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## Messages
Subject:
Re: Wheel Making
Response To:
Don Stephan
Someone please double check my memory and calculations . . .
The length of the circumference of a circle is 2 times 3.14159 times the radius, or 3.14159 times the diameter, here 20", so the circumference would be 62.83 inches. If the circumference is divided into 5 equal segments each would be 12.57" along the curve.
There are 360 degrees in a circle, so I would think five equal pie shaped pieces would each have an interior angle of 360 divided by 5, which my calculator says is 72 degrees.
A triangle with one vertex at the center of the circle and the other two at opposite ends of one of the segments on the circumference would be an isosolese (how is this dad burned word spelled anyway) triangle, meaning the outer two angles are the same. There are 180 degrees in a triangle, and if the center angle is 72 degrees then each of the outer angles would be one half of 180 minus 72, or 54 degrees. The angle setting on a miter saw may be the complement of this angle, 90-54, or 36 degrees.
Finding the straight line length between the two ends of each segment would require a trig calculator or table, and I have neither here.
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Tambahkan ke favorit Tautkan di sini
Beranda
MatematikaRia.com
Transformations
The three main Transformations are:
After any of those transformations (turn, flip or slide), the shape still has the same size, area, angles and line lengths. If one shape can become another using Turns, Flips and/or Slides, then the two shapes are called Congruent.
Resizing
The other important Transformation is Resizing (also called dilation, contraction, compression, enlargement or even expansion). The shape becomes bigger or smaller:
If you have to use Resizing to make one shape become another then the shapes are not Congruent, they are Similar.
Congruent or Similar
So, if one shape can become another using transformation, the two shapes might be Congruent or just Similar
If you ... Then the shapes are ...
... only Rotate, Reflect and/or Translate
Congruent
... need to Resize
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# Integer tutorial¶
integer was the first class implemented in mp++, and as a result it is currently the most featureful and optimised among mp++’s multiprecision classes.
One of the first things that can be noticed about integer is that the class is parametrised over an integral value SSize, called the static size. This compile-time constant represents the number of limbs that will be stored directly within an integer object before resorting to dynamic memory allocation. A limb is the part of a multiprecision integer that fits in a single word-sized unsigned integral type: in the same way an integral value in base 10 is represented as a sequence of digits in the $$\left[0,9\right]$$ range, an integer object is represented internally by an array of 32-bit or 64-bit unsigned C++ integral values. Thus, for instance, if SSize is set to 2 on a 64-bit system, integer is able to represent values in the $$\left(-2^{128},2^{128}\right)$$ range without resorting to dynamic memory allocation. In general, an SSize value of 1 is a good default choice for many use cases.
In addition to the common operators available for all of mp++’s multiprecision classes, integer supports the following additional operators:
• the modulo operator %, which computes the remainder of integer division,
• the bitshifting operators << and >>,
• the bitwise logical operators ~, &, | and ^.
In addition to the binary versions of these operators, the in-place versions are also available. Lower level ternary primitives (e.g., mul_2exp(), tdiv_q_2exp(), bitwise_ior(), etc.) are also provided for those situations in which it is desirable to pass the return value as a function parameter, rather than creating a new return value (as explained earlier in the API overview). For consistency with C++11, the % operator returns a remainder with the same sign as the dividend. The bit-shifting operators << and >> correspond respectively to multiplication and division by a power of 2. The bitwise logical operators behave as-if integer used a two’s complement representation (even if internally a sign-magnitude representation is used instead).
Among the features specific to integer we find:
Many of these features, which are documented in detail in the integer reference, are available in multiple overloads, often both as free and member functions.
## Interacting with the GMP API¶
integer provides a variety of ways for interfacing with the GMP library. There are a few reasons why one would want to use integer in conjunction with the GMP API, such as:
• the necessity of using functions from the GMP API which have not (yet) been wrapped/implemented by mp++,
• passing data from/to mp++ to/from another GMP-based multiprecision library.
mpz_t m;
mpz_init_set_si(m, -4); // Init an mpz_t with the value -4.
int_t n1{m}; // Init an int_t from the mpz_t.
assert(n1 == -4); // Verify that the value is correct.
int_t n2;
n2 = m; // Assign the mpz_t to another int_t.
assert(n2 == -4); // Verify that the value is correct.
mpz_clear(m); // Clear the mpz_t.
Second, it is possible to get a reference to an mpz_t from an integer via the get_mpz_t() member function. This member function will first switch the calling integer to dynamic storage (if the calling integer is not already employing dynamic storage), and it will then return a raw non-owning pointer which can be used both as a const and mutable parameter in the GMP API. For example:
mpz_t b;
mpz_init_set_si(b, -4); // Init an mpz_t with the value -4.
int_t a, c{2}; // Init two integers.
mpz_add(a.get_mpz_t(), b, c.get_mpz_t()); // Compute b + c via the GMP API, storing the result in a.
assert(a == -2); // Verify that the result is correct.
mpz_clear(b); // Clear the mpz_t.
It is important to emphasise that get_mpz_t() forces the use of dynamic storage, thus incurring in a potential performance hit. If only const access is needed, a better alternative to get_mpz_t() is the get_mpz_view() member function. get_mpz_view() returns a read-only view of the calling integer which is implicitly convertible to a const mpz_t, and which can thus be used as a non-mutable function parameter in the GMP API. The creation of the read-only view is lightweight, and, crucially, it does not force the use of dynamic storage in the calling integer. We can slightly modify to previous example to use a read-only view as the third parameter in the mpz_add() call, and verify that the creation of the read-only view did not trigger a promotion from static to dynamic storage:
mpz_t b;
mpz_init_set_si(b, -4); // Init an mpz_t with the value -4.
int_t a, c{2}; // Init two integers.
mpz_add(a.get_mpz_t(), b, c.get_mpz_view()); // Compute b + c via the GMP API, storing the result in a.
assert(a == -2); // Verify that the result is correct.
assert(c.is_static()); // Verify that c is still using static storage.
mpz_clear(b); // Clear the mpz_t.
It must be noted that both get_mpz_t() and get_mpz_view() have to be used carefully, as they return non-owning objects which can easily lead to dangling pointers or references, if misused. The documentation of the two functions explains in detail some of the potential pitfalls that users need to be aware of.
## Serialisation¶
New in version 0.7.
mp++ provides a simple API for the (de)serialisation of integer objects into/from memory buffers and C++ streams. Possible uses of the serialisation API include persistence (e.g., saving/loading integer values to/from a file), the transmission of integer objects over the network (e.g., in distributed computing applications), inter-process communication, etc. The API consists of two main overloaded functions, mppp::integer::binary_save() and mppp::integer::binary_load() (plus their free-function counterparts).
Let’s see a few examples of the serialisation API in action:
int_t a{42}, b;
char buffer[1024]; // Provide ample storage for serialisation.
a.binary_save(buffer); // Serialise a into the buffer.
b.binary_load(buffer); // Deserialise the content of the buffer into b.
assert(b == a); // Check that the original value is recovered.
Here we are serialising the value a into a char buffer of size 1024. In this case we know that, since the original value is small, a buffer of 1024 bytes will provide more than enough space to store the serialised representation of a. The content of the buffer is then read back into the b variable, and we verify that b’s value is indeed equal to the value of a.
Additional overloads are available for the save/load functions, targeting std::vector<char>, std::array<char> and C++ streams. The std::vector<char> binary_save() overload takes care of enlarging the target vector as much as needed to store the serialised integer:
int_t a{42}, b;
std::vector<char> buffer; // An initially-empty vector buffer.
a.binary_save(buffer); // Serialise a into the buffer.
assert(buffer.size() != 0); // Check that the buffer was resized.
b.binary_load(buffer); // Deserialise the content of the buffer into b.
assert(b == a); // Check that the original value is recovered.
Both mppp::integer::binary_save() and mppp::integer::binary_load() retun a std::size_t value representing the number of bytes written/read during (de)serialisation, or zero if an error occurred. A return value of zero can occur, for instance, when serialising into a std::array<char> with insufficient storage:
int_t a{42}, b;
std::array<char, 1> buffer_small; // A std::array<char> with insufficient storage.
auto ret = a.binary_save(buffer_small); // Try to serialise a into the small buffer.
assert(ret == 0); // Verify that an error occurred.
std::array<char, 1024> buffer_large; // A std::array<char> with ample storage.
ret = a.binary_save(buffer_large); // Serialise a into the large buffer.
assert(ret != 0); // Check that no error occurred.
b.binary_load(buffer_large); // Deserialise the content of the buffer into b.
assert(b == a); // Check that the original value is recovered.
The amount of bytes of storage necessary to serialise an integer can always be computed via the mppp::integer::binary_size() function.
Overloads for C++ streams work in a (hopefully) unsurprising fashion:
int_t a{42}, b;
std::stringstream ss; // Let's use a string stream for this example.
a.binary_save(ss); // Serialise a into the stream.
b.binary_load(ss); // Deserialise the content of the stream into b.
assert(b == a); // Check that the original value is recovered.
A couple of important points about the serialisation API need to be emphasised:
• although mp++ does run some consistency checks during deserialisation, the API is not built to protect against maliciously-crafted data. Users are thus advised not to load data from untrusted sources;
• the current binary serialisation format is compiler, platform and architecture specific, it is not portable and it might be subject to changes in future versions of mp++. Users are thus advised not to use the binary serialisation format for long-term persistence or as a data exchange format: for such purposes, it is better to use the string representation of integer objects.
## User-defined literals¶
New in version 0.18.
User-defined literals are available to construct mppp::integer instances with 1, 2 and 3 limbs of static storage. The literals are defined within the inline namespace mppp::literals, and, like the builtin C++ integer literals, they support binary, octal, decimal and hexadecimal representations:
using namespace mppp::literals;
auto n1 = 123_z1; // n1 has 1 limb of static storage,
// and it contains the value 123.
auto n2 = -0b10011_z2; // n2 has 2 limbs of static storage,
// and it contains the value -19
// (-10011 in base 2).
auto n3 = 0146_z1; // n3 has 1 limb of static storage,
// and it contains the value 102
// (146 in base 8).
auto n4 = 0xfe45_z3; // n4 has 3 limbs of static storage,
// and it contains the value 65093
// (fe45 in base 16).
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Home / Weight and Mass Conversion / Convert Hectogram to Ton (assay) (US)
# Convert Hectogram to Ton (assay) (US)
Please provide values below to convert hectogram [hg] to ton (assay) (US) [AT (US)], or vice versa.
From: hectogram To: ton (assay) (US)
### Hectogram to Ton (assay) (US) Conversion Table
Hectogram [hg]Ton (assay) (US) [AT (US)]
0.01 hg0.0342857104 AT (US)
0.1 hg0.3428571037 AT (US)
1 hg3.4285710367 AT (US)
2 hg6.8571420735 AT (US)
3 hg10.2857131102 AT (US)
5 hg17.1428551837 AT (US)
10 hg34.2857103673 AT (US)
20 hg68.5714207347 AT (US)
50 hg171.4285518367 AT (US)
100 hg342.8571036735 AT (US)
1000 hg3428.5710367347 AT (US)
### How to Convert Hectogram to Ton (assay) (US)
1 hg = 3.4285710367 AT (US)
1 AT (US) = 0.2916667 hg
Example: convert 15 hg to AT (US):
15 hg = 15 × 3.4285710367 AT (US) = 51.428565551 AT (US)
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We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
www.vustudents.ning.com
www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More
# Mth 401 All Current Final Term Papers Fall 2014 ~ From (01 March , 2014 to 13 March 2014 )
..How to Join Subject Study Groups & Get Helping Material?..
Views: 2309
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+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)
+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
### Replies to This Discussion
hmmm....thank u so much
me na ni parha kuch b khas
paper was easy..
egien values and vector 2marks 3marks and even 5marks
integrating factor malum karna tha equation given the. 5marks
power series sy reletd 2questions thy 2 and 3 marks
worksin nekalna tha 5marks
legendre's polynomial of degree 3marks
X= AX+F(t) ke form me likhna tha 5marks
mcqs me zayda ter egien values and vetcor or kuch power series waly thy
fari thanks for sharing ur paper..best of luck for ur result
Attention Students: You don’t need to go any other site for current papers pattern & questions. Because all sharing data related to current Final term papers of our members are going from here to other sites. You can judge this at other sites yourself. So don’t waste your precious time with different links. Just keep visiting http://vustudents.ning.com/ for all latest updates.
gud . thnks for sharing
Thanks fari
MTH401 today paper 3/10/2014
40 MCQS mostly in past papers
41) Explain series that are identical are identically zero? (2)
42) Define Legendre equation polynomials? (2)
43) Find wronskian of two vectors? (2)
44) Find integrating factor? (3)
45) Example of power series? (3)
46) Orthogonality condition of Legendre polynomials using delta &m, n? (3)
47) Write system in matrix form? (3) Given data.
48) Straight lines x+y=cx c is a arbiter constant? (3)
49) Identical equation and identical roots in belles differential equation (5)
50) Find eigenvectors given eigenvalues? (5)
Thank you so much Jia .......
U WELCOME
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Science Forums
# Physics Question?
## Recommended Posts
I am not so sure our approach will work in this case ron.. well not yet at any rate, we first need to find what C or T is
I am not sure about T, but if c is the force on the Y sphere, can't we use coulomb's law?
F = {Kc* lq1l *lq2l}/r^2
(Kc times absolute value of test charge times absolute value of source charge divided by r squared)
constant: Kc = 9.0x10^9 Nm^2C^-2
BTW...thanks for the replies so far
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well then I think we would have to have more info about the charges, clearly they must be opposite since they repel, but are they equal in magnitude?
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Well, you do know [math]c[/math]. All you need to do now is to take the terms two-at-a-time and find the respective forces.
However you'll need to know the values of the 'X' and 'Y' charges for this to be solvable for numerical values.
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Well, you do know [math]c[/math]. All you need to do now is to take the terms two-at-a-time and find the respective forces.
However you'll need to know the values of the 'X' and 'Y' charges for this to be solvable for numerical values.
Thank you! That is exactly right. After finding "c," all I had to do was sum the forces in the x diretion to find "T" (tension) . Then plug tension into the sum the forces in the y direction, and then solve for mass.
F(x) = c + (-Tsin30) = 0
F(y) = -mg + (T cos30) = 0
m = (Tcos30)/g
It was simple, but the fundamental part I previously overlooked was the fact that the objects were not moving therfore, sum of the forces in the x and y direction equal zero.
Thanks to all who helped!
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Taki,
I love your enthusiasm. Keep up your hard work and engagement with learning. It's the good kind of contagious. :lol:
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J C.
Present value of an amount
John and Marie’s son Sean is saving a deposit for a house and as they would like to help him, so they have decided to give him his funds early (before he turns 30). They want to be fair and ensure he gets the same amount of money as their daughter Michelle which will be \$71,165.59 at age 30.
What is the present value of this amount if Sean was to receive it when he turns 30? He is currently 26.
I.e. what is the Present Value of that amount if received in 4 years’ time at an interest rate of 4% pa compounding?
I need to use the finance calculator on arachnoid.com to calculate this answer but I am not sure what to put into the different areas. I would appreciate some help with this.
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7.2.4.3.3. Board Skew parameters for RLDRAM II and RLDRAM 3
The following paragraphs describe board skew parameters for RLDRAM II and RLDRAM 3 interfaces.
Maximum CK delay to device
The delay of the longest CK trace from the FPGA to any device/DIMM is expressed by the following equation:
where n is the number of memory clocks. For example, the maximum CK delay for two pairs of memory clocks is expressed by the following equation:
Maximum DK delay to device
The delay of the longest DK trace from the FPGA to any device/DIMM is expressed by the following equation:
where n is the number of DK. For example, the maximum DK delay for two DK is expressed by the following equation:
Minimum delay difference between CK and DK
The minimum delay difference between the CK signal and any DK signal when arriving at the memory device(s). The value is equal to the minimum delay of the CK signal minus the maximum delay of the DK signal. The value can be positive or negative.
The minimum delay difference between CK and DK is expressed by the following equations:
where n is the number of memory clocks and m is the number of DK. For example, the minimum delay difference between CK and DK for two pairs of memory clocks and four DK signals (two DK signals for each clock) is expressed by the following equation:
Maximum delay difference between CK and DK
The maximum delay difference between the CK signal and any DK signal when arriving at the memory device(s). The value is equal to the maximum delay of the CK signal minus the minimum delay of the DK signal. The value can be positive or negative.
The maximum delay difference between CK and DK is expressed by the following equations:
where n is the number of memory clocks and m is the number of DK. For example, the maximum delay difference between CK and DK for two pairs of memory clocks and four DK signals (two DK signals for each clock) is expressed by the following equation:
Maximum delay difference between devices
The maximum delay difference of data signals between devices is expressed by the following equation:
For example, in a two-device configuration there is greater propagation delay for data signals going to and returning from the furthest device relative to the nearest device. This parameter is applicable for depth expansion. Set the value to 0 for non-depth expansion design.
Maximum skew within QK group
The maximum skew between the DQ signals referenced by a common QK signal.
Maximum skew between QK groups
The maximum skew between QK signals of different data groups.
The maximum skew between the address/command signals.
Average delay difference between address/command and CK
A value equal to the average of the longest and smallest address/command signal delay values, minus the delay of the CK signal. The value can be positive or negative.
The average delay difference between the address and command and CK is expressed by the following equation:
where n is the number of memory clocks.
Average delay difference between write data signals and DK
A value equal to the average of the longest and smallest write data signal delay values, minus the delay of the DK signal. Write data signals include the DQ and DM signals. The value can be positive or negative.
The average delay difference between DQ and DK is expressed by the following equation:
where n is the number of DK groups.
Average delay difference between read data signals and QK
A value equal to the average of the longest and smallest read data signal delay values, minus the delay of the QK signal. The value can be positive or negative.
The average delay difference between DQ and QK is expressed by the following equation:
where n is the number of QK groups.
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8 Ahson/RFID Handbook: Applications, Technology, Security, and Privacy 54996_C032 Page Proof page :37pm Compositor Name: JGanesan 596 RFID Handbook: Applications, Technology, Security, and Privacy Step 1 Step 2 b 11 b 1n b 21 b 2n a 11 a 1n b 11 b 1n 1 b 1n a n 11 a n 1, n a n 21 a n,1 a nn a n 11 a n 2n 1 a n 1n 1 a n 2n a n 1n FIGURE 32.4 Operation of PISP. Suppose that for any sequence of authentication messages of length n, at least one message is not received by the intruder. In order to compromise the authentication algorithm the intruder has to perform authentication procedure similar to the tag. To do so the intruder has to forge the key message s ji the intruder has to correctly guess the XOR of the corresponding (n (j i 1) (mod(n)))th column elements of the matrix B. Recall that dim (B) ¼ n. Let the intruder be unfamiliar with the authentication message s 1 ¼ (X 1, b 1 ) sent by the tag to the reader during the first authentication session (Figure 32.4, Step 1). That is, the intruder does not know the nth column of B (a 1n, a 2n,...,a nn ) and the appropriate row vector is b 1 ¼ (b 11, b 12,..., b 1n ). After the tag transmits the authentication message s 1 ¼ (X 1, b 1 ), X 1 ¼ (a 1n a 2n a nn ), b 1 ¼ (b 11, b 12,..., b 1n ), both the tag and the reader shift the rows of B down as described earlier. Note that in the next session the tag will send to the reader the XOR of the updated (n 1)th B 0 s column X 2 ¼ (b 1n 1 a 1n 1 a n 1n 1 ) and a new randomly generated vector b 2 ¼ (b 2n, b 2n 1,..., b 21 ) (Figure 32.4, Step 2). Now matrix B differs from the previous one by the newly inserted first row and the appropriate deletion of the last row. The PISP is secure from the information-theoretic standpoint, or unconditionally secure as was defined in [32, Chapter 2]. That is, the probability that the intruder will forge the key message and successfully perform session on behalf of the reader, is negligible for long enough l, where l is the number of bits of the element of B. Note also that the matrix size n factors in the security of PISP. If n is large the probability that the intruder does not overhear at least one authentication session is greater. However, large n and l require greater storage resources on the tag. Note that if the intruder is allowed to eavesdrop n consecutive sessions, it can desynchronize the tag and the reader by forcing the reader to replace a row in B which is unknown to the tag. in any authentication session S th j. For that Proactive Computationally Secure Protocol We now describe the proactive computationally secure protocol (PCSP). It is the extension of PISP that allows us to lift the assumption on limited intruder eavesdropping. The pseudocode for the protocol is shown in Figure As in PISP, the reader and the tag
12 Ahson/RFID Handbook: Applications, Technology, Security, and Privacy 54996_C032 Page Proof page :37pm Compositor Name: JGanesan 600 RFID Handbook: Applications, Technology, Security, and Privacy 1: Initialization 2: constants 3: q : integer {key size} 4: k[1..q]: integer {key} 5: variables 6: collide : boolean {trial outcome} 7: cfront, pfront : integer, initially 0 8: {currently and previously number of growth points in front} 9: cback, pback: integer, initially 0, 10: {currently and previously number of growth points behind} 11: Operation 12: for i := 1 to q do 13: for j := 1 to pfront do 14: collide := trial() 15: cfront := cfront : if collide = true, then cfront := cfront : collide := trial() 18: if collide = true, then 19: if key[i] = 0 then 20: cback := cback : else 22: cfront := cfront : for j := 1 to pback do 24: collide := trial() 25: cback := cback : if collide = true, then cback := cback : pback := cback, cback := 0, pfront := cfront, cfront := 0 FIGURE 32.7 Tag-side singulation algorithm. such as retail item-tagging where such heavyweight solutions are prohibitively expensive, low overhead cryptography may provide a convenient alternative. Acknowledgments This work was partially supported by Rita Altura Trust Chair in Computer Sciences, Lynne and William Frankel Center for Computer Sciences and Deutsche Telekom, DARPA contract OSU-RF #F C-1901, NSF CAREER Award , and a grant from the Wright Center for Sensor System Engineering.
13 Ahson/RFID Handbook: Applications, Technology, Security, and Privacy 54996_C032 Page Proof page :37pm Compositor Name: JGanesan Low Overhead RFID Security 601 References 1. G. Avoine. Bibliography on security and privacy in RFID systems. Available Online, G. Avoine and P. Oechslin. RFID traceability: A multilayer problem. In Financial Cryptography (FC), Roseau, The Commonwealth Of Dominica, February March Lecture Notes in Computer Science, vol. 3570, pp IFCA, Springer-Verlag, L. Batina, J. Guajardo, T. Kerins, N. Mentens, P. Tuyls, and I. Verbauwhede. An elliptic curve processor suitable for RFID-tags. Cryptology eprint Archive, Report 2006=227, M. Burmester, T. van Le, and B. de Medeiros. Provably secure ubiquitous systems: Universally composable RFID authentication protocols. In Conference on Security and Privacy for Emerging Areas in Communication Networks (SecureComm), Baltimore, Maryland, USA, August September IEEE, B. Defend, K. Fu, and A. Juels. Cryptanalysis of two lightweight RFID authentication schemes. In International Workshop on Pervasive Computing and Communication Security (PerSec) 2007, New York, USA, March IEEE Computer Society Press. 6. S. Dolev and M. Kopeetsky. Secure communication for RFIDs proactive information security within computational security. In Stabilization, Safety, and Security of Distributed Systems (SSS), Lecture Notes in Computer Science, vol pp Springer, M. Feldhofer, S. Dominikus, and J. Wolkerstorfer. Strong authentication for RFID systems using the AES algorithm. In Workshop on Cryptographic Hardware and Embedded Systems (CHES), Boston, Massachusetts, USA, August Lecture Notes in Computer Science, vol pp IACR, Springer-Verlag, K. Finkenzeller. RFID Handbook: Fundamentals and Applications in Contactless Smart Cards and Identification. John Wiley, New York, NY, USA, C. Floerkemeier, R. Schneider, and M. Langheinrich. Scanning with a purpose supporting the fair information principles in RFID protocols. In International Symposium on Ubiquitous Computing Systems (UCS), Tokyo, Japan, November Lecture Notes in Computer Science, vol pp Springer-Verlag, G. Hancke. Practical attacks on proximity identification systems (short paper). In IEEE Symposium on Security and Privacy, Oakland, California, USA, May 2006 pp IEEE Computer Society Press, T.S. Heydt-Benjamin, D.V. Bailey, K. Fu, A. Juels, and T. O Hare. Vulnerabilities in firstgeneration RFID-enabled credit cards. Manuscript, October A. Juels. Minimalist cryptography for low-cost RFID tags. In International Conference on Security in Communication Networks (SCN), Amalfi, Italia, September Lecture Notes in Computer Science, vol p Springer-Verlag, A. Juels. RFID security and privacy: A research survey. Manuscript, September A. Juels and J. Brainard. Soft blocking: Flexible blocker tags on the cheap. In Workshop on Privacy in the Electronic Society (WPES), Washington, DC, USA, October 2004, p ACM Press, A. Juels, R. Rivest, and M. Szydlo. The blocker tag: Selective blocking of RFID tags for consumer privacy. In Conference on Computer and Communications Security (CCS), Washington, DC, USA, October 2003, p ACM Press, G. Karjoth and R. Moskowitz. Disabling RFID tags with visible confirmation: Clipped tags are silenced. In Workshop on Privacy in the Electronic Society (WPES), Alexandria, Virginia, USA, November ACM Press, S. Karthikeyan and M. Nesterenko. RFID security without extensive cryptography. In Proceedings of the 3rd ACM workshop on security of ad hoc and sensor networks (SASN), New York, NY, USA, 2005, pp ACM Press, C. Kaufman, R. Perlman, and M. Speciner. Network Security: Private Communication in a Public World. Prentice- Hall, T. Li and R.H. Deng. Vulnerability analysis of EMAP an efficient RFID mutual authentication protocol. In Second International Conference on Availability, Reliability and Security (AReS), Vienna, Austria, April 2007.
14 Ahson/RFID Handbook: Applications, Technology, Security, and Privacy 54996_C032 Page Proof page :37pm Compositor Name: JGanesan 602 RFID Handbook: Applications, Technology, Security, and Privacy 20. T. Li and G. Wang. Security analysis of two ultra-lightweight RFID authentication protocols. In International Information Security Conference (IFIP SEC), Sandton, Gauteng, South Africa, May IFIP, A.J. Menezes, S.A. Vanstone, and P.C. Van Oorschot. Handbook of Applied Cryptography. CRC Press, Boca Raton, FL, USA, D. Molnar and D. Wagner. Privacy and security in library RFID: Issues, practices, and architectures. In ACM Conference on Computer and Communications Security (CCS), Washington, DC, USA, October 2004, pp , ACM Press, M. Ohkubo, K. Suzuki, and S. Kinoshita. Cryptographic approach to privacy-friendly tags. In RFID Privacy Workshop, MIT, MA, USA, November P. Peris-Lopez, J.C. Hernandez-Castro, J. Estevez-Tapiador, and A. Ribagorda. LMAP: A real lightweight mutual authentication protocol for low-cost RFID tags. In Workshop on RFID Security (RFIDSec), Graz, Austria, July Ecrypt. 25. P. Peris-Lopez, J.C. Hernandez-Castro, J. Estevez-Tapiador, and A. Ribagorda. M2AP: A minimalist mutual- authentication protocol for low-cost RFID tags. In International Conference on Ubiquitous Intelligence and Computing (UIC), Lecture Notes in Computer Science, vol pp Springer-Verlag, P. Peris-Lopez, J.C. Hernandez-Castro, J.M. Estevez-Tapiador, and A. Ribagorda. EMAP: An efficient mutual authentication protocol for low-cost RFID tags. In OTM Federated Conferences and Workshop: IS Workshop, Lecture Notes in Computer Science, vol pp Springer- Verlag, A. Poschmann, G. Leander, K. Schramm, and C. Paar. A family of light-weight block ciphers based on DES suited for RFID applications, July M. Rieback, G. Gaydadjiev, B. Crispo, R. Hofman, and A. Tanenbaum. A platform for RFID security and privacy administration. In USENIX=SAGE Large Installation System Administration conference (LISA), Washington DC, USA, December G. Roussos. Enabling RFID in retail. IEEE Computer, 39(3):25 30, K. Sakiyama, L. Batina, N. Mentens, B. Preneel, and I. Verbauwhede. Smallfootprint ALU for public-key processors for pervasive security. In Workshop on RFID Security (RFIDSec), Graz, Austria, July Ecrypt. 31. W. Stallings. Cryptography and Network Security: Principles and Practices, 2nd end. Prentice-Hall, Upper Saddle River, NJ, USA, D. Stinson. Cryptography: Theory and Practice, 3rd end CRC=C&H, G. Tsudik. YA-TRAP: Yet another trivial RFID authentication protocol. In International Conference on Pervasive Computing and Communications (PerCom), Pisa, Italy, March IEEE Computer Society Press, I. Vajda and L. Buttyán. Lightweight authentication protocols for low-cost RFID tags. In Second Workshop on Security in Ubiquitous Computing (Ubicomp), Seattle, WA, USA, October 2003.
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# A certain quantity is measured on two different scales, the
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A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
A. 20
B. 36
C. 48
D. 60
E. 84
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08 Oct 2006, 12:36
48
R) 24 - 6 = 18
S) 60 - 30 = 30
R = 3/5 S
R = 3/5 (100 - 60) = 24
24 + 24 = 48
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08 Oct 2006, 12:36
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Game Development Reference
In-Depth Information
Figure 15.6 Ten-vertex stencil for new vertex assuming end-points have valence six.
calculate the location of the added vertex based solely on this
extraordinary vertex. This seems intuitively strange, but the result is very
effective. One of the authors of the original paper, Denis Zorin, justifies the
choice of weighting on his website at www.mrl.nyu.edu/dzorin. If both end-
points are extraordinary, then we calculate the location of the added
vertex as the average of the position calculated using each extraordinary
vertex.
The scheme used is:
Valence 3:
v : 4 ;
e 0 :
12 ;
5
e 1 :- 12 ;
e 2 :- 12
Valence 4:
v : 4 ;
e 0 : 8 ;
e 1 : 0, e 2 :- 8 ;
e 3 : 0
1
4 + cos(2
i / N )+ 2 cos(4
i / N )
Valence 5 and 7 or more:
v : 4 ;
e i :
N
where N is the valence.
Border vertices are handled differently from other vertices, taking all the
weighting from adjacent border points. The new edge vertex is calculated
using
V =
16 a +
9
16 b -
9
16 c -
1
16 d
1
Using this technique for border vertices ensures that the mesh doesn't
pull away from any boundary edges.
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This thread is privately moderated by Robert Burson, who may elect to delete unwanted replies.
Dec 17, 2012, 01:51 PM AZ Outback United States, AZ, Chino Valley Joined Dec 2000 1,331 Posts FAI F5J 2013 New Years day and Janurary and Feburary Since at some point in time it will necessary to create a team select for WC in F5J, I would like to have at least a RC-Group competition, to demonstrate what is going on in the world and the possible equipment that would be successful on a WC level. At this point in time I am hosting a New Years day, and Janurary competition for F5J. The rules are below and only represent a generalize competition, however realize in almost every round at a World level someone always makes time and a landing of 50 points. Since current rules allow for a 30s motor run, in the 10minute flight window it will be possible to do exactly 10:00 minutes of flight time. So starting height becomes the discrimnator for the final score as you have 1/2pt subtracted from you total score for each meter of vertical height you launch to, and 3pts per meter if your starting height is 200M or more. To normalize your score I will have to do a little math. You will have three values, starting height in meters ( -.5 points per meter), flight ime 1 point per second, and landing points up to 100pts. Example: starting height 150M so -0.5 X 150 = -75.0 pts flight time 9:58s or 9min x 60pts + 58s x 1pt = 598.0 pts landing points say 45pts total 568 pts (raw score) If this where the highest point total flying againt five other competitors then you would receive 1000pts for the round. If the second highest raw score was 550pt then this pilot would receive 550/568 X 1000pt = 968.83pts As you can see only 18pts(raw score) or maybe just a height difference of 36M with each having the same flight and landing seperated the two scores. So in the current rules starting height is a major player in determining final scores. Since I do not see more than 6 to 12 scores on New Years or Janurary I will just normalize for the group once you have entered your raw scores. You may submit only one score for New Years and a new single score of Janurary. At the end of the month, all raw scores will be normalized. The final requirement is you must submit a graph from your flight recording showing starting height and landing time, I will just trust everyone on landing points. Pictures and equipment list are always welcome. I realize this event will only appeal to a very small group of pilots as does F5B and F5D, but at least we will have some pilots looking at a new and very diffcult event on the World level of flying. Please remember anyone with a electric sailplane height recorder and landing tape can fly this event. FAI F5J generalized TASK: Fly one 10min flight. 1. Any airframe, battery, ESC, prop, motor maybe used. To FAI limits. 2. 10:00 minute window, with flight points starting on launch. Over flying window 0 landing points. 3. A device to record flight starting height, after a 30s or less motor run, at the beginning of the window. To establish starting height fly level for 10s after motor run. 4. .5pts subtracted for each meter of starting height, 3pts subtracted for each meter above 200M, from possible 10minute flight time ((600pts or number of flight seconds) - (starting height(.5 or/and 3))). 5. 50pt. landing tape, - 5pts per meter from 50pt end of tape. Landing points will just be added on since you are not flying in a group to normalize scores. Hope you can joins us for the fun. Last edited by Robert Burson; Feb 12, 2013 at 11:57 AM. Reason: units
Dec 28, 2012, 12:01 PM
AZ Outback
United States, AZ, Chino Valley
Joined Dec 2000
1,331 Posts
Some of the equipment used during provisional events in Europe.
Images
Last edited by Robert Burson; Dec 28, 2012 at 01:00 PM.
Dec 30, 2012, 10:47 AM
AZ Outback
United States, AZ, Chino Valley
Joined Dec 2000
1,331 Posts
Finals group of flyers from Eurotour, in 2012
Images
Dec 30, 2012, 10:54 AM Registered User Miamisburg OH Joined Dec 2003 425 Posts Bob Now that you are in the sunny southwest you will need to change your avatar. Or are you missing the snow, ice, wind and cold? John Lueke
Dec 30, 2012, 11:11 AM AZ Outback United States, AZ, Chino Valley Joined Dec 2000 1,331 Posts Hi John, We do have snow here, but it is usually gone by noon. I am two hours from Phoenix and Flagstaff, so warm sun or snow skiing on any day, this time of year. Hope you have a great New Years. The Best Soaring
Dec 30, 2012, 03:21 PM Registered User Madrid, Spain Joined Feb 2005 195 Posts Hello Robert, I hope you are doing well. Quick questions: 1) Is it needed to be registered before the 1st Jan? 2) I understand you are not following FAI rules for altitude over 200m, isn't it? I'm preparing a team meeting for New Years day here in Madrid and wanted to include many people as possible. Thanks, Javier.
Dec 30, 2012, 06:48 PM Registered User Australia, NSW Joined Sep 2006 52 Posts F5J Australia Hi Robert Paul Osmond from Australia We have been flying F5j Comp rules for the past 18 months and took part in the Euro F5J challenge to gage wear we were in the development of the class, the past 12months has seen a definite increase in participation. the choice of equipment has been very diverse from the humble Radian which has equipped its self very well to the hi-tech composite gliders but it is apparent that the use of large 3.2 mt plus gliders do the best overall. I currently use a Supper AVA 4 flap as my light air model and a Suppra Ultra Light Hybrid as my general purpose model . I hope to submit a score for January 1 to aid in the comparison of the Class i will add photos and height extracts . All the best for the New Year Regards Paul
Dec 31, 2012, 10:47 AM AZ Outback United States, AZ, Chino Valley Joined Dec 2000 1,331 Posts Javier, Happy New Years Eve from the US, no you do not have to sign-up before you fly, I am making this as easy as possible for everyone. New Years Day will be the only day as a date, the rest of January is open to when you can fly, so everyone can decide what works best for them. Thank You for the 200M reminder, I did that once and it was so terrible on my score I never when over the height again, but I will fix that in the first post. You always supply great photos, so I am looking forward to even more. Happy New Years Robert
Dec 31, 2012, 10:57 AM AZ Outback United States, AZ, Chino Valley Joined Dec 2000 1,331 Posts Paul, Thank You, for the overview, of what is happening in the land of down under. The more information the better, the US is a little slower to respond to new rule sets, so I am hoping we can start to have a few events, here in the US. The only new requirement is to have a starting height recording on the score sheet, I think this will be a problem for a while until we develop a little more experience. I see some where in the future, our planes will send this wireless to an app which records the starting height of each plane, for each round, but that maybe some time in the future. Happy New Years Robert Last edited by Robert Burson; Dec 31, 2012 at 11:09 AM.
Jan 01, 2013, 04:06 PM
Registered User
Joined Feb 2005
195 Posts
Quote:
Originally Posted by Robert Burson Javier, Happy New Years Eve from the US, no you do not have to sign-up before you fly, I am making this as easy as possible for everyone. New Years Day will be the only day as a date, the rest of January is open to when you can fly, so everyone can decide what works best for them. Thank You for the 200M reminder, I did that once and it was so terrible on my score I never when over the height again, but I will fix that in the first post. You always supply great photos, so I am looking forward to even more. Happy New Years Robert
Thanks Robert, here you have some pics from today's flights.
http://f5j.es/2013/01/01/comenzando-...e-el-ano-2013/
I'll send you the flight logs as soon as we can.
Jan 01, 2013, 06:23 PM Registered User Australia, NSW Joined Sep 2006 52 Posts Flight resault 1-1-13 Hi Robert Hear is my first entry as you can see in the pics the weather was very foggy had to wait a wile befor i could get a flight in there was not much lifit to be had so had to use more motor than than i would like . Recorded flight time 9.53 = 593 Launch hight 174.7 = 87.35 - Landing inside 1 mt = 45 total 550.65 Best Regards Paul
Jan 01, 2013, 09:30 PM
AZ Outback
United States, AZ, Chino Valley
Joined Dec 2000
1,331 Posts
Great New Years weather in Mesa, AZ. Got out to the field around noon, and launched around 12:30 in light winds, and 90% sun with high cirrus taking the rest. Good launch height and time at 9:57.7 and 50pts. As soon as I find the cable to the F5J height recorder, still not unpacked from the move to the West, I will post up my score.
The Best Soaring
Robert
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# SCM Repository
[matrix] Diff of /pkg/src/bCrosstab.c
[matrix] / pkg / src / bCrosstab.c
# Diff of /pkg/src/bCrosstab.c
revision 463, Sat Jan 29 14:11:34 2005 UTC revision 464, Sat Jan 29 14:12:53 2005 UTC
# Line 27 Line 27
27 } }
28
29 /** /**
30 * Calculate the zero-based index in a row-wise packed lower * Calculate the zero-based index in a row-wise packed lower triangular matrix.
31 * triangular matrix. This is used for the arrays of blocked sparse matrices. * This is used for the arrays of blocked sparse matrices.
32 * *
33 * @param i row number (0-based) * @param i column number (zero-based)
34 * @param k column number (0-based) * @param k row number (zero-based)
35 * *
36 * @return The 0-based index of the (i,k) element of a row-wise packed lower * @return The index of the (k,i) element of a packed lower triangular matrix
* triangular matrix.
37 */ */
38 static R_INLINE int static R_INLINE
39 Lind(int i, int k) int Lind(int k, int i)
40 { {
41 return (i * (i + 1))/2 + k; if (k < i) error("Lind(k = %d, i = %d) must have k >= i", k, i);
42 return (k * (k + 1))/2 + i;
43 } }
44
45 /** /**
Legend:
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12 December, 17:37
# What is 6 over 21 in simplest form
0
1. 12 December, 17:43
0
6/21
Divide the numerator and denominator by using 3.
6/3=2
21/3=7
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## Count amount of duplications/COUNTIFS for column
Hello All,
I am trying to find amount of duplications within a database, based on certain columns.
Below is a sample of the data I have:
• ID, VednorCode, ItemName is the raw data.
• Duplications is what I am trying to get (In Yellow)
ID VendorCode ItemName Duplications 1 H90042800 6251 3 2 H90042800 6251 3 3 H90042800 6251 3 4 H90042800 PD-6251-300 2 5 H90042800 PD-6251-300 2 6 H90042800 PD-6252-300 1 7 H03000773 PD-6252-300 1 8 H03000773 PD-6011-300 3 9 H03000773 PD-6011-300 3 10 H03000773 PD-6011-300 3
Duplications should show the total amount of times each combination of VednorCode and ItemName appears.
Duplications amount in each line refers to VendorCode and ItemName in the same line.
Basically, it is the same as if I would use COUNTIFS(B:B,\$B2,C:C,\$C2) in excel.
Any ideas?
1 Solution
Accepted Solutions
MVP
May be like this
Count(TOTAL <VendorCode, ItemName> ID)
6 Replies
Why 1, 1 for Item code of PD-6252-300 ?? I am assuming, It would be 2, 2
Count(TOTAL <ItemName> ItemName)
Before develop something, think If placed (The Right information | To the right people | At the Right time | In the Right place | With the Right context)
MVP
May be like this
Count(TOTAL <VendorCode, ItemName> ID)
Creator
Author
Hi Anil,
It is 1,1 because the VendorCode is different.
I want to count the amount of times each unique combination VednorCode and ItemName appears.
MVP
Sample attached
After that, I came to know
Count(TOTAL <VendorCode,ItemName> ItemName)
Before develop something, think If placed (The Right information | To the right people | At the Right time | In the Right place | With the Right context)
Creator
Author
Many thanks Sunny and Anil
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# Convert 364 inches to mm
## How many mm is 364 inches?
364 Inches is equal to 9245.6 Millimeters
## Explanation of 364 Inches to Millimeters Conversion
Inches to Millimeters Conversion Formula: mm = in × 25.4
According to 'inches to mm' conversion formula if you want to convert 364 (three hundred sixty-four) Inches to Millimeters you have to multiply 364 by 25.4.
Here is the complete solution:
364″ × 25.4
=
9245.6 mm
(nine thousand two hundred forty-five point six millimeters)
## About "Inches to Millimeters" Calculator
This converter will help you to convert Inches to Millimeters (in to mm). For example, it can help you find out how many mm is 364 inches? (The answer is: 9245.6). Enter the number of inches (e.g. '364') and hit the 'Convert' button.
## Inches to Millimeters Conversion Table
InchesMillimeters
8864.6 mm
8890 mm
8915.4 mm
8940.8 mm
8966.2 mm
8991.6 mm
9017 mm
9042.4 mm
9067.8 mm
9093.2 mm
9118.6 mm
9144 mm
9169.4 mm
9194.8 mm
9220.2 mm
9245.6 mm
9271 mm
9296.4 mm
9321.8 mm
9347.2 mm
9372.6 mm
9398 mm
9423.4 mm
9448.8 mm
9474.2 mm
9499.6 mm
9525 mm
9550.4 mm
9575.8 mm
9601.2 mm
## FAQ
### How many mm is 364 inches?
364″ = 9245.6 mm
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Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
Suppose we know the standard deviation of the population is 3. We have a sample size of 625. We also have a sample mean of 46. We also have a...
A. Suppose we know the standard deviation of the population is 3.7. We have a sample size of 625. We also have a sample mean of 46. We also have a population mean of 43. We want to test the hypothesis for the sample mean with 78% confidence level and do a two-tailed test on the hypothesis. State the null and alternate hypothesis for the sample mean and then test the hypothesis to see if you will accept or reject the hypothesis.
B. Suppose we know the sample standard deviation is 10. We have a sample size of 25. We also have a sample mean of 116. We also have a population mean of 118. We want to test the hypothesis for the sample mean with 98% confidence level and do a two-tailed test on the hypothesis. State the null and alternate hypothesis for the sample mean and then test the hypothesis to see if you will accept or reject the hypothesis.
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1. ## Permutation and Combination
Q: A rectangular table has 7 secured seats, 4 being on one side facing the window and 3 being on the opposite side. In how many ways can 7 people be seated at the table if 2 people, X and Y must sit on the same side?
The answer I was given is 2160, but I don't know how to derive it...>.< Thank you for helping!
2. Here is what i thought.
(4P2 x 5P5) + ( 3P2 X 5P5 ) = 2160
Hope it helps.
3. if X picked on four seat side then
4 choices for x* 3 choices for y ( has to be on same side)*P(5,5)=1440
if X picked on three seat side then
3 choices for x* 2 choices for y * p(5,5) = 720
Total arrangements 1440+720 = 2160
4. Hello, Tangera!
A rectangular table has 7 secured seats, 4 on one side, 3 on the other. In how many
ways can they be seated at the table if 2 people, $\displaystyle X$ and $\displaystyle Y$ must sit on the same side?
Suppose $\displaystyle X$ and $\displaystyle Y$ sit on the 4-side of the table.
. . X has 4 choices of seats; Y has 3 choices of seats.
. . The other five can be seated in $\displaystyle 5!$ ways.
There are: .$\displaystyle 4\cdot3\cdot5! \:=\:1440$ ways.
Suppose $\displaystyle X$ and $\displaystyle Y$ sit on the 3-side of the table.
. . X has 3 choices of seats; Y has 2 choices of seats.
. . The other five can be seated in $\displaystyle 5!$ ways.
There are: .$\displaystyle 3\cdot2\cdot5! :=\:720$ ways.
Therefore, there are: .$\displaystyle 1440 + 720 \;=\;{\color{blue}2160}$ seating arrangements.
,
,
,
# a table has 7 seats 4 being on one side
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# semi-continuity
Okay, imagine you have a toy car and you want to know how fast it can go. You press a button and the car starts moving forward. Now, while the car is moving, you start pressing down on the button more and more. Sometimes, when you press down more, the car will go faster, but sometimes it will stay at the same speed.
This is kind of like semi-continuity, which is a way to talk about how things change when you change something else. In the case of the car, you're changing how much you press the button, and you're seeing how the car responds.
In math, we talk about functions, which are kind of like toy cars. A function takes in some number (like how much you're pressing the button), and then it spits out another number (like how fast the car is going). Just like the car, sometimes the function will change a lot when you change the input, and sometimes it will stay the same.
So, semi-continuity is a way to talk about functions and how they behave when you change the input a little bit. If you have a semi-continuous function, it means that if you change the input a little bit, the output will only change a little too. Just like with the car, sometimes a small change in the input will cause a big change in the output, but sometimes it will only cause a small change.
So, that's what semi-continuity means. It's a way to talk about how functions behave when you change the input a little bit, and whether or not they change a lot or just a little.
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Brahmagupta's Formula: Lesson Plans, Articles, Studies and Background Information
Science Experiments For science fair projects, lesson plans, classroom activities and research projects
• Repeat Famous Experiments and Inventions
• The Scientific Method - How to Experiment
• The Display Board
• Brahmagupta's Formula
Lesson Plans, Articles, Studies and Background Information
For Lesson Plans, Class Activities & Science Fair Projects
Experiments Home Mathematics Brahmagupta's Formula
Brahmagupta's Formula Rsources
Definition
Brahmagupta's formula gives the area of a general quadrilateral given the lengths of its sides and two angles.
In its basic form, Brahmagupta's formula gives the area of a cyclic quadrilateral (inscribed in a circle) whose side lengths are: a, b, c, d:
where s is the semiperimeter:
Background Information
• Brahmagupta's formula - MathWorld [View Resource]
• Brahmagupta's formula - Wikipedia [View Resource]
• Who Was Brahmagupta? - Full MacTutor biography [View Resource]
• Proof of Brahmagupta's formula - PlanetMath [View Resource]
• Brahmagupta's Formula Proof [View Resource]
• Brahmagupta's Formula [View Resource]
• Brahmagupta’s formula for area of a cyclic quadrilateral [View Resource]
• Hero’s and Brahmagupta’s Formulas [View Resource]
K-12 Lesson Plans and Science Fair Projects
• Triangular Discoveries: A Look into Heron's Formula and Beyond [View Resource]
• The Quadratic Formula [View Resource]
• Investigate whether it's possible to expand Hero’s Theorem and Brahmagupta’s Formula to find the area of any convex pentagon/hexagon [View Resource]
Articles and Studies
• A Proof of the Pythagorean Theorem From Heron's Formula [View Resource]
• Heron, Brahmagupta, Pythagoras, and the Law of Cosines Expository Paper [View Resource]
• Mechanical Formula Derivation in Elementary Geometries [View Resource]
• Cyclic Polygons with Rational Sides and Area [View Resource]
• Relationship Between Shadow Length and Time of Day [View Resource]
Theses and Dissertations
• Some congruence properties of Pell’s equation [View Resource]
• Solvability of polynomial equations over the rational field [View Resource]
• Early treatises introducing into Europe the Hindu art of reckoning (1914!) [View Resource]
• Heron's Formula [View Resource]
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# Question c4250
Jun 26, 2016
By definition, $\sec \theta = \frac{1}{\cos} \theta$.
#### Explanation:
Hence, $\cos \theta = - \frac{4}{5}$. We know that $\tan \theta > 0$, and the only quadrant where tangent is positive and the other trig ratios are negative is quadrant III.
Since we only know the side adjacent $\theta$ and the hypotenuse, we must find the side opposite $\theta$.
We can do this using Pythagorean theorem. Let a be $- 4$ and $c$ be 5.
${a}^{2} + {b}^{2} = {c}^{2}$
${b}^{2} = {c}^{2} - {a}^{2}$
${b}^{2} = {\left(5\right)}^{2} - {\left(- 4\right)}^{2}$
${b}^{2} = 25 - 16$
$b = \sqrt{9}$
$b = \pm 3$
We will take the $- 3$, because in quadrant three both opposite and adjacent sides to the angle $\theta$ will be negative.
Now that we know that
$\text{adjacent = -4}$
$\text{opposite} = - 3$
$\text{hypotenuse = 5}$
We can define cosine and cotangent. Cosine is adjacent/hypotenuse, and cotangent is 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite"#.
Applying these definitions to the problem at hand, we have:
$\cot \theta = \frac{4}{3}$
$\cos \theta = - \frac{4}{5}$
Now, adding these is simple arithmetic.
$\frac{4}{3} + \left(- \frac{4}{5}\right) = \frac{20}{15} - \frac{12}{15} = \frac{8}{15}$
Thus, $\cos \theta + \cot \theta = \frac{8}{15}$.
Hopefully this helps!
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American Institute of Mathematical Sciences
April 2018, 12(2): 315-330. doi: 10.3934/ipi.2018014
Hölder stability estimate in an inverse source problem for a first and half order time fractional diffusion equation
Department of Mathematical Sciences, The University of Tokyo, Komaba Meguro Tokyo 153-8914, Japan
Received November 2016 Revised November 2017 Published February 2018
We consider the first and half order time fractional equation with the zero initial condition. We investigate an inverse source problem of determining the time-independent source factor by the spatial data at an arbitrarily fixed time and we establish the conditional stability estimate of Hölder type in our inverse problem. Our method is based on the Bukhgeim-Klibanov method by means of the Carleman estimate. We also derive the Carleman estimate for the first and half order time fractional diffusion equation.
Citation: Atsushi Kawamoto. Hölder stability estimate in an inverse source problem for a first and half order time fractional diffusion equation. Inverse Problems and Imaging, 2018, 12 (2) : 315-330. doi: 10.3934/ipi.2018014
References:
[1] E. E. Adams and L. W. Gelhar, Field study of dispersion in a heterogeneous aquifer: 2.Spatial moments analysis, Water Resources Research, 28 (1992), 3293-3307. doi: 10.1029/92WR01757. [2] B. Amaziane, L. Pankratov and A. Piatnitski, Homogenization of a single-phase flow through a porous medium in a thin layer, Math. Models Methods Appl. Sci., 17 (2007), 1317-1349. doi: 10.1142/S0218202507002339. [3] A. Ashyralyev, Well-posedness of the Basset problem in spaces of smooth functions, Appl. Math. Lett., 24 (2011), 1176-1180. doi: 10.1016/j.aml.2011.02.002. [4] E. Bazhlekova and I. Dimovski, Exact solution of two-term time-fractional Thornley's problem by operational method, Integral Transforms Spec. Funct., 25 (2014), 61-74. doi: 10.1080/10652469.2013.815184. [5] A. L. Bukhgeim and M. V. Klibanov, Uniqueness in the large of a class of multidimensional inverse problems, Dokl. Akad. Nauk SSSR, 260 (1981), 269-272 (in Russian). [6] T. Carleman, Sur un problème d'unicité pour les systèmes d'équations aux derivées partielles à deux variables independantes, Ark. Mat. Astr. Fys., 26B (1939), 1-9. [7] J. Cheng, C.-L. Lin and G. Nakamura, Unique continuation property for the anomalous diffusion and its application, J. Differ. Equ., 254 (2013), 3715-3728. doi: 10.1016/j.jde.2013.01.039. [8] M. Eller and V. Isakov, Carleman estimates with two large parameters and applications, Proc. Conf. Differential-Geometric Methods in the Control of Partial Differential Equations (Boulder, CO, July 1999), Contemp. Math., 268 (2000), 117-136. [9] A. V. Fursikov and O. Y. Imanuvilov, Controllability of Evolution Equations, (Lecture Notes Series vol. 34) Seoul National University, Seoul (Korea), 1996. [10] Y. Hatano and N. Hatano, Dispersive transport of ions in column experiments: An explanation of long-tailed profiles, Water Resources Research, 34 (1998), 1027-1033. doi: 10.1029/98WR00214. [11] L. Hörmander, Linear Partial Differential Operators, Springer-Verlag, Berlin, 1963. [12] O. Y. Imanuvilov, Controllability of parabolic equations, Sbornik Math., 186 (1995), 879-900. [13] O. Y. Imanuvilov and M. Yamamoto, Lipschitz stability in inverse parabolic problems by the Carleman estimate, Inverse Problems, 14 (1998), 1229-1245. doi: 10.1088/0266-5611/14/5/009. [14] V. Isakov, Inverse Problems for Partial Differential Equations, 2nd edition, Springer-Verlag, Berlin, 2006. [15] V. Isakov and N. Kim, Carleman estimates with two large parameters for second order operators and applications to elasticity with residual stress, Appl. Math., 35 (2008), 447-465. doi: 10.4064/am35-4-4. [16] V. Isakov and N. Kim, Carleman estimates with second large parameter for second-order operators, Some Applications of Sobolev Spaces to PDEs, International Math. Ser., SpringerVerlag, 10 (2009), 135-159. [17] D. Jiang, Z. Li, Y. Liu and M. Yamamoto, Weak unique continuation property and a related inverse source problem for time-fractional diffusion-advection equations, Inverse Problems, 33 (2017), 055013 (22p). [18] B. Jin and W. Rundell, A tutorial on inverse problems for anomalous diffusion processes, Inverse Problems, 31 (2015), 035003 (40pp). [19] A. Kawamoto, Lipschitz stability estimates in inverse source problems for a fractional diffusion equation of half order in time by Carleman estimates, UTMS Preprint Series, UTMS 2016-3. [20] A. Kawamoto and M. Machida, Global Lipschitz stability for a fractional inverse transport problem by Carleman estimates, arXiv preprint, arXiv: 1608.07914 (2016). [21] M. V. Klibanov, Inverse problems and Carleman estimates, Inverse Problems, 8 (1992), 575-596. doi: 10.1088/0266-5611/8/4/009. [22] M. V. Klibanov, Carleman estimates for global uniqueness, stability and numerical methods for coefficient inverse problems, J. Inverse Ill-Posed Probl., 21 (2013), 477-560. [23] M. V. Klibanov and A. A. Timonov, Carleman Estimates for Coefficient Inverse Problems and Numerical Applications, VSP, Utrecht, 2004. [24] Z. Li, Y. Liu and M. Yamamoto, Initial-boundary value problems for multi-term time-fractional diffusion equations with positive constant coefficients, Appl. Math. Comput., 257 (2015), 381-397. doi: 10.1016/j.amc.2014.11.073. [25] Z. Li, O. Y. Imanuvilov and M. Yamamoto, Uniqueness in inverse boundary value problems for fractional diffusion equations Zhiyuan, Inverse Problems, 32 (2016), 015004 (16pp). [26] Z. Li and M. Yamamoto, Uniqueness for inverse problems of determining orders of multi-term time-fractional derivatives of diffusion equation, Appl. Anal., 94 (2015), 570-579. [27] C.-L. Lin and G. Nakamura, Unique continuation property for anomalous slow diffusion equation, Commun. Partial Differ. Equations, 41 (2016), 749-758. [28] Y. Liu, Strong maximum principle for multi-term time-fractional diffusion equations and its application to an inverse source problem, Comput. Math. Appl., 73 (2017), 96-108. [29] Y. Luchko, Initial-boundary-value problems for the generalized multi-term time-fractional diffusion equation, J. Math. Anal. Appl., 374 (2011), 538-548. [30] M. Machida, The time-fractional radiative transport equation: Continuous-time random walk, diffusion approximation, and Legendre-polynomial expansion, J. Math. Phys. , 58 (2017), 12pp. [31] R. Metzler and J. Klafter, The random walk's guide to anomalous diffusion: A fractional dynamics approach, Phys. Rep., 339 (2000), 1-77. [32] I. Podlubny, Fractional Differential Equations, Academic Press, New York, 1999. [33] X. Xu, J. Cheng and M. Yamamoto, Carleman estimate for fractional diffusion equation with half order and applicatioin, Appl. Anal., 90 (2011), 1355-1371. [34] M. Yamamoto, Carleman estimates for parabolic equations and applications, Inverse Problems, 25 (2009), 123013 (75pp). [35] M. Yamamoto and Y. Zhang, Conditional stability in determining a zeroth-order coefficient in a half-order fractional diffusion equation by a Carleman estimate, Inverse Problems, 28 (2012), 105010 (10pp).
show all references
References:
[1] E. E. Adams and L. W. Gelhar, Field study of dispersion in a heterogeneous aquifer: 2.Spatial moments analysis, Water Resources Research, 28 (1992), 3293-3307. doi: 10.1029/92WR01757. [2] B. Amaziane, L. Pankratov and A. Piatnitski, Homogenization of a single-phase flow through a porous medium in a thin layer, Math. Models Methods Appl. Sci., 17 (2007), 1317-1349. doi: 10.1142/S0218202507002339. [3] A. Ashyralyev, Well-posedness of the Basset problem in spaces of smooth functions, Appl. Math. Lett., 24 (2011), 1176-1180. doi: 10.1016/j.aml.2011.02.002. [4] E. Bazhlekova and I. Dimovski, Exact solution of two-term time-fractional Thornley's problem by operational method, Integral Transforms Spec. Funct., 25 (2014), 61-74. doi: 10.1080/10652469.2013.815184. [5] A. L. Bukhgeim and M. V. Klibanov, Uniqueness in the large of a class of multidimensional inverse problems, Dokl. Akad. Nauk SSSR, 260 (1981), 269-272 (in Russian). [6] T. Carleman, Sur un problème d'unicité pour les systèmes d'équations aux derivées partielles à deux variables independantes, Ark. Mat. Astr. Fys., 26B (1939), 1-9. [7] J. Cheng, C.-L. Lin and G. Nakamura, Unique continuation property for the anomalous diffusion and its application, J. Differ. Equ., 254 (2013), 3715-3728. doi: 10.1016/j.jde.2013.01.039. [8] M. Eller and V. Isakov, Carleman estimates with two large parameters and applications, Proc. Conf. Differential-Geometric Methods in the Control of Partial Differential Equations (Boulder, CO, July 1999), Contemp. Math., 268 (2000), 117-136. [9] A. V. Fursikov and O. Y. Imanuvilov, Controllability of Evolution Equations, (Lecture Notes Series vol. 34) Seoul National University, Seoul (Korea), 1996. [10] Y. Hatano and N. Hatano, Dispersive transport of ions in column experiments: An explanation of long-tailed profiles, Water Resources Research, 34 (1998), 1027-1033. doi: 10.1029/98WR00214. [11] L. Hörmander, Linear Partial Differential Operators, Springer-Verlag, Berlin, 1963. [12] O. Y. Imanuvilov, Controllability of parabolic equations, Sbornik Math., 186 (1995), 879-900. [13] O. Y. Imanuvilov and M. Yamamoto, Lipschitz stability in inverse parabolic problems by the Carleman estimate, Inverse Problems, 14 (1998), 1229-1245. doi: 10.1088/0266-5611/14/5/009. [14] V. Isakov, Inverse Problems for Partial Differential Equations, 2nd edition, Springer-Verlag, Berlin, 2006. [15] V. Isakov and N. Kim, Carleman estimates with two large parameters for second order operators and applications to elasticity with residual stress, Appl. Math., 35 (2008), 447-465. doi: 10.4064/am35-4-4. [16] V. Isakov and N. Kim, Carleman estimates with second large parameter for second-order operators, Some Applications of Sobolev Spaces to PDEs, International Math. Ser., SpringerVerlag, 10 (2009), 135-159. [17] D. Jiang, Z. Li, Y. Liu and M. Yamamoto, Weak unique continuation property and a related inverse source problem for time-fractional diffusion-advection equations, Inverse Problems, 33 (2017), 055013 (22p). [18] B. Jin and W. Rundell, A tutorial on inverse problems for anomalous diffusion processes, Inverse Problems, 31 (2015), 035003 (40pp). [19] A. Kawamoto, Lipschitz stability estimates in inverse source problems for a fractional diffusion equation of half order in time by Carleman estimates, UTMS Preprint Series, UTMS 2016-3. [20] A. Kawamoto and M. Machida, Global Lipschitz stability for a fractional inverse transport problem by Carleman estimates, arXiv preprint, arXiv: 1608.07914 (2016). [21] M. V. Klibanov, Inverse problems and Carleman estimates, Inverse Problems, 8 (1992), 575-596. doi: 10.1088/0266-5611/8/4/009. [22] M. V. Klibanov, Carleman estimates for global uniqueness, stability and numerical methods for coefficient inverse problems, J. Inverse Ill-Posed Probl., 21 (2013), 477-560. [23] M. V. Klibanov and A. A. Timonov, Carleman Estimates for Coefficient Inverse Problems and Numerical Applications, VSP, Utrecht, 2004. [24] Z. Li, Y. Liu and M. Yamamoto, Initial-boundary value problems for multi-term time-fractional diffusion equations with positive constant coefficients, Appl. Math. Comput., 257 (2015), 381-397. doi: 10.1016/j.amc.2014.11.073. [25] Z. Li, O. Y. Imanuvilov and M. Yamamoto, Uniqueness in inverse boundary value problems for fractional diffusion equations Zhiyuan, Inverse Problems, 32 (2016), 015004 (16pp). [26] Z. Li and M. Yamamoto, Uniqueness for inverse problems of determining orders of multi-term time-fractional derivatives of diffusion equation, Appl. Anal., 94 (2015), 570-579. [27] C.-L. Lin and G. Nakamura, Unique continuation property for anomalous slow diffusion equation, Commun. Partial Differ. Equations, 41 (2016), 749-758. [28] Y. Liu, Strong maximum principle for multi-term time-fractional diffusion equations and its application to an inverse source problem, Comput. Math. Appl., 73 (2017), 96-108. [29] Y. Luchko, Initial-boundary-value problems for the generalized multi-term time-fractional diffusion equation, J. Math. Anal. Appl., 374 (2011), 538-548. [30] M. Machida, The time-fractional radiative transport equation: Continuous-time random walk, diffusion approximation, and Legendre-polynomial expansion, J. Math. Phys. , 58 (2017), 12pp. [31] R. Metzler and J. Klafter, The random walk's guide to anomalous diffusion: A fractional dynamics approach, Phys. Rep., 339 (2000), 1-77. [32] I. Podlubny, Fractional Differential Equations, Academic Press, New York, 1999. [33] X. Xu, J. Cheng and M. Yamamoto, Carleman estimate for fractional diffusion equation with half order and applicatioin, Appl. Anal., 90 (2011), 1355-1371. [34] M. Yamamoto, Carleman estimates for parabolic equations and applications, Inverse Problems, 25 (2009), 123013 (75pp). [35] M. Yamamoto and Y. Zhang, Conditional stability in determining a zeroth-order coefficient in a half-order fractional diffusion equation by a Carleman estimate, Inverse Problems, 28 (2012), 105010 (10pp).
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2020 Impact Factor: 1.639
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# Clocks and Calendar Worksheet-2
Read the question carefully and choose the correct option.
(11) A movie started at 3 O' Clock and finished at 4 O' Clock. How long was the movie?
A. 3 hours B. 4 hours C. 1 hour D. 2 hours
(12) What date is the second Tuesday of November?
A. 8th B. 15th C. 7th D. 9th
(13) Megha started her homework at 6:00 and finished it in 30 minutes. What time did she finish her homework?
A. 7:30 B. 6:45 C. 7:15 D. 6:30
(14) What date is the first Monday of November?
A. 1st B. 15th C. 6th D. 7th
(15) Rahul woke up at 7 o'clock. He took 20 minutes to get ready and 10 minutes to walk to the school. What time did he reach the school?
A. 7:20 B. 7:40 C. 7:30 D. 8:00
(16) What is the day on January 12th?
A. Monday B. Tuesday C. Wednesday D. Thursday
(17) 50 minutes + 10 minutes =?
A. two hours B. 60 hours C. 60 seconds D. one hour
(18) What is the day on April 4th?
A. Sunday B. Monday C. Tuesday D. Wednesday
(19) Following are the results of a race.
Jay: 55 min
Mehul: 50 min
Ali: 45 min
Akash: 48 min
Who is the winner of the race?
A. Jay B. Mehul C. Ali D. Akash
(20) The clock is showing the current time. What was the time 2 hours before?
A. 12 O'Clock B. 6 O'Clock C. 2 O'Clock D. 10 O'Clock
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# Re: Many Pasts? Not according to QM...
From: Stathis Papaioannou <stathispapaioannou.domain.name.hidden>
Date: Sat, 28 May 2005 15:26:24 +1000
Saibal Mitra wrote:
>You have to consider the huge number of alternative states you could be in.
>
>1) Consider an observer moment that has experienced a lot of things. These
>experiences are encoded by n bits. Suppose that these experiences were more
>or less random. Then we can conclude that there are 2^n OMs that all have a
>probability proportional to 2^(-n). The probability that you are one of
>these OMs isn't small at all!
>
>2) Considering perforing n suicide experiments, each with 50% survival
>probability. The n bits have registered the fact that you have survived the
>n suicide experiments. The probability of experiencing that is 2^(-n). The
>2^(n) -1 alternate states are all unconscious.
>
>
>So, even though each of the states in 1 is as likely as the single state in
>2, the probability that you'll find yourself alive in 1 is vastly more
>likely than in 2. This is actually similar to why you never see a mixture
>of
>two gases spontaneously unmix. Even though all states are equally likely,
>there are far fewer unmixed states than mixed ones.
I understand your point, but I think you are making an invalid assumption
about the relationship between a random sampling of all the OM's available
to an individual and that individual's experience of living his life.
Suppose a trillion trillion copies of my mind are made today on a computer
and run in lockstep with my biologically implemented mind for the next six
months, at which point the computer is shut down. This means that most of my
measure is now in the latter half of 2005, in the sense that if you pick an
observer moment at random out of all the observer moments which identify
themselves as being me, it is much more likely to be one of the copies on
the computer. But what does this mean for my experience of life? Does it
mean that I am unlikely to experience 2006, being somehow suspended in 2005?
More generally, if a person has N OM's available to him at time t1 and kN at
time t2, does this mean he is k times as likely to find himself experiencing
t2 as t1? I suggest that this is not the right way to look at it. A person
only experiences one OM at a time, so if he has "passed through" t1 and t2
it will appear to him that he has spent just as much time in either interval
(assuming t1 and t2 are the same length). The only significance of the fact
that there are "more" OM's at t2 is that the person can expect a greater
variety of possible experiences at t2 if the OM's are all distinct.
--Stathis Papaioannou
_________________________________________________________________
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Received on Sat May 28 2005 - 01:29:34 PDT
This archive was generated by hypermail 2.3.0 : Fri Feb 16 2018 - 13:20:10 PST
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# Ordering of Large Cardinal Axioms
One of the unexplained phenomena about large cardinal axioms is that they tend to be in a linear order (in terms of consistency strength). So it brings to mind other natural questions about this "order".
So my question is - For any two large cardinal axioms $A1$, $A2$ such that $ZFC + A1$ proves $ZFC + A2$ is consistent, can we define a large cardinal axiom $A3$ such that $ZFC + A1$ proves $ZFC + A3$ is consistent, and $ZFC + A3$ proves $ZFC + A2$ is consistent?
Meaning, is the order known\thought to be densely ordered (assuming the ordering itself is not some random phenomena)? Alternatively, can we build an example (even a highly contrived one) of $A1$, $A2$, where any axiom we add to $ZFC$ is either stronger than $ZFC + A1$, weaker than $ZFC + A2$, or equiconsistent to one of them (meaning the ordering is not dense)?
P.S - I know there is no agreed upon definition of "large cardinal axiom", but if we discuss their linear order we should be able to ask questions like this as well.
• We expect the consistency strength degrees of large cardinals to be pre-well-ordered. Essentially, because we ultimately expect to identify these degrees with certain "mice", and these should be orderable via comparison processes. Commented Mar 15, 2016 at 15:20
• A good place to read about why this leads to a (pre-)well-ordering is MR1720574 (2000k:03101). Löwe, Benedikt(1-CA); Steel, John R.(1-CA) An introduction to core model theory. (English summary) Sets and proofs (Leeds, 1997), 103–157, London Math. Soc. Lecture Note Ser., 258, Cambridge Univ. Press, Cambridge, 1999. Particularly see Section 2.5 on the mouse order. (Note the link above is to a postscript file.) Commented Mar 15, 2016 at 15:20
• @Andrés: Are these mice nice? And if there is just an iterable mouse, is it nouse? And if it's not, does that mean that the iterable mouse is a terrible mouse? Commented Mar 15, 2016 at 15:23
If by large cardinal axioms you mean axioms which directly allude to cardinals, then the answer is negative.
Namely, if a large cardinal axiom is something like "There exists a cardinal $\kappa$ with such and such properties" and these properties imply certain amount of inaccessibility, in which case we rule out things like "$\sf ZFC$ is consistent" or "$0^\#$ exists", then the answer is negative.
To see that, simply note that there will always be a smallest large cardinal. If you want it to be such that $V_\kappa\models\sf ZFC$, then the least worldly cardinal is your smallest large cardinal. Of course this is debatable, and we don't have a mathematical definition of what is a large cardinal axiom. But much like pornography, "we know it when we see it".
So if we accept that worldly cardinals are the smallest large cardinals, then "There are two worldly cardinals" proves the consistency of "There is a worldly cardinal", but there are no intermediate notions to be found.
Of course you might argue that $\sf ZFC$ with "the theory $\sf ZFC+\exists \kappa$ worldly is consistent" proves the consistency of the existence of worldly cardinals, but now this becomes a question of whether or not this is a large cardinal axiom. Some people might argue that it is, others might argue that it's not.
For what it's worth, statement about "this theory is consistent" are arithmetic, so if we only assume $\sf PA$ with "this theory is consistent" then we can prove the consistency of said theory. But this is really not what we mean when we talk about large cardinals, is it now?
• Of course, I am strictly treating the case where the two extensions of ZFC have different consistency strength. One could argue that we cannot find anything intermediate between "There is a weakly compact cardinal" and "There is a successor of a regular cardinal with the tree property", because the two theories have been shown to be equiconsistent. Commented Mar 15, 2016 at 15:26
• Interesting. Though the example with worldly cardinals makes me wonder what would happen if I'd amend the question to demand a large cardinal axiom that states the existence of an unbounded class of a certain type of cardinals (instead of a single cardinal). Commented Mar 15, 2016 at 20:59
• That would usually be the the same. A worldly limit of inaccessible cardinals is the smallest large cardinal above "proper class of inaccessible cardinals" in terms of consistency strength. Commented Mar 15, 2016 at 21:32
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## Chapter 7 - Day 3 - Lesson 7.3
##### Learning Targets
• Check the Random and Large Counts conditions for constructing a confidence interval for a population proportion.
• Determine the critical value for calculating a C% confidence interval for a population proportion using Table A or technology.
• Calculate a C% confidence interval for a population proportion.
##### Activity:
For today’s lesson, you will need A LOT of Hershey’s kisses. Each group needs 50 kisses. If you’re teaching this lesson after February 14, we suggest hitting up stores to buy some clearance Valentine’s Day kisses on February 15. We were able to get them half off! Groups will be tossing the kisses and keeping track of the proportion of kisses that land flat on the bottom, as opposed to on their side. From there they will create a 95% confidence interval for the true proportion of Hershey kisses that land flat.
When students are working through the activity, draw a number line on the board going from to 1, with tick marks every 0.1. After the groups have calculated their confidence interval, have one person from each group come up to the front and draw their interval under the scale. After going over the activity, reveal that the true proportion (as best as we can guess) is p = 0.35. Draw a vertical line through the drawn intervals showing this. Approximately 95% of the class intervals should have captured the true proportion. This is a great time to review the interpretation of a confidence level from lesson 7.2.
The structure and questioning in this activity are scaffolded really nicely. Students have all the information they need since they’ve already learned about margin of error in chapter 3 and about the standard deviation of a sampling distribution of a sample proportion in chapter 6. However, make sure you point out important information to students. When going over questions #4 and 5, add to the margin that these are the RANDOM and NORMAL conditions which must be met in order to create a confidence interval. After question #6, identify that 2 standard deviations is called a CRITICAL VALUE or z*. We add the critical values for 90%, 95% and 99% in the margin.
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# QlikView Scripting
Discussion Board for collaboration on QlikView Scripting.
Announcements
Not applicable
## Need help with expression
Hi all!
Here is my problem: I have two dimensions, A and B, and I have data C: C=AxB (there is C value for each pair (A,B)). Also, I have a formula: KPI=A*1000/(C*10*B). So, I have to get A-values, for which KPI, calcuted with grouping by B field, is MAX.
I know how to calculate MAX KPI grouping by B field:
=aggr(max(aggr(A*1000/(C*9.81*B),A,B)),B),
but how to get B value?
Tags (2)
1 Solution
Accepted Solutions
Not applicable
## Re: Need help with expression
Are you looking for this?
Expression: Max(TOTAL <B> KPI)
8 Replies
Not applicable
## Re: Need help with expression
Can you provide some sample data with expected output to try out what you are looking for?
Not applicable
## Re: Need help with expression
Hi, Sunny T!
First thing I have to edit my question: " but how to get A value?"
Here is the data to load:
A*1000/(C*9.81*B) as KPI
INLINE
[A, B, C
20, 8, 312
20.5, 8, 321
21, 8, 330
20, 8.5, 294
20.5, 8.5, 300.5
21, 8.5, 307
20, 9, 275
20.5, 9, 282
21, 9, 289]
I decided to calculated KPI already in script.
So here is what we have (pivot table for better understanding). I want to get something like this:
newA newB
20 8
21 8,5
20 9
I need newA and newB for my linear diagram.
Not applicable
## Re: Need help with expression
Are you looking for this?
Expression: Max(TOTAL <B> KPI)
Not applicable
## Re: Need help with expression
Exactly!
Thank you very much!
Not applicable
## Re: Need help with expression
If not, please make clear what part of this topic you still want discussion on .
May you live in interesting times!
Not applicable
## Re: Need help with expression
Hi!
Thanks for help once more But I still can't build what I want..
My purpose is a linear diagram: horizontal axis(dimension) - B, and expression should return optimal A (A, where we have max(KPI) grouping by B).
Output for sample data I gave above mentioned to be like this (in numeric view):
B - 8, A - 20
B - 8.5, A - 21
B - 9, A - 20.
Is it possible?
Not applicable
## Re: Need help with expression
This?
Dimension: B
Expressions
1) =FirstSortedValue(A, -KPI)
2) =Max(KPI)
Not applicable
## Re: Need help with expression
Thank you! Now it works!
Community Browser
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# Solving $\sqrt{8+2x-x^2} > 6-3x$.
So I was solving this inequality. $$\sqrt{8+2x-x^2} > 6-3x$$
• First things first, I obtained that the common domain of definition is actually $[-2,4]$.
• Next we would square and solve the quadratic that follows.
But the "solution" seems to have a part, where they took $6-3x \geq 0$,
which gave another restriction for $x$ as $(-\infty,2]$.
Let's say we have an inequality $\sqrt a>b$. We're often interested in getting rid of the square root, so we want to do something along the lines of 'squaring both sides'. But squaring both sides doesn't necessarily preserve the inequality.
Example:
$$\sqrt5>-3$$
But
$$\implies\sqrt5^2>(-3)^2 \implies 5\gt9$$
is clearly false.
If you want a general rule, then you should use
$$|a|\gt|b|\iff a^2>b^2$$
This explains why you need to consider separate cases here
• Basically 6-3x may have a negetive output also as the domain is R . We take 6-3x >=0 ao that ">" still holds . – user33699 Jul 24 '17 at 12:28
Just use these fundamental rules for irrational (in)equalities, valid on the domain $A\ge 0$: \begin{alignat}{2} &\bullet\quad \sqrt A> B\iff A>B^2\quad\text{OR}&&\quad B<0 \\ &\bullet\quad \sqrt A < B\iff A < B^2\quad\text{AND}&&\quad B\ge 0 \\ &\bullet\quad \sqrt A = B\iff A = B^2\quad\text{AND}&&\quad B\ge 0 \end{alignat}
The domain gives $-2\leq x\leq4$.
For $x>2$ with the domain our inequality is true.
But for $x\leq2$ we can use squaring,
which gives $5x^2-19x+14<0$ or $1<x<2.8$ and since $(1,2]\cup(2,4]=(1,4]$, we get the answer: $$(1,4]$$
• I can upvote now :) – user33699 Jul 24 '17 at 12:16
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# AMUSEMENT PARK RID RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS
#### Trending Tags
Terms · Privacy · Contact
## 5 People In A Room
Hint:
2 people remain. Dont forget to count the murderer.
Did you answer this riddle correctly?
YES NO
Solved: 67%
## A Single Candle On A Cake
Hint:
1 of your 7 year cycles! You go through 7 cycles every year. The first cycle starts on your birthday, and each of the 7 cycles lasts 52 days. (7x52=364).
You only have to find your personal cycle numbers once, because it's always the same, year after year.
Did you answer this riddle correctly?
YES NO
Solved: 32%
## The Most Amount Of Money
Hint:
Mary! John 'had' 200k but is past tense. Mark 'will' have 500k but does not current have it as the question asks who 'has' the most amount of money.
Did you answer this riddle correctly?
YES NO
Solved: 81%
## Thirteenth Birthday
Hint:
A teenager!
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YES NO
Solved: 72%
## Starts Off Temporary
Hint:
The letter 'T'
(T)emporary and Permanen(t)
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YES NO
Solved: 52%
## Pongo And Perdita's Litter
Hint:
In the movie 101 Dalmatians, the dalmatian couple Pongo and Perdita originally had 15 puppies of their own.
As the story progresses, they were able to rescue 99 puppies from Cruella de Vil. (Pongo is the 100th and Perdita is the 101st)
It can be a source of confusion for many, especially those who have read the book, because in the book, Perdita was not Pongo's wife. His wife was referred to as Mrs. Pongo. Perdita is another dalmatian parent who lost her litter to Cruella De Vil. Although not specified in the book, it was believed that she had 8 puppies in her litter.
Did you answer this riddle correctly?
YES NO
Solved: 56%
## 4 Girls Are In A Room
Hint:
Playing Chess with Aire, as Chess is a two person game!
Did you answer this riddle correctly?
YES NO
Solved: 78%
## The Big Animal Race
Hint:
She passed the buck!
Did you answer this riddle correctly?
YES NO
Solved: 40%
## A Monster Likes You
Hint:
He takes another bite.
Did you answer this riddle correctly?
YES NO
Solved: 39%
## A Person Wakes Up From His Night Sleep And Got Some Cereal
Hint:
His eyes
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YES NO
## Legs On The Floor
Hint:
There are six legs on the floor. Four legs from the bed and your own two legs as you stand in the room.
All the animals are on the bed and not on the floor.
Did you answer this riddle correctly?
YES NO
Solved: 68%
## Never Goes In And Never Comes Out
Hint:
Keyhole
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YES NO
Solved: 55%
## Seen In The Middle Of March
Hint:
The letter R
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YES NO
Solved: 71%
## NBA 5 Letter Word Puzzle
Hint: Cut the grass and I will show.
Snake
Did you answer this riddle correctly?
YES NO
## I Have 6 Eggs
Hint: The eggs aren't specified.
4. I broke, fried and ate the same 2 eggs.
Did you answer this riddle correctly?
YES NO
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My mother alwyas took a little red counter to the grocery store. The counter was
used to keep tally of the amount of money she would have spent so far on that visit
to the store, if she bought all the items in her basket. There was a four-digit
display, increment buttons for each digit, and a reset button. There was an overflow
indicator that came up red if more money was entered than the \$99.99 it would
register. (This was a long time ago.)
Write and implement the member functions of a class Counter that simulates and
slightly generalizes the behavior of this grocery store counter. The constructor
should create a Counter object that can count up to the constructor's argument.
That is, Counter(9999) should provide a counter that can count up to 9999. A newly
constructed counter displays a reading of 0. The member function void reset();
sets the counter's number to 0. The member function void incr1(); increments the
units digits by 1q, void incr10(); increments the tens digit by 1, the void
incr100(); and void incr1000(); increment the next two digits respectively.
Accounting for any carry when you increment should require no further action
than adding an appropriate number to the private data member. A member function
bool overflow(); detects overflow. (Overflow is the result of incrementing the
counter's private data member beyond the maximum entered at counter construction.)
Use this class to provide a simulation of my mother's little red clicker. Even
though the display is an integer, in the simulation, the rightmost (lower-order)
two digits are always thought of as cents, and tens of cents, the next digit is
dollars, and the fourth digit is tens of dollars.
Provide keys for cents, dimes, dollars, and tens of dollars. Unfortunately, no
choice of keys seems particularly mnemonic. One choice is to use the keys
asdfo: a for cents, followed by a digit 1 to 9; s for dimes, followed by the
digits 1 to 9; d for dollars, followed by a digit 1 to 9; and f for tens of
dollars, again followed by a digit 1 to 9. Each entry (one of asdf followed
by 1 to 9) is followed by pressing the Return key. Any overflow is reported
after each operation. Overflow can be requested by pressing the o key.
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## ››Convert metric mile to decametre
metric mile decameter
Did you mean to convert metric mile metric mile [high school] to decameter
How many metric mile in 1 decameter? The answer is 0.0066666666666667.
We assume you are converting between metric mile and decametre.
You can view more details on each measurement unit:
metric mile or decameter
The SI base unit for length is the metre.
1 metre is equal to 0.00066666666666667 metric mile, or 0.1 decameter.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between metric miles and decameters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of metric mile to decameter
1 metric mile to decameter = 150 decameter
2 metric mile to decameter = 300 decameter
3 metric mile to decameter = 450 decameter
4 metric mile to decameter = 600 decameter
5 metric mile to decameter = 750 decameter
6 metric mile to decameter = 900 decameter
7 metric mile to decameter = 1050 decameter
8 metric mile to decameter = 1200 decameter
9 metric mile to decameter = 1350 decameter
10 metric mile to decameter = 1500 decameter
## ››Want other units?
You can do the reverse unit conversion from decameter to metric mile, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Metric mile
A metric mile is a distance of 1,500 meters. This distance is used when it is desirable to approximate the statute mile, yet also desirable to use a convenient round number in SI or metric units. A statute mile is actually 1,609.347 meters.
## ››Definition: Decameter
A decametre (American spelling: dekameter, symbol: dam) is a measurement of distance equal to ten metres. This measure is included mostly for completeness. It very rarely has any practical application.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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https://www.teacherspayteachers.com/Product/Capacity-and-Weight-2382285
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Capacity and Weight
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
7.81 MB | 36 pages
PRODUCT DESCRIPTION
Capacity and Weight Review Problems! Review measurement concepts including capacity, weight, and mass with this handy slideshow that can also be printed as TASK CARDS!
Who is This For?
3rd or 4th grade teachers looking for engaging review of capacity, weight, or mass concepts!
What is Included?
* Quick Review Slides
* 24 Capacity and Weight Problems (slideshow or printed as task cards)
* Answer Key and Recording Sheet
I created these problems to help my students prepare for the Texas state test (STAAR), which is known for having students apply concepts in a variety of unique and challenging ways.
In these problems, students must determine the approximate and exact measurements of capacity (liquid volume), weight, and mass, and determine appropriate units for measurement.
I love using slide shows for review and having students show their answers using index cards labeled A, B, and C or having them write answers on their desks using dry-erase markers.
The slides can also be turned into task cards by selecting the "multiple" page print in your printing options and printing out the slides 4 to a page.
This set of capacity and weight problems is perfect for test prep, review, guided math, intervention, and independent work!
Thanks so much! I hope you found just what you needed!:)
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > stdpc7 Structured version Visualization version GIF version
Theorem stdpc7 2002
Description: One of the two equality axioms of standard predicate calculus, called substitutivity of equality. (The other one is stdpc6 2001.) Translated to traditional notation, it can be read: "𝑥 = 𝑦 → (𝜑(𝑥, 𝑥) → 𝜑(𝑥, 𝑦)), provided that 𝑦 is free for 𝑥 in 𝜑(𝑥, 𝑥)." Axiom 7 of [Mendelson] p. 95. (Contributed by NM, 15-Feb-2005.)
Assertion
Ref Expression
stdpc7 (𝑥 = 𝑦 → ([𝑥 / 𝑦]𝜑𝜑))
Proof of Theorem stdpc7
StepHypRef Expression
1 sbequ2 1939 . 2 (𝑦 = 𝑥 → ([𝑥 / 𝑦]𝜑𝜑))
21equcoms 1993 1 (𝑥 = 𝑦 → ([𝑥 / 𝑦]𝜑𝜑))
Colors of variables: wff setvar class Syntax hints: → wi 4 [wsb 1937 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 This theorem depends on definitions: df-bi 197 df-an 385 df-ex 1745 df-sb 1938 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator
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https://community.fabric.microsoft.com/t5/DAX-Commands-and-Tips/Calculated-column-with-sum-as-function/m-p/2900198/highlight/true
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Helper II
## Calculated column with sum as function
I'm looking for a formula that uses sums as logic that we know in Excel. I want to add a calculated column (orange) in the table below. That sums values of column value if the column values of date and color are equal.
I once came across a formula with the EARLIER function, but I don't know the exact formula to get it to work.
1 ACCEPTED SOLUTION
Solution Sage
Hint: You should never use EARLIER. There is a much better alternative and it's called VARIABLES.
``````// This is the formula for your column...
[Total Value] =
var CurrentDate = YourTable[Date]
var CurrentColor = YourTable[Color]
var Output =
sumx(
filter(
YourTable,
YourTable[Date] = CurrentDate
&&
YourTable[Color] = CurrentColor
),
YourTable[Value]
)
return
Output``````
5 REPLIES 5
Helper II
I want to thank you both ( @daXtreme & @zerotyper ) for the quick and well-functioning response. I have been helped tremendously.
As a result of @daXtreme 's additional explanation, I naturally opt for his solution.
Thanks and have a nice weekend!
Frequent Visitor
@brief001 Hope this is the result you want.
Solution Sage
A word of caution.
Using CALCULATE in calculated column is a no-no. This slows DAX tremendously down even on moderate sized sets. I've seen many a time formulas like this that brought the engine down to a complete freeze.
Frequent Visitor
@daXtreme Really thanks for your suggestion.
You are right, just use Variable and simply filter and Aggregation is better than Calculate in a calculated column.
The context transition is too heavy.
Solution Sage
Hint: You should never use EARLIER. There is a much better alternative and it's called VARIABLES.
``````// This is the formula for your column...
[Total Value] =
var CurrentDate = YourTable[Date]
var CurrentColor = YourTable[Color]
var Output =
sumx(
filter(
YourTable,
YourTable[Date] = CurrentDate
&&
YourTable[Color] = CurrentColor
),
YourTable[Value]
)
return
Output``````
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https://www.acmicpc.net/problem/3870
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
2 초 128 MB33121139.286%
## 문제
You are given a sequence a0a1···aN−1 of digits and a prime number Q. For each i ≤ j with ai ≠ 0, the subsequence aiai+1···aj can be read as a decimal representation of a positive integer. Subsequences with leading zeros are not considered. Your task is to count the number of pairs (i, j) such that the corresponding subsequence is a multiple of Q.
## 입력
The input consists of at most 50 datasets. Each dataset is represented by a line containing four integers N, S, W, and Q, separated by spaces, where 1 ≤ N ≤ 105, 1 ≤ S ≤ 109, 1 ≤ W ≤ 109, and Q is a prime number less than 108. The sequence a0 ··· aN−1 of length N is generated by the following code, in which ai is written as a[i].
int g = S;
for(int i=0; i<N; i++) {
a[i] = (g/7) % 10;
if( g%2 == 0 ) { g = (g/2); }
else { g = (g/2) ^ W; }
}
Note: the operators /, %, and ^ are the integer division, the modulo, and the bitwise exclusiveor, respectively. The above code is meant to be a random number generator. The intended solution does not rely on the way how the sequence is generated.
The end of the input is indicated by a line containing four zeros separated by spaces.
## 출력
For each dataset, output the answer in a line. You may assume that the answer is less than 230.
## 예제 입력 1
3 32 64 7
4 35 89 5
5 555 442 3
5 777 465 11
100000 666 701622763 65537
0 0 0 0
## 예제 출력 1
2
4
6
3
68530
## 힌트
In the first dataset, the sequence is 421. We can find two multiples of Q = 7, namely, 42 and 21.
In the second dataset, the sequence is 5052, from which we can find 5, 50, 505, and 5 being the multiples of Q = 5. Notice that we don’t count 0 or 05 since they are not a valid representation of positive integers. Also notice that we count 5 twice, because it occurs twice in different positions.
In the third and fourth datasets, the sequences are 95073 and 12221, respectively.
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https://docs.google.com/forms/d/e/1FAIpQLSdSJlNq54I-0R0JT2JAuNCceWf__gJV0SiumWxELhqwpFz5OA/viewform
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Sea Level Rise in Hawaiʻi
We don't have to look far to see the impacts of sea level rise. Throughout Hawaiʻi's shorelines, the impacts are already visible. Scientists predict an increase anywhere from 1 foot to 6 feet by the end of the century.
In this activity, you will analyze sea level data from two locations around Hawaiʻi and look at another location that is being greatly impacted, Florida.
Email *
Please enter your full name. *
Sea Level Rise Data- Honolulu
The city of Honolulu on Oʻahu is expected to be highly impacted by sea level rise. Visit the following link and explore the different current and future scenarios.
Honolulu Sea Level Rise Simulation: http://www.pacioos.hawaii.edu/shoreline/slr-honolulu/
1 Meter = 3.2 feet
Compare the Hurricane Storm Surge Inundation for current sea level vs. one meter sea level rise. What observation can you make when comparing these scenarios? *
2 points
What observation is true when looking at the 1 M Sea Level Rise Inundation. *
2 points
Sea Level Rise Data- Maui
The island of Maui is expected to also be impacted by sea level rise, with some regions already feeling the impact. Visit the following link and explore how different predictions will impact Maui's coastal communities.
Sea Level Rise Viewer: http://www.pacioos.hawaii.edu/shoreline/slr-hawaii/
At 3.2 feet increase in sea level, you can see the majority of Maui's coastline being impacted. Which TWO statements below describe the areas most significantly impacted? *
2 points
Required
One low lying area that will be impacted on Maui is the Pali road to Lāhaina. Zoom into the Olowalu coastline and select "Flooded Highways 1.1 ft." What observation can you make about this scenario? *
1 point
With sea level rise, the economical loss can be detrimental. Compare the "Potential Economic Loss" of Lāhaina and Honolulu for a 3.2 ft scenario and select the correct answer below which best describes the damage in each area. *
2 points
Sea Level Rise Adaptation Strategies
Use the descriptions of the 4 strategies below to answer the following questions about adapting to sea level rise.
Dune Restoration at Kamaʻole I Beach Park
The photo above of sand dune restoration is an example of which kind of sea level rise adaptation? *
1 point
How is restoring sand dunes and vegetation an example of adaptation? *
Hint: what do they prevent along the shoreline
2 points
House on stilts
The photo above of the house is an example of which kind of sea level rise adaptation? *
1 point
Seawalls and sand bags in Kahana, Maui
The photo above of seawalls and sang bags is an example of which kind of sea level rise adaptation? *
1 point
Moving the "pali" road on Maui
Moving the "pali" road inland would be an example of what kind of sea level rise adaptation? *
1 point
What is one potential limitation or disadvantage of one of the adaptation strategy examples above? *
Choose one of the four examples above and explain one potential disadvantage of that strategy.
2 points
Sea Level Rise Viewer Maps of 3.2ft of sea level rise & economic impact in Waikīkī
With 3.2ft of sea level rise, which areas of Waikīkī experience the highest economic loss? *
1 point
Which TWO areas experience a lot of flooding but less economic loss? *
2 points
Required
Explain why you think certain areas result in more economic loss than others even with a similar amount of flooding/damage? *
hint: what are the differences between the areas in your two answers above in terms of what would be lost and what rebuilding/adapting might involve
2 points
Submit
Clear form
Never submit passwords through Google Forms.
This form was created inside of Maui Huliau Foundation.
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https://brainmass.com/physics/heat-thermodynamics/first-second-law-thermodynamics-618955
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Purchase Solution
# First and second law of thermodynamics
Not what you're looking for?
1. For the following processes, state whether the driving force is the first or second law of thermodynamics. The systems are in italics and we are only interested in whether the properties of the system have changed. Explain your choice in 1-2 sentences.
a. Warming up exercises when muscles are worked the cells "burn" more glucose and the oxidation releases energy which is stored as ATP. However only about 50% is stored and the rest increases the temperature of the muscle tissue.
b. You observe the trees in the forest do not grow in straight lines. Forests that were plants by the CCC camps in the Great Depression have the trees plants in long straight rows. Anyone visiting these knows immediately that this is not a natural forest.
2. 12.0L of monoatomic ideal gas at 15◦C and 4atm are expanded to a final pressure of 1.2atm. Calculate ∆U, q, w, and ∆S if the process is:
a. Reversible and isothermal
i. Prove that for the process in part b ∆S is zero.
d. Explain, in 3-4 sentences, why ∆S is different in all three parts
3. The following equation is an equation of state for an imagined gas: . The term accounts for the attractions between the gas molecules. Determine ?
##### Solution Summary
One part of the problem was to consider some example of thermodynamic processes and conclude is it example of the first or the second law of thermodynamics. Other part of the problem was to calculate internal energy, entropy, work and heat for ideal monatomic gas if we have reversible and irreversible thermodynamics processes.
##### Solution Preview
Hi,
1. For the following processes, state whether the driving force is the first or second law of thermodynamics. The systems are in italics and we are only interested in whether the properties of the system have changed. Explain your choice in 1-2 sentences.
a. Warming up exercises when muscles are worked the cells "burn" more glucose and the oxidation releases energy which is stored as ATP. However only about 50% is stored and the rest increases the temperature of the muscle tissue.
This is an example of the first law of thermodynamics. The first law of thermodynamics is a version of the law of conservation of energy applied on thermodynamics systems. The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed. In our example we have energy that is stored in the form of ATP and our body can use part of that energy for doing some work and part of energy is ...
##### Classical Mechanics
This quiz is designed to test and improve your knowledge on Classical Mechanics.
##### Intro to the Physics Waves
Some short-answer questions involving the basic vocabulary of string, sound, and water waves.
##### Variables in Science Experiments
How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.
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https://jenniferelliskampani.com/lh3pb7792/zI27976gB/
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# Worksheet 4th Grade Handwriting Writing Worksheets Printable Practice For
Liliane Constance November 21, 2020 Worksheets
Division Worksheets. Our Division worksheet generator enables you to create many worksheets with a varied range of questions for this Core but often difficult topic. There is an option for including division worksheets with remainders in the answers for when pupils are to progress their learning. Division worksheets, online interactive activities and other resources for Key Stage 1 and Key Stage 2. Snappy Maths (A Primary Topics Website) . Counting & Number Division Worksheets & Printables Division can be tough for any student, but it’s an essential skill for more advanced math concepts. Our skills-based division worksheets and printables help students from third to fifth grade kick their math skills up a notch.
### Parallel Lines And Transversals Practice Worksheet Answers
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#### Exponent Rules Worksheet Algebra 1
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###### Pre K Science Worksheets
Mar 21, 2021
Multiplication and Division Resources. Before beginning times table recall, students must build an understanding of multiplication and division and how they are linked. The resources on this page allow students to represent multiplication and division situations in pictures and complete fact families. Math explained in easy language, plus puzzles, games, quizzes, videos and worksheets. For K-12 kids, teachers and parents. The division worksheet is a great tool for testing their knowledge of division and helping them to revise key maths skills. Great for consolidating their times table knowledge, these maths division worksheets are perfect for a quick and easy start to the lesson or as a simple homework activity. Worksheets > Math > Grade 4 > Long division. Long division worksheets. Long division is a skill which requires a lot of practice with pencil and paper to master. Our grade 4 long division worksheets cover long division with one digit divisors and up to 4 digit dividends.
Learn fourth grade math—arithmetic, measurement, geometry, fractions, and more. This course is aligned with Common Core standards. If you’re seeing this message, it means we’re having trouble loading external resources on our website. Math Sheets for 4th Grade: PDF Worksheets and Answer Keys. The following collection shares over 100 free and easy to print math sheets for 4th grade on topics including fractions, place value, unit conversion, multiplication, division, and more! Plus every worksheet includes a free answer key. By the time kids finish third grade, they have a fundamental understanding of the four tenets of math: addition, subtraction, multiplication, and division. The real challenging work begins in fourth grade, where concepts such as multi-digit multiplication and complex word problems are introduced.
### Photos of 4th Grade Penmanship Worksheets
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https://www.jobilize.com/course/section/getting-and-creating-data-by-openstax?qcr=www.quizover.com
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# 18.1 Writing c functions in matlab (mex-files) (Page 2/3)
Page 2 / 3
## Getting and creating data
The main MATLAB structure used for holding data in MEX-Files is the mxArray. This structure can hold real data, complex data, arrays, matrices, sparse-arrays, strings, and a whole host of other MATLAB data-structures.Using data from some of the basic structures is shown here, but refer to the MATLAB help for using other data structures.
## Get that data
Lets use my example from above (if you forgot: `[z0,z1] = jasonsFunction(x,y,z);` ). Assume that x is a 2-D matrix, y is a string, and z is an integer. Here we wills ee how to extractand use these different types of data.
We have access to the input paramter x by a pointer held in the array prhs. In C, when referencing an array by index, the variable is automatically dereferenced (ie: you dont need to use a star).For clarity, I will copy the variable x over to an mxArray pointer named xData (This does not need to be done for the code to work).
```//---Inside mexFunction--- //DeclarationsmxArray *xData; double *xValues;int i,j; int rowLen, colLen;double avg; //Copy input pointer xxData = prhs[0];//Get matrix x xValues = mxGetPr(xData);rowLen = mxGetN(xData); colLen = mxGetM(xData);//Print the integer avg of each col to matlab console for(i=0;i<rowLen;i++) {avg=0; for(j=0;j<colLen;j++) {avg += xValues[(i*colLen)+j]; //Another Method:// //avg += *xValues++;} avg = avg/colLen;printf("The average of row %d, is %d",i,(int)avg); }```
The function `mxGetPr` is used to get a pointer to the real data xData. This function takes a pointer to an mxArray as the intput paramter, and returns a pointer array of doubles. A similar function `mxGetPi` can be used for complex data. `mxGetN` and `mxGetM` return integers of the lengths of the row and column in the matrix. If this were an array of data, one of thesereturn values would be zero. MATLAB gives the matrix as rows first, then columns (if you were to traverse the matrix linearly) so to jump by position, (x,y) maps to x*colLen+y. MATLAB organizesits arrays this way to reduce cache misses when the row traversal is on the outside loop. It is good to code it this way if you are working for efficiency. `printf()` will print out to the MATLAB command prompt.
Getting a string is very similar, but has its own method. The example below shows the procedure for getting a string. Again, I will copy the input to a pointer called yData.
```//---Inside mexFunction--- //DeclarationsmxArray *yData; int yLength;char *TheString; //Copy input pointer yyData = prhs[1];//Make "TheString" point to the string yLength = mxGetN(yData)+1;TheString = mxCalloc(yLength, sizeof(char)); //mxCalloc is similar to malloc in C mxGetString(yData,TheString,yLength);```
This last example shows how to get a simple integer. This is the method that has always worked for me, but it seems kind of strange so I imagine there is another way to do this.
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
how did you get the value of 2000N.What calculations are needed to arrive at it
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https://freecadweb.org/wiki/Vector_API
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# Vector API
Jump to: navigation, search
Other languages:
Deutsch • English • español • français • italiano • română
(October 2019) Do not edit this page. The information is incomplete and outdated. For the latest API, see the autogenerated API documentation, or generate the documentation yourself, see Source documentation.
Vectors are used everywhere in FreeCAD.
Example:
```v=FreeCAD.Vector()
v=FreeCAD.Vector(1,0,0)
v=FreeCAD.Base.Vector()
v2 = FreeCAD.Vector(3,2,-5)
v3 = v.add(v2)
print v3.Length```
Length
Returns: returns the length of the vector.
add(Vector)
Description: adds another vector to this one.
Returns: vector
cross(Vector)
Description: the crossproduct between this vector and another.
Returns: vector
distanceToLine(Vector1,Vector2)
Description: the distance between the vector and a line through Vector1 in direction Vector2.
Returns: float
distanceToLineSegment(Vector1,Vector2)
Description: a vector to the closest point on a line segment from Vector1 to Vector2.
Returns: vector
distanceToPlane(Vector1,Vector2)
Description: the distance between the vector and a plane defined by a point and a normal.
Returns: float
dot(Vector)
Description: the dot product between 2 vectors.
Returns: float
getAngle(Vector)
Description: the angle in radians between this vector and another.
Returns: float
multiply(Float)
Description: multiplies (uniform scale) a vector by the given factor.
Returns: nothing
normalize( )
Description: normalizes a vector (sets its length to 1.0).
Returns: nothing
projectToLine(Vector1,Vector2)
Description: projects the vector on a line through Vector1 in direction Vector2.
Returns: nothing
projectToPlane(Vector1,Vector2)
Description: projects the vector on a plane defined by a point (Vector1) and a normal (Vector2).
Returns: nothing
scale(Float,Float,Float)
Description: Same as multiply but lets specify different values for x, y and z directions. (non-uniform scale)
Returns: nothing
sub(Vector)
Description: subtracts another vector from this one.
Returns: vector
x
Returns: the x coordinate of a vector.
y
Returns: the y coordinate of a vector.
z
Returns: the z coordinate of a vector.
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https://plainmath.net/high-school-geometry/83844-pyramid-sabc-has-right-triangular-base-a
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Paxton Hoffman
2022-07-23
Pyramid SABC has right triangular base ABC, with $\mathrm{\angle }ABC={90}^{\circ }$. Sides $AB=\sqrt{3},BC=3$. Lateral lengths are equal and are equal to 2. Find the angle created by lateral length and the base.
dtal50
Expert
Step 1
Let K be a middle point of AC.
Just $\measuredangle SBK=\measuredangle SAK=\mathrm{arccos}\frac{\sqrt{3}}{2}={30}^{\circ }.$.
BK is a median of $\mathrm{\Delta }ABC$ and is not perpendicular to AC, otherwise $AB=BC$, which is a contradiction.
By the way, $BK\perp SK$, but we said about it in the first line.
Step 2
Let SK′ be an altitude of the pyramid.
Thus, since $SA=SB=SC$, we obtain: ΔSAK′≅ΔSBK′ and $\mathrm{\Delta }SA{K}^{\prime }\cong \mathrm{\Delta }SC{K}^{\prime }$, which gives $A{K}^{\prime }=B{K}^{\prime }=C{K}^{\prime },$, which says K′ is a center of the circumcircle for $\mathrm{\Delta }ABC$.
Thus, ${K}^{\prime }\equiv K$.
Do you have a similar question?
Recalculate according to your conditions!
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https://www.physicsforums.com/threads/turns-of-a-wire.383165/
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# Turns of a Wire
## Homework Statement
A generator is designed to produce a maximum emf of 180 V while rotating with an angular speed of 3900 rpm. Each coil of the generator has an area of 2.0×10−2 m^2.
a) If the magnetic field used in the generator has a magnitude of 4.1×10−2 T, how many turns of wire are needed?
emf = NABw ???
## The Attempt at a Solution
I'm not sure if this is even the correct equation, but my main problem is that I'm not really sure what the 3900 rpm is (i.e. what symbol it is and what it is in general). Any help would be appreciated.
Related Introductory Physics Homework Help News on Phys.org
rock.freak667
Homework Helper
## Homework Statement
A generator is designed to produce a maximum emf of 180 V while rotating with an angular speed of 3900 rpm. Each coil of the generator has an area of 2.0×10−2 m^2.
a) If the magnetic field used in the generator has a magnitude of 4.1×10−2 T, how many turns of wire are needed?
emf = NABw ???
## The Attempt at a Solution
I'm not sure if this is even the correct equation, but my main problem is that I'm not really sure what the 3900 rpm is (i.e. what symbol it is and what it is in general). Any help would be appreciated.
the 'w' in your relevant equation is the angular velocity in rad/s. So you need to convert rpm to rad/s
how exactly would i do that? but that is the correct equation though?
rock.freak667
Homework Helper
how exactly would i do that? but that is the correct equation though?
yes it is the correct equation.
1 rpm = 1 revolution per minute.
1 rpm = 2π radians per minute.
How many per second now?
Can you now convert 3900 rpm now?
im not sure if i did it right, but i got 234,000. the radians thing is throwing me off
ideasrule
Homework Helper
No, that's not right. How many radians is 3900 rpm? How many seconds is one minute? How many radians does the thing cover per second if it covers that many radians in that many seconds?
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https://discuss.leetcode.com/topic/12546/java-solution-to-with-o-1-space
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Java solution to with O(1)space
• just use the edge thingking to think about this question, while the num in the array equals 0, make the first num of its row and column to 0,so that can not use the extra space
``````public void setZeroes(int[][] matrix) {
boolean isFirstRowZero = false;
boolean isFirstColumnZero = false;
for(int i = 0; i < matrix.length ; i++)
{
if(matrix[i][0] == 0)
{
isFirstColumnZero = true;
}
}
for(int i = 0; i < matrix[0].length; i++)
{
if(matrix[0][i] == 0)
{
isFirstRowZero = true;
}
}
for(int i = 1; i< matrix.length;i++)
{
for(int j = 1; j < matrix[i].length; j++)
{
if(matrix[i][j] == 0)
{
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for(int i = 1; i< matrix.length;i++)
{
if(matrix[i][0] == 0)
{
for(int j = 0; j < matrix[i].length; j++)
{
matrix[i][j] = 0;
}
}
}
for(int i = 1; i< matrix[0].length;i++)
{
if(matrix[0][i] == 0)
{
for(int j = 0; j < matrix.length; j++)
{
matrix[j][i] = 0;
}
}
}
if(isFirstRowZero)
{
for(int i = 0; i< matrix[0].length;i++)
{
matrix[0][i] = 0;
}
}
if(isFirstColumnZero )
{
for(int i = 0; i< matrix.length;i++)
{
matrix[i][0] = 0;
}
}
}``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.
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http://www.slideshare.net/studsplanet/interest-rate-swaps-caps
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# Interest rate swaps, caps.....
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### Interest rate swaps, caps.....
1. 1. Chapter 29 Interest-Rate Swaps, Caps, and FloorsCopyright © 2010Pearson Education, Inc. 29-1
2. 2. Learning ObjectivesAfter reading this chapter, you will understand what an interest-rate swap is the relationship between an interest-rate swap and forward contracts how interest-rate swap terms are quoted in the market how the swap rate is calculated how the value of a swap is determined the primary determinants of the swap rate how a swap can be used by institutional investors for asset/liability management Copyright © 2010 Pearson Education, Inc. 29-2
3. 3. Learning Objectives (continued)After reading this chapter, you will understand how a structured note is created using an interest-rate swap what a swaption is and how it can be used by institutional investors what a rate cap and floor are, and how these agreements can be used by institutional investors the relationship between a cap and floor and options how to value caps and floors how an interest-rate collar can be created Copyright © 2010 Pearson Education, Inc. 29-3
4. 4. Interest-Rate Swaps In an interest-rate swap, two parties (called counterparties) agree to exchange periodic interest payments. The dollar amount of the interest payments exchanged is based on a predetermined dollar principal, which is called the notional principal amount. The dollar amount that each counterparty pays to the other is the agreed-upon periodic interest rate times the notional principal amount. The only dollars that are exchanged between the parties are the interest payments, not the notional principal amount. This party is referred to as the fixed-rate payer or the floating-rate receiver. The other party, who agrees to make interest rate payments that float with some reference rate, is referred to as the floating-rate payer or fixed-rate receiver. The frequency with which the interest rate that the floating-rate payer must pay is called the reset frequency. Copyright © 2010 Pearson Education, Inc. 29-4
5. 5. Interest-Rate Swaps (continued) Entering into a Swap and Counterparty Risk Interest-rate swaps are over-the-counter instruments, which means that they are not traded on an exchange. An institutional investor wishing to enter into a swap transaction can do so through either a securities firm or a commercial bank that transacts in swaps. The risks that parties take on when they enter into a swap are that the other party will fail to fulfill its obligations as set forth in the swap agreement; that is, each party faces default risk. The default risk in a swap agreement is called counterparty risk. Copyright © 2010 Pearson Education, Inc. 29-5
6. 6. Interest-Rate Swaps (continued) Interpreting a Swap Position There are two ways that a swap position can be interpreted:i. as a package of forward/futures contractsii. as a package of cash flows from buying and selling cash market instruments Although an interest-rate swap may be nothing more than a package of forward contracts, it is not a redundant contract, for several reasons.i. Maturities for forward or futures contracts do not extend out as far as those of an interest-rate swap.ii. An interest-rate swap is a more transactionally efficient instrument because in one transaction an entity can effectively establish a payoff equivalent to a package of forward contracts.iii. Interest-rate swaps now provide more liquidity than forward contracts, particularly long-dated (i.e., long-term) forward contracts. Copyright © 2010 Pearson Education, Inc. 29-6
7. 7. Interest-Rate Swaps (continued) Interpreting a Swap Position To understand why a swap can also be interpreted as a package of cash market instruments, consider an investor who enters into the following transaction:o Buy \$50 million par of a five-year floating-rate bond that pays six-month LIBOR every six months; finance the purchase by borrowing \$50 million for five years at 10% annual interest rate paid every six months. The cash flows for this transaction are shown in Exhibit 29-1 (see Overhead 29- 8). The second column shows the cash flow from purchasing the five-year floating-rate bond. There is a \$50 million cash outlay and then 10 cash inflows. The amount of the cash inflows is uncertain because they depend on future LIBOR. The next column shows the cash flow from borrowing \$50 million on a fixed-rate basis. The last column shows the net cash flow from the entire transaction. As the last column indicates, there is no initial cash flow (no cash inflow or cash outlay). In all 10 six-month periods, the net position results in a cash inflow of LIBOR and a cash outlay of \$2.5 million. This net position, however, is identical to the position of a fixed-rate payer/floating-rate receiver. Copyright © 2010 Pearson Education, Inc. 29-7
8. 8. Exhibit 29-1 Cash Flow for the Purchase of a Five-Year Floating-Rate Bond Financed by Borrowing on a Fixed-Rate BasisTransaction: Purchase for \$50 million a five-year floating-rate bond: floating rate = LIBOR,semiannual pay; borrow \$50 million for five years: fixed rate = 10%, semiannual payments Cash Flow (millions of dollars) From: Six-Month Borrowing Period Floating-Rate Bond a Cost Net 0 –\$50.0 +\$50 \$0 1 +(LIBOR1/2)×50 –2.5 + (LIBOR1/2)×50–2.5 2 +(LIBOR2/2)×50 –2.5 + (LIBOR2/2)×50–2.5 3 +(LIBOR3/2)×50 –2.5 + (LIBOR3/2)×50–2.5 4 +(LIBOR4/2)×50 –2.5 + (LIBOR4/2)×50–2.5 5 +(LIBOR5/2)×50 –2.5 + (LIBOR5/2)×50–2.5 6 +(LIBOR6/2)×50 –2.5 + (LIBOR6/2)×50–2.5 7 +(LIBOR7/2)×50 –2.5 + (LIBOR7/2)×50–2.5 8 +(LIBOR8/2)×50 –2.5 + (LIBOR8/2)×50–2.5 9 +(LIBOR9/2)×50 –2.5 + (LIBOR9/2)×50–2.5 10 +(LIBOR10/2)×50+50 –52.5 + (LIBOR10/2)×50–2.5a The subscript for LIBOR indicates the six-month LIBOR as per the terms of the floating-rate bondat time t. Copyright © 2010 Pearson Education, Inc. 29-8
9. 9. Interest-Rate Swaps (continued) Terminology, Conventions, and Market Quotes The date that the counterparties commit to the swap is called the trade date. The date that the swap begins accruing interest is called the effective date, and the date that the swap stops accruing interest is called the maturity date. The convention that has evolved for quoting swaps levels is that a swap dealer sets the floating rate equal to the index and then quotes the fixed-rate that will apply.o The offer price that the dealer would quote the fixed-rate payer would be to pay 8.85% and receive LIBOR “flat” (“flat” meaning with no spread to LIBOR).o The bid price that the dealer would quote the floating-rate payer would be to pay LIBOR flat and receive 8.75%.o The bid-offer spread is 10 basis points. Copyright © 2010 Pearson Education, Inc. 29-9
10. 10. Interest-Rate Swaps (continued) Terminology, Conventions, and Market Quotes Another way to describe the position of the counterparties to a swap is in terms of our discussion of the interpretation of a swap as a package of cash market instruments.o Fixed-rate payer: A position that is exposed to the price sensitivities of a longer-term liability and a floating-rate bond.o Floating-rate payer: A position that is exposed to the price sensitivities of a fixed-rate bond and a floating-rate liability. The convention that has evolved for quoting swaps levels is that a swap dealer sets the floating rate equal to the index and then quotes the fixed rate that will apply. To illustrate this convention, consider a 10-year swap offered by a dealer to market participants shown in Exhibit 29-2 (see Overhead 29-12). Copyright © 2010 Pearson Education, Inc. 29-10
11. 11. Interest-Rate Swaps (continued) Terminology, Conventions, and Market Quotes In our illustration, suppose that the 10-year Treasury yield is 8.35%. Then the offer price that the dealer would quote to the fixed-rate payer is the 10-year Treasury rate plus 50 basis points versus receiving LIBOR flat. For the floating-rate payer, the bid price quoted would be LIBOR flat versus the 10-year Treasury rate plus 40 basis points. The dealer would quote such a swap as 40–50, meaning that the dealer is willing to enter into a swap to receive LIBOR and pay a fixed rate equal to the 10-year Treasury rate plus 40 basis points, and it would be willing to enter into a swap to pay LIBOR and receive a fixed rate equal to the 10-year Treasury rate plus 50 basis points. The difference between the Treasury rate paid and received is the bid-offer spread. Copyright © 2010 Pearson Education, Inc. 29-11
12. 12. Exhibit 29-2 Meaning of a “40–50” Quote fora 10-Year Swap When Treasuries Yield 8.35%(Bid-Offer Spread of 10 Basis Points) Floating-Rate Fixed-Rate Payer Payer Pay Floating rate of Fixed rate of six-month 8.85% LIBOR Receive Fixed rate of Floating rate of 8.75% six-month LIBOR Copyright © 2010 Pearson Education, Inc. 29-12
13. 13. Interest-Rate Swaps (continued) Calculation of the Swap Rate At the initiation of an interest-rate swap, the counterparties are agreeing to exchange future interest-rate payments and no upfront payments by either party are made. While the payments of the fixed-rate payer are known, the floating-rate payments are not known. This is because they depend on the value of the reference rate at the reset dates. For a LIBOR-based swap, the Eurodollar CD futures contract can be used to establish the forward (or future) rate for three-month LIBOR. In general, the floating-rate payment is determined as follows: floating − rate payment = number of days in period notional amount × three − month LIBOR × 360 Copyright © 2010 Pearson Education, Inc. 29-13
14. 14. Interest-Rate Swaps (continued) Calculation of the Swap Rate The equation for determining the dollar amount of the fixed-rate payment for the period is: fixed − rate payment = number of days in period notional amount × swap rate × 360 It is the same equation as for determining the floating-rate payment except that the swap rate is used instead of the reference rate. Exhibit 29-4 (see Overhead 29-15) shows the fixed-rate payments based on an assumed swap rate of 4.9875%.o The first three columns of the exhibit show the beginning and end of the quarter and the number of days in the quarter. Column (4) simply uses the notation for the period.o That is, period 1 means the end of the first quarter, period 2 means the end of the second quarter, and so on.o Column (5) shows the fixed-rate payments for each period based on a swap rate of 4.9875%. Copyright © 2010 Pearson Education, Inc. 29-14
15. 15. Exhibit 29-4 Fixed-Rate Payments Assuming a Swap Rate of 4.9875% Quarter Days in Period = End Fixed-Rate Payment if Swap Quarter Ends Starts Quarter of Quarter Rate Is Assumed to Be 4.9875%Jan 1 year 1 Mar 31 year 1 90 1 1,246,875Apr 1 year 1 June 30 year 1 91 2 1,260,729July 1 year 1 Sept 30 year 1 92 3 1,274,583Oct 1 year 1 Dec 31 year 1 92 4 1,274,583Jan 1 year 2 Mar 31 year 2 90 5 1,246,875Apr 1 year 2 June 30 year 2 91 6 1,260,729July 1 year 2 Sept 30 year 2 92 7 1,274,583Oct 1 year 2 Dec 31 year 2 92 8 1,274,583Jan 1 year 3 Mar 31 year 3 90 9 1,246,875Apr 1 year 3 June 30 year 3 91 10 1,260,729July 1 year 3 Sept 30 year 3 92 11 1,274,583Oct 1 year 3 Dec 31 year 3 92 12 1,274,583 Copyright © 2010 Pearson Education, Inc. 29-15
16. 16. Interest-Rate Swaps (continued) Calculation of the Swap Rate Given the swap payments, we can show how to compute the swap rate. At the initiation of an interest-rate swap, the counterparties are agreeing to exchange future payments and no upfront payments by either party are made. This means that the present value of the payments to be made by the counterparties must be at least equal to the present value of the payments that will be received. To eliminate arbitrage opportunities, the present value of the payments made by a party will be equal to the present value of the payments received by that same party. The equivalence of the present value of the payments is the key principle in calculating the swap rate. Copyright © 2010 Pearson Education, Inc. 29-16
17. 17. Interest-Rate Swaps (continued) Calculation of the Swap Rate The present value of \$1 to be received in period t is the forward discount factor. In calculations involving swaps, we compute the forward discount factor for a period using the forward rates. These are the same forward rates that are used to compute the floating- rate payments—those obtained from the Eurodollar CD futures contract.o We must make just one more adjustment.o We must adjust the forward rates used in the formula for the number of days in the period (i.e., the quarter in our illustrations) in the same way that we made this adjustment to obtain the payments.o Specifically, the forward rate for a period, which we will refer to as the period forward rate, is computed using the following equation: days in period period forward rate = annual forward rate × 360 Copyright © 2010 Pearson Education, Inc. 29-17
18. 18. Interest-Rate Swaps (continued) Calculation of the Swap Rate Given the payment for a period and the forward discount factor for the period, the present value of the payment can be computed. The forward discount factor is used to compute the present value of the both the fixed-rate payments and floating-rate payments. Beginning with the basic relationship for no arbitrage to exist: PV of floating-rate payments = PV of fixed-rate payments The formula for the swap rate is derived as follows. We begin with: fixed-rate payment for period t = days in period notional amount × swap rate × 360 Copyright © 2010 Pearson Education, Inc. 29-18
19. 19. Interest-Rate Swaps (continued) Calculation of the Swap Rate The present value of the fixed-rate payment for period t is found by multiplying the previous expression by the forward discount factor for period t. We have: present value of the fixed-rate payment for period t = days in period tnotional amount × swap rate × × forward discount factor for period t 360 Summing up the present value of the fixed-rate payment for each period gives the present value of the fixed-rate payments. Letting N be the number of periods in the swap, we have: present value of the fixed-rate payment = days in period tswap rate × ∑ notional amount × × forward discount factor for period t 360 Copyright © 2010 Pearson Education, Inc. 29-19
20. 20. Interest-Rate Swaps (continued) Calculation of the Swap Rate Solving for the swap rate gives swap rate = present value of floating-rate payments N days in period t ∑ t =1 notional amount × 360 × forward discount factor for period t Valuing a Swap Once the swap transaction is completed, changes in market interest rates will change the payments of the floating-rate side of the swap. The value of an interest-rate swap is the difference between the present value of the payments of the two sides of the swap. Copyright © 2010 Pearson Education, Inc. 29-20
21. 21. Interest-Rate Swaps (continued) Duration of a Swap As with any fixed-income contract, the value of a swap will change as interest rates change. Dollar duration is a measure of the interest-rate sensitivity of a fixed-income contract. From the perspective of the party who pays floating and receives fixed, the interest-rate swap position can be viewed as follows: long a fixed-rate bond + short a floating-rate bond This means that the dollar duration of an interest-rate swap from the perspective of a floating-rate payer is simply the difference between the dollar duration of the two bond positions that make up the swap; that is,dollar duration of a swap = dollar duration of a fixed-rate bond – dollar duration of a floating-rate bond Copyright © 2010 Pearson Education, Inc. 29-21
22. 22. Interest-Rate Swaps (continued) Application of a Swap to Asset/Liability Management An interest-rate swap can be used to alter the cash flow characteristics of an institution’s assets so as to provide a better match between assets and liabilities. An interest-rate swap allows each party to accomplish its asset/liability objective of locking in a spread. An asset swap permits the two financial institutions to alter the cash flow characteristics of its assets: from fixed to floating or from floating to fixed. A liability swap permits two institutions to change the cash flow nature of their liabilities. Copyright © 2010 Pearson Education, Inc. 29-22
23. 23. Interest-Rate Swaps (continued) Creation of Structured Notes Using Swaps Corporations can customize medium-term notes for institutional investors who want to make a market play on interest rate, currency, and/or stock market movements. That is, the coupon rate on the issue will be based on the movements of these financial variables. A corporation can do so in such a way that it can still synthetically fix the coupon rate. This can be accomplished by issuing an MTN and entering into a swap simultaneously. MTNs created in this way are called structured MTNs. Copyright © 2010 Pearson Education, Inc. 29-23
24. 24. Interest-Rate Swaps (continued) Primary Determinants of Swap Spreads The swap spread is determined by the same factors that influence the spread over Treasuries on financial instruments (futures / forward contracts or cash) that produce a similar return or funding profile. Given that a swap is a package of futures/forward contracts, the swap spread can be determined by looking for futures/forward contracts with the same risk/return profile. A Eurodollar CD futures contract is a swap where a fixed dollar payment (i.e., the futures price) is exchanged for three-month LIBOR. A market participant can synthesize a (synthetic) fixed-rate security or a fixed-rate funding vehicle of up to five years by taking a position in a strip of Eurodollar CD futures contracts (i.e., a position in every three-month Eurodollar CD up to the desired maturity date). Copyright © 2010 Pearson Education, Inc. 29-24
25. 25. Interest-Rate Swaps (continued) Primary Determinants of Swap Spreads For swaps with maturities longer than five years, the spread is determined primarily by the credit spreads in the corporate bond market. Because a swap can be interpreted as a package of long and short positions in a fixed-rate bond and a floating-rate bond, it is the credit spreads in those two market sectors that will be the key determinant of the swap spread. Boundary conditions for swap spreads based on prices for fixed- rate and floating-rate corporate bonds can be determined. Several technical factors, such as the relative supply of fixed-rate and floating-rate corporate bonds and the cost to dealers of hedging their inventory position of swaps, influence where between the boundaries the actual swap spread will be Copyright © 2010 Pearson Education, Inc. 29-25
26. 26. Interest-Rate Swaps (continued) Development of the Interest-Rate Swap Market The initial motivation for the interest-rate-swap market was borrower exploitation of what was perceived to be “credit arbitrage” opportunities.o These opportunities resulted from differences in the quality spread between lower- and higher-rated credits in the U.S. and Eurodollar bond fixed-rate market and the same spread in these two floating-rate markets. Basically, the argument for swaps was based on a well-known economic principle of comparative advantage in international economics.o The argument in the case of swaps is that even though a high credit- rated issuer could borrow at a lower cost in both the fixed- and floating- rate markets (i.e., have an absolute advantage in both), it will have a comparative advantage relative to a lower credit-rated issuer in one of the markets (and a comparative disadvantage in the other). Copyright © 2010 Pearson Education, Inc. 29-26
27. 27. Interest-Rate Swaps (continued) Role of the Intermediary The role of the intermediary in an interest-rate swap sheds some light on the evolution of the market.o Intermediaries in these transactions have been commercial banks and investment banks, who in the early stages of the market sought out end users of swaps.o That is, they found in their client bases those entities that needed the swap to accomplish a funding or investing objective, and they matched the two entities.o In essence, the intermediary in this type of transaction performed the function of a broker.o The only time that the intermediary would take the opposite side of a swap (i.e., would act as a principal) was to balance out the transaction. Copyright © 2010 Pearson Education, Inc. 29-27
28. 28. Interest-Rate Swaps (continued) Beyond the Plain Vanilla Swap In a generic or plain vanilla swap, the notional principal amount does not vary over the life of the swap. Thus it is sometimes referred to as a bullet swap. In contrast, for amortizing, accreting, and roller coaster swaps, the notional principal amount varies over the life of the swap. An amortizing swap is one in which the notional principal amount decreases in a predetermined way over the life of the swap.o Such a swap would be used where the principal of the asset that is being hedged with the swap amortizes over time. Less common than the amortizing swap are the accreting swap and the roller coaster swap. An accreting swap is one in which the notional principal amount increases in a predetermined way over time. In a roller coaster swap, the notional principal amount can rise or fall from period to period. Copyright © 2010 Pearson Education, Inc. 29-28
29. 29. Interest-Rate Swaps (continued) Beyond the Plain Vanilla Swap The terms of a generic interest-rate swap call for the exchange of fixed- and floating-rate payments. In a basis rate swap, both parties exchange floating-rate payments based on a different reference rate.o The risk is that the spread between the prime rate and LIBOR will change. This is referred to as basis risk. Another popular swap is to have the floating leg tied to a longer-term rate such as the two-year Treasury note rather than a money market rate.o Such a swap is called a constant maturity swap. Copyright © 2010 Pearson Education, Inc. 29-29
30. 30. Interest-Rate Swaps (continued) Beyond the Plain Vanilla Swap There are options on interest-rate swaps.o These swap structures are called swaptions and grant the option buyer the right to enter into an interest-rate swap at a future date.o There are two types of swaptions – a payer swaption and a receiver swaption.i. A payer swaption entitles the option buyer to enter into an interest-rate swap in which the buyer of the option pays a fixed-rate and receives a floating rate.ii. In a receiver swaption the buyer of the swaption has the right to enter into an interest-rate swap that requires paying a floating rate and receiving a fixed-rate. Copyright © 2010 Pearson Education, Inc. 29-30
31. 31. Interest-Rate Swaps (continued) Forward Start Swap A forward start swap is a swap wherein the swap does not begin until some future date that is specified in the swap agreement. Thus, there is a beginning date for the swap at some time in the future and a maturity date for the swap. A forward start swap will also specify the swap rate at which the counterparties agree to exchange payments commencing at the start date. Copyright © 2010 Pearson Education, Inc. 29-31
32. 32. Interest-Rate Caps and Floors An interest-rate agreement is an agreement between two parties whereby one party, for an upfront premium, agrees to compensate the other at specific time periods if a designated interest rate, called the reference rate, is different from a predetermined level. When one party agrees to pay the other when the reference rate exceeds a predetermined level, the agreement is referred to as an interest-rate cap or ceiling. The agreement is referred to as an interest-rate floor when one party agrees to pay the other when the reference rate falls below a predetermined level. The predetermined interest-rate level is called the strike rate. Copyright © 2010 Pearson Education, Inc. 29-32
33. 33. Interest-Rate Caps and Floors (continued) Interest-rate caps and floors can be combined to create an interest-rate collar. This is done by buying an interest-rate cap and selling an interest-rate floor. Some commercial banks and investment banking firms write options on interest-rate agreements for customers. Options on caps are captions; options on floors are called flotions. Copyright © 2010 Pearson Education, Inc. 29-33
34. 34. Interest-Rate Caps and Floors (continued) Risk/Return Characteristics In an interest-rate agreement, the buyer pays an upfront fee representing the maximum amount that the buyer can lose and the maximum amount that the writer of the agreement can gain. The only party that is required to perform is the writer of the interest-rate agreement. The buyer of an interest-rate cap benefits if the underlying interest rate rises above the strike rate because the seller (writer) must compensate the buyer. The buyer of an interest rate floor benefits if the interest rate falls below the strike rate, because the seller (writer) must compensate the buyer. Copyright © 2010 Pearson Education, Inc. 29-34
35. 35. Interest-Rate Caps and Floors (continued) Valuing Caps and Floors The arbitrage-free binomial model can be used to value a cap and a floor. This is because a cap and a floor are nothing more than a package or strip of options. More specifically, they are a strip of European options on interest rates. Thus to value a cap the value of each period’s cap, called a caplet, is found and all the caplets are then summed. We refer to this approach to valuing a cap as the caplet method. (The same approach can be used to value a floor.) Once the caplet method is demonstrated, we will show an easier way of valuing a cap. Similarly, an interest rate floor can be valued. The value for the floor for any year is called a floorlet. Copyright © 2010 Pearson Education, Inc. 29-35
36. 36. Interest-Rate Caps and Floors (continued) Valuing Caps and Floors To illustrate the caplet method, we will use the binomial interest-rate tree used in Chapter 18 to value an interest rate option to value a 5.2%, three-year cap with a notional amount of \$10 million. The reference rate is the one-year rates in the binomial tree and the payoff for the cap is annual. There is one wrinkle having to do with the timing of the payments for a cap and floor that requires a modification of the binomial approach presented to value an interest rate option. This is due to the fact that settlement for the typical cap and floor is paid in arrears. Exhibit 29-11 (see Overhead 29-37) shows the binomial interest rate tree with dates and years. Copyright © 2010 Pearson Education, Inc. 29-36
37. 37. Exhibit 29-11 Binomial Interest RateTree with Dates and Years Identified 7.0053% 5.4289% NHH NH 3.500% 5.7354% N NHL 4.4448% NL 4.6958% NLL Dates: 0 1 2 3 Years: One Two Threes Copyright © 2010 Pearson Education, Inc. 29-37
38. 38. Interest-Rate Caps and Floors (continued) Using a Single Binomial Tree to Value a Cap The valuation of a cap can be done by using a single binomial tree. The procedure is easier only in the sense that the number of times discounting is required is reduced. The method is shown in Exhibit 29-13 (see Overhead 29-40). The three values at Date 2 are obtained by simply computing the payoff at Date 3 and discounting back to Date 2. Let’s look at the higher node at Date 1 (interest rate of 5.4289%). The top number, \$104,026, is the present value of the two Date 2 values that branch out from that node. Copyright © 2010 Pearson Education, Inc. 29-38
39. 39. Interest-Rate Caps and Floors (continued) Using a Single Binomial Tree to Value a Cap The number below it, \$21,711, is the payoff of the Year Two caplet on Date 1. The third number down at the top node at Date 1 in Exhibit 29- 13, which is in bold, is the sum of the top two values above it. It is this value that is then used in the backward induction. The same procedure is used to get the values shown in the boxes at the lower node at Date 1. Given the values at the two nodes at Date 1, the bolded values are averaged to obtain (\$125,737 + \$24,241)/2 = \$74,989. Discounting this value at 3.5% gives \$72,453. This is the same value obtained from using the caplet approach. Copyright © 2010 Pearson Education, Inc. 29-39
40. 40. Exhibit 29-13 Valuing a Cap Using a Single Binomial Tree \$168,711 \$180,530 \$104,026 7.0053% NH \$21,711 \$125,737 5.4289% \$72,753 \$50,636 3.500% \$53,540N 5.7354% \$24,241 \$0 \$24,241 NL 4.4448% \$0 \$0 4.6958% Dates: 0 1 2 3 Years: One Two Threes Copyright © 2010 Pearson Education, Inc. 29-40
41. 41. Interest-Rate Caps and Floors (continued) Applications To see how interest-rate agreements can be used for asset/liability management, consider the problems faced by a commercial bank which needs to lock in an interest-rate spread over its cost of funds. Because the bank borrows short term, its cost of funds is uncertain. The bank may be able to purchase a cap, however, so that the cap rate plus the cost of purchasing the cap is less than the rate it is earning on its fixed-rate commercial loans. If short-term rates decline, the bank does not benefit from the cap, but its cost of funds declines. The cap therefore allows the bank to impose a ceiling on its cost of funds while retaining the opportunity to benefit from a decline in rates. Copyright © 2010 Pearson Education, Inc. 29-41
42. 42. Interest-Rate Caps and Floors (continued) Applications The bank can reduce the cost of purchasing the cap by selling a floor. In this case the bank agrees to pay the buyer of the floor if the reference rate falls below the strike rate. The bank receives a fee for selling the floor, but it has sold off its opportunity to benefit from a decline in rates below the strike rate. By buying a cap and selling a floor the bank creates a “collar” with a predetermined range for its cost of funds. Copyright © 2010 Pearson Education, Inc. 29-42
43. 43. All rights reserved. No part of this publication may be reproduced,stored in a retrieval system, or transmitted, in any form or by any means,electronic, mechanical, photocopying, recording, or otherwise, withoutthe prior written permission of the publisher. Printed in the UnitedStates of America. Copyright © 2010 Pearson Education, Inc. 29-43
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https://community.jmp.com/t5/Discussions/Calculate-a-vector-of-miniums-from-a-2-dimentional-matrix/td-p/270751
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Share your ideas for the JMP Scripting Unsession at Discovery Summit by September 17th. We hope to see you there!
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Super User
## Calculate a vector of miniums from a 2 dimentional matrix
I am looking for a simple method to take a 2xn matrix and product a vector where each element in the vector is the minimum value for each row in the 2xn matrix
``````a=[ 1 4, 3 2, 5 6]
// the final result needs to be
[1, 2, 5]``````
Looping through matrix "a" can easily do this, but......I am hoping someone out there is more fluent in matrices that may have a more efficient solution
Jim
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Super User
## Re: Calculate a vector of miniums from a 2 dimentional matrix
Hi Jim,
``````m = [1 4, 3 2, 5 6];
v = Transpose( V Min( Transpose( m ) ) ); // VMin( m`)` ``````
5 REPLIES 5
Highlighted
Super User
## Re: Calculate a vector of miniums from a 2 dimentional matrix
Hi Jim,
``````m = [1 4, 3 2, 5 6];
v = Transpose( V Min( Transpose( m ) ) ); // VMin( m`)` ``````
Highlighted
Super User
excellent...
Jim
Highlighted
Staff (Retired)
## Re: Calculate a vector of miniums from a 2 dimentional matrix
This set of functions all work on the vertical dimension of a matrix and can use the same technique to get a horizontal version.
Craige
Highlighted
Staff
## Re: Calculate a vector of miniums from a 2 dimentional matrix
I knew there was a function but I couldn't find it, so I did it step by step using matrix operations. Touché @gzmorgan0 !
``````Names Default to Here( 1 );
// define 2xn data matrix
a = [ 1 4, 3 2, 5 6];
// compute row-wise differences as a vector
b = a[0,1] - a[0,2];
// determine deficit in first column of data matrix
c = b < 0;
// determine deficit in second column of data matrix
d = !c;
// create conforming matrix of deficits
e = c || d;
// multiply to get minimum value
f = a * e`;
// extract minimum value from diagonal elements
Vec Diag( f );``````
I doubt that my way is more efficient than iterating over the rows of the data matrix. I am sure it is not more efficient than V Min()!
Learn it once, use it forever!
Highlighted
Staff (Retired)
## Re: Calculate a vector of miniums from a 2 dimentional matrix
Wow! I hope you are doing @DonMcCormack 's challenges! I'm still going to need to work through that a bit longer.
Craige
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Welcome! On this website you can find information about each number.
Number 189763199
Number 189763199 basic info
Number 189763199 has 9 digits. Number 189763199 can be formatted as 189,763,199 or 189.763.199 or 189 763 199 or in case this was a phone number 189-763-199 or 18-976-3199 to be easier to read. Number 189763199 in English words is "one hundred and eighty-nine million, seven hundred and sixty-three thousand, one hundred and ninety-nine". Number 189763199 can be read by triplets (groups of 3 digits) as "one hundred and eighty-nine, seven hundred and sixty-three, one hundred and ninety-nine". Number 189763199 can be read digit by digit as "one eight nine seven six three one nine nine". Number 189763199 is odd. Number 189763199 is not divisible by any single digit number. Number 189763199 is likely to be a prime number.
Number 189763199 conversions
Number 189763199 in binary code is 1011010011111000111001111111. Number 189763199 in octal code is: 1323707177. Number 189763199 in hexadecimal (hexa): b4f8e7f.
The sum of all digits of this number is 53. The digital root (repeated digital sum until you get single-digit number) is 8. Number 189763199 divided by two (halved) equals 94881599.5. Number 189763199 multiplied by two (doubled) equals 379526398. Number 189763199 multiplied by ten equals 1897631990. Number 189763199 raised to the power of 2 equals 3.6010071694714E+16. Number 189763199 raised to the power of 3 equals 6.8333864010082E+24. The square root (sqrt) of 189763199 is 13775.456398973. The sine (sin) of 189763199 degree is -0.017452406131445. The cosine (cos) of 189763199 degree is 0.99984769516173. The base-10 logarithm of 189763199 equals 8.2782119930182. The natural logarithm of 189763199 equals 19.061287531768. The number 189763199 can be encoded to characters as AHIGFCAII. The number 189763199 can be encrypted to chemical element names as hydrogen, oxygen, fluorine, nitrogen, carbon, lithium, hydrogen, fluorine, fluorine.
Numbers simmilar to 189763199
Numbers simmilar to number 189763199 (one digit altered): 89763199289763199179763199199763199188763199189663199189863199189753199189773199189762199189764199189763099189763299189763189189763198
Possible variations of 189763199 with a digit pair swapped: 819763199198763199187963199189673199189736199189761399189763919
Number 189763199 typographic errors with one digit missing: 897631991976319918763199189631991897319918976199189763991897631918976319
Number 189763199 typographic errors with one digit doubled: 118976319918897631991899763199189776319918976631991897633199189763119918976319991897631999
Previous number: 189763198
Next number: 189763200
Several randomly selected numbers:
739460328497079648082648443148212734425802728768222806388903338116012979554127884200875747026938026804561828294778308528060300050671615757156435371143323523871081305504263517790472229668451609049521175945101483451567721137211137135051300799932400472732444287713699207170729755605855289970444163847178601347656781159651862235624708656113747899628655787535935230596078750733856906466712607035.
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## Introducing Supply and Demand
Read the sections on Demand, Supply, Market Equilibrium, and Government Intervention and Disequilibrium for a mathematical exposition of the demand and supply model, clicking through to the next when you have finished each page. The chapter also covers price ceilings and price floor analysis as well as quantity regulations.
### 2. Supply - The Law of Supply
The law of supply states that there is a positive relationship between the quantity that suppliers are willing to sell and the price level.
#### LEARNING OBJECTIVES
Explain the Law of Supply
#### KEY TAKEAWAYS
##### Key Points
• Quantity supplied moves in the same direction as price.
• The supply curve is an upward sloping curve.
• Producers are willing to increase production at higher prices to increase profit.
##### Key Terms
• surplus: That which remains when use or need is satisfied, or when a limit is reached; excess; overplus.
• shortage: a lack or deficiency
• equilibrium: The condition of a system in which competing influences are balanced, resulting in no net change.
The law of supply is a fundamental principle of economic theory. It states that an increase in price will result in an increase in the quantity supplied, all else held constant.
An upward sloping supply curve, which is also the standard depiction of the supply curve, is the graphical representation of the law of supply. As the price of a good or service increases, the quantity that suppliers are willing to produce increases and this relationship is captured as a movement along the supply curve to a higher price and quantity combination.
The Law of Supply: Supply has a positive correlation with price. As the market price of a good increases, suppliers of the good will typically seek to increase the quantity supplied to the market.
The rationale for the positive correlation between price and quantity supplied is based on the potential increase in profitability that occurs with an increase in price.
All else held constant, including the costs of production inputs, the supplier will be able to increase his return per unit of a good or service as the price for the item increases. Therefore, the net return to the supplier increases as the spread or difference between the price and the cost of the good or service being sold increases.
The law of supply in conjunction with the law of demand forms the basis for market conditions resulting in a price and quantity relationship at which both the price to quantity relationship of suppliers and demanders (consumers) are equal. This is also referred to as the equilibrium price and quantity and is depicted graphically at the point at which the demand and supply curve intersect or cross one another. It is the point where there is no surplus or shortage in the market.
Law of Supply and Law of Demand: Equilibrium: The law of supply and the law of demand form the foundation for the establishment of an equilibrium–where the price to quantity combination for both suppliers and demanders are the same.
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# Thread: Which Metagem to use
1. New Registrant
Join Date
May 2010
Posts
6
## Which Metagem to use
Hello,
Sorry for my bad english right away.
Today I did some math on the two meta gems available for Tanks.
+81 Stamina +2% Armor
+81 Stamina +5% Block
Since 5% Block is bugged it's only 1% .
The Armor values are just pre-made numbers to make the math a little more handy, but they are quite close to mine in a Raid. The other Stats are exact ones.
So we start with +2% Armor!
We assume:
40000 Armor with armor Gem: ~55,117% -> 44.883% incoming Dmg left
39215 Armor with block Gem: ~54,626% -> 45.374% incoming Dmg left
44.883 / 45.374 = 0.9891
Taking these numbers is a DMG reduction of ~1.08%
12,05% Dodge (pre DR) = ~10,57% Dodge post DR
15,30% Parry (pre DR) = ~12,86% Parry post DR
5,00% Miss
=28,42% Avoidance
Since Bosslvl is 88, you need 102.4% Avoidance to be unhittable.
So a Boss hits to 73.97% with his melee swings.
1.08%*0.7397 = 0.7988% effective DMG reduction from Austere Shadowspirit Diamond.
Now the Eternal Spiritshadow Diamond
30% Block value
52.01% Block Chance +8.33% (Shield Block 33% Uptime)
Based on data on World of Logs I have a Crit. Block Chance of ~37%
30*0.63+60*0.37 = 41.1% Block value average
By inserting the Block Meta Gem I'd gain 1% Block value and 2% critical Block value.
31*0.63+62*0.37 = 42.47% Block value
An increase of 1,37% Block value which in my case happens 60.34% of all swings.
Shield Block used on CD:
1.37*0.6034 = 0,8266% effective DMG reduction from Eternal Spiritshadow Diamond
Conclusion:
Eternal Shadowspirit Diamond provides 0.0278% more Mitigation if Shield Block is used on CD (in my case)
I'm assuming the Block Meta scales better with gear than the armor gem does, but i didn't do the math
If you find some mistakes feel free to reply
Greets
2. Established Registrant
Join Date
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Posts
534
In the example you used you have 28,4% avoidance and 52% block combined 80,4% so you only have 22% to get a full hit, shield block gives only 22% block in that case so it should be 22/3=7,33% block from shieldblock. The rest looks solid
Did the math myself with 30% avoidance and 54% block used a bit different calculations but my conclusion was that with the blockmeta i took 0,77% more damage on average in the the 20 secs without shieldblock and 2,37% less damage while shieldblock was up. Making the blockmeta slightly better on avarage but i found the difference not big enough to switch yet since armor meta is more reliable and helps in situations with bad hit strings.
Blockmeta will get better when we get closer to 102,4% unhittable, since the armormeta is only better on the full hits we receive. Think I'll switch after 5% more block but also depends how i value EH, might be that when i start with hardmodes I'm not willing to sacrifice the EH the armor meta gives.
3. Established Registrant
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You are missing the 3rd meta.
http://www.wowhead.com/item=52295
4. Originally Posted by Selene
You are missing the 3rd meta.
http://www.wowhead.com/item=52295
That is a very encounter specific gem, so specific I cannot really think of one I would use it on right now. There is no "sartharion+3 breath" to survive.
5. New Registrant
Join Date
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http://www.wowhead.com/item=52289
with http://www.wowhead.com/spell=74238 over http://www.wowhead.com/spell=74253
that's 69 mastery vs 81 stam 2% armor
need some maths on this : )
6. Established Registrant
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Originally Posted by Nukous
http://www.wowhead.com/item=52289
with http://www.wowhead.com/spell=74238 over http://www.wowhead.com/spell=74253
that's 69 mastery vs 81 stam 2% armor
need some maths on this : )
Depends on the Tank and Spec.
Prot Pally depending on how they spec, don't need the run increase.
also imo the 81 stam/+2% armor > 54 mastery
7. New Registrant
Join Date
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69 mastery provides something like .57% block.
0.57*0.30=0.171
so it provides 0.171% Mitigation excluding the crit block
Assuming my average block value is like 42%:
0.57*0.42=0.2394
-> 0.2394% Mitigation including crit block
Last edited by Shaizhowfn; 01-09-2011 at 07:24 AM.
8. New Registrant
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thanks Shaizhowfn
9. New Registrant
Join Date
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Posts
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Originally Posted by Shaizhowfn
+81 Stamina +5% Block
Since 5% Block is bugged it's only 1% .
Are you sure this is actually bugged? Normally, we block for 30% of the damage. 105% of 30% is 31.5%, which is what the metagem seems to actually do.
The wording is really vague and seems to make most people think that the 30% goes to 35%, but a lot of WoW +X% adjustments to things that are percentages themselves tend to be adjustments of the actual percentage value, not just added to the actual percentage value. For example, the Abyssal Seahorse mount says "Increases swim speed by 450%". Swim speed is normally 67.5% of running speed, so the Seahorse moves at 67.5% + 450%*67.5% = 371.25% [same speed as a land or air mount that Increased speed by 271.5%]. Notice that it's NOT 67.5% + 450% = 517.5% movement speed increase. Similarly, +5% to our value of 30% is 30% + 5%*30% = 31.5% NOT 30% + 5% = 35%.
10. Established Registrant
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Fairly certain the gem does indeed increase block to 31.5% damage reduction and 63% damage reduction on a critical block. Which would probably mean the gem is better than that math suggests, unless I read something wrong.
11. Sponsor
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Location
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Posts
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Not trying to take anything away from this thread.
But I personally choose the armor meta for the same reason I choose to aim for passive unhittable (get as much Mastery as possible). It might end up being less damage reduction "overall", but it's gauranteed reduction - as where the block meta is not.
There was discussion in the T&M forums about this, with some talk regarding Warrior crit block aswell.
http://www.tankspot.com/showthread.p...hich-is-better.
12. New Registrant
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When i started with my calculations I also assumed the gem would increase my block value to 31.5%, but after some testing regarding a comment from a guild mate I've found it only gives 31% block value.
Last edited by Shaizhowfn; 01-15-2011 at 01:56 PM.
13. Registrant
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Originally Posted by Bigbad
Blockmeta will get better when we get closer to 102,4% unhittable, since the armormeta is only better on the full hits we receive.
Are we sure that this is true? If what you're saying is true, the armor %DR is calculated after parry/dodge/miss/block. I always assumed that it was calculated before, such that Boss Hits -> Damage to be done is calculated based on Armor -> Avoidance Roll -> If Blocked, take 30% off of that mitigated damage.
14. Established Registrant
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Doesn't really matter if you do 0,42(from armor) * 0,7(from block) or 0,7*0,42 answer will be the same.
15. Originally Posted by Nukous
http://www.wowhead.com/item=52289
with http://www.wowhead.com/spell=74238 over http://www.wowhead.com/spell=74253
that's 69 mastery vs 81 stam 2% armor
need some maths on this : )
Does the run speed stack with the "Stam/run speed" boot enchant? If not, I'd take the boot enchant and an 81 Stam meta.
16. Knight-Captain of Obvious
Join Date
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1,900
For warriors it seems the block performs better in better gear for 2 reasons:
1) You get progressively closer to unhittable/are with shield block.
2) Hold the Line uptime increases which means critical block frequency increases.
I can't speak for the other tanks however.
17. New Registrant
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I made a calculator for just this question: http://www.bacondev.com/tankmeta/
At lower gear levels Austere wins out. With a couple pieces of raid level gear it will shift to favor Eternal instead.
Paladins need a little higher avoidance+block before Eternal wins but it's still very reasonable numbers.
18. Registrant
Join Date
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Posts
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Pretty sure on the latest patch notes they have changed the stam/sbv to only have 1% sbv
19. True, they did say that. But it already was 1%. Well, 1.5% rounded down. Instead of 30% blocked damage, you block 31%.
The question is now: did they just make the description go in line with the net effect or did they reduce the net effect to 20% of the previous value. In a comparison between the Austere and the Eternal option, Eternal only won by a narrow margin as it is/was.
I'd guess they only changed the description but left the effect unchanged, because too many people expected the bonus to be additive rather than multiplicative.
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Export file:
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Existence and uniqueness of miscible flow equation through porous media with a non singular fractional derivative
1 Department of Mathematics, Malaviya National Institute of Technology, Jaipur-302017, India
2 Department of Mathematics, Cankaya University, Ankara-06430, Turkey
3 Institute of Space Sciences, Magurele-Bucharest-R 76900, Romania
4 Department of HEAS(Mathematics), Rajasthan Technical University, Kota-324010, India
## Abstract Full Text(HTML) Figure/Table Related pages
In this paper, we discuss the phenomenon of miscible flow with longitudinal dispersion in porous media. This process simultaneously occur because of molecular diffusion and convection. Here, we analyze the governing differential equation involving Caputo-Fabrizio fractional derivative operator having non singular kernel. Fixed point theorem has been used to prove the uniqueness and existence of the solution of governing differential equation. We apply Laplace transform and use technique of iterative method to obtain the solution. Few applications of the main result are discussed by taking different initial conditions to observe the effect on derivatives of different fractional order on the concentration of miscible fluids.
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# References
1. R. Agarwal, Kritika, S. D. Purohit, A mathematical fractional model with non-singular kernel for thrombin receptor activation in calcium signalling, Math. Meth. Appl. Sci., 42 (2019), 7160-7171.
2. R. Agarwal, M. P. Yadav, R. P. Agarwal, Collation analysis of fractional moisture content based model in unsaturated zone using q-homotopy analysis method, Methods of Mathematical Modelling: Fractional Differential Equations, CRC Press, Taylor & Francis, 151, 2019.
3. R. Agarwal, M. P. Yadav, R. P. Agarwal, et al., Analytic solution of fractional advection dispersion equation with decay for contaminant transport in porous media, Matematicki Vesnik, 71 (2019), 5-15.
4. R. Agarwal, M. P. Yadav, R. P. Agarwal, et al., Analytic solution of space time fractional advection dispersion equation with retardation for contaminant transport in porous media, Progress in Fractional Differentiation and Applications, 5 (2019), 283-295.
5. R. Agarwal, M. P. Yadav, R. P. Agarwal, Analytic solution of time fractional Boussinesq equation for groundwater flow in unconfined aquifer, J. Discontinuity, Nonlinearity Complexity, 8 (2019), 341-352.
6. A. Atangana, D. Baleanu, Caputo-Fabrizio Derivative Applied to Groundwater Flow within Confined Aquifer, J Eng. Phys., 143 (2017), Article Number: D4016005.
7. M. S. Aydogan, D. Baleanu, A. Mousalou, et al., On high order fractional integro-differential equations including the Caputo-Fabrizio derivative, Boundary Value Probl., 2018 (2018), 90.
8. D. Baleanu, A. Mousalou, S. Rezapour, The extended fractional Caputo-Fabrizio derivative of order 0 ≤ σ < 1 on $C_\mathbb{R}[0,1]$ and the existence of solutions for two higher-order series-type differential equations, Adv. Differ. Equations, 2018 (2018), 255.
9. N. R. Bastos, Calculus of variations involving Caputo-Fabrizio fractional differentiation, Statistics, Optimization & Information Computing, 6 (2018), 12-21.
10. J. Bear, Dynamics of fluids in porous media, Courier Corporation, 2013.
11. D. Baleanu, S. S. Sajjadi, A, Jajarmi, et al., New features of the fractional Euler-Lagrange equations for a physical system within non-singular derivative operator, Eur. Phys. J. Plus, 134 (2019), 181.
12. D. Baleanu, A. Jajarmi, J. H. Asad, The fractional model of spring pendulum: New features within different kernels, Proc. Rom. Acad., 19 (2018), 447-454.
13. M. Caputo, M. Fabrizio, A new definition of fractional derivative without singular kernel, Prog. Fractional Differ. Appl., 2 (2015), 73-85.
14. G. Dagan, Flow and transport in porous formations, Springer Science & Business Media, 2012.
15. G. De Josselin de Jong, Longitudinal and transverse diffusion in granular deposits, Trans. Am. Geophys. Union, 39 (1958), 67-74.
16. F. A. Dullien, Porous media: Fluid transport and pore structure, Academic press, 2012.
17. R. A. Greenkorn, Steady flow through porous media, AIChE Journal, 27 (1975), 529-545.
18. J. Hristov, Derivatives with non-singular kernels from the Caputo-Fabrizio definition and beyond: Appraising analysis with emphasis on diffusion models, Front. Fract. Calc., 1 (2017), 270-342.
19. A. Jajarmi, B. Ghanbari, D. Baleanu, A new and efficient numerical method for the fractional modeling and optimal control of diabetes and tuberculosis co-existence, Chaos: An Interdiscip. J. Nonlinear Sci., 29 (2019), 093111.
20. A. Jajarmi, S. Arshad, D. Baleanu, A new fractional modelling and control strategy for the outbreak of dengue fever, Physica A: Stat. Mech. Appl., 535 (2019), 122524.
21. J. Losada, J. J. Nieto, Properties of a new fractional derivative without singular kernel, Prog. Fractional Differ. Appl., 1 (2015), 87-92.
22. P. I. Polubarinova-Koch, Theory of ground water movement, Princeton University Press, 2015.
23. P. G. Saffman, A theory of dispersion in a porous medium, J. Fluid Mech., 6 (1959), 321-349.
24. A. Scheidegger, On the theory of flow of miscible phases in porous media, International Union of Geodesy and Geophysics, 1957.
25. F. W. Schwartz, Macroscopic dispersion in porous media: The controlling factors, Water Resour. Res., 13 (1977), 743-752.
26. M. P. Yadav, R. Agarwal, Numerical investigation of fractional-fractal Boussinesq equation, Chaos: An Interdiscip. J. Nonlinear Sci., 29 (2019), 013109.
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#### Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 20.
Answer:$\frac{ 2}{3}\tan^{-1}\left (\frac{1}{13} \right )$
Hint: We use indefinite integral formula then put limits to solve this integral.
Given: $\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx$
Solution: $I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx$
Put $\sin x=\frac{2\tan \frac{x}{2} }{1+\tan ^2\frac{x}{2}}$
$I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}dx$
\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1}{5+\frac{4\left(2 \tan \frac{x}{2}\right)}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x\\ &=\int_{0}^{\frac{\pi}{2}} \frac{1+\tan ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \end{aligned}
$=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1+\tan ^{2} x=\sec ^{2} x\right]$
Put $\tan \frac{x}{2}=t$
$\sec^ 2\frac{x}{2} .\frac{1}{2}dx=dt$
$\sec ^2\frac{x}{2} dx=2t$
When x=0 then t=0 and when $x=\frac{\pi}{2}$ then t=1
\begin{aligned} &l=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{5+5 \tan ^{2} \frac{x}{2}+8 \tan \frac{x}{2}} d x \\ &=\int_{0}^{1} \frac{1}{5+5 t^{2}+8 t} 2 d t \\ &=\int_{0}^{1} \frac{1}{5\left(1+t^{2}+\frac{8 t}{5}\right)} 2 d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{1+t^{2}+\frac{8 t}{5}} d t \end{aligned}
\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{t^{2}+2 t \frac{4}{5}+\left(\frac{4}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}+1} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}-\frac{16}{25}+1} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\frac{25-16}{25}} d t \\ &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\frac{9}{25}} d t \end{aligned}
\begin{aligned} &=\frac{2}{5} \int_{0}^{1} \frac{1}{\left(t+\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} d t \\ &=\frac{2}{5}\left[\frac{1}{\frac{3}{5}} \tan ^{-1}\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \\ &=\frac{2}{5} \times \frac{5}{3}\left[\tan ^{-1}\left(\frac{\frac{5 t+4}{5}}{\frac{3}{5}}\right)\right]_{0}^{1} \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{5 \times 1+4}{3}\right)-\tan ^{-1}\left(\frac{5 \times 0+4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{9}{3}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}(3)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{9}{3}\right)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}(3)-\tan ^{-1}\left(\frac{4}{3}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{3-\frac{4}{3}}{1+3 \times \frac{4}{3}}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} A-\tan ^{-1} B=\tan ^{-1}\left(\frac{A-B}{1+A B}\right)\right] \end{aligned}
\begin{aligned} &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{\frac{9-4}{3}}{1+4}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{5}{3 \times 5}\right)\right] \\ &=\frac{2}{3}\left[\tan ^{-1}\left(\frac{1}{3}\right)\right] \end{aligned}
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > 3adantl2 GIF version
Description: Deduction adding a conjunct to antecedent. (Contributed by NM, 24-Feb-2005.)
Hypothesis
Ref Expression
3adantl.1 (((𝜑𝜓) ∧ 𝜒) → 𝜃)
Assertion
Ref Expression
3adantl2 (((𝜑𝜏𝜓) ∧ 𝜒) → 𝜃)
Proof of Theorem 3adantl2
StepHypRef Expression
1 3simpb 913 . 2 ((𝜑𝜏𝜓) → (𝜑𝜓))
2 3adantl.1 . 2 (((𝜑𝜓) ∧ 𝜒) → 𝜃)
31, 2sylan 271 1 (((𝜑𝜏𝜓) ∧ 𝜒) → 𝜃)
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 101 ∧ w3a 896 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 103 ax-ia2 104 ax-ia3 105 This theorem depends on definitions: df-bi 114 df-3an 898 This theorem is referenced by: 3ad2antl1 1077 nnmord 6121 ltaprg 6775 lediv2a 7936 zdiv 8386
Copyright terms: Public domain W3C validator
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# Category Archives: Calculus
March 18, 2014, 4:34 pm
# What should mathematics majors know about computing, and when should they know it?
Yesterday I got an email from a reader who had read this post called What should math majors know about computing? from 2007. In the original article, I gave a list of what computing skills mathematics majors should learn and when they should learn them. The person emailing me was wondering if I had any updates on that list or any new ideas, seven years on from writing the article.
If anything, over the past seven years, my feelings about the centrality of computing in the mathematics major have gotten even more entrenched. Mostly this is because of two things.
First, I know more computer science and computer programming now than I did in 1997. I’ve learned Python over the last three years along with some of its related systems like NumPy and SciPy, and I’ve successfully used Python as a tool in my research. I’ve taken a MOOC on algorithms and read, in whole or in part, books…
March 11, 2014, 2:34 pm
# Getting off on the right foot in an inverted calculus class
In the previous post about the flipped/inverted calculus class, we looked at getting student buy-in for the flipped concept, so that when they are asked to do Guided Practice and other such assignments, they won’t rebel (much). When you hear people talk about the flipped classroom, much of the time the emphasis is on what happens before class – the videos, how to get students to do the reading, and so on. But the real magic is what happens in class when students come, prepared with some basic knowledge they’ve acquired for themselves, and put it to work with their peers on hard problems.
But before this happens, there’s an oddly complex buffer zone that students and instructors have to cross, and that’s the time when students arrive at the class meeting. Really? you are thinking. How can arrival to class be such a complicated thing? They show up, you get to work, right? Well…
March 6, 2014, 2:25 pm
# Getting student buy-in for the inverted calculus class
So far, regarding the inverted/flipped calculus course, we’ve discussed why I flipped the calculus class in the first place, the role of self-regulated learning as a framework and organizing principle for the class, how to design pre-class activities that support self-regulated learning, and how to make learning objectives that get pre-class activities started on a good note. This is all “design thinking”. Now it’s time to focus on the hard part: Students, and getting them to buy into this notion of a flipped classroom.
I certainly do not have a perfect track record with getting students on board with an inverted/flipped classroom structure. In fact the first time I did it, it was a miserable flop among my students (even though they learned a lot). It took that failure to make me start thinking that getting student buy-in has to be as organized, systematic, and well-planned as…
March 5, 2014, 2:37 pm
# Creating learning objectives, flipped classroom style
In my last post about the inverted/flipped calculus class, I stressed the importance of Guided Practice as a way of structuring students’ pre-class activities and as a means of teaching self-regulated learning behaviors. I mentioned there was one important difference between the way I described Guided Practice and the way I’ve described it before, and it focuses on the learning objectives.
A clear set of learning objectives is at the heart of any successful learning experience, and it’s an essential ingredient for self-regulated learning since self-regulating learners have a clear set of criteria against which to judge their learning progress. And yet, many instructors – myself included in the early years of my career – never map out learning objectives either for themselves or for their students. Or, they do, and they’re so mushy that they can’t be measured – like any…
March 4, 2014, 2:59 pm
# The inverted calculus course: Using Guided Practice to build self-regulation
This post continues the series of posts about the inverted/flipped calculus class that I taught in the Fall. In the previous post, I described the theoretical framework for the design of this course: self-regulated learning, as formulated by Paul Pintrich. In this post, I want to get into some of the design detail of how we (myself, and my colleague Marcia Frobish who also taught a flipped section of calculus) tried to build self-regulated learning into the course structure itself.
We said last time that self-regulated learning is marked by four distinct kinds of behavior:
1. Self-regulating learners are an active participants in the learning process.
2. Self-regulating learners can, and do, monitor and control aspects of their cognition, motivation, and learning behaviors.
3. Self-regulating learners have criteria against which they can judge whether their current learning status is…
March 3, 2014, 9:00 am
# The inverted calculus course and self-regulated learning
A few weeks ago I began a series to review the Calculus course that Marcia Frobish and I taught using the inverted/flipped class design, back in the Fall. I want to pick up the thread here about the unifying principle behind the course, which is the concept of self-regulated learning.
Self-regulated learning is what it sounds like: Learning that is initiated, managed, and assessed by the learners themselves. An instructor can play a role in this process, so it’s not the same thing as teaching yourself a subject (although all successful autodidacts are self-regulating learners), but it refers to how the individual learner approaches learning tasks.
For example, take someone learning about optimization problems in calculus. Four things describe how a self-regulating learner approaches this topic.
1. The learner works actively on optimization problems as the primary form of…
January 27, 2014, 7:55 am
# The inverted calculus course: Overture
As many Casting Out Nines readers know, last semester I undertook to rethink the freshman calculus 1 course here at my institution by converting it to an inverted or “flipped” class model. It’s been two months since the end of that semester, and this blog post is the first in a (lengthy) series that I’ll be rolling out in the coming weeks that lays out how the course was designed, what happened, and how it all turned out.
Let me begin this series with a story about why I even bother with the flipped classroom.
The student in my programming class looked me straight in the eye and said, “I need you to lecture to me.” She said, “I can’t do the work unless someone tells me how to get started and then shows me how, step by step.” I took a moment to listen and think. “Do you mean that you find the work hard and it’s easier if someone tells you how to start and…
December 18, 2013, 1:25 pm
# Dijkstra, radical novelty, and the man on the moon
Over three years ago, I wrote a post to try to address a fallacy that is used to refute the idea of novel ways of teaching mathematics and science. That fallacy basically says that mathematics and the way people learn it have not fundamentally changed in hundreds if not thousands of years, and therefore the methods of teaching that have “worked” up to this point in history don’t need changing. Or more colloquially, “We were able to put a man on the moon with the way we’ve taught math for hundreds of years, so we shouldn’t change it now.” I sometimes refer to this as the “man on the moon” fallacy because of that second interpretation.
To understand why I think this is a fallacy, read the post above – or better yet, read this long quote from a 1988 paper by Edsger Dijkstra, one of the great scientific minds of the last 100 years and one of the authors of modern…
October 7, 2013, 9:19 am
# The biggest lesson from the flipped classroom may not be about math
For the last six weeks, my colleague Marcia Frobish and I have been involved in an audacious project – to “flip” our freshman Calculus 1 class at Grand Valley State University. I started blogging about this a while back and it’s been quiet around the blog since then, mainly because I’ve been pretty busy actually, you know, planning and teaching and managing the actual course. When I say “audacious project” to describe all this, I’m not engaging in hyperbole. It’s definitely a project – there are screencasts to make, activities to write, instruction to differentiate and so on. And it’s definitely audacious because at the core of this project is a goal of nothing less than a complete reinvention of freshman calculus at the university level. So, no pressure.
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Fundamental concepts
# Fundamental conceptsIntroduction to Algorithms
## Basic Concepts
An algorithm is a set of instructions that are followed to solve a problem or complete a task. The basic operations used to describe algorithms include assignment, comparison, arithmetic, and input/output operations. Notations such as Big O, Big Omega, and Big Theta are used to describe the time complexity of algorithms, which is the amount of time required to execute an algorithm as a function of the size of its input data.
Efficiency and complexity analysis are important considerations when designing and selecting algorithms. The time and space complexity of an algorithm can be calculated and used to determine the efficiency of an algorithm. Best-case, worst-case, and average-case scenarios are analyzed to determine the expected run time of an algorithm.
Common algorithm design techniques include divide-and-conquer and dynamic programming. Divide-and-conquer involves breaking down a problem into smaller sub-problems that can be solved independently, and then combining the solutions to solve the original problem. Dynamic programming involves breaking down a problem into sub-problems, solving each sub-problem only once, and storing the solutions in a table to avoid redundant calculations.
## Searching and Sorting Algorithms
Searching and sorting algorithms are two of the most fundamental types of algorithms in computer science. They are used to organize and manipulate data in a variety of applications, from databases to web search engines.
### Searching Algorithms
Searching algorithms are used to find a specific element in a collection of data. There are two main types of searching algorithms: linear search and binary search.
Linear search, also known as sequential search, is a simple algorithm that checks each element of a collection in turn until it finds the desired element or reaches the end of the collection. Linear search is easy to implement, but it can be slow for large collections.
Binary search is a more efficient algorithm that works by dividing the collection in half repeatedly until the desired element is found. Binary search only works on collections that are already sorted, but it is much faster than linear search for large collections.
### Sorting Algorithms
Sorting algorithms are used to arrange data in a specific order. There are many different sorting algorithms, but some of the most common ones include insertion sort, selection sort, bubble sort, merge sort, and quicksort.
Insertion sort works by iterating through the collection and inserting each element in its proper place in a new, sorted collection. Selection sort works by selecting the smallest unsorted element and swapping it with the first unsorted element. Bubble sort works by repeatedly swapping adjacent elements if they are in the wrong order.
Merge sort and quicksort are more efficient sorting algorithms that use divide-and-conquer techniques to sort the collection. Merge sort works by dividing the collection in half repeatedly until each sub-collection contains only one element, and then merging the sub-collections back together in order. Quicksort works by selecting a pivot element, partitioning the collection into two sub-collections based on the pivot, and then recursively sorting the sub-collections.
Sorting algorithms can be compared in terms of their efficiency and complexity. The time complexity of a sorting algorithm is typically measured by the number of comparisons and swaps required to sort a collection of data. The space complexity of a sorting algorithm is the amount of memory required to perform the sort.
## Data Structures
Data structures are used to store and organize data in a way that allows for efficient access and modification. There are many different data structures, each with its own strengths and weaknesses. Some of the most basic data structures include arrays, linked lists, stacks, queues, and trees.
### Arrays
An array is a collection of elements, each identified by an index or a key. Arrays are used to store data that can be accessed and modified quickly, as each element can be accessed in constant time. However, arrays have a fixed size and can be difficult to modify if the size needs to be changed.
A linked list is a collection of elements, each containing a reference to the next element in the list. Linked lists are used to store data that can be easily added or removed, as elements can be inserted or deleted in constant time. However, linked lists can be slower to access than arrays, as each element must be accessed sequentially.
### Stacks and Queues
Stacks and queues are data structures that are used to store collections of elements. A stack is a collection of elements that are stored and accessed in a last-in, first-out (LIFO) order. A queue is a collection of elements that are stored and accessed in a first-in, first-out (FIFO) order. Both stacks and queues can be implemented using arrays or linked lists.
### Trees
A tree is a collection of elements that are organized hierarchically. Each element in a tree is called a node, and each node can have zero or more child nodes. Trees are used to store data that can be easily searched and modified, as elements can be added or removed in logarithmic time. There are many different types of trees, including binary trees, AVL trees, and B-trees.
Choosing the right data structure for a particular algorithm or application is an important consideration. The efficiency and complexity of an algorithm can depend heavily on the choice of data structure, and selecting the right data structure can greatly improve the performance of an algorithm.
## Graph Algorithms
Graph algorithms are used to analyze and manipulate data that is organized as a graph. A graph is a collection of vertices (nodes) and edges that connect the vertices. Graphs are used to model a wide variety of systems, including computer networks, social networks, and transportation networks.
Breadth-first search (BFS) is a simple graph traversal algorithm that visits all the vertices in a graph in breadth-first order. BFS starts at a specific vertex and explores all the vertices at the same level before moving on to the next level. BFS is often used to find the shortest path between two vertices in an unweighted graph.
### Depth-First Search
Depth-first search (DFS) is another graph traversal algorithm that visits all the vertices in a graph. DFS explores the graph by visiting as far as possible along each branch before backtracking. DFS is often used to find cycles in a graph and to perform topological sorting.
### Shortest Path Algorithms
Shortest path algorithms are used to find the shortest path between two vertices in a graph. There are many different algorithms for finding the shortest path, including Dijkstra's algorithm, Bellman-Ford algorithm, and Floyd-Warshall algorithm. Dijkstra's algorithm is often used to find the shortest path in a weighted graph, while Bellman-Ford algorithm is used to find the shortest path in a graph with negative edge weights.
### Minimum Spanning Tree Algorithms
Minimum spanning tree algorithms are used to find the minimum spanning tree of a graph, which is a tree that connects all the vertices in the graph with the minimum possible total edge weight. There are many different algorithms for finding the minimum spanning tree, including Kruskal's algorithm and Prim's algorithm.
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Karlstad Vision 100 000 och dess implementering för Karlstads Elnät AB 2011-2015
Karlstad University, Faculty of Technology and Science.
2011 (Swedish)Independent thesis Basic level (university diploma), 15 credits / 22,5 HE creditsStudent thesis
##### Abstract [en]
The town of Karlstad has a long term project called Vision 100 000 which aim is to increase the population towards 100 000 inhabitants. This will of course put a strain for the town’s energy distribution grid. This is a work about trying to predict the grids power usage would be for the next five years if the city building plan is finalized. Those predictions are mainly going to be done through calculations with a database program called Facilplus and by checking the measured history of power usage per hour on certain selected customers by the program CustCom. The peak for power usage happened the 22 December at 4 am. This date is important because it would give the main measure for how much it the grid is able to deliver. The future prognoses of the power usage are then going to be calculated. This is going to be done by making sure that the total areas of newly constructed buildings are going to be measured. A mean value for area and power usage is then going to be calculated for respective single houses and apartment houses. This would in turn give the important necessary value that would enable to give Karlstad Elnät AB a mean to calculate a new primarily prognosis for the next five years plan. These prognoses would then be used as plans how to newly constructed habitation areas would affect the energy grid in Karlstad.
##### Abstract [sv]
Karlstads Kommun har en vision att få en befolkningsmängd på 100 000 invånare. För att få en bild av det framtida energibehovet bör man först skapa sig en bild över dagens elbehov. Detta sker på två sätt, först genom att summera energiuttaget från alla de elmätare som finns under nätstationerna som hämtas från insamlingssystemet CustCom. Här får man timrapporten för varje kunds energiförbrukning. Man är då särdeles intresserad av den dagen då nätet var som hårdast belastat. Detta skedde under julhelgen 22 december 2010 klockan 16.00. Den andra metoden är att få fram värden genom beräkningar via javaprogrammet Facilplus. Programmet används även för nätdokumentation (var stationerna och kablar befinner sig geografiskt) och projektering av utbyggnaderna. Belastningsberäkningarna i Facilplus använder sig av Velanders formel för att räkna fram effekten från kända årsförbrukningar av Karlstad Elnät AB:s kunder.
När det gäller att prognosera effektanvändningen för nybebyggelserna, beräknas först ett standardvärde för respektive byggnad (specifik energiförbrukning). Detta värde beräknas med hjälp av byggnaders ytor utifrån begärda ritningar från några noga utvalda områden. Därefter kan energianvändningen per kvadratmeter beräknas utifrån uppmätt energiförbrukning för de kunderna i respektive byggnad. Omvandlingen sker sedan från energiförbrukning till effektförbrukning och man får därmed ett bra mått på hur en viss byggnadstyp har för energianvändning. Då antalet bostäder som ska byggas ut under perioden är kända från de byggnadsplaner som är tillgängliga för allmänheten kan man sedermera få fram en översiktlig storlek för varje nybyggnadsområdes framtida energianvändning. Det kommer även sålunda bli möjligt att avgöra om elnätet bör byggas ut eller om stationer ska omplaceras så att de hamnar inom andra mottagningsstationers matningsområden. Befintliga mottagningsstationer har kapacitet för ytterligare utbyggnad av Karlstad i olika riktningar.
2011. , p. 49
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Available from: 2011-10-06 Created: 2011-09-29 Last updated: 2016-04-08Bibliographically approved
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Slopes - Energy
These questions involve an object moving up or down slopes. There may or may not be friction. The object may be free-wheeling, or may use an engine. There are many types of questions. There is a simple energy relationship between all the energies in all such questions.
Going uphill
These questions revolve around Ek. Note the blue arrow linking the 2 Ek values.
The red arrows represent the energy obstacles in the way, and are negative.
Block any one of the numbers in the diagram with a finger, and guess what it is supposed to be.
Energy relationship in an equation form:
100-20-50 = 30
Question 1
A 2kg trolley is free-wheeling towards a slope of height 3m at a constant speed of 9m.s-1. The length of the slope is 5m and the friction that would be experienced along the slope is 2N.
As you calculate the answers, put the energies back into the diagram, using positive only. This makes it easier to grasp the solution.
1.1 Calculate the kinetic energy of the trolley at the bottom.
1.2. Calculate the Ep of the trolley when it reaches the top.
1.3. Calculate the magnitude of work done to overcome friction along the slope.
1.4. Calculate the kinetic energy of the trolley at the top.
1.5. Calculate the velocity of the trolley at the top.
Summary of the energies:
Bonus Questions
Assume that the trolley had a motor, and could thus go up the slope at constant speed.
1.6. What ADDITIONAL energy would be needed from the motor to achieve a constant speed up the slope?
The trolley would have been DEPRIVED (friction + hill) of a total of:
Thus the motor would have to replace this amount of energy.
Additional energy required would be 68,8J
1.7. How long would it take the trolley to reach the top.
1.8. How much power would be required by the motor to achieve this constant speed up the slope?
This energy is the extra energy the motor itself introduces, and the time is how long it took to get to the top.
Going downhill - can you figure the diagram out?
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https://codea.io/talk/discussion/6107/wave-of-enemies
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#### Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
# Wave of Enemies
edited January 2015 in General Posts: 8
I'm trying to make waves of enemies and am stuck on drawing a certain amount/certain enemies. I want to loop 10 of certain enemies in a single wave and then wait a certain amount of time before the next wave(s) are drawn.
``````-- Final Creatures
-- Use this function to perform your initial setup
function setup()
health={5,2,8,10}
wave={"meteor1","meteor2","meteor3","meteor4","meteor5","meteor6","meteor7","meteo8r","meteor9","meteor10"}
rTable={}
timeout=ElapsedTime+1
end
-- This function gets called once every frame
function draw()
background(141, 47, 47, 255)
--floor
strokeWidth(15)
line(0,250,WIDTH,250)
strokeWidth(7)
rect(-5,-5,WIDTH+10,250)
time()
--beast
strokeWidth(0)
for meteor1,meteor10 in pairs(wave) do
for nbr,rock in pairs(rTable) do
ellipse(rock.x,rock.y,30)
rock.x=rock.x+.1
if rock.x>=WIDTH/2-30 then
rock.x=WIDTH/2-30
end
end
end
end
function time()
if (ElapsedTime>timeout) then
table.insert(rTable,vec2(0,300))
timeout=ElapsedTime+1
end
end
``````
Tagged:
• Mod
Posts: 7,903
@ZacharyU Here's a simple example of creating a wave of objects.
``````displayMode(FULLSCREEN)
function setup()
tab={}
tm=0
create()
end
function create()
for z=1,10 do
table.insert(tab,vec2(z*50,HEIGHT))
end
end
function draw()
background(0)
fill(255)
for a,b in pairs(tab) do
ellipse(b.x,b.y,20)
b.y=b.y-5
end
for a,b in pairs(tab) do
if b.y<0 then
table.remove(tab,a)
end
end
tm=tm+1
if tm>60 then
create()
tm=0
end
end
``````
• Posts: 8
@dave1707 What if I wanted to draw, say, only 5 of the circles. What would I have to change to make it only draw 5?
• Mod
Posts: 7,903
@ZacharyU In the function create(), change the 10 in the 'for' loop to 5.
• Posts: 8
Sorry, I meant draw the circles only 5 times and then they stop getting drawn.
• Mod
edited January 2015 Posts: 7,903
@ZacharyU You could add a variable that gets incremented each time create() is called. If that variable is greater than 5 then you stop calling create().
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https://stacks.math.columbia.edu/tag/0GG4
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Remark 84.3.5. In the situation of Lemma 84.3.3 we have
$DQ_ Y(Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K))) = Rf_* DQ_ X(R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)))$
by Derived Categories of Spaces, Lemma 73.19.2. Thus if $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \in D_\mathit{QCoh}(\mathcal{O}_ X)$, then we can “erase” the $DQ_ Y$ on the left hand side of the arrow. On the other hand, if we know that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K) \in D_\mathit{QCoh}(\mathcal{O}_ Y)$, then we can “erase” the $DQ_ Y$ from the right hand side of the arrow. If both are true then we see that (84.3.2.1) is an isomorphism. Combining this with Derived Categories of Spaces, Lemma 73.13.10 we see that $Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(L, a(K)) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(Rf_*L, K)$ is an isomorphism if
1. $L$ and $Rf_*L$ are perfect, or
2. $K$ is bounded below and $L$ and $Rf_*L$ are pseudo-coherent.
For (2) we use that $a(K)$ is bounded below if $K$ is bounded below, see Lemma 84.3.2.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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http://www.lazytweet.com/2019/06/distance-and-displacement-worksheet-answer-key-luxury-distance-displacement-speed-velocity-worksheet-with-answers/
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# Distance And Displacement Worksheet Answer Key Luxury Distance Displacement Speed Velocity Worksheet With Answers
This post categorized under Vector and posted on April 2nd, 2019.
The velocity of an object is the rate of change of its position with respect to a frame of reference and is a function of time. Velocity is equivalent to a specification of an objects speed and direction of motion (e.g. 7001600000000000000 60 kmh to the north).In mathematics physics and engineering a Euclidean vector (sometimes called a geometric or spatial vector oras heresimply a vector) is a geometric object that has magnitude (or length) and direction.Vectors. Components. Vector addition and subtraction. Scalar product and vector product (dot product and cross product). Displacement velocity and acceleration. Physclips provides multimedia education in introductory physics (mechanics) at different levels. Modules may be used by teachers while students may use the whole package for self
Velocity definition Velocity is the speed at which something moves in a particular direction. Meaning pronunciation translations and examplesHello I love your tutorial but i am having a hard time with the motion vector data section i am not getting the same result as yourself with my fume Fx velocity data.The anatomical properties including ring width latewood percentage strand density earlywood and latewood fiber length tangential earlywood and latewood cell wall thickness tangential earlywood and latewood cell wall diameter radial earlywood and latewood cell diameter and radial earlywood and latewood cell wall thickness associated with
Equation 1 predicts that the average translational velocity V of the earth around the Sun is 29.79 Kms. Of course the earth velocity vector changes continuously in direction and completes a full cycle during a one year period while the earth circles the Sun.Single Deflection Grilles & Registers. Single deflection supply grilles and registers are recommended for sidewall applications requiring pattern adjustability in a single horizontal or vertical plane sill or sidewall location at ceiling line or heating application only.
## An Object Is Moving Along A Circular Path Of Radius R W
Abstract With the maturity and wide availability of GPS wireless telecommunication and Web technologies mgraphicive amounts of object movement data [more]
## Which Of The Following Is An Accurate Statement A All Points On A Rotating Disk
In this paper the vectorysis of the aerodynamic performance of ducted wind turbines is carried out by means of a nonlinear and semi-vectorytical ac [more]
## Culvert Outflow Coming Up Not Shooting Out Of Barrel Tpp
Inlet Control Culvert configuration for which the cross sectional area of the barrel and headwater depth are the primary controls on culvert capaci [more]
## Astronomy And Astrophysics Magnetic Field And Radial Velocity Of The Cp Star A Cvn
The magnitude of the effective field is compatible with a variation of the field on a time scale of a little more than 70 years. The temporal behav [more]
## Stock Illustration Motion Speed Linear Icons Slow Fast Vector Signs Velocity Line Icon Illustration Vehicle Animal Image
A novel by Charles Stross. Copyright Charles Stross 2005. Published by. Ace Books New York July 2005 ISBN 0441012841. Orbit Books London August 20 [more]
## Glowing Blue Orange Rays Luminous Blue Orange Rays Many Sparks Linear Velocity Vector Dynamic Background Image
A dynamic seasonal planet - bright and frosty polar caps above a vivid rust-coloured landscape - has been revealed in the latest pictures of Mars. [more]
## Practice Test Vectors D Motion
Blast a car out of a cannon and chalvectorge yourself to hit a target Learn about projectile motion by firing various objects. Set parameters such [more]
## Why Is Parallel Component Of Velocity Along Position Vector Considered Rate Of C
Academia.edu is a platform for academics to share research papers.Momentum is a measurable quanvectory and the measurement depends on the motion of [more]
## Polar Coordinate System Velocity Vector Written Rbi Term Called Ru Transverse Velocit Q
Polar Coordinate System Velocity Vector Written Rbi Term Called Ru Transverse Velocit Q [more] Ignoring Air Resistance The Horizontal Component Of [more]
## The M Velocity Vector
In mathematics physics and engineering a Euclidean vector (sometimes called a geometric or spatial vector oras heresimply a vector) is a geometric [more]
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http://forums.wolfram.com/mathgroup/archive/2005/Sep/msg00181.html
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|
Re: NMinimize InitialPoints BUGREPORT ?
• To: mathgroup at smc.vnet.net
• Subject: [mg60239] Re: NMinimize InitialPoints BUGREPORT ?
• From: "Ray Koopman" <koopman at sfu.ca>
• Date: Thu, 8 Sep 2005 04:53:20 -0400 (EDT)
• References: <dfm7h0\$h8e\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```lupos wrote:
> hi Mathematica support,
> this posting is a Mathematica 5.1
> bug report for NMinimize especially its InitialPoints Option.
>
> As can be seen from the example below the InitialPoints seem to be
> intermixed if the name of the parameters in this case S, R respectivly
> S, T are not in alphabetical order. everything behaves well as long the
> paramteres are in alphabetical sequence such as {S, T}. if we choose
> the paramters as {S, R} the initial values for the parameters get
> erroneously intermixed.
> watch the very first printed line after the invocation of NMinimize.
> once the initial value for S gets 1. (ok)
> and once the initial value for S gets 2. (error)
>
> note that this happens with all NMinimize methods not only the
>
>
> In[1] :=
> f[v1_, v2_] := v1^2 + (v2 - 10)^2
>
>
>
> In[19]:=
> NMinimize[f[S, T], {S, T}, MaxIterations -> 1,
> EvaluationMonitor :> Print["S ", S, " T ", T],
> Method -> {NelderMead, InitialPoints -> {{1, 2}}}]
>
> >From In[19] := S 1. T 2.
> >From In[19] := S - 0.936293 T 0.280416
> >From In[19] := S 0.448605 T 0.161768
> >From In[19] := S - 0.384898 T 2.11865
> >From In[19] := S - 0.801649 T 3.09709
> >From In[19] := S - 0.801649 T 3.09709
> >From In[19] := S - 0.801649 T 3.09709
> >From In[19] := S 0. T 10.
> >From In[19] := S 0. T 10.
> >From In[19] := S - 0.801649 T 3.09709
>
> Out[19] = {0., {S -> 0., T -> 10.}}
>
>
>
> In[21]:=
> NMinimize[f[S, R], {S, R}, MaxIterations -> 1,
> EvaluationMonitor :> Print["S ", S, " R ", R],
> Method -> {NelderMead, InitialPoints -> {{1, 2}}}]
>
> >From In[21] := S 2. R 1.
> >From In[21] := S 0.280416 R - 0.936293
> >From In[21] := S 0.161768 R 0.448605
> >From In[21] := S 1.88135 R 2.3849
> >From In[21] := S 2.68182 R 4.04549
> >From In[21] := S 2.68182 R 4.04549
> >From In[21] := S 2.68182 R 4.04549
> >From In[21] := S 0. R 10.
> >From In[21] := S 0. R 10.
> >From In[21] := S 2.68182 R 4.04549
>
> Out[21] = {0., {R -> 10., S -> 0.}}
>
>
>
> regards robert
5.2 has a similar bug: with {S,R} it appears to get the initial point
correct, but otherwise follows the same solution path as 5.1.
In[1]:= \$Version
Out[1]= 5.2 for Mac OS X (June 20, 2005)
In[2]:= f[v1_, v2_] := v1^2 + (v2 - 10)^2
In[3]:= NMinimize[f[S, T], {S, T}, MaxIterations -> 1,
EvaluationMonitor :> Print["S ", S, " T ", T],
Method -> {NelderMead, InitialPoints -> {{1, 2}}}]
S 1. T 2.
S -0.936293 T 0.280416
S 0.448605 T 0.161768
S -0.384898 T 2.11865
S -0.801649 T 3.09709
S -0.801649 T 3.09709
S -0.801649 T 3.09709
S 0. T 10.
S 0. T 10.
S -0.801649 T 3.09709
S -0.801649 T 3.09709
S 0. T 10.
S 0. T 10.
Out[3]= {0.,{S->0.,T->10.}}
In[4]:= NMinimize[f[S, R], {S, R}, MaxIterations -> 1,
EvaluationMonitor :> Print["S ", S, " R ", R],
Method -> {NelderMead, InitialPoints -> {{1, 2}}}]
S 1. R 2.
S 0.280416 R -0.936293
S 0.161768 R 0.448605
S 0.881352 R 3.3849
S 1.18182 R 5.54549
S 1.18182 R 5.54549
S 1.18182 R 5.54549
S 0. R 10.
S 0. R 10.
S 1.18182 R 5.54549
S 1.18182 R 5.54549
S 0. R 10.
S 0. R 10.
Out[4]= {0.,{R->10.,S->0.}}
```
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| 2.53125
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https://www.convertunits.com/from/sthene/to/micronewton
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|
## ››Convert sthene to micronewton
sthene micronewton
## ››More information from the unit converter
How many sthene in 1 micronewton? The answer is 1.0E-9.
We assume you are converting between sthene and micronewton.
You can view more details on each measurement unit:
sthene or micronewton
The SI derived unit for force is the newton.
1 newton is equal to 0.001 sthene, or 1000000 micronewton.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between sthenes and micronewtons.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of sthene to micronewton
1 sthene to micronewton = 1000000000 micronewton
2 sthene to micronewton = 2000000000 micronewton
3 sthene to micronewton = 3000000000 micronewton
4 sthene to micronewton = 4000000000 micronewton
5 sthene to micronewton = 5000000000 micronewton
6 sthene to micronewton = 6000000000 micronewton
7 sthene to micronewton = 7000000000 micronewton
8 sthene to micronewton = 8000000000 micronewton
9 sthene to micronewton = 9000000000 micronewton
10 sthene to micronewton = 10000000000 micronewton
## ››Want other units?
You can do the reverse unit conversion from micronewton to sthene, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Sthene
The sthene is the unit of force in the former Soviet mts system, 1933-1955. The symbol is sn.
## ››Definition: Micronewton
The SI prefix "micro" represents a factor of 10-6, or in exponential notation, 1E-6.
So 1 micronewton = 10-6 newtons.
The definition of a newton is as follows:
In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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https://numberworld.info/45940416
| 1,701,880,368,000,000,000
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| 4,034
|
# Number 45940416
### Properties of number 45940416
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 2 * 3 * 239273
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
2bcfec0
Base 32:
1bpvm0
sin(45940416)
0.96185942199314
cos(45940416)
-0.27354424198479
tan(45940416)
-3.5162846602585
ln(45940416)
17.642855810474
lg(45940416)
7.6621949235035
sqrt(45940416)
6777.93596901
Square(45940416)
### Number Look Up
Look Up
45940416 (forty-five million nine hundred forty thousand four hundred sixteen) is a unique figure. The cross sum of 45940416 is 33. If you factorisate the number 45940416 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 3 * 239273. The figure 45940416 has 28 divisors ( 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192, 239273, 478546, 717819, 957092, 1435638, 1914184, 2871276, 3828368, 5742552, 7656736, 11485104, 15313472, 22970208, 45940416 ) whith a sum of 121551192. 45940416 is not a prime number. The number 45940416 is not a fibonacci number. 45940416 is not a Bell Number. 45940416 is not a Catalan Number. The convertion of 45940416 to base 2 (Binary) is 10101111001111111011000000. The convertion of 45940416 to base 3 (Ternary) is 10012110000101220. The convertion of 45940416 to base 4 (Quaternary) is 2233033323000. The convertion of 45940416 to base 5 (Quintal) is 43230043131. The convertion of 45940416 to base 8 (Octal) is 257177300. The convertion of 45940416 to base 16 (Hexadecimal) is 2bcfec0. The convertion of 45940416 to base 32 is 1bpvm0. The sine of 45940416 is 0.96185942199314. The cosine of the figure 45940416 is -0.27354424198479. The tangent of the number 45940416 is -3.5162846602585. The root of 45940416 is 6777.93596901.
If you square 45940416 you will get the following result 2110521822253056. The natural logarithm of 45940416 is 17.642855810474 and the decimal logarithm is 7.6621949235035. that 45940416 is amazing figure!
| 778
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http://mizar.uwb.edu.pl/version/current/html/proofs/cardfil4/63
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| 2,399
|
let T be non empty TopSpace; :: thesis: for s being Function of , the carrier of T
for x being Point of T
for cB being basis of () holds
( x in lim_filter (s,) iff for B being Element of cB ex n being Nat st square-uparrow n c= s " B )
let s be Function of , the carrier of T; :: thesis: for x being Point of T
for cB being basis of () holds
( x in lim_filter (s,) iff for B being Element of cB ex n being Nat st square-uparrow n c= s " B )
let x be Point of T; :: thesis: for cB being basis of () holds
( x in lim_filter (s,) iff for B being Element of cB ex n being Nat st square-uparrow n c= s " B )
let cB be basis of (); :: thesis: ( x in lim_filter (s,) iff for B being Element of cB ex n being Nat st square-uparrow n c= s " B )
hereby :: thesis: ( ( for B being Element of cB ex n being Nat st square-uparrow n c= s " B ) implies x in lim_filter (s,) )
assume A1: x in lim_filter (s,) ; :: thesis: for B being Element of cB ex n being Nat st square-uparrow n c= s " B
hereby :: thesis: verum
let B be Element of cB; :: thesis: ex n being Nat st square-uparrow n c= s " B
B is a_neighborhood of x by YELLOW19:2;
hence ex n being Nat st square-uparrow n c= s " B by ; :: thesis: verum
end;
end;
assume A2: for B being Element of cB ex n being Nat st square-uparrow n c= s " B ; :: thesis:
now :: thesis: for A being a_neighborhood of x ex n being Nat st square-uparrow n c= s " A
let A be a_neighborhood of x; :: thesis: ex n being Nat st square-uparrow n c= s " A
A3: A is Element of BOOL2F by YELLOW19:2;
cB is filter_basis ;
then consider B being Element of cB such that
A4: B c= A by A3;
A5: ex n being Nat st square-uparrow n c= s " B by A2;
s " B c= s " A by ;
hence ex n being Nat st square-uparrow n c= s " A by ; :: thesis: verum
end;
hence x in lim_filter (s,) by Th47; :: thesis: verum
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https://www.courses.com/landt-edutech/design-of-transmission-line-modelling-and-performance
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# Design of Transmission Line: Modelling and Performance
L&T EduTech
This course provides a detailed exploration of transmission line design, modeling, and performance assessment, offering a unique blend of theoretical insights and hands-on applications. Participants will gain proficiency in evaluating and modeling transmission line parameters, analyzing the performance of different types of transmission lines, and investigating phenomena like corona discharge and electromagnetic interference.
Through a specialized focus on MATLAB demonstrations and real-world case studies, participants will develop practical skills for transmission line analysis. The course also addresses the environmental impact of transmission lines and emphasizes the importance of minimizing these effects. To excel in this course, a background in basic electrical engineering principles, including knowledge of circuit analysis, electromagnetism, and mathematical modeling, is recommended.
Certificate Available ✔
##### Course Modules
This course comprises three modules: Transmission Line Parameters, Modeling and Performance of AC Transmission Lines - Part I, and Modeling and Performance of AC Transmission Lines - Part II. Each module delves into specific aspects of transmission line engineering, providing theoretical insights, practical demonstrations, and assessments.
#### TRANSMISSION LINE PARAMETERS
This module covers an in-depth exploration of transmission line parameters, including inductance, capacitance, and different configurations. Participants will gain proficiency in using MATLAB for calculating transmission line parameters and analyzing real-world scenarios, with a focus on practical applications and hands-on demonstrations.
#### MODELLING AND PERFORMANCE OF AC TRANSMISSION LINES -PART-I
Participants will study the classification and performance analysis of short, medium, and long transmission lines, including the use of MATLAB for modeling and analysis. Real-world applications and challenges related to short transmission lines will be explored, providing practical insights into the field.
#### MODELLING AND PERFORMANCE OF AC TRANSMISSION LINES -PART-II
This module delves into the intricacies of corona discharge, electromagnetic interference, and environmental impacts in AC transmission lines. Participants will gain an understanding of corona discharge mechanisms, power losses, and measures to mitigate interference. The module emphasizes the practical application of theoretical concepts through case studies and MATLAB demonstrations.
#### FPGA Design for Embedded Systems
This course provides a comprehensive understanding of FPGA design for embedded systems, covering topics such as Verilog, VHDL, soft-core processors, and FPGA development...
#### Batteries and Electric Vehicles
Arizona State University
Batteries and Electric Vehicles: Gain insights into battery performance in zero emission vehicles, EV charger networks, and regulatory requirements in this comprehensive...
#### Fundamentals of Audio and Music Engineering: Part 1 Musical Sound & Electronics
University of Rochester
Fundamentals of Audio and Music Engineering: Part 1 delves into acoustics, electronics, and their applications in creating music with electronic instruments.
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https://math.stackexchange.com/questions/2781158/question-about-proof-in-neukirchs-algebraic-number-theory
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# Question about proof in Neukirch's Algebraic Number Theory
I was reading Proposition 2.2 in chapter I of Neukirch (page 6 in my edition), which states the following for an extension of rings $A\subseteq B$:
(2.2) Proposition. Finitely many elements $b_1,\dots, b_n\in B$ are all integral over $A$ if and only if the ring $A[b_1,\dots,b_n]$ viewed as an $A$-module is finitely generated.
Neukirch begins the proof by showing that if $b\in B$ is integral over $A$ then $A[b]$ is a finitely generated $A$-module. To do this, he notes that $b$ integral means there is some monic $f(x)\in A[x]$ of degree $n\geq 1$ such that $f(b)=0$. The claim is that $\{1,b,\dots,b^{n-1}\}$ form a generating set for $A[b]$. Neukirch proceeds to take a polynomial $g(x)\in A[x]$ (so that $g(b)$ is an arbritary element in $A[b])$ and states that "we may then write $$g(x)=q(x)f(x)+r(x)$$ for some $q(x),r(x)\in A[x]$ with $\deg(r(x))<n$".
Here is my problem: $A[x]$ is not a Euclidean domain in general. If $A$ is a field then sure, but if $A=\mathbb{Z}$ then $\mathbb{Z}[x]$ is not Euclidean so it would seem this step in the proof is not justified. What am I missing here?
• You can always perform polynomial division as long as the leading coefficient of the divisor is a unit in the ring. In this example, your $f$ is monic so you can divide as usual. – André 3000 May 14 '18 at 18:15
• @Quasicoherent, this is news to me (nice news!). If you'd like to expand this a bit more I'd be willing to accept it as a solution. – Arbutus May 16 '18 at 13:17
This has nothing to do with $\mathbf A[x]$ being Euclidean, nor even $A$$being a domain. By induction, you can suppose$B=A[b]$for a single integral element$b\in B$. Indeed, if$\;b^n+a_{n-1}b^{n-1}+\dots +a_1b+a_0=0$is a monic equation for$b$, then$\;b^n\in \langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle$. We'll prove$b^m\in \langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle$for all$m\ge n$. To set the inductive step, suppose$b^n,\dots,b^m\in \langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle$for some$m. Then \begin{align} b^{m+1}&=b\cdot b^m\in b\,\langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle =\langle \mkern1.5mu b,b^2,\dots b^{n-1}, b^n\mkern 1.5mu\rangle =\langle \mkern1.5mu b,b^2,\dots b^{n-1}\mkern 1.5mu\rangle+\langle \mkern1.5mu b^n\mkern 1.5mu\rangle \\ &\subseteq\langle \mkern1.5mu b,b^2,\dots b^{n-1}\mkern 1.5mu\rangle+\langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle =\langle \mkern1.5mu1,b,\dots b^{n-1}\mkern 1.5mu\rangle. \end{align} • Correct me if I'm mistaken, but it looks like you've provided a separate proof of the fact that ifb$is integral over$A$then$A[b]\$ is finitely generated. Although I appreciate this, it doesn't quite answer my question as to what Neukirch was doing in his proof. I guess I'm looking for something more along the lines of Quasicoherent's comment to my original post. – Arbutus May 16 '18 at 13:13
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http://mytestbook.com/PrintableWorksheets/Worksheet_Grade4_Math_FractionFractiontoDecimal_491.aspx
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go to: myTestBook.com
print help! Use this to print without Ads and Toolbar. dotted fields in the header are editable. Report an error
Instruction:
Question 1
There were 20 students going to a field trip. Five students were not wearing the red field trip T-shirt given to all the students. What fraction of the students was not wearing red T-shirt?
A.
2 3
B.
1 4
C.
2 5
D.
1 3
Question 2
Which choice shows Â
3 6
in simplest form?
A.
1 2
B.
1 3
C.
2 3
D.
1 4
Question 3 Which option shows the following fractions in order from least to greatest? , , A. , , B. , , C. , , D. , ,
Question 4 Which option describes the shaded part in decimal? A. 0.25 B. 0.5 C. 0.75 D. 0.33
Question 5
Which fraction is greater than ?
A.
2 7
B.
1 4
C.
1 5
D.
3 4
Question 6 Jade ordered one extra large pizza with 12 slices, for his birthday party. Nate ate of the slices, Karen ate of the slices, Chris ate of the slices and William ate the remaining slices. Who ate largest part of the pizza? A. Karen B. Chris C. Nate D. William
-
Question 7
Find the sum.
A.
7 6
B.
1 2
C.
1 6
D.
5 6
Question 8
Find the difference.
A.
2 9
B.
5 12
C.
1 10
D.
5 24
Question 9 Stacy, Judy and Lisa are knitting sweaters. Stacy's sweater is completed, Judy's sweater is completed and Lisa's sweater is completed. Who has a greater part finished, Stacy, Judy, or Lisa? A. Judy B. Stacy C. Lisa D. Judy and Lisa
Question 10 How do you write as a decimal? A. 0.1 B. 0.5 C. 0.02 D. 0.2
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https://answers.yahoo.com/question/index?qid=20081217051701AAf90Gd
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# can anybody tell me how to calculate circumference of sector?
area of sector and circumference
Relevance
The area of a circle is πr2 (2 means squared since we can't put it on superscript) and the circumference is 2πr.
Let's start with the area first. Since a sector is a portion of the circle,
Area of Sector= Angle of Sector/360 degrees x πr2
The circumference is more difficult.
Circumference= (Angle of Sector/360 degrees x 2πr) + 2r
This is because 2πr only calculates the outer part but not the radius itself.
To sum it up,
Area of Sector= Angle of Sector/360 degrees x πr2
Circumference= (Angle of Sector/360 degrees x 2πr) + 2r
Note: π here means 22/7 or 3.14, but it depends on what the question wants.
Circumference of a sector = (x/360)*2*(22/7)*r
where x = angle of sector
• Anonymous
5 years ago
Distance from O to S is length of radius = 2 cm Area of circle = πr² = 4π cm² A whole circle is 2π radians Sector is π/5 radians So area of sector is (π/5) / (2π) = 1/10 of area of circle Area of sector = 1/10 *4π cm² = 2π/5 cm²
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http://intrologic.stanford.edu/exercises/exercise_02_04.html
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Introduction to Logic ToolsforThought
Exercise 2.4 - Satisfaction
Consider the sentences shown below.
p ∨ q ∨ r p ⇒ q ∧ r q ⇒ ¬r
There are three proposition constants here, meaning that there are eight possible truth assignments. How many of these assignments satisfy all of these sentences?
0 1 2 3 4 5 6 7 8
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http://perplexus.info/show.php?pid=11162&cid=59411
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All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Square pairs (Posted on 2018-02-07)
Let's denote by p(n) the greatest square less than or equal to n,
n being a positive integer.
List all pairs (m,n), for which:(2m+1)*p(2n+1) = 400; m being less than n.
See The Solution Submitted by Ady TZIDON Rating: 5.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
my solution (no spoiler) | Comment 1 of 3
(0,200)... to ...(0,219)
Posted by armando on 2018-02-07 09:02:31
Search: Search body:
Forums (0)
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https://www.datacamp.com/courses/case-studies-manipulating-time-series-data-in-r
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# Case Studies: Manipulating Time Series Data in R
Strengthen your knowledge of the topics covered in Manipulating Time Series in R using real case study data.
4 Hours12 Videos50 Exercises11,059 Learners3950 XPQuantitative Analyst TrackTime Series Track
or
By continuing, you accept our Terms of Use, our Privacy Policy and that your data is stored in the USA. You confirm you are at least 16 years old (13 if you are an authorized Classrooms user).
## Course Description
In this course, you will strengthen your knowledge of time series topics through interactive exercises and interesting datasets. You’ll explore a variety of datasets about Boston, including data on flights, weather, economic trends, and local sports teams.
1. 1
### Flight Data
Free
You've been hired to understand the travel needs of tourists visiting the Boston area. As your first assignment on the job, you'll practice the skills you've learned for time series data manipulation in R by exploring data on flights arriving at Boston's Logan International Airport (BOS) using xts & zoo.
Review xts fundamentals
50 xp
Identify the time series
50 xp
Flight data
100 xp
Pick out the xts object
50 xp
100 xp
50 xp
100 xp
Visualize flight data
100 xp
Calculate time series trends
100 xp
Saving and exporting xts objects
50 xp
Assessing flight trends
50 xp
Saving time - I
100 xp
Saving time - II
100 xp
2. 2
### Weather Data
In this chapter, you'll expand your time series data library to include weather data in the Boston area. Before you can conduct any analysis, you'll need to do some data manipulation, including merging multiple xts objects and isolating certain periods of the data. It's a great opportunity for more practice!
3. 3
### Economic Data
Now it's time to go further afield. In addition to flight delays, your client is interested in how Boston's tourism industry is affected by economic trends. You'll need to manipulate some time series data on economic indicators, including GDP per capita and unemployment in the United States in general and Massachusetts (MA) in particular.
4. 4
### Sports Data
Having exhausted other options, your client now believes Boston's tourism industry must be related to the success of local sports teams. In your final task on this project, your supervisor has asked you to assemble some time series data on Boston's sports teams over the past few years.
In the following tracks
Quantitative AnalystTime Series
#### Lore Dirick
Director of Data Science Education at Flatiron School
Lore is a data scientist with expertise in applied finance. She obtained her PhD in Business Economics and Statistics at KU Leuven, Belgium. During her PhD, she collaborated with several banks working on advanced methods for the analysis of credit risk data. Lore formerly worked as a Data Science Curriculum Lead at DataCamp, and is and is now Director of Data Science Education at Flatiron School, a coding school with branches in 8 cities and online programs.
#### Matt Isaacs
Political Science PhD interested in data science in defense, security, and international relations
Matt Isaacs is a former Course Development Intern at DataCamp . Matt holds a PhD in Political Science from Brandeis University and has extensive experience in applied data science across the public sector with a focus on analytics in defense, security, and international relations.
## What do other learners have to say?
I've used other sites—Coursera, Udacity, things like that—but DataCamp's been the one that I've stuck with.
Devon Edwards Joseph
Lloyds Banking Group
DataCamp is the top resource I recommend for learning data science.
Louis Maiden
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# Metamath Proof Explorer
## Theorem fneq2d
Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011)
Ref Expression
Hypothesis fneq2d.1 ${⊢}{\phi }\to {A}={B}$
Assertion fneq2d ${⊢}{\phi }\to \left({F}Fn{A}↔{F}Fn{B}\right)$
### Proof
Step Hyp Ref Expression
1 fneq2d.1 ${⊢}{\phi }\to {A}={B}$
2 fneq2 ${⊢}{A}={B}\to \left({F}Fn{A}↔{F}Fn{B}\right)$
3 1 2 syl ${⊢}{\phi }\to \left({F}Fn{A}↔{F}Fn{B}\right)$
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# Worksheet on Calculating Time
Practice the questions given in the worksheet on calculating time. Learn how to solve different problems on calculating time.
We know, the formula to calculate time = distance/speed.
1. Mohan drives a car at a uniform speed of 60 km/hr; find the time taken by him to cover 280 km?
2. The speed of sound in air is 330 m/sec. how long will the sound take to travel 198 km?
3. Sam walks 2 km in 30 minutes. How much time will he take to cover 400m?
4. Mohan cycle and covers a certain distance at the rate of 6 km/hr in 1 1/2 hours. How much time will be taken by him if he covers the same distance by scooter at the rate of 10 km/hr?
5. A man is walking at a speed of 12 km/hr after every km , he takes rest for 2 minutes. How much time will it take to cover a distance of 6 km?
6. A car covers a distance of 30 km in 12 minutes. Find the time taken by it to cover the same distance if its speed is increased by 10 km/hr.
7. A car is running at 56 km/hr. what time will it take to cover 560 meters?
Answers for the worksheet on calculating time are given below to check the exact answers of the above questions on different problems.
1. 4 hours 40 minutes
2. 10 minutes
3. 6 minutes
4. 54 minutes
5. 40 minutes
6. 11 min 15 sec
7. 36 sec
Worksheet on Conversion of Units of Speed
Worksheet on Calculating Time
Worksheet on Calculating Speed
Worksheet on Calculating Distance
Worksheet on Train Passes through a Pole
Worksheet on Relative Speed
Worksheet on Decimal into Percentage
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
## Recent Articles
1. ### Thousandths Place in Decimals | Decimal Place Value | Decimal Numbers
Jul 20, 24 03:45 PM
When we write a decimal number with three places, we are representing the thousandths place. Each part in the given figure represents one-thousandth of the whole. It is written as 1/1000. In the decim…
2. ### Hundredths Place in Decimals | Decimal Place Value | Decimal Number
Jul 20, 24 02:30 PM
When we write a decimal number with two places, we are representing the hundredths place. Let us take plane sheet which represents one whole. Now, we divide the sheet into 100 equal parts. Each part r…
3. ### Tenths Place in Decimals | Decimal Place Value | Decimal Numbers
Jul 20, 24 12:03 PM
The first place after the decimal point is tenths place which represents how many tenths are there in a number. Let us take a plane sheet which represents one whole. Now, divide the sheet into ten equ…
4. ### Representing Decimals on Number Line | Concept on Formation of Decimal
Jul 20, 24 10:38 AM
Representing decimals on number line shows the intervals between two integers which will help us to increase the basic concept on formation of decimal numbers.
5. ### Decimal Place Value Chart |Tenths Place |Hundredths Place |Thousandths
Jul 20, 24 01:11 AM
Decimal place value chart are discussed here: The first place after the decimal is got by dividing the number by 10; it is called the tenths place.
Speed of Train
Relationship between Speed, Distance and Time
Conversion of Units of Speed
Problems on Calculating Speed
Problems on Calculating Distance
Problems on Calculating Time
Two Objects Move in Same Direction
Two Objects Move in Opposite Direction
Train Passes a Moving Object in the Opposite Direction
Train Passes through a Pole
Train Passes through a Bridge
Two Trains Passes in the Same Direction
Two Trains Passes in the Opposite Direction
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## A very long uniform line of charge has a charge per unit length of 4.82 uC/m
1. The problem statement, all variables and given/known data
A very long uniform line of charge has charge per unit length 4.82 uC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length -2.46 uC/m and is parallel to the x-axis at y1= 0.418
1-)What is the magnitude of the net electric field at point y2= 0.202 m on the y-axis?
2-)What is the direction of the net electric field at point y2= 0.202 m on the y-axis?
-y-axis
+y-axis
3-)What is the magnitude of the net electric field at point y3= 0.608 m on the y-axis?
4-)What is the direction of the net electric field at point y3 = 0.608 m on the y-axis?
-y-axis
+y-axis
2. Relevant equations
EA=PA/epsilon-nought
E=kq/r2
3. The attempt at a solution
I don't really know where to start.
At first I tried to do E=(k*dq)/r^2 = k(4.82e-6)/(x^2+.202^2)
But then realized that I don't have a definite integral. Then I tried to use E=(kq1/r1)+(kq2/r2)... but the answer was incorrect. Anyone know how to start this problem..?
PhysOrg.com science news on PhysOrg.com >> Intel's Haswell to extend battery life, set for Taipei launch>> Galaxies fed by funnels of fuel>> The better to see you with: Scientists build record-setting metamaterial flat lens
Recognitions: Homework Help Try looking at this: http://hyperphysics.phy-astr.gsu.edu...elecyl.html#c1
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https://www.transtutors.com/questions/the-probability-that-samantha-will-be-accepted-by-the-college-of-her-choice-and-obta-3438499.htm
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# The probability that Samantha will be accepted by the college of her choice and obtain a scholarship 1 answer below »
The probability that Samantha will be accepted by the college of her choice and obtain a scholarship is 0.35. If the probability that she is accepted by the college is 0.65, find the probability that she will obtain a scholarship given that she is accepted by the college.
VIJAYAKUMAR G
The solution is attached...
## Plagiarism Checker
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# [R] mstate with multiple initial states?
Lucy Leigh lucy.leigh at newcastle.edu.au
Tue Jan 10 05:46:40 CET 2017
```Hi,
I have a multi-state model that I would like to estimate using the 'mstate' package - but I am not sure how best to approach it and was wondering
if anyone could provide some insight for me.
Basically, I have a group of kids who have been randomized to one of 2 treatments, A or B. If they do well on either of these, they
are discharged (A -->D or B--> D). If they do poorly, kids who were receiving treatment A are moved onto treatment B (A-->B), but kids
on B can't go to A, they go straight into ICU (A-->C). If the kids who went from A-->B do well they are discharged (A-->B-->D), and if they
do poorly they go to ICU (A-->B-->C), and then finally after ICU they are discharged (C-->D). There are also a small number of kids who were sent
straight to ICU (so...in fact there are 3 initial states). So the transition matrix looks like:
A B C D
A NA 1 NA 1
B NA NA 1 1
C NA NA NA 1
D NA NA NA NA
However, I am not sure whether I can do this in 'mstate', as there are three initial states? The program requires that we start with
a wide data set, in which there is an event indicator for each state, and a time of entry into each state. If they don't enter a state, then
the time for that state is set to last follow-up (page 4 of https://www.jstatsoft.org/article/view/v038i07 ).
So, in the case where a kid starts in A, and goes to D; the data would look like:
A.status A.time B.status B.time C.Status C.Time D.Status D.time
1 0 0 Final 0 Final 1 Final
And I think this is OK in terms of what is specified for B, because technically the kid is at risk of transitioning into B
until they are discharged.
But what about a kid who starts in B and goes to D? The corresponding data would possibly be?:
A.status A.time B.status B.time C.Status C.Time D.Status D.time
0 Final 1 0 0 Final 1 Final
However, this to me doesn't look right, as technically they are never at risk of going to A if they started in B.
I tried setting the A.time to 0 to reflect the fact that they are never at risk of going back to A, but the numbers I got
from the model (events\$model) were incorrect - basically all the events were going to A, and no one was in B.
And for the kids that started in C...similarly.
So, I was wondering, would it be valid to create a new initial state (e.g. P = pre-treatment), from which the child
then transitions immediately (at say, time = 0.1) to either A or B (or C). So the matrix would be:
P A B C D
P NA 1 1 1 NA
A NA NA 1 NA 1
B NA NA NA 1 1
C NA NA NA NA 1
D NA NA NA NA NA
And then the data for someone who went from A--D would be
A.status A.time B.status B.time C.Status C.Time D.Status D.time
1 0.1 0 0.1 0 Final 1 Final
And then the data for someone who went from B--D would be
A.status A.time B.status B.time C.Status C.Time D.Status D.time
0 0.1 1 0 .1 0 Final 1 Final
When I code the model like this, then the number of events I get from the model is correct.
However, I am not sure whether adding this extra initial state is a valid option?
Thanks in advance for anyone who can help me out,
Lucy Leigh
mstate: An R Package for the Analysis of Competing Risks ...<https://www.jstatsoft.org/article/view/v038i07>
www.jstatsoft.org
Authors: Liesbeth C. de Wreede, Marta Fiocco, Hein Putter: Title: mstate: An R Package for the Analysis of Competing Risks and Multi-State Models
[[alternative HTML version deleted]]
```
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## Monday, July 30, 2012
There is two hour parking all around University of Saskatchewan. I once went to move my car (to avoid a ticket) and found that the parking attendant had marked--in chalk--the top of my tire. I wanted to erase the mark so began driving through as many puddles as possible.
I then convinced myself to find a puddle longer than the circumference of my tire--to guarantee a clean slate and a fresh two hours.
As I walked back to campus, I got thinking about the pattern left behind by my tires. For simplicity, let's take the case of a smaller vehicle--a bike.
If you were to ride a bike through a puddle of a certain width, the trail would look like this:
Is this model correct? Evenly spaced iterations of puddle-width splotches.
Assume that:
width(puddle) < circumference(tire)
and consider the following bike-ish contraptions. Can you predict the pattern? Better yet, can you draw an accurate prediction on graph paper? Assume a six-inch puddle (why not?)
That is the task I present to the students. The emerging patterns are interesting.
Unicycles--one wheel; one pattern.
But now combine them. (Of course, the bike goes in a perfectly straight line...)
A standard bicycle-- two wheels; same size.
Alter it slightly. (You may want to encourage colour coding for overlapping paths...)
Old school--two wheels; different sizes.
Exaggerate the difference.
Crazy old school--two wheels; way different sizes.
How does the pattern change? Is it important to know how far apart the wheels are? (Experiment...)
Just for fun--4 wheels; 3 tracks; 2 sizes.
What do you notice about certain radii? What causes certain patterns to "line-up"?
An interesting task to give a class working on circles, algebraic manipulation, factors, etc.
NatBanting
## Friday, July 27, 2012
### Painting Tape
I came across the following situation while shopping for paint at a local home improvement store:
Admittedly, the three varieties were not positioned like this, but this positioning does raise an interesting question.
"We can see the packages are the same height, what is that height?"
I see this question going one of two ways:
1. The students realize that really any conceivable measurement is possible. (Barring, of course, zero and the negatives) One could make the argument that it also cannot be irrational, but this would be nit-picking. Can a roll of tape have a width of pi/6? Exactly?
2. The students fall prey to their subconscious affinity toward the integers and begin constructing common multiples.
In fairness to the problem, both are very teachable moments, but there is nothing scarier for a teacher--under considerable time constraints--than to see a problem steer students in an alternate, but useful, direction. We know they should explore their curiosity, but can we as teachers shut-up long enough to let them?
Situation (1) leads into an explanation of unit analysis. The height can be any "x" because each roll would simply subsume the thickness of x/6, x/4, x/3 respectfully. This demonstrates great number sense. If the class immediately goes that way, I would show this picture.
Revisit the question:
"What is the height of the packaging?"
"How thick is the individual roll in each package?"
Ideally, students dwell on situation (2) long enough to draw out the ideas of factors, common factors, multiples, divisibility, and lowest common multiples. After which I would drop the unit analysis bomb on them anyway.
Just a thought. Yet another way that mathematics proves to be an inseparable mass despite what neatly organized curricula dictates.
NatBanting
## Tuesday, July 24, 2012
I am frustratingly mathematical. Ask my wife. I see the world as a combination of, in the words of David Berlinski, absolutely elementary mathematics.(AEM). The path of a yo-yo, the tiles in the mall, and the trail of wetness after a bike rides through a puddle are all dissected with simple, mathematical phenomenon. The nice part about AEM is that I can talk about it to almost anyone. People are (vaguely) familiar with graphs, geometric patterns, and circles even if they can't decipher what practical implications they have on their city block. Unfortunately, people (and students) don't often want to hear about them--they need to see them.
I can remember the look on my mother's face when I broke out the silverware to show her that the restaurant table corner was not square. Without a ruler, I showed her that trigonometry allows us to rely on ratio rather than set measurements. As I was in the midst of showing her that the 3-4-5 knife-length rule was breached, the waitress came. Mom was horrified; I was thrilled.
AEM has a visual nature; school mathematics often destroys that nature--reducing it to a simple diagram of a rope hanging from a drainpipe or train chugging its way through the prairie. I am guilty of the same thing in my class. I am a tactile learner so spend the majority of my time teaching with things. Students play with triangles to learn trigonometry and we build models to slope specifications. I often describe problems--good problems--to my students and have them struggle through them. Great learning occurs, but I have robbed them of the ability to find their own problems--the very problems of AEM that exist all around us.
I have been watching the work of many educators for a while now. I love the way they use simple, visual elements to create extremely intriguing problems. This past week I was particularly inspired by Timon Piccini (@MrPicc112) and Andrew Stadel (@mr_stadel). They create video problems that not only test the AEM underpinning, but the curiosity and problem solving of the students. It is under this inspiration that I created my first video-based task. It contains a strong visual component and is based on a natural phenomenon that I observed during housework.
Sprinkler Task (V.2) from Nat Banting on Vimeo.
The video shows two takes of me using my "circular" sprinkler on my oddly-shaped piece of lawn. The natural question that came to my mind was, "Where do I place my sprinkler to minimize the amount of water that is wasted?"
The question is broad. The situation is organic. A simple curiosity can be cultivated in a novel way. That is my favorite part of the problem. I will simply play the video for my students. I will pause after the first take and allow students to absorb the situation. Hopefully cynics will point out that the pattern is not circular; this will lead to a great conversation about perspective and spatial reasoning.
I want students to notice that the first pattern touches a corner of the yard. What if the edge of the circle didn't touch any edge of the yard? Could this possibly be the most efficient watering method?
I will clear up any variables that a good mathematician would. We assume the spray is uniform. We assume that the pressure can be turned up or down to any desired radius. Wasted water is considered water that lands outside of the grass. Wind is not a factor. After our initial conversation, I will re-play the video and see the second case.
Which one wastes more? Have students discuss. Ask the students what information they need to solve the problem. Measurements will undoubtedly come up. If students are done theorizing with the problem, I provide them with a picture of the yard complete with measurements.
The lawn has been modeled as a rectangle and two semi-circles. What error has occurred? Can we refine the model? (Possibly by placing a quarter circle on the bottom right-hand corner). Do the measurements help you calculate how much water is wasted?
To aid in their work, I created a scale diagram of the yard. I then created three different worksheets--each has a different size scale drawing on it. This creates three unique scales within the classroom. As a group, we will try the first placement together. We'll draw the sprinkler and create triangles to calculate the distances to the furthest points. The longest of these must be the radius of the sprinkler circle.
I will then send the students out into groups to discover a more efficient placement. If they are going to communicate with one-another, they will have to convert using their scales. For example, "We placed ours 3 inches down and 2 from the left" won't work if a group has a different scale factor on their diagram.
At the end, I will construct a table of results to see which group indeed maximized the efficiency.
To follow up, I created two other "yards" complete with measurements. I will ask the same question. One involves the possibility of introducing trigonometry and the other has students explore the idea of circumcircles. Both of these diagrams along with three worksheets can be downloaded here.
This is hopefully the first video embodiment of my thinking. Tasks like these not only "real-world", they cater to multiple learning intelligences. The visual, spatial, kinesthetic, and auditory are all engaged. Some of the best lessons in the classroom mirror experiences that students may have outside the classroom.
NatBanting
## Thursday, July 12, 2012
### Creating PBL 3.0
I have been on my project-based learning journey for a while now. This blog has served as the main receptacle for my inspirations, ideas, successes, failures, and reflections. It is now time to document my next step: wide scale revision.
This post will be divided into two main sections:
1. A look back at the posts that brought me to this point. (Reading them may provide some context, but not reading them will provide you with more free time...your call)
2. A look ahead into my revisions and their rationale. I will describe the new administrative and assessment framework around the projects and provide links to the first completed framework online.
Now that we have that out of the way, I guess we should start with section one.
1. My views on PBL have varied drastically as I have experienced it first hand. My initial vision for the course was one of infinite possibilities. Students would develop their own projects and follow them out to fruition. I would provide the supports for them to do so. Only after I tried to do this myself did I find that good projects are hard to find, and even harder to create. My initial (naive) vision can be read here:
http://musingmathematically.blogspot.ca/2011/10/proper-workspace-for-workplace.html
As I re-shaped this initial vision, I discovered that there was a lot more support for these teaching ideas than I originally thought. Every time I found an excuse not to pursue the goal, it was addressed. My skills at project creation were growing and views of prominent educators worldwide began to solidify my belief that a completely project-based class was possible. My solidified vision can be read here:
http://musingmathematically.blogspot.ca/2011/11/more-inspiration-for-math-projects.html
I gathered support and launched two courses that were project-based. I included good problems and tasks for students to learn the basic skills and they were then solidified and utilized in their projects. During the semester, I had a number of roaring successes--both with problems and projects. Students were buying into the deep learning available to them. My largest project success can be read here:
http://musingmathematically.blogspot.ca/2012/04/soft-drink-project-part-1-framework.html
The semester ended and I had a chance to reflect. I knew the students had learned on a deep level, and I left most days surprised with the complexity of their thought and initiative. Upon reflection, I highlighted areas that the class needed to improve on. These included the technology, group work, and assessment. I wanted the course to get stronger in all three areas. My rationale and reflection can be read here:
http://musingmathematically.blogspot.ca/2012/07/project-based-pitfalls.html
That brings us to the next step along the path.
2. My solution to the three major issues addressed above was to implement a continuous feedback assessment structure. That would keep groups accountable as well as improve the formative and summative assessments on the projects. I dubbed the framework, "Project Binders".
Each group has a unique project binder for each project. The projects are no longer allotted a clump of time in which groups are required to produce the final product. I found this approach left quite a few students lost along the way. Every group would come up with a product, but many would be missing key developmental stages along the way. The project binder clearly truncates the project time into "stages". The students are responsible for a certain sub-section of the project during that stage. Each stage is discussed orally, worked on within the group, and assessed by a "stage rubric". A binder includes two copies of each stage rubric--one for the group to use and one for the teacher.
Along with the stage rubrics, students fill out a daily log to infuse the process with self-evaluation. Students are asked who was present, what they accomplished, what their next steps are, and if there were any issues they needed to report. Issues could be anything from a lack of white glue to a slacking team member.
Also included in the binder is a cover page, a calendar page, and a group contract. The calendar page will be put into a clear sheet protector. That way students can write deadlines, stage assessment days, and teacher-group meeting days right into their binder. Each binder comes equipped with a fine-tip dry-erase marker.
The group contract outlines the responsibilities of each member for the duration of the project time. If a group takes issue with a member's conduct, they can fill out a "issue" on the daily log form. A meeting with me decides a future course of action. If the problem persists, that student can be found in violation of the contract and will be forced to form their own group. If this occurs, the student and I will negotiate the amount of appropriate overlap between their new project and their previous group's.
I plan on setting this all up with a set of dividers and leaving plenty of room for the students to hole punch work from the stages and place it right in the binder. (Calculations, geometric drawings, brainstorming, etc.)
My goal, for this year, is to take a step back from technology. I want to refocus my efforts; I think it became a distraction at times last year. I also want to make these project binders accessible to a large number of teachers. I have abandoned the class wiki in favour of a paper calendar and physical progress sheets. It is my hope that this method appeals to more educators.
I have developed a set of templates for each page in a "project binder". I have also developed the specific contents of the "Pop Box Project" project binder. I believe that the new structure will not kill the innovation from the original project. (Linked above). I have posted all the files I have to date on my personal wiki page, and will continue to post the binders as I develop them.
All I ask is that you use the material and provide feedback so we can make this process continually better. I am sure that this is not the last chapter of my PBL story.
NatBanting
## Wednesday, July 4, 2012
### Project-Based Pitfalls
Those of you who follow me on twitter or read this blog regularly know I have been struggling to implement wide scale Project-based Learning (PBL) into my Workplace and Apprenticeship mathematics courses. This strand of classes is probably unfamiliar to those outside of Western Canada. I have included a link to our provincial curriculum below. You can skip to the outcomes and indicators to view which topics need to be addressed. (Page 33)
Let me start out by saying that I think this is an excellent direction for high school mathematics. Some powers-that-be in Saskatchewan would like to see this pathway die out or become analogous with a modified course. I disagree strongly on both counts. This course is an exercise in teacher flexibility. (That's probably why it is hated so much).
I designed my class around an infusion of technology, a large amount of responsibility, peer collaboration, and large-scale projects. I was very happy how it went (for the most part) but there were a few glaring problems that need to be addressed.
You may read this post as a warning if you are planning to implement projects into math class. You can also see it as an encouragement. It has been done, students did learn, and the teacher didn't collapse from administrative stresses.
The Three Biggest Struggles: A Rookie's Guide
1) Using Technology as an Enabler
I was graciously given numerous supports to set-out on my journey. First, and foremost, I was given a schedule where three of my four classes were Workplace and Apprenticeship classes. (Two Gr.10 and one Gr.11). This allowed me to focus on the institution of PBL. My principal bought me new tables to make over the physical appearance of the room. This made peer collaboration more accessible. I was given a document scanner to create digital archives, as well as sixteen laptops. The laptops ran Microsoft Office 2003. That puts them in perspective. Throughout the semester, they became more of a hindrance than a support. I was grateful to have them, but they soon started to crash, lose work, and even lose keys!
After the fact, I reflected upon my use of a wiki-centered class. I wanted the class to seem modern. I wanted the website to be our central hub of communication. The fact was, students rarely accessed the wiki outside of class and the computers acted as a barrier to the central hub. I have been denied new laptops for next year, but am choosing to see it as a blessing. (after an initial period of rage). The re-designed course will be organized in binders with more focus on neat construction. In this case, the technology I needed came in the three-ringed variety.
Make sure to ask yourself, "What does this technology enable my students to do that they could not do before?" Scanners, graphing software, and collaborative structures all proved useful. Archaic laptops became a barrier.
2) Creating Continuous Assessment
I was in constant communication with my students. The beginning of the year was scaffolded to acclimatize them to a system filled with freedom and creativity. As I weened them off of smaller projects onto ones with larger scope, I noticed a drop in commitment. The class evaluations revealed that numerous students felt lost or confused. They became directionless.
Projects that resulted in a creative product were not the problem. Students knew that the creativity of design unlocked diversity. I found that I lacked assessment (and subsequent guidance) on large projects where the product was designated, but pathway was not. For example, one class was designing a moving out plan complete with budget, housing and employment plans, a expense chart, and tax assessment. They knew what they were to hand in, but didn't grasp the possible pathways to get there.
To combat this, I am going to use a series of project checkpoints--flexible due-dates to keep groups on task. Each one includes a self-assessment, a teacher progress report, and a face-to-face meeting. It will keep both students and teacher accountable. In PBL, confusion is your biggest opponent. Students will shut down if they feel like they are on the wrong path. There needs to be scheduled times of encouragement and, if necessary, re-direction throughout larger project phases.
3) Group Accountability
This is every teacher's biggest fear with group work. One student does it all while others play 'Draw Something' with each other. Again, I felt this happening with the larger scope projects. The new assessment plan will help, but students will always have a natural tendency to wander intellectually. Not all off-task time has malicious intent.
At the beginning of the semester, I had students keep a log of what they did each day. The technology limitations halted the process. Having students create a group road map (so to speak) will cut down on time off task. One student realized he needed to get to work when his daily log included:
March 15th
-Filled out my bracket for March Madness. It's the winner, I can tell. Banting's has nothing on mine, he's gonna lose!
March 14th
-Sick, couldn't come into work.
March 13th
-Finished comparisons and graphs.
-Started making PowerPoint
March 12th
-Made all of my bar graphs and started putting them into a presentation.
-Put up my Andre Iguodala poster in my cubical.
This student has obvious interests that distract him from his work. A quick review of his log revealed a lot to him. This brings me full-circle to the technology piece I began with. I didn't have adequate means of self-tracking. The teacher needs to provide students the technologies to learn this important skill.
It is worth noting that I placed 3rd in the school pool--handily beating this student.
I have begun to re-work my classes with these three important reflections in mind. I want to develop step-by-step project binders for the students to keep them on pace. As the project descriptions, rubrics, and exemplars are created, I will post them (in full) on my personal wiki page.
It is ironic how much of a project designing a project-based class has become for myself.
NatBanting
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# On the Unique Games Conjecture Subhash Khot Georgia Inst. Of Technology. At FOCS 2005.
## Presentation on theme: "On the Unique Games Conjecture Subhash Khot Georgia Inst. Of Technology. At FOCS 2005."— Presentation transcript:
On the Unique Games Conjecture Subhash Khot Georgia Inst. Of Technology. At FOCS 2005
NP-hard Problems Vertex Cover MAX-3SAT Bin-Packing Set Cover Clique MAX-CUT ……………..
Approximability : Algorithms A C-approximation algorithm computes (C > 1), for problem instance I, solution A(I) s.t. Minimization problems : A(I) C OPT(I) Maximization problems : A(I) OPT(I) / C
Some Known Approximation Algorithms Vertex Cover 2 - approx. MAX-3SAT 8/7 - approx. Random assignment. Packing/Scheduling (1+ ) – approx. > 0 (PTAS) Set Cover ln n approx. Clique n/log n [Boppana Halldorsson’92] Many more, ref. [Vazirani’01]
PCP Theorem [B’85, GMR’89, BFL’91, LFKN’92, S’92,……] [PY’91] [FGLSS’91, AS’92 ALMSS’92] Theorem : It is NP-hard to tell whether a MAX-3SAT instance is * satisfiable (i.e. OPT = 1) or * no assignment satisfies more than 99% clauses (i.e. OPT 0.99). i.e. MAX-3SAT is 1/0.99 = 1.01 hard to approximate. i.e. MAX-3SAT and MAX-SNP-complete problems [PY’91] have no PTAS.
Approximability : Towards Tight Hardness Results [Hastad’96] Clique n 1- [Hastad’97] MAX-3SAT 8/7 - [Feige’98] Set Cover (1- ) ln n [Dinur’05] Combinatorial Proof of PCP Theorem !
Open Problems in Approximability –Vertex Cover (1.36 vs. 2) [DinurSafra’02] –Coloring 3-colorable graphs (5 vs. n 3/14 ) [ KhannaLinialSafra’93, BlumKarger’97 ] –Sparsest Cut (1 vs. (logn) 1/2 ) [ AroraRaoVazirani’04 ] –Max Cut (17/16 vs 1/0.878… ) [ Håstad’97, GoemansWilliamson’94] ………………………..
Unique Games Conjecture [Khot’02] Implies these hardness results : Vertex Cover 2- [KR’03] Coloring 3-colorable (1) [DMR’05] graphs (variant of UGC) MAX-CUT 1/0.878.. - [KKMO’04] Sparsest Cut, Multi-cut [KV’05, (1) CKKRS’04] Min-2SAT-Deletion [K’02, CKKRS’04]
Unique Games Conjecture Led to … [MOO’05] Majority Is Stablest Theorem [KV’05] “Negative type” metrics do not embed into L 1 with O(1) “distortion”. Optimal “integrality gap” for MAX-CUT SDP with “Triangle Inequality”.
Integrality Gap : Definition Given : Maximization Problem + Specific SDP relaxation. For every problem instance G, SDP(G) OPT(G) Integrality Gap = Max G SDP(G) / OPT(G) Constructing gap instance = negative result.
Overview of the talk The UGC Hardness of Approximation Results I hope UGC is true Attempts to Disprove : Algorithms Connections/applications : Fourier Analysis Integrality Gaps Metric Embeddings
Unique Games Conjecture A maximization problem called “Unique Game” is hard to approximate. “Gap-preserving” reductions from Unique Game Hardness results for Vertex Cover, MAX-CUT, Graph-Coloring, …..
Example of Unique Game OPT = max fraction of equations that can be satisfied by any assignment. x 1 + x 3 = 2 (mod k) 3 x 5 - x 2 = -1 (mod k) x 2 + 5 x 1 = 0 (mod k) UGC For large k, it is NP-hard to tell whether OPT 99% or OPT 1%
2-Prover-1-Round Game (Constraint Satisfaction Problem ) variables constraints
2-Prover-1-Round Game (Constraint Satisfaction Problem ) variables k labels Here k=4 constraints
2-Prover-1-Round Game (Constraint Satisfaction Problem ) variables k labels Here k=4 Constraints = Bipartite graphs or Relations [k] [k]
2-Prover-1-Round Game (Constraint Satisfaction Problem ) variables k labels Here k=4 OPT(G) = 7/7 Find a labeling that satisfies max # constraints
Hardness of Finding OPT(G) Given a 2P1R game G, how hard is it to find OPT(G) ? PCP Theorem + Raz’s Parallel Repetition Theorem : For every , there is integer k( ), s.t. it is NP-hard to tell whether a 2P1R game with k = k( ) labels has OPT = 1 or OPT In fact k = 1/poly( )
Reductions from 2P1R Game Almost all known hardness results (e.g. Clique, MAX-3SAT, Set Cover, SVP, …. ) are reductions from 2P1R games. Many special cases of 2P1R games are known to be hard, e.g. Multipartite graphs, Expander graphs, Smoothness property, …. What about unique games ?
Unique Game = 2P1R Game with Permutations variable k labels Here k=4
Unique Game = 2P1R Game with Permutations variable k labels Here k=4 Permutations or matchings : [k] [k]
OPT(G) = 6/7 Find a labeling that satisfies max # constraints Unique Game = 2P1R Game with Permutations
Unique Games Considered before …… [Feige Lovasz’92] Parallel Repetition of UG reduces OPT(G). How hard is approximating OPT(G) for a unique game G ? Observation : Easy to decide whether OPT(G) = 1.
MAX-CUT is Special Case of Unique Game Vertices : Binary variables x, y, z, w, ……. Edges : Equations x + y = 1 (mod 2) [Hastad’97] NP-hard to tell whether OPT(MAX-CUT) 17/21 or OPT(MAX-CUT) 16/21
Unique Games Conjecture For any , , there is integer k( , ), s.t. it is NP-hard to tell whether a Unique Game with k = k( , ) labels has OPT 1- or OPT i.e. Gap-Unique Game (1- , ) is NP-hard.
Overview of the talk The UGC Hardness of Approximation Results I hope UGC is true Attempts to Disprove : Algorithms Connections/applications : Fourier Analysis Integrality Gaps Metric Embeddings
Case Study : MAX-CUT Given a graph, find a cut that maximizes fraction of edges cut. Random cut : 2-approximation. [GW’94] SDP-relaxation and rounding. min 0 < < 1 / (arccos (1-2 ) / ) = 1/0.878 … approximation. [KKMO’04] Assuming UGC, MAX-CUT is 1/0.878… - hard to approximate.
Reduction to MAX-CUT Unique Game Graph H Completeness : OPT(UG) > 1-o(1) - o(1) cut. Soundness : OPT(UG) < o(1) No cut with size arccos (1-2 ) / + o(1) Hardness factor = / (arccos (1-2 ) / ) - o(1) Choose best to get 1/0.878 … (= [GW’94])
Reduction from Unique Game Gadget constructed via Fourier theorem + Connecting gadgets via Unique Game instance [DMR’05] “UGC reduces the analysis of the entire construction to the analysis of the gadget”. Gadget = Basic gadget ---> Bipartite gadget ---> Bipartite gadget with permutation
Basic Gadget A graph on {0,1} k with specific properties (e.g. cuts, vertex covers, colorability) {0,1} k k = # labels x = 011 Y = 110
Basic Gadget : MAX-CUT Weighted graph, total edge weight = 1. Picking random edge : x R {0,1} k y <-- flip every co-ordinate of x with probability ( 0.8) x {0,1} k y
MAX-CUT Gadget : Co-ordinate Cut Along Dimension i Fraction of edges cut = Pr (x,y) [x i y i ] = Observation : These are the maximum cuts. x i = 0 x i = 1
Bipartite Gadget A graph on {0,1} k {0,1} k (double cover of basic gadget) x = 011 y’ = 110
Cuts in Bipartite Gadget {0,1} k Matching co-ordinate cuts have size =
Bipartite Gadget with Permutation : [k] -> [k] Co-ordinates in second hypercube permuted via . x = 011 Y ’ = 110 (y’) = 011 Example : = reversal of co-ordinates.
Reduction from Unique Game Variables k labels OPT 1 – o(1) or OPT o(1) Permutations : [k] [k]
Instance H of MAX-CUT {0,1} k Vertices Edges Bipartite Gadget via
Proving Completeness Unique Game Graph H (Completeness) : OPT(UG) > 1-o(1) H has - o(1) cut.
Completeness : OPT(UG) 1-o(1) label = 2 label = 1 label = 3 label = 1 label = 3 label = 2 Labels = [1,2,3]
Completeness : OPT(UG) 1-o(1) {0,1} k Vertices Edges Hypercubes are cut along dimensions = labels. MAX-CUT - o(1)
Proving Soundness Unique Game Graph H (Soundness) : OPT(UG) < o(1) H has no cut of size arccos (1-2 ) / + o(1)
MAX-CUT Gadget Cuts = Boolean functions f : {0,1} k {0,1} Compare boolean functions * that depend only on single co-ordinate vs * where every co-ordinate has negligible “influence” (i.e. “non-junta” functions) {0,1} k x y f(x 1 x 2 …….. x k ) = x i f(x 1 x 2 …….. x k ) = MAJORITY Influence (i, f) = Pr x [ f(x) f(x+e i ) ]
Gadget : “Non-junta” Cuts How large can non-junta cuts be ? i.e. cuts with all influences negligible ? Random Cut : ½ Majority Cut : arccos (1-2 ) / > ½ [MOO’05] Majority Is Stablest (Best) Any cut slightly better than Majority Cut must have “influential” co-ordinate.
Non-junta Cuts in Bipartite Gadget [MOO’05] Any “special” cut with value arccos (1-2 ) / + must define a matching pair of influential co-ordinates. {0,1} k
Non-junta Cuts in Bipartite Gadget {0,1} k f : {0,1} k --> {0, 1} g : {0,1} k --> {0, 1} i Infl (i, f), Infl (i, g) > (1) cut > arccos (1-2 ) / +
Instance H of MAX-CUT {0,1} k Vertices Edges Bipartite Gadget via
Proving Soundness Assume arccos (1-2 ) / + cut exists. On /2 fraction of constraints, the bipartite gadget has arccos (1-2 ) / + /2 cut. matching pair of labels on this constraint. This is impossible since OPT(UG) = o(1). Done !
Other Hardness Results Vertex Cover Friedgut’s Theorem Every boolean function with low “average sensitivity” is a junta. Sparsest Cut, Min-2SAT Deletion KahnKalaiLinial Every balanced boolean function has a co-ordinate with influence log n/n. Bourgain’s Theorem (inspired by Hastad-Sudan’s 2-bit Long Code test) Every boolean function with low “noise sensitivity” is a junta. Coloring 3-Colorable [MOO’05] inspired. Graphs
Basic Paradigm by [BGS’95, Hastad’97] Hardness results for Clique, MAX-3SAT, ……. Instead of Unique Games, use reduction from general 2P1R Games (PCP Theorem + Raz). Hypercube = Bits in the Long Code [Bellare Goldreich Sudan’95] PCPs with 3 or more queries (testing Long Code). Not enough to construct 2-query PCPs.
Why UGC and not 2P1R Games? Power in simplicity. “Obvious” way of encoding a permutation constraint. Basic Gadget ----> Bipartite Gadget with permutation.
Overview of the talk The UGC Hardness of Approximation Results I hope UGC is true Attempts to Disprove : Algorithms Connections/applications : Fourier Analysis Integrality Gaps Metric Embeddings
I Hope UGC is True Implies all the “right” hardness results in a unifying way. Neat applications of Fourier theorems [Bourgain’02, KKL’88, Friedgut’98, MOO’05] Surprising application to theory of metric embeddings and SDP-relaxations [KV’05]. Mere coincidence ?
Supporting Evidence [Feige Reichman’04] Gap-Unique Game (C , ) is NP-hard. i.e. For every constant C, there is s.t. it is NP-hard to tell if a UG has OPT > C or OPT < . However C --> 0 as --> 0.
Supporting Evidence [Khot Vishnoi’05] SDP relaxation for Unique Game has integrality gap (1- , ).
Overview of the talk The UGC Hardness of Approximation Results I hope UGC is true Attempts to Disprove : Algorithms Connections/applications : Fourier Analysis Integrality Gaps Metric Embeddings
Disproving UGC means.. For small enough (constant) , given a UG with optimum 1- , algorithm that finds a labeling satisfying (say) 50% constraints.
Algorithmic Results Algorithm that finds a labeling satisfying f( , k, n) fraction of constraints. [Khot’02] 1- 1/5 k 2 [Trevisan’05] 1- 1/3 log 1/3 n [Gupta Talwar’05] 1- log n [CMM’05] 1/k , 1- 1/2 log 1/2 k None of these disproves UGC.
Quadratic Integer Program For Unique Game [Feige Lovasz’92] variable k labels : [k] [k] u 1, u 2, …, u k {0,1} v 1, v 2, …, v k {0,1} u v v i = 1 if Label(v) = i = 0 otherwise
Quadratic Program for Unique Games Constraints on edge-set E. Maximize u i v π(i) (u, v) E i=1,2,..,k u i [k], u i {0,1} u u i 2 = 1 i u i ≠ j, u i u j = 0
SDP Relaxation for Unique Games Maximize u i, v π(i) (u, v) E i=1,2,..,k u i [k], u i is a vector. u || u i || 2 = 1 i=1,2,..,k u i≠j [k], u i, u j = 0
[Feige Lovasz’92] OPT(G) SDP(G) 1. If OPT(G) < 1, then SDP(G) < 1. SDP(G m ) = (SDP(G)) m Parallel Repetition Theorem for UG : OPT(G) < 1 OPT(G m ) 0
[Khot’02] Rounding Algorithm u1u1 ukuk u2u2 vkvk v2v2 v1v1 r r Label(u) = 2, Label(v) = 2 Pr [ Label(u) = Label(v) ] > 1 - 1/5 k 2 Labeling satisfies 1 - 1/5 k 2 fraction of constraints in expected sense. Random r u v
[CMM’05] Algorithm Labeling that satisfies 1/k fraction of constraints. ( Optimal [KV’05]) vkvk v2v2 v1v1 r u1u1 ukuk u2u2 r All i s.t. u i is “close” to r are taken as candidate labels to u. Pick one of them at random.
[Trevisan’05] Algorithm Given a unique game with optimum 1- 1/log n, algorithm finds a labeling that satisfies 50% of constraints. Limit on hardness factors achievable via UGC (e.g. loglog n for Sparsest Cut).
[Trevisan’05] Algorithm [Leighton Rao’88] Delete a few constraints and remaining graph has connected components of low diameter. Variables and constraints
[Trevisan’05] Algorithm A good algorithm for graphs with low diameter.
Overview of the talk The UGC Hardness of Approximation Results I hope UGC is true Attempts to Disprove : Algorithms Connections/applications : Fourier Analysis Integrality Gaps Metric Embeddings
Already Covered Let’s move on ….
Overview of the talk The UGC Hardness of Approximation Results I hope UGC is true Attempts to Disprove : Algorithms Connections/applications : Fourier Analysis Integrality Gaps Metric Embeddings
[KV’05] Integrality Gaps for SDP-relaxations MAX-CUT Sparsest Cut Unique Game Gaps hold for SDPs with “Triangle Inequality”.
Integer Program for MAX-CUT Given G(V,E) Maximize ¼ |v i - v j | 2 (i, j) E i, v i {-1,1} Triangle Inequality (Optional) : i, j, k, |v i - v j | 2 + |v j - v k | 2 |v i - v k | 2
Goemans-Williamson’s SDP Relaxation for MAX-CUT Maximize ¼ || v i - v j || 2 (i, j) E i, v i R n, || v i || = 1 Triangle Inequality (Optional) : i, j, k, || v i - v j || 2 + || v j - v k || 2 || v i - v k || 2
Integrality Gap for MAX-CUT [Goemans Williamson’94] Integrality gap 1/0.878.. [Karloff’99] [Feige Schetchman ’01] Integrality gap 1/0.878.. - SDP solution does not satisfy Triangle Inequality. Does Triangle Inequality make the SDP tighter ? NO if Unique Games Conj. is true !
Integrality Gap for Unique Games SDP Unique Game G with OPT(G) = o(1) SDP(G) = 1-o(1) Orthonormal Bases for R k u 1, u 2, …, u k v 1, v 2, …, v k variables k labels Matchings [k] [k] u v
Integrality Gap for MAX-CUT with Triangle Inequality {-1,1} k u 1, u 2, …, u k u 1 u 2 u 3 ……… u k-1 u k PCP Reduction OPT(G) = o(1) No large cut Good SDP solution
Overview of the talk The UGC Hardness of Approximation Results I hope UGC is true Attempts to Disprove : Algorithms Connections/applications : Fourier Analysis Integrality Gaps Metric Embeddings
Metrics and Embeddings Metric is a distance function on [n] such that d(i, j) + d(j, k) d(i, k). Metric d embeds into metric with distortion 1 if i, j d(i, j) (i, j) d(i, j).
Negative Type Metrics Given a set of vectors satisfying Triangle Inequality : i, j, k, || v i - v j || 2 + || v j - v k || 2 || v i - v k || 2 d(i, j) = || v i - v j || 2 defines a metric. These are called “negative type metrics”. L 1 NEG METRICS
NEG vs L 1 Question [Goemans, Linial’ 95] Conjecture : NEG metrics embed into L 1 with O(1) distortion. Sparsest Cut O(1) Integrality Gap O(1) Approximation [Linial London Rabinovich’94] [Aumann Rabani’98] Unique Games Conjecture [Chawla Krauthgamer Kumar Rabani Sivakumar ’05] [KV’05] (1) hardness result
NEG vs L 1 Lower Bound ( loglog n) integrality gap for Sparsest Cut SDP. [KhotVishnoi’05, KrauthgamerRabani’05] A negative type metric that needs distortion ( loglog n) to embed into L 1.
Open Problems (Dis)Prove Unique Games Conjecture. Prove hardness results bypassing UGC. NEG vs L 1, Close the gap. (log log n) vs ( log n loglog n) [Arora Lee Naor’04]
Open Problems Prove hardness of Min-Deletion version of Unique Games. (log n approx. [GT’05]) Integrality gaps with “k-gonal” inequalities. Is hypercube (Long Code) necessary ?
Open Problems More hardness results, integrality gaps, embedding lower bounds, Fourier Analysis, …… [Samorodnitsky Trevisan’05] “Gowers Uniformity, Influence of Variables, and PCPs”. UGC Boolean k-CSP is hard to approximate within 2 k- log k Independent Set on degree D graphs is hard to approximate within D/poly(log D).
Open Problems in Approximability Traveling Salesperson Steiner Tree Max Acyclic Subgraph, Feedback Arc Set Bin-packing (additive approximation) …………………… Recent progress on Edge Disjoint Paths Network Congestion Shortest Vector Problem Asymmetric k-center (log * n) Group Steiner Tree (log 2 n) Hypergraph Vertex Cover ………………
Linear Unique Games System of linear equations mod k. x 1 + x 3 = 2 3 x 5 - x 2 = -1 x 2 + 5 x 1 = 0 [KKMO’04] UGC UGC in the special case of linear equations mod k.
Variations of Conjecture 2-to-1 Conjecture [K’02] -Conjecture [DMR’05] NP-hard to color 3-colorable graphs with O(1) colors. [k] [k]
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## What Is Marginal Average Cost?
To calculate the marginal average cost, we need to first calculate the average cost,
$\displaystyle \overline{TC}\left( Q \right)=\frac{TC\left( Q \right)}{Q}$
where TC(Q) is the total cost to produce Q units. Once we have the the average cost function, the marginal average cost is simply its derivative,
$\displaystyle \text{Marginal Average Cost}=\overline{TC}{{\,}^{\prime }}\left( Q \right)$
Example Suppose the total cost (in thousands of dollars) to produce Q units is
$\displaystyle TC\left( Q \right)=\frac{9Q-5}{7Q+2}$
a. Find the the average cost of producing 40 units.
b. Find the average cost function $\displaystyle \overline{TC}\left( Q \right)$.
c. Find the marginal average cost function $\displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$.
d. Find and interpret $\displaystyle \overline{TC}{{\,}^{\prime }}\left( 40 \right)$.
The marginal average cost is simply the slope of the tangent line to the average cost$\displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$. The slope has vertical units of thousands of dollars per unit and horizontal units of units. So the rate has units of thousands of dollars per unit per unit. This means that if the production were to increase by one, the average cost would drop by 0.0007701 thousand dollars per unit or 0.7701 dollars per unit. This means that the average cost is decreasing…probably a good thing for the bottom line.
## How Do You Maximize Revenue Using Elasticity?
There are several ways we can define elasticity E. Each indicates how the quantity demanded changes as the price is changed. In the examples below, we’ll utilize elasticity defined as
$\displaystyle \text{E}\approx \frac{P}{Q}\,\frac{dQ}{dP}$
With this definition, the revenue is maximized when E = -1.
Example 1 Suppose the demand function is defined by
$\displaystyle Q=30-\frac{P}{5}$
where Q units are demanded at a price of P dollars each.
a. Find an expression for the elasticity in terms of P.
b. Find the quantity at which the revenue is maximized.
Example 2 Suppose the demand function is defined by
$\displaystyle Q=40-\frac{P}{4}$
where Q units are demanded at a price of P dollars each.
a. Find an expression for the elasticity in terms of P.
b. Find the quantity at which the revenue is maximized.
Example 3 Suppose the demand function is defined by
$\displaystyle Q=39300-7{{P}^{2}}$
where Q units are demanded at a price of P dollars each.
a. Find an expression for the elasticity in terms of P.
b. Find the quantity at which the revenue is maximized.
## How Do You Find the Economic Order Quantity?
Although many textbooks use a restrictive formula to find economic order quantity or economic lots size, you can use tables to come up with more general formulas.
Here are more examples from class.
Problem 1 A restaurant has an annual demand for 1200 bottle of wine. It costs $1 to store one bottle for a year and$5 to place an order. Orders are made when the inventory of wine is zero. If each bottle costs $15, find the optimum number of bottles per order. The strategy above is correct, but there is a math mistake in the last board…do you see their error? Problem 2 If the restaurant in Problem 3 orders wine when the inventory is half of an order size, find the optimum number of bottles per order. This problem can’t be done with the formula given in Help Me Solve This. The change in when the order is made (when the inventory drops to half the order size) leads to a different expression for the storage costs. However, once this pattern is established the cost function can be maximized as before. ## What Is Marginal Average Cost? In each problem below, the average cost function by dividing the cost function by the variable representing the quantity. For a cost function C(Q), the average cost function is $\displaystyle \overline{C}(Q)=\frac{C(Q)}{Q}$ The marginal average cost function is the derivative of the average cost function. Problem 1 Suppose the total cost function for a product is $\displaystyle TC(Q)=\frac{3Q+1}{Q+2}\text{ hundred dollars}$ where Q is the number of units produced. 1. Find the average cost of producing 20 units. 2. Find the average cost function. 3. Find the marginal average cost function. 4. Find and interpret the marginal average cost when 20 units are produced. This means that each of the 20 units costs an average of .1386 hundred dollars or$13.86.
In this board they have used the fact that dividing by Q is the same as multiplying by 1/Q.
Although it is OK to leave the derivative unsimplified, they need to put in 20. So it is best to do some algebra before putting in the value. Since -0.006 is the slope of the tangent line on the average cost function, the units on it is hundreds of dollars per unit per unit:
$\displaystyle \frac{-0.006}{1}\frac{\frac{\text{hundreds of dollars}}{\text{unit}}}{\text{unit}}$
This means that if production is increased by 1 unit, the average cost will drop by 0.006 hundred dollars per unit.
Problem 2 Suppose the total cost function for a product is
$\displaystyle C(x)=\frac{30x^{2}+500}{x+2}\text{ thousand dollars}$
where x is the number of units produced.
1. Find the average cost of producing 10 units.
2. Find the average cost function.
3. Find the marginal average cost function.
4. Find and interpret the marginal average cost when 10 units are produced.
This value tells us that if production is increased by 1 unit, the average cost will drop by 0.3472 thousand dollars per unit or \$347.2 per unit. Had they rounded one more decimal place, we would have had this number to the nearest penny.
Problem 3 Suppose the total cost (in thousands of dollars) to produce Q units is
$\displaystyle TC\left( Q \right)=\frac{9Q-5}{7Q+2}$
a. Find the the average cost of producing 40 units.
b. Find the average cost function $\displaystyle \overline{TC}\left( Q \right)$.
c. Find the marginal average cost function $\displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$.
d. Find and interpret $\displaystyle \overline{TC}{{\,}^{\prime }}\left( 40 \right)$.
The marginal average cost is simply the slope of the tangent line to the average cost$\displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$. The slope has vertical units of thousands of dollars per unit and horizontal units of units. So the rate has units of thousands of dollars per unit per unit. This means that if the production were to increase by one, the average cost would drop by 0.0007701 thousand dollars per unit or 0.7701 dollars per unit. This means that the average cost is decreasing…probably a good thing for the bottom line.
## What is Elasticity?
Here are several problems classes have worked out to calculate the elasticity E,
$\displaystyle E=\frac{P}{Q} \frac{dQ}{dP}$
In this formula, $\displaystyle \frac{dQ}{dP}$ is the derivative of the demand function when it is given as a function of P. Here are two examples the class worked.
Problem 1 Suppose the quantity demanded by consumers in units is given by $Q=5000-5P$ where P is the unit price in dollars.
1. Find the elasticity of demand with respect to price when P = 200.
2. Find the quantity at which revenue is maximized.
This means that a 1% increase in price results in a 0.25% drop in the quantity demanded. The demand is inelastic and the price increase results in an increase in revenue.
Problem 2 Suppose the quantity demanded by consumers in units is given by $Q=100-\frac{P}{2}$ where P is the unit price in dollars.
1. Find the elasticity of demand with respect to price when P = 110.
2. Find the quantity at which revenue is maximized.
This means that a price increase of 1% will lead to a 1.22% drop in demand, demand is elastic and the price increase results in a drop in revenue.
Problem 3 Suppose the quantity demanded by consumers in units is given by $Q=500-0.1{{P}^{2}}$ where P is the unit price in dollars.
1. Find the elasticity of demand with respect to price when P = 100.
2. Find the quantity at which revenue is maximized.
An increase of 1% in price results in a drop in demand of 0.041%…demand is inelastic so the increase will result in an increase in revenue.
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Nrich: Math Puzzles and More
Nrich recently updated their amazing website. I love exploring their backlog of puzzles and games — what a mother lode of resources for math club or a homeschool co-op class!
Want to help your kids learn math? Claim your free 24-page problem-solving booklet, and sign up to hear about new books, revisions, and sales or other promotions.
6 thoughts on “Nrich: Math Puzzles and More”
1. jay says:
There are 55 connections, since there are curved connections too.
1. Paul J. Jacobs says:
Only 46 connections. The outer circle is a single curved line, not 10 curved lines.
2. jay says:
For a 100 pointed rose, 5050 connection including the curves.
3. So basically, you guys are saying that the answer is the sum of all integers between 1 and n, inclusive, where n = the number of points. Gauss gives us the equation cnt = (n*(n-1))/2 to determine the solution. It’s worth noting that you’d still have to subtract n from this value if the circle were a polygon.
4. Yes, Holosim, the answer is the sum of the first n whole numbers. If you look at the lines, you will see the outermost lines form an n-sided polygon. So if you posed the puzzle within the boundary of a polygon, those lines would already be covered, which explains why you would have to subtract n from the formula.
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Question
# A flip coil consists of N turns of circular coils which lie in a uniform magnetic field. Plane of the coils is perpendicular to the magnetic field as shown in figure. The coil is connected to a current integrator which measures the total charge passing through it. The coil is turned through 180o about the diameter. The charge passing through the coil is
A
NBAR
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B
3NBA2R
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C
NBA2R
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D
2NBAR
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Solution
## The correct option is D 2NBARThe emf induced in the coil is given by dϕdt.The current passing through the circuit thus becomes i=dϕdtR.Hence the total charge that passed through the current integrator=∫idt=1R∫dϕdtdt=ΔϕR=NBA−(−NBA)R=2NBAR
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# Maths - Reorthogonalising a matrix - Attempt at an Alternative Method
Another approach to try might be to split the matrices into there individual elements:
The inverse of a 2x2 matrix [O] is:
[O]-1 = 1/(o00 o11 * o01 o10)
o11 -o01 -o10 o00
so expanding out [O]T = [O]-1 gives:
o00 = o11/(o00 o11 * o01 o10)
o10 = -o01/(o00 o11 * o01 o10)
o01 = -o10/(o00 o11 * o01 o10)
o11 = o00/(o00 o11 * o01 o10)
If we constrain the determinant to be 1 then,
(o00 o11 * o01 o10) = 1
o00 = o11
o10 = -o01
we are looking for C where [O] = [C]*[M]
c00 * m00 + c01 * m10 c00 * m01 + c01 * m11 c10 * m00 + c11 * m10 c10 * m01 + c11 * m11
=
c00 c01 c10 c11
m00 m01 m10 m11
so combining these gives:
c00 * m00 + c01 * m10 = c10 * m01 + c11 * m11
c00 * m01 + c01 * m11 = -c10 * m00 - c11 * m10
There seem to be more unknowns than equations so its difficult to see where to take this?
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Question
# Solve:xdydx=y(logy−logx−1).
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Solution
## xdydx=y(lny−lnx−1)⇒1ydydx=1x(lny−lnx−1)⇒1y=dydx=lnyx−(1+lnxx)⇒1ydydx−lnyx=−(1+lnxx).......(1)Let lny=tDifferential w.r.t.x both sideswe get,1ydydx=dtdxSo, eqn (1) becomes,⇒dtdx−tx=−(1+lnxx).......(2)Above differential eqn is a linear differential equation of the form:-dydx+py=Q∴I.F=e∫Pdx=e∫−1xdx(∵P=−1x in eqn(2))Q=(−(1+lnx)x)we know, =e−(lnx)=elnx−1=eln1x=1x(∵elnx=x)∴ Solution is given byy×I.F=∫(I.F.×Q)dx+CSo, solution for eqn (2) is given byt.1x=∫−(1+lnx)x×1xdx+C=∫−1x2dx−∫lnxx2dxt.1x=1x−I1.........(3)Continued : →Let I1=∫lnxx2dxLet lnx=u⇒x=e4Diff we get 1xdx=du∴I1=∫u(eu)2du=∫u.e−2uduusing ILATE Rule.I1=u.∫e−2udu−∫(dudv.∫e−2udu)du=u×e−2u2−∫1×e−2u−2du=u×e−2u2+12∫e−2udu=u×e−2u2+12×e−2u−2+c=12[ue−2u−12e−2u]+cputting value of u, we getI1=12[lnxe−2lnx−12elnx]+c=12[lnx×elnx−2−12elnx−2]+c=12[lnx×x−2−12×x−2]+c=12[lnxx2−12x2]+c=12[2lnx−12]+cI1=14x2[lnx2−1]+cNow Putting value of I1in eqn (3), we gett.1x=1x−14x2[lnx2−1]+cPutting value of t we get.lnyx=1x−14x2[lnx2−lne]+c(∵lne=1)lny=4x−ln(x2×e)4x+cfinal solution for d eqn.
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The Quadratic Programming Solver -- Experimental
# Overview
The OPTMODEL procedure provides a framework for specifying and solving quadratic programs.
Mathematically, a quadratic programming (QP) problem can be stated as follows:
where
is the quadratic (also known as Hessian) matrix is the constraints matrix is the vector of decision variables is the vector of linear objective function coefficients is the vector of constraints right-hand sides (RHS) is the vector of lower bounds on the decision variables is the vector of upper bounds on the decision variables
The quadratic matrix is assumed to be symmetric; i.e.,
Indeed, it is easy to show that even if , then the simple modification
produces an equivalent formulation hence symmetry is assumed. When specifying a quadratic matrix it suffices to list only lower triangular coefficients.
In addition to being symmetric, is also required to be positive semidefinite:
for minimization type of models; it is required to be negative semidefinite for maximization type of models. Convexity can come as a result of a matrix-matrix multiplication
or as a consequence of physical laws, etc. See Figure 12.1 for examples of convex, concave, and nonconvex objective functions.
Figure 12.1: Examples of Convex, Concave, and Nonconvex Objective Functions
The order of constraints is insignificant. Some or all components of or (lower/upper bounds) can be omitted.
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# Evaluate expressions with grouping symbols
## Presentation on theme: "Evaluate expressions with grouping symbols"— Presentation transcript:
Evaluate expressions with grouping symbols
EXAMPLE 2 Evaluate expressions with grouping symbols Evaluate the expression. a. 7(13 – 8) = 7(5) Subtract within parentheses. = 35 Multiply. b. 24 – (32 + 1) = 24 – (9 + 1) Evaluate power. = 24 – 10 Add within parentheses. = 14 Subtract. c. 2[30 – (8 + 13)] = 2[30 – 21] c. Add within parentheses. = 2[9] Subtract within brackets. = 18 Multiply.
Evaluate an algebraic expression
EXAMPLE 3 Evaluate an algebraic expression Evaluate the expression when x = 4. 9x 3(x + 2) = 3(4 + 2) 9 4 Substitute 4 for x. 3 6 9 4 = Add within parentheses. 18 36 = Multiply. = 2 Divide.
GUIDED PRACTICE for Examples 2 and 3 Evaluate the expression. (3 + 9) = 48 6. 3(8 – 22) = 12 7. 2[( 9 + 3) ] = 6
GUIDED PRACTICE for Examples 2 and 3 Evaluate the expression when y = 8. y2 – 3 8. = 61 12 – y – 1 9. = 3 10y + 1 y + 1 10. = 9
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# zbMATH — the first resource for mathematics
An objective approach of balanced cricket team selection using binary integer programming method. (English) Zbl 1360.90158
Summary: Selecting a balanced playing XI in the game of cricket with the right mix of players of different specialization is a difficult decision making problem for the team management. To make the process more objective, optimization techniques can be applied to the process of selection of players from a given squad. Such an exercise has two dimensions. First, a suitable tool for performance measurement of cricketers needs to be defined. Secondly, for selecting a balanced team of XI players, an appropriate objective function and some constraints need to be formulated. Since the captain gets an obvious inclusion in the cricket team, the area specialization of the captain influences the selection of other ten positions in the playing XI. This study attempts to select the optimum balanced playing XI from a squad of players given the specialization of the captain using binary integer programming. To validate the exercise, data from the fifth season of the Indian Premier League has been used.
##### MSC:
90B90 Case-oriented studies in operations research 90C10 Integer programming 90B70 Theory of organizations, manpower planning in operations research
Full Text:
##### References:
[1] Saikia, H; Bhattacharjee, D, On classification of all-rounders of the indian premier league (IPL): a Bayesian approach, Vikalpa: J. Decis. Mak., 36, 51-66, (2011) [2] Lemmer, HH, Team selection after a short cricket series, Eur. J. Sport Sci., (2011) [3] Salas, E; Rosen, MA; Burke, CS; Goodwin, GF; Fiore, SM; Ericsson, KA (ed.); Charness, N (ed.); Feltovich, PJ (ed.); Hoffman, RR (ed.), The making of a dream team: when expert teams do best, 439-453, (2006), New York [4] Kamble, A; Rao, R; Kale, A; Samant, S, Selection of cricket players using analytical hierarchy process, Int. J. Sports Sci. Eng., 5, 207-212, (2011) [5] Ahmed, F; Jindal, A; Deb, K, Multi-objective optimization and decision making approaches to cricket team selection, Appl. Soft Comput., (2012) [6] Barr, GDI; Kantor, BS, A criterion for comparing and selecting batsmen in limited overs cricket, J. Oper. Res. Soc., 55, 1266-1274, (2004) · Zbl 1088.90545 [7] Gerber, H; Sharp, G, Selecting a limited overs cricket squad using an integer programming model, S. Afr. J. Res. Sport Phys. Educ. Recreat., 28, 81-90, (2006) [8] Lourens, M.: Integer optimization for the selection of a twenty20 cricket team$$.$$ Nelson Mandela Metropolitan University. http://statssa.gov.za/isi2009/ScientificProgramme/IPMS/0738.pdf (2010). Accessed 21 June 2010 [9] Brettenny, W.: Integer optimisation of the selection of a fantasy league cricket team$$.$$ Nelson Mandela Metropolitan University. http://dspace.nmmu.ac.za:8080/jspui/bitstream/10948/…/W20%B.pdf (2010). Accessed 21 December 2010 [10] Staden, PJ, Comparison of cricketers’ bowling and batting performances using graphical displays, Curr. Sci., 96, 764-766, (2009) [11] Beaudoin, D; Swartz, T, The best batsmen and bowlers in one-day cricket, S. Afr. Stat. J., 37, 203-222, (2003) · Zbl 1139.62344 [12] Singh, G; Bhatia, N; Singh, S, Fuzzy cognitive maps based cricket player performance evaluator, Int. J. Enterp. Comput. Bus. Syst., 1, 1-15, (2011) [13] Damodaran, U, Stochastic dominance and analysis of ODI batting performance: the Indian cricket team, 1989-2005, J. Sports Sci. Med., 5, 503-508, (2006) [14] Boon, BH; Sierksma, G, Team formation: matching quality supply and quality demand, Eur. J. Oper. Res., 148, 277-292, (2003) · Zbl 1036.90503 [15] Das, D.: Moneyballer: an integer optimization framework for fantasy cricket league selection and substitution. http://debarghyadas.com/files/IPLpaper.pdf (2014). Accessed 11 July 2015 [16] Lewis, A, Towards fairer measures of player performance in one-day cricket, J. Oper. Res. Soc., 56, 804-815, (2005) · Zbl 1084.90513 [17] Iyenger, NS; Sudarshan, P, A method of classifying regions from multivariate data, Econ. Pol. Wkly, 5, 2048-2052, (1982) · Zbl 0505.35075 [18] Lemmer, HH, An analysis of players’ performances in the first cricket twenty20 world cup series, S. Afr. J. Res. Sport Phys. Educ. Recreat., 30, 73-79, (2008) [19] Lemmer, HH, The combined bowling rate as a measure of bowling performance in cricket, S. Afr. J. Res. Sport Phys. Educ. Recreat., 24, 37-44, (2002) [20] Lemmer, HH, A method for the comparison of the bowling performances of bowlers in a match or a series of matches, S. Afr. J. Res. Sport Phys. Educ. Recreat., 27, 91-103, (2005) [21] Narayanan, A.: Test wicket keepers - an analysis. The ESPNcricinfo Blog. http://blogs.cricinfo.com/itfigures/archives/wicketkeepers/ (2008). Accessed 4 March 2010 · Zbl 1139.62344 [22] Lemmer, HH, A measure for the batting performance of cricket players, S. Afr. J. Res. Sport Phys. Educ. Recreat., 26, 55-64, (2004) [23] Chakrabarty, N; Bhattacharjee, D, Aggregation and weightage issues concerning composite index development: experience with digital divide in Asian countries, Asia Pac. J. Libr. Inf. Sci., 2, 24-37, (2012) [24] Bracewell, PJ; Ruggiero, K, A parametric control chart for monitoring individual batting performances in cricket, J. Quant. Anal. Sports, 5, 1-19, (2009)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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## Which cement is used in oil wells?
Oil-well cements are used for cementing work in the drilling of oil wells where they are subject to high temperatures and pressures. They usually consist of portland or pozzolanic cement (see below) with special organic retarders to prevent the cement from setting too quickly.
What are cement slurries?
Cement slurry is a mixture of Portland cement, water, and additives. The cement slurry used in the drilling industry is a cement suspension with a high water content and a relatively low viscosity. Cement slurry is a mixture of Portland cement, water, and additives.
How do you calculate the volume of a concrete slurry?
Calculate Water Requirement and Cement Yield as per Cement Slurry
1. The cement formula given from town is listed below:
2. Determine requirement (gal/sack), yield of cement.
3. Volume = Weight x Absolute Volume.
4. Total weight per sack of cement =94+32.9+1.0267+5.825+8.33k = 133.7517 + 8.33k.
### What is the best bonding agent for concrete?
Epoxy Resin is known for being the most versatile of all concrete bonding agents. This bonding agent is ideal for high performance and lightweight parts. Characterized as being strong but brittle, Epoxy Resin can be formulated to become more flexible without losing its tensile strength.
How do they cement an oil-well?
Cementing is the process of mixing a slurry of cement, cement additives and water and pumping it down through casing to critical points in the annulus around the casing or in the open hole below the casing string.
Oil-well cements are manufactured using the same raw materials and processes11,23,24 as for ordinary, moderate or high sulfate-resistant Portland cement. The composition of the raw meal is designed to produce a clinker of suitable reactivity for oil-well cement usage.
## What is slurry mix?
A slurry is a mixture of solids denser than water suspended in liquid, usually water. The particles may settle below a certain transport velocity and the mixture can behave like a Newtonian or non-Newtonian fluid. Depending on the mixture, the slurry may be abrasive and/or corrosive.
How do you make pourable cement?
The proportions are as follows: 3 parts cement, 9 parts fine sand, ~5 parts water.
What are the performance properties of slurry cement?
The slurry performance properties that are measured include: Cement manufactured to American Petroleum Institute (API) depth and temperature requirements can be purchased in most oil-producing areas of the world. Any properly made Portland cement (consistent from batch to batch) can be used at temperatures up to 570°F.
### Do not add dispersants or retarders to slurry cement?
Do not add dispersants or retarders in excess of the amounts indicated by wellbore conditions, and provide just enough fluid-loss control to place the cement before it gels. Slurry design is affected by the following criteria:
How is volume reduction compensated for in cement slurry?
Normally a slurry volume reduction would be compensated for by contraction and downward movement of fluids from above, thus maintaining its hydrostatic head. However, during the setting process the cement slurry develops a gel structure causing it to stick to the wall thus hindering compensation of the shrinkage.
What is chemical shrinkage of oil well cement?
chemical shrinkage is the sum of the external chemical shrinkage and the contraction pores of the slurry. The formation of contrac-tion pores contribute to the connectivity between pores in a set cement and hence to its permeability. Chemical shrinkage of oil well cements have been investigated by several authors.
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# How do solve 3/(x+1)<=3 graphically?
May 25, 2018
The solution is $x \in \left(- \infty , - 1\right) \cup \left[0 , + \infty\right)$
#### Explanation:
Plot the curve
$f \left(x\right) = \frac{3}{x + 1} - 3$
The function is undefined for $x = - 1$
graph{3/(x+1)-3 [-18.02, 18.03, -9.01, 9.01]}
Then, look where $f \left(x\right) \le 0$
The solution is $x \in \left(- \infty , - 1\right) \cup \left[0 , + \infty\right)$
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Wellcome!
Question 1. Asked on :13 July 2019:04:06:42 PM
#### Find a quadratic polynomial whose sum of zero is 14 and product of zero is -1.
Master Mind
Given that:
α + β = 14 and αβ = -1
α + β = 14 = -ba ..........(i)
αβ = -1 or αβ = -44 = ca ....... (ii)
On comparing both equations we have,
a = 4, b = -1 and c = - 4
ax2 + bx + c
⇒ 4x2 + (-1)x + (-4)
⇒ 4x2 - x - 4 Answer
-Answered by Master Mind On 13 July 2019:04:50:14 PM
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Solution to puzzle 111: Trigonometric progression
Complex Arithmetic Solution
We will make use of Euler's formula, which states that, for any real number t, eit = cos t + i sin t.
It follows that
cos t = ½ (eit + e−it), and
i sin t = ½ (eit − e−it).
Let C = (cos a + cos 2a + ... + cos na), and S = (sin a + sin 2a + ... + sin na).
Then consider P = eia + e2ia + ... + eina = C + iS.
Evaluating P as a geometric series, we have
Recall that P = C + iS. That is, C is the real part of the above expression for P; S is the imaginary part.
Trigonometric Solution
We will make use of the following product-to-sum and sum-to-product trigonometric identities.
sin x · sin y = ½ [cos(x − y) − cos(x + y)] (1) sin x · cos y = ½ [sin(x + y) + sin(x − y)] (2) cos x − cos y = −2 sin ½(x + y) · sin ½(x − y) (3) sin x − sin y = 2 cos ½(x + y) · sin ½(x − y) (4)
Let S = (sin a + sin 2a + ... + sin na).
Then S · sin(a/2) = ½ [cos(a/2) − cos(3a/2) + cos(3a/2) − cos(5a/2) + ... + cos[(2n−1)a/2] − cos[(2n+1)a/2]], by (1). = ½ [cos(a/2) − cos[(2n+1)a/2]]. = sin[(n+1)a/2] · sin(na/2), by (3).
Similarly, let C = (cos a + cos 2a + ... + cos na).
Then C · sin(a/2) = ½ [−sin(a/2) + sin(3a/2) − sin(3a/2) + sin(5a/2) − ... − sin[(2n−1)a/2] + sin[(2n+1)a/2]], by (2). = ½ [−sin(a/2) + sin[(2n+1)a/2]]. = cos[(n+1)a/2] · sin(na/2), by (4).
Six tangents
Show that tan (/13) · tan (2/13) · tan (3/13) · tan (4/13) · tan (5/13) · tan (6/13) = .
`Hint - Solution`
Center of mass
Pennies are placed around the circumference of a circle as follows. Using polar coordinates, and with the center of the circle at the pole, place 1 penny with its center of mass vertically above the point (1,0), 2 pennies piled vertically at (1,a), 3 pennies at (1,2a), ... , n pennies at (1,(n−1)a), where a = 2/n radians. Find the x and y coordinates of the center of mass of the coins.
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## Writing a paper about Cellular Automata
For the discussion of math. Duh.
Moderators: gmalivuk, Moderators General, Prelates
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Writing a paper about Cellular Automata
Hello!
This is my first post in this forum. It seems to have alot of activity and some interesting topics!
First Okay, tell a little about my story:
I kinda droped out of college, I liked math a bit, but i hated school.
I still i have alot of computational experience and I experiment with cellular automata and the like,
i started programming at early age.
I have fragmented knowledge about math, but sometimes i look at a problem and solve it very easily.
but I am really sucky at writing anything down into readable form (for others to understand).
I what i want to ask you about is what does a paper contain?
Does it have mathematical proofs, solutions to problems etc. but what does a proof actually mean?
does someone have a specifical definition of a proof? let's say i have found a solution to a problem that I think
can be usefull for researchers, professors, mathematicians and physicists and the like..
how do i prove that it is correct? by coding a computer program i see that it is correct, but i dont necessary
have the mathematical knowledge (formal) to write a proof down.
I want to learn it, but i dont have access to school. my bachelor-studies in Computer Science was boring and so my
grades reflected that. i dont study any mastering either.
im studying stuff at home, not at an academy. and have many ideas and solutions to problems.
studying the cellular automaton rules i found a way to reverse rule 30, that is to go backwards without much to know about the history.
only last configuration.
i heard it was impossible to do, i read on the web, and looked at seminar by Stephen Wolfram about his book.
but either i have misunderstood or i have the solution on how to go backwards (atleast for this specific rule).
I dont know if it has been done before.
I wanna write a paper about it, but i dont know how. or where to start.
Thanks
Rudi
tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC
### Re: Writing a paper about Cellular Au
It is impossible to do, every configuration has four possibilities of what could lead to it, so it is impossible.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Yakk
Poster with most posts but no title.
Posts: 11077
Joined: Sat Jan 27, 2007 7:27 pm UTC
Location: E pur si muove
### Re: Writing a paper about Cellular Au
What is a proof?
A proof is a justification for an answer to a question.
The proof is typically written out in a sequence of steps such that checking each step is (much) easier than finding the answer to the original proof, and that if the steps are valid then the reader will agree the answer is correct.
Now, that checking is highly context dependent. What kind of steps are acceptable and which are not depends on what is well known about the subject area, and the skill of the reader.
In an academic sense, this means understanding enough about the subject area to understand what kind of steps are easy to verify and are known, and which are not. Creating an invalid proof is really, really easy, where at some point you miss some random edge case. You also have to really specifically describe the question in such a way that there is very little ambiguity, and the same with the answer: doing so is a hard problem, and is one of the things you learn when you study a subject area academically.
For specific examples of proofs, you could pick up any textbook on computer science and automata theory. Or you could search online paper repositories.
"Seeing that a computer program is correct" is not, generally, a reliable form of proof.
On over-estimation and the Dunning–Kruger effect:
Spoiler:
Most people who are not academically inclined seriously over estimate their abilities at the subject area in question. One of the things that happens as you advance up the academic slope is that there are serious amounts of culling, and at each stage (mostly) the people who where really good at the previous stage are the ones left standing. By the time you even get your first tenure track job, you have the high-school cull, the early university cull, the finish university cull, starting grad school cull, the finishing your PhD culls, the getting a good post-doc cull, then the getting a tenure track job cull. Many of these culls are not solely based on intelligence or aptitude at the subject (self motivation ends up being rather important), but the output tends to be pretty expert cookies, who have spent sometimes north of 10,000 hours thinking (with help!) about an increasingly specialized part of science.
Ie: the odds are that what you are thinking isn't all that extraordinary. If you have an inclination to keep it secret, that inclination is almost certainly wrong. The world is full of people who have a "great idea" that they thought about and worked on really hard and, due to the effort in figuring it out, they don't want to share it: in my experience, such people very rarely actually have a particularly "great idea". Academia gets around this problem (of hording ideas) by rating people by what they share, rather than what they keep secret.
My advice to you is to accept that your idea almost certainly isn't all that revolutionary, and instead of going about it in a secret backwards way, try sharing it openly and seeing if people can help you understand what it means. Ie, instead of asking "what is a proof", you could ask for help proving your particular algorithm does what you think it does. The alternative approach is pretty much to start reading introductory textbooks on proving things in your domain of interest (like an undergrad would), and start with spending a few 100 hours learning how to prove easy things and how to read proofs, then move on up to intermediate texts, or even easy to read journal articles. You aren't going to become an expert at "writing a proof" from much less effort than this.
In short, don't try to keep your technique secret. Tell people, and ask for help with the flaws. There will almost certainly be flaws. The worth of your idea is not a function of how hard it was to come up with, and you aren't an expert at coming up with high-worth ideas.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Au
Yakk:
thank you for an interesting and educable answer. i will have to read it a couple of times more to really grasp it all and let it in. I ordered a book last year, its called "Theory And Applications of Cellular Automata" by Martinez and le others, that contains papers with different solutions to problems and papers that introduces applications on some topics in CA. I just recently started looking into it and I find it difficult to understand, but I get some clues by reading the Abstract sections etc.
What I do is much simpler than all these things. Because I like simple solutions! and my solution to this specific problem was a very logical thinking, like injecting something into the automaton and just giving it the right directions.
I know that seeing a computer program that is correct, is not a reliable form of proof. thats why I am looking into actually writing a proof by dividing the different program codes into small sections, and getting it down as some form on a paper or something, hopefully. Ill ask here for help if i need it, if thats okay. i find this forum the best one considering the issues and problems I have, atleast it seems like a forum where one can learn, get hints about things, and talk about interresting subjects etc. But I would have to write down some stuff first, so that it doesnt look like gibberish. I talked with someone who work with patents here in my country - Just to hear if something like this would be worth to patent if it really was a solution to something. Really I dont want to do it, if it happens to be a solution! But I need a meeting with him before actually revealing anything about this, even if it might not be anything. I cant afford to patent it, but i would like some info from someone about it anyway. The only reason why I actually think this way is because "Rule 30" is allready patented. And, what would be a cooler idea to actually patent a reversible solution to that rule? I dont think its extraordinary solution to all worlds problems what i have here, hehe. It took me one day to solve the problem i was facing. And when I looked on the web etc: I found out that no others had done this before. And some people say it's plain impossible. Even Stephen Wolfram himself. I dont understand what he and they mean by that, but anyway. How can something be impossible when I think I have the solution right here. Not that I care. What I care about is my work, and what i did that day was a very concentration to do things right. If I can prove that what I have is actually right, then I would be glad. I dont say those people are wrong, but I want to show how to solve a problem, by using a / some method(s), rather than wasting time on thinking that what I have here is not right, when I can't see it yet, because my program is showing the proof. I dont know how many iterations I would have to do and check to find that it is not correct, a million? or infinite many times? i think thats why i want it down on paper. im sure if i look closer I can find wrong things about it. But if I follow the logical methods I use, it doesnt seem like that. There you have some thoughts about what kind of problems i am in right now. hehe.
I tried to share my idea first at some site, but I didnt find any great answers back. I just got: "No, its impossible to do" etc.. but.. so i am trying to find an environment where knowledge is freely accepted and where people: mathematicians, physicists dont look at things that are impossible, but look on how things can be possible. if you know what i mean.
anyway, i could have written so much more here, i just have so much to say, but it would be too much to read hehe, so ill just wait and see what people say. i will look more into proofs etc. also its almost like i am not at the time being in the mood of putting my notes into something bigger and better. I think i need a break from that, so I will take it out later and look at it then. It's been a very strange past few days, thinking around this kind of thing. I will not have my technique secret after a few talks and when ive gotten a few answers to some questions that i have. i hope that it is okay! i will get back to this, when i have gotten the answers! I want information to be free! But sometimes I wonder, whats the reason why Rule 30 is already patented? Because its a good random generator or is used in public-key cryptography? and used in processes in nature (not only rule 30). but anyway. if one can reverse engineer things, nothing would be impossible! not even finding the cellular automaton for our universe! (although that requires alot of work ).
Meem1029
Posts: 379
Joined: Wed Jul 21, 2010 1:11 am UTC
### Re: Writing a paper about Cellular Au
First of all, tomtom already told you that it's impossible to do and why. Do you have a reason that the point he brought up is invalid?
Second of all, remember that code speaks louder than words. If you have code that does something people say is impossible, post it somewhere (ex github) and make it known so people can tell you what's wrong with it. And approach it in a way that does not seem arrogant and "I'm smarter than the people who spend their lives studying this stuff and proved it impossible but I did it anyway." All that will accomplish is annoying people and making them tear your code apart. The type of people who look at it will generally provide constructive feedback as to what was wrong about it or else be very impressed if your code is right.
cjmcjmcjmcjm wrote:If it can't be done in an 80x24 terminal, it's not worth doing
++\$_
Mo' Money
Posts: 2370
Joined: Thu Nov 01, 2007 4:06 am UTC
### Re: Writing a paper about Cellular Au
Do not ever keep your ideas secret. Out of all the ideas you come up with in your life, almost none will be worth keeping secret. But many will be worth sharing.
snowyowl
Posts: 464
Joined: Tue Jun 23, 2009 7:36 pm UTC
### Re: Writing a paper about Cellular Au
Meem is right. If you have code that does something previously thought impossible - or even pseudocode, that is human-readable but not actually written in a programming language - then post it so we can give it the once-over. Probably there'll be some flaw in it, since some very clever people have already looked into this and found conclusive 100% proof that it's impossible, but you never know.
The preceding comment is an automated response.
lightvector
Posts: 224
Joined: Tue Jun 17, 2008 11:04 pm UTC
### Re: Writing a paper about Cellular Au
Definition: A cellular automaton is reversible if every configuration has a unique predecessor, that is, a unique configuration that transitions to it. Where "configuration" refers to the global state - the state of every cell in the automaton.
Theorem: Rule 30 is not reversible.
Proof: The configuration where every cell is 0 has two predecessors. Itself and the configuration where every cell is 1. QED.
On the other hand, there is an interesting paper that shows (among other things) that rule 30 is reversible if you impose the additional constraint that the right half of any configuration must eventually be all zero. Given this paper together with the above disproof, it's unlikely what you've discovered is new. But it is still good to experiment and learn about such things.
ahammel
My Little Cabbage
Posts: 2135
Joined: Mon Jan 30, 2012 12:46 am UTC
Location: Vancouver BC
Contact:
### Re: Writing a paper about Cellular Au
If you're interested in learning what goes in to a math or CS paper, why not read some? Since you've got some post-secondary education in the topic, you should be able to find some that you can get through without too much difficulty. If I understand you correctly, you don't have access to a university library anymore, but there are plenty of journals that make at least some of their material available for free on the interenet. This is probably a good start. A good public library will have a few scholarly journals available as well.
Edit: P.S. The title of your thread is cut off. I was very confused to learn that you're not writing a paper about cellular gold.
He/Him/His/Alex
God damn these electric sex pants!
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Au
ahammel: thanks for the links! sorry about that topic, i dunno what happened with it during my first post. It should of course have said Cellular Automata. so its an error from my side or something. that paper, i allready have but thanks for telling me anyway, i didnt read and understand it all the way through yet, but i will do it again! I was actually going to send a email to Eric Rowland, but i choosed not too since i just have these messy notes.
okay, thanks, and too all of you: I will try and write some human readable paper about this (because now it's just very chaotic, probably most wouldnt understand it since it is in c++ code) and then I will come back and post it so you can read it and come with suggestions, error-corrections and the like. i will try and find some info on how mathematicians write pseudo-code so i can include it in the paper/article for you to see or whatever.
: another thing (which i forgot) that i formally was going to ask this time, was if someone knows about a good application for windows that can be used to write math symbols and text, draw things like grids, figures and so on to include in an article for example
Rudi
Meem1029
Posts: 379
Joined: Wed Jul 21, 2010 1:11 am UTC
### Re: Writing a paper about Cellular Au
If you're going to write a paper on it at all, you need to learn LaTeX. There are many tutorials on the internet for learning it. A good one for the basics is at Art of Problem Solving (http://www.artofproblemsolving.com/Wiki ... aTeX:About).
cjmcjmcjmcjm wrote:If it can't be done in an 80x24 terminal, it's not worth doing
ahammel
My Little Cabbage
Posts: 2135
Joined: Mon Jan 30, 2012 12:46 am UTC
Location: Vancouver BC
Contact:
### Re: Writing a paper about Cellular Au
rudi wrote:okay, thanks, and too all of you: I will try and write some human readable paper about this (because now it's just very chaotic, probably most wouldnt understand it since it is in c++ code) and then I will come back and post it so you can read it and come with suggestions, error-corrections and the like.
Trust me, anybody who is in a position to make corrections will understand c++ code.
rudi wrote:: another thing (which i forgot) that i formally was going to ask this time, was if someone knows about a good application for windows that can be used to write math symbols and text, draw things like grids, figures and so on to include in an article for example
I strongly support using LaTeX. MiKTeX is the most popular Windows frontend, but emacs and vim make good environments with a few additional packages.
He/Him/His/Alex
God damn these electric sex pants!
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Au
thanks.
im downloading protext, i dunno if it includes latex or miktex. i think i tried miktex without luck, or something else. cant remember.
also im in the middle of a inheritance. my dad passed away last year. and its so much else takes time to do than working on this.
hopefully ill get back when all this is inheritance is over. takes alot of energy.
Rudi
ahammel
My Little Cabbage
Posts: 2135
Joined: Mon Jan 30, 2012 12:46 am UTC
Location: Vancouver BC
Contact:
### Re: Writing a paper about Cellular Au
Protext looks like it's based on MiKTeX, so it should be sufficient. I'm not a windows user, so I can't be of much direct help.
I'm sorry for your loss. Take your time.
He/Him/His/Alex
God damn these electric sex pants!
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Au
ahammel: thanks
Rudi
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Automata
Hey! Im back! im very very tired right now, but i decided to post my results. if anyone can check it, (i dont know if it has some spelling-errors right now), then please tell me how it turned out! also if there are any questions, please post them here, and i will answer. tell me why Rule 30 is not reversible after reading this:
http://www.home.no/rudibs/Rule 30 reversibility.pdf
i believe it is reversible, i just can't believe anything else, or i've missed something essential about the definition they use for reversibility.
i wrote the document fast because i decided to post it before i go to sleep.
i will write a more comprehensive document that might be easier to understand the things i wrote, but they should be easy after some studying.
its really simple. because rule 30 is simple to produce, it should be simple to reverse.
remember the current configurations is all one needs.
please understand that the states of Rule 86 is the key here after reading the 3-cells to determine the left one (a(i-1)).
now i want feedback, questions, bugs, error-corrections and the like. i really want to hear what you say about this.
Rudi
Rudi
tomtom2357
Posts: 563
Joined: Tue Jul 27, 2010 8:48 am UTC
### Re: Writing a paper about Cellular Automata
Okay, if your algorithm does what it is supposed to do, then it should do one of these three things with an infinite string:
1. It outputs all the strings which, when run through rule 30, produce the original string.
2. It outputs one of the strings which, when run through rule 30, produce the original string.
3. It outputs the only string which, when run through rule 30, produce the original string.
Option 1 is possible, and it would be interesting, but it is not reversibility. Reversibility is one to one.
Option 2 is possible, but it is less interesting than option 1, as it is less powerful. However it is not reversibility, because it is weaker than option 1, and just because it is one to one, doesn't mean rule 30 is reversible, as for some strings it will omit a different string which precedes the original string.
Option 3 is impossible because, as we have shown, some strings (if not all) have at least 2 strings which could precede it in the iteration, for example the string of all zeroes.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
rudi
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### Re: Writing a paper about Cellular Automata
tomtom2357: please explain what you mean by this.
i believe that you talk about the definition of string the same as configuration, or?
commonly: the original rule 30 with black cell or with initial configurations (random states of cells) intrinsincly generates consecutive configurations with time steps on forward.
what my algorithm are capable of is to intrinsincly generate the consecutive reverse states of cells with reverse time steps of the current/last configuration. the length of the current configuration is the amount of time steps you need to go back to the original configuration (your random string).
[edit: computational irreducibility has nothing to do with reversibility]
Rudi
OverBored
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### Re: Writing a paper about Cellular Automata
Okay, how about this. I generated a random sequence of digits, and it turned out the at the next step my string looks like this:
.....0000000000000000.....
Or if you prefer,
....Black Black Black Black.......
What did I have previously? Note that this is precisely the question you claim to be answering, but I am confident that if you tell me just one answer, you will be wrong. If you give me more than one answer, you might be right, but this won't be new, as reversible claims there is a unique sequence leading to this.
G4!!
Grob FTW,
Hello. Smithers. You're. Quite good. At. Turning. Me. On.
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Automata
OverBored:
your question is based on what you define being zero or one in your rule 30 pattern. black and white has two different solutions, you need to specify which one it is. a sequence of 1,1,1,1,1,1,1,1's had previously a sequence of ...0,0,1,0,0,1,0,0,1.... (because 1 would have generated infinite zeros), and a sequence of zeroes would have had previous sequence of ...0,0,0,0,0,0,.... you ask me which one of these there is without explicitly tell me if your sequences are inifinite or not - has borders or not. if you go backwards with one black cell you see that there are infinite history yes. but when you know that you can use another dimension to reverse lets say a defined discrete latitce of cells with some random initial configuration, it i dont see the reason why one cannot say that it is?
first of all, the algorithm was not in the first place designed for cyclic borders, but it would work there too if the width of the wrapping was equal to a modulus 8 digit in decimal and if you knew the amount of time steps. one can go backwards as much as one wants anyhow. my algorithm was designed for non-cyclic borders. but, even that the lattice was infinite you would be able to go backwards at specific points of times.
since the rule 30 intrinsinctly generates numbers based on ONE rule, the rule to reverse it is exactly the same, only by opposite states of cells - Rule 86 is this solution. it intrinsincly generates the state of the past a(i-1)discrete cell based on one extra dimension in time, the past - sort of the same dimension as when one generates a rule 30 pattern forward in time.
if rule 30 generates histories based on previous histories. i dont see any reason why generating the histories on past histories can be used as a reversible solution?
Rudi
OverBored
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### Re: Writing a paper about Cellular Automata
rudi wrote:OverBored:
your question is based on what you define being zero or one in your rule 30 pattern. black and white has two different solutions, you need to specify which one it is. a sequence of 1,1,1,1,1,1,1,1's had previously a sequence of ...0,0,1,0,0,1,0,0,1.... (because 1 would have generated infinite zeros), and a sequence of zeroes would have had previous sequence of ...0,0,0,0,0,0,.... you ask me which one of these there is without explicitly tell me if your sequences are inifinite or not - has borders or not. if you go backwards with one black cell you see that there are infinite history yes. but when you know that you can use another dimension to reverse lets say a defined discrete latitce of cells with some random initial configuration, it i dont see the reason why one cannot say that it is?
Sorry, I should have been clearer. I meant the latter. You correctly observe that
....0000000....
is a possible previous state. However, there is another possible previous state:
.....11111111.....
Just to clarify, my sequences are infinite in both directions. I am pretty sure this is a standard definition. Some definitions might require that the number of 1's be bounded, in which case it might be conceivable that the process is reversible, but this isn't what we normally mean.
G4!!
Grob FTW,
Hello. Smithers. You're. Quite good. At. Turning. Me. On.
rudi
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### Re: Writing a paper about Cellular Automata
1) ...111111111111111111... and ...0000000000000... doesnt contain any
information at all.
2) they dont produce any complex patterns.
based on this, it doesnt even have to be generated by Rule 30. its unfortunately a contradiction.
Rudi
rudi
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### Re: Writing a paper about Cellular Automata
thanks for the replies anyhow
Rudi
Qlexander
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### Re: Writing a paper about Cellular Automata
Rudi, I have looked at your solution, and while I haven't checked if it "works", I have found a problem with it.
I understand that you're determining the previous cells gradually, from right to left. For this reason, you allow yourself to use a(i) and a(i+1) in your formula to determine a(i-1). This would be okay, BUT... it means that your algorithm can never start! In order to compute a(0), you need a(1) and a(2). But before you can compute a(1) and a(2), your formula needs a(3) and a(4). And before that, you need a(5) and a(6)! And so on... Your algorithm has no place to start, because it requires at least some information about generation t before it can compute any of the cells in generation t.
Does that make sense?
Shadowfish
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### Re: Writing a paper about Cellular Automata
It sounds like rudi has noticed that some subset of initial conditions yield reversible dynamics for some number of iterations, and that the evolution starting from these initial conditions can be reversed in a straightforward way. This leads to the questions:
1) Will the CA evolve from a reversible state to an irreversible one?
2) Can we specify what subset of initial conditions yields reversible dynamics?
3) Is this result well-known, or weaker than a well-known result?
As an outsider to the CA world, it seems like rudi's algorithm might be interesting if these questions can be answered.
I'm mostly a lurker. I lurk. Kind of like a fish, in the shadows.
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Automata
Qlexander:
Sorry about not have written to much background-information, i will do that in a more comprehensive document like i said earlier, just havent gotten around making the document complete yet.
You have the information - let's say you have only one configuration, but the past are gone, there is just a white background except for your current configuration. Let's say that this lies on t=0.
I will refer to as space-steps as when i decreases by one (i=i-1) which has nothing to do with time-steps (t=t-1).
We will calculate the new cell at t-1, a(i-1).
At the first decrypting sequence a'(i) the configuration lies (the one you got) are on time t=0. a(i-1), a(i) and a(i+1) are all 0 before the first space-step. on the next space-step (that is i-1) the a(i) might be black or white depending on the a'(i-1) state after the first space-step. we know that the first and last few cells are black because of the borders on rule 30, but in general when reading a'(i) where our configuration lies there might be 0 or 1 depending on the rule 86 truth table (or the formula used, which is kind of a mirror of the truth table). combining these values that gets fed into the a(i-1), a(i) and a(i+1) with the previous configuration a'(i) one determines the output a(i-1). so yes, its one extra dimension a'(i). when the configuration-length is complete time decreases by one. all this happens in discrete time anyway. almost like a mobile automaton or turring machine. so in some sense we are reading the current time as well as the past to create a new past.
does this makes sense?
thanks for the questions.
[edit: maybe its much simpler to understand if i say that it is almost similar a sideways evolution, like the (b) one on page 604 in Wolfram's NKS book. the difference that the evolution goes in the i-direction rather than time-direction t and the formula used is not wolfram's typical Rule 30 formula: p xor (p|q) but the reversible-formula in my document - i might have to test it a few times, because i am not sure if i got all the notation right in that doc. however it behaves similar to rule 89 because ive testet that in my simulation. and besides the one in page 604, it's of course just one string of information initially, no more.]
Last edited by rudi on Wed May 30, 2012 12:26 am UTC, edited 2 times in total.
Rudi
rudi
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### Re: Writing a paper about Cellular Automata
Shadowfish wrote:It sounds like rudi has noticed that some subset of initial conditions yield reversible dynamics for some number of iterations, and that the evolution starting from these initial conditions can be reversed in a straightforward way. This leads to the questions:
1) Will the CA evolve from a reversible state to an irreversible one?
2) Can we specify what subset of initial conditions yields reversible dynamics?
3) Is this result well-known, or weaker than a well-known result?
As an outsider to the CA world, it seems like rudi's algorithm might be interesting if these questions can be answered.
I hope someone could answer these quesions, because i dont want my algorithm to go to waste if they are usable in any way for other mathematicians and researchers.
Rudi
rudi
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### Re: Writing a paper about Cellular Automata
sorry people. i knew it, there's something missing in that doc. i forgot to say that i wrote this at night when i was very tired.
now i will correct the formulas in the doc.
a(i-1) is calculated like this:
a(i-1) = (a(i) AND (a(i+1)) xor a(i) xor a(i+1) xor a'(i)
if you look up that logic formula up in lets say wolfram alpha, then you will see that you get rule 86 as the rule to reverse the rule 30 pattern.
and if you look up the common rule 30 logic expression: "a xor (b or c)" you see that the algebraic normal form is exactly the same.
i've updated the doc: http://www.home.no/rudibs/Rule%2030%20reversibility.pdf
hope it's easier for you mathematicians to understand. also if its really bad to have use one extra dimension to produce past states, please tell me why.
hope that helps.
thanks for your patience.
Rudi
gmalivuk
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### Re: Writing a paper about Cellular Automata
Please don't post three times in a row. If you have something more to say, and your last post is still the latest in the thread, just edit it to include the new information.
Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.
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Sizik
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### Re: Writing a paper about Cellular Automata
One thing that should tell you that it's not reversible is the inverted triangles of 0s that rule 30 forms. They form when a section of the line is all 1s, which turns into all 0s on the next line. The sides then shrink over time (since both 100 and 001 map to 1), forming a triangle. In order to reverse that, since the triangles can be upwards of 20 cells wide at the top, your rule would have to map 001, 100, and 000 to 0, in order to have them stay at least the same width as time "decreases". Then, at some point, you'd have to have your line of 0s all turn into 1s, completing the triangle. This requires having 000 map to 111, but you can't do that since you need it to map to 0 in order to form the triangle in the first place.
gmalivuk wrote:
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Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.
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### Re: Writing a paper about Cellular Automata
rudi wrote:if you look up that logic formula up in lets say wolfram alpha, then you will see that you get rule 86 as the rule to reverse the rule 30 pattern.
Rule 86 is a left-right mirror of Rule 30. This is not the same thing as a time-reversal.
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rudi
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### Re: Writing a paper about Cellular Automata
sizik:
look at the information flow on fig 2 in the document. im not really a good explanator, but try and figure out how information flows from a(i-1) to a(i) and a(i+1) at each space-step. when a(i-1) is determined to be for example 1, then at next space-step a(i) is 1 and after the second space step a(i+1) is 1. based on the formula, they all produce 1's at at your particular problem. when they reach a border the formula behaves differently since a(i-1) = (1 ^ 1) xor 1 xor 1 xor 1 = 0
dont know if that answered your question.
you could give me some last configuration of your Rule 30 pattern. I will try with my algorithm to reverse (produce the whole pyramid) and get back to your random/or defined initial configuration. I don't see any flaws at this time with my current formula. the pdf is updated once more. i found another small bug!
gmalivuk:
you cannot prove that it is not a time-reversal algorithm. this formula intrinsincly generates the past reverse states of rule 30 states. rule 30 intrinsicly generates complex patterns forward in time, this formula does the opposite.
Rudi
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### Re: Writing a paper about Cellular Automata
rudi wrote:you cannot prove that it is not a time-reversal algorithm.
If you apply 86 to the following, you definitely don't get the same thing I started with and applied rule 30 to:
..., 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, ...
Like I said, 86 simply flips the result left-to-right. Try it with just a single central 1, and this is obvious.
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rudi
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### Re: Writing a paper about Cellular Automata
gmalivuk:
im not sure if you're doing it wrong.
this is what i get:
so your initial configuration would be, afaik:
....0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,....
(zero in this case is always background).
pdf is updated. maybe its easier to understand what i do now.
Rudi
Sizik
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### Re: Writing a paper about Cellular Automata
I'm going to guess gmalivuk started with ...0,0,1,1,1,1,1,1,1,1,0,0...
Also,
rudi wrote:
Is definitely NOT what you get if you apply rule 86 to ...0,1,1,0,1,1,1,0,1,0,1,1,0,1,1,1,1,0...
gmalivuk wrote:
King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.
Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.
rudi
Posts: 24
Joined: Sun Mar 18, 2012 7:06 pm UTC
### Re: Writing a paper about Cellular Automata
Sizik:
You don't just apply rule 86 on the automaton like any other elementary ca. its a more dynamic algorithm than that.
But yea, i get your point. but sequences like ...0000000... or ...1111111... doesnt make any sense at all. for a rule 30 to produce these sequences, if they do at all.. the background would be the final result. the background is not possible to reverse, because it contains just one 1/2 bit state. but complex patterns are possible to reverse.
Rudi
gmalivuk
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### Re: Writing a paper about Cellular Automata
rudi wrote:But yea, i get your point. but sequences like ...0000000... or ...1111111... doesnt make any sense at all. for a rule 30 to produce these sequences, if they do at all.. the background would be the final result. the background is not possible to reverse, because it contains just one 1/2 bit state. but complex patterns are possible to reverse.
So it looks like you can't reverse the "background" (i.e. infinite strings of 0), nor can you reverse states with an infinite number of non-background cells (i.e. ones that aren't bordered by zeroes on both sides).
It seems you can only reverse when you start with the assumptions that a'(j)=a(j)=0 for j>i, and a(i)=0. Otherwise how do you know where to start?
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rudi
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### Re: Writing a paper about Cellular Automata
So it looks like you can't reverse the "background" (i.e. infinite strings of 0), nor can you reverse states with an infinite number of non-background cells (i.e. ones that aren't bordered by zeroes on both sides).
you can reverse infinite non-background cells. here is for example a history of a infinite cyclic system:
i drawn a yellow line because it seems that things happen and that there is some kind of hidden "border" between the background noise and the regular structures. it's tricky to know what this means at the moment, but those borders seem to increase/decrease by 3 because of the neighborhood length, r=1 perhaps. someone with more analytical knowledge than me should look into this area and see if there is something interesting to find.
It seems you can only reverse when you start with the assumptions that a'(j)=a(j)=0 for j>i, and a(i)=0. Otherwise how do you know where to start?
a'(i) can be anything. this is where the current configuration lies, for example t=0. a(i-1) the determined cell, and a(i) and a(i+1) are t-1. but t-1 are initially background. after a space step, this background may either have 1 or 0 based on a'(i)'s state, but also a(i) and a(i+1)'s states of course, but since initially their states are all 0 a'(i)'s state are most important at first space step. [i use the word space step because i found it easy to use, what i mean is that i decreases].
[edit: you should know where to start, when a'(i) changes state. for example from 0 to 1].
I just first of all would like to thank everybody for their responses. and please make contact if you perhaps would like to contribute into a larger research project, if that even is something that can be done. ive allready thought about building a website where i should put this up. or even write a more comprehensive paper in the future, if that is necessary.
Rudi
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### Re: Writing a paper about Cellular Automata
rudi wrote:a(i) and a(i+1) are t-1. but t-1 are initially background
Like I said: you assume those must be zero. Which means you're not actually able to uniquely reverse all possible configurations.
rudi wrote:you can reverse infinite non-background cells. here is for example a history of a infinite cyclic system:
Sorry, I meant infinite in both directions, since you have to pick a rightmost position to start with.
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rudi
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### Re: Writing a paper about Cellular Automata
Like I said: you assume those must be zero. Which means you're not actually able to uniquely reverse all possible configurations.
They are zero (background). Because there are no history until you actually compute that history based on a'(i).
Sorry, I meant infinite in both directions, since you have to pick a rightmost position to start with.
The question is do you really know where to start when something is infinite, can you pick somewhere on the infinite string, does it really matter?
Rudi
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# Deriving the FEA formulation using triangular elements
• A
• maistral
#### maistral
Hi, it's been a while since I last posted. Anyway, so I went through the trouble of enrolling in two finite element analyses classes and yet, they still don't teach how the 2D formulation has been made. I'll list the things that 'I know' already to get some things clear.
1. I know how to derive the 1D finite element formulation using the Galerkin linear weighting functions (in the end I studied this myself as the classes I took did not teach this).
2. I know that you can mechanically solve the 1D formulation using the generic force = stiffness matrix × displacement method (this is what I initially knew before taking the undergrad and graduate classes, and yet this is the one they teach here. Yep, waste of money).
3. I also know the shape functions for 2D (the triangle thing, where T(x,y) = Ti(xi,yi)Ni(x,y) + Tj(xj,yj)Nj(x,y) + Tk(xk,yk)Nk(x,y) then the weighting functions are Ni(x,y), Nj(x,y), and Nk(x,y), which is integrated to the differential equations as in 1D).
So I'm trying to figure out how did everything get derived since I don't want to blindly follow formulations without any information of how they arrived... there. So I took some resources from the internet and upon checking them, my nose instantly bled . They were talking about integrations along volumes and there was even an integral with a small circle in the middle! Apparently some deep knowledge in mathematics courses are needed in order to understand these, and most of them involve integrating an entire vector which just made me more dizzy and confused.
I actually wanted to avoid those things as I don't know how they would be relevant to me as I'm not really a mathematics major/graduate, I'm an engineering graduate. I am truly sorry for asking help here if it's too much to ask as I really wanted to know where did these things come from, or at least guide me until I achieve the 'unsimplified' but 'clear' solution I wanted - what I'm hoping for is someone be able to guide me with the pertinent theorems in order to arrive with the default, unsimplified formulation similar to the 1D case; something like:
Ti × (insert lengthy integral) + Tj × (insert lengthy integral) + Tk × (insert lengthy integral) = 0.
Anyway, I started off by using the Laplace equation. This is what I have now:
Anyway, these are the questions.
1. I know I am supposed to multiply the entire PDE with the weighting function. Which weighting function do I use? I remember in the 1D case you have to integrate the differential equations twice, one for the ith case and one for the jth case. Am I correct to assume that I integrate this differential equation thrice, seeing there are three vertices in the triangular element? So that would mean I integrate the PDE; one for the ith, one for the jth case, and one for the kth case?
2. I am assuming that you have to use double integration along the x-axis and the y-axis. My problem is about the limits. Which limits do I use? If the bottom limit of the integrands are xi and yi, what would be the upper limit?
I hope, and pray very dearly, that someone helps me. I even intend to spend some money just to learn this (I already did and they just taught me stuff that I know already which is extremely disheartening). Please, help me. Thank you.
Here's a book that is low on math and high on worked out examples that you can follow by hand:
https://www.amazon.com/dp/0070087148/?tag=pfamazon01-20
You might also want to study this in case you haven't done so already:
https://www.amazon.com/dp/0070552215/?tag=pfamazon01-20
You probably know chapters 1 and 2, study chapters 3 and 6 for 2D stuff. If you want to know why it works, you have to know about the variational form though.
The first thing to learn is how to transform any triangle and rectangle to the unit triangle or rectangle. Then learn how to integrate a function on any triangle by transforming it to the unit triangle.
Then realize that solving the pde in variational form just means discretizing the domain into a lot of triangles, transforming the problem from the unit triangle (because all these neat integration methods only work from -1..+1) to the actual triangle, and construct the global mass matrix and solve the system Ax=b
It helps to write an actual solver in e.g. matlab, scilab, python or whatever you have available that has some basic matrix-vector stuff.
1. no, the integral is in 2D. The number of nodes does increase the size of the matrix. Try to see this by just integrating a function in 2D on some rectangle (with known solution so you can check the outcome).
2. After discretization, the integration becomes the sum of the integration over the unit triangle/rectangle.
Hope this helps.
Here's a book that is low on math and high on worked out examples that you can follow by hand:
https://www.amazon.com/dp/0070087148/?tag=pfamazon01-20
You might also want to study this in case you haven't done so already:
https://www.amazon.com/dp/0070552215/?tag=pfamazon01-20
You probably know chapters 1 and 2, study chapters 3 and 6 for 2D stuff. If you want to know why it works, you have to know about the variational form though.
The first thing to learn is how to transform any triangle and rectangle to the unit triangle or rectangle. Then learn how to integrate a function on any triangle by transforming it to the unit triangle.
Then realize that solving the pde in variational form just means discretizing the domain into a lot of triangles, transforming the problem from the unit triangle (because all these neat integration methods only work from -1..+1) to the actual triangle, and construct the global mass matrix and solve the system Ax=b
It helps to write an actual solver in e.g. matlab, scilab, python or whatever you have available that has some basic matrix-vector stuff.
1. no, the integral is in 2D. The number of nodes does increase the size of the matrix. Try to see this by just integrating a function in 2D on some rectangle (with known solution so you can check the outcome).
2. After discretization, the integration becomes the sum of the integration over the unit triangle/rectangle.
Hope this helps.
Hi! Thanks for replying. I'll try and read the resources you pointed out. Thank you very much!
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##### Notes
This supplemental quiz accompanies the Grade 1 Daily Math Operations Review printable. Use on the fifth day of Daily Math Practice - Geometry Review for progress monitoring.
##### Print Instructions
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# Geometry Review Quiz - Grade 1 (Grade 1)
Print Test (Only the test content will print)
## Geometry Review Quiz - Grade 1
1.
How many equal parts of the rectangle?
1. 1
2. 2
3. 3
4. 4
2.
How many equal parts of the square?
1. 1
2. 2
3. 3
4. 4
3.
Which figure shows halves?
4.
What two shapes make up the figure?
1. circle and square
2. circle and rectangle
3. half-circle and square
4. half-circle and rectangle
5.
Which of the following would change the shape of the rectangle?
1. color it blue
2. make it smaller
3. remove one side
4. turn it clockwise
6.
Does this shape have a circle for a face?
1. yes
2. no
7.
Skylar has 4 of the shape pieces shown.
Which shape could Skylar make with the 4 shape pieces?
1. circle
2. square
3. triangle
8.
Which would change the shape so it was no longer a triangle?
1. make it bigger
2. color it yellow
3. flip it upwards
4. remove a side
9.
Which figure shows quarters?
10.
Mateo is sorting shape pieces. He puts all the 3-sided shapes in one group. Which shape could go in the group?
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > lssvacl Structured version Visualization version GIF version
Theorem lssvacl 19717
Description: Closure of vector addition in a subspace. (Contributed by NM, 11-Jan-2014.) (Revised by Mario Carneiro, 19-Jun-2014.)
Hypotheses
Ref Expression
lssvacl.p + = (+g𝑊)
lssvacl.s 𝑆 = (LSubSp‘𝑊)
Assertion
Ref Expression
lssvacl (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → (𝑋 + 𝑌) ∈ 𝑈)
Proof of Theorem lssvacl
StepHypRef Expression
1 simpll 766 . . . 4 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → 𝑊 ∈ LMod)
2 eqid 2822 . . . . . 6 (Base‘𝑊) = (Base‘𝑊)
3 lssvacl.s . . . . . 6 𝑆 = (LSubSp‘𝑊)
42, 3lssel 19700 . . . . 5 ((𝑈𝑆𝑋𝑈) → 𝑋 ∈ (Base‘𝑊))
54ad2ant2lr 747 . . . 4 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → 𝑋 ∈ (Base‘𝑊))
6 eqid 2822 . . . . 5 (Scalar‘𝑊) = (Scalar‘𝑊)
7 eqid 2822 . . . . 5 ( ·𝑠𝑊) = ( ·𝑠𝑊)
8 eqid 2822 . . . . 5 (1r‘(Scalar‘𝑊)) = (1r‘(Scalar‘𝑊))
92, 6, 7, 8lmodvs1 19653 . . . 4 ((𝑊 ∈ LMod ∧ 𝑋 ∈ (Base‘𝑊)) → ((1r‘(Scalar‘𝑊))( ·𝑠𝑊)𝑋) = 𝑋)
101, 5, 9syl2anc 587 . . 3 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → ((1r‘(Scalar‘𝑊))( ·𝑠𝑊)𝑋) = 𝑋)
1110oveq1d 7155 . 2 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → (((1r‘(Scalar‘𝑊))( ·𝑠𝑊)𝑋) + 𝑌) = (𝑋 + 𝑌))
12 simplr 768 . . 3 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → 𝑈𝑆)
13 eqid 2822 . . . . 5 (Base‘(Scalar‘𝑊)) = (Base‘(Scalar‘𝑊))
146, 13, 8lmod1cl 19652 . . . 4 (𝑊 ∈ LMod → (1r‘(Scalar‘𝑊)) ∈ (Base‘(Scalar‘𝑊)))
1514ad2antrr 725 . . 3 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → (1r‘(Scalar‘𝑊)) ∈ (Base‘(Scalar‘𝑊)))
16 simprl 770 . . 3 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → 𝑋𝑈)
17 simprr 772 . . 3 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → 𝑌𝑈)
18 lssvacl.p . . . 4 + = (+g𝑊)
196, 13, 18, 7, 3lsscl 19705 . . 3 ((𝑈𝑆 ∧ ((1r‘(Scalar‘𝑊)) ∈ (Base‘(Scalar‘𝑊)) ∧ 𝑋𝑈𝑌𝑈)) → (((1r‘(Scalar‘𝑊))( ·𝑠𝑊)𝑋) + 𝑌) ∈ 𝑈)
2012, 15, 16, 17, 19syl13anc 1369 . 2 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → (((1r‘(Scalar‘𝑊))( ·𝑠𝑊)𝑋) + 𝑌) ∈ 𝑈)
2111, 20eqeltrrd 2915 1 (((𝑊 ∈ LMod ∧ 𝑈𝑆) ∧ (𝑋𝑈𝑌𝑈)) → (𝑋 + 𝑌) ∈ 𝑈)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 = wceq 1538 ∈ wcel 2114 ‘cfv 6334 (class class class)co 7140 Basecbs 16474 +gcplusg 16556 Scalarcsca 16559 ·𝑠 cvsca 16560 1rcur 19242 LModclmod 19625 LSubSpclss 19694 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2178 ax-ext 2794 ax-sep 5179 ax-nul 5186 ax-pow 5243 ax-pr 5307 ax-un 7446 ax-cnex 10582 ax-resscn 10583 ax-1cn 10584 ax-icn 10585 ax-addcl 10586 ax-addrcl 10587 ax-mulcl 10588 ax-mulrcl 10589 ax-mulcom 10590 ax-addass 10591 ax-mulass 10592 ax-distr 10593 ax-i2m1 10594 ax-1ne0 10595 ax-1rid 10596 ax-rnegex 10597 ax-rrecex 10598 ax-cnre 10599 ax-pre-lttri 10600 ax-pre-lttrn 10601 ax-pre-ltadd 10602 ax-pre-mulgt0 10603 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2622 df-eu 2653 df-clab 2801 df-cleq 2815 df-clel 2894 df-nfc 2962 df-ne 3012 df-nel 3116 df-ral 3135 df-rex 3136 df-reu 3137 df-rmo 3138 df-rab 3139 df-v 3471 df-sbc 3748 df-csb 3856 df-dif 3911 df-un 3913 df-in 3915 df-ss 3925 df-pss 3927 df-nul 4266 df-if 4440 df-pw 4513 df-sn 4540 df-pr 4542 df-tp 4544 df-op 4546 df-uni 4814 df-iun 4896 df-br 5043 df-opab 5105 df-mpt 5123 df-tr 5149 df-id 5437 df-eprel 5442 df-po 5451 df-so 5452 df-fr 5491 df-we 5493 df-xp 5538 df-rel 5539 df-cnv 5540 df-co 5541 df-dm 5542 df-rn 5543 df-res 5544 df-ima 5545 df-pred 6126 df-ord 6172 df-on 6173 df-lim 6174 df-suc 6175 df-iota 6293 df-fun 6336 df-fn 6337 df-f 6338 df-f1 6339 df-fo 6340 df-f1o 6341 df-fv 6342 df-riota 7098 df-ov 7143 df-oprab 7144 df-mpo 7145 df-om 7566 df-wrecs 7934 df-recs 7995 df-rdg 8033 df-er 8276 df-en 8497 df-dom 8498 df-sdom 8499 df-pnf 10666 df-mnf 10667 df-xr 10668 df-ltxr 10669 df-le 10670 df-sub 10861 df-neg 10862 df-nn 11626 df-2 11688 df-ndx 16477 df-slot 16478 df-base 16480 df-sets 16481 df-plusg 16569 df-0g 16706 df-mgm 17843 df-sgrp 17892 df-mnd 17903 df-mgp 19231 df-ur 19243 df-ring 19290 df-lmod 19627 df-lss 19695 This theorem is referenced by: lsssubg 19720 lspprvacl 19762 lspvadd 19859 lidlacl 19977 minveclem2 24028 pjthlem2 24040 lshpkrlem5 36369 lcfrlem6 38802 lcfrlem19 38816 mapdpglem9 38935 mapdpglem14 38940
Copyright terms: Public domain W3C validator
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Symmetry And Group
By Steinke G. F.
Best symmetry and group books
20th Fighter Group
The 20 th Fighter workforce joined the eighth Air strength Command in Dec of '43, flying the P-38 in lengthy diversity bomber escort function. the gang later switched over to the P-51 in July of '44. the gang destroyed a complete of 449 enemy airplane in the course of its wrestle journey. Over one hundred fifty images, eight pages of colour, eighty pages.
Multiplizieren von Quantengruppen
Inhaltsangabe:Einleitung: Quantengruppen als quantisierte Universelle Einhüllende von Lie-Algebren sind Gegenstand der vorliegenden Arbeit. Sie bietet eine Einführung in die Thematik, setzt lediglich Grundkenntnisse der Darstellungstheorie Halbeinfacher Lie-Algebren voraus, wie sie etwa bei Humpfreys, Jacobsen, Serre oder Bourbaki vermittelt werden, und ordnet die Darstellungstheorie der Quantengruppen in die Physik konformer Feldtheorien ein.
Extra info for 4-Dimensional Elation Laguerre Planes Admitting Non-Solvable Automorphism Groups
Sample text
Ht„x1 is the group „ n t 1Ht . ; ; ^-^ 2. Show that the map a^C(a)—^ -0 1 -1/ 1 defines a representation of the cyclic group gp{a I =1 }. Prove that this representation is irreducible over the field of real numbers. 3. Let E be a linear map of the m-dimensional vector space V into itself such that E2 = E, e # i (the identity map). E # 0, Define W = {wiwe = 0}. U= {ulue = u}, Show that U and W are subspaces of V which are invariant under UEGU, WEcW. Prove that E, that is V= UO+W. Deduce that if E is an m x m matrix such that E 2 =E, E#0, E #I, there exists an integer r satisfying 1 r < m and a non-singular matrix T such that T- 'ET = (0 34 0 ) = i.
E m (X). 37) - where the right-hand side is an abbreviation for 4)(x). This enables us to recast the expression for the inner product of two characters 4(x) and 0(x). When x lies in Ca , then 4(x) = O a , i (x} _ 111a , {x ') = t1ra . 38) In particular, if x and x' are simple characters, the character relations (of the first kind, p. 39) state that 1 g ^ haXaX►a--{0 ifX#X' ►. 39) (i, j = 1, 2, ... , k). 1(11„1 g)). 39) states that the rows of U form a system of unitary-orthogonal k-tuples.
Let w=Nlx1 +fl2x2+ ... + flaxa be another element of G. Then y = w if and only if a ; = /3 (i = 1, 2, ... 20) 1v=v. Furthermore, G c possesses a multiplicative structure. 21) where j (i, j) is a well-defined integer lying between 1 and g. i)• This multiplication is associative by virtue of the associative law for G. Thus, if u, y, w E G e , then (uv)w = u(vw). As we have already remarked (p. 27), a vector space which is endowed with an associative multiplication is called an algebra. Accordingly, we call G c the group algebra of G over C .
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# How Many Teaspoons Is 7g Of Yeast?
## How many teaspoons is 7g dried yeast?
2 teaspoonsDry yeast in small packs has most universal weight.
One packet, one sachet or one envelope weighs 7 grams (0.25 oz or 2 teaspoons).
1 teaspoon (5 ml) of dry yeast equals 3.5 grams..
## How do I measure 7g of yeast?
1 1/2 spoons will be fine. … Baking needs to be accurate. … Two level teaspoons equals 7g of yeast. … Use 2 teaspoons (5ml) for 7g of yeast. … I just use 1 and a half 5ml measures and accept its half a gram out…. … TBH, the only totally accurate way of measuring would be with scales or a half teaspoon measure.More items…
## How much is 3 grams to teaspoon?
Common Grams to Teaspoon ConversionsGramsTeaspoons3 g0.6 tsp4 g0.8 tsp5 g1 tsp6 g1.2 tsp6 more rows•Aug 17, 2020
## How much is a gram of yeast in teaspoons?
One gram of active dry yeast converted to teaspoon equals to 0.35 tsp. How many teaspoons of active dry yeast are in 1 gram? The answer is: The change of 1 g ( gram ) unit in a active dry yeast measure equals = into 0.35 tsp ( teaspoon ) as per the equivalent measure and for the same active dry yeast type.
## How much is 7g dry yeast?
1 Answer. (1) A packet of yeast is typically 7g exactly. So if you’re buying yeast by the packet, use one packet. But assuming you will be measuring from bulk yeast, the correct measure by volume would be 2 1/4 tsp instead of 2 1/2 tsp.
## How many eggs is 57 grams?
Large eggs are about 57 grams or 3 1/4 tablespoons of egg. The mess begins before you’ve even cracked an egg. Of this about 1/3 is the yolk and 2/3 is the white so one egg white will weigh approximately 40g.
## How many teaspoons is 7g?
Common Grams to Teaspoon ConversionsGramsTeaspoons4 g0.8 tsp5 g1 tsp6 g1.2 tsp7 g1.4 tsp6 more rows•Aug 17, 2020
## What happens if you add too much yeast?
Too much yeast could cause the dough to go flat by releasing gas before the flour is ready to expand. If you let the dough rise too long, it will start having a yeast or beer smell and taste and ultimately deflate or rise poorly in the oven and have a light crust.
## How much is a tablespoon of yeast?
Baking Conversion TableU.S.Metric1 tablespoon17.07 grams1 teaspoon instant dry yeast3.1 grams2 1/4 teaspoons instant dry yeast7 grams1 tablespoon instant dry yeast9.3 grams82 more rows
## How do I convert dry yeast to fresh?
ConversionsTo convert from fresh yeast to active dry yeast, multiply the fresh quantity by 0.4. Active dry yeast must be hydrated in warm water before being incorporated into a dough.To convert from fresh yeast to instant dry yeast, multiply the fresh quantity by 0.33.
## How much dry yeast do I use for bread?
You can increase the size of most bread recipes by simply doubling, tripling, etc. all of the ingredients, including the yeast. Depending on the recipe and rising time, you may use as little as 1 teaspoon, or up to 2 1/4 teaspoons (sometimes more) of instant yeast per pound (about 4 cups) of flour.
## How many teaspoons is 4g?
Sliding down the label to the total carbohydrates it reads sugars “4g,” or “4 grams.” This important bit of information is your key to converting grams into teaspoons. Four grams of sugar is equal to one teaspoon. To be precise, 4.2 grams equals a teaspoon, but the nutrition facts rounds this number down to four grams.
## How many teaspoons is 4g yeast?
0.32 tspOne gram of instant yeast converted to teaspoon equals to 0.32 tsp.
## How do you use dry yeast?
Active dry yeast needs to be activated before use. To do this, mix together lukewarm water, sugar and yeast, stirring vigorously to ensure the yeast is fully dissolved. Cover, and set aside in a warm place for 5 to 10 minutes. Once froth forms on top, your yeast is activated and ready to use.
## How many teaspoons is 5g?
1.18 tspHow Many Teaspoons Are in a Gram?GramsTeaspoons5 grams1.18 tsp15 grams3.6 tsp25 grams6 tsp35 grams8.4 tsp3 more rows
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https://fairmodel.econ.yale.edu/vote2020/inrankn7.htm
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# President 2012
## Post Mortem after the 2012 Presidential Election
The discussion below was on the website before the 2012 election. The last data collection is at the end, dated November 6, 2012. These data are used for the test of the ranking assumption. Obama took FL and all the states ranked above it. He lost NC and all the states ranked below it. So the ranking assumption was perfect. Note that Intrade was not perfect because it predicted that Obama would lose FL.
## Background
The paper,
Interpreting the Predictive Uncertainty of Elections, Journal of Politics, April 2009,
provides an interpretation of the uncertainty that exists on election morning as to who will win. The interpretation is based on the theory that there are a number of possible conditions of nature than can exist on election day, of which one is drawn. Political betting markets like Intrade provide a way of trying to estimate this uncertainty. (Polling standard errors do not provide estimates of this type of uncertainty. They estimate sample-size uncertainty, which can be driven close to zero with a large enough sample.)
This paper also introduces a "ranking assumption," which puts restrictions on the possible conditions of nature that can exist on election day. Take as an example the vote in each state for the Democratic candidate for president. Rank the states by the probability that the candidate wins the state. The ranking assumption says that if the candidate wins state i, he or she wins every state ranked above state i.
Given some ranking, the ranking assumption can be tested by simply looking to see after the fact if the candidate won a state ranked lower than one he or she lost. In the paper the assumption was tested for the 2004 and 2008 presidential elections using Intrade probabilities at 6am Eastern standard time on the day of the election. The test is thus a test of the joint hypothesis that the Intrade probabilities are right and the ranking assumption is right. Using the Intrade probabilities, the ranking assumption was perfect in 2004 and off by one in 2008. In 2008 Missouri was ranked above Indiana, and Obama won Indiana and lost Missouri. Both of these elections were very close. Obama won Indiana with 50.477 percent of the two-party vote and lost Missouri with 49.937 percent of the two-party vote. Otherwise, 2008 was perfect.
Evidence is also presented in the paper, although this is not a test of the ranking assumption, that Intrade traders use the ranking assumption to price various contracts.
## 2012 Election
I plan to collect Intrade probabilities at 6am Eastern standard time on six days: September 11, 25, October 9, 23, 30, and November 6. The November 6 data will be used to test the ranking assumption. The data as they are collected are presented in Table 1.
## September 11, 2012
All but 11 states on Intrade either have probabilities close to 0 or 1 or have essentially no trading. The 11 states, ranked by probabilities for the Democratic candidate, are:
state prob votes sumvotes MN 82.5 10 217 PA 80.1 20 237 NH 71.0 4 241 WI 68.0 10 251 NV 67.9 6 257 CO 64.8 9 266 OH 61.9 18 284 pivot IA 59.7 6 VA 58.9 13 FL 47.0 29 NC 28.3 15
61.0 = Intrade probability that the Democratic candidate wins the presidential election.
"sumvotes" is the sum of the electoral votes of all the states ranked above the state plus the state's vote. 270 votes are needed to win. You can see that Ohio is the pivot state. If Obama takes Ohio and all the states ranked above it, he gets 284 votes. Of the states ranked below Ohio, he could also win by not taking Ohio and taking any one of the others.
There is evidence that the Intrade traders are using the ranking assumption. According to the ranking assumption, the probability that Obama wins overall is the probability that he wins the pivot state, Ohio, which is 61.9. The price of the contract on Intrade that the Democratic candidate wins the presidential election (in the electoral college) is 61.0, close to 61.9. If the ranking assumption were not being used, the overall probablilty would be much higher if, say, the probabilities were thought to be independent. The ranking assumption says that Obama is not going to take, say, Iowa or Virginia unless he also takes Ohio, so the probabilities for Iowa and Virginia do not matter. The probability he wins overall is just the probability he wins Ohio. If the probabilities were independent, the probability of an overall win would be 82.3 percent (computed numerically).
## September 25, 2012
The results are:
state prob votes sumvotes MN 92.2 10 217 PA 87.0 20 237 NV 79.8 6 243 WI 79.0 10 253 NH 75.6 4 257 OH 72.6 18 275 pivot IA 65.1 6 VA 64.3 13 CO 61.1 9 FL 55.0 29 NC 42.6 15
72.4 = Intrade probability that the Democratic candidate wins the presidential election.
Ohio is still pivot, and it seems clear that the Intrade traders are using the ranking assumption. If the probabilities were thought to be independent, the overall probability would be 93.1 percent (computed numerically).
## October 9, 2012
The results are:
state prob votes sumvotes MN 90.0 10 217 PA 87.4 20 237 NV 76.7 6 243 NH 74.8 4 247 WI 70.0 10 257 IA 67.7 6 263 OH 66.6 18 281 pivot VA 63.0 13 CO 61.0 9 FL 49.9 29 NC 30.0 15
62.7 = Intrade probability that the Democratic candidate wins the presidential election.
Ohio is still pivot. If the ranking assumption were being used by the Intrade traders, the probability of winning overall should be 66.6, the probability for Ohio. It is in fact 62.7, so a discrepancy of 3.9. In the two earlier samples above the discrepancies were 0.9 and 0.2. If the probabilities were thought to be independent, the overall probability would be 89.0 percent (computed numerically). It is thus surprising that the overall probability of 62.7 is smaller than both the probability for Ohio and the overall probability if the state probabilities were independent. Could be a thin market problem or market manipulation of the overall probability contract.
## October 23, 2012
The results are:
state prob votes sumvotes NV 85.0 6 213 MN 81.0 10 223 PA 75.2 20 243 WI 71.0 10 253 IA 57.8 6 259 OH 57.0 18 277 pivot NH 53.5 4 VA 48.0 13 CO 45.5 9 FL 30.0 29 NC 18.4 15
61.7 = Intrade probability that the Democratic candidate wins the presidential election.
Ohio is still pivot. If the ranking assumption were being used by the Intrade traders, the probability of winning overall should be 57.0, the probability for Ohio. It is in fact 61.7, so a discrepancy of -4.7. This compares to a positive 3.9 on October 9, 2012. If the probabilities were thought to be independent, the overall probability would be 68.1 percent (computed numerically). So in this case the overall probability is in between what the ranking assumption would imply and what the independence assumption would imply.
## October 30, 2012
The results are:
state prob votes sumvotes MN 92.0 10 217 PA 82.0 20 237 NV 80.0 6 243 WI 71.2 10 253 IA 61.6 6 259 OH 58.8 18 277 pivot NH 58.3 4 CO 47.8 9 VA 45.1 13 FL 30.1 29 NC 22.0 15
62.0 = Intrade probability that the Democratic candidate wins the presidential election.
Ohio is still pivot. If the ranking assumption were being used by the Intrade traders, the probability of winning overall should be 58.8, the probability for Ohio. It is in fact 62.0, so a discrepancy of -3.2. If the probabilities were thought to be independent, the overall probability would be 73.9 percent (computed numerically).
## November 6, 2012
The results are:
state prob votes sumvotes MN 90.6 10 217 NV 88.1 6 223 PA 79.0 20 243 WI 77.5 10 253 IA 70.0 6 259 OH 68.8 18 277 pivot NH 68.0 4 VA 55.8 13 CO 54.1 9 FL 35.0 29 NC 20.5 15
67.9 = Intrade probability that the Democratic candidate wins the presidential election.
Ohio is still pivot. If the ranking assumption were being used by the Intrade traders, the probability of winning overall should be 68.8, the probability for Ohio. It is in fact 67.9, so a discrepancy of 0.9. If the probabilities were thought to be independent, the overall probability would be 82.7 percent (computed numerically).
This ranking will be used to test the ranking assumption. The probabilities are at 6am on the day of the election. The ranking assumption (combined with the Intrade probabilities) says, for example, that if Obama wins Colorado, Romney will not win Virginia, New Hampshire, or Ohio. If, on the other hand, Romney wins New Hampshire, the ranking assumption says that he will also win Virginia and Colorado.
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# GATE | GATE CS 2008 | Question 5
• Last Updated : 27 Oct, 2020
In the Karnaugh map shown below, X denotes a don’t care term. What is the minimal form of the function represented by the Karnaugh map?
```
A)
B)
C)
D) ```
(A) A
(B) B
(C) C
(D) D
Explanation:
One group consists of (0000, 0010, 1000, 1010) which gives b’d’
Other group is (0000, 0001, 1000, 1001) which gives a’d’
So, solution is b’d’ + a’d’
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×
×
# Solutions for Chapter 8: One-Factor Experiments: General
## Full solutions for Essentials of Probability & Statistics for Engineers & Scientists | 1st Edition
ISBN: 9780321783738
Solutions for Chapter 8: One-Factor Experiments: General
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##### ISBN: 9780321783738
This expansive textbook survival guide covers the following chapters and their solutions. Essentials of Probability & Statistics for Engineers & Scientists was written by and is associated to the ISBN: 9780321783738. Since 47 problems in chapter 8: One-Factor Experiments: General have been answered, more than 22567 students have viewed full step-by-step solutions from this chapter. Chapter 8: One-Factor Experiments: General includes 47 full step-by-step solutions. This textbook survival guide was created for the textbook: Essentials of Probability & Statistics for Engineers & Scientists, edition: 1.
Key Statistics Terms and definitions covered in this textbook
• 2 k factorial experiment.
A full factorial experiment with k factors and all factors tested at only two levels (settings) each.
• `-error (or `-risk)
In hypothesis testing, an error incurred by rejecting a null hypothesis when it is actually true (also called a type I error).
• Asymptotic relative eficiency (ARE)
Used to compare hypothesis tests. The ARE of one test relative to another is the limiting ratio of the sample sizes necessary to obtain identical error probabilities for the two procedures.
• Backward elimination
A method of variable selection in regression that begins with all of the candidate regressor variables in the model and eliminates the insigniicant regressors one at a time until only signiicant regressors remain
• Bayes’ estimator
An estimator for a parameter obtained from a Bayesian method that uses a prior distribution for the parameter along with the conditional distribution of the data given the parameter to obtain the posterior distribution of the parameter. The estimator is obtained from the posterior distribution.
• Bernoulli trials
Sequences of independent trials with only two outcomes, generally called “success” and “failure,” in which the probability of success remains constant.
• Bimodal distribution.
A distribution with two modes
• Binomial random variable
A discrete random variable that equals the number of successes in a ixed number of Bernoulli trials.
• C chart
An attribute control chart that plots the total number of defects per unit in a subgroup. Similar to a defects-per-unit or U chart.
• Categorical data
Data consisting of counts or observations that can be classiied into categories. The categories may be descriptive.
• Central composite design (CCD)
A second-order response surface design in k variables consisting of a two-level factorial, 2k axial runs, and one or more center points. The two-level factorial portion of a CCD can be a fractional factorial design when k is large. The CCD is the most widely used design for itting a second-order model.
• Control limits
See Control chart.
• Critical region
In hypothesis testing, this is the portion of the sample space of a test statistic that will lead to rejection of the null hypothesis.
• Discrete distribution
A probability distribution for a discrete random variable
• Distribution free method(s)
Any method of inference (hypothesis testing or conidence interval construction) that does not depend on the form of the underlying distribution of the observations. Sometimes called nonparametric method(s).
• Eficiency
A concept in parameter estimation that uses the variances of different estimators; essentially, an estimator is more eficient than another estimator if it has smaller variance. When estimators are biased, the concept requires modiication.
• Erlang random variable
A continuous random variable that is the sum of a ixed number of independent, exponential random variables.
• Error propagation
An analysis of how the variance of the random variable that represents that output of a system depends on the variances of the inputs. A formula exists when the output is a linear function of the inputs and the formula is simpliied if the inputs are assumed to be independent.
• Finite population correction factor
A term in the formula for the variance of a hypergeometric random variable.
• Gamma random variable
A random variable that generalizes an Erlang random variable to noninteger values of the parameter r
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How to use Classification Metrics in Python
Classification Metrics are a set of techniques used to evaluate the performance of a classifier. These metrics provide a way to measure the accuracy, precision, recall and other aspects of a classifier’s performance. They are widely used in machine learning to evaluate the effectiveness of a model and make informed decisions about which model to use for a given problem.
In Python, the popular library scikit-learn provides a number of built-in functions for calculating classification metrics such as accuracy_score, precision_score, recall_score, f1_score, among others. Each of these functions calculates a different metric, and they can all be used to evaluate the performance of a classifier on a dataset.
Accuracy is the most common metric used to evaluate the performance of a classifier, it is the ratio of the number of correct predictions to the total number of predictions made.
Precision is the proportion of true positive predictions in relation to the total number of positive predictions made by the classifier.
Recall is the proportion of true positive predictions in relation to all the positive instances in the dataset.
F1 score is the harmonic mean of precision and recall.
In addition to these metrics, other useful classification metrics include AUC-ROC(Area Under Receiver Operating Characteristic) curve, confusion matrix and cross validation score.
AUC-ROC curve is a graphical representation of the performance of a classifier, which plots the true positive rate against the false positive rate.
Confusion Matrix is a table that is used to define the performance of a classification algorithm, it gives the number of true positives, false positives, true negatives and false negatives.
Cross-validation score is a technique for measuring the performance of a classifier by dividing the dataset into training and test subsets, and then training the classifier on the training set and evaluating it on the test set.
In conclusion, classification metrics are a set of techniques used to evaluate the performance of a classifier. Scikit-learn, a python library provide us with several built-in functions for calculating various classification metrics such as accuracy, precision, recall, f1-score, AUC-ROC and so on. These metrics allow us to evaluate the effectiveness of a model and make informed decisions about which model to use for a given problem.
In this Machine Learning Recipe, you will learn: How to use Classification Metrics in Python.
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