A 100-pound propane tank holds 23.6 pounds of propane when its full. If your fireplace is 20,000 BTU and you use it 12 hours a day, the 100-gallon propane tank will last you around nine days.
Can I transport a 20lb propane tank on its side?
A standard 20 lb grill tank should never be on its side for use or transport for all of the above reasons. In addition they are equipped with an OPD (overfill protection device) that shuts the fill
off at the legal 80%. This is basically a float valve that can be damaged or jammed if the tank is on its side. | {"url":"http://blog.pristine.io/suwiloba60564.html","timestamp":"2024-11-14T13:58:07Z","content_type":"text/html","content_length":"27635","record_id":"<urn:uuid:e882f78f-8f6a-48c8-8558-e4b1aebf27a7>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00132.warc.gz"} |
Improving Your Meshing with Partitioning
Often, the most tedious step of finite element modeling is subdividing your CAD geometry into a finite element mesh. This step, usually just called meshing, can sometimes be fully automated. More
often, however, the careful finite element analyst will want to semi-manually create their meshes. Although this does require more work, sometimes there are significant advantages in doing so. In
this blog entry, we will look at one of the key manual meshing techniques: the concept of geometric partitioning.
How Does the Meshing Algorithm Work?
Let’s start by giving a very conceptual introduction to how a 3D CAD geometry is meshed when you use the default mesh settings in COMSOL Multiphysics. The default mesh settings will always use a Free
Tetrahedral mesh to discretize an arbitrary volume into smaller elements. Tetrahedral elements (tets) are the default element type because any geometry, no matter how topologically complex, can be
subdivided and approximated as tets. Within this article, we will only discuss free tetrahedral meshing, although there are situations when other types of meshes can be more appropriate, as discussed
A cylinder (left) is meshed with triangular elements (grey) on the surface and the tetrahedral meshing algorithm subdivides the volume with tets (cyan). The ends are omitted for clarity.
At a conceptual level, the tetrahedral meshing algorithm begins by applying a triangular mesh to all of the faces of the volume that you want to mesh. The volume is then subdivided into tetrahedra,
such that each triangle on the boundary is respected and the size and shape of the tetrahedra inside the volume meets the specified size and growth criteria. If you get the error message “Failed to
respect boundary element edge on geometry face” or similar, it is because the shape of the tetrahedra became too distorted during this process.
Of course, the true algorithm can only be stated mathematically, not in words. There are, however, cases that can cause this algorithm some difficulties, and these cases can be understood without
resorting to any equations. The free tetrahedral meshing algorithm can have difficulties if:
1. The part is extremely complex with very detailed regions mixed with coarse mesh.
2. The aspect ratios of the edges and boundaries defining the domain are very large.
Let’s take a look at some examples of each case and how partitioning can help us.
Simplifying Complex Geometries
To get us started, let us consider a modestly complex geometry: the Helix geometry primitive. You can certainly think of more complex geometries than this, but we can illustrate many concepts
starting with this case.
Go ahead and open a new COMSOL Model file and create a helix with ten turns, and then mesh it with the default settings, as shown below.
A ten-turn helix primitive with the corresponding default tetrahedral mesh.
When you were meshing this relatively simple part, you may have noticed that the meshing step took a relatively long time. So let’s look at how partitioning can simplify this geometry. Add a Work
plane to your geometry sequence that bisects the length of the helix and then add a Partition feature, using the Work plane as the partitioning object.
A Work plane is used to partition the helix.
As you can see from the image above, the resultant ten-turn helix object is now composed of twenty different domains, each representing a half-turn of the helix. When you re-mesh this model, you will
find that the meshing time is reduced, which is good. Each domain represents a much easier meshing problem than the original problem, and, furthermore, the domains can be meshed in parallel on a
multi-core computer.
However, you’re probably also thinking to yourself that we now have twenty different domains, and that we’ve subdivided the six surfaces of this helix into one hundred two surfaces, including the
internal boundaries, which are dividing up the domain. Although this geometry now meshes a lot faster, we have added many more domains and boundaries that can be a distraction as we apply material
properties and boundary conditions. What we actually want is to use the partitioned geometry for the mesh, but ignore the partitioning during the set-up of the physics.
What you’ll want to do next is to add a Virtual Operation, the Mesh Control Domains operation. This feature will take, as input, all twenty domains defining the helix. The output will appear to be
our original helix, and when we apply material properties and physics settings, there will be only one domain and six boundaries.
The Mesh Control Domains will specify that these are different domains only for the purposes of meshing.
When you now mesh this geometry, you’ll observe that you have the best of both. The meshing takes relatively little time, and the physics settings will be easy to apply. If you haven’t already, try
this out on your own!
We have only looked at one example geometry here, but there are many other cases where you’ll want to use this type of partitioning. Domains that look like combs or serpentines or objects that have
many holes, cutouts, or domains embedded within them all present situations in which you should consider partitioning. Also, keep in mind that you don’t need to partition with planes; you can create
and use other objects for partitioning. We’ll take a look at such an example next.
Geometries with High Aspect Ratios
The CAD geometries you are working with can often contain some edges or surfaces that have vastly different sizes relative to the other edges and surfaces defining a domain. We often want to avoid
such situations, since small features on a large domain may not be that important for our analysis objectives.
We’ve already looked at how we can ignore these small features using Virtual Operations to Simplify the Geometry, but what if these small features are important? Let’s examine how partitioning can
help us in terms of the example geometry shown below.
A flow domain to be meshed. Three small inlets, with even smaller fillets, protrude from the main pipe.
The geometry that you see above has a large pipe with three smaller pipes protruding from it. The small fillets that round the transition between the two have dimensions that are over one hundred
times smaller than the pipe volume. If we mesh this domain with the default mesh settings, the same settings will be used throughout. However, we will almost certainly want to have smaller mesh sizes
around the inlets.
The default mesh will use one setting for all elements within the model. That will not be very useful here. We could just add additional Size features to the mesh, and apply these features to all of
the faces around the small pipes to adjust the element sizes at these boundaries, but this is not quite optimal. It’s a lot of work and might not give us exactly what we want.
We can also use partitioning to define a small volume within which we will want to have different mesh settings. In the figure below, additional cylinders have been included that surround each of the
smaller pipes and extend some distance into the pipe.
Additional domains (wireframe) which will be used for partitioning of the blue domain.
Results of the partitioning operation.
These additional cylinder objects can be used to partition the original modeling domain, as shown above. Using the Mesh Control Domains, it will again be possible to simplify this geometry down to a
single domain for the purposes of physics and materials settings. Once you get to the meshing step, however, it is possible to add a Size feature to the Mesh sequence that will set the element size
settings of these newly partitioned domains. This gives us control over the element sizes in these domains and makes things a little bit easier for the mesher.
Different size features can be applied to each partitioned geometry.
What About When Automatic Meshing Fails?
The geometries that we have looked at here can be meshed with minimal effort or modification to the default meshing settings, but this is not always the case. It is relatively easy to come up with a
geometry that no meshing algorithm will ever be able to mesh in a reasonable amount of time. What can we do in that situation?
The answer (as I’m sure you’ve already guessed) is partitioning along with one other concept: divide and conquer. When confronted with a domain that does not mesh, use partitioning to divide it into
two domains. Try to individually mesh each one. If one of the domains does not mesh, keep partitioning each half. Using this approach, you’ll very quickly zoom in on the problematic region of the
original domain. You can then decide if you want to simplify the problematic parts of the geometry via the usage of Virtual Operations, or you can use the techniques we’ve outlined here and mesh
sub-domain by sub-domain, or you can even use some combination of the two.
Another technique that you can use is to apply a Free Triangular mesh on all of the boundaries of the imported geometry. Surface meshing is much faster than volume meshing and will almost always
succeed. Visually inspect the resultant surface mesh. It will then often be immediately apparent where in the model the small features and problematic areas are. Once you know where the issues are,
delete the Free Triangular mesh, since the free tetrahedral meshing algorithm will typically want to adjust the mesh on the boundaries, but will not do so if there is already a surface mesh defined.
Along with the Virtual Operations which we have already mentioned for simplifying the geometry for meshing, you can also use the Repair and Defeaturing functionality to clean up CAD data originating
from another source. The Virtual Operations will simply create an abstraction of the CAD geometry which can only be used inside of the COMSOL software, as compared to the Repair and Defeaturing
operations which will modify the CAD directly, and will create a modified CAD representation that can be written out from COMSOL Multiphysics to other software packages.
Summary of Meshing with Partitioning
We have now looked at two different representative cases where the default mesh settings are not optimal — a domain that is very complex as well as a domain with extreme aspect ratios. In both cases,
we can use partitioning along with the Mesh Control Domains Virtual Operations feature to simplify the meshing operations.
We have also presented some strategies for handling cases in which your geometry will not mesh with the default settings. It is also worth saying that such situations arise most often when working
with imported CAD geometry that was meant for manufacturing, rather than analysis purposes. If you are given a CAD file with many features that are cosmetic rather than functional or that you are
reasonably certain will not affect the physics of the problem, consider removing these features in the originating CAD package, before they even get to COMSOL Multiphysics.
In future blog posts, we will also look at combining partitioning with swept meshing, which is another powerful technique in your toolkit as you use COMSOL Multiphysics. Stay tuned!
Comments (10)
Saleha Quadssia
May 5, 2015
Its a great post and I have been trying to implement the partitioning using work-plane in my 3D imported geometry but I just end up with a line in the geometry but it still is not divided. So I
decided to try the simple helix model you proposed here and its the same issue. I end up with the geometry with one line at the place of work plane but no separate domains. I think there is something
very basic I am missing. Can you please help?
Walter Frei
May 5, 2015 COMSOL Employee
Hello Saleha,
Such questions would be more appropriate to pose to your COMSOL Support Team: http://www.comsol.com/support/
Aditya Date
July 16, 2015
Waka Waka
Mohammad Almajhadi
October 5, 2015
Once you create the partition, change the “PARTITION WITH” option into Workplan, and press “Build All” icon.
Nicola Young
February 17, 2016
This meshing technique looks very useful. I am trying to implement the latter part where you partition the mesh around the inlets.
It is not quite clear how you do this. I have revolved a cylindrical work plane, partitioned it and set it as a mesh control domain. However any objects within this disappear at the last build stage
in geometry.
Could you please explain how you implement the mesh partitioning while keeping domains that already exist (in your case, the inlet), from disappearing?
Many thanks,
Stephan Piotrowski
August 7, 2020
I think there are some details missing in this post, to the extent that it is difficult to apply the suggested strategies in my opinion. Setting a Mesh Control Domain seems to remove the geometry
completely. I am playing around with this right now and I believe you have to check the “Keep objects to partition” option so that after partitioning you have the partitioned domains plus the
original objects. Curiously, the “Partition Domains” operation does not appear to have this option.
Ravinder Banyal
March 30, 2016
I have a 3D geometry with two objects. Two objects are in tangential contact. One object is a cylinder which is supported by another object having V-shape. In 2D (x-y) plane you can think of it as
letter ‘o’ dropped inside the two inclined lines of letter ‘V’. Letter ‘o’ makes point contact with each inclined lines of letter ‘V’. I get my 3D geometry (cylinder+V-block) by extruding the 2D
geometry along z-direction. Now the point contacts in 2D become line contact in 3D. When I do the meshing, I get the error at the error message: “Failed to generate mesh for domain. Failed to respect
boundary element edge on geometry face.” The mesh size along the contact lines between cylinder and V-block becomes diminishingly small. I do not really need small mesh along the contact lines. How
can I overcome such mesh related issues where two objects/domains are in only in tangential contact?
Bridget Cunningham
April 1, 2016 COMSOL Employee
Hello Ravinder,
Thank you for your comment.
For questions specific to your modeling work, please contact our support team.
Online support center: https://www.comsol.com/support
Email: support@comsol.com
Mahendar Kumbham
March 9, 2017
Very useful post. Thank you so much for your time.
Kamuran Turksoy
February 26, 2020
When I use Mesh Control Domains option, it combines all the domains in my geometry and creates a single one. This happens even though I specify the domains to be combined not all. | {"url":"https://www.comsol.com/blogs/improving-your-meshing-with-partitioning?setlang=1","timestamp":"2024-11-07T20:43:08Z","content_type":"text/html","content_length":"123478","record_id":"<urn:uuid:c1336b80-a6d2-4da2-a298-f51f0ba41a19>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00306.warc.gz"} |
Mathematical operations on grids in the spectral domain
gmt grdfft ingrid [ ingrid2 ] -Goutfile|table [ -Aazimuth ] [ -Czlevel ] [ -D[scale|g] ] [ -E[r|x|y][+n][+w[k]] ] [ -F[r|x|y]params ] [ -I[scale|g] ] [ -Nparams ] [ -Q] [ -Sscale|d ] [ -V[level] ] [
-fflags ] [ --PAR=value ]
Note: No space is allowed between the option flag and the associated arguments.
grdfft will take the 2-D forward Fast Fourier Transform and perform one or more mathematical operations in the frequency domain before transforming back to the space domain. An option is provided to
scale the data before writing the new values to an output file. The horizontal dimensions of the grid are assumed to be in meters. Geographical grids may be used by specifying the -fflags option that
scales degrees to meters. If you have grids with dimensions in km, you could change this to meters using grdedit or scale the output with grdmath.
Required Arguments
Optionally, append =ID for reading a specific file format [Default is =nf] or ?varname for a specific netCDF variable [Default is the first 2-D grid found by GMT]. The following modifiers are
☆ +b - Select a band [Default is 0].
☆ +d - Divide data values by the given divisor [Default is 1].
☆ +n - Replace data values matching invalid with NaN.
☆ +o - Offset data values by the given offset [Default is 0].
☆ +s - Scale data values by the given scale [Default is 1].
Note: Any offset is added after any scaling.
Specify the name of the output grid file (see Grid File Formats) or the 1-D spectrum table (see -E).
Optional Arguments
Take the directional derivative in the azimuth direction measured in degrees CW from north.
Upward (for zlevel > 0) or downward (for zlevel < 0) continue the field zlevel meters.
Differentiate the field, i.e., take \(\frac{\partial}{\partial z}\) of the grid z. This is equivalent to multiplying by \(k_r\) in the frequency domain (\(k_r\) is radial wave number). Append a
scale to multiply by \(k_r \cdot\)scale) instead. Alternatively, append g to indicate that your data are geoid heights in meters and output should be gravity anomalies in mGal. Repeatable.
[Default is no scale].
Estimate power spectrum in the radial or a horizontal direction. No grid file is created. If one grid is given then f (i.e., frequency or wave number), power[f], and 1 standard deviation in power
[f] are written to the file set by -G [standard output]. If two grids are given we write f and 8 quantities: Xpower[f], Ypower[f], coherent power[f], noise power[f], phase[f], admittance[f], gain
[f], coherency[f]. Each quantity is followed by its own 1-std dev error estimate, hence the output is 17 columns wide. Select your spectrum by choosing one of these directives:
□ r - Choose a radial spectrum [Default].
□ x - Compute the spectrum in the x-direction instead.
□ y - Compute the spectrum in the y-direction instead.
Two modifiers are available the adjust the output further:
□ +w - Write wavelength w instead of frequency f, and if your grid is geographic you may further append k to scale wavelengths from meter [Default] to km.
□ +n - Normalize spectrum so that the mean spectral values per frequency are reported [By default the spectrum is obtained by summing over several frequencies.
Filter the data. Place x or y immediately after -F to filter x or y direction only; default is isotropic [r]. Choose between a cosine-tapered band-pass, a Gaussian band-pass filter, or a
Butterworth band-pass filter.
Specify four wavelengths lc/lp/hp/hc in correct units (see -fflags) to design a bandpass filter: wavelengths greater than lc or less than hc will be cut, wavelengths greater than lp and less
than hp will be passed, and wavelengths in between will be cosine-tapered. E.g., -F1000000/250000/50000/10000 -fflags will bandpass, cutting wavelengths > 1000 km and < 10 km, passing
wavelengths between 250 km and 50 km. To make a highpass or lowpass filter, give hyphens (-) for hp/hc or lc/lp. E.g., -Fx-/-/50/10 will lowpass x, passing wavelengths > 50 and rejecting
wavelengths < 10. -Fy1000/250/-/- will highpass y, passing wavelengths < 250 and rejecting wavelengths > 1000.
Gaussian band-pass:
Append lo/hi, the two wavelengths in correct units (see -fflags) to design a bandpass filter. At the given wavelengths the Gaussian filter weights will be 0.5. To make a highpass or lowpass
filter, give a hyphen (-) for the hi or lo wavelength, respectively. E.g., -F-/30 will lowpass the data using a Gaussian filter with half-weight at 30, while -F400/- will highpass the data.
Butterworth band-pass:
Append lo/hi/order, the two wavelengths in correct units (see -fflags) and the filter order (an integer) to design a bandpass filter. At the given cut-off wavelengths the Butterworth filter
weights will be 0.707 (i.e., the power spectrum will therefore be reduced by 0.5). To make a highpass or lowpass filter, give a hyphen (-) for the hi or lo wavelength, respectively. E.g., -F-
/30/2 will lowpass the data using a 2nd-order Butterworth filter, with half-weight at 30, while -F400/-/2 will highpass the data.
Note: For filtering in the time (or space) domain instead, see grdfilter.
Filename for output netCDF grid file OR 1-D data table (see -E). This is optional for -E (spectrum written to standard output) but mandatory for all other options that require a grid output.
Integrate the field, i.e., compute \(\int z(x,y) dz\). This is equivalent to divide by \(k_r\) in the frequency domain (\(k_r\) is radial wave number). Append a scale to divide by \(k_r \cdot\)
scale instead. Alternatively, append g to indicate that your data set is gravity anomalies in mGal and output should be geoid heights in meters. Repeatable. [Default is no scale].
Choose or inquire about suitable grid dimensions for FFT and set optional parameters. Control the FFT dimension via these directives:
□ a - Let the FFT select dimensions yielding the most accurate result.
□ f - Force the FFT to use the actual dimensions of the data.
□ m - Let the FFT select dimensions using the least work memory.
□ r - Let the FFT select dimensions yielding the most rapid calculation.
□ s - Just present a list of optional dimensions, then exit.
Without a directive we expect -Nnx/ny which will do FFT on array size nx/ny (must be >= grid file size). Default chooses dimensions >= data which optimize speed and accuracy of FFT. If FFT
dimensions > grid file dimensions, data are extended and tapered to zero.
Control detrending of data by appending a modifier for removing a linear trend. Consult module documentation for the default action:
□ +d - Detrend data, i.e. remove best-fitting linear trend.
□ +a - Only remove the mean value.
□ +h - Only remove the mid value, i.e. 0.5 * (max + min).
□ +l - Leave data alone.
Control extension and tapering of data by appending a modifier to control how the extension and tapering are to be performed:
□ +e - Extend the grid by imposing edge-point symmetry [Default].
□ +m - Extends the grid by imposing edge mirror symmetry.
□ +n - Turns off data extension.
Tapering is performed from the data edge to the FFT grid edge [100%]. Change this percentage via modifier +twidth. When +n is in effect, the tapering is applied instead to the data margins as no
extension is available [0%].
Control messages being reported:
□ +v - Report suitable dimensions during processing.
Control writing of temporary results: For detailed investigation you can write the intermediate grid being passed to the forward FFT; this is likely to have been detrended, extended by
point-symmetry along all edges, and tapered. Use these modifiers to ave such grids:
□ +w - Set the suffix from which output file name(s) will be created (i.e., ingrid_prefix.ext) [Default is “tapered”], where ext is your file extension
□ +z - Save the complex grid produced by the forward FFT. By default we write the real and imaginary components to ingrid_real.ext and ingrid_imag.ext. Append p to instead use the polar form of
magnitude and phase to files ingrid_mag.ext and ingrid_phase.ext.
Selects no wavenumber operations. Useful in conjunction with -N modifiers when you wish to write out the 2-D spectrum (or other intermediate grid products) only.
Multiply each element by scale in the space domain (after the frequency domain operations). [Default is 1.0]. Alternatively, append d to convert deflection of vertical to micro-radians.
Select verbosity level [w]. (See full description) (See cookbook information).
Geographic grids (dimensions of longitude, latitude) will be converted to meters via a “Flat Earth” approximation using the current ellipsoid parameters.
-^ or just -
Print a short message about the syntax of the command, then exit (Note: on Windows just use -).
-+ or just +
Print an extensive usage (help) message, including the explanation of any module-specific option (but not the GMT common options), then exit.
-? or no arguments
Print a complete usage (help) message, including the explanation of all options, then exit.
Temporarily override a GMT default setting; repeatable. See gmt.conf for parameters.
Grid Distance Units
If the grid does not have meter as the horizontal unit, append +uunit to the input file name to convert from the specified unit to meter. If your grid is geographic, convert distances to meters by
supplying -fflags instead.
netCDF COARDS grids will automatically be recognized as geographic. For other grids geographical grids were you want to convert degrees into meters, select -fflags. If the data are close to either
pole, you should consider projecting the grid file onto a rectangular coordinate system using grdproject
Data Detrending
The default detrending mode is to remove a best-fitting linear plane (+d). Consult and use -N to select other modes.
Normalization of Spectrum
By default, the power spectrum returned by -E simply sums the contributions from frequencies that are part of the output frequency. For x- or y-spectra this means summing the power across the other
frequency dimension, while for the radial spectrum it means summing up power within each annulus of width delta_q, the radial frequency (q) spacing. A consequence of this summing is that the radial
spectrum of a white noise process will give a linear radial power spectrum that is proportional to q. Appending n will instead compute the mean power per output frequency and in this case the white
noise process will have a white radial spectrum as well.
Note: Below are some examples of valid syntax for this module. The examples that use remote files (file names starting with @) can be cut and pasted into your terminal for testing. Other commands
requiring input files are just dummy examples of the types of uses that are common but cannot be run verbatim as written.
To obtain the normalized radial spectrum from the remote data grid @white_noise.nc, after removing the mean, let us try:
gmt grdfft @white_noise.nc -Er+n -N+a > spectrum.txt
To upward continue the sea-level magnetic anomalies in the file mag_0.nc to a level 800 m above sealevel:
gmt grdfft mag_0.nc -C800 -V -Gmag_800.nc
To transform geoid heights in m (geoid.nc) on a geographical grid to free-air gravity anomalies in mGal:
gmt grdfft geoid.nc -Dg -V -Ggrav.nc
To transform gravity anomalies in mGal (faa.nc) to deflections of the vertical (in micro-radians) in the 038 direction, we must first integrate gravity to get geoid, then take the directional
derivative, and finally scale radians to micro-radians:
gmt grdfft faa.nc -Ig -A38 -S1e6 -V -Gdefl_38.nc
Second vertical derivatives of gravity anomalies are related to the curvature of the field. We can compute these as mGal/m^2 by:
gmt grdfft gravity.nc -D -D -V -Ggrav_2nd_derivative.nc
To compute cross-spectral estimates for co-registered bathymetry and gravity grids, and report result as functions of wavelengths in km, try:
gmt grdfft bathymetry.nc gravity.grd -E+wk -fg -V > cross_spectra.txt
To examine the pre-FFT grid after detrending, point-symmetry reflection, and tapering has been applied, as well as saving the real and imaginary components of the raw spectrum of the data in topo.nc,
gmt grdfft topo.nc -N+w+z -fg -V -Q
You can now make plots of the data in topo_taper.nc, topo_real.nc, and topo_imag.nc. | {"url":"http://docs.generic-mapping-tools.org/latest/grdfft.html","timestamp":"2024-11-04T01:30:29Z","content_type":"text/html","content_length":"35001","record_id":"<urn:uuid:b37eafc4-ab7c-4a58-bdb8-6fac36a8ba5e>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00391.warc.gz"} |
Cost of Equity Formula (Quick Method to Calculate)
In this article, we will look at the introduction of the cost of equity formula. The introduction of cost of equity is the first step in calculating the required return on capital employed (ROCE) for
an investment decision. The cost of equity is a key element of valuation because it reflects the opportunity costs of the capital employed in a project.
There are several factors that go into calculating the cost of equity. For example, if a company has a beta higher than 1.0, it must provide a higher risk premium than a company with a lower beta. In
addition, a high-beta stock is more volatile than a low-beta stock. The risk premium is calculated by dividing a stock’s beta by its expected growth rate.
Introduction of Cost of Equity
The cost of equity is the ratio of a company’s share price to the company’s book value, which is equal to shareholders’ equity plus the market value of the company’s long-term debt.
This ratio indicates how much investors are willing to pay for company ownership. If a company has a higher cost of equity, investors will pay more for its shares than if it has a lower cost of
equity. When calculating this ratio, the first step is to calculate shareholders’ equity.
How To Calculate the Cost of Equity Formula
The cost of EquityFormula is a method for determining the cost of an equity investment. It involves analysing the risks involved in holding equity investments and comparing the risk to the expected
returns. A high beta means higher risk, while a low beta means lower risk. The formula takes several factors into account, including beta, which is a measure of volatility in a company’s stock price.
Several online resources offer beta values for various companies.
Investors and companies use the cost of equity formula to evaluate the benefits of investing in a company. The formula measures the risk associated with a company’s stock, the dividends it will pay
in the future, and the historical rates of return for the stock. It is also used to compare the risks of an investment against the risks of the broader market.
It has remained largely unchanged over the past four decades and is based on the capital asset pricing model. According to the CAPM, the cost of equity is the required return on equity divided by the
risk factor (usually the yield of a ten-year treasury bond). A premium is then added to reflect the risk associated with the investment.
Finding the Cost of Equity Using Different Types
1. GGM variable
The GGM variable in the cost of equity formula can be useful in calculating the fair value of a stock. It can also be useful in determining whether the stock is a good investment or not. Generally,
it is best to purchase undervalued stocks and sell overvalued stocks. The GGM can also be used to compare multiple investment opportunities.
Professor Myron J. Gordon developed the GGM model in the late 1950s. It uses three variables to value a company’s shares: the dividend per share (DPS), the required rate of return (r), and the
required dividend growth rate (g). These factors are obtained from the financial reports of the company.
The DGM model extends the DDM by incorporating an aspect of growth into the cost of equity formula. It includes a variable – the dividend growth rate – to measure dividend growth. The equations can
be rearranged to generate an expression as a function of the GGM. This model is part of the second proposition of the Modigliani and Miller model, which looks at firms’ value and capital structure.
2. Dividend Discount Model
The Dividend Discount Model for Cost of Equity (DDM) estimates the cost of equity capital by comparing the current price to a company’s expected future cash flows. It is an alternative to the Capital
Asset Pricing Model. It is a simple and logical formula that allows for different growth phases and applies to many companies.
Using the DDM in estimating the cost of equity can help investors understand the true cost of a particular stock. This method is most applicable to firms that distribute dividends to shareholders.
However, it can also be used on firms that do not distribute profits. Both models make assumptions about the number of dividends a company can pay, and these assumptions are dependent on the
company’s earnings.
3. The Capital Asset Pricing Model
The Capital Asset Pricing Model (CAPM) is an investment formula that computes an investor’s expected return on a security. The expected return is measured using the risk-free rate, typically equal to
the yield on a 10-year U.S. government bond. However, a security’s risk premium may be lower or higher depending on the market.
The CAPM has several applications in the finance industry. It is often used in financial modeling and calculating the equity cost. It is also used to estimate a firm’s net present value and determine
its enterprise value. In short, the CAPM is a very useful tool for evaluating investment opportunities and estimating the cost of equity.
This CAPM formula is a popular way to determine the cost of equity for a particular company. The CAPM model calculates the return based on the company’s risk and the cost of raising funds from equity
investors. However, determining the cost of equity is difficult because it relies on market expectations.
You need to know the cost of equity formula to determine each new asset’s ROI. This means knowing how much money you must spend before your project is profitable. You will also be able to compare
your ROI with other projects and see how your investment compares.
This cost of EquityFormula is the basis for calculating the cost of capital. Moreover, this is how much your investors must pay to make your company successful. The better your cost of equity, the
cheaper the capital. | {"url":"https://stoneoakbusiness.com/cost-of-equity-formula/","timestamp":"2024-11-13T01:38:43Z","content_type":"text/html","content_length":"85554","record_id":"<urn:uuid:6a28de57-a8ba-423e-984c-d0fab4891861>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00023.warc.gz"} |
Year 7 Lucky Dip
Gradient of a Line
Practise the skill of finding the gradients of straight lines by counting squares and dividing rise by run.
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Time Totals
Ten questions requiring a time calculation based on popular films.
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How Many of Each?
Work out how many items were bought from the information given.
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Express as a Fraction
Express one quantity as a fraction of another, where the fraction could be less than 1 or a mixed number.
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Area and Perimeter
Show that you know the area and perimeter formulas of basic shapes.
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A series of self-marking exercises on adding, subtracting, multiplying and dividing fractions.
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Players decide where to place the cards to make an equation with the largest possible solution.
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Boxed In Fractions
The classic dots and boxes two-player game with the addition of some fractions which determine your score.
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Yohaku Puzzles
Fill in the blank spaces so that the cells give the sum or the product shown in each row and column.
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Cracked Clock Quiz
A self marking set of ten mathematical questions about a clock which cracked!
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Calculate the missing fractions in these partly completed arithmagon puzzles.
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Months of the Year
Match the month name with the month number in this drag and drop activity
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Magic Square Jigsaws
Interactive jigsaw puzzles of four by four magic squares.
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Coordinates Picture
Plot the coordinates and join them with straight lines to produce a picture.
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Grid Arithmetic
Fill in a multiplication grid with the answers to simple multiplication and division questions.
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A self marking exercise on using ratio notation, reducing a ratio to its simplest form and dividing a given quantity into a number of parts in proportion to a given ratio.
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Know Your Place
Without a calculator perform some calculations requiring a knowledge of place value.
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Fickle Fractions
Compare pairs of fractions to identify the largest or smallest in order to move through the maze.
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Double, Double, Halve and Treble
Questions about scaling up the ingredients in the correct proportion for the witch's brew at Hallowe'en.
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Arrange the given digits to make six 3-digit numbers that combine in an awesome way.
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Fraction of ...
Practise your ability to find a fraction of a given amount with this self marking exercise.
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Old Equations
Solve these linear equations that appeared in a book called A Graduated Series of Exercises in Elementary Algebra by Rev George Farncomb Wright published in 1857.
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Largest Product
A drag and drop activity challenging you to arrange the digits to produce the largest possible product.
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Angles in a Triangle
A self marking exercise involving calculating the unknown angle in a triangle.
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Xmas Consonants
After working out which vowels are missing from the Christmas words do some basic calculations.
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Negative Numbers
Use negative numbers in basic arithmetic and algebraic calculations and word problems.
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Clock Times Pairs
The traditional pairs or Pelmanism game adapted to test the ability to compare analogue and digital times.
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Mix and Math
Determine the nature of adding, subtracting and multiplying numbers with specific properties.
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Letter Sums
Use your mental arithmetic skills to add up the values of the letters in these mathematical words.
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Area Maze
Use your knowledge of rectangle areas to calculate the missing measurement of these composite diagrams.
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Missing Terms
Can you work out which numbers are missing from these number sequences?
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Fraction Percentage
Match the fraction with the equivalent percentage. A drag and drop self marking exercise.
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Arrange the given numbers on the cross so that the sum of the numbers in both diagonals is the same.
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Fraction Decimal Pairs
The traditional pairs or Pelmanism game adapted to test knowledge of simple fractions and their equivalent decimals.
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A self marking exercise on indices (powers or exponents) including evaluating expressions and solving equations.
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Basic probability questions in an online exercise.
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Time Sort
Arrange the analogue and digital clock faces in order from earliest to latest
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Rounding DP
A self marking exercise requiring students to round numbers to a given number of decimal places.
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Magic Square
Each row, column and diagonal should produce the same sum.
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Fraction Percentage Pairs
The traditional pairs or Pelmanism game adapted to test knowledge of simple fractions and their equivalent percentages.
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Particular Pipes
Construct the pipes using a set number of pieces with lengths given as fractions, decimals or percentages.
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Make 1000
Use the numbers on the strange calculator to make a total of 1000
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Triside Totals
Arrange the digits 1 to 9 on the triangle so that the sum of the numbers along each side is equal to the given total.
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Pick The Primes
Pick the prime fruit from the tree as quickly as possible. Practise to improve your personal best time.
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Calculator Workout
An animated guide to using a scientific calculator for Secondary and High School students.
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Hi-Low Predictions
A version of the Play Your Cards Right TV show. Calculate the probabilities of cards being higher or lower.
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Indices Pairs
The traditional pairs or pelmanism game adapted to test knowledge of indices.
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Birthday Days
Compute the name of the day of the week from the given birthday clues.
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Rounding SF
A self marking exercise requiring students to round numbers to a given number of significant figures.
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Make an expression
Use the digits given to form an expression equivalent to the given total.
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Pairs 240
Find the pairs of numbers that multiply together to give a product of 240 in this collection of matching games.
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Area Two
How many different shapes with an area of 2 square units can you make by joining dots on this grid with straight lines?
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Twelve Days
How many gifts did my true love send to me according to the traditional Christmas song 'Twelve Days of Christmas'.
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For each pair of numbers subtract the sum from the product then divide the result by 20 without a calculator.
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A self marking exercise on identifying coordinates culminating in finding the mid point of two given points.
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Not Too Close
The students numbered 1 to 8 should sit on the chairs so that no two consecutively numbered students sit next to each other.
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Name the polygons and show the number of lines and order of rotational symmetry.
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HCF and LCM
Practise finding the highest common factor (H.C.F), sometimes called the greatest common divisor, and the lowest common multiple (L.C.M) of two numbers.
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How Many Squares? 2
How many different sets of four dots can be joined to form a square?
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Angle Parallels
Understand and use the relationship between parallel lines and alternate and corresponding angles.
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Test your understanding of averages with this self marking quiz about mean, median and range.
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Mystery Numbers
If '7 D in a W' stands for 7 days in a week, what do you think these mystery numbers are?
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T Puzzle
Use the pieces of the T puzzle to fit into the outlines provided. A drag, rotate and drop interactive challenge.
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Missing Operations Exercise
Each box represents a missing operation (add, subtract, multiply or divide). What are they?
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The Transum version of the traditional sliding tile puzzle.
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Area Wall Puzzles
Divide the grid into rectangular pieces so that the area of each piece is the same as the number it contains.
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Without Lifting The Pencil
Can you draw these diagrams without lifting your pencil from the paper? This is an interactive version of the traditional puzzle.
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Vector Connectors
Exercises about vectors and coordinates; using one to find the other.
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Don't Shoot The Square
You will need to be quick on the draw to shoot all of the numbers except the square numbers.
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An investigation of the minimum number of moves required to make the blue and green frogs swap places.
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Arrange the cards to create a valid mathematical statement.
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Expand algebraic expressions containing brackets and simplify the resulting expression in this self marking exercise.
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Eleven In Your Head
Multiply numbers by eleven in your head.
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Improper Fractions
A self-marking online exercise on converting improper fractions to mixed numbers and vice versa.
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Practise converting between miles and kilometres with this self marking quiz.
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Alpha Twist
Develop your skills and understanding of rotation in this fast-paced challenge.
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Angle Chase
Find all of the angles on the geometrical diagrams.
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Multi-step Problems
Solve multi-step problems in contexts, deciding which operations and methods to use and why.
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Number Palindromes
A collection of activities based around the theme of palindromic numbers.
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Plotting Graphs
Complete a table of values then plot the corresponding points to create a graph.
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Factor Trees Challenge
Can you determine the unique digits that will complete these factor trees?
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Changing The Subject
Rearrange a formula in order to find a new subject in this self marking exercise.
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Mixed Numbers
A self marking quiz about the application of the four operations to mixed numbers.
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How Many Triangles?
A self marking step by step approach to calculating the number of triangles in a design.
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Venn Diagram
Place each of the numbers 1 to 16 on the correct regions on the Venn diagram.
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Show that you know how many whatsits are in a thingamabob and other 'Numbers In A ...' answers.
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Transformation Tetris
Develop your skills translating and rotating shapes in this fast paced classic game.
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Words and Concepts
Fill in the missing words to show an understanding of the vocabulary of equations, inequalities, terms and factors.
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Quad Areas
Calculate the areas of all the possible quadrilaterals that can be constructed by joining together dots on this grid.
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Ratios vs Fractions
Relate the language of ratios and the associated calculations to the arithmetic of fractions.
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Time Arithmetic
Practise adding, subtracting, multiplying and dividing times with these self-marking exercises.
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Estimation is a very important skill. Use this activity to practise and improve your skills.
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Scale Drawings
Measure line segments and angles in geometric figures, including interpreting scale drawings.
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An online board game for two players evaluating algebraic equations and inequalities.
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A logic challenge requiring a strategy to update each of the numbers in a grid.
Try It! | {"url":"https://transum.org/Go/Pick.asp?YG=7&Title=Year+7+Lucky+Dip","timestamp":"2024-11-14T16:54:07Z","content_type":"text/html","content_length":"55719","record_id":"<urn:uuid:591c9048-f06b-4e36-8a7b-7aa2504c099c>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00897.warc.gz"} |
A contractor 2 - math word problem (83832)
A contractor 2
A contractor has to complete his contract in 92 days. 234 men were set to do the work, each of them works for 16 hours/day. After 66 days, 4/7 of the work is completed. How many additional men may be
employed, so that the work may be completed in time, each of additional men work 18h/day.
Correct answer:
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Showing 1 comment:
Dr Math
contract period =92 days.
234 men work for 16 hours/day.
In 66 days they did 4/7 of the work .
addl men work 18h/day.
Work left 1-4/7=3/7
Days left 92-66=26
66 days 4/7 work
In 26 days will do 26*4/7*66=52/231
Work left = 3/7-52/231 =(99-52)/231=47/231
4/7 in 66*234*16 hours
1 work in 66*234*16*7/4 hours
47/231 work will do in 66*234*16*7/4*47/231 hours
= By 66*234*16*7/4*47/231/(26*18) men
=(66*234*16*7*47)/(4*231*26*18)= 188 men
Tips for related online calculators
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Units of physical quantities:
Themes, topics:
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