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https://inv.alid.pw/posts/kleisli-lifting-distributive/ | 1,685,720,189,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00282.warc.gz | 360,041,977 | 4,045 | # Lifting monads to Kleisli categories with distributive laws
It seems my first non-trivial post on my new website is going to be something fairly technical and likely not of any interest to anyone in particular. However, it’s related to something I’m working on at the moment, and a little perfunctory searching seems to indicate that there’s some sort of hole in the literature here, at least as far as my understanding goes. This is partly also a means to justify my faffing about to get MathJax working on this site. I assume familiarity with functors and natural transformations, but not much category theory beyond that.
A monad (on some category $$\mathcal{C}$$ ), historically also known as a triple, consists of three pieces of data:
1. a functor $$T : \mathcal{C} \rightarrow \mathcal{C}$$;
2. a natural transformation $$\eta = \eta^T : 1 \rightarrow T$$ called the unit; and
3. a natural transformation $$\mu = \mu^T: TT \rightarrow T$$ called multiplication
such that $$\mu \circ T\mu = \mu \circ \mu T$$ (associativity) and $$\mu \circ T \eta = \mu \circ \eta T = 1$$ (unit law), famously summarised as ‘a monad is a monoid in the category of endofunctors’1. Given a monad $$T$$, we may form its Kleisli category $$\mathbf{Kl}(T)$$, which has the same objects as the underlying category and where an arrow $$X \nrightarrow Y$$ is a $$\mathcal{C}$$-arrow $$X \rightarrow TY$$. I’ll use this different notation for Kleisli arrows because it’s important to keep straight the distinction between them and arrows in the underlying category. The identities are given by the unit maps $$\eta_X$$ and composition of $$f: X \nrightarrow Y$$ and $$g: Y \nrightarrow Z$$ uses the multiplication:
$X \xrightarrow{f} TY \xrightarrow{Tg} TTZ \xrightarrow{\mu_Z} TZ$
This construction is well-known from its use in functional programming to model effects.
Distributive laws come up a lot in the world of string diagrams and SMTs, where we use them to combine two theories together subject to some laws governing their interaction. The most familiar example of a distributive law is multiplication over addition: $a(b + c) = ab + ac$ Beck (p95) extended the notion to monads, defining a law distributing a monad $$S$$ over a monad $$T$$ to be a natural transformation $$\lambda: ST \rightarrow TS$$ satisfying certain coherence laws. In the paper, he in fact gives a way to recover the multiplication/addition law back from this construction, which is pretty cool.
The main theorem related to distributive laws of monads is the equivalence of
1. a distributive law of $$S$$ over $$T$$;
2. a coherent multiplication on $$TS$$;
3. a lifting of $$T$$ to the category of $$S$$-algebras (which I haven’t introduced, since it isn’t really relevant to this post); and
4. a lifting of $$S$$ to the Kleisli category of $$T$$.
Now, this certainly is true, but this is where the hole appears. Beck’s original paper omits this last condition (perhaps because Kleisli categories weren’t widespread back then?). On the other hand, Cheng implicitly assumes the result for Kleisli categories. I have a feeling that it possibly follows out of the treatment of adjoint functors in Section 3 in the Beck paper, but I can’t quite get my head around it, and in any case I needed to write the construction down from first principles for my application.
To warm up, let’s look at the most well-known use for a distributive law $$\lambda : ST \to TS$$: giving the composite functor $$TS$$ a monad structure. Simply composing the units $$\eta^T \circ \eta^S$$ gives a natural transformation which will be the unit of the composite monad, but the bit which actually requires a distributive law is defining a multiplication $$\mu^{TS} : TSTS \rightarrow TS$$. It’s one of those times when there’s essentially only one thing you can write down, and that is exactly what you need:
$TSTS \xrightarrow{T \lambda S} TTSS \xrightarrow{\mu^T \mu^S} TS$
and all the relevant coherence conditions match up to satisfy the monad laws.
For the Kleisli category, we need to define a monad $$\tilde{S}$$ on $$\mathbf{Kl}(T)$$. Even just to define the functor part we already need the distributive law: the objects are the same, but for a map $$f: X \nrightarrow Y$$, we need to define
$\tilde{S}f: SX \rightarrow TSY = SX \xrightarrow{Sf} STY \xrightarrow{\lambda_Y} TSY$
So handwaving over the proofs that this is a proper functor, we now just need a unit and multiplication. The unit is again just a composite of the constituent units: $\eta_X: X \nrightarrow \tilde{S}X = X \xrightarrow{\eta^S_X} SX \xrightarrow{\eta^T_{SX}} TSX$ and the multiplication is a little more fiddly, but still straightforward: $\mu_X: \tilde{S}\tilde{S}X \nrightarrow \tilde{S}X = SSX \xrightarrow{\mu^S_X} SX \xrightarrow{S\eta^T_X} STX \xrightarrow{\lambda_X} TSX$ And, again glossing over all the annoying coherence proofs, we’re done!
There are quite a few pieces I’ve left out of all of this, notably going back the other way, from a lifting of $$S$$ to $$\mathbf{Kl}(T)$$ to a distributive law $$ST \rightarrow TS$$, and most of the verifications that all these things satisfy the laws they are supposed to, but this is mostly what I wanted to figure out. Hopefully it’s useful to someone other than me. Any mistakes or comments, send me an email or tweet and I’d be happy to discuss. | 1,378 | 5,345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-23 | latest | en | 0.914634 |
http://docplayer.net/3392961-Network-security-cs-5490-6490-fall-2015-lecture-notes-8-26-2015.html | 1,529,929,330,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00337.warc.gz | 83,995,037 | 25,326 | # Network Security CS 5490/6490 Fall 2015 Lecture Notes 8/26/2015
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1 Network Security CS 5490/6490 Fall 2015 Lecture Notes 8/26/2015 Chapter 2: Introduction to Cryptography What is cryptography? It is a process/art of mangling information in such a way so as to make it very hard to decipher but allowing a secret method of unmangling. Should cryptographic algorithms be secret? No, if we want to check whether these are robust against vulnerabilities or not. The more these are used and tested, the more are the chances of pitfalls getting exposed. The longer these are kept secret, the longer it will take to know the vulnerabilities fully and hence longer it will take to test the robustness against various types of attacks. [E.g., 2G GSM security algorithms were easily reverse engineered.] Though there are certain exceptions (Government and military sometimes like to not disclose their cryptographic algorithms). Might make business sense to keep the algorithm secret in some cases. Good cryptographic algorithms: Easy to use by good guy Hard for bad guy to break Who benefits from increasing processing power? Good guys can benefit from increasing processing power as long as the key sizes are increased appropriately. Increasing a bit only increases the work for the good guys by a little bit but doubles the amount of work for the bad guys (for brute force attacks). Comments in the class the good guys could guys could perform more computations for higher security good guys could perform faster crytanalysis Monoalphabetic cipher: replacing an alphabet with another alphabet. Brute force attack with 26! trials
2 Vulnerable to analysis attacks which letters occur most commonly, etc., newspaper puzzles Very easy to break with chosen plaintext attack Breaking an encryption (what is available to an adversary): Ciphertext only: Trudy has access to some cipher text Known Plaintext : has some <plaintext, ciphertext> pairs these can be used for an offline dictionary attack (most of the time the secret keys are based on passwords so one can try to figure out the password by trying out a dictionary of passwords, for every guess try to see if the encrypted plaintext matches the ciphertext) Chosen plaintext: Trudy is able to get cipher text for a chosen plaintext, i.e. has < chosen plaintext, cipher text> pair or pairs. She uses this to figure out the secret key (e.g., by choosing The quick brown fox jumps over the lazy dog in a telegraph office that encrypts the messages that it sends out using a monoalphabetic cipher). When designing encryption methods and protocols, we must be cognizant of the above Cryptofunctions: Secret key functions: same key used for encryption and decryption Public key functions: 2 different keys used (public key for encryption, private key for decryption) Hash functions/message digests (MD*) Secret Key Encryption 1. Encrypting data 2. Securely storing data o You can store the encrypted data securely and decrypt it when you want with the knowledge of the secret key (do not loose the secret key) 3. Authentication challengeresponse authentication (r A and r B are nonces) Alice Bob r A > < K AB {r A } < r B K AB {r B } > The challenges must not be repeated (at least until the keys are changed). The above protocol is vulnerable to different types of attacks. One such attack is called the reflection attack where Trudy impersonates Alice by starting two connections to Bob. For the first connection she sends r A to Bob as her first message. For the second connection, she sends r B and then uses the response of Bob K AB {r B } in her response over the first connection and hence successfully impersonates Alice. Trudy ignore K AB {r A }. We will look at this attack in more details later.
3 Public Key Encryption 1. Encrypting data 2. Agreeing upon a shared secret key Encrypting data with public key is orders of magnitude slower than with a secret key and hence generally avoided. Public Key Encryption is used primarily for encrypting a secret key rather than data. It is used to send a key to other party in a secure way. 3. Digital signatures One cannot create a digital signature scheme with secret key functions because at least two parties know the shared secret. In case of digital signatures using private keys, the private key is (expected to be) known only to one party. You cannot achieve nonrepudiation using secret keys. This must be achieved through public/private keys. 4. Authentication Alice will send a nonce encrypted with Bob s public key key, Bob will decrypt it using his private key and send it back to Alice. Alice Bob K B +{r A } > < r A < K A +{r B } r B >
4 Alice can have Bob decrypt any message that was in the past encrypted using Bob s public key that Alice had recorded. In the above diagram such a message is represented by r A. Hash Functions Characteristics: Always produces same length output (whatever be the input length) it should be computationally infeasible to find 2 messages M1 and M2 such that H(M1) = H(M2) It should be computationally infeasible to find M, given H(M) Manytoone, multiple inputs maps to a single output Integrity protection: Alice sends {M,H(M)} over the network, Bob computes H(M) from M and checks if its same as received H(M) from Alice. Problem: Person in the middle > Trudy replaces M,H(M)with his own message M and hash H(M ). o Better to send M, H(M, S) where S is the shared secret between A and B, or o sign H(m) with Alice s private key and send M and the signed hash. Data integrity: Store hash of data D and store it along with it. Every time data is accessed, find hash of data D and compare it with stored H(D). Problem: Trudy may change the program to always produce the right hash and hence fool Alice into believing that data is always right (or never tempered with). Chapter 3: Secret Key Cryptography We will first learn about block encryption. Two key questions what is the right key size? What is the right block size? Key size cannot be too small. Small keys could be bruteforced. The Perlman book suggests that 64bit security is pretty good. So a 64bit key must be strong enough against brute force attacks (requires the adversary to try 2^64 keys in the worst case). Block size a small block size (say k bits) allows an adversary to create and effectively search through a table of <plain text, cipher text> pairs. Here the cipher text output for a plain text input is obtained through random substitution that corresponds to tossing a coin k times with a head outcome represented by a 1 and a tail outcome represented by a 0. A block size of 64 bits would be good because it will be computationally infeasible for an adversary to create and maintain a <plain text, cipher text> table with 2^64 entries. The goal is to obtain an output that appears to be uncorrelated to the input. Any change in any bit location in the input should not just appear in specific locations of the output. The change in the input should be reflected on all bits of the output. A block size of 64 bits and a substitution table created for all possible 64 bit inputs by tossing coins will be perfect in terms of meeting this goal. However, this huge table is also the shared key. This cannot be efficiently or realistically exchanged between two parties.
5 Substitution memory requirement: for each of the 2 k inputs, there is a kbit output obtained by tossing a coin ktimes. Total entries are 2 k, and size of the table = k.2 k On the other hand, a permutation can be very effectively represented and can be agreed upon between two parties. The problem is that, a permutation is not very secure. One can figure it out by seeing which bits of the output change as a result of changes in the specific input bit patterns. Permutation memory requirement: each input bit is placed at specific location in output. Input= kbits, permutation mapping memory = k.log 2 k bits. Some mix of substitutions and permutations perhaps could help?
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### CUNSHENG DING HKUST, Hong Kong. Computer Security. Computer Security. Cunsheng DING, HKUST COMP4631
Cunsheng DING, HKUST Lecture 08: Key Management for One-key Ciphers Topics of this Lecture 1. The generation and distribution of secret keys. 2. A key distribution protocol with a key distribution center. | 7,848 | 35,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-26 | longest | en | 0.920695 |
https://testbook.com/question-answer/peak-inverse-voltage-for-a-full-wave-rectifier-is--5811e6db995a2d70d5f67f9d | 1,632,848,014,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00171.warc.gz | 593,384,926 | 29,433 | # Peak inverse voltage for a full wave rectifier is
This question was previously asked in
ALP CBT 2 Electrician Previous Paper: Held on 23 Jan 2019 Shift 3
View all RRB ALP Papers >
1. Em
2. 2Em
3. 3Em
4. 4Em
Option 2 : 2Em
## Detailed Solution
Peak Inverse Voltage is the maximum voltage that appears across a reverse biased diode:
If Em is the maximum voltage of the sinusoidal input signal
Then
For Half Wave rectifier $${\rm{PIV\;}} = {\rm{\;}}{{\rm{E}}_{\rm{m}}}$$
For Full Wave rectifier $${\rm{PIV\;}} = {\rm{\;}}2{{\rm{E}}_{\rm{m}}}$$
For Bridge Wave rectifier $${\rm{PIV\;}} = {\rm{\;}}{{\rm{E}}_{\rm{m}}}$$ | 225 | 625 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-39 | latest | en | 0.581711 |
https://projecteuler.chat/viewtopic.php?f=22&t=900&sid=beaf1a7c0ff50e5fb19eb8a41dfe24a9 | 1,624,513,156,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488551052.94/warc/CC-MAIN-20210624045834-20210624075834-00319.warc.gz | 404,489,118 | 10,622 | ## Twelve Knights
Decision and strategy, simulations and probability models, card and board games, ...
arif
Posts: 111
Joined: Tue Oct 09, 2007 4:35 am
Location: Lahore, Pakistan
### Twelve Knights
You must be familiar with the eight queens problem, where you have to place eight queens on a board so that no queen can capture another.
If we place knights in the following formation
n n - - n n - -
n n - - n n - -
- - - - - - - -
- - - - - - - -
n n - - n n - -
n n - - n n - -
- - - - - - - -
- - - - - - - -
We have twelve knights such that none can capture any other.
What is the maximum number of knights that can be placed on the chessboard with the same conditions.
[Very easy once you see the answer]
The i-th root of i = 4.810477381
stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 11:57 pm
Location: Netherlands
### Re: Twelve Knights
32 ?
jaap
Posts: 559
Joined: Tue Mar 25, 2008 3:57 pm
Contact:
### Re: Twelve Knights
arif wrote:We have twelve knights such that none can capture any other.
I think you'll find your example has 16 knights.
uws8505
Posts: 58
Joined: Tue Sep 30, 2008 3:13 pm
Location: South Korea
### Re: Twelve Knights
I get it. If we fill the dark squares on the chessboard(as you know) with knights, there are no knights capturing each other.
Math and Programming are complements
Eigenray
Posts: 62
Joined: Mon Jul 14, 2008 5:20 pm
### Re: Twelve Knights
For an mxn board, that gives a lower bound of mn/2 or (mn+1)/2. If there exists a knight's tour, we get an upper bound of mn/2. But if not?
uws8505
Posts: 58
Joined: Tue Sep 30, 2008 3:13 pm
Location: South Korea
### Re: Twelve Knights
Eigenray wrote:If there exists a knight's tour, we get an upper bound of mn/2.
A knight's tour may exist on a board with odd number of cells(such as 5*5). Then the upper bound should be (mn+1)/2, not mn/2.
Math and Programming are complements
jaap
Posts: 559
Joined: Tue Mar 25, 2008 3:57 pm
Contact:
### Re: Twelve Knights
uws8505 wrote:
Eigenray wrote:If there exists a knight's tour, we get an upper bound of mn/2.
A knight's tour may exist on a board with odd number of cells(such as 5*5). Then the upper bound should be (mn+1)/2, not mn/2.
Yes. Maybe Eigenray was thinking only of closed knight's tours.
Long, complete, and I hope correct analysis follows:
Expand
uws8505
Posts: 58
Joined: Tue Sep 30, 2008 3:13 pm
Location: South Korea
### Re: Twelve Knights
That's an interesting result
Math and Programming are complements
jaap
Posts: 559
Joined: Tue Mar 25, 2008 3:57 pm
Contact:
### Re: Twelve Knights
There was a slight oversight in my analysis, though the result stays the same.
Expand
Maddox
Posts: 1
Joined: Wed Aug 29, 2012 2:08 pm
Location: USA
Contact:
### Re: Twelve Knights
Hey interesting result. I like that type of game which belong with math. I will play this game in this Sunday then i will post my result on it. | 858 | 2,880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-25 | latest | en | 0.854087 |
https://www.gizmocrazed.com/2018/04/usc-wide-receiver-josh-imatorbhebhe039s-crazy-floating-jump/ | 1,642,677,327,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301737.47/warc/CC-MAIN-20220120100127-20220120130127-00472.warc.gz | 842,771,995 | 46,314 | Tuesday, January 11, 2022
# USC Wide Receiver Josh Imatorbhebhe's Crazy Floating Jump
Josh Imatorbhebhe is a wide receiver for the University of Southern California football team. And, um, he can jump crazy high. Take a close look at his vertical jump in the video above. What looks weird about it? It almost looks like he just hangs there at the top of his jump.
Is this normal physics, some type of optical illusion, or just plain magic?
In order to explore this epic jump, the first thing that needs to be done is to plot Imatorbhebhe’s position as a function of time. I can scale this video using Josh’s height (1.88 meters). After that, I can use my favorite free software Tracker Video Analysis to get the position of his feet in each frame. Here’s what that looks like. Oh, just a note—I tried to scale the video using the scale in the background, but something wasn’t right.
When a ball is tossed into the air, its position-time graph should be a parabola. Clearly, this isn’t a parabola since it looks flat at the top (that’s the part where he magically stays in the air). So it’s true—something weird is going on here.
Of course, there is a difference between a tossed ball and a jumping human. The motion of a human can’t be represented by just one point, because humans can move their body. While in the air, a human could pull their feet up which would change the position of the center of mass. This means that although the center of mass for the human moves in a parabola, the feet (or any point) by itself could move in some weirder fashion. A great example of this is the Invisible Box Challenge—where a person appears to step on a box that isn’t there by moving the body in a special way.
So, you can see that fitting a parabola to this data for the motion of the foot doesn’t quite work. The fitting function deviates from the data—especially at the highest point. However, it is somewhat surprising that the parabolic fit gives a t-squared coefficient of -4.9 m/s2 and this would make the overall acceleration have a value of -9.8 m/s2. This is what you would expect for a ball accelerating due to gravity alone.
How about a quick comparison? What would the motion of a tossed ball look like right next to the feet of Josh Imatorbhebhe? I can model this in python (you can see the code here). If the ball starts in the same position with an initial velocity of 0.88 m/s, I get the following.
The red curve is a parabola since it’s an object with a constant acceleration. Notice how the data from Josh’s jump sort of oscillates around the parabolic plot? That’s because the video data is just marking his feet, which does not represent the whole body. As Josh rotates his arms around, it also moves his feet closer and then farther from the center of mass.
It is this oscillation that make it appear that he floats at the top of the jump!
However, this comes at a cost. Notice that the highest the ball travels (red curve) is 1.146 meters. The highest the foot gets is only 1.113 meters. That’s a difference of 3.3 centimeters (1.3 inches). Waving his arms makes a cool effect, but he doesn’t jump as high (as measured by his feet). If he timed his arm swinging correctly, he could get his feet even higher at the cost of a lower apparent “hang” at the top.
Still, in the end it’s a super impressive jump.
76,236Fans
18,129Followers | 796 | 3,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-05 | latest | en | 0.959264 |
https://planetmath.org/IncircleRadiusDeterminedByPythagoreanTriple | 1,716,817,577,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00550.warc.gz | 385,474,582 | 3,912 | # incircle radius determined by Pythagorean triple
If the sides of a right triangle are integers, then so is the radius of the incircle of this triangle.
For example, the incircle radius of the Egyptian triangle is 1.
Proof. The sides of such a right triangle may be expressed by the integer parametres $m,\,n$ with $m>n>0$ as
$\displaystyle a\;=\;2mn,\quad b\;=\;m^{2}\!-\!n^{2},\quad c\;=\;m^{2}\!+\!n^{2};$ (1)
the radius of the incircle (http://planetmath.org/Incircle) is
$\displaystyle r\;=\;\frac{2A}{a\!+\!b\!+\!c},$ (2)
where $A$ is the area of the triangle. Using (1) and (2) we obtain
$r\;=\;\frac{2\cdot 2mn\cdot(m^{2}\!-\!n^{2})/2}{2mn\!+\!(m^{2}\!-\!n^{2})\!+\!% (m^{2}\!+\!n^{2})}\;=\;\frac{2mn(m\!+\!n)(m\!-\!n)}{2m(m\!+\!n)}\;=\;(m\!-\!n)n,$
which is a positive integer.
Remark. The corresponding radius of the circumcircle need not to be integer, since by Thales’ theorem, the radius is always half of the hypotenuse which may be odd (e.g. 5).
Title incircle radius determined by Pythagorean triple Canonical name IncircleRadiusDeterminedByPythagoreanTriple Date of creation 2013-03-22 17:45:46 Last modified on 2013-03-22 17:45:46 Owner pahio (2872) Last modified by pahio (2872) Numerical id 14 Author pahio (2872) Entry type Feature Classification msc 11A05 Synonym incircle radius of right triangle Related topic Triangle Related topic PythagoreanTriple Related topic DifferenceOfSquares Related topic FirstPrimitivePythagoreanTriplets Related topic X4Y4z2HasNoSolutionsInPositiveIntegers | 507 | 1,529 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 6, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-22 | latest | en | 0.647425 |
http://openmusictheory.com/popRockHarmony | 1,656,415,039,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00477.warc.gz | 42,630,365 | 5,081 | Harmony in pop/rock music does not always follow the same norms and patterns of classical-era music. Thus, functional-bass notation does not work for all situations. Instead, we will primarily use Roman numerals for our analysis of harmony in pop/rock music, occasionally using functional bass as a supplement to help us make sense of a particular harmonic pattern or relate it to what we’ve studied in classical harmony.
One key difference between rock and classical harmony is that chords in pop/rock music are almost always root-position triads or seventh chords. This affects the “rules” of harmonic syntax, as 6/3 chords in classical progressions are replaced by 5/3 chords in pop/rock progressions. (For instance, the classical progression IV–IV6 becomes IV–VI. The same bass line does the same work, but by using a 5/3 chord instead of a 6/3 chord, the functional progression changes to something that would “break the rules” of classical syntax.) This can make harmonic analysis a little tricky in pop/rock music, especially since there is no published theory of rock harmony that is equal to Quinn’s functional theory of classical harmony. However, it makes chord labeling and harmonic dictation simpler. Most of the time, all you need is the scale degree of the bass.
Following is a chart of bass scale degrees and the roots/Roman numerals most typically associated with them. Keep this chart handy when transcribing and dictating rock harmonic progressions. As you can see, most bass notes typically go with a single chord.
bass 5/3 or 7 6/3 or 6/5
do I
re II
mi/me [III] I
fa IV
sol V
la/le VI [IV]
te VII
ti V
Bass scale degrees and commonly associated harmonies in pop/rock music. Less common chords are enclosed in square brackets.
## Harmonic functions in minor
Harmonic functions in minor keys of pop/rock songs, particularly in the last couple decades, tend to deviate from classical syntax in different ways than major-key songs. See the Harmonic functions in minor resource for more details about this.
## Schemata
There are a number of common stock chord progressions that recur in many pop/rock songs. Typically, these stock progressions, or schemata, will occur in cyclical patterns; that is, the same progression will repeat multiple times in a row. This is particularly common in choruses of verse-chorus songs, but also happens in verses, strophes, and bridges. This is helpful for identifying harmonies by ear, since in addition to listening for bass scale degrees and considering whether the harmonies are chords of the fifth (5/3 or 7) or chords of the sixth (typically 6/3 or 6/5), we can listen for common patterns that we’ve heard in other songs. Following are a number of common schemata for pop/rock harmonic progressions.
The “50s doo-wop” progression
||: I – VI – IV – V :||
or
||: I – VI – II – V :||
The “Singer/Songwriter” progression
||: VI – IV – I – V :|| (in major)
||: I – VI – III – VII :|| (in minor)
||: I – V – VI – IV :|| (“With or Without You” variant)
||: IV – I – V – VI :|| (“deceptive” variant)
The “Puff” progression
I – III – IV . . . (to begin a phrase)
The blues progression
||: I - - - | IV - I - | V IV I - :|| (12-bar blues)
||: I - - - | I - - - | IV - I - | V IV I -:|| (16-bar blues)
The Pachelbel progression
||: I – V – VI – III – IV – I – IV – V :||
||: I – V6 – VI – III6 – IV – I6 – IV – V :|| (stepwise bass version)
I – V – VI – III . . . (to begin a phrase; “truncated” version)
The lament progression
||: I – VII – VI – V :||
The circle-of-fifths progression (in minor)
||: I – IV – VII – III :||
Plagal progressions
||: I - IV :||
||: I - bVII - IV - I :||
||: bVI - bIII - bVII - IV - I :||
## Rock’s “tonal systems”
Walter Everett has posited six tonal systems to which most pop/rock songs belong. The following sections summarize four of these with examples (all but 1 and 6—the most and least tonal). See Everett’s 2004 article in Music Theory Online for a more detailed exploration of these “systems.”
System 1: Pop/rock tonality that strongly resembles common-practice classical tonality (major and minor keys, with parallel-key borrowing).
System 2: Pop/rock tonality that draws on diatonic modes (such as Dorian, Mixolydian, Aeolian).
System 3: Pop/rock tonality that strongly resembles common-practice classical tonality, but which borrows from diatonic modes.
System 4: Blues-based pop/rock tonality.
System 5: Minor-pentatonic-based pop/rock tonality.
System 6: Chromatically inflected minor-pentatonic-based pop/rock tonality. | 1,172 | 4,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | latest | en | 0.924724 |
https://www.edaboard.com/threads/discrete-components-low-distortion-variable-with-a-single-resistor-audio-oscillator.397099/ | 1,619,157,052,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039601956.95/warc/CC-MAIN-20210423041014-20210423071014-00499.warc.gz | 847,357,624 | 35,753 | # Discrete components, low distortion, variable with a single resistor, audio oscillator
#### neazoi
##### Advanced Member level 5
Hi,
I would like to built a simple audio oscillator in the 480Hz-2700Hz range with the next requirements:
1. Discrete components
2. Variable in the 480Hz-2700Hz range
3. Frequency to be varied with a single variable resistor preferably, not a stereo potentiometer or other dual controls, if possible
4. Single range, i.e not switching to be required for the whole 480-2700Hz
4. Low distortion, so as to avoid harmonics in the passband
Can you propose me any circuits, or present any simulations to try?
Thank you
#### barry
##### Advanced Member level 5
sine wave? square wave? triangle wave? Low distortion: 1%? 0.00001%?
Hey, here's an idea:
1) Go to google.com , bing.com, or some other search engine.
2) Type: "Audio oscillator circuit".
3) Hit "ENTER".
3) Gasp in amazement at the number of resources available.
#### neazoi
##### Advanced Member level 5
sine wave? square wave? triangle wave? Low distortion: 1%? 0.00001%?
Hey, here's an idea:
1) Go to google.com , bing.com, or some other search engine.
2) Type: "Audio oscillator circuit".
3) Hit "ENTER".
3) Gasp in amazement at the number of resources available.
Amazing answer! Thank you I will do so...
#### betwixt
##### Super Moderator
Staff member
I presume this has to be a sine wave or the 'avoid harmonics in the passband' wouldn't make sense.
I think the only practical solution would be a Wien bridge oscillator but for low distortion you need a dual gang resistor (ideally linear so not a 'stereo' control).
The only other solution I can think of is to generate a square wave first but at four times the frequency you want, divide it by two in a bistable circuit to ensure it has equal mark-space (a 'perfect' square wave), the divide by two again but use constant current sources to alternately charge and discharge a capacitor, this give a 'perfect' triangle wave at the frequency you want. Finally, use a non-linear amplifier or clamp to progressively compress the peaks of the triangle wave to make it close to a sine shape. This is the technique used in some 'universal' wave generators because it produces simultaneous square, triangle and sine outputs.
It isn't simple though.
Brian.
### neazoi
Points: 2
#### neazoi
##### Advanced Member level 5
I presume this has to be a sine wave or the 'avoid harmonics in the passband' wouldn't make sense.
I think the only practical solution would be a Wien bridge oscillator but for low distortion you need a dual gang resistor (ideally linear so not a 'stereo' control).
The only other solution I can think of is to generate a square wave first but at four times the frequency you want, divide it by two in a bistable circuit to ensure it has equal mark-space (a 'perfect' square wave), the divide by two again but use constant current sources to alternately charge and discharge a capacitor, this give a 'perfect' triangle wave at the frequency you want. Finally, use a non-linear amplifier or clamp to progressively compress the peaks of the triangle wave to make it close to a sine shape. This is the technique used in some 'universal' wave generators because it produces simultaneous square, triangle and sine outputs.
It isn't simple though.
Brian.
Indeed it sounds too complicated.
I was thinking that one could also use a variable crystal oscillator on HF (vxo) and mix it down to audio. Then use a simple LPF to keep only the audio.
This should work in theory and should provide a very clean sinewave.
I am not sure if I will go that way, I still investigate oscillators such as the one attached. A forum member had replied this a few months ago if I remember well. But as far as I understand it uses a dual potentiometer isn't it?
So maybe the VXO solution is the best and the simplest?
#### Attachments
• wienAFosc.PNG
95.6 KB · Views: 9
#### betwixt
##### Super Moderator
Staff member
R7 and R8 have to be ganged in that design. It is a Wien bridge with a JFET as active gain controller.
Yes, you can mix a high frequency crystal oscillator with a VCO but it may not be as simple as it sounds. Frequency stability is the main issue, you would have a high frequency RF generator with frequency set by a potentiometer, that sound like a recipe for drift with temperature. There is also the point that unless you keep both oscillator levels well within the mixer dynamic range it will distort the output waveform and you have to be careful what frequencies you use if you are mixing them to get a low frequency result. Without sufficient isolation between the signals you might get injection locking when they are only a few hundred Hz apart.
Brian.
#### neazoi
##### Advanced Member level 5
Without sufficient isolation between the signals you might get injection locking when they are only a few hundred Hz apart.
This is in fact what I was thinking the most.
You see you can have two crystal oscillators, the one fixed say at 10MHz the other variable (vxo) 1-3KHz (crystal pulling), but you then have to have a network at the output to isolate their outputs.
It more reminds me of a 2-tone generator like the one presented in this page Inband IMD Immunity Testing (ab4oj.com) but you can also find such designs on the net. These use a bridge at the output that isolates the two closely-operated oscillators. Perhaps I should search more for a two-tone generator circuits, if I want to go that way.
But there are some other ideas, for example one can have 2 oscillators, one of them operating at half the frequency and mix it's harmonics with the other. This should still produce an audible done from one of the first oscillator harmonics, mixed with the main signal of the other oscillator without interlocking. A simple audio LPF after the mixing of these oscillators should be enough for keeping only the audio tone. See for example the polyakov mixer attached. The 2 signals should not interlock.
What do you think?
#### Attachments
• Beat_Crystal_Freq_Meter_W5JSN.gif
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• PolyakovPlus2133.jpg
245.8 KB · Views: 7
Last edited:
#### schmitt trigger
##### Advanced Member level 5
This is one of those times, when one really misses the XR2206.
#### betwixt
##### Super Moderator
Staff member
XR2206 was really useful, it used a clamp circuit to 'round off' the triangle wave as I described earlier. Not perfect, but acceptable.
I don't think it would be practical to construct an XR2206 out of discrete components though.
Brian.
#### Audioguru
##### Advanced Member level 5
The retro project was needed to use discrete components. Vacuum tubes?
#### neazoi
##### Advanced Member level 5
The retro project was needed to use discrete components. Vacuum tubes?
I was thinking about transistors only.
I have tried this circuit but I could not keep the level constant at the edges of the frequency spectrum. I think it has to do with the incandescent properties. I only had a 500r stereo pot so I increased the caps to 1uF. Anyway I a not very satisfied and after all it uses a stereo pot.
Now I could replace the stereo pot with two LDRs, driven with leds, which in turn would be driven by a mono potentiometer. Even if ldrs are not very matched, it should provide acceptable distortion I think. I believe that to some point the ldrs couls be replaced by a set of transistors used as variable resistance devices.
In that sense, maybe a phase shift oscillator can be used, but could it provide a good sinewave if set for low gain?
I have not yet tried the simulated circuit above (a bit more complex)
The RF mixers idea, driving a polyakov mixer could be derived from this circuit https://www.qsl.net/dl1gsj/html/qrssrx30.html
#### Attachments
• invwein.gif
7.7 KB · Views: 1
#### Easy peasy
##### Advanced Member level 5
A Triangle wave is only a few % different to a sine wave - so make a triangle wave generator and filter it - too easy ...
#### Audioguru
##### Advanced Member level 5
A Triangle wave is only a few % different to a sine wave - so make a triangle wave generator and filter it - too easy ...
A filtered triangle wave generator will work if the frequency is not changed because then all the RC filters would also need changing.
You might as well make a Wien Bridge or Phase shift oscillator that does not need filtering.
##### Super Moderator
Staff member
When designing the XR2206 the makers looked for the most efficient method to create a sinewave. They decided on the diode shaping technique. I imagine it's because the effect is independent of frequency.
Once you have a regulated triangle wave, a pair of zener diodes back-to-back is sufficient to round the peaks so you get a semblance of a sine. (If you wish you can shape it further by adding another set of diodes with a different volt rating.)
The rightmost XY figure plots the output against a pure sine. The aim is to carefully adjust the potentiometer until the trace is a straight line, or as close as you can get.
### neazoi
Points: 2
#### neazoi
##### Advanced Member level 5
When designing the XR2206 the makers looked for the most efficient method to create a sinewave. They decided on the diode shaping technique. I imagine it's because the effect is independent of frequency.
Once you have a regulated triangle wave, a pair of zener diodes back-to-back is sufficient to round the peaks so you get a semblance of a sine. (If you wish you can shape it further by adding another set of diodes with a different volt rating.)
The rightmost XY figure plots the output against a pure sine. The aim is to carefully adjust the potentiometer until the trace is a straight line, or as close as you can get.
View attachment 167880
The output is by far not a sinewave though. Maybe more like something that approximates a sinewave only...
I think I am going to give a try to the post #5 circuit, as my options start to decrease. I will let you know about the results.
I will also give a try to this transistor used as a resistance device thing. Maybe for the limited range I am interested in, this is a very viable solution.
#### neazoi
##### Advanced Member level 5
Ok this circuit worked. The BJTs were more critical, I could not use 2n2222, it needed to be bc547. Now, at 12v the gain is a bit more so there is some clipping at the bottom of the sine. I think the ALC does not work correctly, cause even if I remove C6 I see no signal change. However if O remove the jfet, there is no oscillation anymore
Update: I replaced R10 with a higher value resistor (33k) and the distortion (and the signal level a bit) went down
Also I would like to avoid 1uF capacitors (can I use electrolytics?) and use 470nF maximum. What resistors I need to increase for this change?
#### Attachments
• wienAFosc.PNG
95.6 KB · Views: 5
Last edited:
#### LvW
##### Advanced Member level 5
A filtered triangle wave generator will work if the frequency is not changed because then all the RC filters would also need changing.
When you feed the output of a triangle generator to a classical difference amplifier (long-tailed pair) you get a pretty good sine wave with a THD of lower than 1% - indpendent on frequency (because this is not a filtering but a wave shape).
However, it is important that the triangle amplitude has a certain (fixed) value of ap.p 60mVolts.
#### Easy peasy
##### Advanced Member level 5
A ganged potentiometer will vary the freq and the R's of the filter to keep the filtering in the same ratio ... making it easy to produce a sine from a triangle .... you can do similar with digital pots ...
#### mtwieg
##### Advanced Member level 5
Back in my undergrad circuit design course, we were tasked with making a sine wave generator whose output amplitude was independent of frequency, without special ICs. Most students took the approach of using resistor/diode networks to shape a triangle wave into a sine wave. But one student took the approach of creating a switched capacitor filter whose clock frequency tracked with the output frequency. Results were very nice, assuming you didn't adjust the frequency faster than his PLL could track.
#### Audioguru
##### Advanced Member level 5
My very low distortion variable frequency sinewave circuit uses a CD4018 counter to produce a stepped levels "sinewave" that is also filtered with a switched-capacitor Butterworth lowpass filter IC. The clock tor the counter and for the switched capacitor filter is the same so it adjusts the frequency with a single pot with perfect immediate tracking..
The Wien Bridge oscillator with the obsolete Jfet (others are still available) uses a dual linear pot. Its 1uF capacitors are used for low frequencies that you do not need, use 330nF instead.
Transistors have a wide range of current gain and a high output resistance, use an opamp instead.
The AGC does not work if the diode is backwards.
#### Attachments
• Wien bridge oscillator with FET AGC.PNG
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2K | 3,049 | 13,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-17 | latest | en | 0.833734 |
https://www.sophia.org/tutorials/measures-of-dispersion--2 | 1,511,115,264,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805708.41/warc/CC-MAIN-20171119172232-20171119192232-00176.warc.gz | 864,952,711 | 15,150 | +
# Measures of Dispersion
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Author: Sara Gorsuch
##### Description:
Define and explain these new terms:
+Range
+Deviation from the mean
+Mean absolute deviation
+Variance
Bringing you through the ideas behind range, deviation from the mean, mean absolute deviation, variance and showing you how it works, showing examples of how it all is put together into one idea.
(more)
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Tutorial
Range: It’s the difference between the highest (largest) value (H) - the lowest (smallest) value (L) pieces of data.
so range = H(highest value) -L(lowest value)
-----> some people also look at this as range= max(x) - min(x)
Variance: Is a sample measure of the spread of data about the mean.
Mean absolute deviation: tells us the average distance that a piece of data is from the mean.
What are the formula for these?
To find the formulas for these parts of standard deviation look through this great pdf, find the glossary section on the far right hand side and look
under measures of dispersion.
## Measures of Dispersion
Source: Elementary Statistics : Sixth Edition: Robert Johnson.
## Example of Finding the Sample of Variance.
This video describes what the variance is and shows us examples of how to calculate it. (It does briefly go into standard deviation) | 428 | 1,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-47 | longest | en | 0.891446 |
http://www.vbforums.com/showthread.php?674833-RESOLVED-Concentration-game-2D-picturebox-array | 1,529,272,434,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859817.15/warc/CC-MAIN-20180617213237-20180617233237-00062.warc.gz | 507,305,481 | 22,768 | Thread: [RESOLVED] Concentration game 2D picturebox array
1. [RESOLVED] Concentration game 2D picturebox array
Hello,
I am required to make a game of concentration, whereby cards are shown for x amount of seconds and then truned over. The user then needs to remember where the cards were and pick two at a time in order to win points - whilst being timed.
Any pointers on where to start, or how to make a 2D control array of pictureboxes that can display the cards into a grid of (10x10) pictureboxes randomly?
Thanks
PMB
2. Re: Concentration game 2D picturebox array
Add a TableLayoutPanel to your form with the appropriate numbers of rows and columns, then add a PictureBox to each cell.Handle the Click event of each control to change the Image and use a Timer to change it back.
3. Re: Concentration game 2D picturebox array
Will this method allow me to somehow shuffle the cards, randomly, so that they're not in the same lcoation each time a New Game is selected?
I need to have a deck of 52 cards, in some sort of type that then related to an integer in the pictureboxes, or coordinates?, but I am unsure as to the best way to go about it. It uses to randomize function, or so I believe, as no number can be repeated.
Thanks
PMB
4. Re: Concentration game 2D picturebox array
If you have a list of 52 items then you just randomise that first and then place them into position one by one. It's just like dealing a pack of cards. You shuffle the cards and then deal from the top. You might like to follow the CodeBank link in my signature and check out my thread on Randomising A List.
5. Re: Concentration game 2D picturebox array
I've successfully managed to randomize the cards, using this code:
vb Code:
`Dim Cards() As Integer Private Sub shuffle() Dim numberOfCards = 52 Cards = Enumerable.Repeat(0, numberOfCards ).ToArray() Dim Tmp As Integer Randomize() Dim i As Integer = 0 Do While i < numberOfCards Tmp = Int((numberOfCards + 1) * Rnd()) If Array.IndexOf(Cards, Tmp) < 0 Then Cards(i) = Tmp ListBox1.Items.Add(Tmp) i += 1 End If Loop End Sub`
This puts the numbers 1 to 52 in a litsbox randomly - which will be removed, it was simply for confiration. How would I now convert this to randomize the display of images on PictureBoxes that were added into a Panel? It's a 10x10 grid.
Also, at the minute I have them all showing red back using this code:
vb Code:
`PictureBox(Row, Column).Image = Image.FromFile(path & "cards\Red.png") PictureBox(Row, Column).SizeMode = PictureBoxSizeMode.StretchImage`
So, what I am trying to do is randomize the face cards that will show upon clicking each PictureBox. I need two of each at any one time so that there's a pair.
Thanks
PMB
6. Re: Concentration game 2D picturebox array
Well, if you have a 10x10 grid, then you need 100 cards. However, depending on how you want to represent those 100 cards, your shuffling code may or may not work. If you represent them as the numbers 1-100 it's fine, but then you have to match up the pairs. If you represent them as the numbers 1-50 with each value appearing twice, it's simpler to represent and match each card, but your shuffling algorithm won't work.
Then it's simply a case of mapping each card's integer value to the relevant picture to display as its face. (e.g. {1,2} map to card face 1, {3,4} map to card face 2, ... (if you decide to go with numbers 1-100))
Oh, well there is of course mapping index of the collection to the relevant picture box as well. That is, for the first element in your shuffled collection, you need to take the value, work out the picture to show, and work out which picture box to show it in, and show that picture in that picture box. Repeat for every element in your collection.
7. Re: Concentration game 2D picturebox array
Or you could do as suggested in the first place and randomise like this:
Code:
```Dim rng As New Random
myArray = myArray.OrderBy(Function(x) rng.NextDouble()).ToArray()```
That will randomise the list no matter what it contains. The original list could be sequential numbers generated by Enumerable.Range or it might be an array of instances of a Card class or whatever you like.
8. Re: Concentration game 2D picturebox array
Apologies for the slow understanding but I am not an avid user of VB.NET or Visual Studio 2010 - this task is an assignment at College.
I have located the 100 pictureboxes to the panel also - using the AddHandler and AddressOf. Now I need 50 pairs, out of the 52 to be displayed in any of the pictureboxes at random. I've had a good play with it and don't seem to be able to get it right.
Also, the images for the cards are in the 'Resources' part of the program, would is be better to have these imported from file using 'Image.FromFile()' ?
Any code to acheive the random deposition of card in the pictureboxes would be massively advantageous.
Thanks
PMB
9. Re: Concentration game 2D picturebox array
You've alreay got the code. Put your PictureBoxes into an array or collection and randomise it as you have alreay been shown. You can then use a loop to place your randomised cards into the randomised PictureBoxes.
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Featured | 1,330 | 5,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.853832 |
https://byjus.com/questions/the-length-breadth-and-height-of-a-room-are-in-the-ratio-3-2-1-if-the-breadth-and-height-are-halved-while-the-length-is-doubled-then-total-area-of-the-four-walls-of-the-room-will-1-remain-the-same/ | 1,656,731,487,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103983398.56/warc/CC-MAIN-20220702010252-20220702040252-00179.warc.gz | 195,443,534 | 41,930 | # The length breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height are halved while the length is doubled, then total area of the four walls of the room will (1) Remain the same (2) Decrease by 30% (3) Decrease by 15% (4) Decrease by 18.75%
Let us consider the room with
Length (l) = 3x
Breadth (b) = 2x
Height = 1x
Area of the walls (A) = 2(l + b) × h
A = 2(3x + 2x) × 1x
A = 10x2 sq. unit.
Now
l1 = 6x
b1 = x
h1 = x/2
Nwe area (A1) = 2(6x + x) (x/2)
A1 = 7x2
$$\begin{array}{l}% decrease in the area of walls = \frac{10x^{2}-7x^{2}}{10x^{2}}\end{array}$$
% decrease in the area of walls = 30%
0 (0)
(3)
(3)
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Suppose we have a square ABCD measuring sides be (x+1). If we draw a diagonal BC calculate the diagonal by using Pythagoras theorem. | 342 | 937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-27 | latest | en | 0.82958 |
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# Probability and Statistics
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Code Completion Credits Range Language
BIE-PST Z,ZK 5 2P+1R+1C English
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Synopsis:
The students will learn the basics of probabilistic thinking, the ability to synthesize prior and posterior information and learn to work with random variables. They will be able to to apply basic models of random variable distributions and solve applied probabilistic problems in informatics and computer science. Using the statistical induction they will be able to perform estimations of unknown distributional parameters from random sample characteristics. They will also be introduced to the methods of determining the statistical dependence of two or more random variables.
Requirements:
Basics of combinatorics and mathematical analysis.
Syllabus of lectures:
1. Probability: Random events, event space structure, probability of a random event and its basic properties.
2. Conditional probability: Dependent and independent events, Bayes theorem.
3. Random variables: Distribution function of a random variable, continuous and discrete distributions, quantiles, median.
4. Characteristics of random variables: Expected value, variance, general moments, kurtosis and skewness.
5. Overview of basic distributions: binomial, geometric, Poisson, uniform, normal, exponential. Their basic properties.
6. Random vectors: Joint and marginal statistics, correlation coefficient, dependence and independence of random variables.
7. Random vectors: Conditional distributions, sums of random variables.
8. Limit theorems: Laws of large numbers, central limit theorem
9. Statistical estimation: Classification and processing of data sets, characteristics of position, variance and shape, sampling moments, graphical representation of data.
10. Point estimation: Random sample, basic sample statistics, sample mean and variance, distributions (t-distribution, F-distribution, chi square).
11. Interval estimation: Confidence intervals for expectation and variance.
12. Hypothesis testing: Testing strategy, tests for expectation and variance, their modifications. Application of statistical testing in CS.
13. Correlation and regression analysis: Linear and quadratic regression, sample correlation.
Syllabus of tutorials:
1. Basics of probability.
2. Conditional probability.
3. Random variables.
4. Basic characteristics of random variables.
5. Using basic distributions.
6. Random vectors - independence, covariance.
7. Random vectors - conditional distributions and sums.
8. Limit theorems
9. Processing of sets of data.
10. Statistical point estimation.
11. Interval estimation.
12. Hypotheses testing.
13. Regression and correlation analysis.
Study Objective:
The goal of the module is to introduce the students to basics of probability theory and mathematical statistics while focusing on applications in informatics.
Study materials:
1. Johnson, J. L. ''Probability and Statistics for Computer Science''. Wiley-Interscience, 2008. ISBN 0470383429.
2. Li, X. R. ''Probability, Random Signals, and Statistics''. CRC, 1999. ISBN 0849304334.
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Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet1448106.html | 741 | 3,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-30 | latest | en | 0.774356 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Units_of_Measure/Physical_Quantities | 1,675,352,064,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500028.12/warc/CC-MAIN-20230202133541-20230202163541-00570.warc.gz | 182,266,446 | 28,607 | # Physical Quantities
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Chemistry is a quantitative science. Amounts of substances and energies must always be expressed in numbers and units (in order to make some sense of what you are talking about). You should also develop a sensation about quantities every time you encounter them; you should be familiar with the name, prefix, and symbol used for various quantities.
However, due to the many different units we use, expression of quantities is rather complicated. We will deal with the number part of quantities on this page, using SI Units.
are expressed by eXXX or EXXX.
Words Number Prefix Symbol Exponent
----- --------------------- ------ ------ -of 10--
Quindrillion 1,000,000,000,000,000 e15
Trillion 1,000,000,000,000 Tera- T e12
Billion 1,000,000,000 Giga- G e9
Million 1,000,000 Mega- M e6
Thousand 1,000 Kilo- k e3
Hundred 100
Ten 10
One 1
Tenth 0.1 Deci- d
Hundredth 0.01 Centi- c
Thousandth 0.001 Milli- m e-3
Millionth 0.000001 Micro- u (mu) e-6
Billionth 0.000000001 Nano- n e-9
Trillionth 0.000000000001 Pico- p e-12
----- --------------------- ------ ------ -------
By now, you probably realized that every time the number increases by a factor of a thousand, we give a new name, a new prefix, and a new symbol in its expression.
After you are familiar with the words associated with these numbers, you should be able to communicate numbers with ease. Consider the following number:
123,456,789,101,234,567
In words, this 18-digit number takes up a few lines:
One hundred twenty three quindrillions, four hundred fifty six trillions, seven hundred eighty nine billions, one hundred and one millions, two hundred and thirty four thousands, five hundred and sixty seven.
If a quantity makes use of this number, the quantity has been measured precisely. Most quantities do not have a precise measurement to warrant so many significant figures. The above number may often be expressed as 123e15 or read as one hundred twenty three quindrillions.
There are seven basic quantities in science, and these quantities, their symbols, names of their units, and unit symbols are listed below:
== Basic Quantity == ==== Unit =====
Name Symbol Symbol Name
============= ====== ====== ========
Length l m meter
Mass m kg Kilogram
Time t s Second
Electric current I A Ampere (C/s)
Temperature T K Kelvin
Amount of substance n mol Mole
Luminous intensity Iv cd Candela
============= ====== ====== ========
*The unit ampere, A, is equal to Coulombs per second, (A = C/s).
Physical Quantities is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. | 1,114 | 4,102 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-06 | latest | en | 0.503816 |
http://www.chegg.com/homework-help/university-physics-plus-modern-physics-plus-masteringphysics-with-etext-access-card-package-13th-edition-chapter-19-solutions-9780321675460 | 1,474,973,686,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661023.80/warc/CC-MAIN-20160924173741-00199-ip-10-143-35-109.ec2.internal.warc.gz | 381,184,210 | 20,029 | University Physics Plus Modern Physics Plus MasteringPhysics with eText -- Access Card Package (13th Edition) View more editions Solutions for Chapter 19
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Chapter: Problem:
For the following processes, is the work done by the system (defined as the expanding or contracting gas) on the environment positive or negative? (a) expansion of the burned gasoline–air mixture in the cylinder of an automobile engine; (b) opening a bottle of champagne; (c) filling a scuba tank with compressed air; (d) partial crumpling of a sealed, empty water bottle as you drive from the mountains down to sea level.
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 5
The work done of the system is,
Here, is the pressure, is the initial volume, and is the final volume of the substance.
• Step 2 of 5
(a)
During the expansion of the burned gasoline-air mixture in the cylinder of an automobile engine, the final volume of the gasoline and air mixture in the cylinder increase. Therefore, the change in the volume of the cylinder has a positive value.
Therefore, the work done by the system is as follows:
Therefore, the work done by the burned gasoline-air mixture system on the environment is positive.
• Step 3 of 5
(b)
In opening a bottle of champagne, then the final volume of the champagne will increases.
Therefore, the change in the volume of the champagne system has a positive value.
Therefore, the work done by the system is as follows:
Therefore, the work done by the champagne system on the environment is positive.
• Step 4 of 5
(c)
In filling a scuba tank with compressed air, the final volume of the scuba tank will decrease.
Therefore, the change in the volume of the scuba tank with compressed air has a negative value.
Therefore, the work done by the system is as follows:
Therefore, the work done by the scuba tank with compressed air on the environment is negative.
• Step 5 of 5
(d)
In partial crumpling of a sealed, empty water bottle, as you drive from the mountains to sea level, the volume of the gas in the bottle decreases due to the increases in the pressure.
Therefore, the change in the volume of the gas in the empty bottle has a negative value.
Therefore, the work done by the system is as follows:
Therefore, the work done by the gas in the empty water bottle on the environment is negative.
Corresponding Textbook
University Physics Plus Modern Physics Plus MasteringPhysics with eText -- Access Card Package | 13th Edition
9780321675460ISBN-13: 0321675460ISBN: Authors:
Alternate ISBN: 9780321696595, 9780321696656, 9780321696670, 9780321696687, 9780321696694, 9780321736116, 9780321850027 | 660 | 2,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-40 | latest | en | 0.904706 |
https://www.ask.com/news/streamlining-workflow-advanced-excel-formulas | 1,721,074,257,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00050.warc.gz | 582,199,232 | 20,907 | In today’s data-driven world, Excel has become an indispensable tool for professionals across industries. While most people are familiar with basic Excel functions, there is a whole world of advanced formulas that can help streamline your workflow and increase efficiency. In this article, we will explore some of the most powerful and time-saving advanced Excel formulas that can take your data analysis to the next level.
## VLOOKUP: Finding Data Made Easy
One of the most widely used advanced Excel formulas is VLOOKUP. This function allows you to search for a specific value in a table and return a corresponding value from another column. VLOOKUP is particularly useful when dealing with large datasets where finding specific information manually would be time-consuming.
To use VLOOKUP, you need to specify four arguments: lookup_value (the value you want to find), table_array (the range of cells containing the data), col_index_num (the column number from which you want to retrieve the result), and range_lookup (a logical value indicating whether you want an exact match or an approximate match).
For example, if you have a sales data table and want to find the total sales for a particular product, you can use VLOOKUP to quickly retrieve that information without manually searching through rows of data.
## INDEX MATCH: A Dynamic Duo
While VLOOKUP is powerful, it does have limitations. One major limitation is that it only searches for values in the leftmost column of a table. This is where INDEX MATCH comes in handy.
INDEX MATCH allows you to search for values in any column within a table and returns a corresponding value from another column. The formula consists of two parts: INDEX returns the value at a specified row and column within an array, while MATCH searches for a specified value within an array and returns its position.
The flexibility of INDEX MATCH makes it extremely useful in situations where you need to perform advanced lookups or when the column you want to search for is not the leftmost column.
## SUMIFS: Summing Up with Conditions
When dealing with large datasets, you often need to sum up values that meet certain criteria. SUMIFS is an advanced Excel formula that allows you to add up values based on multiple conditions.
With SUMIFS, you can specify multiple criteria and corresponding ranges, and Excel will only consider the values that meet all of the specified conditions. This function is particularly useful when you want to analyze data based on specific parameters, such as summing up sales for a particular region within a certain time frame.
By using SUMIFS, you can avoid manual calculations and quickly obtain accurate results even from complex datasets.
## CONCATENATE: Combining Text Strings
In some cases, you may need to combine text strings from different cells into a single cell. This is where CONCATENATE comes in handy. CONCATENATE allows you to join multiple text strings together.
To use CONCATENATE, simply specify the cells or text strings you want to combine within the formula. You can also add additional text or symbols between the strings by including them within quotation marks.
For example, if you have a list of first names and last names in separate columns and want to create a full name column, CONCATENATE can help you quickly merge these two columns into one.
In conclusion, advanced Excel formulas like VLOOKUP, INDEX MATCH, SUMIFS, and CONCATENATE can greatly enhance your data analysis capabilities and streamline your workflow. By familiarizing yourself with these formulas and incorporating them into your daily tasks, you can save time and effort while ensuring accurate results. So why not take advantage of these powerful tools and become an Excel expert?
This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability. | 750 | 3,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.854433 |
https://www.edumple.com/cbse-class-6/mathmatics/chapter-9-data-handling5/notes | 1,722,729,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00808.warc.gz | 614,313,107 | 16,016 | ## 1. Data Handling
Introduction
Data :
Data is the collection of information or facts . It may be a collection of numbers , figures ,words ,or symbols.
Raw Data :
The data that has been collected in the original form is called raw data.
For example :
The marks obtained by 10 students in the Maths exam are 60, 40, 35, 27, 67, 72, 40, 37, 45, 50. This collection of data is raw data.
Type of Data
Thus, primary data are original whereas secondary data are not original.
Organization of Data
To draw a meaningful conclusion from the data, the data must be organized. Understanding and analyzing the data becomes easier when it is represented by pictures or tables.
Ex.: A teacher wants to know the choice of food of each student as part of the mid-day meal program. The teacher assigns the task of collecting this information to Maria. Maria does so using paper and a pencil. After arranging the choices in a column, she puts against a choice of food one ( | ) mark for every student making that choice.
Umesh, after seeing the table suggested a better method to count the students. He asked Maria to organize the marks ( | ) in a group of ten as shown below :
Rajan made it simpler by asking her to make groups of five instead of ten, as shown below:
Frequency Distribution
The number of times a particular observation occurs is called its frequency. In a class marks obtained by 20 students are as follows : 1, 2, 3, 4, 5, 4, 6, 6, 8, 9, 10, 2, 3, 3, 5, 3, 6, 8, 8, 7
This row data can be represented in meaning full manner.
## 1. Data Handling
Chapter 9
Data Handling
Introduction to data,
Data handling is appertained to the procedure done to organize the information handed in order to perform fine operations on them.
Scoring or marks, measuring height/ weight, calculating distance travelled are all data for different situations. In our day-to-day life, raw information or data, in terms of volume or numbers is used regularly.
Below are a many examples.
Classification of data
Tally mark,
Pictographs,
It is a representation in terms of pictures/symbols.
.
Bar graph
Information is displayed in terms of bars of equal width and equal gap in between. The heights are proportional to the values they represent.
Classification of bar graphs,
Double Bar Graph
Information is displayed for two sets of data in terms of bars together. This is used for comparison of data.
Multiple Bar Graphs
Multiple bar graphs are a bar graph, which is used for comparing more than one kind of information.
Example:
Construct simple bar graph
In the bar graph, the information represented by the bars of the same width with equal gaps but the height of the bars represents the respective values.
## Representation of Data
Representation of Data
We have learnt how to collect the information and how to organize it using a frequency table. To represent the study of this data, we can draw graphs for this data.
Pictographs and bar graphs are simple ways of representing this data.
1. Pictograph : A pictograph represents data through pictures of objects. It helps answer the questions on the data at a glance. We represent the data using pictures as following example :
Ex. A cupboard has five compartments. In each compartment a row of books is arranged. The details are indicated in the adjoining table :
By seeing above pictograph we can easily get information that Row 4 has the greatest number of books. Row 3 has the least number of books and no Row which does not have books.
Interpretation of a Pictograph :
Ex.1 The following pictograph shows the number of absentees in a class of 30 students during the previous week :
(a) On which day were the maximum number of students absent?
(b) Which day had full attendance?
(c) What was the total number of absentees in that week?
Sol : (a)Maximum absentees were on saturday. (There are 8 pictures in the row for saturday; on all other days, the number of pictures are less).
(b) Against thursday, there is no picture, i.e. no one is absent. Thus, on that day the class had full attendance.
(c) There are 20 pictures in all. So, the total number of absentees in that week was 20.
Ex.2 A survey was carried out on 30 students of class VI in a school. Data about the different modes of transport used by them to travel to school was displayed as pictograph. What can you conclude from the pictograph?
Sol. From the pictograph we find that:
(a) The number of students coming by private car is 4.
(b) Maximum number of students use the school bus. This is the most popular way.
(c) Cycle is used by only three students.
(d) The number of students using the other modes can be similarly found.
Drawing a Pictograph :
Ex.3 The following are the number of electric bulbs purchased for a lodging house during the first four months of a year.
2. Bar Graph: Sometimes representing data by pictograph is difficult due to large data. There is another way of representing data visually is Bar Graph. Bars of uniform width can be drawn horizontally or vertically with equal spacing between them and then the length of each bar represents the given number. Such a method of representing data is called a bar diagram or a bar graph.
## Construction of a bar graph
Construction of a bar graph :
Steps :
1. On the graph paper draw a horizontal line (x-axis) and a vertical line (y-axis)
2. Along the horizontal line, mark points at equal distances and write the names of the items for which the data is to be represented.
3. Choose a suitable scale which shall determine the height of the bars.
4. Locate the heights of different bars according to scale.
5. On x-axis draw bars of equal width according to the required height.
6. Keep the distance between the bars the same.
Drawing a bar graph : Let us draw a bar graph for this data.
First of all draw a horizontal line and a vertical line. On the horizontal line we will draw bars representing each fruit and on vertical line we will write numerals representing number of students.
Let us choose a scale. It means we first decide how many students will be represented by unit length of a bar.
Here, we take 1 unit length to represent 1 student only. We get a bar graph as shown in adjoining figure.
Ex. 4 : Read the bar graph given below and answer the following questions.
(i) What is the total number of students in all the clubs together ?
Sol. 20 + 30 +40 + 20 = 110 students
(ii) Which club has more students and how many ?
Sol. Drama club, 40 students
(iii) Which club have equal number of students ?
Sol. Cricket club and music club
(iv) How any students are there in hockey club ?
Sol. 30 students.
(v) What information is given in the bar graph ?
Sol. The bar graph represetns the number of members of different sports clubs.
## Average
Average
Average is equal to the sum of observation dividing by the number of observation.
Ex.1 Find the average of all prime numbers between 30 and 50.
Sol. There are five prime numbers between 30 and 50.
They are 31, 37, 41, 43, and 47
Ex.2 Find the average of first 5 multiples of 7. | 1,682 | 7,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-33 | latest | en | 0.923219 |
https://vikramlearning.com/jntuh/notes/ic-applications-and-hdl-simulation-lab/ic-565-pll-applications/302 | 1,716,592,522,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00374.warc.gz | 517,807,024 | 8,944 | # IC 565 - PLL Applications
Prerequisite: Basic operation of PLL and Pin diagram of IC 565.
Objective: To study the operation of NE565 PLL – with a given free running frequency.
Apparatuses:
• DC power supply
• CRO
• Function Generator
• Resistors 2.2kΩ, 10KΩ, 18KΩ – 1No. each
• Capacitor 10µF – 1No
• Capacitor 0.001µF – 2Nos
• IC565 - 1No
Theory:
The 565 IC is available as a14-pin DIP package. The output frequency of the VCO is fo = 1.2 /4 RTCT, where RT and CT are the external Resistor and Capacitor connected to pin 8 and pin 9. A value between 2k and 20k is recommended for RT. The VCO free running frequency is adjusted with RT and CT, so that it is at the centre of the input frequency range.
Circuit diagram:
Procedure:
1. Connect the circuit as shown in the figure.
2. Measure the free running frequency of VCO at pin 4 with the input signal Vin = zero. Compare it with the calculated value = 1.2/4RTCT
3. Now apply the input signal of 1Vpp square wave of 100Hz to pin 2. Connect channel 1 of the CRO to pin 2 and display this signal on the scope.
4. Gradually increase the input frequency till the PLL is locked to the input frequency.
5. This frequency fCL gives the lower end of the capture range.
6. Go on increase the input frequency, till PLL stops tracking the input signal. This frequency fLH gives the upper end of the lock range.
7. If the input frequency is increased further the loop will be in unlocked condition only.
8. Now gradually decrease the input frequency till the PLL is again get locked. This is the frequency fCH, the upper end of the capture range. Keep on decreasing the input frequency until the loop is unlocked. This frequency fLL gives the lower end of the lock range.
Expected Wave Forms:
Result:
fO Free running frequency:
fLL Lower Locking frequency:
fCL Lower capture frequency:
fCH Higher capture frequency:
fLH Higher Locking frequency:
Outcome: After conducting this experiment, students are able to understand the applications of PLL.
VIVA Questions:
1. What are the basic blocks of a PLL?
Ans: A phase detector, low pass filter, amplifier and a VCO in feedback loop.
2. Define VCO?
Ans: A voltage controlled oscillator is an oscillator circuit in which the frequency of oscillations is controlled by an externally applied voltage.
3. What is the formula for the free running frequency Fo of 565 PLL?
Ans:
4. What is PLL?
Ans: A PLL is a closed loop system designed to lock output frequency and phase to the frequency and phase of an input signal.
5. Define lock range?
Ans: When PLL is in lock, it can track frequency changes in the incoming signal. The range of frequencies over which the PLL can maintain lock with the incoming signal is called as lock range.
6. Define pull-in time?
Ans: The total time taken by the PLL to establish lock is called pull-in time.
7. Define capture range?
Ans: The range of frequencies over which the PLL can acquire lock with the input signal is called as capture range.
8. What is the function of the LPF in PLL?
Ans: Controls the capture range and lock range of PLL.
9. What are the applications of PLL?
Ans: Frequency multiplication / Division, Frequency translation, AM detection, FM Demodulation, FSK Demodulator etc.
10. Which is greater – lock in range or capture range?
Ans: Lock in Range.
• Updated
Oct 22, 2016
• Views
21,708
You May Like | 831 | 3,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.825071 |
https://brainly.com/question/355820 | 1,484,573,480,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00231-ip-10-171-10-70.ec2.internal.warc.gz | 808,259,773 | 8,586 | • Brainly User
2015-03-16T13:15:27-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
1/4 I think it is.
I hope I helped you out
That is incorrect, it is 1 1/6
• Brainly User
2015-03-16T13:18:51-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
1/2 + 2/3 = 3/6 + 4/6 = 7/6 = 1 1/6
Or
1/2 + 2/3= 7/6
You could simplify that to 1 1/6 | 247 | 861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-04 | latest | en | 0.952229 |
http://antoine.frostburg.edu/chem/senese/101/thermo/faq/print-energy-required-for-temperature-rise.shtml | 1,537,347,597,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156096.0/warc/CC-MAIN-20180919083126-20180919103126-00003.warc.gz | 18,038,118 | 2,244 | # How can I calculate the energy required to cause a temperature rise?
The specific heat of water is 1.0 cal/g°C. How much energy is required to raise the temperature of 25.1 g of water by 72.0 degree Celsius?
Becky
1. What is the unknown? Your problem asks "How much energy...?" The units aren't specified, but since you spot 'calories' in the given information, it will probably be easiest to get your answer in calories.
2. What information is given? You know that:
• the specific heat is 1.0 cal/g°C
• the mass of water is 25.1 g
• the temperature change is 72.0°C
3. How is the information in the problem related to the unknown? The relationship between heat and temperature change is
heat = mass × specific heat × temperature change
4. Do the math. Now you're ready to do the calculation itself:
heat = (25.1 g)(1.0 cal/g°C)(72.0°C)
5. Is the final answer in the correct units? You want only calories. All other units should 'cancel' in the calculation somewhere.
6. Is the size of the answer reasonable? Your reasoning should go something like this: The specific heat means that it takes 1.0 calories of heat to warm 1 g of water up by 1°C. So it should take 25.1 calories to warm up 25.1 g of water by 1°C. And it should take 72.0×25.1 calories to warm up 25.1 g of water by 72.0°C.
You can also check the answer by working the problem backwards. For example, take the calories just calculated and divide them by the temperature change and the grams of water; do you get a specific heat of 1.0 cal/g°C?
Author: Fred Senese senese@antoine.frostburg.edu
General Chemistry Online! How can I calculate the energy required to cause a temperature rise? | 427 | 1,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-39 | latest | en | 0.912245 |
https://discuss.codecademy.com/t/i-am-not-getting-output-one-of-the-following-is-missing-or-broken-when-we-tried-to-use-it-alice-lloyd-tyler-students-get-class-average-get-letter-grade/116880 | 1,537,596,603,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158045.57/warc/CC-MAIN-20180922044853-20180922065253-00129.warc.gz | 495,279,166 | 4,675 | # I am not getting output "One of the following is missing or broken when we tried to use it: alice, lloyd, tyler, students, get_class_average, get_letter_grade"
#1
lloyd = {
"name": "Lloyd",
"homework": [90.0, 97.0, 75.0, 92.0],
"quizzes": [88.0, 40.0, 94.0],
"tests": [75.0, 90.0]
}
alice = {
"name": "Alice",
"homework": [100.0, 92.0, 98.0, 100.0],
"quizzes": [82.0, 83.0, 91.0],
"tests": [89.0, 97.0]
}
tyler = {
"name": "Tyler",
"homework": [0.0, 87.0, 75.0, 22.0],
"quizzes": [0.0, 75.0, 78.0],
"tests": [100.0, 100.0]
}
def average(numbers):
total=sum(numbers)
total=float(total)/len(numbers)
def get_average(student):
homework=average(student["homework"])
quizzes=average(student["quizzes"])
tests=average(student["tests"])
``````return .1*homework + \
.3*quizzes + \
.6*tests``````
if score>=90:
return "A"
elif score>=80:
return "B"
elif score>=70:
return "C"
elif score>=60:
return "D"
else:
return "F"
students = ["lloyd", "alice", "tyler"]
def get_class_average(students):
results=[]
for student in students:
results.append(get_average(student))
return average(results)
q = get_class_average(students)
# printing class average
print q
#2
``students = ["lloyd", "alice", "tyler"]``
it contains strings, why? calling get_class_average:
``q = get_class_average(students)``
with the list containing strings results in an error
the list should contain the students like you build it in 3. put it together
#3
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 500 | 1,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-39 | latest | en | 0.757576 |
https://uk.mathworks.com/matlabcentral/answers/462560-iterate-simultaneous-equations-till-it-reach-convergence?s_tid=prof_contriblnk | 1,606,769,823,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00456.warc.gz | 519,877,999 | 22,901 | # Iterate simultaneous equations till it reach convergence
1 view (last 30 days)
Senthil on 17 May 2019
Answered: Senthil on 17 May 2019
Question: How to iterate T_R, T_f and T_a for multiple times till T_f reaches convergence.
Code below:
Q_R=935.9432834;
h_f=2500;
A_f=0.017671459;
A_a=0.017671459;
h_cs=13.3;
L_c=0.15;
h_rs=8.8;
R_c=0.075;
T_amb=35;
E_bm=0.05;
SIGMA=5.67E-08;
T_sky=29;
h_a=10;
h_c=10;
h_ct=10;
m_f=2;
c_f=4186;
A_s=0.070685835;
A_top=0.017671459;
hct=8.8;
t_lid=0.01;
k_lid=237;
m_a=0.000797131;
c_a=718;
DELTAt=1;
m_R=0.5;
c_R=910;
T_R0=25; %initial condition"
T_f0=25; %initial condition"
T_a0=25; %initial condition"
T_R=(Q_R-(h_f*A_f)*(T_R0-T_f0)-((h_cs*2*L_c)+(h_rs*R_c))*3.14*R_c*(T_R0-T_amb)-(2*L_c+R_c)*(3.14*R_c*E_bm*SIGMA*(T_R0^4-T_sky^4))/(T_R0-T_amb))*(DELTAt)/(m_R*c_R)+T_R0;
T_f=(((h_f*A_f*(T_R0-T_f0))-(h_c*A_a*(T_f0-T_a0)))*DELTAt/(m_f*c_f))+T_f0;
T_a=(((h_a*A_a)*(T_f0-T_a0))+((h_a*A_s)*(T_R0-T_a0))-(A_top*(T_a0-T_amb))/(1/(h_ct+((E_bm*SIGMA*(T_R0^4-T_sky^4))/(T_a0-T_amb)))+(t_lid/k_lid)))*(DELTAt)/(m_a*c_a)+T_a0;
Expected Output Required:
Here the output is generated for 20 time steps. But we need to iterate till the difference between T_f upto 0.01 or 0.02.
#### 1 Comment
KSSV on 17 May 2019
Where is iteration in the code?
Senthil on 17 May 2019
Actually i need to integrate the iteration code in the above program. | 598 | 1,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-50 | latest | en | 0.57382 |
https://essaychimp.com/2017/10/16/a-certain-type-of-bacteria-has-been-incubating-within-infect-67/ | 1,586,474,551,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00498.warc.gz | 444,332,836 | 13,708 | # A certain type of bacteria has been incubating within infect
A certain type of bacteria has been incubating within infected tissues. Let P(t) denote the bacte… Show more A certain type of bacteria has been incubating within infected tissues. Let P(t) denote the bacteria population, P>or equal to 0, as a function of time(measured in hours), per unit volume of tissues. Let t=0 to be the time that the patient’s immune system begins responding to the antigens. Suppose the bacteria’s growth is continuous with a rate of growth equals twice of its present polulation size, and that the patient’s immune system is able to reduce P by 9000, per hour. THe progress of the infection can be modeled by the first order differential equation for (t>0): P’=2P-9000 (a) Does the equation have an equilibrium solution? If it does, find it and classify its stability. (b) Suppose P(0)=6000, per hour. what is the patient’s prognosis? (c)Proper vaccination would provide the host with a stockpile of antibodies keyed to the specific antigens, thereby speeds up the immune system’s response time. The immune system would, therefore, responds while P(t) is at a lower count than otherwise. Suppose the immune system of an infected, but vaccinated, patient responds at P(0)=3000, what is this patient’s prognosis? (d) A proplerly applied antibiotics regimen will increase the rate of reduction of P to 15000per hour. Under such treatment, the equation describing P becomes P’=2P-15000. Once again, suppose P(0) = 6000, what is the patient’s prognosis? • Show less
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or | 1,028 | 4,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-16 | latest | en | 0.916029 |
http://math.stackexchange.com/questions/62600/how-to-solve-this-problem-using-calculus | 1,469,455,889,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824230.71/warc/CC-MAIN-20160723071024-00123-ip-10-185-27-174.ec2.internal.warc.gz | 156,901,035 | 18,934 | # How to solve this problem using Calculus?
During a homework problem I was supposed to solve this using Calculus. Here's the problem:
A cow car is traveling at $50 \ \text{mi}/\text{h}$ when the brakes are fully applied, producing a constant deceleration of $26 \ \text{ft}/\text{s}^2$. What is the distance traveled before the car comes to a stop?
So, I just solved this with Physics equations and came up with $103.4 \ \text{ft}$. However, I guess I'm supposed to use Calculus, seeing this is a Calculus class. ;) Any ideas how to do this?
I guess that $s''(t) = -26 \ \text{ft}/\text{s}^2$.
-
Cows have brakes? :) – J. M. Sep 7 '11 at 16:29
Clarification to @all: ft/s is the unit feet per sec. I.e., the s is not the same as the distance $s$. – Srivatsan Sep 7 '11 at 16:30
@Ramp: The second derivative denotes acceleration, not velocity; the units of $s''(t)$ would be feet per second square, not feet per second. – Arturo Magidin Sep 7 '11 at 16:31
Can you show that applying your calculus knowledge to the equation you have quoted gives you the physics equation you have used to solve the problem [integrate twice and be careful with constants] – Mark Bennet Sep 7 '11 at 16:31
@Mark: Wouldn't s(t)=-13/3 t^3 ft/s^2 + cx + d? – Ramp Sep 7 '11 at 16:37
Let $v(t)$ be the velocity. You are told that $v(0)=50$ mi/h; you are also told that when brakes are applied, acceleration, $v'(t)$, is $-26\ \mathrm{ft}/\mathrm{sec}^2$. It's better to put everything in the same units, so let's change velocity to feet per second: $$50\frac{\mathrm{mi}}{\mathrm{hr}} = 50\left(\frac{5280\text{ feet}}{1\text{ mi}}\right)\left(\frac{1\text{ hr}}{3600\text{ sec}}\right) = \frac{220}{3}\text{ft}/\text{sec}.$$
So $v(0) = \frac{220}{3}$ ft/sec. $v'(t) = -26$ ft/sec${}^2$.
By the First Fundamental Theorem of Calculus, we have: $$v(t) -v(0) = \int_0^t v'(x)\,dx = \int_0^t(-26)\,dx = -26x\Bigm|_0^t = -26t.$$ Therefore, $v(t) = v(0) - 26t = \frac{220}{3}-26t$.
The car comes to a stop when $v(t)=0$. Solving $v(t)=0$ we get $$t = \frac{220}{78} = \frac{110}{39}\text{ seconds.}$$ How far did it travel? The distance traveled is the integral of the velocity, so \begin{align*} \text{distance traveled} &= \int_0^{110/39}v(t)\,dt\\ &= \int_0^{110/39}\left(\frac{220}{3} - 26t\right)\,dt \\ &= \frac{220}{3}t - 13t^2\Bigm|_0^{110/39}\\ &= \frac{24200}{117} - \frac{157300}{1521}\\ &\approx 103.4188\text{ feet.} \end{align*}
-
Oh snap I haven't gotten to the fundamental theorem of Calculus yet. Thanks for your help, though. – Ramp Sep 7 '11 at 16:47
For constant acceleration, calculus only justifies the usual equation $s=x_0+v_0t+\frac{1}{2}at^2$. With $a=-26, v_0=\frac{220}{3}, x_0=0$ you get $s=\frac{v_0^2}{2|a|}$. I agree with your answer.
-
Thanks for agreeing. But I need to know how to solve it with Calculus. ;) – Ramp Sep 7 '11 at 16:50 | 997 | 2,850 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-30 | latest | en | 0.873213 |
http://www.mathworks.com/help/stats/fitlmematrix.html?s_tid=gn_loc_drop&requestedDomain=www.mathworks.com&nocookie=true | 1,500,628,479,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423764.11/warc/CC-MAIN-20170721082219-20170721102219-00115.warc.gz | 481,567,838 | 29,605 | # Documentation
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# fitlmematrix
Fit linear mixed-effects model
## Syntax
lme = fitlmematrix(X,y,Z,[])
lme = fitlmematrix(X,y,Z,G)
lme = fitlmematrix(___,Name,Value)
## Description
example
lme = fitlmematrix(X,y,Z,[]) creates a linear mixed-effects model of the responses y using the fixed-effects design matrix X and random-effects design matrix or matrices in Z. [] implies that there is one group. That is, the grouping variable G is ones(n,1), where n is the number of observations. Using fitlmematrix(X,Y,Z,[]) without a specified covariance pattern most likely results in a nonidentifiable model. This syntax is recommended only if you build the grouping information into the random effects design Z and specify a covariance pattern for the random effects using the 'CovariancePattern' name-value pair argument.
example
lme = fitlmematrix(X,y,Z,G) creates a linear mixed-effects model of the responses y using the fixed-effects design matrix X and random-effects design matrix Z or matrices in Z, and the grouping variable or variables in G.
example
lme = fitlmematrix(___,Name,Value) also creates a linear mixed-effects model with additional options specified by one or more Name,Value pair arguments, using any of the previous input arguments. For example, you can specify the names of the response, predictor, and grouping variables. You can also specify the covariance pattern, fitting method, or the optimization algorithm.
## Examples
collapse all
Load the sample data.
load carsmall
Fit a linear mixed-effects model, where miles per gallon (MPG) is the response, weight is the predictor variable, and the intercept varies by model year. First, define the design matrices. Then, fit the model using the specified design matrices.
y = MPG; X = [ones(size(Weight)), Weight]; Z = ones(size(y)); lme = fitlmematrix(X,y,Z,Model_Year)
lme = Linear mixed-effects model fit by ML Model information: Number of observations 94 Fixed effects coefficients 2 Random effects coefficients 3 Covariance parameters 2 Formula: y ~ x1 + x2 + (z11 | g1) Model fit statistics: AIC BIC LogLikelihood Deviance 486.09 496.26 -239.04 478.09 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'x1' 43.575 2.3038 18.915 92 1.8371e-33 'x2' -0.0067097 0.0004242 -15.817 92 5.5373e-28 Lower Upper 39 48.151 -0.0075522 -0.0058672 Random effects covariance parameters (95% CIs): Group: g1 (3 Levels) Name1 Name2 Type Estimate Lower Upper 'z11' 'z11' 'std' 3.301 1.4448 7.5421 Group: Error Name Estimate Lower Upper 'Res Std' 2.8997 2.5075 3.3532
Now, fit the same model by building the grouping into the Z matrix.
Z = double([Model_Year==70, Model_Year==76, Model_Year==82]); lme = fitlmematrix(X,y,Z,[],'Covariancepattern','Isotropic')
lme = Linear mixed-effects model fit by ML Model information: Number of observations 94 Fixed effects coefficients 2 Random effects coefficients 3 Covariance parameters 2 Formula: y ~ x1 + x2 + (z11 + z12 + z13 | g1) Model fit statistics: AIC BIC LogLikelihood Deviance 486.09 496.26 -239.04 478.09 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'x1' 43.575 2.3038 18.915 92 1.8371e-33 'x2' -0.0067097 0.0004242 -15.817 92 5.5373e-28 Lower Upper 39 48.151 -0.0075522 -0.0058672 Random effects covariance parameters (95% CIs): Group: g1 (1 Levels) Name1 Name2 Type Estimate Lower Upper 'z11' 'z11' 'std' 3.301 1.4448 7.5421 Group: Error Name Estimate Lower Upper 'Res Std' 2.8997 2.5075 3.3532
Load the sample data.
load(fullfile(matlabroot,'examples','stats','weight.mat'));
weight contains data from a longitudinal study, where 20 subjects are randomly assigned 4 exercise programs (A, B, C, D) and their weight loss is recorded over six 2-week time periods. This is simulated data.
Define Subject and Program as categorical variables. Create the design matrices for a linear mixed-effects model, with the initial weight, type of program, week, and the interaction between the week and type of program as the fixed effects. The intercept and coefficient of week vary by subject.
This model corresponds to
where = 1, 2, ..., 120, and = 1, 2, ..., 20. are the fixed-effects coefficients, = 0, 1, ..., 8, and and are random effects. stands for initial weight and is a dummy variable representing a type of program. For example, is the dummy variable representing program type B. The random effects and observation error have the following prior distributions:
Subject = nominal(Subject); Program = nominal(Program); D = dummyvar(Program); % Create dummy variables for Program X = [ones(120,1), InitialWeight, D(:,2:4), Week,... D(:,2).*Week, D(:,3).*Week, D(:,4).*Week]; Z = [ones(120,1), Week]; G = Subject;
Since the model has an intercept, you only need the dummy variables for programs B, C, and D. This is also known as the 'reference' method of coding dummy variables.
Fit the model using fitlmematrix with the defined design matrices and grouping variables.
lme = fitlmematrix(X,y,Z,G,'FixedEffectPredictors',... {'Intercept','InitWeight','PrgB','PrgC','PrgD','Week','Week_PrgB','Week_PrgC','Week_PrgD'},... 'RandomEffectPredictors',{{'Intercept','Week'}},'RandomEffectGroups',{'Subject'})
lme = Linear mixed-effects model fit by ML Model information: Number of observations 120 Fixed effects coefficients 9 Random effects coefficients 40 Covariance parameters 4 Formula: Linear Mixed Formula with 10 predictors. Model fit statistics: AIC BIC LogLikelihood Deviance -22.981 13.257 24.49 -48.981 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'Intercept' 0.66105 0.25892 2.5531 111 0.012034 'InitWeight' 0.0031879 0.0013814 2.3078 111 0.022863 'PrgB' 0.36079 0.13139 2.746 111 0.0070394 'PrgC' -0.033263 0.13117 -0.25358 111 0.80029 'PrgD' 0.11317 0.13132 0.86175 111 0.39068 'Week' 0.1732 0.067454 2.5677 111 0.011567 'Week_PrgB' 0.038771 0.095394 0.40644 111 0.68521 'Week_PrgC' 0.030543 0.095394 0.32018 111 0.74944 'Week_PrgD' 0.033114 0.095394 0.34713 111 0.72915 Lower Upper 0.14798 1.1741 0.00045067 0.0059252 0.10044 0.62113 -0.29319 0.22666 -0.14706 0.3734 0.039536 0.30686 -0.15026 0.2278 -0.15849 0.21957 -0.15592 0.22214 Random effects covariance parameters (95% CIs): Group: Subject (20 Levels) Name1 Name2 Type Estimate Lower 'Intercept' 'Intercept' 'std' 0.18407 0.12281 'Week' 'Intercept' 'corr' 0.66841 0.21077 'Week' 'Week' 'std' 0.15033 0.11004 Upper 0.27587 0.88573 0.20537 Group: Error Name Estimate Lower Upper 'Res Std' 0.10261 0.087882 0.11981
Examine the fixed effects coefficients table. The row labeled 'InitWeight' has a -value of 0.0228, and the row labeled 'Week' has a -value of 0.0115. These -values indicate significant effects of the initial weights of the subjects and the time factor in the amount of weight lost. The weight loss of subjects who are in program B is significantly different relative to the weight loss of subjects who are in program A. The lower and upper limits of the covariance parameters for the random effects do not include zero, thus they seem significant. You can also test the significance of the random-effects using the compare method.
Load the sample data.
load flu
The flu dataset array has a Date variable, and 10 variables for estimated influenza rates (in 9 different regions, estimated from Google® searches, plus a nationwide estimate from the Centers for Disease Control and Prevention, CDC).
To fit a linear-mixed effects model, where the influenza rates are the responses, combine the nine columns corresponding to the regions into an array that has a single response variable, FluRate, and a nominal variable, Region, the nationwide estimate WtdILI, that shows which region each estimate is from, and the grouping variable Date.
flu2 = stack(flu,2:10,'NewDataVarName','FluRate',... 'IndVarName','Region'); flu2.Date = nominal(flu2.Date);
Define the design matrices for a random-intercept linear mixed-effects model, where the intercept varies by Date. The corresponding model is
where is the observation for level of grouping variable Date, is the random effect for level of the grouping variable Date, and is the observation error for observation . The random effect has the prior distribution,
and the error term has the distribution,
y = flu2.FluRate; X = [ones(468,1) flu2.WtdILI]; Z = [ones(468,1)]; G = flu2.Date;
Fit the linear mixed-effects model.
lme = fitlmematrix(X,y,Z,G,'FixedEffectPredictors',{'Intercept','NationalRate'},... 'RandomEffectPredictors',{{'Intercept'}},'RandomEffectGroups',{'Date'})
lme = Linear mixed-effects model fit by ML Model information: Number of observations 468 Fixed effects coefficients 2 Random effects coefficients 52 Covariance parameters 2 Formula: y ~ Intercept + NationalRate + (Intercept | Date) Model fit statistics: AIC BIC LogLikelihood Deviance 286.24 302.83 -139.12 278.24 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'Intercept' 0.16385 0.057525 2.8484 466 0.0045885 'NationalRate' 0.7236 0.032219 22.459 466 3.0502e-76 Lower Upper 0.050813 0.27689 0.66028 0.78691 Random effects covariance parameters (95% CIs): Group: Date (52 Levels) Name1 Name2 Type Estimate Lower 'Intercept' 'Intercept' 'std' 0.17146 0.13227 Upper 0.22226 Group: Error Name Estimate Lower Upper 'Res Std' 0.30201 0.28217 0.32324
The confidence limits of the standard deviation of the random-effects term , do not include zero (0.13227, 0.22226), which indicates that the random-effects term is significant. You can also test the significance of the random-effects using compare method.
The estimated value of an observation is the sum of the fixed-effects values and value of the random effect at the grouping variable level corresponding to that observation. For example, the estimated flu rate for observation 28
where is the best linear unbiased predictor (BLUP) of the random effects for the intercept. You can compute this value as follows.
beta = fixedEffects(lme); [~,~,STATS] = randomEffects(lme); % compute the random effects statistics STATS STATS.Level = nominal(STATS.Level); y_hat = beta(1) + beta(2)*flu2.WtdILI(28) + STATS.Estimate(STATS.Level=='10/30/2005')
y_hat = 1.4674
You can simply display the fitted value using the fitted(lme) method.
F = fitted(lme); F(28)
ans = 1.4674
Load the sample data.
load(fullfile(matlabroot,'examples','stats','shift.mat'));
The data shows the deviations from the target quality characteristic measured from the products that five operators manufacture during three shifts: morning, evening, and night. This is a randomized block design, where the operators are the blocks. The experiment is designed to study the impact of the time of shift on the performance. The performance measure is the deviations of the quality characteristics from the target value. This is simulated data.
Define the design matrices for a linear mixed-effects model with a random intercept grouped by operator, and shift as the fixed effects. Use the 'effects' contrasts. 'effects' contrasts mean that the coefficients sum to 0. You need to create two contrast coded variables in the fixed-effects design matrix, X1 and X2, where
The model corresponds to
where represents the observations, and represents the operators, = 1, 2, ..., 15, and = 1, 2, ..., 5. The random effects and the observation error have the following distributions:
and
S = shift.Shift; X1 = (S=='Morning') - (S=='Night'); X2 = (S=='Evening') - (S=='Night'); X = [ones(15,1), X1, X2]; y = shift.QCDev; Z = ones(15,1); G = shift.Operator;
Fit a linear mixed-effects model using the specified design matrices and restricted maximum likelihood method.
lme = fitlmematrix(X,y,Z,G,'FitMethod','REML','FixedEffectPredictors',.... {'Intercept','S_Morning','S_Evening'},'RandomEffectPredictors',{{'Intercept'}},... 'RandomEffectGroups',{'Operator'},'DummyVarCoding','effects')
lme = Linear mixed-effects model fit by REML Model information: Number of observations 15 Fixed effects coefficients 3 Random effects coefficients 5 Covariance parameters 2 Formula: y ~ Intercept + S_Morning + S_Evening + (Intercept | Operator) Model fit statistics: AIC BIC LogLikelihood Deviance 58.913 61.337 -24.456 48.913 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'Intercept' 3.6525 0.94109 3.8812 12 0.0021832 'S_Morning' -0.91973 0.31206 -2.9473 12 0.012206 'S_Evening' -0.53293 0.31206 -1.7078 12 0.11339 Lower Upper 1.6021 5.703 -1.5997 -0.23981 -1.2129 0.14699 Random effects covariance parameters (95% CIs): Group: Operator (5 Levels) Name1 Name2 Type Estimate Lower 'Intercept' 'Intercept' 'std' 2.0457 0.98207 Upper 4.2612 Group: Error Name Estimate Lower Upper 'Res Std' 0.85462 0.52357 1.395
Compute the best linear unbiased predictor (BLUP) estimates of random effects.
B = randomEffects(lme)
B = 0.5775 1.1757 -2.1715 2.3655 -1.9472
The estimated deviation from the target quality characteristics for the third operator working the evening shift is
You can also display this value as follows.
F = fitted(lme); F(shift.Shift=='Evening' & shift.Operator=='3')
ans = 0.9481
Load the sample data.
load carbig
Fit a linear mixed-effects model for miles per gallon (MPG), with fixed effects for acceleration and horsepower, and uncorrelated random effect for intercept and acceleration grouped by the model year. This model corresponds to
with the random-effects terms having the following prior distributions:
where represents the model year.
First, prepare the design matrices for fitting the linear mixed-effects model.
X = [ones(406,1) Acceleration Horsepower]; Z = {ones(406,1),Acceleration}; G = {Model_Year,Model_Year}; Model_Year = nominal(Model_Year);
Now, fit the model using fitlmematrix with the defined design matrices and grouping variables.
lme = fitlmematrix(X,MPG,Z,G,'FixedEffectPredictors',.... {'Intercept','Acceleration','Horsepower'},'RandomEffectPredictors',... {{'Intercept'},{'Acceleration'}},'RandomEffectGroups',{'Model_Year','Model_Year'})
lme = Linear mixed-effects model fit by ML Model information: Number of observations 392 Fixed effects coefficients 3 Random effects coefficients 26 Covariance parameters 3 Formula: Linear Mixed Formula with 4 predictors. Model fit statistics: AIC BIC LogLikelihood Deviance 2194.5 2218.3 -1091.3 2182.5 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'Intercept' 49.839 2.0518 24.291 389 5.6168e-80 'Acceleration' -0.58565 0.10846 -5.3995 389 1.1652e-07 'Horsepower' -0.16534 0.0071227 -23.213 389 1.9755e-75 Lower Upper 45.806 53.873 -0.7989 -0.3724 -0.17934 -0.15133 Random effects covariance parameters (95% CIs): Group: Model_Year (13 Levels) Name1 Name2 Type Estimate Lower 'Intercept' 'Intercept' 'std' 8.6657e-07 NaN Upper NaN Group: Model_Year (13 Levels) Name1 Name2 Type Estimate Lower 'Acceleration' 'Acceleration' 'std' 0.18783 0.12523 Upper 0.28172 Group: Error Name Estimate Lower Upper 'Res Std' 3.7258 3.4698 4.0007
Note that the random effects covariance parameters for intercept and acceleration are separate in the display. The standard deviation of the random effect for the intercept does not seem significant.
Refit the model with potentially correlated random effects for intercept and acceleration. In this case, the random-effects terms has this prior distribution
where represents the model year.
First, prepare the random-effects design matrix and grouping variable.
Z = [ones(406,1) Acceleration]; G = Model_Year; lme = fitlmematrix(X,MPG,Z,G,'FixedEffectPredictors',.... {'Intercept','Acceleration','Horsepower'},'RandomEffectPredictors',... {{'Intercept','Acceleration'}},'RandomEffectGroups',{'Model_Year'})
lme = Linear mixed-effects model fit by ML Model information: Number of observations 392 Fixed effects coefficients 3 Random effects coefficients 26 Covariance parameters 4 Formula: Linear Mixed Formula with 4 predictors. Model fit statistics: AIC BIC LogLikelihood Deviance 2193.5 2221.3 -1089.7 2179.5 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'Intercept' 50.133 2.2652 22.132 389 7.7727e-71 'Acceleration' -0.58327 0.13394 -4.3545 389 1.7075e-05 'Horsepower' -0.16954 0.0072609 -23.35 389 5.188e-76 Lower Upper 45.679 54.586 -0.84661 -0.31992 -0.18382 -0.15527 Random effects covariance parameters (95% CIs): Group: Model_Year (13 Levels) Name1 Name2 Type Estimate 'Intercept' 'Intercept' 'std' 3.3475 'Acceleration' 'Intercept' 'corr' -0.87971 'Acceleration' 'Acceleration' 'std' 0.33789 Lower Upper 1.2862 8.7119 -0.98501 -0.29675 0.1825 0.62558 Group: Error Name Estimate Lower Upper 'Res Std' 3.6874 3.4298 3.9644
Note that the random effects covariance parameters for intercept and acceleration are together in the display, with an addition of the correlation between the intercept and acceleration. The confidence intervals for the standard deviations and the correlation between the random effects for intercept and acceleration do not include 0s, hence they seem significant. You can compare these two models using the LinearMixedModel.compare method.
Load the sample data.
load(fullfile(matlabroot,'examples','stats','weight.mat'));
weight contains data from a longitudinal study, where 20 subjects are randomly assigned 4 exercise programs, and their weight loss is recorded over six 2-week time periods. This is simulated data.
Define Subject and Program as categorical variables.
Subject = nominal(Subject); Program = nominal(Program);
Create the design matrices for a linear mixed-effects model, with the initial weight, type of program, and week as the fixed effects.
D = dummyvar(Program); X = [ones(120,1), InitialWeight, D(:,2:4), Week]; Z = [ones(120,1) Week]; G = Subject;
This model corresponds to
where = 1, 2, ..., 120, and = 1, 2, ..., 20.
are the fixed-effects coefficients, = 0, 1, ...,8, and and are random effects. stands for initial weight and is a dummy variable representing a type of program. For example, is the dummy variable representing program type B. The random effects and observation error have the following prior distributions:
Fit the model using fitlmematrix with the defined design matrices and grouping variables. Assume the repeated observations collected on a subject have common variance along diagonals.
lme = fitlmematrix(X,y,Z,G,'FixedEffectPredictors',... {'Intercept','InitWeight','PrgB','PrgC','PrgD','Week'},... 'RandomEffectPredictors',{{'Intercept','Week'}},... 'RandomEffectGroups',{'Subject'},'CovariancePattern','Isotropic')
lme = Linear mixed-effects model fit by ML Model information: Number of observations 120 Fixed effects coefficients 6 Random effects coefficients 40 Covariance parameters 2 Formula: Linear Mixed Formula with 7 predictors. Model fit statistics: AIC BIC LogLikelihood Deviance -24.783 -2.483 20.391 -40.783 Fixed effects coefficients (95% CIs): Name Estimate SE tStat DF pValue 'Intercept' 0.4208 0.28169 1.4938 114 0.13799 'InitWeight' 0.0045552 0.0015338 2.9699 114 0.0036324 'PrgB' 0.36993 0.12119 3.0525 114 0.0028242 'PrgC' -0.034009 0.1209 -0.28129 114 0.77899 'PrgD' 0.121 0.12111 0.99911 114 0.31986 'Week' 0.19881 0.037134 5.3538 114 4.5191e-07 Lower Upper -0.13723 0.97883 0.0015168 0.0075935 0.12986 0.61 -0.27351 0.2055 -0.11891 0.36091 0.12525 0.27237 Random effects covariance parameters (95% CIs): Group: Subject (20 Levels) Name1 Name2 Type Estimate Lower 'Intercept' 'Intercept' 'std' 0.16561 0.12896 Upper 0.21269 Group: Error Name Estimate Lower Upper 'Res Std' 0.10272 0.088014 0.11987
## Input Arguments
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Fixed-effects design matrix, specified as an n-by-p matrix, where n is the number of observations, and p is the number of fixed-effects predictor variables. Each row of X corresponds to one observation, and each column of X corresponds to one variable.
Data Types: single | double
Response values, specified as an n-by-1 vector, where n is the number of observations.
Data Types: single | double
Random-effects design, specified as either of the following.
• If there is one random-effects term in the model, then Z must be an n-by-q matrix, where n is the number of observations and q is the number of variables in the random-effects term.
• If there are R random-effects terms, then Z must be a cell array of length R. Each cell of Z contains an n-by-q(r) design matrix Z{r}, r = 1, 2, ..., R, corresponding to each random-effects term. Here, q(r) is the number of random effects term in the rth random effects design matrix, Z{r}.
Data Types: single | double | cell
Grouping variable or variables, specified as either of the following.
• If there is one random-effects term, then G must be an n-by-1 vector corresponding to a single grouping variable with M levels or groups.
G can be a categorical vector, numeric vector, character array, or cell array of character vectors.
• If there are multiple random-effects terms, then G must be a cell array of length R. Each cell of G contains a grouping variable G{r}, r = 1, 2, ..., R, with M(r) levels.
G{r} can be a categorical vector, numeric vector, character array, or cell array of character vectors.
Data Types: single | double | char | cell
### Name-Value Pair Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside single quotes (' '). You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'CovariancePattern','Diagonal','DummyVarCoding','full','Optimizer','fminunc' specifies a random-effects covariance pattern with zero off-diagonal elements, creates a dummy variable for each level of a categorical variable, and uses the fminunc optimization algorithm.
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Names of columns in the fixed-effects design matrix X, specified as the comma-separated pair consisting of 'FixedEffectPredictors' and a cell array of length p.
For example, if you have a constant term and two predictors, say TimeSpent and Gender, where Female is the reference level for Gender, as the fixed effects, then you can specify the names of your fixed effects in the following way. Gender_Male represents the dummy variable you must create for category Male. You can choose different names for these variables.
Example: 'FixedEffectPredictors',{'Intercept','TimeSpent','Gender_Male'},
Data Types: cell
Names of columns in the random-effects design matrix or cell array Z, specified as the comma-separated pair consisting of 'RandomEffectPredictors' and either of the following:
• A cell array of length q when Z is an n-by-q design matrix. In this case, the default is {'z1','z2',...,'zQ'}.
• A cell array of length R, when Z is a cell array of length R with each element Z{r} of length q(r), r = 1, 2, ..., R. In this case, the default is {'z11','z12',...,'z1Q(1)'},...,{'zr1','zr2',...,'zrQ(r)'}.
For example, suppose you have correlated random effects for intercept and a variable named Acceleration. Then, you can specify the random-effects predictor names as follows.
Example: 'RandomEffectPredictors',{'Intercept','Acceleration'}
If you have two random effects terms, one for the intercept and the variable Acceleration grouped by variable g1, and the second for the intercept, grouped by the variable g2, then you specify the random-effects predictor names as follows.
Example: 'RandomEffectPredictors',{{'Intercept','Acceleration'},{'Intercept'}}
Data Types: cell
Name of response variable, specified as the comma-separated pair consisting of 'ResponseVarName' and a character vector.
For example, if your response variable name is score, then you can specify it as follows.
Example: 'ResponseVarName','score'
Data Types: char
Names of random effects grouping variables, specified as the comma-separated pair 'RandomEffectGroups' and either of the following:
• Character vector — If there is only one random-effects term, that is, if G is a vector, then the value of 'RandomEffectGroups' is a character vector containing the name for the grouping variable G. The default is 'g'.
• Cell array of character vectors — If there are multiple random-effects terms, that is, if G is a cell array of length R, then the value of 'RandomEffectGroups' is a cell array of length R, where each cell contains the name for the grouping variable G{r}. The default is {'g1','g2',...,'gR'}.
For example, if you have two random-effects terms, z1 and z2, grouped by the grouping variables sex and subject, then you can specify the names of your grouping variables as follows.
Example: 'RandomEffectGroups',{'sex','subject'}
Data Types: char | cell
Pattern of the covariance matrix of the random effects, specified as the comma-separated pair consisting of 'CovariancePattern' and a character vector, a square symmetric logical matrix, or a cell array of character vectors or logical matrices.
If there are R random-effects terms, then the value of 'CovariancePattern' must be a cell array of length R, where each element r of this cell array specifies the pattern of the covariance matrix of the random-effects vector associated with the rth random-effects term. The options for each element follow.
'FullCholesky' Default. Full covariance matrix using the Cholesky parameterization. fitlme estimates all elements of the covariance matrix. 'Full' Full covariance matrix, using the log-Cholesky parameterization. fitlme estimates all elements of the covariance matrix. 'Diagonal' Diagonal covariance matrix. That is, off-diagonal elements of the covariance matrix are constrained to be 0. $\left(\begin{array}{ccc}{\sigma }_{b1}^{2}& 0& 0\\ 0& {\sigma }_{b2}^{2}& 0\\ 0& 0& {\sigma }_{b3}^{2}\end{array}\right)$ 'Isotropic' Diagonal covariance matrix with equal variances. That is, off-diagonal elements of the covariance matrix are constrained to be 0, and the diagonal elements are constrained to be equal. For example, if there are three random-effects terms with an isotropic covariance structure, this covariance matrix looks like$\left(\begin{array}{ccc}{\sigma }_{b}^{2}& 0& 0\\ 0& {\sigma }_{b}^{2}& 0\\ 0& 0& {\sigma }_{b}^{2}\end{array}\right)$where σ2b is the common variance of the random-effects terms. 'CompSymm' Compound symmetry structure. That is, common variance along diagonals and equal correlation between all random effects. For example, if there are three random-effects terms with a covariance matrix having a compound symmetry structure, this covariance matrix looks like$\left(\begin{array}{ccc}{\sigma }_{b1}^{2}& {\sigma }_{b1,b2}& {\sigma }_{b1,b2}\\ {\sigma }_{b1,b2}& {\sigma }_{b1}^{2}& {\sigma }_{b1,b2}\\ {\sigma }_{b1,b2}& {\sigma }_{b1,b2}& {\sigma }_{b1}^{2}\end{array}\right)$where σ2b1 is the common variance of the random-effects terms and σb1,b2 is the common covariance between any two random-effects term . PAT Square symmetric logical matrix. If 'CovariancePattern' is defined by the matrix PAT, and if PAT(a,b) = false, then the (a,b) element of the corresponding covariance matrix is constrained to be 0.
Example: 'CovariancePattern','Diagonal'
Example: 'CovariancePattern',{'Full','Diagonal'}
Method for estimating parameters of the linear mixed-effects model, specified as the comma-separated pair consisting of 'FitMethod' and either of the following.
'ML' Default. Maximum likelihood estimation 'REML' Restricted maximum likelihood estimation
Example: 'FitMethod','REML'
Observation weights, specified as the comma-separated pair consisting of 'Weights' and a vector of length n, where n is the number of observations.
Data Types: single | double
Indices for rows to exclude from the linear mixed-effects model in the data, specified as the comma-separated pair consisting of 'Exclude' and a vector of integer or logical values.
For example, you can exclude the 13th and 67th rows from the fit as follows.
Example: 'Exclude',[13,67]
Data Types: single | double | logical
Coding to use for dummy variables created from the categorical variables, specified as the comma-separated pair consisting of 'DummyVarCoding' and one of the following.
ValueDescription
'reference'Default. Coefficient for first category set to 0.
'effects'Coefficients sum to 0.
'full'One dummy variable for each category.
Example: 'DummyVarCoding','effects'
Optimization algorithm, specified as the comma-separated pair consisting of 'Optimizer' and either of the following.
'quasinewton' Default. Uses a trust region based quasi-Newton optimizer. Change the options of the algorithm using statset('LinearMixedModel'). If you don't specify the options, then LinearMixedModel uses the default options of statset('LinearMixedModel'). 'fminunc' You must have Optimization Toolbox™ to specify this option. Change the options of the algorithm using optimoptions('fminunc'). If you don't specify the options, then LinearMixedModel uses the default options of optimoptions('fminunc') with 'Algorithm' set to 'quasi-newton'.
Example: 'Optimizer','fminunc'
Options for the optimization algorithm, specified as the comma-separated pair consisting of 'OptimizerOptions' and a structure returned by statset('LinearMixedModel') or an object returned by optimoptions('fminunc').
• If 'Optimizer' is 'fminunc', then use optimoptions('fminunc') to change the options of the optimization algorithm. See optimoptions for the options 'fminunc' uses. If 'Optimizer' is 'fminunc' and you do not supply 'OptimizerOptions', then the default for LinearMixedModel is the default options created by optimoptions('fminunc') with 'Algorithm' set to 'quasi-newton'.
• If 'Optimizer' is 'quasinewton', then use statset('LinearMixedModel') to change the optimization parameters. If you don't change the optimization parameters, then LinearMixedModel uses the default options created by statset('LinearMixedModel'):
The 'quasinewton' optimizer uses the following fields in the structure created by statset('LinearMixedModel').
Relative tolerance on the gradient of the objective function, specified as a positive scalar value.
Absolute tolerance on the step size, specified as a positive scalar value.
Maximum number of iterations allowed, specified as a positive scalar value.
Level of display, specified as one of 'off', 'iter', or 'final'.
Method to start iterative optimization, specified as the comma-separated pair consisting of 'StartMethod' and either of the following.
ValueDescription
'default'An internally defined default value
'random'A random initial value
Example: 'StartMethod','random'
Indicator to display the optimization process on screen, specified as the comma-separated pair consisting of 'Verbose' and either false or true. Default is false.
The setting for 'Verbose' overrides the field 'Display' in 'OptimizerOptions'.
Example: 'Verbose',true
Indicator to check the positive definiteness of the Hessian of the objective function with respect to unconstrained parameters at convergence, specified as the comma-separated pair consisting of 'CheckHessian' and either false or true. Default is false.
Specify 'CheckHessian' as true to verify optimality of the solution or to determine if the model is overparameterized in the number of covariance parameters.
Example: 'CheckHessian',true
## Output Arguments
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Linear mixed-effects model, returned as a LinearMixedModel object.
For properties and methods of this object, see LinearMixedModel.
## More About
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### Cholesky Parameterization
One of the assumptions of linear mixed-effects models is that the random effects have the following prior distribution.
$b~N\left(0,{\sigma }^{2}D\left(\theta \right)\right),$
where D is a q-by-q symmetric and positive semidefinite matrix, parameterized by a variance component vector θ, q is the number of variables in the random-effects term, and σ2 is the observation error variance. Since the covariance matrix of the random effects, D, is symmetric, it has q(q+1)/2 free parameters. Suppose L is the lower triangular Cholesky factor of D(θ) such that
$D\left(\theta \right)=L\left(\theta \right)L{\left(\theta \right)}^{T},$
then the q*(q+1)/2-by-1 unconstrained parameter vector θ is formed from elements in the lower triangular part of L.
For example, if
$L=\left[\begin{array}{ccc}{L}_{11}& 0& 0\\ {L}_{21}& {L}_{22}& 0\\ {L}_{31}& {L}_{32}& {L}_{33}\end{array}\right],$
then
$\theta =\left[\begin{array}{c}{L}_{11}\\ {L}_{21}\\ {L}_{31}\\ {L}_{22}\\ {L}_{32}\\ {L}_{33}\end{array}\right].$
### Log-Cholesky Parameterization
When the diagonal elements of L in Cholesky parameterization are constrained to be positive, then the solution for L is unique. Log-Cholesky parameterization is the same as Cholesky parameterization except that the logarithm of the diagonal elements of L are used to guarantee unique parameterization.
For example, for the 3-by-3 example in Cholesky parameterization, enforcing Lii ≥ 0,
$\theta =\left[\begin{array}{c}\mathrm{log}\left({L}_{11}\right)\\ {L}_{21}\\ {L}_{31}\\ \mathrm{log}\left({L}_{22}\right)\\ {L}_{32}\\ \mathrm{log}\left({L}_{33}\right)\end{array}\right].$
## Alternative Functionality
You can also fit a linear mixed-effects model using fitlme(tbl,formula), where tbl is a table or dataset array containing the response y, the predictor variables X, and the grouping variables, and formula is of the form 'y ~ fixed + (random1|g1) + ... + (randomR|gR)'.
## See Also
#### Introduced in R2013b
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https://everything2.com/user/Superunknown_GP/writeups/IP+subnetting?lastnode_id= | 1,611,415,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538082.57/warc/CC-MAIN-20210123125715-20210123155715-00596.warc.gz | 320,590,100 | 7,653 | IP subnetting is the art of taking a given address range and dividing it into several smaller networks.
How to subnet an IP address range: The first thing you need to figure out is how many bits are required for the network and hosts portions of the address. So if you had a class B (yeah, I know about CIDR, but I'm starting slow), and you needed, say, 200 subnets:
Class B: 143.17.0.0/16
By default, A class B has 16 bits in the network portion. You need enough bits to be above 200 in decimal. The first power of 2 that's above 200 is 2 to the 8th, which equals 256 (although you don't get two of those subnets if you're taking an MCP exam, because Microsoft is dumb). So you shift the network portion to the right 8 bits, giving you:
143.17.0.0/24
This gives you 254 usable subnets (per Microsoft's definiton- 256 anywhere else). You can determine the number of hosts by taking 2 to the power of however many bits are left, so (2^(32-24)) comes out as 256. Now, you need to subtract 2 from the network address and broadcast address. This gives you 254.
The next step is to figure out the subnet mask. To do this, convert everything in the network portion of the address to 1s (in binary), and everything in the host portion to 0s.
11111111.11111111.11111111.00000000
Now convert back to decimal: 255.255.255.0
Look familiar? Yeah, it's the subnet mask for a class C.
Now, to find the individual networks:
The first 16 bits were assigned by your ISP, so you can't touch those. The last 8 bits are the host portion, so you can't use those to determine networks, either. This leaves you with the third octet. Your usable networks are all possible combinations of those 8 bits:
143.17.00000001.00000000
143.17.00000010.00000000
143.17.00000011.00000000
143.17.00000100.00000000
and so on.
I left off the bottom range because Microsoft's exams say you can't use it- on their exams you can't use the top one either.
The network address (the address that defines a given network) is the lowest address in a subnet. So for your subnet:
143.17.00000100.00000000-143.17.00000100.11111111
Similarly, the highest address in that subnet is the broadcast address, so in that same range:
143.17.00000100.00000000-143.17.00000100.11111111
143.17.8.255 (143.17.00000100.11111111) is your broadcast address. Pretty simple, huh? All you need now is to figure out the default gateway. The default gateway is the near side of a router- the router interface that is on the same network as you. Typically, a default gateway is one number higher than the network address, but this is not always the case- it depends what the netadmin assigned to it.
To tackle a tougher one: Your address range is 14.192.0.0/11 (a range so large you'll never be able to buy one these days). Converting this to binary gives us:
00001101.11000000.00000000.00000000
The section you get to play with is after the first 11 bits (the first octect and the first 3 from the second). So let's say you needed 1,000 subnets of this gigantic network. The next higher power of 2 is 2^10th, which comes out to 1024. Again, Microsoft doesn't let you use the top or bottom networks (assuming you're taking one of their exams), so you're left with 1022. Taking your IP range, we move the network portion 10 bits to the right, leaving you with:
14.192.0.0/21
So the dividing line between the hosts and network portions are:
00001101.11000000.00000|000.00000000
So to find your subnet mask, take everything to the left of that dividing line and turn it into a 1. Take everything to the right and turn it into zeroes. This gives you:
11111111.11111111.11111000.00000000, which, in decimal, is 255.255.248.0.
To figure out the number of hosts per subnet, merely count the digits to the right of the dividing line and take 2 to that number. In this case, we get 2^11, which is 2048. Subtract 2 (that network address and broadcast address again!), and you're left with 2046.
Finding the individual networks is again nice and easy. Your ISP assigned the top 11 bits, the bottom 11 bits are for hosts, so you're left with the remaining (middle) 10. So your addresses look like this:
00001101.110|00000.00001|000.00000000
00001101.110|00000.00010|000.00000000
00001101.110|00000.00011|000.00000000
I segmented it out so you can see what are we're working with here. All you have to do now is convert it back to binary, giving you:
14.192.8.0
14.192.16.0
14.192.24.0
Et Cetera. Again, the gateway is the address of the router interface on your network.
I hope this is helpful to someone.
Oh, and I'm looking for a job as a netadmin's assistant while I learn the ropes. Any offers? | 1,273 | 4,649 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2021-04 | latest | en | 0.935979 |
https://ch.mathworks.com/matlabcentral/cody/problems/1159-coin-tossing-probability-of-same-heads-for-n-tosses/solutions/273887 | 1,585,481,679,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00166.warc.gz | 405,441,120 | 16,444 | Cody
Problem 1159. Coin Tossing: Probability of Same Heads for N tosses
Solution 273887
Submitted on 5 Jul 2013 by Dan
• Size: 13
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
%% assert(isequal(.5, round(1e6*coin_head_match(1))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.5000
2 Pass
%% assert(isequal(.375, round(1e6*coin_head_match(2))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.3750
3 Pass
%% assert(isequal(.3125, round(1e6*coin_head_match(3))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.3125
4 Pass
%% assert(isequal(.273438, round(1e6*coin_head_match(4))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.2734
5 Pass
%% assert(isequal(.246094, round(1e6*coin_head_match(5))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.2461
6 Pass
%% assert(isequal(.225586, round(1e6*coin_head_match(6))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.2256
7 Pass
%% assert(isequal(.139950, round(1e6*coin_head_match(16))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.1399
8 Pass
%% assert(isequal(.125371, round(1e6*coin_head_match(20))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.1254
9 Pass
%% assert(isequal(.114567, round(1e6*coin_head_match(24))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.1146
10 Pass
%% assert(~isequal(1,2))
11 Pass
%% assert(isequal(.099347, round(1e6*coin_head_match(32))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.0993
12 Pass
%% assert(isequal(.070386, round(1e6*coin_head_match(64))/1e6))
ans = Columns 1 through 10 0.5000 0.3750 0.3125 0.2734 0.2461 0.2256 0 0 0 0 Columns 11 through 20 0 0 0 0 0 0.1399 0 0 0 0.1254 Columns 21 through 30 0 0 0 0.1146 0 0 0 0 0 0 Columns 31 through 40 0 0.0993 0 0 0 0 0 0 0 0 Columns 41 through 50 0 0 0 0 0 0 0 0 0 0 Columns 51 through 60 0 0 0 0 0 0 0 0 0 0 Columns 61 through 64 0 0 0 0.0704 ans = 0.0704 | 2,758 | 5,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-16 | latest | en | 0.58556 |
http://interferencetheory.com/Research/ResearchVenusBlueprintDatabase/files/f9233b24e839bfd0bd37cfa37d562f96-88.html | 1,553,237,991,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202635.43/warc/CC-MAIN-20190322054710-20190322080710-00463.warc.gz | 103,424,762 | 4,732 | # StarKey as Birth of Venus
In the diagram below, the Earth, Moon and Venus are sized proportional to one another and compared to the StarKey Mt. Meru design and Rosslyn central pillar to determine:
1) if the StarKey could represent a geometric harmony of proportions between the Earth, Moon and Venus within the intersecting Vesica Piscis region of two circles,
2) if the StarKey overlaid on the Vesica Piscis could be meant to symbolize the female reproductive organ, and
3) if the StarKey overlaid the central Shekinah Pillar was intended to symbolize the male reproductive organ.
Altogether, it could constitute a microcosm Birth of Venus symbolism inside the already discussed StarKey floor plan (see here).
The most impressive thing revealed by this study is the fact that all spheres fit into the StarKey design when the large pentagrams are scaled to equal Phi-squared x Moon_Diameter. This is an extraordinarily simple relationship and suggests something more than coincidence. The variance from this simple calculation to the actual sum of the diameters is only 1.1%.
The entire StarKey fits into a double gothic vesica piscis arch, resembling those either side of the central pillar in Rosslyn chapel. This, in turn, fits within an Egyptian Triangle identical to the one used to build the Great Pyramid of Khufu.
The size of these three spheres closely align with the Phi proportions represented by the StarKey design. Whether this was understood or not by Rosslyn designer Sinclair and Hay is hard to tell, but it is interesting that the musical angels are located exactly where the Moon is located, sitting on the (yellow) Venus pentagram.
Proof method: We could determine if they knew this by taking the diagonal of one of the pentagram carvings and apply the same formula I used to see if it gives us the correct height of the pillar. If it does, then it proves they used the same Meru StarKey design on the pillars as they did on the floor plan. All I need is for someone to measure one of the star's diagonals and then the pillar up to the star's horizontal crossbar for comparison. A proportional match between the pillar geometry and the StarKey would strongly suggest the interpretation proposed here.
Outside of the cosmological symbolism, the male / female symbolism seems undeniable. The Vesica Piscis and StarKey together are very suggestive of the female sex organ while the pillar itself is phallic. The two together represent a union and insemination, most probably representing Venus fertility and birth of the cherbim (Aphrodite’s mythological children by Zeus). This specific pillar location has already been discussed as the birth location inside the birth canal of the chapel (see here) with the StarKey overlaid. We can only conclude that the pillar location represents the insemination of Venus in the microcosm and the chapel Venus herself in the macrocosm.
Cross reference: When we divide Height of StarKey by Earth Diameter, the result is 2.2. When you then take a look at this theoretical analysis of the StarKey physics I did several days ago, you'll see the same number. Earth makes for a perfect egg geometry when combined with the Moon. The size of Venus is the only thing keeping this stack from becoming a 100% perfect match to the StarKey. | 696 | 3,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-13 | latest | en | 0.918498 |
https://www.reddit.com/user/fishify/ | 1,521,736,447,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647892.89/warc/CC-MAIN-20180322151300-20180322171300-00024.warc.gz | 867,157,375 | 27,369 | ×
[–] 3 points4 points (0 children)
It appears that your confusion lies in understanding the index notation. For simplicity, we'll consider a simple scenario in which we don't need to worry about upper and lower indices. We'll assume repeated indices are summed over.
What is the ij component of the matrix product MN?
It is (MN)ij = MikNkj.
Notice because of the way we combine rows and columns in matrix multiplication, the index we sum over is the second index of M and the first index of N. This matters; we would get something different if computed MikNjk.
So to recast an index expression into a matrix product, the summed on indices must appear in the correct positions.
For example, MkiNkj = MTikNkj = (MTN)ij
The transpose is needed to get the index being summed over into the correct position so that we can interpret the expression as a matrix product.
Edit: fixed formatting.
[–] 5 points6 points (0 children)
If you don't have research, don't have a research section. I've written letters for students in such a situation, and one of the things I will say, after characterizing the strength of the student, is that while it's the case that this student doesn't have research experience, one has to start somewhere.
The point is to show what your skills/talents are and what your record is that should make you a good choice for the program.
[–] 2 points3 points (0 children)
The new tax bill doesn't give favorable treatment to all LLC and S-corporations. My understanding is that if the business's main asset is the skill or reputation of one of the corporation's employees, the favored tax rate does not apply.
That said, the law was written quickly, and the full implications/interpretations are still being sorted out. I think that part was pretty clear, but how it's going to be interpreted is probably still somewhat open.
[–] 0 points1 point (0 children)
This is probably more suited to /r/asksciencediscussion.
[–] 9 points10 points (0 children)
From one LIGO's own web pages w/various LIGO facts :
Curvature of the Earth: LIGO’s arms are so long that the curvature of the Earth is a measurable 1 meter (vertical) over the 4 km length of each arm. The most precise concrete pouring and leveling imaginable was required to counteract this curvature and ensure that LIGO’s vacuum chambers were "flat" and level. Without this work, LIGO's lasers would hit the end of each arm 1 m above the mirrors it is supposed to bounce off of!
[–] 4 points5 points (0 children)
It is easy to measure from Earth the period with which the Cepheid variable's brightness oscillates.
For Cepheid variables, there's an additional property, namely that once you know the period of oscillation, you know how intrinsically bright they are.
Now for the final step: We also know how bright they appear to be from Earth. Once we know this, since we also know how intrinsically bright they are, we can determine how far away they are.
[–] 1 point2 points (0 children)
Mermin's book Quantum Computer Science is a great place to start. It's not exhaustive, but it covers the core and will be a good way to see if you want to go further into the subject. The book was designed around a course he taught for people in computer science and math in particular, so I think it would be well-suited for you.
[–] 2 points3 points (0 children)
LIGO's black hole merger signals have been viewed as very reliable. Not only have they had multiple events, but the most recent black hole merger was detected by VIRGO as well.
That said, this latest observation lets us actually see the event that produced the gravitational wave via entirely separate channels, which is yet additional confirmation that they know what they are seeing: their interpretation of the signal was a neutron star merger, and the corresponding electromagnetic signal was observed!
[–] 0 points1 point (0 children)
LIGO detects a signal by seeing something other than the noise. To determine what that signal corresponds to, LIGO matches the signal against lots of templates, calculated versions of what kind of signal different events would produce (neutron star mergers, black hole mergers, different initial masses, etc.).
[–] 0 points1 point (0 children)
No, they travel at the speed of light.
[–] 0 points1 point (0 children)
Yes, gravitational waves travel at the speed of light. This is something that we could determine indirectly before, but now with this observation, we have direct evidence: the gravitational wave signal and the gamma ray burst (photons, i.e., light) arrived together.
[–] 9 points10 points (0 children)
You need to distinguish functions over the reals (functions that take real numbers as inputs) and real-valued functions (functions whose outputs are real numbers).
If by real function you mean what is conventionally meant -- a real-valued function -- then the input need not be a real number.
For example, the absolute value function over the complex numbers has complex-valued inputs and real-valued outputs. You can define the determinant function on nxn matrices that has nxn matrices for its inputs and real numbers for its outputs.
Knowing that a function produces real numbers does not tell you that its domain consists of real numbers.
[–] 9 points10 points (0 children)
Due to the properties of rotations, the only possible values for the spin of a particle are 0, 1/2, 1, 3/2, 2, 5/2, etc.
But there's an extra feature. If an electron had spin 0, 1, 2, or another whole number, there'd be no exclusion principle; you could fit as many electrons as you wish in an orbital. (Of course, if this happened, chemistry wouldn't be remotely like chemistry as we know it.)
If a particle has spin 1/2, 3/2, 5/2, etc., you can only have a finite number of those particles in an orbital, but the rule is different from you what you proposed.
For spin 1/2, you get two electrons per orbital, as you know. For spin 3/2, you would get four electrons per orbital. For spin 5/2, you would get six electrons per orbital. In general, if you have a half-integral spin, the number of electrons per orbital is twice the value of spin plus one.
[–] 2 points3 points (0 children)
At 1 AU, the strain would have been about 5 x 10-7, so still pretty tiny. You wouldn't see or feel anything, but you might hear the event at that distance; a deformation of 1 part per 107 of the eardrum is the edge of what you can hear.
[–] 0 points1 point (0 children)
Thanks!
[–] 1 point2 points (0 children)
The neutron stars wound up close to each other and started orbiting each other. As they did, they lost energy to gravitational waves, and eventually spiraled in till they collided with each other and merged.
[–] 2 points3 points (0 children)
Just to add : Here's a paper on the likelihood of neutrino detections down the line:
http://iopscience.iop.org/article/10.3847/2041-8213/aa8d14
[–] 21 points22 points (0 children)
At the press conference, the team reported that the signal here was much larger than for the black hole mergers, because the objects were so much closer.
Scanning through the material on LIGO's site, I found the strain from the first black hole merger to be 1 x 10-21 and from this merger to be 6 x 10-20.
Edit: Fixed a minus sign!
[–] 17 points18 points (0 children)
Spoke to a LIGO physicist -- neutrino observations from this event at this distance were not expected, so no surprise. I've yet to see the estimates of how close it would have to be for neutrinos to have been observed, however.
[–] 24 points25 points (0 children)
Also, to add: LIGO just measures the masses of progenitor objects. It is only astrophysics that tells us what sort of objects those are or could be.
[–] 0 points1 point (0 children)
The reason is that we now have a detector sensitive enough to see events.
[–] 1 point2 points (0 children)
Realistically, there is no need for this. When I'm reading letters and I encounter a letter from someone I know, I know that I know that person. On the other hand, if your advisor or one of your letter writers chooses to ping someone, that's their business, but it is not something you should ask. I would only ask someone to do this if there was a particular place you were especially invested in (other than it's the most prestigious).
[–] 0 points1 point (0 children)
Yes.
[–] 16 points17 points (0 children)
It's typical at many institutions. It doesn't mean they're looking for someone that speaks to that department. The outside person is there just to give a broader institutional perspective. Even if the outside person is from an allied field, that person is generally just functioning as an outsider looking in.
[–] 37 points38 points (0 children)
Here's a symmetry argument: Which way would the beam get re-directed? There is no preferred direction, so it wouldn't be consistent for the light to be deflected one way or another.
More precisely: If the situation is rotationally symmetric along the axis along which the incoming light travels, the behavior of the outgoing light must be unchanged under rotations about that axis. Consequently, the beam of light going into the new medium has to go along the same direction as the incident light was traveling in the situation you describe. | 2,186 | 9,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-13 | latest | en | 0.958298 |
https://everything2.com/user/cntrational/writeups/Calculus | 1,548,012,116,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583730728.68/warc/CC-MAIN-20190120184253-20190120210253-00401.warc.gz | 504,503,627 | 5,679 | You use geometry for manipulating shapes. You use algebra for doing math with unknown numbers. You use trigonometry for studying triangles, and by extension, waves. So what's calculus for?
Calculus is the study of change.
When you differentiate something, you get the derivative, which represents the rate at which something increases or decreases.
Integration is combining derivatives, or minute changes, into a whole result.
Using these, you can handle changing values in anything: acceleration, electromagnetism, money... calculus can handle them all.
Useful, eh? | 112 | 571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-04 | latest | en | 0.905532 |
https://www.palnet.io/puzzle/@quantumdeveloper/math-contest-8-1-sbi-prize | 1,590,828,845,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00221.warc.gz | 838,349,629 | 34,181 | # Math Contest #8 [1 SBI prize]
in puzzle • 9 months ago
Here you can keep your brain fit by solving math related problems and also earn SBI by doing so.
The problems usually contain a mathematical equation that in my opinion is fun to solve or has an interesting solution.
I will also only choose problems that can be solved without additional tools(at least not if you can calculate basic stuff in your head), so don't grab your calculator, you won't need it.
↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓
#### Every friday and tuesday!
↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓
# Rules
## Some additional rules for the case that the problem is an equation or a collection of equations.
#### And one more thing: I only want integer solutions!
↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓
# Problem
Today I want to give you something different(ial):
I don't know how much you now about differential equation( if you know nothing it would be time to learn it). Because of that I want to give you an easy and a hard equation. You'll get 1 point for solving the easy one and 2 for solving the hard one:
In case you don't know:
f'(x) is the first derivative of f(x) with respect to x.
g''(x) is the second derivative of g(x) with respect to x.
### Solve for f(x)[1 point] or g(x)[2 points].
I don't know if you all know about differential equations. Please tell me if this is maybe too hard right now.
↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓
You can take a look at the results of the last contest here. There you can also find out how I evaluate your entries.
To everyone who already participated in a past contest, come back today and try a new problem(tell me if you don't want to be tagged):
@addax @ajayyy @athunderstruck @bwar @contrabourdon @heraclio @hokkaido @mytechtrail @onecent @sidekickmatt @sparkesy43 @syalla @tonimontana @vote-transfer
In case no one gets a result(which I doubt), I will give away the prize to anyone who comments.
↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓
@contrabourdon sponsors this contest with 2 STEEM weekly.
You can support him by using a witness vote on untersatz, so he can further support this and other contests.
Sort Order:
· 9 months ago
g(x)=c1e^x+c2e^(-x)+x^2+2
· 9 months ago
Ugly formatting, but I accept it.
By the way if you want to write *something* without the word in between getting italic you can use `\*` instead of `*`
And if you want to write indexes, but your keyboard (-layout) doesnt allow it, you could at least use c_1 to show that you actually wanted to use a subscript there.
· 9 months ago
Yes somehow it looked way better in the preview:D
· 9 months ago
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https://community.qlik.com/thread/251505 | 1,540,021,827,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512592.60/warc/CC-MAIN-20181020055317-20181020080817-00007.warc.gz | 659,407,082 | 25,312 | 12 Replies Latest reply: Feb 27, 2017 6:49 PM by Aar Kay
# Doubt in Interval Match
I have the below dataset :-
And I want this Output :-
can anyone help me to do this....
• ###### Re: Doubt in Interval Match
Try this:
Table:
F1, F2,F3
1,12/12/2016,90
1,01/01/2017,92
2,01/02/2017,56
2,22/02/2017,59
];
F1,
F2 as StartDate,
If(F1=peek(F1),Date(peek(StartDate)-1,'DD/MM/YYYY'),Date(today())) as [End Date]
Resident
Table
Order By
F1,F2 Desc;
Drop Table
Table;
• ###### Re: Doubt in Interval Match
Thanks Aar Kay...
Load F1,Date#(F2,'DD/MM/YYYY') as F2; // why we are using this preceeding Load
and
If(F1=peek(F1),Date(peek(StartDate)-1,'DD/MM/YYYY'),Date(today())) as [End Date]
Can you explain me the above line
• ###### Re: Doubt in Interval Match
preceeding load is just to tell qlikview that F2 is date field in 'DD/MM/YYYY' format for the sample data.
If(F1=peek(F1) -this is if field value of F1 is equal to previous row value of F1,
then Date(peek(StartDate)-1,'DD/MM/YYYY') i.e take value of startdate "minus" 1day from the previousrow
if field value of F1 is "not" equal to previous F1 then return today()
• ###### Re: Doubt in Interval Match
if I want F3 Column also in the resultant table. for this what I need to do..
• ###### Re: Doubt in Interval Match
Try this:
Table:
F1, F2,F3
1,12/12/2016,90
1,01/01/2017,92
2,01/02/2017,56
2,22/02/2017,59
];
F1,
F2 as StartDate,
If(F1=peek(F1),Date(peek(StartDate)-1,'DD/MM/YYYY'),Date(today())) as [End Date],
F3
Resident
Table
Order By
F1,F2 Desc;
Drop Table
Table;
• ###### Re: Doubt in Interval Match
Did you change F1 to * in the preceding load
Table:
F1, F2,F3
1,12/12/2016,90
1,01/01/2017,92
2,01/02/2017,56
2,22/02/2017,59
];
F1,
F2 as StartDate,
If(F1=peek(F1),Date(peek(StartDate)-1,'DD/MM/YYYY'),Date(today())) as [End Date],
F3
Resident
Table
Order By
F1,F2 Desc;
Drop Table
Table;
• ###### Re: Doubt in Interval Match
Below is the code which i tried to execute but gives error :-
Gives the following error :-
• ###### Re: Doubt in Interval Match
Sorry About that because of load * -F2 field is being loaded twice that is why it is giving you error
Use this Load F1,F3,Date#(F2,'DD/MM/YYYY') as F2;
• ###### Re: Doubt in Interval Match
Thank You So Much Aar Kay. Now It is Working
• ###### Re: Doubt in Interval Match
I have this dataset :-
In this I have to match the OrderDate with the previous table columns of start date and end date and Derive a new field Of Amount which is equals to Price * Qty {Price From the previous table}
and i want this output :-
• ###### Re: Doubt in Interval Match
Using Intervalmatch like below:
Table:
ProductID, F2,F3
1,12/12/2016,90
1,01/01/2017,92
2,01/02/2017,56
2,22/02/2017,59
];
Temp:
ProductID,
F2 as StartDate,
If(ProductID=peek(ProductID),Date(peek(StartDate)-1,'DD/MM/YYYY'),Date(today())) as [End Date],
F3
Resident
Table
Order By
ProductID,F2 Desc;
Drop Table
Table;
A:
OrderID,Order Date,ProductID,Salesrep,Qty
9123,15/12/2016,1,Suresh,6
6635,01/02/2017,2,Ramesh,4
5589,25/02/2017,2,Shubham,3
4698,02/02/2017,1,Ankit,2
];
Inner Join(A)
IntervalMatch([Order Date],ProductID)
StartDate,
[End Date],
ProductID
Resident Temp;
Left Join (A)
StartDate,
[End Date],
ProductID,
F3
Resident Temp;
Drop Table Temp;
Final: | 1,128 | 3,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-43 | latest | en | 0.713873 |
https://pubs.nctm.org/search?q=%22CCSS.Math.Content.HSA-CED.A.1%22 | 1,675,010,534,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00726.warc.gz | 496,480,870 | 31,467 | # Search Results
## You are looking at 1 - 10 of 47 items for :
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## Media Clips: Lumbering Along
Students analyze items from the media to answer mathematical questions related to the article. Some media pieces about lumber provide an opportunity for work in graphing, volume, and restricting domains to real-world settings.
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## Mathematical Lens: Pumped Up about Mathematics
“Mathematical Lens” uses photographs as a springboard for mathematical inquiry and appears in every issue of the Mathematics Teacher. All submissions should be sent to the department editors. For more background information on “Mathematical Lens” and guidelines for submitting a photograph and questions, please visit http://www.nctm.org/publications/content.aspx?id=10440#lens.
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## Multiple Representations and Connections with the Sierpinski Triangle
A task using multiple representations helps students write explicit algebraic equations.
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## Calendar and Solutions to Calendar – March 2011
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## Parabolic Sound Reflectors–History and Modeling
In first- and second-year algebra classrooms, the all-too-familiar whine of “when are we ever going to use this in real life?” challenges mathematics teachers to find new, engaging ways to present mathematical concepts. The introduction of quadratic equations is typically modeled by describing the motion of a moving object with respect to time, and typical lessons include uninspiring textbook practice problems that portray dropping or shooting objects from given distances or at particular time intervals. For a novel approach to exploring quadratics, we chose to step outside the classroom to look at some phenomena in the field of acoustics. Our activity incorporates mathematical modeling to provide a multirepresentational view of the math behind the physics and to provide a conceptual basis for analyzing and understanding a real-world quadratic situation.
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## Media Clips: How Many People Can Fit on the National Mall? – Zombie Attack
Students analyze items from the media to answer mathematical questions related to the article. Estimating the size of the crowd at the Obama inauguration leads to estimating skills and finding areas, whereas the Zombie epidemic leads to modeling infectious disease.
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## Six Principles for Quantitative Reasoning and Modeling
Modeling the motion of a speeding car or the growth of a Jactus plant, teachers can use six practical tips to help students develop quantitative reasoning.
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## Extreme Word Problems: Interweave Student Imagination with Math
Let students find the connecting thread to create, illustrate, and share word problems to bridge school math and real-life math.
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## Isle of Cats
What happens with cat populations when they are not controlled? Consider the case of Aoshima Island in Japan. Aoshima Island is called a cat island: Its cat population is 130 and growing; its human population is 13. The cats live in colonies and are fed and cared for by people who live on the islands.
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## Angry Birds Mathematics: Parabolas and Vectors
Vector properties and the birds' frictionless environment help students understand the mathematics behind the game. | 657 | 3,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.875509 |
http://www.mathprotutoring.com/blog/?tag=act-math&paged=6 | 1,685,909,628,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00275.warc.gz | 82,253,602 | 11,772 | # Algebra 1, 2 – systems of equations
Q: 100 people are to be seated at 22 tables. Some tables seat 4 people and some seat 6 people. If no seats are empty, how many tables seating 4 people are there?
Explanation: Solve the system: x + y = 22; 4x + 6y = 100. My ‘CRAMER’ program will solve this for you! The number of 4-person tables is the x-value of the solution.
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# Algebra 1 – distance
Q: To the nearest tenth, what is the distance between the points (-6,2) and (6,5)?
(Use the distance formula, and round to the nearest tenth.)
My TI-84 Plus program DISTANCE will find this value.
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# SAT – Counting and Ordering
Q: Bob has ten different books, but must choose only three to bring along on his vacation. In how many ways can Bob choose three books?
Explanation: This is a combination (not permutation) problem, because the order of the books in Bob’s bag does not matter. Find the nCr function on your calculator and enter ’10 nCr 3′. On the TI-84 Plus, nCr is in MATH, PRB, 3. Or, do this by hand using the formula for nCr.
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# Geometry – Volume
Q: What is the maximum number of boxes with dimensions 4”x5”x8” that can be packed into one large box with dimensions 20”x20”x24”?
Explanation: First, decide that a whole number of small boxes will entirely fill the large box. To do this, look at the dimensions and match them. 4″ fits into 20″ five times. 5″ fits into 20″ four times. 8″ fits into 24″ three times. That means, you’ll get five rows of small boxes, four boxes per row, and three layers deep. So, the quickest way to get the answer is to multiply 5x4x3 = 60.
BUT if you don’t notice that, just do (large volume)/(small volume).
Large volume = 20x20x24 = 9600
Small volume = 4x5x8 = 160
(large volume)/(small volume) = 9600/160 = 60
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# Geometry – Volume
Q: Two rectangular prisms have the same volume. Prism 1 has dimensions 4 ft, 5 ft, and 7 ft. Prism 2 has dimensions 14 ft, 4 ft, and x ft. What is the value of x?
Explanation: Volume of a rectangular prism = (L)(W)(H)
Volume 1 = (4)(5)(7) = 140
Volume 2 = (14)(4)(x) = 56x
Since the volumes are equal, 56x = 140. x = 2.5
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# SAT – remainders
If summer begins on a Tuesday and ends 100 days later, on which day of the week does summer end?
Explanation: A week has 7 days. Dividing 100 by 7, you get 14 remainder 2. That’s 14 weeks and 2 days. Beginning on a Tuesday, 14 weeks later would land on a Tuesday too. Two more days brings you to Thursday.
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# Geometry – circles
Q: The area of a 45 degree sector of circle M is 100 square units. What is the area of circle M?
Explanation: The 45° sector is exactly 1/8 of the circle. (45/360 = 1/8)
100 = 1/8 of the area of the whole circle.
The area of the whole circle is 800.
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# Algebra 1, 2 – systems of equations
Two cubes and one pyramid weigh 9 lb. Seven cubes and two pyramids weigh 24 lb. All cubes weigh the same, and all pyramids weigh the same. What is the weight of one pyramid?
Explanation: Let x be the weight of a cube. Let y be the weight of a pyramid.
The first sentence becomes: 2x + y = 9.
The second sentence is: 7x + 2y = 24.
Solve this as a system of equations by any method you choose… substitution, elimination, Cramer’s Rule, etc. Or use my TI-84 Plus program ‘CRAMER’.
x = 2, y = 5
‘y’ is the weight of one pyramid, so the answer is 5.
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# Geometry – prism
How many vertices does a rectangular prism have?
If you have trouble drawing it or visualizing it, consider this instead: A prism has two identical bases. All of the vertices are along the edges of the two bases. Since a rectangular prism has rectangles for bases, each base will have 4 vertices. Two rectangular bases will have 8 vertices total. | 1,429 | 5,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-23 | latest | en | 0.667008 |
https://www.crazy-numbers.com/en/35893 | 1,550,604,531,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247491141.23/warc/CC-MAIN-20190219183054-20190219205054-00511.warc.gz | 799,266,384 | 4,348 | Discover a lot of information on the number 35893: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 35893
Is 35893 a prime number? No
Is 35893 a perfect number? No
Number of divisors 8
List of dividers 1, 11, 13, 143, 251, 2761, 3263, 35893
Sum of divisors 42336
## How to write / spell 35893 in letters?
In letters, the number 35893 is written as: Thirty-five thousand eight hundred and ninety-three. And in other languages? how does it spell?
35893 in other languages
Write 35893 in english Thirty-five thousand eight hundred and ninety-three
Write 35893 in french Trente-cinq mille huit cent quatre-vingt-treize
Write 35893 in spanish Treinta y cinco mil ochocientos noventa y tres
Write 35893 in portuguese Trinta e cinco mil oitocentos noventa e trĂªs
## Decomposition of the number 35893
The number 35893 is composed of:
2 iterations of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
1 iteration of the number 9 : The number 9 (nine) represents humanity, altruism. It symbolizes generosity, idealism and humanitarian vocations.... Find out more about the number 9
Other ways to write 35893
In letter Thirty-five thousand eight hundred and ninety-three
In roman numeral
In binary 1000110000110101
In octal 106065
In US dollars USD 35,893.00 (\$)
In euros 35 893,00 EUR (€)
Some related numbers
Previous number 35892
Next number 35894
Next prime number 35897
## Mathematical operations
Operations and solutions
35893*2 = 71786 The double of 35893 is 71786
35893*3 = 107679 The triple of 35893 is 107679
35893/2 = 17946.5 The half of 35893 is 17946.500000
35893/3 = 11964.333333333 The third of 35893 is 11964.333333
358932 = 1288307449 The square of 35893 is 1288307449.000000
358933 = 46241219266957 The cube of 35893 is 46241219266957.000000
√35893 = 189.45448002093 The square root of 35893 is 189.454480
log(35893) = 10.488297569392 The natural (Neperian) logarithm of 35893 is 10.488298
log10(35893) = 4.5550097589461 The decimal logarithm (base 10) of 35893 is 4.555010
sin(35893) = -0.2992749984479 The sine of 35893 is -0.299275
cos(35893) = -0.9541669011782 The cosine of 35893 is -0.954167
tan(35893) = 0.31365057630731 The tangent of 35893 is 0.313651 | 821 | 2,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-09 | latest | en | 0.749106 |
https://affordablehomework.com/2020/12/29/mgmt-650-fall-2017-week-2-homework/ | 1,610,940,088,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00159.warc.gz | 203,492,647 | 10,245 | # Mgmt 650 fall 2017 week 2 homework
MGMT 650 Fall 2017 Week 2 Homework
(Last updated 7/3 /2017)
1. Basic Excel Operation:
Use a new sheet called “Grade Calculator” in the same Excel file that you are submitting for this assignment.
After this week you will receive the solution from your professor and you will be able to use your solution or our solution for future classes.
2. Basic Descriptive statistics
A quality characteristic of interest for a tea-bag-filling process is the weight of the tea in the individual bags. If the bags are under filled, two problems arise. First, customers may not be able to brew the tea to be as strong as they wish. Second, the company may be in violation of the truth-in-labeling laws. In this example, the label weight on the package indicates that, on average, there are 5.5 grams of tea in a bag. If the average amount of tea in a bag exceeds the label weight, the company is giving away product. Getting an exact amount of tea in a bag is problematic because of variation in the temperature and humidity inside the factory, differences in the density of the tea, and the extremely fast filling operation of the machine (approximately 170 bags a minute). The following table provides the weight in grams of a sample of 50 bags produced in one hour by a single machine:
5.85
5.74
5.42
5.44
5.53
5.37
5.54
5.45
5.52
5.41
5.57
5.5
5.53
5.54
5.55
5.61
5.66
5.48
5.64
5.51
5.67
5.42
5.49
5.61
5.53
5.42
5.56
5.3
5.49
5.55
5.67
5.53
5.42
5.58
5.59
5.5
5.38
5.5
5.53
5.58
5.69
5.47
5.43
5.25
5.56
5.62
5.5
5.56
5.61
5.38
a. Compute the arithmetic mean and median.
b. Compute the first quartile and third quartile.
c. Compute the range, interquartile range, variance, standard deviation, and coefficient of variation.
Parts d and e are your chance to show your understanding. You are a consultant brought in by the company and given the sample. What are your recommendations to the company, based on your reading for this week?
d. Interpret the measures of central tendency within the context of this problem. Why should the company producing the tea bags be concerned about the central tendency?
e. Interpret the measures of variation within the context of this problem. Why should the company producing the tea bags be concerned about variation?
3. Built In Excel Functions
Review Excel help files (and other sources if you wish) and explain in your own words:
a. The difference between the functions QUARTILE.INC() and QUARTILE.EXC(). Which of the two will you get if you use QUARTILE()?
b. The difference between STDEV.P() and STDEV.S() and the difference between VAR.P() and VAR.S().
c. What is the Excel function COVARIANCE (This has two forms: COVARIANCE.P() and COVARIANCE.S())?
d. How will you calculate Coefficient of Variation of one sample using Excel?
4. Frequency Distribution
According to Zillow.com, here is a list of the prices of the property sold (in thousands) in Washington, DC June 29 – 30, 2016. Use Excel’s capability (=FREQUENCY()) to create a frequency distribution using 5 bins.
316
680
292
440
1860
2333
1775
844
843
785
265
1315
765
368
763
536
466
1261
1875
1113
299
205
830
1618
1451
625
1473
2093
449
1879
869
920
1700
2450
1460
1562
1380
888
1750
495
240
2250
1420
785
1556
949
1435
1475
945
645
Hint: To score the full points for this question, set it up so that Excel calculate the bins for you. This way if you change the data the bins are changed automatically.
5. Histogram, bar chart, and pie chart
Use the house price data and the bins data from problem 4 to create:
a. A histogram using the data analysis tool.
b. A bar chart
c. A pie chart
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Total price:
\$26
The price is based on these factors: | 1,514 | 5,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-04 | latest | en | 0.891518 |
https://mathsmartinthomas.wordpress.com/2017/10/08/finding-square-roots-of-complex-numbers/ | 1,540,157,709,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514355.90/warc/CC-MAIN-20181021203102-20181021224602-00075.warc.gz | 722,405,775 | 15,225 | # Finding square roots of complex numbers
Example: find w so that $w^2 = -16 + 30i$
Method 1: calculator
Find the modulus of the number to be square-rooted, by mental arithmetic or the calculate: |w| = 34
Use the calculator to find the argument of w
calculate $\mathrm{Ans}\div 2$
calculate $\sqrt{\mathrm{mod}} \, \angle \, \mathrm{Ans}$
convert back to cartesian form: in this case you get 3+5i
and check (just to make sure no rounding error on the calculator)
and write the negative of that answer (in this case -3-5i) as the other square root
Convert back to a+bi form: w ≈ 3+5i (round your answer back to a neat value, to correct the calculator error introduced when halving a decimal approximation of the argument)
See below for calculator images.
Method 2: Equate modulus and real part
Let square root = w = a + bi
Then $|w^2| = |w|^2 = a^2 + b^2 = 34$
$\mathrm{Re}(w^2) = a^2 - b^2 = -16$
Solve those simultaneous equations for a2 and b2
a2=9 and b2=25
Only problem now is to work out the valid signs for a and b. $\mathrm{Im}(w^2) = 2ab$. In this case a and b must have the same sign (both negative or both positive) because $\mathrm{Im}(w^2) > 0$. (If $\mathrm{Im}(w^2) < 0$ then a and b would have to have opposite signs, one positive, one negative).
So w = ± (3+5i)
Method 3: Equate real and imaginary parts (the method in the old Edexcel textbook)
Let square root = w = a + bi
Then $\mathrm{Re}(w^2) = a^2 - b^2 = -16$
$\mathrm{Im}(w^2) = 2ab = 30$
Substitute $b = \frac{15}{a}$ from the second equation into the first
$a^2 - \frac{225}{a^2} = -16$
Multiply up to get a quadratic equation in $a^2$
$a^4 +16a^2 - 225 = 0$
Solve: a2 = 9, so a = ± 3, and b = ± 5
So w = ± (3+5i)
Method 1 (calculator) – images
This will usually give you the exact answer in the exact form you want it. The likely exception is if you’re finding something like √(−1 + 2√2 i), where the answer is (1 + √2 i).
The calculator will give you (1 + 1.414213562 i), and it won’t convert that to (1 + √2 i) even if you press the S-D key.
Workaround: if you get an answer for real or imaginary part of the square root which isn’t a whole number, square your answer to see if it’s an exact square root of a whole number (in A level, it pretty much always will be).
It’s worth memorising the facts that
√2 ≈ 1.414
√3 ≈ 1.732
√5 ≈ 2.236
Often come in useful, like it comes in useful to be able to remember your way home from school, or to a friend’s house, without having to look at Google Maps. | 776 | 2,511 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 14, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2018-43 | latest | en | 0.792405 |
https://kr.mathworks.com/matlabcentral/cody/problems/751-implement-simple-rotation-cypher/solutions/364430 | 1,606,181,924,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141169606.2/warc/CC-MAIN-20201124000351-20201124030351-00388.warc.gz | 367,228,943 | 17,070 | Cody
# Problem 751. Implement simple rotation cypher
Solution 364430
Submitted on 4 Dec 2013 by Hussain
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% inStr = 'doug'; shift = 10; outStr = 'nyeq'; assert(isequal(cypher(inStr, shift),outStr))
2 Pass
%% inStr = 'thequickbrownfox'; shift = 5; outStr = 'ymjvznhpgwtbsktc'; assert(isequal(cypher(inStr, shift),outStr))
3 Pass
%% inStr = 'thecrowfliesatmidnight'; shift = 22; outStr = 'pdaynksbheaowpiezjecdp'; assert(isequal(cypher(inStr, shift),outStr))
### Community Treasure Hunt
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Start Hunting! | 230 | 760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | latest | en | 0.512462 |
https://testbook.com/question-answer/find-the-value-of-cosec2-x-1-cos-x--60ed452b5822dabf4c7dc092 | 1,638,436,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00182.warc.gz | 604,716,475 | 30,326 | # Find the value of ∫ cosec2 x (1 - cos x) dx
This question was previously asked in
Airforce Group X 13 Jul 2021 Shift 1 Memory Based Paper
View all Airforce Group X Papers >
1. - cot x + cosec x + c
2. cot x - cosec x + c
3. cot x + cosec x + c
4. -cosec x - cot x + c
Option 1 : - cot x + cosec x + c
## Detailed Solution
Concept:
Some useful formulas are:
∫ cosec x. cot x dx = - cosec x + c
∫ cosec2x dx = - cot x + c
Calculation:
∫ cosec2 x (1 - cos x) dx
= ∫ cosec2x dx - ∫ cosec x. cot x dx
= (-cot x) - (-cosec x) + c
= -cot x + cosec x + c | 223 | 560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-49 | latest | en | 0.675972 |
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# Working with cash discounts on inventory purchases: how much, what interest rate?
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
You place an order for 1,000 units of inventory at a price of \$350 per unit. The supplier offers terms of 2/10, net 30. Answer the following questions about payment.
a. If you take the full period to pay, how much should you remit?
b. How soon must you pay to get the discount?
c. What will be the amount of the discount if you take it?
d. If you don't take the discount, how much interest are you paying implicitly?
#### Solution Preview
(a) You will not get any discount and you remit = 350 * 1,000 = \$350,000.
(b) If you pay within 10 days (tenth ...
#### Solution Summary
The solution is very understandable as it explains and calculates answers to the problem.
\$2.49 | 233 | 982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-23 | longest | en | 0.915635 |
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1. ## Combinations of numbers that add up to a total.
I looked around for a solution but could not find one. I am trying to find combinations of numbers that can add up to a total. The numbers available and the total are different each time.
For example, I have an array that contains the following numbers:
[16,12,8,6,4,2]
I want to find the combinations that add up to 140, like this:
8x16,1x12
8x16,1x8,1x4
11x12,1x8
Notice I do not do:
8x16,2x6
8x16,3x4
This is because as a rule, only the first number can be used in multiple of more than one. All other number pairs can only be a 1x.
Output string could be formatted like this in a single string containing all combinations:
8x16,1x12|8x16,1x8,1x4|11x12,1x8
• Looks like homework to me...
• Its not homework, but thanks for looking.
I'm stuck on this problem and am looking for an elegant way to do it. I gave a simple starting array, but it could be far more complex, such as [64, 48, 40, 36, 32, 28, 24, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2], which would produce a larger number of combinations. Even larger arrays would be expected when testing with larger totals, so something fast and efficient is needed.
• So what do you use this for then?
• If I understand you correctly only the first number pair is actually a pair (i.e. to items from the array) - all other are just one item from the array?
Or are each pair always only one from the array and the multiple can be arbitrary?
• Not to ask a dumb question, but what is the point in displaying the 1x values?
That is, why
8x16,1x12
8x16,1x8,1x4
11x12,1x8
8x16,12
8x16,8,4
11x12,8
And why is it listed in that order instead of larger number first?
16x8,12
16x8,8,4
12x11,8
It is the kind of requirements you put on it that make it feel like homework. Doesn't feel like a real world problem.
Oh...and where did 11x12 come from? There's no 11 in your original array.
Anyway, there's really no answer for this except recursion. Probably could be done without it, but so much easier to conceive of using recursion.
• This is being used for an imposition model (how to layout things on a bigger area).
Your right, there is no point in displaying the 1x notation, other than for readability. It could very well be:
8x16,8,4
11x12,8
4x32,12
Regardless, the final output contains all the possible combinations in a single string separated by a delimiter, for example:
8x16,8,4|11x12,8|4x32,12|etc.
8 x 16 + 8 + 4 = 140
11 x 12 + 8 = 140
4 x 32 + 12 = 140
First number is arbitrary, but the second is always from the array. Only the first number can be used more than once (8x16, all others are 1x).
Hope that helps clarify it.
• Okay, almost have it. But one more thing:
For example, I have an array that contains the following numbers:
[16,12,8,6,4,2]
I want to find the combinations that add up to 140, like this:
8x16,1x12
8x16,1x8,1x4
11x12,1x8
Ummm... once again, you do *NOT* have 11 in your array. So will ignore that one.
But...
*ARE* you allowed to use the same number twice???
8x16,1x8,1x4
??? That seems inconsistent with your other rules.
And your list there is wrong, as I read it. You are missing possibilities.
My result:
Code:
```[16,12,8,6,4,2]
16x8,1x12|16x8,1x8,1x4|16x8,1x6,1x4,1x2|16x6,1x16,1x12,1x8,1x6,1x2|12x8,1x16,1x12,1x8,1x6,1x2```
Again, that assumes I can re-use a number in a "1x" when it has already been used in the first product.
• Here, rather than waiting for you:
Code:
```<html>
function targetFromArray( arr, target )
{
// sort descending, just to be safe
arr = arr.sort( function(n1,n2) { return n2-n1; } );
// first, find factors:
for ( var i = 0; i < arr.length; ++i )
{
var n1 = arr[i];
var ar2 = arr.concat([]); // clone the array
ar2.splice(i,1); // remove n1 from it
for ( var j = i+1; j < arr.length; ++j )
{
var n2 = arr[j];
var ar3 = ar2.concat([]); // clone the array
// then remove n2 from it
for ( var n = 0; n < ar3.length; ++n )
{
if ( ar3[n] == n2 )
{
ar3.splice( n, 1 );
break;
}
}
var product = n1 * n2;
var list = n1 + "x" + n2;
if ( product == target )
{
} else if ( product < target ) {
sumFromArray( list, ar3, target - product );
}
}
}
}
function sumFromArray( list, arx, target )
{
for ( var i = 0; i < arx.length; ++i )
{
var n = arx[i];
var nlist = list + ",1x" + n;
if ( n == target )
{
} else if ( n < target ) {
var ary = arx.concat([]);
ary.splice( 0, i+1 );
if ( ary.length > 0 ) sumFromArray( nlist, ary, target - n );
}
}
}
var test = [8,2,16,4,12,6];
function testit() { document.getElementById("result").innerHTML = targetFromArray( test, 140 ); }
</script>
<body>
<input type="button" value="try it" onclick="testit()"/>
<div id="result"></div>
</body>
</html>```
That code does *NOT* use "1xN" where "N" is one of the number in the first product.
So for
Code:
```test = [8,2,16,4,12,6];
result is 16x8,1x12|16x8,1x6,1x4,1x2```
If you want to have "1xN" where "N" is already in the first product, make one tiny change:
Code:
``` } else if ( product < target ) {
sumFromArray( list, ar3, target - product );
}```
to
Code:
``` } else if ( product < target ) {
sumFromArray( list, arr, target - product );
}```
Then for
Code:
```test = [8,2,16,4,12,6];
result is 16x8,1x12|16x8,1x8,1x4|16x8,1x6,1x4,1x2|16x6,1x16,1x12,1x8,1x6,1x2|12x8,1x16,1x12,1x8,1x6,1x2```
Happy?
• Wow, that looks complicated How about:
Code:
```//possibleParts(numbers:array,target:number)
var possibleParts = (function(){
var r;
var a = function(numbers,target,partial){
var i, s;
for(i=0,s=0;i<partial.length;s+=partial[i++]);
if(s > target){return};
partial.s = s;
(s == target || 0 == (target-s)%partial[0]) && r.push(partial);
for (i = 0; i < numbers.length; i++){
a(numbers.slice(i+1),target,partial.concat(numbers[i]));
};
};
var b = function(x,target){
for(var i = 0; i < x.length; i++){
for(var j = 0; j < x[i].length; j++){
x[i][j] = j == 0 && x[i].s < target
? ((target-x[i].s+x[i][0])/x[i][0]) + 'x' + x[i][0]
: x[i][j] = '1x' + x[i][j];
};
x[i] = x[i].join(',');
};
x = x.join('|');
return x
};
return function(numbers,target){
r = [];
a(numbers,target,[]);
return b(r,target)
}
})();```
• Test:
possibleParts([16,12,8,6,4,2],140)
Result:
8x16,1x12|7x16,1x12,1x8,1x6,1x2|8x16,1x8,1x4|8x16,1x6,1x4,1x2|11x12,1x8|10x12,1x8,1x6,1x4,1x2|11x12, 1x6,1x2|16x8,1x6,1x4,1x2|17x8,1x4|23x6,1x2|35x4|70x2
• Doesn't seem to work.
Calling possibleParts([16,12,8,6,4,2],140)
Produced this result:
8x16,1x12|7x16,1x12,1x8,1x6,1x2|8x16,1x8,1x4|8x16,1x6,1x4,1x2|11x12,1x8|10x12,1x8,1x6,1x4,1x2|11x12, 1x6,1x2|16x8,1x6,1x4,1x2|17x8,1x4|23x6,1x2|35x4|70x2
??? It seems to ignore the array of numbers.
EDIT: So we agree. And it doesn't fulfill the conditions he asked for, at all.
• I think it does exactly what he asks for ;-)
(Took me a while to understand what was asked for:
in each result the first item can be multiplied several times)
• <shrug> I guess we will find out, if he ever comes back.
I do note that adding that condition to your code would be easy, so not a big deal.
•
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# The Lambda Calculus and The JavaScript
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### The Lambda Calculus and The JavaScript
1. 1. orb@nostacktrace.com The Lambda Calculus λ and The JavascriptNorman Richards
2. 2. orb@nostacktrace.com Why lambda calculus? Alan TUring Alonzo CHURCh Invented the turing machine, an Created lambda calculus, the imperative computing model basis for functional programming. We spend all our time studying this guy’s work (because he’s awesome)Norman Richards http://en.wikipedia.org/wiki/Alonzo_Church But if we want to think functionally, http://en.wikipedia.org/wiki/Alan_Turing we should pay attention to this guy too (because he’s also awesome)
3. 3. orb@nostacktrace.com The premise Functional programming is good Using functional techniques will make your code better Understanding lambda calculus will improve your functionalNorman Richards
4. 4. orb@nostacktrace.com A Lambda Expression bound variable This is the argument of the function λx.x body A lambda expression that uses the bound variable and represents the function valueNorman Richards
5. 5. orb@nostacktrace.com A Lambda Expression function(x) { λx.x return x }Norman Richards
6. 6. orb@nostacktrace.com Not a Lambda Expression λy.x free variable This variable isn’t bound, so this expression is meaninglessNorman Richards
7. 7. orb@nostacktrace.com A Lambda Expression function(y) { λy.x return x }Norman Richards These are meaningless
8. 8. orb@nostacktrace.com A Lambda Expression parenthesis Not needed, but used for clarity here λx.(λ.y.x) bound or free? x is bound in the outer expression and free in inner expression.Norman Richards
9. 9. orb@nostacktrace.com A Lambda Expression function(x) { return function(y) { λx.λ.y.x return x } }Norman Richards
10. 10. orb@nostacktrace.com Lambda Application unicorns don’t exist There are no assignments or external references. We’ll use upper-case words to stand for some valid lambda expression, but it must mean something. λx.x UNICORN application by substitution Apply UNICORN to the expression, substituting UNICORN for every occurrence of the bound valueNorman Richards
11. 11. orb@nostacktrace.com Lambda Application The I (identity) combinator Combinatory logic predates lambda calculus, but the models are quite similar. Many lambda expressions are referred to by their equivalent combinator names. function(x) { λx.x UNICORN return x ! UNICORN }(UNICORN); ! UNICORNNorman Richards
12. 12. orb@nostacktrace.com Lambda Application the K (constant) combinator This function takes an input argument and returns a function that returns that same value λx.λy.x UNICORN λy.UNICORN MITT ! λy.UNICORN ! UNICORN always a UNICORNNorman Richards No matter what argument is passed in, the result is always the same. http://leftaction.com/action/ask-kansas-mitt-romney-unicorn
13. 13. orb@nostacktrace.com Lambda Application self-application This function takes a function and applies it to itself. function(f) { λs.(s s) return f(f); }Norman Richards
14. 14. orb@nostacktrace.com Lambda Application Substitute λx.x for s λs.(s s) λx.x function(f) { ! λx.x λx.x return f(f); }(function (x) { ! λx1.x1 λx.x return x; For clarity, rename x as x1 ! λx.x });Norman Richards Substitute λx.x for x1.
15. 15. orb@nostacktrace.com Infinite application λs.(s s) λs.(s s) ! λs.(s s) λs.(s s) ! λs.(s s) λs.(s s) ! … the halting problemNorman Richards Just as you’d expect, there’s no algorithm to determine whether or not a given lambda expression will terminate
16. 16. orb@nostacktrace.com Turing Complete I: λx.x K: λx.λy.x S: λx.λy.λz.x z (y z)Norman Richards http://en.wikipedia.org/wiki/SKI_combinator_calculus
17. 17. orb@nostacktrace.com Actually, just two lambdas S K K UNICORN ! λx.λy.λz.(x z (y z)) K K UNICORN ! λy.λz.(K z (y z)) K UNICORN ! λz.(K z (K z)) UNICORN ! K UNICORN (K UNICORN) ! λx.λy.x UNICORN (K UNICORN) ! λy.UNICORN (K UNICORN)Norman Richards ! UNICORN Thus, S K K is computationally equivalent to identity
18. 18. orb@nostacktrace.com Other Turing Complete things Magic: the Gathering http://www.toothycat.net/ ~hologram/Turing/ C++ Templates The number 110 http://citeseerx.ist.psu.edu/Norman Richards viewdoc/summary? http://mathworld.wolfram.com/ doi=10.1.1.14.3670 Rule110.html
19. 19. orb@nostacktrace.com TURING TARPIT Beware of the Turing tar-pit in which everything is possible but nothing of interest is easy. Alan PerlisNorman Richards http://pu.inf.uni-tuebingen.de/users/klaeren/epigrams.html
20. 20. orb@nostacktrace.com Currying function(x) { return function(y) { return function (z) { A function that returns a FUNCTION return UNICORN; .. that returns a function that returns a function. You can think of this as a function of three arguments. } λx.λy.λz.UNICORN X Y Z } }(X)(Y)(Z);Norman Richards function(x,y,z) { return UNICORN; }(X,Y,Z);
21. 21. orb@nostacktrace.com Currying Sometimes abbreviated Sometimes, you’ll see lambda expressions written like this for clarity. The arguments are still curried. λx y z.UNICORN X Y Z HASKELL CURRY ! λy z.UNICORN Y Z Curry studied combinatory logic, a variant of lambda calculus. ! λz.UNICORN Z ! UNICORNNorman Richards http://www.haskell.org/haskellwiki/Haskell_Brooks_Curry
22. 22. orb@nostacktrace.com data structures CHURCH PAIR ENCODING PAIR takes two arguments, x and y and returns a function that takes another function and applies x and y to it. PAIR: λx.λy.λf.(f x y) BuILDING UP STATE PAIR takes two arguments, x and y and returns a function that takes another function and applies x and y to it.Norman Richards
23. 23. orb@nostacktrace.com data structures PAIR: λx.λy.λf.(f x y) FIRST: λx.λy.x SECOND: λx.λy.y ACCESSOR FUNCTIONS FIRST and SECOND are functions that can beNorman Richards passed to pair. They take two arguments and return the first and second, respectively.
24. 24. orb@nostacktrace.com data structures PAIR: λx.λy.λf.(f x y) DOGCAT: PAIR DOG CAT ! λx.λy.λf.(f x y) DOG CAT ! λy.λf.(f DOG y) CAT ! λf.(f DOG CAT)Norman Richards
25. 25. orb@nostacktrace.com data structures DOGCAT: λf.(f DOG CAT) FIRST: λx.λy.x SECOND: λx.λy.y DOGCAT FIRST ! λf.(f DOG CAT) λx.λy.x ! λx.λy.x DOG CATNorman Richards ! λy.DOG CAT ! DOG
26. 26. orb@nostacktrace.com data structures DOGCAT: λf.(f DOG CAT) FIRST: λx.λy.x SECOND: λx.λy.y DOGCAT SECOND ! λf.(f DOG CAT) λx.λy.x ! λx.λy.x DOG CATNorman Richards ! λy.y CAT ! CAT
27. 27. orb@nostacktrace.com data structures RGB: λr.λg.λb.λf.(f r g b) RED: λr.λg.λb.r GREEN: λr.λg.λb.g BLUE: λr.λg.λb.b BLACK_COLOR: RGB 255 255 255Norman Richards WHITE_COLOR: RGB 0 0 0 NuMBERS? We haven’t seen how to construct numbers yet, but for the moment imagine that we did.
28. 28. orb@nostacktrace.com data structures AVERAGE? AVG: λx.λy.??? If we did have numbers, we could probably do math like simple averaging MIX_RGB: λc1.λc2.λf. (f (AVG (c1 RED) (c2 RED)) (AVG (c1 GREEN)(c2 GREEN)) (AVG (c1 BLUE) (c2 BLUE)))Norman Richards OO LAMBDA? MIX_RGB takes two colors and returns a new color that is the mix of the two
29. 29. orb@nostacktrace.com CONDITIONALS COND: λe1.λe2.λc.(c e1 e2) TRUE: λx.λy.x FALSE: λx.λy.y Look familiar? Think of a boolean condition as a PAIR of values. TRUE selects the first one and FALSE selects the second one.Norman Richards
31. 31. orb@nostacktrace.com BOOLEAN LOGIC NOT: λx.(COND FALSE TRUE x) AND: λx.λy.(COND y FALSE x) OR: λx.λy.(COND TRUE y x)Norman Richards
32. 32. orb@nostacktrace.com NUMBERS ZERO: λx.x SUCC: λn.λs.(s false n) ONE: SUCC ZERO ! λs.(s false ZERO) TWO: SUCC ONE ! λs.(s false ONE) ! λs.(s false λs.(s false ZERO))Norman Richards PEANO NUMBERS PEANO numbers are based on a zero function and a successor function
33. 33. orb@nostacktrace.com NUMBERS ISZERO ZERO ZERO: λx.x ! λn.(n FIRST) λx.x ISZERO: λn.(n FIRST) ! λx.x FIRST ! FIRSTNorman Richards ! TRUE
34. 34. orb@nostacktrace.com NUMBERS ISZERO: λn.(n FIRST) ONE: λs.(s FALSE ZERO) ISZERO ONE ! λn.(n FIRST) λs.(s FALSE ZERO) ! λs.(s FALSE ZERO) FIRST ! (FIRST FALSE ZERO)Norman Richards ! FALSE
35. 35. orb@nostacktrace.com NUMBERS PRED: λn.(n SECOND) PRED (SUCC NUM) ! λn.(n SECOND) λs.(s FALSE NUM) ! λs.(s FALSE ZERO) SECOND ! SECOND FALSE NUM ! NUMNorman Richards but... ZERO isn’t SUCC of a number
36. 36. orb@nostacktrace.com NUMBERS PRED: λn.(ISZERO n) ZERO (n SECOND) What is Pred of zero? We can at least test for zero and return another number.Norman Richards
37. 37. orb@nostacktrace.com NUMBERS ADD:λx.λy.(ISZERO Y) function add(x,y) { x if (y==0) { (ADD (SUCC X) return x (PRED Y)) } else { return add(x+1,y-1) NOT VALID Recursive definitions aren’t possible . }Norman Richards Remember, names are just syntactics sugar. All definitions must be finite. How do we do this? }
38. 38. orb@nostacktrace.com RECURSION DIVERSION If we could pass the function in Now we’ll make a function of three arguments, the first being the function to recurse on ADD 1:λf.λx.λy.(ISZERO Y) x (f f (SUCC X) (PRED Y)) Then we could call it RECURSIVELY Just remember that the function takes threeNorman Richards arguments, so we should pass the function to itself.
39. 39. orb@nostacktrace.com RECURSION DIVERSION ADD 1:λf.λx.λy.(ISZERO Y) x (f f (SUCC X) (PRED Y)) ADD: ADD 1 ADD 1 ! λx.λy.(ISZERO Y) xNorman Richards (ADD 1 ADD1 (SUCC X) (PRED Y)) FINITE DEFINiTION This definition of ADD is a bit repetitive, but it doesn’t recurse infinitely. If only we could clean this up a bit... | 3,003 | 9,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-34 | latest | en | 0.626316 |
https://forum.industrial-craft.net/thread/10998-ic2-experimental-reactor-planner-open-beta/?postID=215439 | 1,537,974,892,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267165261.94/warc/CC-MAIN-20180926140948-20180926161348-00135.warc.gz | 514,424,430 | 15,752 | # IC2 Experimental Reactor Planner open beta
• I guess my issue is not understanding why he used 2 for single as a base rather then 4 when they say one uranium cell produces 4 on its own.
I know this reply is a few months late, but look at the formula for [Triangular Numbers](https://en.wikipedia.org/wiki/Triangular_Numbers). The formula is:
0.5 * (n * (n+1)
So, if 4 is the base, then the amount for single uranium is likewise:
4 * (0.5 * (n * (n+1)))
Since 4 * 0.5 is 2, that's why you see 2 in the code.
-----
Speaking of code, I noticed that placing a single Heat Vent increases causes the tool to say "10 iron" for the costs, but from what I can tell, it only costs 8.5 iron, and always has only cost 8.5 iron. You can see a recipe for it [here](http://crafting-guide.com/brow…strial_craft_2/heat_vent/), or by using JEI in the game. I looked in the code to see how it is calculated, and I just see that it is hardcoded to 10:
Code
1. public static final MaterialsList MATERIALS = new MaterialsList(
2. 10, java.util.ResourceBundle.getBundle("Ic2ExpReactorPlanner/Bundle").getString("IRON"),
3. 1, java.util.ResourceBundle.getBundle("Ic2ExpReactorPlanner/Bundle").getString("TIN"),
4. 16.0 / 3, "Copper");
As far as I count, you need 1.5 iron for the iron bars, 4 iron for the plates, and 3 iron for the electric motor, making a total of 8.5.
Since all the other types of heat vents use the basic heat vent in their material calculations, and since every commonly used reactor contains heat vents of some type, the iron calculations in the official list of reactor designs are wrong on every single reactor listed. Am I just misunderstanding something about the crafting here?
I also noticed that the heat exchanger code for transfer() looks extremely buggy. Your notes say that you adapted the code from IC2 code, so then I went and checked it, and wow, it matches. I find that to be a bit, um, unsettling. Is there a way that we can file bugs against the IC2 code?
Likewise, your code for adjustCurrentHeat() returns a negative value in cases where the component overflows and in cases where it underflows. I checked IC2 decompiled code and saw the same thing. Since you are trying to mimic what we see in the game, you did the right thing, but wow, that is really interesting. I'm wondering if it is possible to abuse these bugs and make a reactor that performs far better than it should.
• I also noticed that the heat exchanger code for transfer() looks extremely buggy. Your notes say that you adapted the code from IC2 code, so then I went and checked it, and wow, it matches. I find that to be a bit, um, unsettling. Is there a way that we can file bugs against the IC2 code?
Likewise, your code for adjustCurrentHeat() returns a negative value in cases where the component overflows and in cases where it underflows. I checked IC2 decompiled code and saw the same thing. Since you are trying to mimic what we see in the game, you did the right thing, but wow, that is really interesting. I'm wondering if it is possible to abuse these bugs and make a reactor that performs far better than it should.
Bug reports can go on Mantis if there's something awry, if you don't especially want to create an account there you can just post in here) . I doubt there is any normal scenario where overflowing could happen, at least to any kind of advantage, but a lot of the reactor logic is quite old so imperfections are possible.
145 Mods isn't too many. 9 types of copper and 8 types of tin aren't too many. 3 types of coffee though?
I know that you believe that you understood what you think I said, but I am not sure you realise that what you read was not what I meant.
---- Minecraft Crash Report ----
// I just don't know what went wrong
I see this too much.
• Speaking of code, I noticed that placing a single Heat Vent increases causes the tool to say "10 iron" for the costs, but from what I can tell, it only costs 8.5 iron, and always has only cost 8.5 iron.
...
As far as I count, you need 1.5 iron for the iron bars, 4 iron for the plates, and 3 iron for the electric motor, making a total of 8.5.
I agree that 10 iron is incorrect, and I'm not sure quite where that came from - in August 2015 I changed the planner to show raw materials instead of components (as requested earlier in this thread), and if I had goofed and thought iron bars were 1:1 with iron ingots, that would have led to a recipe of 11 iron instead of 10. However, your total of "8.5 iron" is ignoring the fact that iron bars have to be crafted in batches of 16 (from 6 iron). If I changed the planner to calculate the iron cost of a heat vent as 8.5, designs with heat vent counts of 1, 5, 9, 13, etc. would undercount the effective iron requirements by 4.5. I haven't figured out how to deal with this yet.
• In my opinion, if you want to only give one number for the heat vent, it should be 8.5, not 9. If you round up for every heat vent, then the total is going to be off by a lot.
People who are competing to come up with the most effective design are probably looking at this as "I want to make 5-7 reactors, and get the most EU for the resources I spend". To most of us, we don't mind making a few extra iron bars for the first reactor, because we know that they'll eventually get use somewhere else. We just don't want to spend more on that first reactor than need.
Also, we may have some iron bars already in our inventory before our first reactor! I don't think that there's any way to accommodate that.
If you do want to do something far more complicated, you could output something at the end that says "Iron cost: 112-115 ingots". That would indicate to people that if they have 115 ingots, they can build this entire reactor without any more mining. If they are making multiple reactors, the average cost will be 112. Also note that although this concept is pretty easy, it could be somewhat difficult to implement. I'm not saying it's a necessary feature, haha!
• New
I looked at the other components that use iron bars in their recipes, and they had the iron counts double what they should be for those iron bars. The only explanation I can think of is that maybe I was looking at the GregTech 5 crafting recipe for iron bars (6 iron rods to 8 iron bars, and if I make the rods by using a file instead of in the lathe or extruder, they take a full ingot each, but that's wasting iron, and by the time a GT5 player is ready to build a nuclear reactor, more material-efficient ways to make iron rods (and iron bars) should be available.
Also, we may have some iron bars already in our inventory before our first reactor! I don't think that there's any way to accommodate that.
Considering that iron bars are required for the scuba helmet that is part of the complete hazmat suit, I'd be surprised if you didn't, come to think of it.
Anyway, material counts should be fixed in v2.2.1. | 1,699 | 6,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-39 | latest | en | 0.895547 |
https://johncarlosbaez.wordpress.com/2011/09/12/fools-gold/ | 1,656,382,741,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103347800.25/warc/CC-MAIN-20220628020322-20220628050322-00476.warc.gz | 370,686,235 | 35,020 | ## Fool’s Gold
My favorite Platonic solids are the regular dodecahedron, with 12 faces and 20 corners:
and the regular icosahedron, with 12 corners and 20 faces:
They are close relatives, with all the same symmetries… but what excites me most is that they have 5-fold symmetry. It’s a theorem that no crystal can have 5-fold symmetry. So, we might wonder whether these shapes occur in nature… and if they don’t, how people dreamt up these shapes in the first place.
It’s widely believed that the Pythagoreans dreamt up the regular dodecahedron after seeing crystals of iron pyrite—the mineral also known as ‘fool’s gold’. Nobody has any proof of this. However, there were a lot of Pythagoreans in Sicily back around 500 BC, and also a lot of pyrite. And, it’s fairly common for pyrite to form crystals like this:
This crystal is a ‘pyritohedron’. It looks similar to regular dodecahedron—but it’s not! At the molecular level, iron pyrite has little crystal cells with cubic symmetry. But these cubes can form a pyritohedron:
(By the way, you can click on any of these pictures for more information.)
You’ll notice that the front face of this pyritohedron is like a staircase with steps that go up 2 cubes for each step forwards. In other words, it’s a staircase with slope 2. That’s the key to the math here! By definition, the pyritohedron has faces formed by planes at right angles to these 12 vectors:
$\begin{array}{cccc} (2,1,0) & (2,-1,0) & (-2,1,0) & (-2,-1,0) \\ (1,0,2) & (-1,0,2) & (1,0,-2) & (-1,0,-2) \\ (0,2,1) & (0,2,-1) & (0,-2,1) & (0,-2,-1) \\ \end{array}$
On the other hand, a regular dodecahedron has faces formed by the planes at right angles to some very similar vectors, where the number 2 has been replaced by this number, called the golden ratio:
$\displaystyle {\Phi = \frac{\sqrt{5} + 1}{2}}$
Namely, these vectors:
$\begin{array}{cccc} (\Phi,1,0) & (\Phi,-1,0) & (-\Phi,1,0) & (-\Phi,-1,0) \\ (1,0,\Phi) & (-1,0,\Phi) & (1,0,-\Phi) & (-1,0,-\Phi) \\ (0,\Phi,1) & (0,\Phi,-1) & (0,-\Phi,1) & (0,-\Phi,-1) \\ \end{array}$
Since
$\Phi \approx 1.618...$
the golden ratio is not terribly far from 2. So, the pyritohedron is a passable attempt at a regular dodecahedron. Perhaps it was even good enough to trick the Pythagoreans into inventing the real thing.
If so, we can say: fool’s gold made a fool’s golden ratio good enough to fool the Greeks.
At this point I can’t resist a digression. You get the Fibonacci numbers by starting with two 1’s and then generating a list of numbers where each is the sum of the previous two:
1, 1, 2, 3, 5, 8, 13, …
The ratios of consecutive Fibonacci numbers get closer and closer to the golden ratio. For example:
$\begin{array}{ccl} 1/1 &=& 1 \\ 2/1 &=& 2 \\ 3/2 &=& 1.5 \\ 5/3 &=& 1.6666.. \\ 8/5 &=& 1.6 \\ \end{array}$
and so on. So, in theory, we could use these ratios to make cubical crystals that come closer and closer to a regular dodecahedron!
And in fact, pyrite doesn’t just form the 2/1 pyritohedron I showed you earlier. Sometimes it forms a 3/2 pyritohedron! This is noticeably better. The 2/1 version looks like this:
while the 3/2 version looks like this:
Has anyone ever seen a 5/3 pyritohedron? That would be even better. It would be quite hard to distinguish by eye from a true regular dodecahedron. Unfortunately, I don’t think iron pyrite forms such subtle crystals.
Okay. End of digression. But there’s another trick we can play!
These 12 vectors I mentioned:
$\begin{array}{cccc} (\Phi,1,0) & (\Phi,-1,0) & (-\Phi,1,0) & (-\Phi,-1,0) \\ (1,0,\Phi) & (-1,0,\Phi) & (1,0,-\Phi) & (-1,0,-\Phi) \\ (0,\Phi,1) & (0,\Phi,-1) & (0,-\Phi,1) & (0,-\Phi,-1) \\ \end{array}$
besides being at right angles to the faces of the dodecahedron, are also the corners of the icosahedron!
And if we use the number 2 here instead of the number $\Phi$, we get the vertices of a so-called pseudoicosahedron. Again, this can be made out of cubes:
However, nobody seems to think the Greeks ever saw a crystal shaped like a pseudoicosahedron! The icosahedron is first mentioned in Book XIII of Euclid’s Elements, which speaks of:
the five so-called Platonic figures which, however, do not belong to Plato, three of the five being due to the Pythagoreans, namely the cube, the pyramid, and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus.
So, maybe Theaetetus discovered the icosahedron. Indeed, Benno Artmann has argued that this shape was the first mathematical object that was a pure creation of human thought, not inspired by anything people saw!
That idea is controversial. It leads to some fascinating puzzles, like: did the Scots make stone balls shaped like Platonic solids back in 2000 BC? For more on these puzzles, try this:
• John Baez, Who discovered the icosahedron?
But right now I want to head in another direction. It turns out iron pyrite can form a crystal shaped like a pseudoicosahedron! And as Johan Kjellman pointed out to me, one of these crystals was recently auctioned off… for only 47 dollars!
It’s beautiful:
So: did the Greeks ever seen one of these? Alas, we may never know.
For more on these ideas, see:
• John Baez, My favorite numbers: 5.
• John Baez, This Week’s Finds in Mathematical Physics, "week241" and "week283".
• Ian O. Angell and Moreton Moore, Projections of cubic crystals, International Union of Crystallography.
To wrap up, I should admit that icosahedra and dodecahedra show up in many other places in nature—but probably too small for the ancient Greeks to see. Here are some sea creatures magnified 50 times:
And here’s a virus:
The gray bar on top is 10 nanometers long, while the bar on bottom is just 5 nanometers long.
The mathematics of viruses with 5-fold symmetry is fascinating. Just today, I learned of Reidun Twarock‘s recent discoveries in this area:
• Reidun Twarock, Mathematical virology: a novel approach to the structure and assembly of viruses, Phil. Trans. R. Soc. A 364 (2006), 3357-3373.
Most viruses with 5-fold symmetry have protein shells in patterns based on the same math as geodesic domes:
But some more unusual viruses, like polyomavirus and simian virus 40, use more sophisticated patterns made of two kinds of tiles:
They still have 5-fold symmetry, but these patterns are spherical versions of Penrose tilings! A Penrose tiling is a nonrepeating pattern, typically with approximate 5-fold symmetry, made out of two kinds of tiles:
To understand these more unusual viruses, Twarock needed to use some very clever math:
• Thomas Keef and Reidun Twarock, Affine extensions of the icosahedral group with applications to the three-dimensional organisation of simple viruses, J. Math. Biol. 59 (2009), 287-313.
But that’s another story for another day!
### 27 Responses to Fool’s Gold
1. Arrow says:
Typo in picture: 2/1=1 should be 1/1=1
2. Vasileios Anagnostopoulos says:
Dear John,
I am not a physics graduate but I have a big interest in physics. Recently I was reading some Group Theory for Physics texts and as it is usual the text starts with crystallographic groups. Unfortunately the analysis of the said text was not good or thorough. Can you recommend a text(s) for crystallographic groups with all the hard core analysis? Classification, defintions and so on.
Regards
Vasileios Anagnostopoulos.
• John Baez says:
Dear Vasileios – I don’t really have a favorite book on this subject. Maybe someone else can recommend one. And maybe you can tell use which books you were reading, so we can know what you consider insufficient.
Back home in California I have Group Theory and Its Application to Physical Problems by Morton Hamermesh. This has a lot of material on crystallographic groups. But I never really loved that book.
I am very fond of Conway, Doyle, Gilman and Thurston’s classification of wallpaper groups using orbifolds, which I discussed in "week267" of This Week’s Finds. Wallpaper groups are the 2d analogue of crystallographic groups.
Conway, Friedrichs, Huson, and Thurston have generalized this orbifold argument to crystallographic groups. But that’s probably not what you want! Their work seems to assume we already know the statement of the result to be proved.
3. The buckyball (C60) is even smaller (I mean much smaller than a virus) and its symmetry group is the same as that of a dodecahedron or an icosahedron (alternating group of degree 5).
4. Beauty is in the pyrite of the beholder.
5. John Baez says:
Berényi Péter mentioned buckyballs. Buckyballs are really cool!
First chemists speculated that carbon might form C60 molecules shaped like soccer balls, and later they managed to produce them. In took a bit of work to get ahold of at first… but later they realized that lowly soot contains lots of this stuff!
Since Buckminster Fuller was fond of this shape in his geodesic domes—with each pentagon or hexagon filled in by triangles—C60 and its relatives are called fullerenes, and the shape is affectionately called a buckyball. For more about this stuff, try:
• P. W. Fowler and D. E. Manolpoulos, An Atlas of Fullerenes, Oxford University Press, 1995.
• M. S. Dresselhaus, G. Dresselhaus, and P. C. Eklund, Science of Fullerenes and Carbon Nanotubules, Academic Press, New York, 1994.
• G. Chung, B. Kostant and S. Sternberg, Groups and the buckyball, in Lie Theory and Geometry, eds. J.-L. Brylinski, R. Brylinski, V. Guillemin and V. Kac, Birkhauser, 1994.
In fact, for the person who has everything: you can now buy 99.5% pure C60 at the following site:
Mathematically, we get the shape of a buckyball by
taking an icosahedron and chop off all 12 corners. The result is called a ‘truncated icosahedron’. It has 12 regular pentagonal faces, one for each corner of the original icosahedron. It has 20 regular hexagonal faces, one for each face of the original icosahedron. All its edges are the same length.
It’s called an Archimedean solid because, while not quite Platonic in its beauty, every face is a regular polygon and every vertex looks alike: two pentagons abutting one hexagon.
And now, some more fancy stuff!
Here’s a story about icosahedra, buckyballs, and last letter Galois wrote before his famous duel, taken from "week79" of This Week’s Finds, but based on this marvelous paper:
• Bertram Kostant, The graph of the truncated icosahedron and the last letter of Galois, Notices of the AMS 42 (September 1995), 959-968.
When I was a graduate student at MIT I realized that Kostant (who teaches there) was deeply in love with symmetry, and deeply knowledgeable about some of its more mysterious byways. Unfortunately I didn’t dig too deeply into group theory at the time, so I spent years later trying to catch up.
Let’s start with the Platonic solids. Note that the cube and the octahedron are dual—putting a vertex in the center of each of the cube’s faces gives you an octahedron, and vice versa. So every rotational symmetry of the cube can be reinterpreted as a symmetry of the octahedron, and vice versa. Similarly, the dodecahedron and the icosahedron are dual, while the tetrahedron is self-dual. So while there are 5 Platonic solids, there are really only 3 different symmetry groups here.
These 3 "Platonic groups" are very interesting. The symmetry group of the tetrahedron is the group A4 of all even permutations of 4 things, since by rotating the tetrahedron we can achieve any even permutation of its 4 vertices. The symmetry group of the cube is S4, the group of all permutations of 4 things. What are the 4 things here? Well, we can draw 4 line segments connecting opposite vertices of the cube; these are the 4 things! The symmetry group of the icosahedron is A5, the group of even permutations of 5 things. What are the 5 things? It we take all the line segments connecting opposite vertices we get 6 things, not 5, but we can’t get all even permutations of those by rotating the icosahedron.
To find the 5 things is a bit trickier; I leave it as a puzzle here. See:
• John Baez, Some thoughts on the number 6.
Once we convince ourselves that the rotational symmetry group of the icosahedron is A5, it follows that it has 5!/2 = 60 elements. But there is a much easier way to see this.
The truncated icosahedron has 5 × 12 = 60 vertices. Every symmetry of the icosahedron is a symmetry of the truncated icosahedron, so A5 acts to permute these 60 vertices. Moreover, we can find an element of A5 that moves a given vertex of the truncated icosahedron to any other one, since "every vertex looks alike". Also, there is a unique element of A5 that does the job. So there must be precisely as many elements of A5 as there are vertices of the truncated icosahedron, namely 60.
Kostant’s paper requires more knowledge of group theory to understand. He focuses on a fact buried in Galois’ last letter, written to the mathematician Chevalier on the night before Galois’ fatal duel. He was thinking about some groups we’d now call PSL(2,F). Here F is a field (for example, the real numbers, the complex numbers, or Zp, the integers mod p where p is prime). PSL(2,F) is a "projective special linear group over F."
What does that mean? Well, first of all, SL(2,F) is the 2×2 matrices with entries in F having determinant equal to 1. These form a group under good old matrix multiplication. The matrices in SL(2,F) that are scalar multiples of the identity matrix form the "center" Z of SL(2,F)—the group of guys who commute with everyone else. We can form the quotient group SL(2,F)/Z, and get a new group called PSL(2,F).
Now Galois was thinking about PSL(2,Zp) where p is prime. There’s an obvious way to get this group to act as permutations of p+1 things. Here’s how! For any field F, the group SL(2,F) acts as linear transformations of the 2-dimensional vector space over F, and it thus acts on the set of lines through the origin in this vector space… which is called the "projective line" over F. But anything in SL(2,F) that’s a scalar multiple of the identity doesn’t move lines around, so we can mod out by the center and think of the quotient group PSL(2,F) as acting on projective line. (By the way, this explains the point of working with PSL instead of plain old SL.)
Now, an element of the projective line is just a line through the origin in F2. We can specify such a line by taking any nonzero vector (x,y) in F2 and drawing the line through the origin and this vector. However, (x’,y’) and (x,y) determine the same line if (x’,y’) is a scalar multiple of (x,y). Thus lines are in 1-1 correspondence with vectors of the form (1,y) or (x,1). When our field F is Zp, there are just p+1 of these. So PSL(2,Zp) acts naturally on a set of p+1 things.
What Galois told Chevalier is that PSL(2,Zp) doesn’t act nontrivially as permutation of any set with fewer than p+1 elements if p > 11. This presumably means he knew that PSL(2,Zp) does act nontrivially on a set with only p elements if p = 5,7, or 11. For example, PSL(2,5) turns out to be isomorphic to A5, which acts on a set of 5 elements in an obvious way. PSL(2,7) and PSL(2,11) act on a 7-element set and an 11-element set, respectively, in sneaky ways which Kostant describes.
These cases, p = 5, 7 and 11, are the the only cases where this happens and PSL(2,Zp) is "simple" in the technical sense of group theorists. In each case it is very amusing to look at how PSL(2,Zp) acts nontrivially on a set with p elements and consider the subgroup that doesn’t move a particular element of this set. For example, when p = 5 we have PSL(2,5) = A5, and if we look at the subgroup of even permutations of 5 things that leaves a particular thing alone, we get A4. Kostant explains how if we play this game with PSL(2,7) we get S4, and if we play this game with PSL(2,11) we get A5. These are the 3 Platonic groups again!!
But notice an extra curious coincidence. A5 is both PSL(2,5) and the subgroup of PSL(2,11) that fixes a point of an 11-element set. This gives a lot of relationships between A5, PSL(2,5), and PSL(2,11). What Kostant does is take this and milk it for all it’s worth! In particular, it turns out that one can think of A5 as the vertices of the buckyball, and describe which vertices are connected by an edge using the embedding of A5 in PSL(2,11). I won’t say how this goes… read his paper!
• Platonic solids, or at least their symmetries, may come back at a fm scale, some six orders of magnitude smaller than that of a buckyball (~1nm).
I guess as long as a 3D Euclidean space is a reasonable local approximation, it keeps happening.
• John Baez says:
I don’t believe Robert Moon’s model of nuclear physics, but there is something smaller than a buckyball with the same symmetry group: dodecahedrane!
It’s formed from 20 carbons at the vertices of a dodecahedron and 20 hydrogens bonded to these.
Unfortunately this molecule hasn’t been found in nature yet, but chemists can synthesize it. It’s a real tour de force! These use reactions like these:
These pictures come from this review article:
• Leo A. Paquette, Dodecahedrane – the chemical transliteration of Plato’s universe (a review), Proc. Nat. Acad. Sci. USA 14 part 2 (1982), 4495-4500.
• It may well be the case that Moon’s model is flawed, still, I love the old man. The point is he enjoyed himself tremendously while having quite some indisputable achievements.
Anyway, the idea that preferred configurations of finite clusters of densely packed spheres has some relation to the polyhedral groups, is not inconceivable.
Just ponder on this 10\$ problem for example.
ps. This dodecahedrane thing is fascinating indeed, I have not known its synthesis is so complicated!
• John Baez says:
I hadn’t heard of Moon. If he did interesting work on clusters of densely packed spheres, perhaps it can be applied to gold atom nanoclusters and their kin.
The ability of chemists to do these complicated syntheses seems amazing to me. I have enough trouble making a really good omelette.
• John wrote
The matrices in SL(2,F) that are scalar multiples of the identity matrix…
in other words, I or -I. :)
• John Baez says:
Right. But my definition of PSL was designed to work for PSL(n,F), not just PSL(2,F). I was trying to give you the most for your money… which I can only afford because you’re not paying anything.
6. Boris Borcic says:
Keef and Twarock’s is a fascinating paper, thanks for passing it on. Although I am not up to otherwise comment on it, it prompts me to document a fun fact that I haven’t seen documented anywhere, and that has at least in common with their work to call for looking into 4D analogues (in turn a fun idea by itself, that mathematics would now allow to dream up 4D viruses in non-trivial logical detail).
I believe the right concise wording for my fun fact – and I would welcome rectification – is to say that the surface of the regular platonic dodecahedron
is isometric to the outer surface or shell of both Poinsot’s great dodecahedron
and to that of the first stellation of the icosahedron (aka small triambic icosahedron).
All three have icosahedral symmetry. The latter two are self-intersecting polyhedra so that a distinction needs to be made between their surface and the part of it that’s apparent to the outside; what I call the outer shell and what is usually the only surface considered when making paper models. To restate my fun fact in simpler words, if you imagine these surfaces made out of paper, it is in principle possible to refold any one of the three without any cutting, to obtain any of the other two. Easier in practice would be to cut up the surface enough to lay it out flat and then refold it and glue together the lips of the cuts.
This property has been well known to paper modelists for a long time in the case of the two self-intersecting polyhedra, but AFAIK it hadn’t yet been remarked that both are similarly related in turn to the regular dodecahedron. The reason this is not obvious is that the mapping from or to the regular dodecahedron preserves no edge or face, while in the case of the other two polyhedra the mapping relates both faces and edges one-to-one (when considering the outer shells as polyhedra in their own right, or else we probably need to speak of “folds” for edges and “flat polygons” for faces).
On the left is a flat surface from which either one of the three polyhedra displayed can be constructed. A single pentagon from the regular dodecahedron is shown in red, while in green is shown a single triangular face of the outer shell of either a great dodecahedron or a small triambic icosahedron.
The property I expose here has been well known to paper modelists for a long time in the case of the two self-intersecting polyhedra, but AFAIK the remark that both are similarly related to the regular dodecahedron, is new. The reason this was not obvious is that starting from the regular dodecahedron the mapping preserves no edge or face, while in the case of the self-intersectors the relationship relates the faces and edges one-to-one (when considering the outer shells as polyhedra in their own right, or else we probably need to speak of “folds” for edges and “flat polygons” for faces).
7. […] This is a response to a recent Azimuth blog post by John Baez. […]
8. John Baez says:
Nice stuff, Boris! I’m sorry the wonderful images in your comment did not appear at first. Because this is a free WordPress blog, it’s impossible for anyone but me to include images in their comments, and it’s also impossible for people to preview their comments. Sorry!
However, I routinely fix mistakes in people’s comments. And
if someone wants to include an image, they can just include the URL, e.g.
and I’ll do the rest. (Of course, this takes me a bit of work, so I’d appreciate it if people include images sparingly.)
To me the most interesting thing about Keef and Twarock’s work is that it suggests a relation between the more sophisticated patterns for viral protein coats and the higher-dimensional geometry underlying quasiperiodic tilings and quasicrystals.
For a great introduction to the higher-dimensional geometry behind quasiperiodic tilings, see:
• Greg Egan, De Bruijn and Tübingen.
In a nutshell, the idea is this. This shape has 5-fold symmetry:
but it’s really a creature from the 4th dimension! It’s just a particular view of a 4-simplex, which is the 4d analogue of a tetrahedron.
So, tilings with approximate 5-fold symmetry can be generated as slices of regular lattices in 4 dimensions!
Using this idea, Egan can generate pictures like this:
Click for large versions and go here for more sophisticated patterns. Everyone is free to use these patterns as wallpaper or computer art, but if you do, please credit Greg!
For more of the underlying math, see "week271".
9. Stephan says:
Small typos in the golden ratio part:
5/3 ≈ 1.67, not 1.6
8/5 = 1.6, not 1.625
13/8 = 1.625 (missing)
10. […] blogs, Libres pensées d’un mathématicien ordinaire remembered John Michael Hammersley while Azimuth studied Fool’s Gold. Flavors and Seasons shared a reflection on discussions and Regularize […]
11. John Baez says:
Johan Kjelmann, who works in the mineral collection at Uppsala Universitet, recently found some interesting crystals made of cobaltite (CoAsS). They’re from Tunaberg, Sweden. They’re quite small, but they’re shaped like pseudoicosahedra!
Johan notes that you can see more about cobaltite, including a rotating picture of a pseudoicosahedron, here:
Cobaltite, mindat.org.
12. John Baez says:
Johan Hjellman sent me an email that’s too interesting to keep to myself. So, I’ll post it here with a few small improvements. The most interesting part to me is that the pyrite crystals forming pyritohedra of type 5/3 and 13/8 have also been seen! The latter would be visually indistinguishable from a regular dodecahedron.
Hi, I browsed your blog. Interesting and informative stuff.
First comment is that pyrite crystals are not built up of small cubic cells, but the symmetry of the “building block”—the unit cell—is cubic, Alas, it is OK as an image. Thus, your sentence “At the molecular level, iron pyrite has little cubic crystal cells” would be better as follows: “At a molecular level, iron pyrite has a unit cell with (of?) cubic symmetry” or similar.
The concept that crystals are built up of polyhedra was introduced by the french mineralogist Rene Just Haüy around 1800 and his images of stacked crystal models became standard in the crystallography parts of the mineralogy books of the 19th c. I was quite surprised and happy to see them on your site in a modern context. Haüy was also heading a small manufacture of real wooden models that were sold from Paris in the first decades of the 19th c. We have a couple in our collection; my attachment shows “structure de dodecahedre pentagonal”, but there are some great collections in Berlin, Vienna, etc comprising several hundred models.
Second comment, as I am not a mathematician, it’s all a bit tricky—anyhow one concept I understood is your discussion on the pentagonal dodecahedra and their relation to Fibonacci numbers. I understand that the numbers are getting closer to the golden ratio, and you mentioned the 3/2 pyritohedron that is even a better approximation to the regular dodecahedron. So I went to a mineralogical classic- Hinte’s Handbuch der Mineralogie and checked out what Miller indices have actually been observed on pyrite and it turns out that the forms (5.3.0) and (13.8.0) have also been observed.
• Johan Kjellman says:
John, I haven’t read the original references for the forms (5.3.0) and (13.8.0) – for sure if any of these forms were the only one present on the crystals they would for sure be fantastic near perfect regular dodecahedra – but most likely they only appeared as tiny faces modifying crystals of more common forms.
• John Baez says:
Thanks – I over-optimistically imagined full dodecahedral crystals.
• Wolfgang says:
Hello,
Since you mention Penrose and quasiperiodic patterns I assume that you are possibly aware of perfect (!) dodecahedral quasicrystals found in an Ho-Mg-Zn alloy (see
http://en.wikipedia.org/wiki/File:Ho-Mg-ZnQuasicrystal.jpg ).
By the way, talking as a devils advocate, I admire Twarocks virus tiling, but did anyone calculate the actual (spherical) angles of the tiles on the sphere? They cannot be very Penrose-tile-like, since such a tiling would be mostly planar, given its connectivity as shown for the virus. Thus, they have to be distorted? And another thought: Twarock combines kites and large rhombs, taking tiles from both distinct Penrose tilings that would not match in the plane, but what about the other choice, darts and small rhombs, would they tile a spherical surface, too? I wonder why mathematicians most often still do not look at “less-ideal” cases, there would be an awful amount of discoveries waiting.
Finally, back to pyrite. I found it most interesting that this substace grows in crystals with curved (!) faces. One may think of flattening out the pyritohedron in order to obtain them. See
http://onlinelibrary.wiley.com/doi/10.1002/zaac.200400404/abstract or the corresponding article on http://www.sandforsk.se .
Greetings
13. Wolfgang says:
I forgot another substance forming near-miss pentagon-dodecahedral crystals…
Click to access JCrystalGrowth206%281999%2993.pdf
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 7,111 | 27,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2022-27 | latest | en | 0.891977 |
https://www.cagednomoremovie.com/what-are-the-different-types-of-differential-manometer/ | 1,675,300,295,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499954.21/warc/CC-MAIN-20230202003408-20230202033408-00554.warc.gz | 702,623,897 | 12,587 | # What are the different types of differential manometer?
## What are the different types of differential manometer?
The differential manometer is further classified into three types:
• Two piezometer manometer.
• U-tube differential manometer.
• Inverted differential manometer.
## What are manometers and how are they classified?
It is also known as low pressure type manometer. In this type, one side of transparent tube or column is attached the pipe where the pressure of the liquid is to be measured and the other end is open into the atmosphere. This type of manometer only measures low positive pressure it cannot measure negative pressure.
What is the difference between simple manometers and differential manometers?
A simple manometer consists of a glass tube having one of its ends connected to a point where pressure is to be measured and other end remains open to atmosphere. A differential manometer is a device that measures the difference in pressure between two places.
What is a manometer What are the two types of manometers?
A Manometer by definition is a device used for measuring the pressure of a fluid by balancing it against a pre-determined column of fluid. The most common type of manometer which almost all us might have seen is sphygmomanometer. It is the device which doctors use to measure your blood pressure.
### What is the difference between U-tube differential manometer and inverted U-tube differential manometers?
The U-tube Inverted differential manometer can be used to measure a very small pressure difference between two points. This U-tube Inverted differential manometer is just the inverted version of the U-tube upright differential manometer.
### Which of the following is limitation of manometers?
They are fragile and hence offer less portability. The manometric fluids density depends on temperature. Hence errors may result due to change in the temperature. It has a slow response which makes it unsuitable for measuring fluctuating pressures.
What is the difference between vacuum pressure and gauge pressure?
Gauge pressure and vacuum pressure are both measured relative to the atmospheric pressure. The difference is that gauge pressure is higher than atmospheric pressure, while vacuum pressure is lower than the atmospheric pressure.
What are the 3 types of U-tube manometer?
Types of Manometer
• U-Tube Manometer.
• Differential U-Tube Manometer.
• Inverted U-Tube Manometer.
• small Manometer.
• Inclined Manometer.
#### Where are manometers used?
What are the applications of a manometer?
• It can be used for pressure monitoring applications.
• It can also be used to monitor the air and gas pressure for the compressor.
• A manometer can be used to measure the static pressure and vacuum.
• Mercury absolute manometers are used in power plants.
#### What are the advantages and limitations of manometers?
• It has a low cost.
• The manometer is suitable for low-pressure applications.
• It is simple in construction.
• It has better sensitivity.
• It has good accuracy.
• It has simple operation and construction.
Which sensor measures pressure below than atmospheric pressure?
vacuum gauge
A vacuum gauge is a pressure gauge used to measure pressures lower than the ambient atmospheric pressure, which is set as the zero point, in negative values (for instance, −15 psig or −760 mmHg equals total vacuum).
How many PSI is full vacuum?
14.7 psi
features a psi of 0, that means that the ducting in question is at 0% vacuum pressure (otherwise known as “standard atmosphere”). -14.7 psi, on the other hand, is absolute full vacuum pressure, meaning that the duct hose is as empty as can ever be possible. | 753 | 3,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-06 | latest | en | 0.953591 |
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(d) $$\frac{1}{9}$$ Go back to Class 9 Maths Chapter 1 Solutions to see other exercises. Also, we know that an irrational number is non-terminating non-recurring. Here you can download NCERT Exemplar Problems for Class 9 Maths Solutions PDF, all solutions are listed chapter wise.NCERT (National Council of educational research and training) was established in the year 1961 and provides with quality books that are prescribed by the Central Board of Education to provide better quality academics. Learning of fundamentals in maths is always essential. An irrational number between 1/3 and 1/2 i.e., between 0-3 and 0.5 is 0.4141141114…………. Students must be familiar with the questions. With this article, students can download the complete PDF. Now, draw an arc with centre B and radius BD, meeting AC produced at E, such that NCERT Solutions for Class 12 Maths Chapter 1 is the best & smart way to master in relations and functions chapter.All the solutions are well-structured by the subject experts based on the NCERT Syllabus prescribed by CBSE Board. NCERT Exemplar Class 9 Mathematics Chapter 13 Surface Areas and Volumes are provided here for students to prepare for exams and make them score well. NCERT Class 9 Mathematics Exemplar Solution Chapter Unit 1 Number Systems are the complete study materials for any student studying in class 12. = 99x = 40 => y is a rational number. Which of the following is equal to x? You will also learn about discount prices and different formulas related to these terms. As we know positive numbers lies on right hand side of 0 and negative numbers lies on left hand side of 0. 1Ox – x = (8.888…) – (0.888…) ⇒ 9x = 8 NCERT Exemplar Class 8 (PDF) Download, MCQs 2021. Chapter 4 - Linear Equations in Two Variables. Thus x = 40/99. NCERT Solutions for Class 9th: Ch 14 Statistics Maths Page No: 239 Exercise 14.1 1. Linear Equations in Two Variables. NCERT Exemplar Class 9 Maths is officially provided by the National Council of Education Research and Training, on its official website, ncert.nic.in. NCERT Exemplar 9th Maths Ch 10 … Let us draw BD perpendicular to AC passing through a point B, intersecting the semi-circle at a point D. The NCERT Solutions class 9 maths is solved keeping various parameters in mind such as stepwise marks, formulas, mark distribution, etc., This in turn, helps you not to lose even a single mark. Go through Maths answers by experts to figure out whether the given decimal is non-terminating repeating or terminating. (c) Between two rational numbers, there are infinitely many rational numbers. NCERT Solutions for Class 9 Maths. Practice in Google Form. (vii) A rational number between 2.357 and 3.121 is 3. Latest News. Question 1. Menu. NCERT Exemplar for Class 9 Maths Chapter 2 With Solution | Polynomials. 10x = 2.5555… … (2) Now, draw an arc with centre B and radius BD, meeting AC produced at E, such that Solution: Find a number which is non-terminating non-recurring lying between them. 100x = 0.3232… …(2) Chapter 4 - Linear Equations in Two Variables Chapter 8 - Quadrilaterals. (b) there is exactly one rational number NCERT Solutions for Class 9 Maths Chapter 1 Number Systems. NCERT Exemplar Class 9 for Maths Chapter 1 – Number Systems. Rational numbers, Irrational numbers, Irrational numbers between two number are the topics covered under this chapter. ⇒ 99x = 13.3 (v) 0.15 and 0.16 NCERT Solutions for Class 9 Maths Chapter 1 Number Systems available in Hindi Medium and English medium PDF to download. The product of any two irrational numbers is Justify your answer by an example. The chapter has a total of 96 Exemplar Problems. Thus, 1.010010001… is an irrational number. If we want to measure our height, we would use feet or meters. 100x = 40.4040… …(2) Give an example in support of your answer. Is x+y necessarily an irrational number? The value of 1.999… in the form of p/q, where p and q are integers and q≠0, is (ii) Amount of rainfall. Let x and y be rational and irrational numbers, respectively. (d) a whole number. Solution: These maths NCERT solutions of class 9 chapter 1 not only help with studying for … Since quotient of two non-zero rational number is a rational number. (iv) -2/5 and -1/2 NCERT Exemplar Class 7 Maths Solutions Chapter 1: Integers. Books. An irrational number between 0 and 0.1 is 0.007000700007………. CBSE Class 12 Syllabus: 2020-21 (All Subjects) - Download PDF just … Chapter 2 - Polynomials Chapter 3 - Coordinate Geometry. On multiplying both sides of (1) by 10, we get Thus 23/90, (vi) Let x = $$0.1\bar { 34 }$$ or x = 0.13434… …(1) You can opt for Chapter 13 - Surface Areas and Volumes NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below. Chapter 2 - Polynomials Chapter 3 - Coordinate Geometry. (viii) .404040… Taking OB = √17 as radius and point line at point P. Thus, the point P represents √17 on the number line. 1. (iv) True, because there are infinitely many numbers which cannot be written in the form p/q, q ≠0. (a) there is no rational number Class 12. (iii) 5.2 Let O be the mid-point of AC. (iii) 1/3 and 1/2 Solution: Solving Latest year 2021 Exemplar Problems Solutions for Class 9 Mathematics is the best option to understand the concepts given in NCERT books and do advanced level preparations for Class 9 exams. (a) a natural number. Represent the given numbers on the number line as shown below: Question5. Solution: Exercise: 1.1. (d) sometimes rational, sometimes irrational (d) We know that, the product of any two irrational numbers is sometimes rational and sometimes irrational. But, this contradicts the fact that y is an irrational number. This solution contains questions, answers, images, explanations of the complete chapter 1 titled Number Systems of Maths taught in class 9. The textbook of Maths Exemplar class 9 can be very useful for the students to score well in the exams. On subtracting (1) from (2), we get If possible, let xy be a rational number. p q, where p and q are integers and q ≠0. Class 9 Maths solutions for this chapter helps you confidently answer the NCERT textbook questions on topics such as parallel lines, co-interior angles, transversals, angle sum property of a triangle etc. (ii) 0 and 0.1 Chapter 9 - Areas of Parallelogram and Triangles. Thus x = 32/9900 = 8/2475, (viii) Let x = 0.404040… …(1) NCERT Solutions For Class 9 Maths: Chapter 1 Number Systems. we have 2√3 + √3 = √3(2+1) = 3√3, Question 11. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE … An irrational number between 3.623623 and 0.484848 is 1.909009000………. Questions from very important … Practise with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 1 Number Systems to understand the real number system. (i) x2 =5 Students can also Download FREE PDF of NCERT Exemplar for Class 9 Maths Chapter 10 with Solutions. (i) We have, 5 = 2² +1 Now, draw a right angled ΔOAB, in which OA = 2 units and AB = 1 unit and ∠OAB = 90°. The students will be able to cover their CBSE syllabus. Download Toppr – Best Learning App for Class 5 to 12. Answer Five examples from day-to-day life: (i) Daily expenditures of household. Now, we will draw a semi-circle with centre O and radius OA. These are type of questions can be asked in the examination. Questions from very important … These solutions are prepared according to the NCERT and CBSE Syllabus. Subjects: Science & Mathematics. (vi) √2 and √3 (c) a real number. (vii) .00323232… (iii) Bill of electricity. NIOS Books. Solution: (iii) √8.1 Chapters covered in class 9 ncert exemplar maths book are given below Chapter 1- Number Systems This chapter contains problems based on topics like representation of real numbers on number line, real numbers and their decimal expansion, irrational number problems, exponent laws for real numbers, etc. Now, we will draw a semi-circle with centre O and radius OA. Multiply (i) by 10, we get (i) False, here √2 is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number. Extra questions with answers. (ii) There are infinitely many integers between any two integers. p,q both are integers and these numbers are called irrational numbers. By using Pythagoras theorem, we get Free Download NCERT Solutions for Class 9 Maths Chapter 14 Exercise 14.1 & Exercise 14.2 statistics in PDF form. (a) 6√5 (b) an integer. NCERT exemplar problems and solutions for CBSE Class 10 Mathematics chapter 8 , Introduction to Trigonometry and its Applications (Exercise 8.1), is available here. (viii) .0001 and .001 Rationalise the denominator of the following Toppr Nation is providing NCERT exemplar solutions for class IX Maths and Science. (iv) √2.3 Download NCERT solutions for this chapter here. NCERT Solutions based on latest NCERT Books as well as NCERT Exemplar Problems chapters are being updated for new academic year 2020-21. NCERT Solutions for Class 9 Maths Chapter 1- Number Systems As the Number System is one of the important topics in Maths, it has a weightage of 6 marks in Class 9 Maths exams. NCERT Exemplar for class 9 for Maths Chapter 1 – Number Systems. Chapter 3 carries 3 exercises. Decimal representation of a rational number cannot be In this chapter, we will learn. (c) 2 Back of Chapter Questions . Question -1 Is zero a rational number? Solution: Question 1. BE = BD = √2.3 units. Thus, 10.124124 is a rational number. These solutions will help students to frame a better understanding of all the topics. NCERT Solutions for Class 9 Maths. Exemplar questions have been designed by NCERT to include practice questions of all levels and students capabilities. 10x = 19.999… (ii) Let us draw BD perpendicular to AC passing through a point B, intersecting the semi-circle at a point D. (iv) A rational number between -2/5 and 1/2 is 0. NCERT Exemplar Problems textbook is very helpful to understand the basic concepts of Mathematics. On multiplying both sides, of (1) by 10, we get Can you write it in the form . These exemplars problems and solutions are designed by experts in accordance with the CBSE latest syllabus (2020-2021). Each chapter has been divided into various exercises which include problems based on different concepts and topics. CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 9 Maths Chapter 12 Heron’s Formula solved by expert teachers as per NCERT (CBSE) Book guidelines. 1. Solution: Question 10. 10x-x = (2.555…)-(0.255…) ⇒ 9x = 2.3 Important features of Class 10 Mathematics NCERT Exemplar are: ... CBSE Class 10 Mathematics NCERT Solutions: Chapter 12, Areas Related to Circles; CBSE Class 10 Mathematics NCERT Solutions: Chapter 11, Constructions; Top Ranking CBSE Sample Paper Books for 2021 Boards. (c) a real number On subtracting (1) from (2), we get Question 1. Thus, 0.5918 is a rational number. as radius and point O as centre, draw an arc which meets the number line at point P. (c) non-terminating recurring This is one subject that will be useful for every student irrespective of their branch. Let x = 1.999… (i) Practise with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 1 Number Systems to understand the real number system. NCERT Solutions For Class 9 Maths: Chapter 1 Number Systems. We have prepared chapter wise solutions for all characters are given below. Now, draw an arc with centre B and radius BD, meeting AC produced at E, such that On multiplying both sides of (1) by 10, we get 1/7 solution: Question 9 of Education Research and Training, on its official,. Chapter 2 – Polynomials the ratio 3: 5: 9: Comparing Quantities number! 1 Solutions to help you to get good marks in your emails given decimal is non-terminating but recurring 7.2! Two parallels it in the Class 9 Maths with Solutions to help you to revise Syllabus! 3.121 is 3 Exemplar Class 8 ( PDF ) download, MCQs.... A line segment AB = 2.3 units and extend it to c that. The form p/q, q≠0 p, q ≠0 Maths Chapter-14Statistics and Probability are solved by faculty. Of concepts is also made for your easy understanding images, explanations of the complete PDF of following! Simple methods and shortcut by using important formulas solution for easy understanding of students and. 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The Exemplar book a and b in each of the Chapter wise Solutions for Class,!: Comparing Quantities balls she plays can improve their Maths skills with the help of NCERT Exemplar for Class to. And get all Exercise Solutions in your emails you can collect from your day-to-day life (! Base and between the same base and between the same parallels the help of NCERT with... Chapter 2 with solution | Polynomials 0.484848 is 1.909009000……… when x ≠0 will surely your! Go through Maths answers by experts to figure out whether the following statements are true or False the end this. Guidance of well qualified teachers the Exemplar book is done properly under the of... Of Education Research and Training, on its official website, ncert.nic.in, and integers, this the! Following statements are true or False score well in the Exercise of textbooks there are numbers can. Times out of 30 balls she plays 14.1 & Exercise 14.2 statistics in PDF form the that... 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Also made for your easy understanding be available till May, 2020 0, then =... 9 Mathematics ( Chapter 1 number Systems are the topics covered under this Chapter types of,! Base and the two parallels the denominator of the complete Chapter 1 number Systems to understand real! Probability that she did not hit a boundary 6 times out of 30 balls she plays of household does. Is 1.585585558……… out main Page materials for any student studying in Class 12 the following solution Exercise. A = 2 + √3, which is rational non-terminating non-recurring 14 chapters in the Exercise of there! Q are integers an average three questions are asked from this unit irrational number between 2.357 and is. 9 for Maths given here comes with step by step Solutions to help you to get good in... For CBSE ncert exemplar class 9 maths solutions chapter 1 9th: Ch 14 statistics Maths Page No: 239 Exercise 14.1 & Exercise 14.2 in! 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For Chapter wise Solutions for Class 9 Maths Chapter 1 with solution | number Systems – here all! Written in the NCERT and CBSE sample papers can be downloaded in.. 15 Probability Ex 15.1 that √3 is solution: ( i ) a rational between... Convenient & fun, 8, 9, 10, 11 and 12 for FREE PDF! Solution in each of the complete CBSE Syllabus Formula Ex 12.1 help you to get marks! Per the CBSEClass 9th Maths Syllabus hand side of 0 learn about discount and. There are different type of questions like MCQs, MCQ, Short and long Answer questions.: integers Exemplar for Class 9 Maths solution check out main Page that. Syllabus ( 2020-2021 ), 2020, xy is necessarily irrational only when x ≠0 a... Per ncert exemplar class 9 maths solutions chapter 1 CBSEClass 9th Maths Syllabus Log in ; Search for: Search regarding! Such that BC = 1 unit of their branch 2.040040004………….. ( ix ) a number... Chapters or other subjects NCERT Solutions for Class 9 Maths Chapter 1 number... | 7,574 | 30,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-31 | latest | en | 0.900763 |
http://oeis.org/A062405 | 1,591,452,805,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513321.91/warc/CC-MAIN-20200606124655-20200606154655-00222.warc.gz | 83,579,187 | 3,978 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A062405 Sums of specific numbers when generating a type of triangle based on pi(x). 0
2, 2, 5, 5, 8, 16, 11, 13, 18, 19, 20, 28, 30, 37, 30, 39, 39, 57, 46, 44, 52, 64, 62, 75, 60, 71, 79, 85, 74, 83, 90, 88, 95, 100, 111, 96, 104, 105, 115, 144, 117, 125, 148, 126, 132, 143, 165, 165, 144, 160, 172, 161, 174, 194, 173, 194, 198, 174, 212, 200, 205 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS Chris Caldwell, How Many Primes Are There? FORMULA Pi(x) denotes the number of prime numbers less than or equal to x. Starting with x = 2 take Pi(2x) - Pi(x). The numbers 1, 1, 2, 1, 2, 2, 2, 3, 4, 3, ... will be found. Now create a new sequence based on the number of repeats. 2, 1, 1, 3, 1, 1, 1, ... Arrange these in a triangle such that 2 is the first row; 1, 1 is the second row; 3, 1, 1 is the third row; 1, 1, 2, 1 is the fourth row; etc. taking one more term each time. Now take the sum of the numbers on each row and this interesting sequence is generated. EXAMPLE Example: The first number on the triangle is 2 because the number 1 is repeated twice in the sequence of Pi(2x) - Pi(x). CROSSREFS Cf. A000040, A000720. Sequence in context: A240495 A304393 A325535 * A071181 A213675 A321185 Adjacent sequences: A062402 A062403 A062404 * A062406 A062407 A062408 KEYWORD easy,nonn AUTHOR Rory Kulz (entropix(AT)amnaria.com), Jul 08 2001 EXTENSIONS More terms from David Wasserman, Jun 27 2002 STATUS approved
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Last modified June 6 10:12 EDT 2020. Contains 334859 sequences. (Running on oeis4.) | 687 | 1,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-24 | latest | en | 0.762056 |
http://www.sixsigmacertificationcourse.com/prioritization-matrix/ | 1,725,989,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00562.warc.gz | 54,358,168 | 13,657 | # Prioritization Matrix
In the previous topic, you created a tree diagram. Key issues identified must also be sorted in the order of their importance. A prioritization matrix helps determine the order for dealing with different issues or selecting solutions according to their relative importance. In this topic, you will create a prioritization matrix to identify crucial issues that require immediate attention.
You must prioritize key issues identified according to their importance to determine which issues or solutions are the most critical and need to be addressed first. Using a prioritization matrix, you can determine such crucial areas where Six Sigma projects need be implemented to improve your business processes.
#### What are Prioritization Matrices?
A prioritization matrix is a tool used for determining the most important issues or solutions. This tool can be used for any prioritization activity. In a Six Sigma project, it can be used for filtering or prioritizing either causes or solutions. For example, the cause-and-effect matrix is a prioritization matrix used for prioritizing causes, while the criteria-based matrix is used for prioritizing solutions.
Whatever the matrix, the procedure used for prioritization is the same. The Six Sigma team identifies different criteria to be used for measuring different solutions. The relative importance of each criterion is determined and a numerical value indicating the weight of each criterion is entered in a column. Ratings are then assigned to the solutions against different criteria. The different solutions are scored against the identified criteria. Each rating is then multiplied by the criteria weights to obtain the weighted scores. The weighted scores are then added to find the cumulative value. The option with the highest total value is regarded as the option with the highest priority.
Fig: A sample prioritization matrix
#### What are the use of Prioritization Matrices?
• Prioritizing complex issues involving several criteria against which the issues are assessed.
• Assigning scores to the criteria or issues when data is available.
• Choosing key areas to be focused upon immediately.
• And, garnering team support and approval of crucial issues.
#### How to Create a Prioritization Matrix?
To create a prioritization matrix for determining key areas where Six Sigma projects need be implemented first:
1. Gather the team members required to participate in the exercise.
1. It is not necessary to limit the team to only the project team members.
2. Involve all stakeholders who would have influence on the items to be prioritized.
3. Ideally, the size of the team should be around eight.
4. Agree upon the scope and the duration of the exercise because there is a tendency to drift from the main subject of interest, which may consume more time.
2. Identify criteria against which the items to be prioritized can be weighed. Select the criteria according to your business and the nature of the process.
3. Assign weights to the criteria on a scale of 1 to 10.
1. Use an ascending scale while assigning the weight. Usually, the higher the better.
2. Draw consensus within the team on the weights.
4. Assign ratings for each item to be prioritized against different criteria on a scale of 1 to 10. Alternatively, the team can also pick solutions one by one and gather votes for each criterion. Members who agree that a solution fits a particular criterion vote for it. Ensure that necessary information is available to the team before voting.
5. Compute the total score for each item.
1. Multiply the rating for an item by the weight.
2. Add the total scores for each item.
6. Continue doing this for every criterion of all the items.
7. Shortlist the items with higher scores for implementation. | 729 | 3,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.923435 |
https://mathematica.stackexchange.com/questions/213480/need-help-solving-ode-with-parametricndsolvevalue-solver?noredirect=1 | 1,600,623,670,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198287.23/warc/CC-MAIN-20200920161009-20200920191009-00258.warc.gz | 517,185,683 | 35,890 | # Need help solving ODE with ParametricNDSolveValue solver
I am trying to apply @bbgodfrey method for finding the right initial guess for the following problem but I get an error message and I don't understand where it comes from.
I would also like to know why FindRoot in Do loop doesn't work in this case.
Below is my minimal working code:
l1 = 0.81;
Z = 500;
x0 = 10;
v0 = 0.02;
\[Epsilon] = $MachineEpsilon; yl = -12; yu = 0; l0 = 0.071420.; ps = ParametricNDSolveValue[{y''[r] + 2 y'[r]/r == -4 \[Pi] l k Exp[-y[r]], y[\[Epsilon]] == y0, y'[\[Epsilon]] == 0, WhenEvent[r == 1, y'[r] -> y'[r] + Z l]}, {y, y'}, {r, \[Epsilon], R}, {k, l}, Method -> {"StiffnessSwitching"}, AccuracyGoal -> 5, PrecisionGoal -> 4, WorkingPrecision -> 15]; Do[x = i x0; v = i^3 v0; yl = -12; yu = 0; R = Rationalize[v^(-1/3), 0]; l = Rationalize[l1/(i x0), 0]; fy := (Do[yc = (yl + yu)/2; (* guess finder function *) test = First[ps[yc]]["Domain"][[1, 2]]; If[test == 1, Throw[yc]]; If[Last[ps[yc]][test] > 0, yu = yc, yl = yc], {i, 50}]; yc); yint = Which[1 == First[ps[yl]]["Domain"][[1, 2]], yl, 1 == First[ps[yu]]["Domain"][[1, 2]], yu, True, Catch[fy]]; nn = FindRoot[Last[ps[y0]][R], {y0, yint}, Evaluated -> False][[1, 2]]; Tot = 4 \[Pi] nn NIntegrate[ r^2 Exp[-First[ps[nn, l]][r]], {r, \[Epsilon], R}, PrecisionGoal -> 4]; Print[NumberForm[i*1., 5], " ", NumberForm[Tot, 5]];, {i, 292/100, 31/10, 1/100}] • The first error message your code gives me says that ps was called with one parameter having a value of -12. That would be in the line ψint = ... ps[ yl ] ... Your definition of ps expects 2 parameters. However, the definition of ps contains y0 and R, which are undefined. Perhaps y0 and R should be a third and fourth parameters of ps ? If they are not parameters, they need numeric values. Also, to debug loops, start with With[ {i=3}, ... ] and add statements one at a time, making sure each one works for the value of i. Then try to replace the With with Table instead of Do. – LouisB Jan 25 at 3:25 • @LouisB First, I need to work with Do. And in Do statements R is numerical value. For y0, I belived that the code should take yint and pass it to y0. Also, there is a typo ψint should be yint. – aluuzz Jan 25 at 3:37 • Just to understand the syntax, try this immediately after your ps = statement, y0 = 1; R = 1.2; Plot[ ps[0.5, 0.027] [[1]] [r], {r, 0, R}, PlotRange -> {{0, R}, {0, 4}}]. This indicates we must set values of y0 and R, then evaluate ps with 2 parameters, followed by a subscript of 1 or 2 followed by [ r ]. Of course we could use First and Last instead of the [[ ... ]] notation. – LouisB Jan 25 at 5:00 • @LouisB OK. I just want to let know that the code that I am using is taken from here [mathematica.stackexchange.com/questions/183148/…, and it works fine. What I am asking here is to implemented bbgodfrey method for finding the right initial guess to the above code. – aluuzz Jan 25 at 5:12 • @aluuzz 1) what problem do you want to solve? 2) it looks so that you have syntax errors there. – Alex Trounev Jan 28 at 20:53 ## 1 Answer We can use my code without changing for Z<=15000 l1 = 0.81; Z = 15000; x0 = 10; v0 = 0.02; \[Epsilon] =$MachineEpsilon;
l0 = 0.071420.;
ps = ParametricNDSolveValue[{y''[r] +
2 y'[r]/r == -4 \[Pi] l k Exp[-y[r]], y[\[Epsilon]] == y0,
y'[\[Epsilon]] == 0, WhenEvent[r == 1, y'[r] -> y'[r] + Z l]}, {y,
y'}, {r, \[Epsilon], R}, {k, l},
Method -> {"StiffnessSwitching"}, AccuracyGoal -> 5,
PrecisionGoal -> 4, WorkingPrecision -> 15];
Do[x = i x0;
v = i^3 v0;
R = Rationalize[v^(-1/3), 0];
l = Rationalize[l1/(i x0), 0];
nn = FindRoot[Last[ps[y0, l]][R], {y0, -1}, Evaluated -> False][[1,
2]];
Tot = 4 \[Pi] nn NIntegrate[
r^2 Exp[-First[ps[nn, l]][r]], {r, \[Epsilon], R},
PrecisionGoal -> 4];
Print[NumberForm[i*1., 5], " ", NumberForm[Tot, 5]];, {i, 292/100,
31/10, 1/100}] // Quiet
Out:
2.92 15007.
2.93 15007.
2.94 15006.
2.95 15006.
2.96 15006.
2.97 15006.
2.98 15006.
2.99 15006.
3. 15006.
3.01 15006.
3.02 15006.
3.03 15006.
3.04 15006.
3.05 15006.
3.06 15006.
3.07 15006.
3.08 15005.
3.09 15005.
3.1 15005.
I checked code @bbgodfrey (question as it is). The result was much worse than with my code, but the code works for some k up to Z = 20000:
p[Z0_, g0_, k0_, R0_] :=
Block[{Z = Z0, g = Rationalize[g0, 0], k2 = Rationalize[k0, 0],
yl = -8, yu = 0, ps, fy, y00, sol}, \[Epsilon] =
10^-4; R = Rationalize[R0, 0];
ps = ParametricNDSolveValue[{y''[r] +
2 y'[r]/r == -4 Pi k2 Exp[-y[r]], y[\[Epsilon]] == y0,
y'[\[Epsilon]] == 0,
WhenEvent[r == 1, y'[r] -> y'[r] + Z g]}, {y,
y'}, {r, \[Epsilon], R}, {y0}, Method -> "StiffnessSwitching",
WorkingPrecision -> 20];
fy := (Do[yc = (yl + yu)/2;
tst = First[ps[yc]]["Domain"][[1, 2]];
If[tst == R, Throw[yc]];
If[Last[ps[yc]][tst] > 0, yu = yc, yl = yc], {i, 50}]; yc);
y00 = Which[R == First[ps[yl]]["Domain"][[1, 2]], yl,
R == First[ps[yu]]["Domain"][[1, 2]], yu, True, Catch[fy]];
sol = FindRoot[Last[ps[y0]][R], {y0, y00}, Evaluated -> False][[1,
2]]; L =
4 \[Pi] k2/g0 NIntegrate[
r^2 Exp[-First[ps[sol]][r]], {r, \[Epsilon], R}]]
Using of p[]
l1 = 0.81;
Z0 = 20000;
x0 = 10;
v0 = 0.02;
l0 = 0.071420.; k = 1/29;
Do[x = i x0;
v = i^3 v0;
R = Rationalize[v^(-1/3), 0];
l = Rationalize[l1/(i x0), 0];
Print[NumberForm[i*1., 5], " ",
NumberForm[p[Z0, l, k l, R] // Quiet, 5]];, {i, 292/100, 31/10,
1/100}]
Out
2.92 20010.
2.93 20010.
2.94 20010.
2.95 20009.
2.96 20009.
2.97 20009.
2.98 20009.
2.99 20008.
3. 20008.
3.01 8.3397*10^(21)
3.02 1.309*10^(22)
3.03 2.0575*10^(22)
3.04 3.2386*10^(22)
3.05 5.1048*10^(22)
3.06 8.0576*10^(22)
3.07 1.2736*10^(23)
3.08 2.0159*10^(23)
3.09 3.1952*10^(23)
3.1 5.0714*10^(23)
` | 2,308 | 5,727 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-40 | latest | en | 0.80461 |
https://www.maths.tcd.ie/~dwilkins/Courses/MA232A/Euclid_ETexts/EuclidBookI_Lardner1855/EuclidBookIProp08_Lardner1855.php | 1,544,867,137,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826842.56/warc/CC-MAIN-20181215083318-20181215105318-00423.warc.gz | 969,870,687 | 3,822 | Euclid, Elements of Geometry, Book I, Proposition 8 (Edited by Dionysius Lardner, 1855)
Proposition VIII. Theorem.
[Euclid, ed. Lardner, 1855, on Google Books]
(68) If two triangles (A B C and E F D) have two sides of the one respectively equal to two sides of the other (A B to E F and C B to D F), and also have the base (A C) equal to the base (E D), then the angles (B and F) contained by the equal sides are equal.
For if the equal bases A C, E D be conceived to be placed one upon the other, so that the triangles shall lie at the same side of them, and that the equal sides A B and E F, C B and D F be conterminous, the vertex B must fall on the vertex F; for to suppose them not coincident would contradict the seventh proposition. The sides B A and B C being therefore coincident with F E and F D, the angles B and F are equal.
(69) It is evident that in this case all the angles and sides of the triangles are respectively equal each to each, and that the triangles themselves are equal. This appears immediately by the eighth axiom.
In order to remove from the threshold of the Elements a proposition so useless, and, to the younger students, so embarrassing as the seventh, it would be desirable that the eighth should be established independently of it. There are several ways in which this might be effected. The following proof seems liable to no objection, and establishes the eighth by the fifth.
Let the two equal bases be so applied one upon the other that the equal sides shall be conterminous, and that the triangles shall lie at opposite sides of them, and let a right line be conceived to be drawn joining the vertices.
1° Let this line intersect the base.
Let the vertex F fall at G, the side E F in the position A G, and D F in the position C G. Hence B A and A G being equal, the angles G B A and B G A are equal (V). Also C B and C G being equal, the angles C G B and C B G are equal (V). Adding these equals to the former, the angles A B C and A G C are equal; that is, the angles E F D and A B C are equal.
2° Let the line G B fall outside the coincident bases.
The angles G B A and B G A, and also B G C and G B C are proved equal as before; and taking the latter from the former, the remainders, which are the angles A G C and A B C, are equal, but A G C is the angle F.
3° Let the line B G pass through either extremity of the base.
In this case it follows immediately (V) that the angles A B C and A G C are equal; for the lines B C and C G must coincide with B G, since each has two points upon it (52).
Hence in every case the angles B and F are equal.
This proposition is also sometimes demonstrated as follows.
Conceive the triangle E F D to be applied to A B C, as in Euclid's proof. Then because E F is equal to A B, the point F must be in the circumference of a circle described with A as centre, and A B as radius. And for the same reason, F must be on a circumference with the centre C, and the radius C B. The vertex must therefore be at the point where these circles meet. But the vertex B must be also at that point; wherefore &c.
Next: Proposition 9
Previous: Proposition 7
This proposition in other editions: | 791 | 3,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-51 | latest | en | 0.924934 |
https://corrfunc.readthedocs.io/en/master/modules/custom_weighting.html | 1,642,467,001,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00219.warc.gz | 250,267,639 | 5,998 | # Implementing Custom Weight Functions¶
Corrfunc supports custom weight functions. On this page we describe the recommended procedure for writing your own. When in doubt, follow the example of pair_product.
First, see Computing Weighted Correlation Functions for basic usage of Corrfunc’s weight features.
The steps are:
1. Add a type to the weight_method_t enum in utils/defs.h (something like MY_WEIGHT_SCHEME=1).
2. Determine how many weights per particle your scheme needs, and add a case to the switch-case block in get_num_weights_by_method() in utils/defs.h. Corrfunc supports up to MAX_NUM_WEIGHTS=10 weights per particle; most schemes will simply need 1. To provide multiple weights per particle via the Python interface, simply pass a weights array of shape (N_WEIGHTS_PER_PARTICLE, N_PARTICLES).
3. Add an if statement that maps a string name (like “my_weight_scheme”) to the weight_method_t (which you created above) in get_weight_method_by_name() in utils/defs.h.
4. Write a function in utils/weight_functions.h.src that returns the weight for a particle pair, given the weights for the two particles. The weights for each particle are packed in a const pair_struct_DOUBLE struct, which also contains the pair separation. You must write one function for every instruction set you wish to support. This can be quite easy for simple weight schemes; the four functions for pair_product are:
/*
* The pair weight is the product of the particle weights
*/
static inline DOUBLE pair_product_DOUBLE(const pair_struct_DOUBLE *pair){
return pair->weights0[0].d*pair->weights1[0].d;
}
#ifdef __AVX512F__
static inline AVX512_FLOATS avx512_pair_product_DOUBLE(const pair_struct_DOUBLE *pair){
return AVX512_MULTIPLY_FLOATS(pair->weights0[0].a512, pair->weights1[0].a512);
}
#endif
#ifdef __AVX__
static inline AVX_FLOATS avx_pair_product_DOUBLE(const pair_struct_DOUBLE *pair){
return AVX_MULTIPLY_FLOATS(pair->weights0[0].a, pair->weights1[0].a);
}
#endif
#ifdef __SSE4_2__
static inline SSE_FLOATS sse_pair_product_DOUBLE(const pair_struct_DOUBLE *pair){
return SSE_MULTIPLY_FLOATS(pair->weights0[0].s, pair->weights1[0].s);
}
#endif
See utils/avx512_calls.h, utils/avx_calls.h and utils/sse_calls.h for the lists of available vector instructions.
1. For each function you wrote in the last step, add a case to the switch-case block in the appropriate dispatch function in utils/weight_functions.h.src. If you wrote a weighting function for all four instruction sets, then you’ll need to add the corresponding function to get_weight_func_by_method_DOUBLE(), get_avx512_weight_func_by_method_DOUBLE, get_avx_weight_func_by_method_DOUBLE(), and get_sse_weight_func_by_method_DOUBLE().
2. Done! Your weight scheme should now be accessible through the Python and C interfaces via the name (“my_weight_scheme”) that you specified above. The output will be accessible in the weightavg field of the results array.
Pair counts (i.e. the npairs field in the results array) are never affected by weights. For theory functions like Corrfunc.theory.xi and Corrfunc.theory.wp that actually return a clustering statistic, the statistic is weighted. For pair_product, the random distribution used to compute the expected bin weight from an unclustered particle set (the RR term) is taken to be a spatially uniform particle set where every particle has the mean weight. See RR in Weighted Clustering Statistics for more discussion. This behavior (automatically returning weighted clustering statistics) is only implemented for pair_product, since that is the only weighting method for which we know the desired equivalent random distribution. Custom weighting methods can implement similar behavior by modifying countpairs_xi_DOUBLE() in theory/xi/countpairs_xi_impl.c.src and countpairs_wp_DOUBLE() in theory/wp/countpairs_wp_impl.c.src. | 860 | 3,837 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-05 | longest | en | 0.643086 |
https://de.mathworks.com/matlabcentral/cody/problems/2674-generalised-hamming-number/solutions/735668 | 1,590,375,232,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00337.warc.gz | 328,791,597 | 15,887 | Cody
# Problem 2674. Generalised Hamming Number
Solution 735668
Submitted on 15 Sep 2015 by Mehmet OZC
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% X = 5; n=5; y_correct = 5; assert(isequal(hamming2(X,n),y_correct))
2 Pass
%% X = 10; n=5; y_correct = 9; assert(isequal(hamming2(X,n),y_correct))
3 Pass
%% X = 100; n=5; y_correct = 34; assert(isequal(hamming2(X,n),y_correct))
4 Pass
%% X = 100; n=7; y_correct = 46; assert(isequal(hamming2(X,n),y_correct))
5 Pass
%% X = 100; n=100; y_correct = 100; assert(isequal(hamming2(X,n),y_correct))
6 Pass
%% X = 100; n=13; y_correct = 62; assert(isequal(hamming2(X,n),y_correct))
7 Pass
%% X = 10^13; n=7; y_correct = 19674; assert(isequal(hamming2(X,n),y_correct))
8 Pass
%% X = 10^5; n=13; y_correct = 1848; assert(isequal(hamming2(X,n),y_correct))
9 Pass
%% X = 10^5; n=5; y_correct = 313; assert(isequal(hamming2(X,n),y_correct))
10 Pass
%% X = 123456; n=5; y_correct = 327; assert(isequal(hamming2(X,n),y_correct))
11 Pass
%% X = 10^13; n=3; y_correct = 624; assert(isequal(hamming2(X,n),y_correct))
12 Pass
%% X = 123456; n=2; y_correct = 17; assert(isequal(hamming2(X,n),y_correct)) | 489 | 1,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-24 | latest | en | 0.474176 |
https://www.physicsforums.com/threads/limit-proof.414498/ | 1,521,335,829,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645405.20/warc/CC-MAIN-20180317233618-20180318013618-00647.warc.gz | 830,932,895 | 14,212 | # Limit Proof
1. Jul 6, 2010
### jgens
So, I'm trying to prove that $\lim_{x \to a} f(x) = \lim_{h \to 0} f(a+h)$ and I'm wondering if this "proof" is correct:
Suppose that $\lim_{x \to a} f(x) = l$ and $\lim_{h \to 0} f(a+h) = m$ with $l \neq m$. Then, for any $\varepsilon > 0$ there must be some $\delta > 0$ such that $0 < |x-a| < \delta$ implies that $|f(x) - l| < \varepsilon$ and $0< |h| < \delta$ that $|f(a+h) - m| < \varepsilon$.
Clearly $a - \delta < x < a + \delta$ and $a - \delta < a + h < a + \delta$, so there is a number $x_1$ such that $|f(x_1) - l| < \varepsilon$ and $|f(x_1) - m| < \varepsilon$.
Choosing $\varepsilon = |l-m|/2$, it follows that $|l-m| = |l - f(x_1) + f(x_1) -m| \leq |f(x_1) - l| + |f(x_1) - m| < |l-m|/2 + |l-m|/2 < |l-m|$, which is a contradiction. Therefore, $l=m$.
Does this work alright? Or is there another way that I should approach this problem? I appreciate any feedback! Thanks!
2. Jul 6, 2010
### snipez90
Your approach seems to work, but I think it would have been much easier to write out what one of the limits means and then later set x = a + h. This approach also supports the intuitive idea behind this equality. There is also a way to prove this without resorting to the definition of the limit via some manipulations.
3. Jul 6, 2010
### losiu99
Looks ok. It is possible to do it more directly. Suppose $$\lim_{x\to a}f(x)=L$$. Fix $$\epsilon > 0$$. For some $$\delta$$-neighbourhood of $$a$$, $$|f(x)-L|<\epsilon$$. Hence, if $$|h|<\delta$$, $$|f(a+h)-L|<\epsilon$$ holds, and therefore $$\lim_{h\to 0}f(a+h)=L$$. It's essentially the same, though.
Edit: I'm too slow
Edit2: Wow, 3 minutes...way too slow | 609 | 1,676 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-13 | longest | en | 0.82982 |
demux.in | 1,659,900,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00396.warc.gz | 215,841,059 | 9,529 | Student Interview Experience
# Round 1 (Coding Round, 1.5 hours):
Question-1:
N strings of varying lengths are given. Each character used in the strings has some value. The value of a string is the sum of all characters used in that string. Given a length L, we have to use a subset of the given strings so that the total length of the strings used is L and the sum of values of all strings is maximised.
Hint: Knapsack
Question-2:
There are N ships, each with some seats S[i] (1≤i≤N). There are M passengers who want to buy the tickets for ships standing in a queue. The price of the ticket for a particular ship is always equal to the vacant seats in the ship. We have to find the maximum and minimum costs possible if all M passengers buy tickets.
Hint: Sorting
Question-3:
There are N nodes in a graph. They are connected using directed edges. For each vertex i (1≤i≤N), we have to determine whether there exists a path such that starting from vertex i, we reach vertex i again.
Hint: Topological Sort, O(N)
# Round 2 (Coding Round, 1.5 hours):
Question-1:
Given a 2D array of size N*M, we have to return the maximum area from the array such that the elements in that area increase as we go right in the row and down the column.
Hint: DP, Stacks, O(max(M,N)²)
Question-2:
There is a dictionary which contains some words. We are given a start word A and an end word B, both belonging to the dictionary. We are to determine the minimum number of moves so that we reach B starting from A, provided the intermediate words are all in the dictionary. Allowed operations are insertion of a character, deletion of a character, modification of a character.
Hint: BFS
Question-3:
A vehicle has to reach x = D starting from x = 0. The vehicle can store infinite amount of petrol. The vehicle has P_initial units of petrol initially. 1 unit distance can be travelled using 1 unit of petrol in 1 min, or by dragging the vehicle in 5 min. There are N petrol pumps in between the source and destination. The amount of petrol at station i is P[i] (1≤i≤N). Filling the vehicle’s petrol tank takes K mins. Find the minimum time in which we can reach B.
Hint: 2D DP
# Round 3 (Technical Interview, 1.25 hours):
The interviewer took a look at my resume and asked some questions from it. Then he gave me a DSA question.
Question:
Given an array with N elements, with some values, we’ve to select exactly X elements such that the sum of values is maximised. The constraint was that each subarray of size K should have atleast one element which is there in our selection of X items.
Solution Approach 1:
I started off with the brute force solution — iterate over all X sized subsets of the array and check whether atleast one element is used from each K sized subarray and return the required answer.
Time Complexity: O(2^N*N²)
Solution Approach 2:
Soon I realised that if I had the solution for it’s subproblems, I could formulate the complete solution. So I thought of a 2D DP where dp[i][j] denotes the maximum sum of values if I choose to select the ith index in my subset and have exactly j elements in my subset. Then my final answer would be max(dp[i][X], (1≤i≤N).
Time Complexity: O(N*X*K)
I discussed the solution with the interviewer and then I was given 10 minutes to code up the solution.
# Round 4 (HR + Technical Interview, 1.25 hours):
>Basic 4 concepts of OOP and their examples
>Run-time polymorphism and virtual functions
>Use of abstract classes
>Indexing and it’s use
>B-Trees and B+Trees and their implementation
Then, I was given another DSA question:
Question:
There are N bananas in city A, and another city B at a distance of D from city A. We’ve to transfer maximum number of bananas from A to B. The only mode of transport is an elephant which has a max capacity C. The elephant eats CR bananas per unit distance. (The elephant can leave the leave the bananas mid-way and pick them up later)
Solution Approach:
I used a sample test to get the basic idea of how to approach the question and then generalised the result. The idea was to let the elephant carry C bananas for every journey and put them at an intermediate distance x, return back and carry the rest of the bananas to x. Since the elephant needs bananas to move, 2*CR*x ≤ C for all trips where he has to make a round trip and CR*x ≤ C for the last trip. Using this information, we can find x and thus the number of bananas (say N’) the elephant can carry till point x. The problem f(N, D) is then reduced to f(N’, D-x) and we can use recursion to solve the problem.
And that was it! After an hour or so, I was told that I was hired as a summer intern in Media.net.
A huge credit of bagging this internship goes to Team Demux — a group of dedicated and knowledgable mentors who helped me strengthen my concepts throughout the 2019 summer break. My family and friends also played an important role as they supported me throughout the internship process.
Overall, the interview experience was a great one and I learned a lot from it. | 1,184 | 5,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-33 | latest | en | 0.941236 |
https://notes.dzackgarza.com/Projects/2022%20Algebraic%20Geometry%20Oral%20Exam/500_Problems/Rising%20Sea%20Exercises/sections/2020-06-14.html | 1,660,741,756,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00443.warc.gz | 417,459,258 | 10,682 | # Week 1
## Exercise 1.3H: Right Exactness of Tensoring
Show that the following endofunctor \begin{align*}\begin{align*} F: {\mathsf{R}{\hbox{-}}\mathsf{Mod}}&\to {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\\ X &\mapsto X\otimes_R N \\ (X\xrightarrow{f} Y) &\mapsto (X\otimes_R N \xrightarrow{f \otimes\operatorname{id}_N} Y\otimes_R N) \end{align*}\end{align*} is exact.
Solution:
Note: to make sense of the functor, we may need to show that there is an isomorphism \begin{align*}\hom_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(X, Y) \otimes_R \hom_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(A, B) \to \hom_{\mathsf{R}{\hbox{-}}\mathsf{Mod}}(X\otimes_R A, Y\otimes_R B).\end{align*} This is what makes taking $$f:X\to Y$$ and $$g:A\to B$$ and forming $$f\otimes g: X\otimes A \to Y\otimes B$$ well-defined?
Let $$A\xrightarrow{f} B \xrightarrow{g} C \to 0$$ be an exact sequence, so
• $$\operatorname{im}f = \ker g$$ by exactness at $$B$$
• $$\operatorname{im}g = C$$ by exactness at $$C$$.
Applying the above $$F$$ yields \begin{align*} A\otimes_R N \xrightarrow{f\otimes\operatorname{id}_N} B\otimes_R N \xrightarrow{g\otimes\operatorname{id}_N} C\otimes_R N \to 0 \end{align*}
We thus need to show
• Exactness as $$C\otimes_R N$$: $$\operatorname{im}(g\otimes\operatorname{id}_N) = C\otimes_R N$$, i.e. this is surjective.
• Exactness at $$B\otimes_R N$$: $$\operatorname{im}(f\otimes\operatorname{id}_N) = \ker(g\otimes id_N)$$.
We’ll use the fact that every element in a tensor product is a finite sum of elementary tensors.
• Claim: $$\operatorname{im}(g\otimes\operatorname{id}_N) \subseteq C\otimes_R N$$.
• Let $$b\otimes n \in B\otimes_R N$$ be an elementary tensor
• Then $$(g\otimes\operatorname{id}_N)(b\otimes n) \coloneqq g(b) \otimes\operatorname{id}_N (n) = g(b) \otimes n$$
• Since $$\operatorname{im}(g) = C$$, there exists a $$c\in C$$ such that $$g(b) = c$$, so $$g(b) \otimes n = c \otimes n \in C\otimes_R N$$
• Extend by linearity:
\begin{align*}
\qty{g\otimes_R \operatorname{id}_N}\qty{\sum_{i=1}^m r_i \cdot b_i \otimes n_i} =\sum_{i=1}^m (g\otimes\operatorname{id}_N)(r_i\cdot b_i \otimes n_i) \coloneqq\sum_{i=1}^m g(r_i\cdot b_i) \otimes\operatorname{id}_N(n_i) =_H \sum_{i=1}^m r_i\cdot c_i \otimes n_i \in C\otimes_R N
\end{align*}
where we’ve used bilinearity for the first equality, and the equality marked with $$H$$ uses above the proof for elementary tensors, and noted that we can pull ring scalars $$r_i\in R$$ through $${\mathsf{R}{\hbox{-}}\mathsf{Mod}}$$ morphisms. - Claim: $$C \otimes_R N \subseteq \operatorname{im}(g\otimes\operatorname{id}_N)$$. - Let $$c\otimes n \in C\otimes_R N$$ be an elementary tensor. - Then $$c\in C = \operatorname{im}(g)$$ implies $$c = g(b)$$ for some $$b\in B$$. - So $$c\otimes n = g(b) \otimes n = (g\otimes\operatorname{id}_N)(b\otimes n) \in B\otimes_R N$$. - Extend by linearity:
\begin{align*}
\sum_{i=1}^m r_i\cdot c_i \otimes n_i =_H \sum_{i=1}^m g(r_i\cdot b_i) \otimes n_i = \sum_{i=1}^m (g\otimes\operatorname{id}_N)(r_i\cdot b_i \otimes n_i) = (g\otimes\operatorname{id}_N)\qty{\sum_{i=1}^m r_i\cdot b_i \otimes n_i}
.\end{align*}
This proves (1).
• Claim: $$\operatorname{im}(f\otimes\operatorname{id}_N) \subseteq \ker(g\otimes\operatorname{id}_N)$$.
• Let $$b\otimes n \in \operatorname{im}(f\otimes\operatorname{id}_N)$$, we want to show $$(g\otimes\operatorname{id}_N)(b\otimes n) = 0 \in C\otimes_R N$$.
• Then $$b\otimes n = f(a)\otimes n$$ for some $$a\in A$$.
• By exactness of the original sequence, $$\operatorname{im}f \subseteq \ker g$$, so $$g(f(a)) = 0 \in C$$
• Then
\begin{align*}
(g\otimes\operatorname{id}_N)\qty{ b \otimes n} = (g\otimes\operatorname{id}_N)(f(a)\otimes n) \coloneqq g(f(a)) \otimes n = 0\otimes n = 0\in C\otimes_R N
\end{align*}
where we’ve used the fact that $$0\otimes x = 0$$ in any tensor product.
• Extend by linearity.
• Claim (nontrivial part): $$\ker(g\otimes\operatorname{id}_N) \subseteq \operatorname{im}(f\otimes\operatorname{id}_N)$$.
Note: the problem is that
\begin{align*}
x\in \ker(g\otimes\operatorname{id}_N) \implies x = \sum_{i=1}^m r_i\cdot b_i \otimes n_i \implies (g\otimes\operatorname{id}_N)\qty{\sum_{i=1}^m r_i\cdot b_i \otimes n_i} = \sum_{i=1}^m r_i\cdot g(b_i) \otimes n_i = 0\in C\otimes_R N
\end{align*}
but this does not imply that $$g(b_i) = 0\in C$$ for all $$i$$, which is what you would need to use $$\operatorname{im}f = \ker g$$ to write $$g(b_i) = 0\implies \exists a_i, f(a_i) = b_i$$ and pull everything back to $$A\otimes_R N$$.
• Strategy: use the first claim and the first isomorphism theorem to obtain this situation:
\begin{center}
\begin{tikzcd}
{B\otimes_R N \over \operatorname{im}(f\otimes_R \operatorname{id}_N)} \ar[r, hook, "i"]\ar[rrr, bend left, dotted, "\alpha"] & {B\otimes_R N \over \ker(g \otimes_R \operatorname{id}_N)} \ar[r, "\cong"] & \operatorname{im}(g\otimes_R \operatorname{id}_N) \ar[equal]{r} & C\otimes_R N
\end{tikzcd}
\end{center}
• The first injection $$i$$ will exist because $$\operatorname{im}(g\otimes_R \operatorname{id}_N) \subseteq \ker(g\otimes_R \operatorname{id}_N)$$ by the first claim.
• The middle isomorphism is the first isomorphism theorem.
• The RHS equality follows from surjectivity of $$g\otimes_R \operatorname{id}_N$$
• We then apply a strengthened version of the 1st isomorphism theorem for modules:
Hungerford Ch.4 Thm 1.7: If $$f:A\to B$$ is a $$R{\hbox{-}}$$module morphism and $$C\leq \ker f$$ then there is a unique map $$\tilde f: A/C\to B$$ which is an isomorphism iff $$f$$ is an epimorphism and $$C = \ker f$$.
Following Hungerford Ch.4 Prop. 5.4, p.210.
• Since $$\operatorname{im}(g\otimes_R \operatorname{id}_N)\subseteq \ker(g\otimes_R \operatorname{id}_N)$$, by the theorem the map $$\alpha$$ exists and satisfies the same formula, i.e. $$\alpha = \tilde g \otimes\tilde \operatorname{id}_N$$ where the tilde denotes the induced map on quotients, so $$\alpha([b\otimes n]) = g(b)\otimes n$$.
• We will show it is an isomorphism, which forces $$\operatorname{im}(g\otimes_R \operatorname{id}_N) \cong \ker(g\otimes_R \operatorname{id}_N)$$ by the above theorem.
• Constructing the inverse map: define
\begin{align*}
\tilde \alpha^{-1}: C\times N &\to {B\otimes_R N \over \operatorname{im}(g\otimes_R \operatorname{id}_N) } \\
(c, n) &\mapsto (b \otimes n) \operatorname{mod}\operatorname{im}(g\otimes_R \operatorname{id}_N) {\quad \operatorname{where} \quad} b \in g^{-1}(c)
,\end{align*}
which we will show well-defined (i.e. independent of choice of $$b$$) and $$R{\hbox{-}}$$linear, lifting to a map $$\alpha^{-1}$$ out of the tensor product by the universal property which is a two-sided inverse for $$\alpha$$.
• Well-defined:
• $$g^{-1}(b)$$ exists because $$g$$ is surjective.
• If $$b\neq b'$$ and $$g(b') = 0$$, then $$0 = g(b) - g(b') = g(b-b')$$ so $$b-b' \in \ker g$$.
• By the original exactness, $$b-b' \in \operatorname{im}f$$ so $$b-b' = f(a)$$ for some $$a\in A$$.
• Then $$f(a) \otimes n \in \operatorname{im}(f\otimes\operatorname{id})$$ implies $$f(a)\otimes n \equiv 0 \operatorname{mod}\operatorname{im}(f\otimes\operatorname{id})$$.
• Then noting that $$b-b' = f(a) \implies b = f(a) + b'$$, working mod $$\operatorname{im}(g\otimes_R \operatorname{id}_N)$$ we have
\begin{align*}
b \otimes n \equiv (f(a) + b') \otimes n \equiv \qty{f(a) \otimes n} + \qty{b' \otimes n} \equiv b'\otimes n
.\end{align*}
• $$R{\hbox{-}}$$linear:
• ?
• Two-sided identity:
• $$(\alpha \circ \alpha^{-1})(c\otimes n) = \alpha(b\otimes n) = g(b)\otimes n = c\otimes n$$, so $$\alpha\circ \alpha^{-1}= \operatorname{id}$$.
• $$(\alpha^{-1}\circ \alpha)([b\otimes n]) = \alpha^{-1}(g(b) \otimes n) = [b'\otimes n]$$ where $$b'\in g^{-1}(g(b))$$ implies $$b'=b$$, so $$\alpha\circ\alpha^{-1}= \operatorname{id}$$.
$$\hfill\blacksquare$$ | 2,944 | 7,785 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-33 | latest | en | 0.452589 |
https://www.scala-exercises.org/cats/functor | 1,721,516,283,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00605.warc.gz | 843,913,981 | 8,708 | ## Functor
A `Functor` is a ubiquitous type class involving types that have one "hole", i.e. types which have the shape `F[*]`, such as `Option`, `List` and `Future`. (This is in contrast to a type like `Int` which has no hole, or `Tuple2` which has two holes (`Tuple2[*,*]`)).
The `Functor` category involves a single operation, named `map`:
``def map[A, B](fa: F[A])(f: A => B): F[B]``
This method takes a function `A => B` and turns an `F[A]` into an `F[B]`. The name of the method `map` should remind you of the `map` method that exists on many classes in the Scala standard library, for example:
``````Option(1).map(_ + 1)
List(1, 2, 3).map(_ + 1)
Vector(1, 2, 3).map(_.toString)``````
### Creating Functor instances
We can trivially create a `Functor` instance for a type which has a well behaved `map` method:
``````import cats._
implicit val optionFunctor: Functor[Option] = new Functor[Option] {
def map[A, B](fa: Option[A])(f: A => B) = fa map f
}
implicit val listFunctor: Functor[List] = new Functor[List] {
def map[A, B](fa: List[A])(f: A => B) = fa map f
}``````
However, functors can also be created for types which don't have a `map` method. For example, if we create a `Functor` for `Function1[In, *]` we can use `andThen` to implement `map`:
``````implicit def function1Functor[In]: Functor[Function1[In, *]] =
new Functor[Function1[In, *]] {
def map[A, B](fa: In => A)(f: A => B): Function1[In, B] = fa andThen f
}``````
This example demonstrates the use of the kind-projector compiler plugin This compiler plugin can help us when we need to change the number of type holes. In the example above, we took a type which normally has two type holes, `Function1[*,*]` and constrained one of the holes to be the `In` type, leaving just one hole for the return type, resulting in `Function1[In,*]`. Without kind-projector, we'd have to write this as something like `({type F[A] = Function1[In,A]})#F`, which is much harder to read and understand.
### Using Functor
#### map
`List` is a functor which applies the function to each element of the list:
``Functor[List].map(List("qwer", "adsfg"))(_.length)``
`Option` is a functor which only applies the function when the `Option` value is a `Some`:
``````Functor[Option].map(Option("Hello"))(_.length) should be(res0)
Functor[Option].map(None: Option[String])(_.length) should be(res1)``````
### Derived methods
#### lift
We can use `Functor` to "lift" a function from `A => B` to `F[A] => F[B]`:
``````val lenOption: Option[String] => Option[Int] = Functor[Option].lift(_.length)
lenOption(Some("abcd"))``````
We can now apply the `lenOption` function to `Option` instances.
``````val lenOption: Option[String] => Option[Int] = Functor[Option].lift(_.length)
lenOption(Some("Hello")) should be(res0)``````
#### fproduct
`Functor` provides an `fproduct` function which pairs a value with the result of applying a function to that value.
``````val source = List("Cats", "is", "awesome")
val product = Functor[List].fproduct(source)(_.length).toMap
product.get("Cats").getOrElse(0) should be(res0)
product.get("is").getOrElse(0) should be(res1)
product.get("awesome").getOrElse(0) should be(res2)``````
#### compose
Functors compose! Given any functor `F[_]` and any functor `G[_]` we can create a new functor `F[G[_]]` by composing them:
``val listOpt = Functor[List] compose Functor[Option]``
In the previous example the resulting functor will apply the `map` operation through the two type constructors: `List` and `Option`.
``````val listOpt = Functor[List] compose Functor[Option]
listOpt.map(List(Some(1), None, Some(3)))(_ + 1) should be(res0)`````` | 1,058 | 3,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.836216 |
vitsyrovatka.info | 1,590,830,538,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00485.warc.gz | 603,155,819 | 6,313 | Vít Syrovátka
Clustering
Let's continue with clustering sites according to their distances calculated in the previous excercise.
1. Cluster sites based on their community dissimilarities. Consecutively make 6 clusterings using the combination of two distances (hellinger and bray-curtis) and three clustering algorithms:
1. single (Nearest neighbour)
2. complete (Farthest neighbour)
3. average (UPGMA) hint
2. Plot the 6 dendrograms. hint
3. To validate how well do clusterings preserve original distances, calculate correlations between cophenetic and original distances and plot them against each other. Which clustering preserves best the original distances? hint
4. Since simple linkage clustering lead to a strong chaining of sites preventing an easy interpretation of clusters we will stick to the results of the other two clustering algorithms: complete linkage a UPGMA. Find the number of clusters for each clustering to obtain approximately 4 reasonable clusters. I.e. find the distance level at which about 4 clusters are developed. Ignor simple clusters containing just one sample.
5. Identify which samples belong to which clusters using cutree().
6. Use table() to compare the groupings of sites.
7. Which groupings are the most similar and which are the most different? | 269 | 1,287 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-24 | latest | en | 0.889893 |
https://oeis.org/A095719 | 1,632,334,823,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057371.69/warc/CC-MAIN-20210922163121-20210922193121-00283.warc.gz | 459,737,603 | 3,884 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A095719 Sum(floor(C(n-k,k)/(k+1)),k=0..n/2). 0
1, 1, 2, 2, 4, 5, 8, 11, 18, 25, 40, 59, 90, 137, 210, 319, 492, 754, 1164, 1798, 2786, 4317, 6710, 10438, 16266, 25377, 39650, 62013, 97108, 152212, 238822, 375058, 589520, 927365, 1459960, 2300097, 3626211, 5720649, 9030450, 14263675 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Sums of diagonal entries in A011847 LINKS FORMULA a(n) = sum(floor(C(n-k, k)/(k+1)), k=0..n/2); MAPLE a:=n->add(floor(C(n-k, k)/(k+1)), k=0..n/2); MATHEMATICA Table[Sum[Floor[Binomial[n-k, k]/(k+1)], {k, 0, n/2}], {n, 40}] (* Harvey P. Dale, Apr 02 2019 *) CROSSREFS Cf. A011847. Sequence in context: A274142 A006206 A060280 * A153952 A050364 A078465 Adjacent sequences: A095716 A095717 A095718 * A095720 A095721 A095722 KEYWORD nonn AUTHOR Mike Zabrocki, Jul 08 2004 STATUS approved
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Last modified September 22 13:48 EDT 2021. Contains 347607 sequences. (Running on oeis4.) | 512 | 1,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | latest | en | 0.555743 |
https://web2.0calc.com/questions/help-please-thanks-for-everyone-who-help | 1,606,857,754,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00202.warc.gz | 553,894,922 | 5,961 | +0
# Help please! thanks for everyone who help!
+1
116
3
+154
2.3
Apr 2, 2020
#1
+154
+2
Thanks Again for the help!!
Apr 2, 2020
#2
+112413
+1
First one
Area = (1/2) (WX) (WY) sin W = (1/2) (10.3)(12.8) sin ( 63°) ≈ 58.7 in ^2
Second one
arc length = r * central angle in radians
(7/15)pi = 14 * central angle divide both sides by 14
(1/14) (7 / 15) pi = central angle = (7 / 210) pi = (1/30)pi
Last one
sin θ = - sqrt [ 1 - cos^2 theta ] = - sqrt [ 1 - 2/9 ] = - sqrt [ 7] / 3
sin B = 4/5 cos B = 3/5
sin ( θ + B) = sinθ cos B + sin B cos θ = [ - sqrt (7)/3 ] [ 3/5] + [ 4/5] [ -sqrt (2) / 3 ] =
[ -3sqrt (7) - 4sqrt (2) ] / 15
Apr 2, 2020
#3
+154
+1
Thanks Can you help on my other qustions Plzz
https://web2.0calc.com/questions/help-plzz_9#r2
whatisthis1010 Apr 2, 2020 | 429 | 864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-50 | latest | en | 0.207643 |
https://physics.stackexchange.com/questions/131122/modeling-a-stool-with-three-legs-pushed-on-one-leg | 1,632,682,281,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00188.warc.gz | 488,523,260 | 37,888 | # Modeling a stool with three legs pushed on one leg
If a stool with three legs is pushed (horizontally) on one leg in the direction of the axis passing through the other two legs, it starts both moving and rotating.
Is it correct to assume 1/3 of force on each of the legs? This comes a bit in contrast with intuition. If we imagine the legs flexible, then the pushed leg should deform more (i.e. stronger friction force).
What is the correct modeling of the friction forces?
p.s. this seems a simplified version of Push a box in a plane with friction. How to deal with the rotation? but there's no clear answer in that thread.
If you are pushing exactly at ground level you can expect 1/3 of the vertical force on each leg - but if you are pushing any higher, the torque balance will make the pushed leg "lighter".
Key to understanding the problem is that this is an unstable equilibrium - you cannot aim exactly between the other two legs. Even if you did, there will be small random fluctuations in the coefficient of friction between the stool legs and the floor. As soon as a tiny deviation from equilibrium occurs, you impart a rotational torque on the stool; and as soon as the stool begins to rotate, you have moved away from equilibrium and the rotational torque will be amplified because the distance of the two far legs to the line of force wil no longer be equal.
Summary:
1. Push exactly along "line of equilibrium"
2. Fluctuation in coefficient of friction
3. Unequal forces on two legs
4. Net torque rotates stool slightly
5. Distance of legs to line of force no longer equal
6. Rotation established
I will try to draw a picture when I have a bit more time | 363 | 1,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-39 | longest | en | 0.937885 |
https://rightbankwarsaw.com/football-game/are-high-school-soccer-fields-the-same-size-as-professional.html | 1,639,015,068,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00063.warc.gz | 575,381,084 | 19,620 | # Are high school soccer fields the same size as professional?
Contents
High school soccer fields are not the same size as professional fields, and will typically fall into the U-14 category sized soccer field. The high school soccer field regulation size is set by the National Federation of State High School Associations (NFHS) and range from 55 to 80 yards wide and 100 to 120 yards long.
## How much bigger is a professional soccer field?
The field dimensions are within the range found optimal by FIFA: 110–120 yards (100–110 m) long by 70–80 yards (64–75 m) wide. These soccer field dimensions are wider than the regulation American football field width of 53 1⁄3 yards (48.8 m), or the 65-yard (59 m) width of a Canadian football field.
## Are professional soccer fields different sizes?
Ultimately, Premier League pitches are all roughly the same size. However, the Laws of the Game state pitches can be between 90 and 120 metres long, and drastically different in width – between 50 and 100 metres, although they must be longer than they are wide.
IT IS INTERESTING: Question: Will Ireland ever win the World Cup?
## Are high school and college soccer fields the same size?
Here’s how the two soccer fields compare. High school soccer fields can be the same size as a professional soccer field. The length and width requirements at both levels of soccer are similar. However, due to space restrictions, many high school soccer fields are smaller than professional soccer fields.
## What is the length of a high school soccer field?
U.S. high school soccer field dimensions are set according to NFHS (the National Federation of State High School Associations) rules. High school fields range in size from 55 yards wide x 100 yards long to 80 yards wide x 120 yards long.
## How big is a full size soccer field?
A full-size pitch may be anywhere from 50-100 yards in width and 100-130 yards in length. For the Bundesliga, Article 37, Section 6 of Appendix VI of the DFL Statutes states that a pitch must be 105 meters (115 yards) long and 68 meters (74 yards) wide.
## Is MLS field smaller than FIFA?
No, an MLS field is not smaller than FIFA regulation. A Major League Soccer (MLS) fields dimensions are in the acceptable range of FIFA international standard sizes. FIFA allows for a range of dimensions for a soccer field, and every MLS field fits within those dimensions.
## Is Camp Nou pitch bigger than Old Trafford?
To many people the field looks a lot wider but it is the same size as many other football fields in Europe. There are even bigger fields like that on Old Trafford for example. This stadium is often used for other purposes like concerts, international sports events etc.
## How big is Tottenham’s pitch?
The Desso GrassMaster hybrid grass football pitch at the Tottenham Hotspur Stadium is 105 metres by 68 metres, with an area of over 7,000 square metres.
## How big is Arsenal’s pitch?
The pitch is 105 by 68 metres (115 by 74 yd) in size and the total grass area at Emirates is 113 by 76 metres (124 by 83 yd). Like Highbury, it runs north–south, with the players’ tunnel and the dugouts on the west side of the pitch underneath the main TV camera.
## What is the size of a youth soccer field?
Field & Goal Dimensions
AGE Field Width Field Length
U11 (40 – 50 yards) (70 – 80 yards)
U12 (40 – 55 yards) (100 – 105 yards)
U13 (50 – 60 yards) (100 – 110 yards)
Adult (65 – 80 yards) (110 – 120 yards)
## What does FIFA stand for?
FIFA — Soccer’s World Governing Body
Founded in 1904 to provide unity among national soccer associations, the Federation Internationale de Football Association (FIFA) boasts 209 members, rivaling that of the United Nations, and is arguably the most prestigious sports organization in the world.
## How tall is the crossbar in soccer?
Goals are placed at the centre of each goal-line. These consist of two upright posts placed equidistant from the corner flagposts, joined at the top by a horizontal crossbar. The inner edges of the posts must be 7.32 metres (8 yd) apart, and the lower edge of the crossbar must be 2.44 metres (8 ft) above the ground.
IT IS INTERESTING: What does competitive soccer mean?
## How big is a U15 soccer field?
U15 / U16 Age Group:
The pitch again increases in size at U15/U16 and at this age group the soccer field dimensions should be 91 metres for the length and 55 metres for the width. After this age group, youth soccer progresses to full size adult pitches which are 90 to 120 metres in length and 45 to 90 metres in width.
## What is a soccer field called?
In soccer, a pitch is another name for the playing surface on which the game of soccer is played. A soccer pitch is also known as a soccer field. Using the word “pitch” to describe the area of ground where a soccer game is played originated in England during the 1800s. | 1,126 | 4,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-49 | latest | en | 0.960918 |
http://ncatlab.org/nlab/show/observing+black+holes | 1,397,799,197,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00166-ip-10-147-4-33.ec2.internal.warc.gz | 156,150,802 | 8,153 | # Contents
## Idea
The typical ways of – indirectly – observing black holes in Nature – are
• 1. via X-ray emission. The nearby gas or nearby stars get acreted about a black hole and accelerate in the vicinity producing radiation emission;
• 2. in star clusters with no gas, such as globular clusters, the only way is to see how the black hole affects the motion of nearby stars.
## X-ray emission
Consider first stars. Radiation pressure outwards, gravity inwards. If the radiation stronger the star dissolves. Eddington’s luminosity limit
$L_{Edd} = \frac{G M m_0 4\pi c}{\sigma}$
In astrophysical units $L_{Edd} = \frac{M}{M_{sun}}\cdot 1.26 \cdot 10^38 erg/s$. For about ten solar masses this is about $10^{39} erg/s$. So something radiating above this limit is heavier of ten solar masses, under the assumption that the emission is isotropic and there are no other mechanisms which can violate the Eddington’s limit (e.g. no beaming). So the sources above $10^{39} erg/s$ (and under the assmptions above) are called the ultraluminous X-ray sources. Candidates for IMBH.
Spectrum observation in the acretion disk. Get temperature as a function of $R$ by various means, this now into black body radiation formulas, and then model the spectrum. Hoter toward the inside, fit to the mass of the black hole supposed around a center.
This method for fitting with the spectrum is called multicolor disk model.
## Effects on nearby stars
Effects of a black hole on the motion of nearby stars can be observed directly via black hole influence sphere or as global evidence (the whole cluster is affected). If the black hole has mass $M$, we can look at the escape velocity $\sqrt{\frac{2GM}{R}_{inf}}\sim \langle v \rangle$.
$R_{inf}$ – is the influence sphere radius. $GM_{BH}/R_{inf} = GM_{GC}/R_{GC}$. Thus
Typically, for $R_{GC} = 1 pc$, $M_{BH} \sim 1 \% M_{GC}$, $R_{inf}\sim 1/100 pc\sim 1 arc sec$. $D_{sun GC} \sim Kpc$.
Very good estimates of black hole mass. But there is a problem here. The number of stars which is in the black hole influence region, it depends on the volume of the black hole influence region. The ratio of volumes
$\frac{V_{inf}}{V_{GC}} \cong 10^{-6}$
while the number of stars in the globular cluster is $10^4 - 10^6$.
so even less than one star in the influence sphere unless there is much bigger density of stars. In the center of the globular cluster there is $10^2$-factor of higher density. That is the place where the black holes would typically occur (heavier than the average star, they tend in the thermalization limit to be found about in the center of the star).
Crowding is the issue here, sometimes lucky to see the individual star. How angularly distant from the center you find the star. At least 100 solar masses black hole. Size around square root of the mass. Another problem is low completeness – we will observe at least two stars, usually $02-0.5$ sun masses each star in the cluster. All this is about the observation of globular clusters in our galaxy (more difficult for other galaxies of course) – about 200 globular clusters in the halo? of Milky Way.
It is important to compare the properties like density and velocity dispersion in the cluster (the corresponding graphs are in the school notes, versus $log R$). Velocity dispersion bigger in the center; actually a peak is near the center: Cusp phenomenon in the presence of intermediate mass black hole?.
• see Noyola and Gebherdt 2006, 2007, 2008 for observations
For example in the cluster $\Omega$-centauri.
Small number of stars then also lots of fluctuations. So one needs to be careful about the statistics. The presence of the cusp has been predicted by Bahcall and Wolf in 1970-s (analytic calculation, gives the shape of the density cusp as well; later simulations have been found).
More recently it has been shown by Trenti, Vesperini, Pasquato 2010, even if you do not put a black hole in the cluster after a while when the relaxation takes place the cusp cn appear anyway. So other phenomena possible.
If we look outside of the black hole influence sphere, it has been shown by simulations that if you take the ratio between the radius $R_c$ at which the projected density declines and compare with $R_h$, then we can see the energy loss in the cluster. It can be supported by the presence of energy sources in the cluster like binary stars or by a black hole. This can be seen on the profile of the projected density – the larger density when energy source.
IMBH – hosting GCs has a large core.
Another signature of the presence of black hole, acceleration of the global shrinking the cluster puff the core and mass segregation accelerates in the cluster (infall of heavy stars).
Heavier main sequence stars (0.8 solar mass) compare with smaller ones (0.2 solar masses). Distribution tells if there is an energy source near the core. If you know how many binary stars there in principle one could be able to tell if there is a black hole there or not.
## References
In parts taken from Mario’s 3rd lecture at the Croatian black hole school, 2010.
Revised on June 22, 2010 18:09:03 by Urs Schreiber (87.212.203.135) | 1,257 | 5,164 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2014-15 | longest | en | 0.856965 |
https://wiki.octave.org/wiki/index.php?title=Special:MobileDiff/6407&mobileaction=toggle_view_mobile | 1,579,377,259,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00320.warc.gz | 736,651,860 | 4,856 | # Changes
## TISEAN package
, 13:04, 9 June 2015
updated with plotting code
{{Code|Finding unstable periodic orbits|<syntaxhighlight lang="octave" style="font-size:13px">
# Create maps
hen = henon (1000);hen_noisy hen = hen + std (hen) * 0.02 .* (-6 + sum (rand ([size(hen), 12]), 3));
# Find orbits
up [lengths, data] = upo(hen(:,1), 2, 'p',6,'v',0.1, 'n', 100);
</syntaxhighlight>}}
The vector {{Codeline|lengths}} contains the size of each of the found orbits and {{Codeline|data}} contains their coordinates. To obtain delay coordinates and plot these orbits onto the original data we will use {{Codeline|upoembed}}.
{{Code|Plotting unstable periodic orbits|<syntaxhighlight lang="octave" style="font-size:13px">
# Create delay coordinates for the orbits and data
up = upoembed (lengths, data, 1);
delay_hen = delay (hen(:,1));
# Plot all of the data
plot (delay_hen(:,1), delay_hen(:,2), 'r.;Noisy Henon;','markersize',2,...
up{4}(:,1), up{4}(:,2),'gx;Fixed Point;','markersize',20,'linewidth',1, ...
up{3}(:,1), up{3}(:,2),'b+;Period 2;','markersize',20,'linewidth',1, ...
up{2}(:,1), up{2}(:,2),'ms;Period 6;','markersize',20,'linewidth',1, ...
up{1}(:,1), up{1}(:,2),'cs;Period 6;','markersize',20,'linewidth',1);
</syntaxhighlight>}}
The plotting options are passed to make the orbits more visible.
=== Nonlinear Prediction ===
156
edits | 441 | 1,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-05 | latest | en | 0.51986 |
http://www.chegg.com/homework-help/questions-and-answers/1-find-f-s-cos-t-b-sin-t-u-t-plot-poles-zeros-s-plane-2-find-f-s-2d-t-4e-2t-cos-2t-u-t-plo-q2083633 | 1,454,857,989,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701149548.13/warc/CC-MAIN-20160205193909-00197-ip-10-236-182-209.ec2.internal.warc.gz | 332,432,171 | 12,822 | 1- Find F(s) for = A[ a cos(ßt) + b sin(ßt)] u(t) and plot and the poles and zeros in the S- plane
2- find f(s) for = 2d(t) +4e^-2t cos(2t) u(t) and plot and the poles and zeros in the S- plane
3- find f(s) for = 10[3-6t-4e^-2t] u(t) and plot and the poles and zeros in the S- plane
4- find f(s) for = [10e^-3t - 5e^-5t ] u(t) and plot and the poles and zeros in the S- plane | 155 | 378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2016-07 | latest | en | 0.849163 |
https://www.oddsmonkey.com/blog/matched-betting/football-accumulators-what-they-are-and-how-to-do-them/ | 1,709,276,475,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00067.warc.gz | 913,613,495 | 40,219 | # Football accumulators: what they are and how to do them
Football accumulators are now firmly embedded in Lad culture. There used to be a time when betting was seen as a silly thing to do, but now that it’s been given a mainstream platform, more people are placing football accumulators than ever before.
The thing with football accumulators, is that they allow people to dream big. We always see the bet slips shared on social media of that lucky punter who placed a £1 bet and won tens of thousands!
This of course, creates a situation where people try to emulate that success. Everyone classes themselves as knowledgeable about football, so it should be easy, right? Well, not quite…
## What are football accumulators?
A football accumulator is a bet that involves multiple selections. You choose a number of teams to bet on and add them all to your bet slip. You need them ALL to win, to be able to win your bet.
Find out more about accumulators in this post.
Because you require every selection to win, it also means the odds are much bigger. Take the following accumulator as an example:
We’ve chosen 4 different selections, all with quite high odds. The accumulative odds for this bet are 357.33. Should we place a £1 bet, we’ll get a huge payout of £357.33!
It’s simple to calculate the returns, too. All you need to do is multiply the price of each selection by the next one. So in the above example, it is:
5.5 x 2.75 x 4.50 x 5.25 = 357.33
## Can betting on football accumulators be profitable?
Bookmakers absolutely love football accumulators, which should perhaps tell its own story. They are one of the most profitable income streams for a sportsbook, and act as a great promotional tool when happy customers share their bet slips which go viral. Bookmakers are all-too-happy to share these bet slips themselves, and profess how great accumulators are.
But did you know that behind the scenes, every bookmaker has sophisticated algorithms judging the betting patterns of each customer?
They literally rate customers on a scale ranging from ‘great punter’ to ‘mug punter’.
The usefulness of this is that they can limit the amount of money that ‘great punters’ can bet, and offer more and more promotions and higher limits to ‘mug punters’.
Betting on an accumulator weights the algorithm heavily towards judging you as a ‘mug punter’; someone they love to have as a customer.
Football accumulators are fun but to win on a regular basis, you have to be very lucky. Which is why bookmakers make a lot of money from them.
Take our earlier accumulator as an example. The odds were just over 356/1. That means the chances of your accumulator winning is just once in every 356 times. Ouch!
## How can you improve your chances?
One of the keys to successful accumulators is to not back the favourites each week. This is exactly what the bookmakers love.
Every weekend, thousands of people bet on the top 4 or 6 teams in the Premier League to win. This strategy has the bookmakers laughing all the way to the bank, as quite often, one or two teams slip up, at very short prices!
Research is key. If you focus on finding bigger prices and ‘outside the box’ selections, you should see your success increase. This also allows you to keep the accumulator as small as possible, while still keeping the price big.
Use stat sites and look for trends in more obscure markets and you can find juicy prices. If you go for the lazy route and back the short priced favourites then you will keep making the same mistakes.
Also, consider removing the number of things that can go wrong by backing selections in markets with fewer outcomes.
Most people back selections in the traditional Win-Draw-Win market. But consider backing bigger priced teams in a Double Chance market and you may find yourself pleasantly surprised by the outcome.
## Where can you find the best football accumulators?
Here at OddsMonkey, we have some fantastic resources geared towards helping you find the best value football accumulators around. We can show you how to win consistently and generate real profit without worrying about striking it lucky.
Use our AccaMatcher to search bookmakers and build your own accumulators. Choose your favourite acca type and guarantee yourself a profit if you use the Lock-In method. For more information, take a look at the AccaMatcher and see how you could profit from placing football accumulators.
Check out the AccaMatcher
#### Peter OddsMonkey
If you have a maths or Matched Betting question, Peter is your man! An ex-maths & P.E. teacher, Peter has been matched betting for over 10 years.
Looking for ways to make money online?
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Episode 1 Kate has just signed up for OddsMonkey and has joined both of our buzzing online communities, our Facebook and Discord! Not only that, | 1,199 | 5,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-10 | latest | en | 0.95882 |
https://textbook.nipraxis.org/numpy_random.html | 1,721,212,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00439.warc.gz | 506,358,171 | 8,546 | # Numpy random number generators#
#: standard imports
import numpy as np
# print arrays to 4 decimal places
np.set_printoptions(precision=4, suppress=True)
We often need random numbers, for tests and for taking random samples, and for other things. np.random is a submodule within numpy:
type(np.random)
module
It contains function that will create a random number generator.
# Make a random number generator.
rng = np.random.default_rng()
type(rng)
numpy.random._generator.Generator
This generator is an object that has a set of methods for returning random numbers of various sorts. For example, to return a single random number from the default normal distribution (mean 0, variance 1):
rng.normal()
-0.34226728081257085
You can set the mean and variance with the first two input parameters:
# Random number from distribution with mean 15, variance 2
rng.normal(15, 2)
14.57283980956615
To return a 8 by 5 array of random numbers from the same distribution:
rng.normal(15, 2, size=(8, 5))
array([[12.6106, 15.1636, 12.9252, 13.8115, 14.0583],
[ 7.6281, 17.1641, 14.0648, 12.1612, 16.4149],
[15.4589, 15.7429, 15.3513, 14.5002, 16.4679],
[14.0262, 13.3321, 18.0072, 17.213 , 15.3075],
[15.2396, 14.2613, 15.1604, 15.1503, 14.811 ],
[11.9148, 16.5794, 14.5032, 17.9572, 14.4331],
[13.0585, 17.0516, 12.121 , 17.6757, 11.4749],
[18.4293, 12.2325, 18.3925, 16.5487, 15.9709]])
A 5 by 3 array of random numbers from the standard normal distribution with mean 1 and variance 1:
rng.normal(size=(5, 3))
array([[-0.9556, -2.6117, 1.5814],
[-1.1922, -0.2481, 1.6778],
[-1.8005, 0.9453, -1.4027],
[ 1.8498, 0.1025, -1.0869],
[-0.182 , 0.2813, -2.5928]])
## Making random numbers predictable#
Sometimes you want to make sure that the random numbers are predictable, in that you will always get the same set of random numbers from a series of calls to the rng methods. You can achieve this by giving the random number generator a seed when you create it. This is an integer that sets the random number generator into a predictable state, such that it will always return the same sequence of random numbers from this point:
# Set the state of the random number generator on creation.
new_rng = np.random.default_rng(seed=42)
# One set of random numbers
first_random_arr = new_rng.normal(size=(4, 2))
first_random_arr
array([[ 0.3047, -1.04 ],
[ 0.7505, 0.9406],
[-1.951 , -1.3022],
[ 0.1278, -0.3162]])
# Another set
second_random_arr = new_rng.normal(size=(4, 2))
second_random_arr
array([[-0.0168, -0.853 ],
[ 0.8794, 0.7778],
[ 0.066 , 1.1272],
[ 0.4675, -0.8593]])
# Make another random number generator with the same seed.
new_rng2 = np.random.default_rng(seed=42)
# The same as "first_random_arr" above.
new_rng2.normal(size=(4, 2))
array([[ 0.3047, -1.04 ],
[ 0.7505, 0.9406],
[-1.951 , -1.3022],
[ 0.1278, -0.3162]])
# The same as "second_random_arr" above.
new_rng2.normal(size=(4, 2))
array([[-0.0168, -0.853 ],
[ 0.8794, 0.7778],
[ 0.066 , 1.1272],
[ 0.4675, -0.8593]]) | 1,044 | 3,019 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.716088 |
http://mathhelpforum.com/calculus/121312-newton-s-method.html | 1,513,638,885,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948627628.96/warc/CC-MAIN-20171218215655-20171219001655-00189.warc.gz | 181,645,920 | 10,623 | 1. ## Newton's Method
approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
2. Hi there, have a look at this post.
http://www.mathhelpforum.com/math-he...ngly-easy.html
make your starting guess $x_1 = 1.5$
3. Originally Posted by wisezeta
approximate the root of h(x)=x^3 -3 using Newton's Method (correct to hundrehts)
you should have a valid rough estimate of the root , $x_0$ , to start the iterations.
note that the cube root of 3 is somewhere between 1 and 2
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
4. The actual root is 1.442, but the answer I get using 1.5 as my estimation gives me 1.5556. Is this right?
5. Are you converging on this figure? how many iterations have you done?
6. sorry for the typo ... the iteration equation should be a minus
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
try it again. | 270 | 847 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-51 | longest | en | 0.833615 |
https://elo.mastermath.nl/course/info.php?id=823 | 1,709,580,788,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00633.warc.gz | 229,945,251 | 9,690 | ### Coding Theory - M1 - 8EC
Prerequisites
-- Linear algebra (vector spaces, linear maps, kernel, matrices). For example, chapters I--IV of "Linear Algebra" by S. Lang, 3rd ed., Springer Undergraduate Texts in Mathematics.
-- Basic algebra (polynomials with coefficients in a field, basics of finite fields). For example, sections A.3--A.5 of https://www.win.tue.nl/~henkvt/images/CODING.pdf (in the Appendix).
-- Elementary combinatorics (double counting, the binomial coefficient, etc).
-- Elementary graph theory (the definitions).
Aim of the course
In the context of digital communication, error-correcting codes are mathematical objects that correct errors in noisy and lossy channels. They find several applications in satellite communications, digital audio systems, wireless communications, error correction chips, deep space exploration probes, flash memories, networking, and in many other contexts.
This course offers an introduction to the mathematical theory of error correction (encoding and decoding information using mathematics). The main aims are:
-- to become familiar with the mathematical methods used to protect digital information from noise, damage, and jamming attacks;
-- to become familiar with the mathematical structure of error-correcting objects;
-- to illustrate how error-correcting codes can be used to construct cryptosystems;
-- to understand the connections between coding theory and other branches of mathematics (algebra, combinatorics, probability, graph theory)
-- to become familiar with the mathematical modeling of network communications.
Lecturer
Alberto Ravagnani (TU/e)
https://a.ravagnani.win.tue.nl/ | 343 | 1,660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.805756 |
http://abrajbank.com/u9la44z32/cW24952iA/ | 1,582,823,546,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146744.74/warc/CC-MAIN-20200227160355-20200227190355-00333.warc.gz | 5,370,352 | 12,473 | # Math Digit Addition Worksheets Problems For Grade Preschool Homework Sheets Free Activity 1st Graders Printable High School Students Best Kids Worksheet Websites Horizontal Lines
By SiMs4lab1m. Addition Worksheets. At Monday, October 07th 2019, 20:55:28 PM.
There are basic skills every kindergarten needs to know before starting the school year. While the first portion of many school years is used to review basic concepts, teachers quickly move on to new subjects, assuming most children have had plenty of exposure to the basic concepts, including letter and number recognition. The ultimate goal in kindergarten is to get children ready to read, which takes a great deal of time. Teachers typically move on when a majority of the class is ready to learn the basics of reading. If your child is left behind, it could leave him behind for the rest of his school career.
Many children are being left behind due to lack of math skills. Schools today seem to do a poor job of preparing students for math at the middle and high school level. Here are 5 tips that parents can use to help their child be successful at math. Start early. Before your child goes to preschool, they need to be familiar with small numbers, up to 10. Two is easy to teach and point out. Pair of socks, shoes, etc. Five fingers on a hand and toes on feet. Ten total fingers and toes. At the preschool level, start counting up to 20. Add small numbers, 1 plus 1 is 2. 2 plus 1 is 3. You can even begin the fraction of one half. Half a sandwich, and other food items are a great start. When finishing kindergarten, your child needs to be able to count past 20 and know what larger numbers mean as well. Not working with them, just be familiar. | 377 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-10 | latest | en | 0.960533 |
http://sepwww.stanford.edu/data/media/public/docs/sep121/paper_html/node24.html | 1,545,185,753,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376830305.92/warc/CC-MAIN-20181219005231-20181219031220-00029.warc.gz | 256,125,404 | 3,377 | Next: Results on a simple Up: Principles of norm and Previous: Shaping filters and the
Shaping filters and the norm
The strong primary in Figure can be seen as an outlier that attracts much of the solver's efforts during the filter estimation.
filterl2 Figure 3 Shaping filter estimated for the 1-D problem with the norm. This filter is not a single spike at lag=0.
filterl1 Figure 4 Shaping filter estimated for the 1-D problem with the norm. This filter is a single spike at lag=0.
Consequently, some of the signal leaks into the noise. Because the norm is robust to outliers, I propose estimating the filter coefficients with it. This insensitivity to large noise'' has a statistical interpretation: robust measures are related to long-tailed density functions in the same way that is related to the short-tailed gaussian density function Tarantola (1987). In this section, I show that the norm solves the problem highlighted in the preceding section.
Now our goal is to estimate one shaping filter that minimizes the objective function
(10)
The function in equation () is singular where any residual component vanishes implying that the derivative of is not continuous everywhere. Unfortunately, most of our optimization techniques, e.g., conjugate-gradient or Newton methods, assume that the first derivative of the objective function is continuous in order to find its minimum. Therefore, specialized techniques have been developed to either minimize or approximate the norm. For instance, various approaches based on a linear programming viewpoint have been used with success, e.g., Barrodale and Roberts 1980. Another idea is to minimize a hybrid norm with an iteratively reweighted least-squares (IRLS) method Bube and Langan (1997); Gersztenkorn et al. (1986); Scales and Gersztenkorn (1988).
Alternatively, the Huber norm is utilized. This technique yields a good approximation of the norm. Therefore, the objective function we minimize becomes
(11)
The minimization of the Huber norm is performed with a quasi-Newton method (Appendix ). The only parameter that needs to be set is in equation (). Consistent with the strategy of Chapter , I set
(12)
From now on, I call norm the Huber norm. In text section, I show that the norm leads to the desired result for the filter estimation problem.
Next: Results on a simple Up: Principles of norm and Previous: Shaping filters and the
Stanford Exploration Project
5/5/2005 | 526 | 2,440 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-51 | longest | en | 0.916071 |
http://lists.openscad.org/pipermail/discuss_lists.openscad.org/2019-November/028916.html | 1,586,176,102,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371624083.66/warc/CC-MAIN-20200406102322-20200406132822-00365.warc.gz | 99,871,722 | 2,984 | # [OpenSCAD] Discuss manifoldness, co-incident faces edges etc
Fri Nov 15 03:15:43 EST 2019
```If you can partition the self-overlapping object into separate sections
that don't self overlap but do overlap each other than they can be unioned
back together. So perhaps the solution lies that way.
On Fri, 15 Nov 2019 at 07:19, <arnholm at arnholm.org> wrote:
> On 2019-11-15 01:25, Jordan Brown wrote:
> > Good question, but you don't need self-intersection to have to answer
> > that question. What color do you get if you have two independent
> > strokes of different color that overlap?
>
> Since this is an analogy of booleans the answer is that you will have to
> apply the proper colour model (additive or subtractive or something
> else) to find the answer. This is possible in a union of two bodies,
> see below.
>
> > Yes. (It need not be vertices. Strictly, I think it's that one or
> > more edges penetrate one or more faces.)
>
> Agreed, I just simplified it.
>
> > Indeed, two material objects cannot occupy the same space. But again
> > you don't have to get self-intersection to have that problem; you
> > can't have two independent steel rods overlap either. We *do* allow
> > two cylinders to overlap, via union, so our cylinders are not steel
> > rods. Two independent cylinders are allowed to occupy the same space,
> > so why isn't a bent cylinder allowed to have two parts that occupy the
> > same space?
>
> The answer is that when you union two different cylinders we have a well
> understood process (boolean union) to resolve overlaps that typically
> also offer a way to handle a way to handle properties, at least I
> believe that is the case in Carve even if I have not used those
> features. I don't know if CGAL has something in that area.
>
> However, there is no union operator that I know of that is able to
> handle "self-unions", and thus no way to handle the properties that may
> not necessarily be constant throughout the solid.
>
> Since this is about manifoldness, it is worth mentioning that solids
> with different subvolumes consisting of different materials (e.g.
> different plastic colour) is a good example where something else than
> 2-manifold solids is both useful and even required. The internal
> boundary between the materials causes the body to have N-manifold edges,
> and faces with no "outside". This is is perfectly meaningful both
> topologically and physically, such solids are all around us.
>
> Returning to the bent cylinder case, if it was all one material you
> could argue that selecting the material in the resulting overlap would
> not matter since it would be the same either way (assuming the property
> was constant that is, but you still have to figure out how to do self
> union). If the bar was N-manifold with different materials you would
> have to resolve the property issue.
>
> overlapping cylinders vs. one self-overlapping bent cylinder is that
> union is defined for the first case, but I am not so sure about the
> other. Perhaps a self-union can be defined that resolves self overlaps,
> I have not studied that.
>
> Carsten Arnholm
>
>
>
> _______________________________________________ | 762 | 3,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.973198 |
https://www.transtutors.com/questions/need-help-with-an-intermediate-accounting-problem-this-is-the-info-and-solution-i-ha-1188227.htm | 1,550,841,455,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247517815.83/warc/CC-MAIN-20190222114817-20190222140817-00009.warc.gz | 989,324,964 | 18,300 | # Need help with an Intermediate Accounting problem: This is the info and solution I have so far -...
Need help with an Intermediate Accounting problem: This is the info and solution I have so far - Percentage-of-Completion Method Newberg Construction Corporation contracted to construct a building for $400,000. Construction began in 2007 and was completed in 2008. Data relating to the contract are as follows: Year Ended December 31 2007 2008 Cost incurred$200,000 $110,000 Estimated costs to complete$100,000 Newberg uses the percentage-of-completion method as the basis for income recognition. For the years ended December 31, 2013 and 2014, respectively, how much gross profit should Newberg report? If required, round percentage completed to one decimal place. When required, round dollar amounts to the nearest dollar. This what I have: Newburg Construction Corporation Gross Profit/Loss Reported 2007 and 2008 2007 2008 Cost incurred to date $200,000$310,000 Estimated Costs to complete 100,000 -------- Percent Completed ?% 100% Estimated Profit on Contract 100,000 90,000 Gross Profit to be recognized ? ? I need help on the steps on how to calculate the three with the question marks. I figured out the others but could not get those three to calculate correctly. Thanks | 285 | 1,283 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-09 | latest | en | 0.914697 |
https://1library.net/subject/teaching-of-statistics | 1,638,570,391,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00258.warc.gz | 153,036,335 | 28,845 | # Teaching of Statistics
## Top PDF Teaching of Statistics:
### Cultivation of College Students’ Digital Economic Honesty in the Teaching of Statistics
3.3. Developing the Behavioral Habit of “Maintaining Integrity” The formation of credit requires a process of accumulation. It is necessary to build a good credit image through continuous accumulation. The establishment of statistical credit requires the efforts of the vast number of statisticians through hard work, so that every teacher and every student can do their best to maintain statistical integrity. In this way, there is hope in the establishment of integrity. In the traditional statistical classroom, however, the students mainly listen to the teacher’s analysis and observe their operation, while rarely do their own hands to verify and question the teacher’s point of view. To develop the behavior of “maintaining integrity” in the statistical classroom, students need to participate in all aspects. They should experience every aspect of teaching, from the class- room scene, on-machine experiments and on-campus and off-campus research to experience the formation of theory, data, results, so as to expand their ability to apply statistical knowledge, and cultivate their digital economic integrity ha- bits.
### Creating artificial data for teaching of statistics
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### XLR: A Free Excel Add In for Introductory Business Statistics
other statistical software to improve students’ learning curve. However, in endorsement of the Mathematical Association of America Guidelines for Programs and Departments in Undergraduate Mathematical Sciences [4], the American Statistical Association commented that “Generic packages such as Excel are not sufficient even for the teaching of statistics, let alone for research and consulting.” Numerous studies have highlighted the defi- ciencies and dangers of using Excel as a statistical pack- age for teaching and research. Reference [5-9] performs extensive studies to reveal that several statistical algo- rithms used in Excel yield erroneous results. Reference [10] demonstrates the danger of using Excel to perform regression analysis that involves correlated independent variables. Reference [11,12] discuss additional problems in using Excel while [13] recommends researchers against using Excel for any scientific purpose.
### Environmental Interfaces in Teaching Economic Statistics
Mainly developed since the 1990s, Statistics Education was conceived in an unease context, trying to question and reflect over problems related with the teaching and learning of this discipline. This educational perspective was ignited by the difficulties that students have in thinking or reasoning statistically even when they show calculation skills. Seeking to differentiate the pedagogical problems presented by Statistics from those presented by the teaching of Mathematics, several authors, such as Gal, Chance, Garfield and Ben-Zvi, among others, converged on the idea that the teaching of Statistics should focus on the development of three specific skills: statistical thinking, statistical reasoning, and statistical literacy.
### Teaching statistics to engineering students – an Australian experience of using educational technologies
The survey revealed consistent opinions regarding the two main directions for the future statistics module of the course. The majority of the students think that the course allowed less than sufficient time to teach statistics. They felt that the duration of 4 weeks is not enough to learn the statistical content of the course. Interestingly, most of the students suggested a stand-alone statistics course for Engineering. Some students were surprised that this was the only statistical component that they had seen in the whole engineering program. Many of the USQ students are of mature age and have long industrial experience and value statistical knowledge more than the fresh graduates who enter university directly from the colleges do. Giving due weight to statistics within the engineering curriculum would benefit the profession enormously, including augmenting the employability of our new graduates. Statistical methods are becoming an indispensable part of many scientific disciplines and engineering is no different. The views of the teaching team and the survey outcomes reinforce the need for increased formal teaching of statistics for engineering students. Therefore, this emphasizes the necessity of offering statistics as a stand-alone course and to include it as a learning outcome for the program. The idea is to include statistical concepts and methods in as many engineering courses as possible within the program, including setting and conducting experiments and hands-on analysis of experimental data using computing software and group projects to solve real engineering problems. Active participation of students and interaction among fellow colleagues [8] are essential to enhance students’ learning, and ensure hands-on statistical support in the process of teaching statistics to help solving diverse engineering problems.
### Teaching statistics to medical students using problem-based learning: the Australian experience
I visited the following universities with medical schools: the University of Western Australia, Perth, Flinders Uni- versity, Adelaide, Monash University, Melbourne, the University of Melbourne, the Australian National Univer- sity, Canberra (medical school about to commence), the University of Sydney, the University of Newcastle, and the University of Queensland, Brisbane. I omitted the Univer- sity of Adelaide, the University of New South Wales, and James Cook University, Townsville, but was able to get information from them by email. I identified potential informants from the university website and emailed them as follows: 'I am interested in the teaching of statistics to medical students. Last year I visited several medical schools in Australia, but unfortunately I did not manage to visit [your university] on this trip, a serious omission. I have written a report on my experiences, which is availa- ble on one of my websites at http:// www.mbland.sghms.ac.uk/ozpbl.htm. I am particularly interested in how statistics is taught in problem-based learning programmes. I am presenting this material at the forthcoming International Biometrics Conference in Cairns and preparing a paper for possible publication. I would be very interested to learn how these matters are ordered in [your university] . . . Are you the right person to ask? If not, could you suggest someone who would be?' I obtained helpful replies from all three universities. Results
### Teaching statistics to medical students using problem-based learning: the Australian experience
I visited the following universities with medical schools: the University of Western Australia, Perth, Flinders Uni- versity, Adelaide, Monash University, Melbourne, the University of Melbourne, the Australian National Univer- sity, Canberra (medical school about to commence), the University of Sydney, the University of Newcastle, and the University of Queensland, Brisbane. I omitted the Univer- sity of Adelaide, the University of New South Wales, and James Cook University, Townsville, but was able to get information from them by email. I identified potential informants from the university website and emailed them as follows: 'I am interested in the teaching of statistics to medical students. Last year I visited several medical schools in Australia, but unfortunately I did not manage to visit [your university] on this trip, a serious omission. I have written a report on my experiences, which is availa- ble on one of my websites at http:// www.mbland.sghms.ac.uk/ozpbl.htm. I am particularly interested in how statistics is taught in problem-based learning programmes. I am presenting this material at the forthcoming International Biometrics Conference in Cairns and preparing a paper for possible publication. I would be very interested to learn how these matters are ordered in [your university] . . . Are you the right person to ask? If not, could you suggest someone who would be?' I obtained helpful replies from all three universities. Results
### Vulva Lesions In A Tertiary Institution In Nigeria
The vulva is the area adjacent to the vagina in female reproductive region. This area is laden with both neoplastic and non-neoplastic lesions. This is a retrospective study of the entire female genital tract specimen received at the Department of Pathology, University of Calabar Teaching Hospital Calabar over a period of 10 years. A total of 6779 samples were received during the study period and 1602 were gynaecological lesions, out of which 175 were malignant. 13 malignant vulva tumors were seen with predominant squamous cell carcinoma (10 out of 13) and the rest were choriocarcinoma, leiomyosarcoma and embryonal rhabdomyosarcoma which were one case each. Adenomyoma was the predominant benign lesion seen with 26 cases seen out of 33 benign lesions seen in the vulva.
### A study of the effectiveness of the contextual lab activity in the teaching and learning engineering statistics at the Universiti Tun Hussein Onn Malaysia (UTHM)
The conceptual framework of this research was adapted from [3]. The dependent variable of this study was the achievement test. The researchers investigated whether the use of the contextual concept in the lab activities influenced the students’ achievement in the statistics test as compared to the use of the non-contextual concept in the lab activities. Both groups were taught by the same lecturer to eliminate the extraneous variable.
### Making Sense of Basic Education Statistics in Tanzania: Emerging Issues and Implications for Future Practice
The government introduced the education policy in 2014 with the aim to improve teaching and learning facilities, including the availability of textbooks, laboratories and technological devices such as computers [5]. In achieving this objective, the Tanzanian government has ensured that one third of the required books are available in primary schools [16]. There has been an increase in enrolment since the declaration of fee free basic education without a clear correspondence to the availability of the textbooks. In the most recent statistics, primary school enrolment of pupils in Standard I-VII has increased by 8.5% from 9,317,791 pupils in the year 2017 to pupils 10,111,671 in the year 2018. In 2017 the average Pupils Book Ratio (PBR) ranged from 1: 5 and 1:6 while in 2018 there was a critical shortage of textbooks for history 1:6, ‘Najifunza Afya na mazingira’ 1:5 and ‘Najifunza Kuendeleza Sanaa na Michezo’ 1:5, [19]. It is possible to conclude that the prevalence of high PBR is attributed to the introduction of the fee free education policy. On the other hand, it is possible to agree that the PBR is affected by an increase of pupils’ enrolment but there has been a concern from educational stakeholders on the quality of the text-books. The major concern is on the textbooks whose content lack proper connection with the curriculum [27]. It is noted with concerns that the language used in English primary school textbooks is complex, in the sense that the majority of pupils could not read [28]. One would say, although there must be a significant PBR, still the need to maintain quality is inevitable. Reflecting on the SDG number four, it states that by 2030, member states should build and upgrade education facilities that are child, disability and gender sensitive and provide safe, non
### Vol 15, No 2 (2016)
students who exit out of introductory statistics and enter their classes to use statistics. Therefore, a symbiotic relationship exists between statistics departments and client disciplines whereby statistics courses prepare students to be successful in their major area of study. Faculty attitudes and perceptions are an important area of research, particularly outside of mathematics and statistics departments, since statistics is widely taught and utilized by faculty across many client disciplines (Sterling, Rosenbaum, & Weinkam, 1995; Carlson, 2002; Switzer & Horton, 2007; Doehler, Taylor, & Smith, 2013). Indeed, Eichler (2010) states that future research should ―investigate the teachers’ attitudes towards statistics in more countries involving more representative samples of ordinary teachers‖ (p. 4). The research presented here considers individuals teaching statistics within many other disciplines. Therefore, it helps to fill a gap in the literature. Since statistics is utilized and taught by faculty in a large number of disciplines, it is imperative to also consider how faculty perceive statistics. If instructors have a poor attitude towards statistics and its usefulness, they could prevent statistics from being a positive experience for their students. Garfield, Hogg, Schau, and Whittinghill (2002) state that ―our courses should attempt to build strong positive attitudes towards statistics […] to increase their chances of using statistics after they leave our courses‖ (p. 3). It seems logical that if this is a goal, then instructors and other individuals who influence students should also have ―strong positive attitudes towards statistics.‖ Zieffler et al. (2008) recognize the need for instructors to help students have a positive learning experience when studying statistics. They also state that learning of statistics could increase if students’ attitudes toward the discipline improved. Therefore, student learning of statistics may increase when positive attitudes towards the discipline are displayed by faculty. Although this likely applies more so to faculty teaching statistics within any discipline, this may also apply to faculty who do not actually teach statistics. For example, if a student taking an introductory statistics class overhears a professor in another discipline saying that doing statistics is too hard for him/her, this could negatively impact the student’s learning.
### Teaching medical statistics to undergraduate medical students: what is taught and what is really useful for a medical professional? A report of the Education Committee of the Italian Society of Medical Statistics and Clinical Epidemiology (SISMEC)
institutional representatives of the academic medical world: the Presidents ‘pro tempore’ of the 50 academic medical disciplines and 56 Deans for Education of the medical courses. Six main domains were investigated: (1) epidemiology and demography, (2) biostatistics, (3) prognosis, (4) diagnosis, (5) efficacy, and (6) EBM. We chose to use a simpler form with a reduced number of more general topics in the hope to increase compliance. For every topic, the sole question was ‘How much is this skill useful for a medical professional’, allowing an answer on a 5-point Likert scale, ranging from ‘not at all’ to ‘essential’. Further to every topic addressed in the questionnaire, we associated one or more appropriate elementary teaching units of the core curriculum defined by the Conference of Deans for Education of medical courses, in order to suggest the medical framework for which it could be applicable (http://presidenti-medicina.it/core- curriculum/). The English translation of the questionnaire is reported in supplementary Table S2.
### The Preparedness of Preservice Secondary Mathematics Teachers to Teach Statistics: A Cross-Institutional Mixed Method Study.
A teacher’s preparedness not only relies on cognitive aspects, but also affective constructs such as beliefs, attitudes, and self-efficacy (e.g., Ball et al., 2008). Teachers’ beliefs play a crucial role in the pedagogical approaches they use, the time spent on a subject, and impact students’ learning (e.g., Wilkins, 2008). However there is a lack of research on secondary teachers’ beliefs in regards to statistics and teaching statistics (Batanero et al., 2011). This study focuses on PSMTs’ self-efficacy for teaching high school students the statistical standards in CCSSM. Bandura (1986) defines self-efficacy as “people’s judgments of their capabilities to organize and execute courses of action required to attain designated types of performance” (p.391). Judgments of one’s own self-efficacy are task-specific and change over time (Bandura, 1997; Pajares, 1997). A teacher has two types of self-efficacy for each content area they teach: self-efficacy to know the content themselves, and self-efficacy to teach the topic to students, known as teaching efficacy.
### The effect of explicit teaching of discourse markers on Iranian efl learners’ pragmatic fluency
4.53333, respectively and the significance level is 0.001; thus, the participants in the experimental group who were exposed to the treatment (explicit teaching of discourse markers) in contrast of control group in post-test had better performance. Also based on Figure 4.2 the frequency of using DMs among the experimental group was more than control group. The findings of this investigation were in line with Hays (1992) have a great investigation in DMs in classroom oral discourse. He asserted that discourse markers had a great influence on students' oral interaction. The results also were in agreement withPierrehumbert and Hirschberg (1986; 1990), Redeker (1991), Fraser (1998), Schiffrin (1987), and others findings about the impact of implicit teaching of discourse markers on learners' achievements. The findings reviled that the participants in the experimental group who received explicit instruction on discourse markers made more frequent use of them in their oral production, in contrast the learners who were in the control group and received implicit instruction did not use discourse markers frequently in their aural production.
### Teaching Practices and the Management of Student Motivation, Effort and Achievement
Teaching Practices and the Management of Student Motivation, Effort and Achievement Gunnes, Trude and Donze, Jocelyn Statistics Norway, Strasbourg University and Toulouse School of Econo[r]
### Collaborating to Teach Research Methods in Education
describes the challenge of teaching research methods to a diverse and multi-disciplinary group of students, who may or may not conduct scholarly research in practice. Through this collaboration in both teaching and writing, we were able to address for ourselves some of the paradigmatic differences between qualitative and quantitative research, and offer evidence for our students that perhaps this dichotomy is not always a useful one. In addition, we have documented an example of simple collaboration at the post-secondary level from which we hope other instructors might benefit. We feel that, based upon our own experiences and a (albeit non-random) sampling of comments made by students at the end of the classes, that we were successful in these areas. We believe that through our collaboration, students were supported in and confident enough to pursue their unique research agendas (as either consumers or producers of research), even when these agendas were outside of each author’s individual comfort zone. The diversity of student needs was manageable when the authors worked together. Collaboration within the context of a research methods class allows both the teachers and students the opportunity to better understand the variety of tools available to answer questions which pertain to educational problems and issues. If not for this collaboration we believe we would have had a lesser experience, as would the students. When we return to our initial question, what can a statistician and an arts-based researcher learn from one another? the answer is, indeed, plenty.
### (Game Based) Student Response Systems Engage Students with Research Teaching Nexus Activities and Support Their Skills Development
Research-teaching nexus framework assists teachers to support activities which are focused on the current research in the disciplines promoting dis- cussions and skills development. The challenges for teachers are related to student research skills development, knowledge application across contexts and student experience on statistics. The aim of this investigation was to compare student views following two approaches: one which integrated technology and one which did not into the teaching delivery process. Forty first-year Psychology students were split into two groups (A and B) and fol- lowed both teaching approaches in a different attending order (one approach without the use of technology and another one with technology or vice versa). (Game-based) student response systems (PollEveryWhere and Kahoot) were integrated into the teaching process, whilst the learning content was the same for both cases. Students evaluated the two approaches by completing two on- line surveys with items relating to research-teaching nexus activities and skills development. The integration of (game-based) student response systems into teaching process increased student engagement in learning, improved the in- teractions between students and teachers and allowed them to develop the re- levant research skills. Students received the technology intervention as a way to work on a creative learning environment which allowed them to develop skills and knowledge/experience around research methods and statistics. Al- though this investigation took place in a Psychology School, the design of a research methods and statistics module based on the combination of colla- borative problem-based learning with the use of (game-based) student re- sponse systems can be applied in other disciplines.
### Theory of Statistics - Free Computer, Programming, Mathematics, Technical Books, Lecture Notes and Tutorials
began writing text around the bullets, and I put the notes on the class website. Later I decided that a single document with a fairly extensive subject index (see pages ?? through ??) would be useful to serve as a companion for the study of mathematical statistics. The other big deal about this document is the internal links, which of course is not something that can be done with a hardcopy book. (One thing I must warn you about is that there is a (known) bug in the L A TEX package hyperref; if the referenced point happens to occur | 4,352 | 22,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-49 | latest | en | 0.927564 |
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# The study undertaken by Department of the Interior will
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The study undertaken by Department of the Interior will [#permalink]
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25 Mar 2005, 10:33
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The study undertaken by Department of the Interior will involve examination and tagging of the California condor in order to obtain information about their daily movements, foraging habits, and sites where they nest.
(A) their daily movements, foraging habits, and sites where they nest
(B) their daily movements, foraging habits, and their nesting sites
(C) its daily movements, foraging habits, and nesting sites
(D) its daily movements, foraging habits, and about nesting sites
(E) daily movements, foraging habits, and sites in which there are nests
If you have any questions
New!
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25 Mar 2005, 10:53
C.
Quote:
The study undertaken by Department of the Interior will involve examination and tagging of the California condor in order to obtain information about their daily movements, foraging habits, and sites where they nest.
(A) their daily movements, foraging habits, and sites where they nest
(B) their daily movements, foraging habits, and their nesting sites
(C) its daily movements, foraging habits, and nesting sites
(D) its daily movements, foraging habits, and about nesting sites
(E) daily movements, foraging habits, and sites in which there are nests
A - their daily movements - their is wrong in this
B - same as A
C - Correct
D - and about nesting sites - is wordy
E - unnecessarily long
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25 Mar 2005, 10:56
OA is C.
But isn't condor plural, and hence require their?
Example:
These condor are under the protection of the department of the interior.
We wouldn't say:
These condor is under the protection of the department of the interior.
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31 Mar 2005, 11:19
OA is C.
But isn't condor plural, and hence require their?
Example:
These condor are under the protection of the department of the interior.
We wouldn't say:
These condor is under the protection of the department of the interior.
in the stem--it metion about california condor,which in this case we take as 1 group/family,,,,,,whereas in yr example these codor refers to a particluar number of codor (2,3,4'....)which are under consideration.......hece these 'are' is uders.........
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i hate when people do'nt post the OA, it leaves in guessing!!!!
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31 Mar 2005, 19:51
'their' is the wrong pronoun. California condor is singular as it's a collective noun that refers to a particular family of condors. So (A), (B) are out.
(E) is also out as there is a problem with the sentence when it joins the non-underline portion of the sentece. It would only be correct if the sentence was "condor's daily movements, ...."
Between C and D, C is better as it has parallelism.
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Display posts from previous: Sort by | 1,216 | 4,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2016-50 | longest | en | 0.919055 |
https://convertoctopus.com/340-cubic-centimeters-to-milliliters | 1,638,743,416,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00397.warc.gz | 249,534,414 | 7,813 | ## Conversion formula
The conversion factor from cubic centimeters to milliliters is 1, which means that 1 cubic centimeter is equal to 1 milliliters:
1 cm3 = 1 ml
To convert 340 cubic centimeters into milliliters we have to multiply 340 by the conversion factor in order to get the volume amount from cubic centimeters to milliliters. We can also form a simple proportion to calculate the result:
1 cm3 → 1 ml
340 cm3 → V(ml)
Solve the above proportion to obtain the volume V in milliliters:
V(ml) = 340 cm3 × 1 ml
V(ml) = 340 ml
The final result is:
340 cm3 → 340 ml
We conclude that 340 cubic centimeters is equivalent to 340 milliliters:
340 cubic centimeters = 340 milliliters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 milliliter is equal to 0.0029411764705882 × 340 cubic centimeters.
Another way is saying that 340 cubic centimeters is equal to 1 ÷ 0.0029411764705882 milliliters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred forty cubic centimeters is approximately three hundred forty milliliters:
340 cm3 ≅ 340 ml
An alternative is also that one milliliter is approximately zero point zero zero three times three hundred forty cubic centimeters.
## Conversion table
### cubic centimeters to milliliters chart
For quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to milliliters
cubic centimeters (cm3) milliliters (ml)
341 cubic centimeters 341 milliliters
342 cubic centimeters 342 milliliters
343 cubic centimeters 343 milliliters
344 cubic centimeters 344 milliliters
345 cubic centimeters 345 milliliters
346 cubic centimeters 346 milliliters
347 cubic centimeters 347 milliliters
348 cubic centimeters 348 milliliters
349 cubic centimeters 349 milliliters
350 cubic centimeters 350 milliliters | 484 | 1,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-49 | latest | en | 0.690142 |
https://www.convertunits.com/from/cubic+mile/to/cubic+centimeter | 1,632,465,513,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057504.60/warc/CC-MAIN-20210924050055-20210924080055-00512.warc.gz | 750,600,321 | 23,464 | ## ››Convert cubic mile to cubic centimetre
cubic mile cubic centimeter
How many cubic mile in 1 cubic centimeter? The answer is 2.3991275756488E-16.
We assume you are converting between cubic mile and cubic centimetre.
You can view more details on each measurement unit:
cubic mile or cubic centimeter
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 2.3991275756488E-10 cubic mile, or 1000000 cubic centimeter.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between cubic miles and cubic centimeters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of cubic mile to cubic centimeter
1 cubic mile to cubic centimeter = 4.1681818430585E+15 cubic centimeter
2 cubic mile to cubic centimeter = 8.3363636861169E+15 cubic centimeter
3 cubic mile to cubic centimeter = 1.2504545529175E+16 cubic centimeter
4 cubic mile to cubic centimeter = 1.6672727372234E+16 cubic centimeter
5 cubic mile to cubic centimeter = 2.0840909215292E+16 cubic centimeter
6 cubic mile to cubic centimeter = 2.5009091058351E+16 cubic centimeter
7 cubic mile to cubic centimeter = 2.9177272901409E+16 cubic centimeter
8 cubic mile to cubic centimeter = 3.3345454744468E+16 cubic centimeter
9 cubic mile to cubic centimeter = 3.7513636587526E+16 cubic centimeter
10 cubic mile to cubic centimeter = 4.1681818430585E+16 cubic centimeter
## ››Want other units?
You can do the reverse unit conversion from cubic centimeter to cubic mile, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Cubic mile
A cubic mile is an Imperial / U.S. customary (non-SI non-metric) unit of volume, used in the United States. It is defined as the volume of a cube with sides of 1 mile (5280 feet, 1760 yards, ≈1.609 kilometre) in length.
## ››Definition: Cubic centimeter
A cubic centimetre (cm3) is equal to the volume of a cube with side length of 1 centimetre. It was the base unit of volume of the CGS system of units, and is a legitimate SI unit. It is equal to a millilitre (ml).
The colloquial abbreviations cc and ccm are not SI but are common in some contexts. It is a verbal shorthand for "cubic centimetre". For example 'cc' is commonly used for denoting displacement of car and motorbike engines "the Mini Cooper had a 1275 cc engine". In medicine 'cc' is also common, for example "100 cc of blood loss".
## ››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! | 782 | 2,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-39 | latest | en | 0.72036 |
http://en.wikibooks.org/wiki/Statistics:Numerical_Methods/Basic_Linear_Algebra_and_Gram-Schmidt_Orthogonalization | 1,432,742,852,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929012.53/warc/CC-MAIN-20150521113209-00288-ip-10-180-206-219.ec2.internal.warc.gz | 77,858,126 | 20,049 | # Statistics/Numerical Methods/Basic Linear Algebra and Gram-Schmidt Orthogonalization
## Introduction
Basically, all the sections found here can be also found in a linear algebra book. However, the Gram-Schmidt Orthogonalization is used in statistical algorithm and in the solution of statistical problems. Therefore, we briefly jump into the linear algebra theory which is necessary to understand Gram-Schmidt Orthogonalization.
The following subsections also contain examples. It is very important for further understanding that the concepts presented here are not only valid for typical vectors as tuple of real numbers, but also functions that can be considered vectors.
## Fields
### Definition
A set $R$ with two operations $+$ and $*$ on its elements is called a field (or short $(R,+,*)$), if the following conditions hold:
1. For all $\alpha, \beta \in R$ holds $\alpha+\beta \in R$
2. For all $\alpha, \beta \in R$ holds $\alpha+\beta = \beta+\alpha$ (commutativity)
3. For all $\alpha, \beta,\gamma \in R$ holds $\alpha+(\beta+\gamma) = (\alpha+\beta)+\gamma$ (associativity)
4. It exist a unique element $0$, called zero, such that for all $\alpha \in R$ holds $\alpha+0 = \alpha$
5. For all $\alpha \in R$ a unique element $-\alpha$, such that holds $\alpha+ (-\alpha) = 0$
6. For all $\alpha, \beta \in R$ holds $\alpha*\beta \in R$
7. For all $\alpha, \beta \in R$ holds $\alpha*\beta = \beta*\alpha$ (commutativity)
8. For all $\alpha, \beta,\gamma \in R$ holds $\alpha*(\beta*\gamma) = (\alpha*\beta)*\gamma$ (associativity)
9. It exist a unique element $1$, called one, such that for all $\alpha \in R$ holds $\alpha*1 = \alpha$
10. For all non-zero $\alpha \in R$ a unique element $\alpha^{-1}$, such that holds $\alpha * \alpha^{-1} = 1$
11. For all $\alpha, \beta,\gamma \in R$ holds $\alpha*(\beta+\gamma) = \alpha*\beta+\alpha*\gamma$ (distributivity)
The elements of $R$ are also called scalars.
### Examples
It can easily be proven that real numbers with the well known addition and multiplication $(IR, +, *)$ are a field. The same holds for complex numbers with the addition and multiplication. Actually, there are not many more sets with two operations which fulfill all of these conditions.
For statistics, only the real and complex numbers with the addition and multiplication are important.
## Vector spaces
### Definition
A set $V$ with two operations $+$ and $*$ on its elements is called a vector space over R, if the following conditions hold:
1. For all $x, y\in V$ holds $x+y \in V$
2. For all $x, y \in V$ holds $x+y = y+x$ (commutativity)
3. For all $x, y, z \in V$ holds $x+(y+z) = (x+y)+z$ (associativity)
4. It exist a unique element $\mathbb{O}$, called origin, such that for all $x \in V$ holds $x+\mathbb{O} = x$
5. For all $x \in V$ exists a unique element $-v$, such that holds $x+ (-x) = \mathbb{O}$
6. For all $\alpha\in R$ and $x\in V$ holds $\alpha*x \in V$
7. For all $\alpha, \beta \in R$ and $x\in V$ holds $\alpha*(\beta*x) = (\alpha*\beta)*x$ (associativity)
8. For all $x\in V$ and $1 \in R$ holds $1*x = x$
9. For all $\alpha\in R$ and for all $x, y\in V$holds $\alpha*(x+y) = \alpha*x+\alpha*y$ (distributivity wrt. vector addition)
10. For all $\alpha, \beta \in R$ and for all $x\in V$holds $(\alpha+\beta)*x = \alpha*x+\beta*x$ (distributivity wrt. scalar addition)
Note that we used the same symbols $+$ and $*$ for different operations in $R$ and $V$. The elements of $V$ are also called vectors.
Examples:
1. The set $IR^p$ with the real-valued vectors $(x_1,...,x_p)$ with elementwise addition $x+y=(x_1+y_1,...,x_p+y_p)$ and the elementwise multiplication $\alpha*x = (\alpha x_1,...,\alpha x_p)$ is a vector space over $IR$.
2. The set of polynomials of degree $p$, $P(x) = b_0 + b_1 x + b_2 x^2 + ... + b_p x^p$, with usual addition and multiplication is a vector space over $IR$.
### Linear combinations
A vector $x$ can be written as a linear combination of vectors $x_1,...x_n$, if
$x = \sum_{i=1}^n \alpha_i x_i$
with $\alpha_i \in R$.
Examples:
• $(1,2,3)$ is a linear combination of $(1,0,0),\,(0,1,0), \,(0,0,1)$ since $(1,2,3)=1*(1,0,0)+2*(0,1,0)+3*(0,0,1)$
• $1+2*x+3*x^2$ is a linear combination of $1+x+x^2,\, x+x^2,\, x^2$ since $1+2*x+3*x^2=1*(1+x+x^2)+1*(x+x^2)+1*(x^2)$
### Basis of a vector space
A set of vectors $x_1, ..., x_n$ is called a basis of the vector space $V$, if
1. for each vector $x \in V$ exist scalars $\alpha_1,...,\alpha_n \in R$ such that $x = \sum_i \alpha_i x_i$ 2. there is no subset of $\{ x_1, ..., x_n \}$ such that 1. is fulfilled.
Note, that a vector space can have several bases.
Examples:
• Each vector $(\alpha_1, \alpha_2, \alpha_3) \in IR^3$ can be written as $\alpha_1 * (1,0,0) + \alpha_2 *(0,1,0) + \alpha_3 * (0,0,1)$. Therefore is $\{(1,0,0), (0,1,0), (0,0,1) \}$ a basis of $IR^3$.
• Each polynomial of degree $p$ can be written as linear combination of $\{ 1, x, x^2, ..., x^p\}$ and therefore forms a basis for this vector space.
Actually, for both examples we would have to prove condition 2., but it is clear that it holds.
### Dimension of a vector space
A dimension of a vector space is the number of vectors which are necessary for a basis. A vector space has infinitely many number of basis, but the dimension is uniquely determined. Note that the vector space may have a dimension of infinity, e.g. consider the space of continuous functions.
Examples:
• The dimension of $IR^3$ is three, the dimension of $IR^p$ is $p$ .
• The dimension of the polynomials of degree $p$ is $p+1$.
### Scalar products
A mapping $<.,.>: V\times V \rightarrow R$ is called a scalar product if the following holds for all $x,x_1,x_2,y,y_1,y_2 \in V$ and $\alpha_1, \alpha_2 \in R$ :
1. $<\alpha_1 x_1 + \alpha_2 x_2, y> = \alpha_1 + \alpha_2 $
2. $ = \alpha_1 + \alpha_2 $
3. $ = \overline{}$ with $\overline{\alpha + \imath \beta} = \alpha - \imath \beta$
4. $ \geq 0$ with $=0 \Leftrightarrow x = \mathbb{O}$
Examples:
• The typical scalar product in $IR^p$ is $ = \sum_i x_i y_i$.
• $ = \int_a^b f(x)*g(x) dx$ is a scalar product on the vector space of polynomials of degree $p$.
### Norm
A norm of a vector is a mapping $\|.\|: V \rightarrow R$, if holds
1. $\| x \| \geq 0$ for all $x \in V$ and $\| x \| =0 \Leftrightarrow x=\mathbb{O}$ (positive definiteness)
2. $\| \alpha v \|= \mid \alpha \mid \| x \|$ for all $x\in V$ and all $\alpha\in R$
3. $\| x+y \| \leq \| x \| + \| y \|$ for all $x,y\in V$ (triangle inequality)
Examples:
• The $L_q$ norm of a vector in $IR^p$ is defined as $\|x\|_q = \sqrt[q]{\sum_{i=1}^p x_i^q}$.
• Each scalar product generates a norm by $\|x\| = \sqrt{}$, therefore $\|f\| = \sqrt{\int_a^b f^2(x) dx}$ is a norm for the polynomials of degree $p$.
### Orthogonality
Two vectors $x$ and $y$ are orthogonal to each other if $=0$. In $IR^p$ it holds that the cosine of the angle between two vectors can expressed as
$\cos(\angle(x,y)) = \frac{}{\|x\|\|y\|}$.
If the angle between $x$ and $y$ is ninety degree (orthogonal) then the cosine is zero and it follows that $=0$.
A set of vectors $x_1, ..., x_p$ is called orthonormal, if
$ = \begin{cases} 0 & \mbox{ if } i\neq j \\ 1 & \mbox{ if } i=j \end{cases}$.
If we consider a basis $e_1,..., e_p$ of a vector space then we would like to have a orthonormal basis. Why ?
Since we have a basis, each vector $x$ and $y$ can be expressed by $x=\alpha_1 e_1 + ... +\alpha_p e_p$ and $y=\beta_1 e_1 + ... +\beta_p e_p$. Therefore the scalar product of $x$ and $y$ reduces to
$\$ $= <\alpha_1 e_1 + ... +\alpha_p e_p, \beta_1 e_1 + ... +\beta_p e_p>\$ $= \sum_{i=1}^p \sum_{j=1}^p \alpha_i \beta_j $ $= \sum_{i=1}^p \alpha_i \beta_i $ $= \alpha_1 \beta_1 + ... + \alpha_p \beta_p.\$
Consequently, the computation of a scalar product is reduced to simple multiplication and addition if the coefficients are known. Remember that for our polynomials we would have to solve an integral!
## Gram-Schmidt orthogonalization
### Algorithm
The aim of the Gram-Schmidt orthogonalization is to find for a set of vectors $x_1, ..., x_p$ an equivalent set of orthonormal vectors $o_1,...,o_p$ such that any vector which can be expressed as linear combination of $x_1, ..., x_p$ can also be expressed as linear combination of $o_1,...,o_p$:
1. Set $b_1 = x_1$ and $o_1 = b_1 / \|b_1\|$
2. For each $i>1$ set $b_i = x_i - \sum_{j=1}^{i-1} \frac{}{} b_j$ and $o_i = b_i / \|b_i\|$, in each step the vector $x_i$ is projected on $b_j$ and the result is subtracted from $x_i$.
### Example
Consider the polynomials of degree two in the interval$[-1,1]$ with the scalar product $ = \int_{-1}^1 f(x) g(x) dx$ and the norm $\|f\| = \sqrt{}$. We know that $f_1(x)=1, f_2(x)=x$ and $f_3(x)=x^2$ are a basis for this vector space. Let us now construct an orthonormal basis:
Step 1a: $b_1(x) = f_1(x) = 1$
Step 1b: $o_1(x) = \frac{b_1(x)}{\|b_1(x)\|} = \frac{1}{\sqrt{}} = \frac{1}{\sqrt{\int_{-1}^1 1 dx}} = \frac{1}{\sqrt{2}}$
Step 2a: $b_2(x) = f_2(x) - \frac{}{} b_1(x) = x - \frac{\int_{-1}^1 x\ 1 dx}{2} 1 = x - \frac{0}{2} 1 = x$
Step 2b: $o_2(x) = \frac{b_2(x)}{\|b_2(x)\|} = \frac{x}{\sqrt{}} = \frac{x}{\sqrt{\int_{-1}^1 x^2 dx}} = \frac{x}{\sqrt{2/3}} = x\sqrt{3/2}$
Step 3a: $b_3(x) = f_3(x) - \frac{}{} b_1(x) - \frac{}{} b_2(x) = x^2 - \frac{\int_{-1}^1 x^2 1\ dx}{2} 1 - \frac{\int_{-1}^1 x^2 x\ dx}{2/3} x = x^2 - \frac{2/3}{2} 1 - \frac{0}{2/3} x = x^2 - 1/3$
Step 3b: $o_3(x) = \frac{b_3(x)}{\|b_3(x)\|} = \frac{x^2-1/3}{\sqrt{}} = \frac{x^2-1/3}{\sqrt{\int_{-1}^1 (x^2-1/3)^2 dx}} = \frac{x^2-1/3}{\sqrt{\int_{-1}^1 x^4 - 2/3 x^2 + 1/9\ dx}} = \frac{x^2-1/3}{\sqrt{8/45}} = \sqrt{\frac{5}{8}} (3x^2-1)$
It can be proven that $1/\sqrt{2}, x\sqrt{3/2}$ and $\sqrt{\frac{5}{8}} (3x^2-1)$ form a orthonormal basis with the above scalarproduct and norm.
### Numerical instability
Consider the vectors $x_1 = (1,\epsilon, 0, 0), x_2 = (1,0,\epsilon,0)$ and $x_3 = (1,0,0,\epsilon)$. Assume that $\epsilon$ is so small that computing $1+\epsilon = 1$ holds on a computer (see http://en.wikipedia.org/wiki/Machine_epsilon). Let compute a orthonormal basis for this vectors in $IR^4$ with the standard scalar product $=x_1y_1+x_2y_2+x_3y_3+x_4y_4$ and the norm $\|x\| = \sqrt{x_1^2+x_2^2+x_3^2+x_4^2}$.
Step 1a. $b_1 = x_1 = (1,\epsilon,0,0)$
Step 1b. $o_1 = \frac{b_1}{\|b_1\|} = \frac{b_1}{\sqrt{1+\epsilon^2}} = b_1$ with $1+\epsilon^2=1$
Step 2a. $b_2 = x_2 - \frac{}{} b_1 = (1,0,\epsilon,0) - \frac{1}{1+\epsilon^2} (1,\epsilon,0,0) = (0,-\epsilon,\epsilon,0)$
Step 2b. $o_2 = \frac{b_2}{\|b_2\|} = \frac{b_2}{\sqrt{2\epsilon^2}} = (0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$
Step 3a. $b_3 = x_3 - \frac{}{} b_1 - \frac{}{} b_2 = (1,0,0,\epsilon) - \frac{1}{1+\epsilon^2} (1,\epsilon,0,0) - \frac{0}{2\epsilon^2} (0,-\epsilon,\epsilon,0) = (0, -\epsilon, 0,\epsilon)$
Step 3b. $o_3 = \frac{b_3}{\|b_3\|} = \frac{b_3}{\sqrt{2\epsilon^2}} = (0,-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$
It obvious that for the vectors
- $o_1 = (1, \epsilon, 0, 0)\$
- $o_2 = (0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$
- $o_3 = (0,-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$
the scalarproduct $ = 1/2 \neq 0$. All other pairs are also not zero, but they are multiplied with $\epsilon$ such that we get a result near zero.
### Modified Gram-Schmidt
To solve the problem a modified Gram-Schmidt algorithm is used:
1. Set $b_i = x_i$ for all $i$
2. for each $i$ from $1$ to $n$ compute
1. $o_i = \frac{b_i}{\|b_i\|}$
2. for each $j$ from $i+1$ to $n$ compute $b_j = b_j - o_i\$
The difference is that we compute first our new $b_i$ and subtract it from all other $b_j$. We apply the wrongly computed vector to all vectors instead of computing each $b_i$ separately.
### Example (recomputed)
Step 1. $b_1 = (1,\epsilon,0,0)$, $b_2 = (1,0,\epsilon,0)$, $b_3 = (1,0,0,\epsilon)$
Step 2a. $o_1 = \frac{b_1}{\|b_1\|} = \frac{b_1}{\sqrt{1+\epsilon^2}} = b_1 = (1,\epsilon,0,0)$ with $1+\epsilon^2=1$
Step 2b. $b_2 = b_2 - o_1 = (1,0,\epsilon,0) - (1,\epsilon,0,0) = (0, -\epsilon, \epsilon, 0)\$
Step 2c. $b_3 = b_3 - o_1 = (1,0,0,\epsilon) - (1,\epsilon,0,0) = (0, -\epsilon, 0, \epsilon)\$
Step 3a. $o_2 = \frac{b_2}{\|b_2\|} = \frac{b_2}{\sqrt{2\epsilon^2}} = (0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)$
Step 3b. $b_3 = b_3 - o_2 = (0, -\epsilon, 0, \epsilon) - \frac{\epsilon}{\sqrt{2}} (0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0) = (0, -\epsilon/2, -\epsilon/2, \epsilon)$
Step 4a. $o_3 = \frac{b_3}{\|b_3\|} = \frac{b_3}{\sqrt{3/2\epsilon^2}} = (0,-\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}})$
We can easily verify that $ = 0$.
## Application
### Exploratory Project Pursuit
In the analysis of high-dimensional data we usually analyze projections of the data. The approach results from the Theorem of Cramer-Wold that states that the multidimensional distribution is fixed if we know all one-dimensional projections. Another theorem states that most (one-dimensional) projections of multivariate data are looking normal, even if the multivariate distribution of the data is highly non-normal.
Therefore in Exploratory Projection Pursuit we jugde the interestingness of a projection by comparison with a (standard) normal distribution. If we assume that the one-dimensional data $x$ are standard normal distributed then after the transformation $z=2\Phi^{-1}(x)-1$ with $\Phi(x)$ the cumulative distribution function of the standard normal distribution then $z$ is uniformly distributed in the interval $[-1;1]$.
Thus the interesting can measured by $\int_{-1}^1 (f(z)-1/2)^2 dx$ with $f(z)$ a density estimated from the data. If the density $f(z)$ is equal to $1/2$ in the interval $[-1;1]$ then the integral becomes zero and we have found that our projected data are normally distributed. An value larger than zero indicates a deviation from the normal distribution of the projected data and hopefully an interesting distribution.
### Expansion with orthonormal polynomials
Let $L_i(z)$ a set of orthonormal polynomials with the scalar product $=\int_{-1}^1 f(z)g(z) dz$ and the norm $\|f\| = \sqrt{}$. What can we derive about a densities $f(z)$ in the interval $[-1;1]$ ?
If $f(z)=\sum_{i=0}^I a_i L_i(z)$ for some maximal degree $I$ then it holds
$\int_{-1}^1 f(z) L_j(z) dz = \int_{-1}^1 \sum_{i=0}^I a_i L_i(z) L_j(z) dz = a_j \int_{-1}^1 L_j(z) L_j(z) dz = a_j$
We can also write $\int_{-1}^1 f(z) L_j(z) dz = E(L_j(z))$ or empirically we get an estimator $\hat{a}_j = \frac{1}{n} \sum_{k=1}^n L_j(z_k)$.
We describe the term $1/2 = \sum_{i=1}^I b_i L_i(z)$ and get for our integral
$\int_{-1}^1 (f(z)-1/2)^2 dz = \int_{-1}^1 \left(\sum_{i=0}^I (a_i-b_i) L_i(z)\right)^2 dz = \sum_{i,j=0}^I \int_{-1}^1 (a_i-b_i)(a_j-b_j) L_i(z) L_j(z) dz = \sum_{i=0}^I (a_i-b_i)^2.$
So using a orthonormal function set allows us to reduce the integral to a summation of coefficient which can be estimated from the data by plugging $\hat{a}_j$ in the formula above. The coefficients $b_i$ can be precomputed in advance.
### Normalized Legendre polynomials
The only problem left is to find the set of orthonormal polynomials $L_i(z)$ upto degree $I$. We know that $1, x, x^2, ..., x^I$ form a basis for this space. We have to apply the Gram-Schmidt orthogonalization to find the orthonormal polynomials. This has been started in the first example.
The resulting polynomials are called normalized Legendre polynomials. Up to a sacling factor the normalized Legendre polynomials are identical to Legendre polynomials. The Legendre polynomials have a recursive expression of the form
$L_i(z) = \frac{(2i-1) L_{i-1}(z) - (i-1) L_{i-2}(z)}{i}$
So computing our integral reduces to computing $L_0(z_k)$ and $L_1(z_k)$ and using the recursive relationship to compute the $\hat{a}_j$'s. Please note that the recursion can be numerically unstable! | 5,691 | 15,832 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 251, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2015-22 | latest | en | 0.844186 |
https://www.lmfdb.org/EllipticCurve/Q/115920dv2/ | 1,643,410,067,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00638.warc.gz | 885,758,110 | 25,882 | # Properties
Label 115920dv2 Conductor $115920$ Discriminant $1.161\times 10^{27}$ j-invariant $$\frac{16077778198622525072705635801}{388799208512064000000}$$ CM no Rank $1$ Torsion structure $$\Z/{2}\Z \times \Z/{2}\Z$$
# Related objects
Show commands: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([0, 0, 0, -7571755803, -253591059413302])
gp: E = ellinit([0, 0, 0, -7571755803, -253591059413302])
magma: E := EllipticCurve([0, 0, 0, -7571755803, -253591059413302]);
$$y^2=x^3-7571755803x-253591059413302$$
## Mordell-Weil group structure
$\Z\times \Z/{2}\Z \times \Z/{2}\Z$
### Infinite order Mordell-Weil generator and height
sage: E.gens()
magma: Generators(E);
$P$ = $$\left(-\frac{1264049044808593}{25067438929}, \frac{7614016989860802598}{3968852403311783}\right)$$ $\hat{h}(P)$ ≈ $30.717914854766757803637767854$
## Torsion generators
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
$$\left(-50051, 0\right)$$, $$\left(100477, 0\right)$$
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
$$\left(-50426, 0\right)$$, $$\left(-50051, 0\right)$$, $$\left(100477, 0\right)$$
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$115920$$ = $2^{4} \cdot 3^{2} \cdot 5 \cdot 7 \cdot 23$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $1160948215829686910976000000$ = $2^{24} \cdot 3^{20} \cdot 5^{6} \cdot 7^{4} \cdot 23^{2}$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$\frac{16077778198622525072705635801}{388799208512064000000}$$ = $2^{-12} \cdot 3^{-14} \cdot 5^{-6} \cdot 7^{-4} \cdot 23^{-2} \cdot 631^{3} \cdot 3999871^{3}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $4.3036087471460685572436676520\dots$ Stable Faltings height: $3.0611554222520684021288129121\dots$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $1$ sage: E.regulator() magma: Regulator(E); Regulator: $30.717914854766757803637767854\dots$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $0.016184565145139906683080677129\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $256$ = $2^{2}\cdot2^{2}\cdot2\cdot2^{2}\cdot2$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $4$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L'(E,1)$ ≈ $7.9544975054373351525103623810513579947$
## Modular invariants
Modular form 115920.2.a.ch
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q - q^{5} + q^{7} + 4 q^{11} + 6 q^{13} - 2 q^{17} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 123863040 $\Gamma_0(N)$-optimal: no Manin constant: 1
## Local data
This elliptic curve is not semistable. There are 5 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $4$ $I_{16}^{*}$ Additive -1 4 24 12
$3$ $4$ $I_{14}^{*}$ Additive -1 2 20 14
$5$ $2$ $I_{6}$ Non-split multiplicative 1 1 6 6
$7$ $4$ $I_{4}$ Split multiplicative -1 1 4 4
$23$ $2$ $I_{2}$ Split multiplicative -1 1 2 2
## Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2Cs 2.6.0.1
## $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
$p$-adic regulators are not yet computed for curves that are not $\Gamma_0$-optimal.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 add add nonsplit split ordinary ordinary ordinary ss split ordinary ss ordinary ordinary ordinary ordinary - - 5 6 1 1 1 1,1 2 1 1,1 1 1 1 1 - - 0 0 0 0 0 0,0 0 0 0,0 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2.
Its isogeny class 115920dv consists of 4 curves linked by isogenies of degrees dividing 4.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z \times \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $4$ $$\Q(\sqrt{-15}, \sqrt{-23})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{3}, \sqrt{23})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{-3}, \sqrt{-5})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $8$ Deg 8 $$\Z/2\Z \times \Z/6\Z$$ Not in database $16$ 16.0.13153238385373209600000000.1 $$\Z/4\Z \times \Z/4\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 2,262 | 5,892 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-05 | latest | en | 0.265706 |
https://infinitylearn.com/surge/question/mathematics/state-whether-the-following-statement-is-true-or-falseif-ab/ | 1,716,459,847,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058614.40/warc/CC-MAIN-20240523080929-20240523110929-00854.warc.gz | 278,908,020 | 31,258 | State whether the following statement is true or false.If ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure), then ∆ABE ≅ ∆ACF.
# State whether the following statement is true or false.If ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure), then ∆ABE ≅ ∆ACF.
1. A
True
2. B
False
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### Solution:
Given that ∆ABC is a triangle, where BE & CF are altitudes to the respective sides AC and AB such that BE=CF.
Consider
Thus, by AAS congruence criteria,
Hence, by AAS congruence criteria, we proved thatTherefore, the given statement is true.
So, option (1) is correct.
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Verify OTP Code (required) | 246 | 967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.818462 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=VRTX | 1,490,314,398,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187227.84/warc/CC-MAIN-20170322212947-00380-ip-10-233-31-227.ec2.internal.warc.gz | 542,519,495 | 111,203 | Back to list of Stocks See Also: Seasonal Analysis of VRTXGenetic Algorithms Stock Portfolio Generator, and Fourier Calculator
# Fourier Analysis of VRTX (Vertex Pharmaceuticals Incorpor)
VRTX (Vertex Pharmaceuticals Incorpor) appears to have interesting cyclic behaviour every 96 weeks (3.7095*cosine), 84 weeks (3.6802*sine), and 134 weeks (2.7566*sine).
VRTX (Vertex Pharmaceuticals Incorpor) has an average price of 34.44 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
Right click on the graph above to see the menu of operations (download, full screen, etc.)
## Fourier Analysis
Using data from 7/24/1991 to 3/13/2017 for VRTX (Vertex Pharmaceuticals Incorpor), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
034.44278 0
114.4027 -19.00926 (1*2π)/13381,338 weeks
24.26197 -22.46857 (2*2π)/1338669 weeks
36.00474 -14.63279 (3*2π)/1338446 weeks
4-7.28212 -10.71228 (4*2π)/1338335 weeks
51.3864 -11.69936 (5*2π)/1338268 weeks
6-5.01868 -.88576 (6*2π)/1338223 weeks
7-5.04737 -3.32142 (7*2π)/1338191 weeks
8.12196 -5.3052 (8*2π)/1338167 weeks
9-1.48784 2.34754 (9*2π)/1338149 weeks
10-1.96527 -2.75657 (10*2π)/1338134 weeks
111.61693 -.04522 (11*2π)/1338122 weeks
12.12571 1.58968 (12*2π)/1338112 weeks
131.06323 -2.28844 (13*2π)/1338103 weeks
143.70951 -1.44435 (14*2π)/133896 weeks
15-.74426 -1.97729 (15*2π)/133889 weeks
16.23588 -3.68021 (16*2π)/133884 weeks
17.79334 -1.40883 (17*2π)/133879 weeks
18-1.27442 -1.7525 (18*2π)/133874 weeks
19-.40692 -2.82254 (19*2π)/133870 weeks
20-.4524 .78381 (20*2π)/133867 weeks
21.44529 -.82451 (21*2π)/133864 weeks
221.53385 -2.65133 (22*2π)/133861 weeks
23-.10577 -1.80286 (23*2π)/133858 weeks
24-2.11734 -1.80603 (24*2π)/133856 weeks
25-.9866 -.70752 (25*2π)/133854 weeks
26-1.06932 .69542 (26*2π)/133851 weeks
27.38172 -1.30008 (27*2π)/133850 weeks
28.37094 -.98977 (28*2π)/133848 weeks
29-1.51879 .04703 (29*2π)/133846 weeks
30-.42129 -1.38204 (30*2π)/133845 weeks
31-.18318 .45013 (31*2π)/133843 weeks
32.37212 .76255 (32*2π)/133842 weeks
33.80652 -1.47228 (33*2π)/133841 weeks
34.36476 .32551 (34*2π)/133839 weeks
35.15262 -1.05002 (35*2π)/133838 weeks
36.07923 -1.47729 (36*2π)/133837 weeks
37-.43178 .44306 (37*2π)/133836 weeks
38.13296 -1.0927 (38*2π)/133835 weeks
39.82516 .44078 (39*2π)/133834 weeks
40.26597 -.30251 (40*2π)/133833 weeks
411.10486 -1.39842 (41*2π)/133833 weeks
42.6019 -.81659 (42*2π)/133832 weeks
43.0232 -.94486 (43*2π)/133831 weeks
44.22516 -1.13076 (44*2π)/133830 weeks
45-.43862 -.86825 (45*2π)/133830 weeks
46-.06018 -.32376 (46*2π)/133829 weeks
47.32268 -.68492 (47*2π)/133828 weeks
48.09033 -.94958 (48*2π)/133828 weeks
49-.23079 -.75939 (49*2π)/133827 weeks
50-.23937 -.9337 (50*2π)/133827 weeks
51.14268 -.89724 (51*2π)/133826 weeks
52-.46841 -.34753 (52*2π)/133826 weeks
53-.57308 -.70172 (53*2π)/133825 weeks
54-.05058 -.28748 (54*2π)/133825 weeks
55-.29235 .04777 (55*2π)/133824 weeks
56.13581 -.40291 (56*2π)/133824 weeks
57.64302 -.20348 (57*2π)/133823 weeks
58-.19341 -.38442 (58*2π)/133823 weeks
59.43932 -.75752 (59*2π)/133823 weeks
60.48316 -.58799 (60*2π)/133822 weeks
61-.60906 -.77278 (61*2π)/133822 weeks
62-.09434 -.71505 (62*2π)/133822 weeks
63-.20427 -.24657 (63*2π)/133821 weeks
64.01101 -.39491 (64*2π)/133821 weeks
65.57662 -.4862 (65*2π)/133821 weeks
66-.1893 -.60628 (66*2π)/133820 weeks
67.34732 -.95478 (67*2π)/133820 weeks
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517-.00403 -.05362 (517*2π)/13383 weeks
518-.14007 -.07151 (518*2π)/13383 weeks
519-.12378 -.04485 (519*2π)/13383 weeks
520-.11073 .16267 (520*2π)/13383 weeks
521.01275 .04752 (521*2π)/13383 weeks
522.079 -.0015 (522*2π)/13383 weeks
523-.00995 -.04235 (523*2π)/13383 weeks
524-.023 -.19155 (524*2π)/13383 weeks
525-.05751 -.05928 (525*2π)/13383 weeks
526-.11905 -.04992 (526*2π)/13383 weeks
527-.15139 -.07834 (527*2π)/13383 weeks
528-.11782 -.03124 (528*2π)/13383 weeks
529-.14546 -.0095 (529*2π)/13383 weeks
530-.0779 .01489 (530*2π)/13383 weeks
531-.0851 .07499 (531*2π)/13383 weeks
532-.14895 .07377 (532*2π)/13383 weeks
533.01724 .03492 (533*2π)/13383 weeks
534.03441 .07744 (534*2π)/13383 weeks
535.03021 .02668 (535*2π)/13383 weeks
536.02817 -.12501 (536*2π)/13382 weeks
537-.0188 -.03661 (537*2π)/13382 weeks
538-.09489 -.10989 (538*2π)/13382 weeks
539-.05925 -.01995 (539*2π)/13382 weeks
540-.05447 .0455 (540*2π)/13382 weeks
541-.10487 -.06718 (541*2π)/13382 weeks
542.06088 .00266 (542*2π)/13382 weeks
543-.03909 -.05042 (543*2π)/13382 weeks
544-.12155 -.14215 (544*2π)/13382 weeks
545-.07331 -.02709 (545*2π)/13382 weeks
546-.21475 .06192 (546*2π)/13382 weeks
547-.06945 -.06913 (547*2π)/13382 weeks
548-.04198 .07104 (548*2π)/13382 weeks
549-.05554 .06961 (549*2π)/13382 weeks
550-.01589 -.06949 (550*2π)/13382 weeks
551-.07054 .06698 (551*2π)/13382 weeks
552.00585 -.01808 (552*2π)/13382 weeks
553.02569 -.02874 (553*2π)/13382 weeks
554-.00288 -.04144 (554*2π)/13382 weeks
555-.10528 -.0855 (555*2π)/13382 weeks
556-.03089 -.03599 (556*2π)/13382 weeks
557-.10661 -.03185 (557*2π)/13382 weeks
558-.05558 .02913 (558*2π)/13382 weeks
559-.04436 -.0169 (559*2π)/13382 weeks
560-.04939 .00599 (560*2π)/13382 weeks
561-.02218 .07167 (561*2π)/13382 weeks
562.00981 -.05478 (562*2π)/13382 weeks
563.05739 -.01915 (563*2π)/13382 weeks
564-.01187 -.07454 (564*2π)/13382 weeks
565-.03326 -.1679 (565*2π)/13382 weeks
566-.10553 -.09595 (566*2π)/13382 weeks
567-.1611 .05268 (567*2π)/13382 weeks
568-.04974 -.03189 (568*2π)/13382 weeks
569.0226 -.03192 (569*2π)/13382 weeks
570-.1091 -.04118 (570*2π)/13382 weeks
571-.10792 -.05386 (571*2π)/13382 weeks
572-.04427 .03112 (572*2π)/13382 weeks
573-.10456 -.02433 (573*2π)/13382 weeks
574.03955 -.05944 (574*2π)/13382 weeks
575-.02856 -.0176 (575*2π)/13382 weeks
576-.14042 -.12674 (576*2π)/13382 weeks
577.00052 -.05649 (577*2π)/13382 weeks
578-.19016 -.01263 (578*2π)/13382 weeks
579-.04839 -.08053 (579*2π)/13382 weeks
580-.07178 .02656 (580*2π)/13382 weeks
581-.16639 -.12224 (581*2π)/13382 weeks
582-.10537 -.08902 (582*2π)/13382 weeks
583-.16721 .10675 (583*2π)/13382 weeks
584-.13156 -.05605 (584*2π)/13382 weeks
585-.10917 .05126 (585*2π)/13382 weeks
586-.118 .11418 (586*2π)/13382 weeks
587-.05921 -.05839 (587*2π)/13382 weeks
588.02016 .12823 (588*2π)/13382 weeks
589-.01337 .01617 (589*2π)/13382 weeks
590.06707 -.11707 (590*2π)/13382 weeks
591-.00938 -.00238 (591*2π)/13382 weeks
592-.13608 -.09839 (592*2π)/13382 weeks
593-.03992 -.15714 (593*2π)/13382 weeks
594-.12518 .02161 (594*2π)/13382 weeks
595-.1952 -.11343 (595*2π)/13382 weeks
596-.07872 -.07201 (596*2π)/13382 weeks
597-.27322 .11867 (597*2π)/13382 weeks
598-.15409 -.1016 (598*2π)/13382 weeks
599-.00358 .10607 (599*2π)/13382 weeks
600-.17663 .09082 (600*2π)/13382 weeks
601-.08578 -.07879 (601*2π)/13382 weeks
602-.08398 .17049 (602*2π)/13382 weeks
603-.19742 .02616 (603*2π)/13382 weeks
604.01751 .03278 | 10,620 | 24,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-13 | longest | en | 0.780886 |
https://www.quizover.com/course/section/educator-section-to-determine-the-equivalence-and-validity-by-openstax | 1,547,802,378,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660020.5/warc/CC-MAIN-20190118090507-20190118112507-00636.warc.gz | 923,194,474 | 19,763 | # 1.23 To determine the equivalence and validity of different
Page 1 / 1
## Memorandum
2. Learner’s own method
3.1 9 080 717
3.2 8 301 883
• 6 485 185
1. 4 2. 1
3 2
1 3
2 3
1 44
8 1
## To use strategies to check solutions [lo 1.11]
1. In the previous activity you had the opportunity to use your own strategies to solve problems. From the feedback in the class you would have realised that there are many ways of adding numbers.
Work in pairs. Discuss the following methods of adding and explain to each other how the answers are calculated:
## Let’s revise!
1.1 845 908 + 25 876 + 343 621
800 000 + 40 000 + 5 000 + 900 + 0 + 8 20 000 + 5 000 + 800 + 70 + 6 + 300 000 + 40 000 + 3 000 + 600 + 20 + 1 1 100 000 + 100 000 + 13 000 + 2 300 + 90 + 15
= 1 000 000 + (100 000 + 100 000) + 10 000 + (3 000 + 5 000) + 300 + (90 + 10) + 5
= 1 215 405
1.2
= 1 215 405
1.3
1 1 1 2 1 8 4 5 9 0 8 2 5 8 7 6 + 3 4 3 6 2 1 1 2 1 5 4 0 5
2. Can you show your friend any other method of calculation?
_____________________________________________________________________
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_____________________________________________________________________
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3. Use any method of your choice (without a calculator) and calculate the sum of:
3.1 4 623 577 + 1 239 246 + 3 217 894
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
3.2 2 851 416 + 4 981 235 + 469 232
_____________________________________________________________________
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_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
3.3 2 153 892 + 264 058 + 4 067 235
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
_____________________________________________________________________
## Brain teasers!
Complete the missing numbers:
1.
1. 2 1 3 4 1 ___ 2. ___ 7 9 1 2 3 8 9 2 ___ 7 4 ___ 2 8 1 6 1 7 5 ___ 9 2 1 1 ___ 2 3 3 ___ 1 4 1 1 9 1 0 0 ___ 7 1 + 1 ___ 9 1 2 3 + 1 4 3 0 ___ ___ ___ 1 ___ 0 8 5 9 5 8 5 8 7
## Time for self-assessment
• Tick the appropriate rectangle:
Yes Mostly Some-times No I can investigate adding methods and evaluate them (determine whether they are correct). (LO 1.11 en LO 2.6) * I can add correctly without a calculator. (LU 1.8) * I need more exercise in addition. *
* Ask your educator to explain again and give you extra exercises.
## Assessment
Learning Outcome 1: The learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems.
Assessment Standard 1.11: We know this when the learner uses a range of strategies to check solutions and judge the reasonableness of solutions.
Learning Outcome 2: The learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems using algebraic language and skills.
Assessment Standard 2.6: We know this when the learner determines, through discussion and comparison, the equivalence of different descriptions of the same relationship or rule presented:
2.6.3: e.g. number sentences.
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
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
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
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
Got questions? Join the online conversation and get instant answers! | 1,631 | 6,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-04 | latest | en | 0.451479 |
https://electronics.stackexchange.com/questions/347491/calculating-base-resistor-value/347494 | 1,627,956,281,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154408.7/warc/CC-MAIN-20210802234539-20210803024539-00708.warc.gz | 223,809,313 | 41,580 | # Calculating base resistor value
I've looked around on the internet, but there's so much conflicting advice on how to find the value for the necessary resistor between the base of the transistor and the microcontroller. I'm completely at a loss at this point.
The microcontroller I'm using is a the huzzah ESP8266 breakout board from adafruit, and the transistors are TIP41C's.
Here's a schematic:
• Could you point me to some of these conflicting information? – pipe Dec 31 '17 at 12:33
• What is the maximum current each transistor must be able to conduct? What is the logic high voltage coming out of the digital part? How much current can each digital output source? Is 12 V really the maximum the transistors must switch? Especially if this LED strip current is reasonably high, why not the more obvious solution of using N channel MOSFETs instead of NPN transistors? – Olin Lathrop Dec 31 '17 at 13:15
First you have to determine what the maximum collector current is that the transistor needs to support when on.
Divide this collector current by the gain of the transistor to get the minimum required base current. The gain of a BJT varies with operating point, so you have to look at the datasheet carefully to see what gain value you can assume at your maximum collector current.
Once you know the minimum required base current, it's just a matter of Ohm's law to find the base resistance. The B-E junction of the transistor looks like a diode to the driving circuit. Figure it will drop around 700 mV when the base current is flowing. Subtract that from the logic high level to find the voltage across the base resistor. By Ohm's law, the resistance is the voltage divided by the current.
Remember that this calculation was based on the minimum required base current. That means the result is the maximum allowed base resistance. It's usually good to round down the resistance one or two standard values to leave some margin.
Now that you've picked a base resistance, work backwards to find the actual base current, and check to make sure the digital output can deliver that. If not, that transistor in that circuit won't work.
That all said, why not use a FET? It seems you only need to switch 12 V. It looks like the IRLML2502, for example, would work nicely here. Then you don't need a base resistor at all. The on-state voltage drop will also be lower.
• you still need a gate resistor though – Vinzent Dec 31 '17 at 13:17
• @Vin: No, you don't. In high speed switching applications a few Ohms of gate resistance is sometimes required. Any normal digital output simply doesn't have the drive capability to cause the problem a gate resistor would solve. – Olin Lathrop Dec 31 '17 at 13:20
• @Cas: First, a FET is a transistor. Second, read the datasheet instead of guessing. The IRLML2502 is specified for 80 mOhms Rdson at 2.5 V gate drive. It does better (45 mOhms) with 4.5 V gate drive, but 80 mOhms is good enough for many applications. At 1 A, it would only drop 80 mV and dissipate 80 mW. That's a lot better than any BJT can do. – Olin Lathrop Dec 31 '17 at 13:27
• @Cas: A FET (field effect transistor) is a transistor, but not all transistors are FETs. – Olin Lathrop Dec 31 '17 at 13:29
• At your low voltage of 12 V, there are a number of MOSFETs that can switch nicely driven from 0 to 3.3 V logic. Just because you selected a inappropriate FET doesn't mean all FETs are inappropriate. If the LED current is higher and you can tolerate 150 uA leakage, then the IRLML6344 is a even better choice. It is guaranteed not to exceed 37 mOhms with 2.5 V gate drive. – Olin Lathrop Dec 31 '17 at 13:33
An NPN transistor is a current amplifier, it allowes a current to flow between collector and emitter that is hFe times greater than the current flowing from base to emitter. this allowes you to controll the current through the collector by controlling the current that flows through the base.
If you want to use an NPN transistor as a current amplifier the base current Ib you need is the collecter current Ic divided by the hFe of the transistor found in the datasheet so
$$I_b=\frac{I_c}{h_{Fe}}$$
If the transistor is used as a current amplifier it will not be able to pull the voltage completely to ground, in order to do that you have to be in the regon of operation of the transistor where you are allowing enough current though relatively, to allow the transistor to switch completely ON. the way you do that is that you raise the base current by a factor of 5, so now the equation for using a bjt transistor as a switch becomes
$$I_b=\frac{5I_c}{h_{Fe}}$$
Now you know the base current needed if you know the current you need to switch on though the collector-emitter path.
now you can calculate the resistor needed if you know the supply voltage it is
$$R_b = \frac{V_{cc}-V_d}{I_b}$$
where Vcc is the supply voltage and Vd is the diode voltage of the base diode typically $V_d=0.66V$.
In conclusion
$$R_b = \frac{h_{Fe}(V_{cc}-V_d)}{5I_c}$$
The scaling by a factor of 5 is just a rule of thumb but you can always analyze from the datasheet how much current you need though the base to switch it on.
• Would this work? (12-0.6)*30/5*0.3=228 228 Ohms? – Casper Dec 31 '17 at 12:47
• If those are your values, eg. 12v, 30gg, 300mA.. well then yearh it would work. but are you really supplying the mcu with 12v? and 228ohm resistor value doesn't sound far off when you considder that you are switching 300mA with a transistor that has a hFe of only 30 which is very low, something else i would wory about is 12v/228ohm=53mA thats quite a large current to be putting out from an mcu output – Vinzent Dec 31 '17 at 12:59
• Unfortunately, your answer is working backwards to an Ib that may not be available, as it isn't here. Would be good to rework your answer to first calculate the required hFE from the required Ic and available Ib, then look up hFE(min) from the datasheet, then show the maximum available Ic if hFE(min) isn't up to it. (If I wasn't on a phone keyboard, I would have posted such an answer :-) ) – TonyM Dec 31 '17 at 13:14
• The question wasn't about how to choose the right transistor the question was how to find the base resistor value, to select a base resistor you will allways need to know the base current. I just showed the general idea/equation – Vinzent Dec 31 '17 at 13:22
• Sorry, Vcc is 3.3V is then. I mistook the Vcc as 12V (for the LED strip) going across the transistor. – Casper Dec 31 '17 at 13:59
I'll risk going off topic - change your NPNs, use a darlington pair or use N-MOSes.
TIP41C has a minimum gain of 15. Low(ish) LED strip are 4.8W per meter - that's 400mA. On the other hand ESP8266 has a maximum GPIO current of 12mA. If you do the calculations 12mA/(400mA/15/3) you end up with a maximum length od 1.35 meter.
I took extreme values - IRL it should work with roughly 2 to 2.5 meters (7 to 8 feet) of a LED strip but that's all.
For your actual question - you already have two good answers, so I'm not gonna repeat them. | 1,839 | 7,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-31 | latest | en | 0.931765 |
https://princeharrymemorial.com/what-is-the-jonswap-spectrum/ | 1,701,990,430,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100705.19/warc/CC-MAIN-20231207221604-20231208011604-00466.warc.gz | 509,002,047 | 12,201 | ## What is the Jonswap spectrum?
The JONSWAP spectrum is similar to the Pierson-Moskowitz spectrum except that waves continues to grow with distance (or time) as specified by the a term, and the peak in the spectrum is more pronounced, as specified by the g term.
### What is wave energy spectrum?
Wave spectra are used by oceanographers to describe the distribution of energy at different frequencies. The form of the spectrum can be related to wind speed and direction and the duration of the storm and the fetch (or distance upwind) over which it has blown,…
What are spectral moments?
Spectral moments (mean, variance, skewness, kurtosis) have been used to classify fricative place of articulation in English and other languages. Logistic regression is used to test the independence of fricative and vowel cues in syllable perception, and to model developmental trends.
What is water shoaling?
Definition of Shoaling: Shoaling is the deformation of incident waves on the lower shoreface that starts when the water depth becomes less than about half of the wavelength, causing the waves to become steeper: increase in amplitude and decrease in wavelength.
## What is 1/3 of the wave spectrum?
1. The significant wave height, H_s, is the mean of the highest third of the waves; instead of H_s the notation H_{1/3} is also often used. H_s represents well the average height of the highest waves in a wave group.
### What is peak wave period?
The peak wave period (in seconds) is defined as the wave period associated with the most energetic waves in the total wave spectrum at a specific point. Peak wave periods are contoured by isolating different swell events with a single line—not contouring at a specific interval as in other forecast graphics.
What is a speed wave?
Summary. Wave speed is the distance a wave travels in a given amount of time, such as the number of meters it travels per second. Wave speed is related to wavelength and wave frequency by the equation: Speed = Wavelength x Frequency.
What are the parameters of the wave spectrum?
The wave spectrum can be summarized to a reasonable accuracy using a small number of parameters. The most important are a measure of average wave height and period, followed by descriptors of directional properties. Wave height and period parameters are defined in terms of moments of the omnidirectional spectrum.
## Why is the Ochi Hubble spectrum called a two peaked spectrum?
This is why the Ochi-Hubble spectrum is often called a two-peaked spectrum; however in practice, the resulting total spectrum typically has only one peak (from the remotely-generated swell) plus a shoulder of energy from the local wind-generated sea.
### Why is there a scatter of wave spectra?
This ‘scatter’ is not due to inaccuracies of observation but to the fact that the sea states observed are at various stages of development and includes swell as well as sea components. The data can be combined in many ways. They can, for example, be averaged over the North Atlantic or world wide. | 625 | 3,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | longest | en | 0.942812 |
http://what-when-how.com/category/3d-animation-using-maya/ | 1,488,267,355,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174154.34/warc/CC-MAIN-20170219104614-00527-ip-10-171-10-108.ec2.internal.warc.gz | 269,978,141 | 4,728 | # 3D Animation Using Maya
## Introduction to 3D Space — Cartesian Coordinate System (Essential Skills)(3D Animation Using Maya)
3D animation in Maya is created by bringing together many parts to form a greater whole. Understanding how these parts relate to each other is one of the keys to learning Maya. Instead of completely focusing on one part at a time, we’ll always keep the bigger picture in mind. So what are these different […]
## Workspace — Getting Around Maya (Essential Skills) (3D Animation Using Maya) Part 1
The Maya workspace presents the tools and user interface ( UT ) elements necessary to create all aspects of your animation (see Figure 1.16). In order to avoid being overwhelmed, some task-specific elements are only there when you need them, and others are ever-present. Let’s look at some of the most commonly used elements. Figure […]
## Workspace — Getting Around Maya (Essential Skills) (3D Animation Using Maya) Part 2
Rotate Tool Another common manipulation is rotation, which is accomplished with the Rotate Tool (see Figures 1.27 and 1.28). To use the rotate tool, select and LMB-drag to increase or decrease the rotational angle. The manipulator handles allow you to rotate on any of the individual axes by using the inner circles, or around the […]
## Modeling — Primitives, Wireframes, Surfaces, and Normals (Essential Skills) (3D Animation Using Maya) Part 1
A geometric primitive is a basic building block of 3D CO geometry, such as a sphere, cube, cylinder, cone, or torus. In the following examples we’ll look at a primitive sphere. In Figure 1.38 you see the wireframe representation of a sphere. The same sphere can be represented in a faceted shaded mode, as in […]
## Modeling — Primitives, Wireframes, Surfaces, and Normals (Essential Skills) (3D Animation Using Maya) Part 2
Modeling Exercise — Barn In this modeling exercise we’ll use the polygon and NURBS primitives as building blocks for a barn and silo. Numeric precision is a bit more important here than in the last exercise, because of the symmetrical construction of the building. We’ll use the Channel Box to enter in exact values, and […]
## Animation — Keyframe Basics (Essential Skills) (3D Animation Using Maya)
This can be as simple as the sky’s color turning from day to night, or as complex as a three-headed dog chasing a busload of screaming babies. The common thread is that over a period of time, things change. In Maya, animation is generally created by specifying the key moments (or most important poses) when […]
## Shading — Color, Specular, and Ambient (Essential Skills) (3D Animation Using Maya)
Shading defines the look of your model when it is rendered. We’ll begin with a com-1 ~V1 mon shading style, called Phong shading, and look at three of its attributes (see Fig ure 1.82): color, specular color, and ambient color. Later we will add textures, texture maps, transparency, glows, and other effects. Figure 1.82: Basic […]
## Lighting — Point Lights: Intensity, Color, and Shadows (Essential Skills) (3D Animation Using Maya)
Lighting in Maya is much like lighting for theater or live action film. To illuminate a VV scene, you must place lights in space, brighten or dim them, color them, and set ^ countless other of their attributes. You can think of lighting in realistic terms: How high should this light be to cast a […]
## Rendering — Outputting a Scene to Frames (Essential Skills) (3D Animation Using Maya)
Your final step in creating a 3D animation will be to produce a sequence of production-quality frames that can be output to video, film, or a compressed video format such as an MPEG-2 QuickTime movie. When you instruct Maya to render your animation, it takes all of the data in the camera’s view and calculates […]
## Project 1 — Ball Rolling Down an Inclined Plane (Essential Skills) (3D Animation Using Maya)
For your first project, create and render a three-second animation of a ball rolling down a tilted board. Be sure to use all the skills you’ve learned so far in the areas of modeling, animation, shading, lighting, and rendering. Include: • Geometric primitives, • Keyframed animation, • Colored Phong shaders, and • Point lights with […] | 951 | 4,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-09 | longest | en | 0.869381 |
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Question:
Let $\mathrm{R}$ be the set of real numbers.
Statement-1: $A=\{(x, y) \in R \quad R: y-x$ is an integer $\}$ is an equivalence relation on $R$.
Statement-2: $B=\{(x, y) \in R \quad R: x=\alpha y$ for some rational number $\alpha\}$ is an equivalence relation on $\mathrm{R}$.
1. Statement- 1 is true, Statement- 2 is false.
2. Statement- 1 is false, Statement- 2 is true
3. Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement- 1
4. Statement- 1 is true, Statement- 2 is true; Statement- 2 is not a correct explanation for Statement- 1
Correct Option: 1
Solution: | 201 | 647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-33 | latest | en | 0.74889 |
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# Punjab University
## Journal of Mathematics (ISSN 1016-2526)
Vol. 40 (2008) pp. 37-48
## Stability Analysis of Predator-Prey Population Model with Time Delay and
Constant Rate of Harvesting
Syamsuddin Toaha
Department of Mathematics
Hasanuddin University
90245, Makassar, Indonesia
E-mail: syamsuddint@yahoo.com
Malik Abu Hassan
University Putra Malaysia
43400, UPM, Serdang Selangor Darul Ehsan, Malaysia
E-mail: malik@fsas.upm.edu.my
Abstract. This paper studies the effect of time delay and harvesting on
the dynamics of the predator - prey model with a time delay in the growth
rate of the prey equation. The predator and prey are then harvested with
constant rates. The constant rates may drive the model to one, two, or
none positive equilibrium points. When there exist two positive equilib-
rium points, one of them is possibly stable. In the case of the constant
rates are quite small and the equilibrium point is not stable, an asymptot-
ically stable limit cycle occurs. The result showed that the time delay can
induce instability of the stable equilibrium point, Hopf bifurcation and
stability switches.
Key Words: Predator-prey, Limit cycle, Time delay, Harvesting rate, Hopf bifurcation.
1. I NTRODUCTION
The Lotka-Volterra model is one of the earliest predator-prey models to be based on
sound mathematical principles. It forms the basis of many models used today in the anal-
ysis of population dynamics and is one of the most popular models in mathematical ecol-
ogy. In both the analysis and experiment, the predator and prey can coexist by reducing
the frequency of contact between them, Luckinbill [13]. In the context of predator-prey
interaction, some studies that treat population can be extended by considering harvesting,
stocking, diffusion, and time delay. In the model with harvesting, some studies relate the
population to the economic problems. The time delay is considered into the population
dynamics when the rate of change of the population is not only a function of the present
population but also depends on the past population.
One predator-one prey system in Hogart et al. [10] where both the predator and prey are
harvested with constant yield has been considered and the stability at maximum sustain-
able yield is established. Martin and Ruan [14] have analyzed generalized Gause predator-
prey models where the prey is harvested with constant rate while Kar [12] considered the
37
38 Syamsuddin Toaha and Malik Abu Hassan
predator-prey model with the predator harvested and suggested that it is ideal to study the
combined harvesting of predator and prey population models. The effect of constant rate
of harvesting has been studied by Holmberg [11] and the results showed that the constant
catch quota can lead to both oscillations and chaos and an increased risk for over exploita-
tion. While the effects on population size and yield of different levels of harvesting of a
predator in a predator-prey system have been explored by Matsuda and Abrams [15] and
showed that the predator may increase in population size with increasing shing effort.
Brauer and Soudack [3] have analyzed the global behavior of a predator-prey system
under constant rate predator harvesting. They showed how to classify the possibilities and
determine the region of stability. They found that if the equilibrium point is asymptotically
stable, which is determined by a local linearization, then every solution whose initial value
is in some neighborhood of the stable equilibrium point tends to it as the time approaches
innity. There exists an asymptotically stable limit cycle when the constant rate is small
and the equilibrium point is unstable. A predator-prey model with Holling type using
harvesting efforts as control has been presented by Srinivasu et al. [17] and showed that
with harvesting, it is possible to break the cyclic behavior of the system and introduces a
globally stable limit cycle in the system.
The effect of constant rate of harvesting on the dynamics of predator-prey systems has
been investigated by many authors, see, for example, Brauer and Soudack [2, 4], Myer-
scough et al. [16], Dai and Tang [7], Xiao and Ruan [18]. Some interesting dynamical
behaviors have been observed such as the stability of the equilibria, existence of Hopf bi-
furcation and limit cycles. It is also observed that in some cases, before a catastrophic
harvest rate is reached the effect of harvesting is to stabilize the equilibrium point of the
population system. In this paper we present a deterministic and continuous model for
predator - prey population based on Lotka - Volterra model which is extended by incor-
porating time delay and constant rates of harvesting of both populations. The objective of
this paper is to study the combined effects of harvesting and time delay on the dynamics
of predator-prey model.
## 2. T HE P REDATOR - P REY P OPULATION M ODEL
We consider a predator - prey model based on Lotka - Volterra model with one predator
and one prey populations. The model for the rate of change of prey population (x) and
predator population (y) is
dx x
= rx(1 ) xy
dt K
dy
= cy + xy. (2.1)
dt
The model includes parameter K, the carrying capacity, for the prey population in the
absence of the predator. The parameter r is the intrinsic growth rate of the prey, c is the
mortality rate if the predator without prey, measures the rate of consumption of prey by
the predator, measures the conversion of prey consumed into the predator reproduction
rate. All the parameters are assumed to be positive.
The equilibrium points of model (2.1) are (0, 0), (K, 0) and E = (x , y ) = ( cb , r(Kc)
K ).
In order to get a positive equilibrium point we assume that K c > 0. The Jacobian
matrix of model (2.1) takes the form
r 2rx
K y x
J= .
y c + bx
Stability Analysis of Predator-Prey Population Model .... 39
## The characteristic equation of the Jacobian matrix J at the equilibrium point E is
cr cr
f () = 2 + K + K (K c) and the eigenvalues have negative real parts. It means that
the equilibrium points E is locally asymptotically stable. Furthermore, since K c > 0
then the equilibrium point E is also globally asymptotically stable, see Ho and Ou [9].
## 3. T HE P REDATOR -P REY M ODEL WITH T IME D ELAY AND C ONSTANT R ATE OF
H ARVESTING
We consider the predator and prey populations of model (2.1) where both populations
are subjected to a constant rate of harvesting. Before we go to the model with time delay,
we need to analyze the stability of the equilibrium point of the model without time delay.
The model without time delay is
dx
= x(r bx y) Hx
dt
dy
= y(c + x) Hy , (3.1)
dt
r
where r, b = K , , c, , Hx , Hy are positive constants. The constants Hx and Hy denote
the rate of harvesting for the populations x and y respectively.
dy
By setting dx
dt = 0 and dt = 0 then we have the relations
x(r bx y) = Hx (3.2)
y(c + x) = Hy . (3.3)
2
From (3.2) we have y = rxbx x
Hx
which follows that r 2 4bHx should be positive in
order to get the equilibrium point in the positive quadrant. Hence we have to assume that
2
Hx < r4b . Since Hy is positive, then from (3.3) we should assume that x > c .
## F IGURE 1. Phase plane and directions of the trajectories
From the phase plane, we know that it is possible to get two, one or no equilibrium
points, Figure 1. There are two positive equilibrium points of the model when
40 Syamsuddin Toaha and Malik Abu Hassan
bx3 (r + bc)x2 + (Hx + Hy + rc)x Hx c < 0, for some positive x > c . Let
the two equilibrium points be E1 = (x1 , y1 ) and E2 = (x2 , y2 ). The equilibrium point E1
is possible to be asymptotically stable, while the equilibrium point E2 is not stable, it is a
To analyze the stability of the equilibrium point E1 we linearize the model around the
equilibrium point E1 . The Jacobian matrix of the model is
r 2bx y x
J= .
y c + x
The characteristic equation of the Jacobian matrix at this point is
2 (P + S) + P S + QR = 0, (3.4)
where
P = r 2bx1 y1 ,
Q = x1 ,
R = y1 , and
S = c + x1 .
P + S < 0.
## Example 1. Consider model (3.1) with parameters r = 1, b = 0.01, = 1, c = 0.3,
= 0.05, Hx = 0.01, and Hy = 0.02. The equilibrium points of the model in the positive
quadrant are E1 = (6.42819, 0.93416) and E2 = (99.56243, 0.00428). The eigenvalues
associated with the equilibrium point E1 are 0.02066 0.54633i and the eigenvalues
associated with equilibrium point E2 are 0.99177 and 4.67437. This reveals that the
equilibrium point E1 is asymptotically stable while the equilibrium point E2 is a saddle
point and unstable.
## Example 2. Consider again model (3.1) with parameters r = 1, b = 0.01, = 1, c = 0.3,
= 0.05, Hx = 0.1, and Hy = 0.2. There are two equilibrium points of the model in the
positive quadrant, they are, E1 = (10.51819, 0.88531) and E2 = (95.42203, 0.04473).
The equilibrium point E1 has eigenvalues 0.06512 0.66313i and the equilibrium point
E2 has eigenvalues 0.91354 and 4.43147. This means that both equilibrium points are
not stable.
From Examples 1 and 2 we know that the equilibrium E1 may be a stable or an unstable
equilibrium point. It depends on the values of the parameters and the level of constant
rate of harvesting. Apparently, the equilibrium point E1 tends to the equilibrium point E
when the harvesting function Hx and Hy approach zero. If the equilibrium point E for the
non-harvesting model is asymptotically stable, then the eigenvalues of the Jacobian matrix
of the linearized system have negative real parts. Since the eigenvalues are continuous in
Hx and Hy , the equilibrium point E1 is asymptotically stable for sufciently small H x > 0
and Hy > 0 . On the other hand, when the equilibrium point E1 is unstable, there exists an
asymptotically stable limit cycle. Theory of perturbation of periodic solutions, Coddington
and Levinson [5], shows that there is an asymptotically stable limit cycle for small H x > 0
and Hy > 0. Thus, the qualitative behavior of the system for Hx = 0 and Hy = 0 carries
over to small Hx > 0 and Hy > 0.
Stability Analysis of Predator-Prey Population Model .... 41
Now we consider the predator - prey population model with time delay and constant
rate of harvesting. Both predator and prey populations are subjected to constant rate of
harvesting. The model is
dx(t)
= rx(t) bx(t)x(t ) x(t)y(t) Hx ,
dt
dy(t)
= cy(t) + x(t)y(t) Hy . (3.5)
dt
A predator-prey model with time delays in the growth rate of the predator population
and the prey harvested with constant rate has been analyzed by Martin and Ruan [14].
They showed that the time delays can induce instability, oscillations via Hopf bifurcation
and switching stability.
To linearize the model about the equilibrium point E1 of model (3.5), let
u(t) = x(t) x1 and v(t) = y(t) y1 . We then obtain the linearized model
u(t) = (r bx1 y1 )u(t) bx1 u(t ) x1 v(t)
v(t) = y1 u(t) + (c + x1 )v(t). (3.6)
From the linearized model we obtain the characteristic equation
(, ) = 2 + a1 e a2 a3 e + a4 , (3.7)
where
a1 = bx1
a2 = r c bx1 + x1 y1
a3 = bcx1 + bx21 , and
a4 = rc + ax1 + bcx1 + bx21 cy1 .
## For = 0, the characteristic equation (3.7) becomes 2 + (a1 a2 ) a3 a4 =
0. This characteristic equation is the same with the characteristic equation (3.7). The
eigenvalues of the characteristic equation are either real and negative or complex conjugate
with negative real parts if and only if
a1 a2 > 0 and a3 + a4 > 0. (3.8)
Hence, in the absence of time delay, the equilibrium point E1 is locally asymptotically
stable if and only if both conditions a1 a2 > 0 and a3 + a4 > 0 are satised.
Now for 6= 0, if = i, > 0 , is a root for the characteristic equation (3.7), then
we have
## Separating the real and imaginary parts, we get
2 + a4 + a1 sin( ) a3 cos( ) = 0
a2 + a1 cos( ) + a3 sin( ) = 0,
or equivalently
2 + a4 = a1 sin( ) + a3 cos( )
a2 = a1 cos( ) + a3 sin( ). (3.9)
42 Syamsuddin Toaha and Malik Abu Hassan
## Squaring both sides gives
4 2a4 2 + a24 = a21 2 sin2 ( ) 2a1 a3 sin( ) cos( ) + a23 cos2 ( )
a22 2 = a21 2 cos2 ( ) + 2a1 a3 sin( ) cos( ) + a23 sin2 ( ).
Adding both equations and regrouping by powers of , we obtain the following fourth
degree polynomial
## 4 (a21 + 2a4 a22 ) 2 + a24 a23 = 0. (3.10)
Then we obtain
1
q
2
{(a21 + 2a4 a22 ) (a21 + 2a4 a22 )2 4(a24 a23 )}.
= (3.11)
2
From the equation (3.11), it follows that if
a22 2a4 a21 > 0 and a24 a23 > 0, (3.12)
then the equation (3.10) does not have any real solutions.
To nd the necessary and sufcient conditions for nonexistence of time delay induced
instability, we now use the following theorem.
Theorem 3. (Kar, [12]). A set of necessary and sufcient conditions for an equilibrium
point (x , y ) to be asymptotically stable for all 0 is
## (1) The real parts of all the roots of (, 0) = 0 are negative,
(2) For all real and 0, (i, ) 6= 0, where i = 1.
Theorem 4. If conditions (3.8), (3.12) and Theorem 3 are satised, then the equilibrium
point E1 is locally asymptotically stable for all 0.
Again, if
a24 a23 > 0, a22 2a4 a21 < 0, and
(a22 2a4 a21 )2> 4(a24 a23 ), (3.13)
2 2
hold, then there are two positive solutions of . Substituting into equation (3.9) and
solving for , we obtain
2
1 (a1 a 1 a4 + a 2 a3 ) 2k
k = arctan{ 2 2 ) } + , k = 0, 1, 2, . (3.14)
a1 a2 + a3 (a4
## Differentiating equation (3.7) with respect to , we obtain
(2 a2 + a1 e (a1 a3 e )) d
d = (a1 a3 )e
,
Stability Analysis of Predator-Prey Population Model .... 43
therefore
d 1 2 a2 a1
( ) = + .
d (a1 a3 )e (a1 a3 )
(2 a2 +a4 )
From equation (3.7), we have e = (a1 a3 ) . Then we obtain
d 1 2 a2 a1
( ) = 2
+ .
d ( a2 a4 ) (a1 a3 )
2a2 a1
( d
d )
1
= (2 a2 a4 ) + (a1 a3 ) .
Thus,
d(Re) d 1
sign = sign{Re( ) }=i
d =i d
2 a2 a1
= sign{Re( )=i + Re( )=i
(2 a2 + a4 ) (a1 a3 )
+ Re( )=i }
2i a2 a1
= sign{Re( 2
) + Re( )
i( a2 i + a4 ) i(a1 i a3 )
+ Re( )}
i
a2 + 2 2 2a4 a21
= sign{ 2 22 }.
a2 + ( 2 + a4 )2 a21 2 + a23
From equation (3.10), we know that
a21 2 + a23 = 4 + (a22 2a4 ) 2 + a24 = a22 2 + ( 2 + a4 )2 ,
then we obtain
d(Re) a22 + 2 2 2a4 a21
sign{ }=i = sign{ 2 2 2 }
d a2 2 2
+ ( + a4 ) 2 a2 + ( 2 + a4 )2
= sign{2 (a21 + 2a4 a22 )}.
2
(3.15)
Theorem 5. Let k be dened by equation (3.14). If the conditions (3.7) and (3.13) are
[0, 0+ ) (S0 , 1+ )
S S
satised,
S then the equilibrium point E1 is stable when
+
) and unstable when [0+ , 0 ) (1+ , 1 ) (m1
+
S S
(m1 , m , m1 ), for
some positive integer m. Therefore there are bifurcations at the equilibrium point E 1 when
= k , k = 0, 1, 2, .
Proof. Since the conditions (3.8) and (3.13) are satised, then to prove the theorem we
need only to verify the transversality conditions, see Cushing [6],
d(Re) d(Re)
| = + > 0 and | = < 0,
d k d k
d(Re) d(Re)
d | =ik > 0 and d | =ik < 0.
+
## From (3.15) and (3.11), it follows that
44 Syamsuddin Toaha and Malik Abu Hassan
d(Re) 2
sign{ }=i+ = sign{2+ (a21 + 2a4 a22 )}
d q
= sign{ (a21 + 2a4 a22 )2 4(a24 a32 )},
therefore,
d(Re)
|=+ , = + > 0,
d k
d(Re)
d |=+ , =k > 0.
+
Again,
d(Re) 2
sign{ }=i = sign{2 (a21 + 2a4 a22 )}
d q
= sign{ (a21 + 2a4 a22 )2 4(a24 a32 )},
therefore,
d(Re)
|= , = < 0.
d k
Hence, the transversality conditions are satised. This completes the proof.
## Example 6. Consider model (3.5) with parameters r = 3.5, b = 0.04, = 1, c = 0.3,
= 0.05, Hx = 0.02, and Hy = 0.01. The equilibrium point of the model is E1 =
(6.06146, 3.25424). For = 0, the Jacobian matrix of the model associated with the
equilibrium point has eigenvalues 0.11804 0.98570i. This means that the equilibrium
point of the model without time delay is stable. The conditions (3.8) and (3.13) are satised.
Some trajectories of x(t) and y(t) with various time delays are given in Figures 2, 3, and
4.
From Figures 2a and 2b with time delay = 1.2, the equilibrium point
(6.06146, 3.25424) is stable. Figures 3a and 3b with time delay = 1.53 show that the
equilibrium point (6.06146, 3.25424) is unstable. The rst critical value of time delay is
= 0+ = 1.37941. When < 1.37941, the equilibrium point (6.06146, 3.25424) is
asymptotically stable; when = 1.37941 the equilibrium point (6.06146, 3.25424) loses
its stability; and when > 1.37941 but less then the second critical value of time delay, the
equilibrium point (6.06146, 3.25424) becomes unstable and there is a bifurcating periodic
solution, see Figure 4. Following Theorem 5 we have
0+ = 1.37941, 0 = 5.39314,
1+ = 6.98104, 1 = 12.53884,
2+ = 12.58266, 2 = 19.68453,
3+ = 18.68453, and 3 = 26.83023.
Then we have 2 stability switches from stability to instability and to stability.
Stability Analysis of Predator-Prey Population Model .... 45
## FIGURE 2b. Trajectory of predator with y(0) = 3.2642 and = 1.2
4. D ISCUSSION
In the analysis of the positive equilibrium point of model (3.1), it is quite difcult to
determine the value of the equilibrium points analytically. We just state that there exists
either one, or two, or none positive equilibrium points by inspection the phase plane of the
model. In the case of two positive equilibrium points occur, one of the equilibrium point
is possibly stable and the other is a saddle point. In this paper, we just analyze the case
of there exist two positive equilibrium points and focus on analyzing the effect of the time
delay on the stable equilibrium point. Actually we may also try to analyze the effect of the
time delay on the stability of the unstable equilibrium point.
46 Syamsuddin Toaha and Malik Abu Hassan
## FIGURE 3b. Trajectory of predator with y(0) = 3.2642 and = 1.53
There is still a lot of work to do in the predator-prey models with time delay and har-
vesting. For example, it would be interesting to consider time delay and harvesting in gen-
eralized Gause-type predator-prey model and in some another generalized predator-prey
models as in Martin and Ruan [14]. It would also be interesting to study the Wangersky-
Cunningham model with some delays in both the predator and prey model as in the Bartletts
model, see Bartlett [1] and Hasting [8].
R EFERENCES
[1] M.S. Bartlett, On theoretical models for competitive and predatory biological systems, Biometrika, 44
(1957), 27-42.
Stability Analysis of Predator-Prey Population Model .... 47
## FIGURE 4. Trajectory of (x(t), y(t)), with x(0) = 6.0715, y(0) = 3.2642
and = 1.37941
[2] F. Brauer and A.C. Soudack, Stability regions and transition phenomena for harvested predator-prey sys-
tems, Journal Math. Biology, 7 (1979), 319-337.
[3] F. Brauer and A.C. Soudack, Stability regions in predator-prey systems with constant-rate prey harvesting,
Journal Math. Biology, 8 (1979), 55-71.
[4] F. Brauer and A.C. Soudack, Coexistence properties of some predator-prey systems under constant rate
harvesting and stocking, Journal Math. Biology, 12 (1981), 101-114.
[5] E.A. Coddington and N. Levinson, Theory of ordinary differential equations, McGraw - Hill, New York,
(1955).
[6] J.M. Cushing, Integrodifferential equations and delay models in population dynamics. Heidelberg: Springer-
Verlag, (1977).
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Appl. Math. 58 (1998), 193-210.
[8] A. Hasting, Delays in recruitment at different trophic levels: effects on stability, J. Math. Biol. 21 (1984),
35-44.
[9] Ho, C.P. and Ou, Y.L. . Inuence of time delay on local stability for a predator-prey system, Journal of
Tunghai Science, 4 (2002), 47-62.
[10] Hogarth, W.L., Norbury, J., Cunning, I. and Sommers, K., Stability of a predator-prey model with harvesting,
Ecological Modelling, 62 (1992), 83-106, ibid. 4 (2002), 47-62.
[11] J. Holmberg, Socio-ecological principles and indicators for sustainability, PhD Thesis, Goteborg Univer-
sity, Sweden, (1995).
[12] T.K. Kar, Selective harvesting in a prey-predator shery with time delay, Mathematical and Computer Mod-
elling, 38 (2003), 449-458.
[13] L.S. Luckinbill, Coexistence in laboratory populations of paramecium aurelia and its predator didinium
nasutum, Journal of Ecology, 54, 6, (1973), 1320-1327.
[14] A. Martin and S. Ruan, Predator-prey models with time delay and prey havesting, J. Math. Biol, 43 (2001),
247-267.
[15] H. Matsuda and P.A. Abrams, Effects of predators-prey interaction and adaptive change on sustainable
yield, Can. J. Fish. Aquat. Sci./J. Can. Sci. Halieut. Aquat., 61, 2, (2004), 175-184.
[16] M.R. Myerscough, B.F. Gray, W.L. Hogarth, J. Norbury . An analysis of an ordinary differential equation
model for a two-species predator-prey system with harvesting and stocking, Journal Math. Biology, 30
(1992), 389-411.
[17] P.D. Srinivasu, S. Ismail and C.R. Naidu, Global dynamics and controllability of a harvested prey-predator
system, J. Biological Systems, 9, 1, (2001), 67-79.
48 Syamsuddin Toaha and Malik Abu Hassan
[18] D. Xiao and S. Ruan, Bogdanov-Takens bifurcations in predator-prey systems with constant rate of harvest-
ing, Fields Institute Communications, 21 (1999), 493-506. | 6,448 | 20,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-35 | latest | en | 0.868165 |
https://cs.stackexchange.com/questions/167999/integrality-gap-of-maximum-coverage-lp | 1,718,774,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00751.warc.gz | 173,270,957 | 38,132 | # Integrality gap of Maximum Coverage LP
Consider Maximum-Coverage problem, which means there are set $$A$$ and $$n$$ subsets of $$A$$, we can choose $$k$$ subsets to cover some elements, and Maximum-Coverage is the assignment that cover most elements in $$A$$. We know by LP relaxation there is $$1-1/e$$ approximation algorithm, however, the integrality gap for it is $$1-1/e+\epsilon$$, which means there is an instance $$I$$ such that $$LP(I)=1$$ and $$Opt(I)<1-1/e+\epsilon$$, how to choose the problem set $$I$$?
• What do you mean by $LP(I) = 1$? Commented May 10 at 7:24
• If the integrality gap is $1-\frac{1}{e} + \epsilon$, then by definition we have $LP(I) \ge \frac{OPT(I)}{1-\frac{1}{e} + \epsilon}$. Commented May 10 at 7:27
• What do you mean by "choose the problem set $I$"? Do you mean you want to find an instance (not problem set) where this bound is tight? Commented May 10 at 7:29
• Yes, I want to find a counterexample. Commented May 10 at 12:25
• I would recommend you modify your question to better reflect your requirements. Commented May 10 at 12:33 | 318 | 1,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.928197 |
https://forums.qrz.com/index.php?threads/push-pull-output-transformers-part-iii-the-final-countdown.582815/page-5 | 1,627,120,912,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150264.90/warc/CC-MAIN-20210724094631-20210724124631-00463.warc.gz | 278,272,761 | 20,864 | # Push-Pull Output Transformers - Part III, The Final Countdown:
Discussion in 'Amateur Radio Amplifiers' started by KD2NCU, Sep 28, 2017.
1. ### KD2NCUHam MemberQRZ Page
No, it's your turn to finish explaining and answering questions first. Ask WW1WW for help if needed.
Answer all of the following questions.
Question 1: Go back and answer the simple questions. (Without getting into a convoluted discussion, of course.)
Figure 1: What does the current coming out of the source look like? Check only one answer below. (Keep it simple now)
•____ 1 amp pulses matching the resistor current exactly?
•____ A constant flat DC current of 0.25 amps?
Figure 2: What will the current coming out of the supply be? Check one. Keep it simple.
•____ A full wave rectified sinewave of current with peak value of 50 mA matching the collector current exactly?
•____ A flat DC current equal to the average value of the collector current or 0.637 x 50 mA = 31.85 mA?
Question 2: Just tell me (no convoluted discussions) how a DC supply works. I'll start you off. A DC supply will work very hard to keep the voltage constant. Now you tell me what a DC supply does regarding current. Keep it simple!
Question 3: Describe what the current looks like just as it enters the center tap of the bifilar feed coil. ie; full wave rectified pulses of current essentially replicas of the individual transistor current pulses, or a dead flat DC current equal to the average of the transistor current pulses.
Question 4: If you say the current entering the center tap is a dead flat DC equal to the average of the transistor current pulses then explain when Q1 first turns on but the collector/drain current has only risen to a few milliamps but the current coming out of the DC supply is a flat DC of about 10 amps, all those extra amps and electrons are going (fill in the blank) ____________________ .
Question 5: Does WW1WW's assertion that all those extra electrons representing the unexplained 10 amps of question 4 are converted into stored energy in the magnetic field sound plausible to you? Keep it simple now, like maybe YES or NO. If so, can you or WW1WW provide a valid physics reference that supports this new "discovery"?
Question 6: Does WW1WW's assertion that all those extra electrons representing the unexplained 10 amps of question 4 are leaking through the off transistor seem plausible to you? Simple now, should be a YES/NO answer. Either you agree with him or you don't.
Question 7: What is "Pure DC" (no convoluted discussions) You and WW1WW throw it around like it has some special meaning. I have been taking it to mean the voltage is constant but the DC supply will provide whatever current the connected load demands or asks for because that's all a DC supply can do. I think you and WW1WW think that "Pure DC" has some extra meaning. So please explain in simple terms what you are assuming when you say at point B you have Pure DC.
2. ### K7JEMHam MemberQRZ Page
Your diagrams are interesting, but not representative of an RF power amplifier. Go back to the simple RF power amplifier I posted and show that "pulsating DC" is being drawn from the PS, at a rate equal to the instantaneous value being drawn at the collector. After we get this simple matter resolved, we can move to other topics. I would submit that the current being provided by the PS is a "pure DC" with a very small trace of RF.
If we can't agree on this, there is no reason to go any further.
3. ### K7JEMHam MemberQRZ Page
1 amp pulses matching the switch closure.
Well, this must be a trick question, because neither answer is correct. We don't know what the waveform is, so hard to predict what the current waveform will look like. The second answer has values provided, but no values are given on the schematic, so who knows? We don't know the voltage, the input voltage or current, the Re value, or anything else. So, neither answer is correct.
The DC supply will supply the current needed for the attached circuit, up to its maximum output.
Assuming standard amplifier circuit with bypass capacitor, it will be dead flat DC input, with a few milliamps of AC ripple. The DC will be equal to the current draw of the amplifier, assuming steady constant power out during measurement.
Charging capacitors and inductors throughout the cycle.
Yes. Energy is stored in inductors and capacitors. This is not a new concept.
I don't think he said that. He mentioned that there was some capacitance in the other transistor.
Pure direct current would be a constant flow of current, not pulsating at RF. An easy way to see this is to insert a non inductive resistor of low resistance, right into the DC feed point. Something that would drop 1 volt or so at full amp power out. Now activate the amp, and use your scope to look at the voltage drop across the resistor. You will see a 1 volt (or whatever) drop. You may see a small amount of RF, but probably a mV or so. Won't be a lot in any normally designed amplifier of any configuration.
I think this is something you need to do, to prove to yourself that the current supplying these amps is not "pulsating" to any degree beyond the feed choke.
5. ### WA1GFZHam MemberQRZ Page
you are confusing a transistor with a tuned tank circuit with broadband amplifier. Apples and oranges
6. ### K7JEMHam MemberQRZ Page
No confusion going on here. There is an inductor between the DC supply and the transistor. It has inductive reactance. That reactance is what isolates the DC from the RF.
7. ### WA1GFZHam MemberQRZ Page
KD2DCU,
The flux would be balanced but there would still be common mode inductance with no offset in the old BH loop, in a perfect world. In the real world at least in all my testing the flux is not canceled and even a 1 turn third winding picks up the offset by the flux transformer action not some leakage mumbo jumbo. I've spent hundreds of hours making measurements and concluded T2 is not necessary if the transformer is designed correctly. We just bought a 2500 watt AR amp at work and the first thing I checked was the output transformer circuit. NO T2s. BTW we have a number of AR amps and none of them have a T2.
T2 was added in commercial hammy hambone rigs because the half ass flux transformer had big offsets and ran hot. I duplicated the effect. I then converted the exact same cores to a balanced TLT and the cores ran stone cold. My case I din't have room for T2 and T3 on my MRI amplifier boards so found a better way. That allowed me consistent performance 160 through 6 meters. (The AR amp runs 10khz to 220 MHz)
I had this same battle with myself a few years ago so more fun to watch gfz
8. ### KD2NCUHam MemberQRZ Page
Now I'm not arguing for or against using the T2/T3 approach and never have been. The start of this entire chain of threads was me asking why the T2/T3 approach is even used and what is T2 doing that T3 can't do by itself. A lot of answers I got were absolute B.S. that defy simple physics and basic circuit analysis. Initially, in gentlemanly fashion, I provided physics and circuit analysis to show which areas I thought were fallacy and why. Ever since then, people have shrieked, "you're wrong", "it's a choke", "but it's pure DC", "this guy doesn't understand" and took it personal asking about my education, hurling insults, etc. But that's OK, because all the shrieking, asking about my education, and hurling insults is the surest sign of weakness, self doubt, and admitting they don't really have an argument. Besides, with my experience, I've been worked on by experts and real engineers, physicists, etc. and these guys with the mouths are amateurs who seem unable to use physics and circuit analysis. I've never lost a technical argument yet because I do my homework first while most people responding on this site think they already know everything and start running their mouths.
The posts I have provided have used solid physics and well established circuit analysis to show which claims I believe are false and why. Prior to you, WA1GFZ, not a single person actually tried to refute ANY of the analysis based on physics and circuit analysis. They just shrieked "you're wrong" etc. etc. and hurled insults. Gee, I wonder what that means. Now you've said a couple of things that lead me to believe you have at least looked at some of my prior analysis. At one point you did say you agree with my analysis but I don't know which analysis you were talking about.
So what about you, WA1GFZ, do you believe in physics and established circuit analysis techniques?
So here are some of the responses I originally got. Let's see what you think about them.
1. There is absolutely no RF of any kind in the coils of T2. What say ye WA1GFZ?
2. It is absolutely impossible for the core of T2 to saturate "because it's not grounded". Well, typically it is AC grounded and the core isn't aware and doesn't give a @#\$% whether it's grounded. Imbalance will staircase up the BH curve until either the core saturates or until something in the external circuit finally limits the current. What do you say about this?
3. The bifilar feed coil keeps RF out of the DC supply. This defies basic physics and basic circuit analysis. In the circuit below I've said that the current entering the center tap has to equal the drain current of the on transistor regardless of whether T2 is providing any inductance at all simply from conservation of charge (Kirchhoff's current law applied to the circuit.) The current enters the center tap, splits into I1 and I2, and regardless of whether they are equal or not, they just wander around a bit and then recombine to become the collector or drain current except for some leakage current through the off transistor. Yes, in the real world there will be some leakage current through the off transistor. But this is a red herring. Regardless of the other real world realities of imbalance, imperfect windings, non-linear BH curve, and so on, other than the small leakage current through the off transistor, there is only one place current gets into the circuit (the center tap) and one way for the current to get out of the circuit and that's the collector or drain of the conducting transistor. Unless, of course, electrons are elastic and can bunch up in the wires or something. Example below. At the very start of Q1's conduction cycle, say the current has only risen to a few milliamps, the current going into the center tap must also be just a few milliamps. K7JEM holds that the current at the center tap will be essentially flat with minor ac ripple and its value will be the average of the collector current pulses. So if the collector currents are positive sine pulses peaking at 16 amps then the current as it enters the center tap will be essentially flat DC current of about 10 amps. I asked where all those extra electrons are going. IE: the collector current is 1 mA or some other very low value, but the current going into the center tap is 10 amps flat DC. What accounts for this difference? Where are all those amps and trons going when the collector current is so low compared to the incoming current. K7JEM and WW1WW think the electrons are being stored in the magnetic field of the T2 coils/cores and maybe even T3's coil/core. So electrons are somehow jumping out of the wires of T2 and T3, somehow transforming to some sort of magnetic ..... goo or maybe magnetic goo-ons of some sort and hanging around drinking covfefe until the collector current rises to 10 amps at which time there is now a shortfall so covfefe break is over, put down your cups, all you goo-ons turn back to electrons and jump back into the coils to make up the shortfall between the constant flat 10 amps and the collector current which is about to peak at 16 amps. But don't worry all you trillions of goo-ons, because you'll be back again shortly when the collector current falls below 10 amps again and there is an excess of amps and trons.
At the last minute, K7JEM threw in that the mismatch of trons and amps also goes into charging capacitors during the normal cycle. Problem is, there are no capacitors present INSIDE the circuit for the trons to get stored in.
What do you think about the electrons turning into magnetic goo-ons theory WA1GFZ?
It's a pretty fantastic theory, but I think he's screwing with me, I don't believe for a minute that anyone would really believe such a fantasy. What do you think?
Below right is K7JEM's theory of magnetic goo-ons. Whaddaya think about this WA1GFZ? Could maybe THIS explain the imbalance you always find? Imbalance in the electron to magnetic goo-on conversion plasma?
9. ### K7JEMHam MemberQRZ Page
I think your image is rather humorous, so I offer a video to rebut:
Yes, inductors can store energy, it is a well known phenomenon. I'll let you do some research on that.
It is interesting that you place a current probe inside the bypass capacitor. But the DC supply does not feed at that point, it feeds to the bypass capacitor, so readings need to be taken at that point, if we are talking about DC that is coming from the supply. So, move the current probe to a place where the DC actually feeds, and explain what the DC current will look like at that location. I have already made my explanation, has to do with inductors and capacitors that store energy, if only for a few nanoseconds. But that's what is going on.
It would be nice to work with schematics and diagrams that were actual circuits with real parts and values listed. That is why I provided the link to the 2M class C amplifier. The main thing is that we know what is going on, rather than try to figure out some main issues, like inductor values, capacitor values, voltage and current, etc.
The diagram and photo that I posted are easy to reference, and a good starting point. If you don't like them, you can post any schematic that you like, as long as it has part values and is actually a working amplifier. I chose the simple class C design with a single transistor, since it eliminates a lot of parts, possible current paths, etc. If we can't agree on how this works, it will be much more difficult on a more complex amplifier design.
KD2NCU, | 3,206 | 14,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-31 | latest | en | 0.8851 |
https://www.thestudentroom.co.uk/showthread.php?t=68606 | 1,547,853,812,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660818.25/warc/CC-MAIN-20190118213433-20190118235433-00636.warc.gz | 978,834,005 | 28,703 | This discussion is closed.
#1
1) Make l the subject of the formula T = 2pi sqrt l/g
(that is t equals 2pi times the sqr root of l over g)
2) simplify x^5/2 x x^1/2
(that is x to the power of 5 over 2 times sqr root of x)
0
14 years ago
#2
(Original post by m66)
1) Make l the subject of the formula T = 2pi sqrt l/g
(that is t equals 2pi times the sqr root of l over g)
2) simplify x^5/2 x x^1/2
(that is x to the power of 5 over 2 times sqr root of x)
1) T = 2pi sqrt l/g
---> T^2 = 4Pi^2l/g
---> l = (T^2g)/(4Pi^2)
2) x^(5/2) . x^(1/2) = x^(5/2 + 1/2) = x^(6/2) = x^3
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http://www.coachingindians.com/articles/secondary-school-level-preparation-strategies-for-maths.html | 1,553,612,431,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205534.99/warc/CC-MAIN-20190326135436-20190326161436-00225.warc.gz | 241,531,750 | 3,220 | Home » Articles » Secondary School level preparation strategies for maths
# Secondary School level preparation strategies for Maths
Students at Secondary level can take up maths little more interestingly. All depends on how the student perceives and feels what is math's all about.
Now proceeding to mathematical concepts, in the secondary level, firstly, as a student let us understand the theory of a glass half full or half empty. You should have confidence that if I get 60% in math's I know 3/4th of maths. Now start feeling that in math's, all I should strive hard is for only 1/4th of maths.
As a student to do your little shopping where you can learn addition, subtraction at the first stage. When you are learning maths through shopping or going to a nearby market, you should excuse yourself if you did little mistakes. Do not get confused by calling yourself a fool or stupid. Do not worry even if others call you so. Everyone learns anything only by doing few mistakes. So, you can convince yourself by saying that you had a loss of Rs.10 and the shop keeper had a gain of Rs.10. Now this gives a way for you to understand what profit and loss is.
As a student try and understand few things in math's and not everything, when you attempt to solve a problem. You should not lose your confidence if you could not successfully complete the sum. Now instead of feeling that you did not understand the concept, you can clarify with your teacher and understand up to a certain level. Now it will be the duty of your teacher to guide you through the next half of the problem.
As a student, you need to change your mindset that in math's subject you need not achieve 100/100. Instead you need to convince yourself if you know 60/100 and the rest left for you to learn is only 40/100. Slowly split this 40 into four portions and make yourself realize it is easy to achieve more marks in maths. Even if it does not happen, you should allow yourself to gain confidence knowing 70% or 60% math's is equally good.
Especially in the secondary level, a student should never start a new concept altogether with a new theory. You should introduce to yourself every new concept with the old and familiar concept, where your confidence increases. In short, as a student learn known topics and move on to unknown topics. Do not start learning from unknown to known topics. Start working and follow the Illustrations and examples and understand them at every step. Once you gain confidence, do the same sum without seeing. Next, go and try a similar problem from exercises. This way, keep changing numbers and solve the same problem or try similar problems. This will boost your confidence in maths. Also try to keep a new note book, put the date and do math's systematically. Go back to your previous day's work and keep gaining confidence.
Slowly adopt this kind of interesting methodology; and learn algebra, geometry, cost concepts, time concepts, and so on. | 622 | 2,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-13 | latest | en | 0.963543 |
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# Measurement.micrometer & vernier caliper
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### Measurement.micrometer & vernier caliper
1. 1. Presented by: Riel M. Casinillo Faculty Measurement and Measuring Devices (Micrometer and Vernier Caliper)
2. 2. What are we going to do today? <ul><li>1) We are going to read and use the following instruments in measuring certain quantities: </li></ul><ul><ul><li>Micrometer </li></ul></ul><ul><ul><li>Vernier Caliper </li></ul></ul>
3. 3. The Micrometer <ul><li>A micrometer allows a measurement of the size of a body. It is one of the most accurate mechanical devices in common use. </li></ul>Parts of a Micrometer
4. 4. How to read and use the Micrometer?
5. 5. <ul><li>The line labeled 0 is the primary pointer. When it lines up with the central horizontal line on the cylinder, as shown, then the distance between the measuring rods is exactly an integral or half-integral number of millimeters. </li></ul><ul><li>The left-hand side of the thimble has markings all around it. </li></ul><ul><li>3) In this case, then, the distance between the measuring rods is 7.000 mm . The upward line on the cylinder corresponding to 7 mm is barely visible under the thimble. </li></ul>Primary pointer Central horizontal line 1 2 3 4 6 7 0.5 1.5 2.5 3.5 4.5 5.5 6.5
6. 6. Remember!!! <ul><li>Report to 3 decimal places. Your instrument is capable of reading up to thousandths of a millimeter. ( That is why it is called a micrometer, 10 -6 of a meter). </li></ul>
7. 7. 1) If we rotate the thimble one complete revolution so it moves to the right, it will look as shown. 2) Now the distance between the measuring rods is 7.500 mm. The downward line on the cylinder corresponding to 7.5 mm is barely visible. 3) Note that this is all consistent with the fact that the markings on the cylinder correspond to fifty divisions for a complete revolution. Thus the numbers on the thimble correspond to hundredths of a millimeter. 7 7.5 7.5+0.000 = 7.500 mm
8. 8. <ul><li>One more complete revolution takes the distance to 8.000 mm, as shown. </li></ul>8 0.000 8.000 mm + 0.000 mm = 8.000 mm
9. 9. <ul><li>Now turn the thimble just a little bit more, and end up with the situation shown. </li></ul><ul><li>We have increased the distance by 0.120 mm from 8.000, so the distance between the measuring rods is now 8.120 mm. </li></ul>8 0.120 mm 8.000 mm + 0.120 mm = 8.120 mm
10. 10. <ul><li>If we give the thimble one complete further rotation, we end up with the micrometer as shown above. </li></ul><ul><li>Now the distance is 0.120 mm greater than 8.500 mm, so the distance is 8.620 mm. </li></ul>8.5 mm 0.120 mm 8.500 mm + 0.120 mm = 8.620 mm
11. 11. <ul><li>Finally, if we turn the thimble a tiny bit further, it might end up like the figure above. </li></ul><ul><li>Now the distance is greater than 8.620 mm, but clearly less than 8.630 mm. We might estimate this reading to be 8.624 mm. </li></ul>8.5 mm 0.120 mm 8.500 mm + 0.12 4 mm = 8.62 4 mm
12. 12. <ul><li>Depending on your eyes and screen resolution, you might be fairly confident that the reading is less than, say, 8.627 mm and similarly confident that it is greater than 8.621 mm. Thus you might assign a Reading Error to this measurement of 0.003 mm </li></ul><ul><li>So, we would report the distance as 8.624 ± 0.003 mm. </li></ul>
13. 13. Let us try!!!
14. 14. 1) Answer: 7.880 mm
15. 15. 2) Answer: 7.721± 0.003 mm
16. 16. 3) Answer: 3.560 mm
17. 17. 4) Answer: 5.801 ± 0.003 mm
18. 18. 5) Answer: 3.090 mm
19. 19. Zero error
20. 20. The Vernier Caliper <ul><li>This instrument may be used to measure outer dimensions of objects (using the main jaws), inside dimensions (using the smaller jaws at the top), and depths (using the stem). </li></ul>
21. 21. How to read and use the Vernier Caliper?
22. 22. The reading here is 3.7 mm or 0.37 cm.
23. 23. The reading here is 15.8 mm or 1.58 cm .
24. 24. In figure 4 above, the first significant figures are taken as the main scale reading to the left of the vernier zero, i.e. 3.7 cm. The remaining digits are taken from the vernier scale reading that lines up with any main scale reading, (i.e. 0.46 mm or 0.046 cm) on the vernier scale. Thus the reading is 3.746 cm.
25. 25. In figure above, the first significant figures are taken as the main scale reading to the left of the vernier zero, i.e. 3.4 cm. The remaining digit is taken from the vernier scale reading that lines up with any main scale reading, (i.e. 0.60 mm or 0.060 cm) on the vernier scale. Therefore the reading is 3.460 cm.
26. 26. In figure 6 the zero and the ten on the vernier scale both line up with main scale readings, therefore the reading is 4.000 cm.
27. 27. Let us try!!!
28. 28. 1) Answer: 3.090 cm
29. 29. Answer: 0.810 cm 2)
30. 30. 3) Answer: 12.168 cm
31. 31. Zero Error
32. 32. Okay, that is enough! Let us try to apply it in real life.
33. 33. Try to measure the quantities of the following objects. Fill-out the table below. Quantities to Measure Micrometer Vernier Caliper 1) Diameter of a 1-peso coin 2) Thickness of a 5-peso coin 3) Thickness of your hair
34. 34. Give me a reading of: <ul><li>5.66 cm </li></ul><ul><li>4.78 cm </li></ul> | 1,828 | 5,705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2016-44 | latest | en | 0.815434 |
https://github.com/dennybritz/reinforcement-learning/tree/master/PolicyGradient | 1,558,548,627,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256887.36/warc/CC-MAIN-20190522163302-20190522185302-00340.warc.gz | 489,586,835 | 17,292 | # dennybritz/reinforcement-learning
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CliffWalk Actor Critic Solution.ipynb Dec 28, 2017
CliffWalk REINFORCE with Baseline Solution.ipynb Jan 24, 2018
Continuous MountainCar Actor Critic Solution.ipynb Dec 28, 2017
### Learning Goals
• Understand the difference between value-based and policy-based Reinforcement Learning
• Understand the REINFORCE Algorithm (Monte Carlo Policy Gradient)
• Understand Actor-Critic (AC) algorithms
• Understand Deterministic Policy Gradients (Optional)
• Understand how to scale up Policy Gradient methods using asynchronous actor-critic and Neural Networks (Optional)
### Summary
• Idea: Instead of parameterizing the value function and doing greedy policy improvement we parameterize the policy and do gradient descent into a direction that improves it.
• Sometimes the policy is easier to approximate than the value function. Also, we need a parameterized policy to deal with continuous action spaces and environments where we need to act stochastically.
• Policy Score Function `J(theta)`: Intuitively, it measures how good our policy is. For example, we can use the average value or average reward under a policy as our objective.
• Common choices for the policy function: Softmax for discrete actions, Gaussian parameters for continuous actions.
• Policy Gradient Theorem: `grad(J(theta)) = Ex[grad(log(pi(s, a))) * Q(s, a)]`. Basically, we move our policy into a direction of more reward.
• REINFORCE (Monte Carlo Policy Gradient): We substitute a samples return `g_t` form an episode for Q(s, a) to make an update. Unbiased but high variance.
• Baseline: Instead of measuring the absolute goodness of an action we want to know how much better than "average" it is to take an action given a state. E.g. some states are naturally bad and always give negative reward. This is called the advantage and is defined as `Q(s, a) - V(s)`. We use that for our policy update, e.g. `g_t - V(s)` for REINFORCE.
• Actor-Critic: Instead of waiting until the end of an episode as in REINFORCE we use bootstrapping and make an update at each step. To do that we also train a Critic Q(theta) that approximates the value function. Now we have two function approximators: One of the policy, one for the critic. This is basically TD, but for Policy Gradients.
• A good estimate of the advantage function in the Actor-Critic algorithm is the td error. Our update then becomes `grad(J(theta)) = Ex[grad(log(pi(s, a))) * td_error]`.
• Can use policy gradients with td-lambda, eligibility traces, and so on.
• Deterministic Policy Gradients: Useful for high-dimensional continuous action spaces where stochastic policy gradients are expensive to compute. The idea is to update the policy in the direction of the gradient of the action-value function. To ensure exploration we can use an off-policy actor-critic algorithm with added noise in action selection.
• Deep Deterministic Policy Gradients: Apply tricks from DQN to Deterministic Policy Gradients ;)
• Asynchronous Advantage Actor-Critic (A3C): Instead of using an experience replay buffer as in DQN use multiple agents on different threads to explore the state spaces and make decorrelated updates to the actor and the critic.
Required:
• David Silver's RL Course Lecture 7 - Policy Gradient Methods (video, slides)
Optional:
### Exercises
• REINFORCE with Baseline
• Actor-Critic with Baseline
• Actor-Critic with Baseline for Continuous Action Spaces
• Deterministic Policy Gradients for Continuous Action Spaces (WIP)
• Deep Deterministic Policy Gradients (WIP) | 832 | 3,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-22 | latest | en | 0.824846 |
https://oeis.org/A027277 | 1,582,871,983,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147054.34/warc/CC-MAIN-20200228043124-20200228073124-00517.warc.gz | 485,825,651 | 4,256 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A027277 a(n) = Sum_{k=0..n} binomial(2*k,k)*binomial(2*n-k,k). 1
1, 3, 13, 67, 375, 2189, 13089, 79479, 487833, 3018355, 18792303, 117589689, 738844719, 4658460165, 29458662005, 186761788579, 1186655988771, 7554520173441, 48176764031385, 307706150625855, 1968040844127793, 12602972755261195, 80798365998084795, 518536437750443773 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Previous name was: a(n) = self-convolution of row n of array T given by A026568. LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 FORMULA From Peter Luschny, May 14 2016: (Start) a(n) = hypergeom([1/2, -n, 1/2-n], [1, -2*n], -16) for n>=1. a(n) = (2*n*(4*n-5)*(-9+4*n)*(-7+4*n)*a(n-3) - (4*n-5)*(50*n^3-175*n^2+152*n-9)* a(n-2) + (80*n^3-260*n^2+198*n-27)*(n-1)*a(n-1)) / (n*(n-1)*(-9+4*n)*(-1+2*n)) for n>=3. (End) a(n) ~ sqrt(5 + 13/sqrt(17)) * ((9 + sqrt(17))/2)^n / (4*sqrt(Pi*n)). - Vaclav Kotesovec, May 14 2016 MAPLE a := n -> add(binomial(2*k, k)*binomial(2*n-k, k), k=0..n): seq(a(n), n=0..23); # Peter Luschny, May 14 2016 MATHEMATICA Table[Sum[Binomial[2k, k] Binomial[2n-k, k], {k, 0, n}], {n, 0, 30}] (* Michael De Vlieger, May 14 2016 *) PROG (PARI) vector(30, n, n--; b=binomial; sum(k=0, n, b(2*k, k)*b(2*n-k, k)) ) \\ G. C. Greubel, May 23 2017, modified Aug 03 2019 (MAGMA) B:=Binomial; [(&+[B(2*k, k)*B(2*n-k, k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Aug 03 2019 (Sage) b=binomial; [sum(b(2*k, k)*b(2*n-k, k) for k in (0..n)) for n in (0..30)] # G. C. Greubel, Aug 03 2019 (GAP) B:=Binomial;; List([0..30], n-> Sum([0..n], k-> B(2*k, k)*B(2*n-k, k) )); # G. C. Greubel, Aug 03 2019 CROSSREFS Cf. A026568. Sequence in context: A298611 A136784 A284717 * A242798 A239198 A234282 Adjacent sequences: A027274 A027275 A027276 * A027278 A027279 A027280 KEYWORD nonn AUTHOR EXTENSIONS New name from Peter Luschny, May 14 2016 STATUS approved
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Last modified February 28 01:38 EST 2020. Contains 332319 sequences. (Running on oeis4.) | 1,000 | 2,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-10 | latest | en | 0.470879 |
https://careertrend.com/how-7985591-calculate-overburden-pressure.html | 1,726,347,781,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00341.warc.gz | 132,890,229 | 40,047 | # How to Calculate Overburden Pressure
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Overburden pressure is the vertical pressure applied on a layer of rock from the rock and soil above it. Overburden pressure is sometimes called overburden stress, as well. If the materials making up a specific area, as well as the porosity and depths at which those materials are found is known, a few calculations can establish an approximation of local overburden pressure. To know more exactly, however, the area in question should undergo density logging.
Take a density log reading of the area you are measuring. This service can be contracted by professionals who possess the necessary equipment. The area to be measured will need to be dug out to create a well bore through to the lowest layer of rock to be measured (the overburden pressure established will be the overburden pressure being acted upon that particular rock layer). The engineers taking the density log measurement will emit gamma rays into the bore in order to establish the density (expressed as kilograms per cube meter) of the sample.
Multiply the density log reading value by the total depth (in meters) of the measured area.
Multiply the sum of Step 1 by .01 to obtain the overburden pressure.
#### Tip
If you can find data of the average rock density in your area, refer to this to cross check your own calculations.
#### Warning
Calculate every data point in either metric or English units, but not a combination. If any measurements exist in English units, convert them to metric.
References
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Darby Stevenson began writing in 1997 for his high-school newspaper, the "Alsea Valley Voice," which won him statewide awards for Best Feature Article and Best Personality Interview. He holds a Bachelor of Arts in international studies and a Bachelor of Arts in religious studies from the University of Oregon.
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https://gmat.kmf.com/question/bdi23k.html | 1,723,262,758,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00046.warc.gz | 227,017,734 | 12,705 | GMAT 考满分题库
Magoosh - - 7
Suppose n children are going to sit, in any order, on a row of n chairs. What is the value of n?
Statement #1: If one child leaves, and one chair is removed from the row, there would be 4320 fewer arrangements of the remaining children.
Statement #2: In this row, there are 5040 possible arrangements of the n children.
• AStatement 1 ALONE is sufficient to answer the question, but statement 2 alone is NOT sufficient.
• BStatement 2 ALONE is sufficient to answer the question, but statement 1 alone is NOT sufficient.
• CBOTH statements 1 and 2 TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
• DEach statement ALONE is sufficient to answer the question.
• EStatement 1 and 2 TOGETHER are NOT sufficient to answer the question.
|
• 按热度
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# A fair dice is thrown ten times one after another. What is
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A fair dice is thrown ten times one after another. What is [#permalink] 21 May 2003, 17:37
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A fair dice is thrown ten times one after another. What is the probability of getting an odd number on an even numbered throw and an even number on an odd numbered throw?
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[#permalink] 22 May 2003, 17:25
i think the problem is worth trying
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[#permalink] 22 May 2003, 22:12
don't quite understand the question? can you explain ?
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[#permalink] 22 May 2003, 22:51
first, third, fifth, seventh & ninth throw.....will give an even number on the dice
second, fourth, sixth, eigth and tenth throw...will give an odd number on the dice
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[#permalink] 23 May 2003, 02:26
Odd throws - five - each throw probability that even turns up is 1/2.
Even throws - five - similar to above probability that odd turns up on each throw is also 1/2
Hence probability that on Every odd throw , we get even = (1/2)^5
And Probability that on Every even throw , we get odd = (1/2)^5
Probability that both these events happen = (1/2)^5 * (1/2)^5
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[#permalink] 23 May 2003, 02:49
Nothing to comment about.........perfect
[#permalink] 23 May 2003, 02:49
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# A fair dice is thrown ten times one after another. What is
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 900 | 2,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2013-48 | longest | en | 0.885044 |
https://www.numbersaplenty.com/175756 | 1,725,998,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00377.warc.gz | 849,942,157 | 3,216 | Search a number
175756 = 2276277
BaseRepresentation
bin101010111010001100
322221002111
4222322030
521111011
63433404
71331260
oct527214
9287074
10175756
11110059
1285864
1361cc9
14480a0
1537121
hex2ae8c
175756 has 12 divisors (see below), whose sum is σ = 351568. Its totient is φ = 75312.
The previous prime is 175753. The next prime is 175757. The reversal of 175756 is 657571.
175756 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
175756 is an admirable number.
It is a hoax number, since the sum of its digits (31) coincides with the sum of the digits of its distinct prime factors.
It is an Ulam number.
It is not an unprimeable number, because it can be changed into a prime (175753) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3111 + ... + 3166.
2175756 is an apocalyptic number.
It is an amenable number.
175756 is a primitive abundant number, since it is smaller than the sum of its proper divisors, none of which is abundant.
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
175756 is a wasteful number, since it uses less digits than its factorization.
175756 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 6288 (or 6286 counting only the distinct ones).
The product of its digits is 7350, while the sum is 31.
The square root of 175756 is about 419.2326323177. The cubic root of 175756 is about 56.0148769998.
It can be divided in two parts, 1 and 75756, that added together give a palindrome (75757).
The spelling of 175756 in words is "one hundred seventy-five thousand, seven hundred fifty-six". | 493 | 1,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.890089 |
https://www.lmfdb.org/EllipticCurve/Q/1045/b/3 | 1,679,688,613,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945288.47/warc/CC-MAIN-20230324180032-20230324210032-00116.warc.gz | 964,359,330 | 28,451 | # Properties
Label 1045.b3 Conductor $1045$ Discriminant $47155625$ j-invariant $$\frac{104857852278310619039721}{47155625}$$ CM no Rank $1$ Torsion structure $$\Z/{2}\Z$$
# Related objects
Show commands: Magma / Oscar / PariGP / SageMath
## Simplified equation
$$y^2+xy=x^3-x^2-982409x-374543312$$ y^2+xy=x^3-x^2-982409x-374543312 (homogenize, simplify) $$y^2z+xyz=x^3-x^2z-982409xz^2-374543312z^3$$ y^2z+xyz=x^3-x^2z-982409xz^2-374543312z^3 (dehomogenize, simplify) $$y^2=x^3-15718547x-23986490514$$ y^2=x^3-15718547x-23986490514 (homogenize, minimize)
comment: Define the curve
sage: E = EllipticCurve([1, -1, 0, -982409, -374543312])
gp: E = ellinit([1, -1, 0, -982409, -374543312])
magma: E := EllipticCurve([1, -1, 0, -982409, -374543312]);
oscar: E = EllipticCurve([1, -1, 0, -982409, -374543312])
sage: E.short_weierstrass_model()
magma: WeierstrassModel(E);
oscar: short_weierstrass_model(E)
## Mordell-Weil group structure
$$\Z \oplus \Z/{2}\Z$$
magma: MordellWeilGroup(E);
### Infinite order Mordell-Weil generator and height
$P$ = $$\left(\frac{13348}{9}, \frac{1000658}{27}\right)$$ (13348/9, 1000658/27) $\hat{h}(P)$ ≈ $6.6552718759833089389679662935$
sage: E.gens()
magma: Generators(E);
gp: E.gen
## Torsion generators
$$\left(-572, 286\right)$$
comment: Torsion subgroup
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
oscar: torsion_structure(E)
## Integral points
$$\left(-572, 286\right)$$
comment: Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
## Invariants
Conductor: $$1045$$ = $5 \cdot 11 \cdot 19$ comment: Conductor sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); oscar: conductor(E) Discriminant: $47155625$ = $5^{4} \cdot 11 \cdot 19^{3}$ comment: Discriminant sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); oscar: discriminant(E) j-invariant: $$\frac{104857852278310619039721}{47155625}$$ = $3^{3} \cdot 5^{-4} \cdot 11^{-1} \cdot 13^{3} \cdot 19^{-3} \cdot 313^{3} \cdot 3863^{3}$ comment: j-invariant sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); oscar: j_invariant(E) Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) sage: E.has_cm() magma: HasComplexMultiplication(E); Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $1.7158240888248362330886097080\dots$ gp: ellheight(E) magma: FaltingsHeight(E); oscar: faltings_height(E) Stable Faltings height: $1.7158240888248362330886097080\dots$ magma: StableFaltingsHeight(E); oscar: stable_faltings_height(E)
## BSD invariants
Analytic rank: $1$ sage: E.analytic_rank() gp: ellanalyticrank(E) magma: AnalyticRank(E); Regulator: $6.6552718759833089389679662935\dots$ comment: Regulator sage: E.regulator() G = E.gen \\ if available matdet(ellheightmatrix(E,G)) magma: Regulator(E); Real period: $0.15164442574914819669892743864\dots$ comment: Real Period sage: E.period_lattice().omega() gp: if(E.disc>0,2,1)*E.omega[1] magma: (Discriminant(E) gt 0 select 2 else 1) * RealPeriod(E); Tamagawa product: $12$ = $2^{2}\cdot1\cdot3$ comment: Tamagawa numbers sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); oscar: tamagawa_numbers(E) Torsion order: $2$ comment: Torsion order sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); oscar: prod(torsion_structure(E)[1]) Analytic order of Ш: $1$ (exact) comment: Order of Sha sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Special value: $L'(E,1)$ ≈ $3.0277046455138353542434232564$ comment: Special L-value r = E.rank(); E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: [r,L1r] = ellanalyticrank(E); L1r/r! magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
## BSD formula
$\displaystyle 3.027704646 \approx L'(E,1) = \frac{\# Ш(E/\Q)\cdot \Omega_E \cdot \mathrm{Reg}(E/\Q) \cdot \prod_p c_p}{\#E(\Q)_{\rm tor}^2} \approx \frac{1 \cdot 0.151644 \cdot 6.655272 \cdot 12}{2^2} \approx 3.027704646$
# self-contained SageMath code snippet for the BSD formula (checks rank, computes analytic sha)
E = EllipticCurve(%s); r = E.rank(); ar = E.analytic_rank(); assert r == ar;
Lr1 = E.lseries().dokchitser().derivative(1,r)/r.factorial(); sha = E.sha().an_numerical();
omega = E.period_lattice().omega(); reg = E.regulator(); tam = E.tamagawa_product(); tor = E.torsion_order();
assert r == ar; print("anayltic sha: " + str(RR(Lr1) * tor^2 / (omega * reg * tam)))
/* self-contained Magma code snippet for the BSD formula (checks rank, computes analyiic sha) */
E := EllipticCurve(%s); r := Rank(E); ar,Lr1 := AnalyticRank(E: Precision := 12); assert r eq ar;
sha := MordellWeilShaInformation(E); omega := RealPeriod(E) * (Discriminant(E) gt 0 select 2 else 1);
reg := Regulator(E); tam := &*TamagawaNumbers(E); tor := #TorsionSubgroup(E);
assert r eq ar; print "analytic sha:", Lr1 * tor^2 / (omega * reg * tam);
## Modular invariants
$$q + q^{2} - q^{4} + q^{5} - 3 q^{8} - 3 q^{9} + q^{10} - q^{11} + 2 q^{13} - q^{16} - 6 q^{17} - 3 q^{18} + q^{19} + O(q^{20})$$
comment: q-expansion of modular form
sage: E.q_eigenform(20)
\\ actual modular form, use for small N
[mf,F] = mffromell(E)
Ser(mfcoefs(mf,20),q)
\\ or just the series
Ser(ellan(E,20),q)*q
magma: ModularForm(E);
Modular degree: 4416
comment: Modular degree
sage: E.modular_degree()
gp: ellmoddegree(E)
magma: ModularDegree(E);
$\Gamma_0(N)$-optimal: yes
Manin constant: 1
comment: Manin constant
magma: ManinConstant(E);
## Local data
This elliptic curve is semistable. There are 3 primes of bad reduction:
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$5$ $4$ $I_{4}$ Split multiplicative -1 1 4 4
$11$ $1$ $I_{1}$ Non-split multiplicative 1 1 1 1
$19$ $3$ $I_{3}$ Split multiplicative -1 1 3 3
comment: Local data
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
oscar: [(p,tamagawa_number(E,p), kodaira_symbol(E,p), reduction_type(E,p)) for p in bad_primes(E)]
## Galois representations
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2B 8.12.0.6
comment: mod p Galois image
sage: rho = E.galois_representation(); [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
gens = [[1665, 8, 1664, 9], [1, 0, 8, 1], [7, 6, 1666, 1667], [1, 8, 0, 1], [1, 4, 4, 17], [217, 212, 214, 1047], [1372, 1, 783, 6], [1467, 1466, 1058, 219], [624, 3, 181, 2]]
GL(2,Integers(1672)).subgroup(gens)
Gens := [[1665, 8, 1664, 9], [1, 0, 8, 1], [7, 6, 1666, 1667], [1, 8, 0, 1], [1, 4, 4, 17], [217, 212, 214, 1047], [1372, 1, 783, 6], [1467, 1466, 1058, 219], [624, 3, 181, 2]];
sub<GL(2,Integers(1672))|Gens>;
The image $H:=\rho_E(\Gal(\overline{\Q}/\Q))$ of the adelic Galois representation has level $$1672 = 2^{3} \cdot 11 \cdot 19$$, index $48$, genus $0$, and generators
$\left(\begin{array}{rr} 1665 & 8 \\ 1664 & 9 \end{array}\right),\left(\begin{array}{rr} 1 & 0 \\ 8 & 1 \end{array}\right),\left(\begin{array}{rr} 7 & 6 \\ 1666 & 1667 \end{array}\right),\left(\begin{array}{rr} 1 & 8 \\ 0 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 4 \\ 4 & 17 \end{array}\right),\left(\begin{array}{rr} 217 & 212 \\ 214 & 1047 \end{array}\right),\left(\begin{array}{rr} 1372 & 1 \\ 783 & 6 \end{array}\right),\left(\begin{array}{rr} 1467 & 1466 \\ 1058 & 219 \end{array}\right),\left(\begin{array}{rr} 624 & 3 \\ 181 & 2 \end{array}\right)$.
The torsion field $K:=\Q(E[1672])$ is a degree-$4815382$ Galois extension of $\Q$ with $\Gal(K/\Q)$ isomorphic to the projection of $H$ to $\GL_2(\Z/1672\Z)$.
## $p$-adic regulators
Note: $p$-adic regulator data only exists for primes $p\ge 5$ of good ordinary reduction.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 ord ss split ss nonsplit ord ord split ord ord ord ord ord ord ss 1 3,3 2 3,1 1 3 1 2 1 1 1 1 1 1 1,1 0 0,0 0 0,0 0 0 0 0 0 0 0 0 0 0 0,0
## Isogenies
gp: ellisomat(E)
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2 and 4.
Its isogeny class 1045.b consists of 4 curves linked by isogenies of degrees dividing 4.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $2$ $$\Q(\sqrt{209})$$ $$\Z/2\Z \oplus \Z/2\Z$$ Not in database $2$ $$\Q(\sqrt{-19})$$ $$\Z/4\Z$$ Not in database $2$ $$\Q(\sqrt{-11})$$ $$\Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{-11}, \sqrt{-19})$$ $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $4$ 4.0.75449.1 $$\Z/8\Z$$ Not in database $8$ 8.4.21336229885501696.2 $$\Z/2\Z \oplus \Z/4\Z$$ Not in database $8$ 8.0.163720581376.3 $$\Z/8\Z$$ Not in database $8$ 8.0.688798743721.1 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database $8$ 8.2.20012416875.1 $$\Z/6\Z$$ Not in database $16$ deg 16 $$\Z/4\Z \oplus \Z/4\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/8\Z$$ Not in database $16$ deg 16 $$\Z/16\Z$$ Not in database $16$ deg 16 $$\Z/2\Z \oplus \Z/6\Z$$ Not in database $16$ deg 16 $$\Z/12\Z$$ Not in database $16$ deg 16 $$\Z/12\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial. | 3,798 | 9,625 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.200708 |
http://www.efunda.com/glossary/units/units--luminous_luminance--lambert.cfm | 1,566,534,342,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317847.79/warc/CC-MAIN-20190823041746-20190823063746-00017.warc.gz | 250,141,426 | 7,355 | Luminous Luminance Units: Lambert
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Glossary » Units » Luminous Luminance » Lambert
Lambert (L) is a unit in the category of Luminous luminance. It is also known as lamberts. This unit is commonly used in the US unit system. Lambert (L) has a dimension of L-2J where L is length, and J is luminous intensity. It can be converted to the corresponding standard SI unit cd/m2 by multiplying its value by a factor of 3183.
Note that the seven base dimensions are M (Mass), L (Length), T (Time), Q (Temperature), N (Aamount of Substance), I (Electric Current), and J (Luminous Intensity).
Other units in the category of Luminous luminance include Apostilb (asb), Candela Per Square Meter (cd/m2), Candle Per Square Centimeter (candle/cm2), Candle Per Square Foot (candle/ft2), Candle Per Square Inch (candle/in2), Candle Per Square Meter (candle/m2), Foot-Lambert (ft-L), Nit (nit), and Stilb (sb). | 288 | 1,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | longest | en | 0.824202 |
https://www.shaalaa.com/question-bank-solutions/x-d-y-d-x-x-cos-2-y-x-y-general-particular-solutions-differential-equation_66749 | 1,619,104,391,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039610090.97/warc/CC-MAIN-20210422130245-20210422160245-00227.warc.gz | 1,092,161,373 | 9,786 | # X D Y D X + X Cos 2 ( Y X ) = Y - Mathematics
Sum
$x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y$
#### Solution
We have,
$x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y$
$\frac{dy}{dx} + \cos^2 \left( \frac{y}{x} \right) = \frac{y}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{y}{x} - \cos^2 \left( \frac{y}{x} \right)$
Putting y = vx, we get
$\frac{dy}{dx} = v + x\frac{dv}{dx}$
$\therefore v + x\frac{dv}{dx} = v - \cos^2 \left( v \right)$
$\Rightarrow x\frac{dv}{dx} = - \cos^2 v$
$\Rightarrow \sec^2 v\ dv = - \frac{1}{x}dx$
Integrating both sides, we get
$\int sec^2 v\ dv = - \int\frac{1}{x}dx$
$\Rightarrow \tan v = - \log x + C$
$\Rightarrow \tan \frac{y}{x} = - \log x + C$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 54 | Page 146 | 354 | 877 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-17 | latest | en | 0.358912 |
http://www.scientificamerican.com/article.cfm?id=dna-sudoku | 1,369,219,065,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701577515/warc/CC-MAIN-20130516105257-00055-ip-10-60-113-184.ec2.internal.warc.gz | 694,636,342 | 29,474 | # DNA Sudoku
Researchers get help from a venerable number theory and a popular puzzle game to solve genetic medical mysteries
FILLING IN THE BLANKS: Could pinpointing genetic mutations be as easy as 1-2-3, er, A-C-G-T? Image: FLICKER/EDUARDO MUESES
• ### The Best Science Writing Online 2012
Showcasing more than fifty of the most provocative, original, and significant online essays from 2011, The Best Science Writing Online 2012 will change the way...
A 2,000-year-old math theorem, along with Sudoku, may soon help researchers untangle DNA at blazing speeds.
Hunting for a particular genetic mutation in hundreds of thousands of specimens can be an expensive and time-consuming process. In the past several years, faster multiplex DNA sequencing machines have sped up the acquisition of data, but researchers have still been hobbled by having to label each sample with a unique molecular identifier (or bar code) for analysis.
Scientists at Cold Spring Harbor Laboratory (CSHL) in Long Island, N.Y., are proposing a new take on a very old idea to tackle large data sets simultaneously. The team is applying the Chinese remainder theorem to pinpoint single samples in larger pools, which are arranged in rows and columns.
Invented about 2,000 years ago, the theorem is a method for mapping information using prime and co-prime numbers. In the case of DNA sequencing and Sudoku, the theorem is used to organize data points with coordinates in a box, but it can also be used to figure out all sorts of missing information in other domains, such as distant points sensed with high-speed radar, pieces of code, and who that attractive person was that you saw at three out of seven parties on a cruise ship.
By using the idea, researchers can deal with whole libraries of genetic information instead of looking at just "one genetic sequence at a time," says Yaniv Erlich, the lead author of the paper, published as the cover story of this month's Genome Research.
In Sudoku players must fill every row and column each with all nine numerals, but in applying this to so many genetic samples to search, the researchers call on state-of-the-art robots, machines and programs to do the specimen placing and searching for them. "Every cell in a Sudoku [puzzle] is like a specimen, and every digit is like a genotype," says Erlich, a doctoral student who had used the Chinese remainder theorem in previous work with radar. He brought the idea to the attention of his CSHL professor Greg Hannon.
View
1. 1. PaoloMagrassi 12:36 PM 7/2/09
Chinese Remainder Theorem [Britannica]: ancient theorem that gives the conditions necessary for multiple equations to have a simultaneous integer solution. [...] It addresses the following type of problem. One is asked to find a number that leaves a remainder of 0 when divided by 5, remainder 6 when divided by 7, and remainder 10 when divided by 12. The simplest solution is 370. Note that this solution is not unique, since any multiple of 5 (= 420) can be added to it and the result will still solve the problem.
2. 2. riverboots 05:11 PM 7/2/09
Does the Chinese Remainder Theorem become case-sensitive in neurological conformity?
3. 3. emueses 10:46 AM 7/6/09
Interesting article. BTW, thanks for using my picture.
4. 4. jwr257 12:50 PM 7/12/09
Here is an easier solution..use the idle time on your computer (Windows, Mac, or Linux) to cure diseases, study global warming, discover pulsars, and do many other types of scientific research. It's safe, secure, and easy by using BOINC at http://boinc.berkeley.edu/. BOINC ( The Berkeley Open Infrastructure for Network Computing) is a non-commercial middleware system for volunteer and grid computing.
5. 5. Benedict9 in reply to riverboots 12:56 AM 7/13/09
Why naturally!
6. 6. ervedge 12:59 AM 7/30/09
The reference/url for "the paper" by Erlich should be:
http://genome.cshlp.org/content/19/7/1243
http://hannonlab.cshl.edu/dna_sudoku/main.html
7. 7. bertwindon 02:35 PM 4/3/10
Baloni
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