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https://www.geeksforgeeks.org/program-to-find-the-value-of-tann/?ref=rp | 1,618,648,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00141.warc.gz | 869,512,762 | 23,821 | Related Articles
Program to find the value of tan(nΘ)
• Last Updated : 23 Aug, 2020
Given a value of tan(Θ) and a variable n <=15. The task is to find the value of tan(nΘ) using property of trignometric functions.
Examples:
Input: tan(Θ) = 0.3, n = 10
Output: -2.15283
Input: tan(Θ) = 0.3, n = 5
Output: 0.37293
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
This problem can be solved using De moivre’s theoremand Binomial theorem
Below is the implementation of above approach:
## C++
// C++ program to find the value // of cos(n-theta) #include #define ll long long int#define MAX 16using namespace std;ll nCr[MAX][MAX] = { 0 }; // This function use to calculate the// binomial cofficient upto 15void binomial(){ // use simple DP to find cofficient for (int i = 0; i < MAX; i++) { for (int j = 0; j <= i; j++) { if (j == 0 || j == i) nCr[i][j] = 1; else nCr[i][j] = nCr[i - 1][j] + nCr[i - 1][j - 1]; } }} // Function to find the value ofdouble findTanNTheta(double tanTheta, ll n){ // store required answer double ans = 0, numerator = 0, denominator = 0; // use to toggle sign in sequence. ll toggle = 1; // calculate numerator for (int i = 1; i <= n; i += 2) { numerator = numerator + nCr[n][i] * pow(tanTheta, i) * toggle; toggle = toggle * -1; } // calculate denominator denominator = 1; toggle = -1; for (int i = 2; i <= n; i += 2) { numerator = numerator + nCr[n][i] * pow(tanTheta, i) * toggle; toggle = toggle * -1; } ans = numerator / denominator; return ans;} // Driver code.int main(){ binomial(); double tanTheta = 0.3; ll n = 10; cout << findTanNTheta(tanTheta, n) << endl; return 0;}
## Java
// Java program to find the value // of cos(n-theta) public class GFG { private static final int MAX = 16 ; static long nCr[][] = new long [MAX][MAX] ; // This function use to calculate the // binomial coefficient upto 15 static void binomial() { // use simple DP to find cofficient for (int i = 0; i < MAX; i++) { for (int j = 0; j <= i; j++) { if (j == 0 || j == i) nCr[i][j] = 1; else nCr[i][j] = nCr[i - 1][j] + nCr[i - 1][j - 1]; } } } // Function to find the value of static double findTanNTheta(double tanTheta, int n) { // store required answer double ans = 0, numerator = 0, denominator = 0; // use to toggle sign in sequence. long toggle = 1; // calculate numerator for (int i = 1; i <= n; i += 2) { numerator = numerator + nCr[n][i] * Math.pow(tanTheta, i) * toggle; toggle = toggle * -1; } // calculate denominator denominator = 1; toggle = -1; for (int i = 2; i <= n; i += 2) { numerator = numerator + nCr[n][i] * Math.pow(tanTheta, i) * toggle; toggle = toggle * -1; } ans = numerator / denominator; return ans; } // Driver code public static void main(String args[]) { binomial(); double tanTheta = 0.3; int n = 10; System.out.println(findTanNTheta(tanTheta, n)); }// This code is contributed by ANKITRAI1 }
## Python3
# Python3 program to find the value # of cos(n-theta) import math MAX=16nCr=[[0 for i in range(MAX)] for i in range(MAX)] # Function to calculate the binomial # cofficient upto 15 def binomial(): # use simple DP to find cofficient for i in range(MAX): for j in range(0,i+1): if j == 0 or j == i: nCr[i][j] = 1 else: nCr[i][j] = nCr[i - 1][j] + nCr[i - 1][j - 1] # Function to find the value of cos(n-theta) def findTanNTheta(tanTheta,n): # store required answer numerator=0 denominator=1 # to store required answer ans = 0 # use to toggle sign in sequence. toggle = 1 # calculate numerator for i in range(1,n+1,2): numerator = (numerator + nCr[n][i]* (tanTheta**(i)) * toggle) toggle = toggle * -1 # calculate denominator toggle=-1 for i in range(2,n+1,2): numerator = (numerator + nCr[n][i]* (tanTheta**i) * toggle) toggle = toggle * -1 ans=numerator/denominator return ans # Driver code if __name__=='__main__': binomial() tanTheta = 0.3 n = 10 print(findTanNTheta(tanTheta, n)) # this code is contributed by sahilshelangia
## C#
// C# program to find the value // of cos(n-theta) using System;public class GFG { private static int MAX = 16 ; static long[,] nCr = new long [MAX,MAX] ; // This function use to calculate the // binomial coefficient upto 15 static void binomial() { // use simple DP to find cofficient for (int i = 0; i < MAX; i++) { for (int j = 0; j <= i; j++) { if (j == 0 || j == i) nCr[i,j] = 1; else nCr[i,j] = nCr[i - 1,j] + nCr[i - 1,j - 1]; } } } // Function to find the value of static double findTanNTheta(double tanTheta, int n) { // store required answer double ans = 0, numerator = 0, denominator = 0; // use to toggle sign in sequence. long toggle = 1; // calculate numerator for (int i = 1; i <= n; i += 2) { numerator = numerator + nCr[n,i] * Math.Pow(tanTheta, i) * toggle; toggle = toggle * -1; } // calculate denominator denominator = 1; toggle = -1; for (int i = 2; i <= n; i += 2) { numerator = numerator + nCr[n,i] * Math.Pow(tanTheta, i) * toggle; toggle = toggle * -1; } ans = numerator / denominator; return ans; } // Driver code public static void Main() { binomial(); double tanTheta = 0.3; int n = 10; Console.Write(findTanNTheta(tanTheta, n)); } }
## PHP
Output:
-2.15283
Want to learn from the best curated videos and practice problems, check out the C++ Foundation Course for Basic to Advanced C++ and C++ STL Course for foundation plus STL.
My Personal Notes arrow_drop_up | 2,054 | 6,602 | {"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} | 3.765625 | 4 | CC-MAIN-2021-17 | latest | en | 0.457412 |
http://exxamm.com/blog/Blog/13660/zxcfghfgvbnm4?Class%2012 | 1,558,681,159,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257553.63/warc/CC-MAIN-20190524064354-20190524090354-00037.warc.gz | 69,088,547 | 16,625 | Mathematics Derivatives of implicit functions and Derivatives of inverse trigonometric functions For CBSE-NCERT
Click for Only Video
### Topic covered
♦ Derivatives of implicit functions
♦ Derivatives of inverse trigonometric functions
### Derivatives of implicit functions :
=> Now we use chain rule to find derivatives of these functions.
Q 3115423369
Find (dy)/(dx) if x – y = π.
Class 12 Chapter 5 Example 24
Solution:
One way is to solve for y and rewrite the above as
y= x - pi
But then (dy)/(dx) =1
Q 3115523460
Find (dy)/(dx), if y + sin y = cos x.
Class 12 Chapter 5 Example 25
Solution:
We differentiate the relationship directly with respect to x, i.e.,
(dy)/(dx) + d/(dx) (sin y) = d/(dx) (cos x)
which implies using chain rule
(dy)/(dx) + cos y * (dy)/(dx) = - sin x
This gives (dy)/(dx) = - (sin x)/(1+cos y)
where y ≠ (2n + 1) π
### Derivatives of inverse trigonometric functions :
● We remark that inverse trigonometric functions are continuous functions, Now we use chain rule to find derivatives of these functions.
● E.g., : Find the derivative of f given by f (x) = tan^(–1) x assuming it exists.
Let y = tan^(–1) x.
Then, x = tan y.
=> Differentiating both sides w.r.t. x, we get
1=sec^2 y (dy)/(dx)
which implies that
(dy)/(dx) = 1/(sec^2 y) = 1/(tan^2 y) =1/(1+ (tan(tan^(-1) x))^2) =1/(1+x^2)
Q 3145623563
Find the derivative of f given by f (x) = sin^(–1) x assuming it exists.
Class 12 Chapter 5 Example 26
Solution:
Let y = sin^(–1) x. Then, x = sin y.
Differentiating both sides w.r.t. x, we get
1 = cos y (dy)/(dx)
which implies that (dy)/(dx) = 1/(cos y) = 1/(cos (sin^(-1) x) )
Observe that this is defined only for cos y ≠ 0, i.e., sin^(–1) x ≠ - pi/2 , pi/2 , i.e., x ≠ – 1, 1,
i.e., x ∈ (– 1, 1).
To make this result a bit more attractive, we carry out the following manipulation.
Recall that for x ∈ (– 1, 1), sin (sin^(–1) x) = x and hence
cos^2 y = 1 – (sin y)^2 = 1 – (sin (sin^(–1) x))^2 = 1 – x^2
Also, since y ∈ ( -pi/2, pi/2) , cos y is positive and hence cos y = sqrt ( 1-x^2)
Thus, for x ∈ (– 1, 1),
(dy)/(dx) = 1/(cos y ) = 1/(sqrt (1-x^2) )
Q 3105623568
Find the derivative of f given by f (x) = tan^(–1) x assuming it exists.
Class 12 Chapter 5 Example 27
Solution:
Let y = tan^(–1 )x. Then, x = tan y.
Differentiating both sides w.r.t. x, we get
1= sec^2 y (dy)/(dx)
which implies that
(dy)/(dx) = 1/(sec^2 y) =1/(1+ tan^2 y) = 1/( 1+(tan( tan^(-1) x ))^2 ) = 1/(1+x^2)
Finding of the derivatives of other inverse trigonometric functions is left as exercise.
The following table gives the derivatives of the remaining inverse trigonometric functions
(Table 5.4): | 916 | 2,659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2019-22 | latest | en | 0.769332 |
https://community.deeplearning.ai/t/batch-norm-reducing-internal-covariate-shift/455263 | 1,716,279,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00126.warc.gz | 147,275,054 | 6,873 | # Batch Norm reducing internal covariate shift
Without batch norm, the mean and variance of Z[l] is determined by a complex interaction between parameters and activations in the layers before l. This distribution fluctuates a lot. However, with batch norm the mean and variance can be controlled with beta and gamma. The learning algo learns good values of beta and gamma and we get a relatively consistent distribution for each mini batch thereby reducing internal covariate shift.
My question is how can we get a consistent distribution if beta and gamma also keep changing in every iteration of Gradient Descent? With fluctuating beta and gamma, the distribution also fluctuates.
I understand that forcing mean=0 and variance=1 for all units in all layers reduces the expressive power of the neural network and it cannot learn a good mapping from input to output. But I don’t see how a constantly fluctuating beta and gamma can give a consistent distribution. What’s to stop beta and gamma to keep changing throughout training and keep changing distribution just like it was before batch norm?
Appreciate any clarification
As with other learning algorithms, the optimizer & learning rate will play a vital role in determining the magnitude of change of the parameters (\beta and \gamma). Changes should be minor if the layer encounters many batches of data are similarly distributed.
Thanks for your response! I have another question.
It goes from a distribution of mean=0 and variance=1 at the beginning of training to whatever the ideal mean and variance are eventually. Does this mean beta and gamma reach roughly optimal values early in the optimization process? So that during the rest of the training they don’t fluctuate much and thereby we have a steady distribution and can learn W,b much faster.
What if the mean (0) and variance(1) we start with are very very far from what is ideal for a particular Z. It might take very long for beta, gamma to reach good values. Is the hope that for the majority of the Zs we reach good values of beta, gamma pretty fast so that for majority of the optimization process we are working with reasonably steady distribution?
With mini batches having similar distributions and W,b updating only slightly at each iteration (controlled by learning rate, etc), why do we even need batch norm to begin with? It seems like there is very little change in distribution from iteration to iteration anyway.
\beta and \gamma are learnt like any other NN problem.
Wild fluctiation of the parameters IMO means that the learning rate is too high or the less possible, dataset isn’t well shuffled (I haven’t dug that deep to observe this problem).
While learning will almost flat out if there are many similarly distributed batches, we want to account for the reality that not all mini batches are similarly distributed.
Consider a simple linear regression problem with 1 input feature. We don’t pick 2 points and draw a line that call it y. The line of best fit is one that has the least error like MSE. It’s the same idea here. We learn the best rescaling parameters from mini batches of data.
1. If we choose to have a small learning rate, then the beta and gamma change only slightly from iteration to iteration thereby giving us a relatively steady distribution to deal with even as beta and gamma march slowly towards their optimum value.
2. If learning rate is on the larger side, beta and gamma update by bigger values. So until they reach near optimum values we pay a price of having more internal covariate shift. However we don’t pay this price for too long as beta, gamma reach their optimal values much sooner and from that point on we have low ICS for the rest of training.
Would this be a fair way to summarize it?
This is not necessarily true since there’s nothing stopping you from providing a really high learning rate for rapid convergence.
Tuning learning rate (which I’ve never done from what tensorflow provides for this layer) has similar meaning to that of a regression problem. If learning rate is too low, it’s takes more iterations for model to reduce loss and arrive at the
local minimum loss. If learning rate is too high, overshooting and bouncing around the local minima is something to be aware of.
Correct. I guess I meant as high learning rare as possible with which we can still converge ideally. Not extremely high where we could have oscillations and potentially have divergence.
Sure. Learning rates have a lot of schedules. So, as long as you’ve got the hang of this intuition, you’re good to go. | 909 | 4,586 | {"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} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.921001 |
http://flaguide.org/tools/math/fault/fault3.htm | 1,723,785,156,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00825.warc.gz | 10,643,970 | 6,455 | ## Tools - Math 'Fault Finding and Fixing' Interpreting and Misinterpreting Data Tasks, Set #3
Percentages, Set #1 (solutions) || Combinations and Chance, Set #2 (solutions)
Interpreting and Misinterpreting Data: Set #3 (solutions) || Set #4 (solutions)
Malcolm Swan
Mathematics Education
University of Nottingham
Malcolm.Swan@nottingham.ac.uk
Jim Ridgway
School of Education
University of Durham
Jim.Ridgway@durham.ac.uk
Each question contains a selection of errors or misleading interpretations of data. The aim of this assessment is to provide the opportunity for you to: explain clearly the source of each error or misinterpretation. rectify the errors and produce correct interpretations.
1. Along a country road This graph shows a car and a motorbike travelling along a country road. What is wrong with the following statement? I think that they are travelling at the same speed after 4 seconds. You can tell that because the graphs cross.
2. Swimming pool The graph above shows the progress of a swimming race. Here is a commentary of the race. Highlight the mistakes in this commentary and write a better one. Sam goes quickly into the lead. He is swimming at 15 metres per second. Janet is swimming at only 10 metres per second. After 22 seconds, Janet overtakes Sam. Janet swims more quickly than Sam from 25 seconds until she turns at 50 seconds. Sam overtakes Janet after 55 seconds, but she catches up again. 5 seconds later, Janet is in the lead until right near the end. Sam swims at a steady 30 metres per second after the turn, until 80 seconds, while Janet is gradually slowing down. Sam wins by 10 seconds.
Explain clearly how you know that an error has been made.
Show how the error should be put right.
3. College magazine Karl is thinking of producing a college magazine. He produces a prototype of the magazine and conducts a small survey to compare male and female opinions of it. He asks the following question among a random sample of students: Would you pay a dollar for this magazine? The results are shown below.
He concludes that females are less likely to buy the magazine than males.
Explain why Karl is wrong and say what a sensible conclusion would be.
4. Car and Bicycle Production The diagram below shows how the world production of cars and bicycles has changed from 1965 to 1995.
Explain, with reasons, whether or not you think that this diagram fairly represents the numerical information given
Percentages, Set #1 (solutions) || Combinations and Chance, Set #2 (solutions)
Interpreting and Misinterpreting Data: Set #3 (solutions) || Set #4 (solutions)
Introduction || Assessment Primer || Matching Goals to CATs || CATs || Tools || Resources
Search || Who We Are || Site Map || Meet the CL-1 Team || WebMaster || Copyright || Download
College Level One (CL-1) Home || Collaborative Learning || FLAG || Learning Through Technology || NISE | 651 | 2,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.941599 |
http://www.teachwithme.com/blogs/artscrafts-a-activities/itemlist/tag/%20Math%20Games | 1,386,409,998,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053921/warc/CC-MAIN-20131204131733-00061-ip-10-33-133-15.ec2.internal.warc.gz | 551,645,354 | 28,209 | ## Dice Games To Reinforce Math Skills
1 2 3 Come Do Some Nice Dice Activities With Me!
MY Y5'S LOVED playing with dice. I did all sorts of fun activities with them to help reinforce math concepts with numbers 1 to 6, so I decided to design a dice packet complete with cards and activities. Click on the link to view/download this fun packet. Dice Activities That Teach Math Skills.
Dice are a wonderful vehicle for teaching your kiddo's to subitize. Subitizing, was coined in 1949 by E.L. Kaufman. The term is derived from the Latin adjective subitus which means "sudden". A person who has affectively mastered this skill immediately knows how many items there are, without having to stop and count them.
According to studies most people can subitize up to 10. Dominoes are also a fun way to get subitizing practice in. Click on the link for my Dominoe Math Packet.
With that in mind, I thought it would be helpful to have a set of big dice flashcards to use for practice. Print, laminate & trim the cards and fasten them together with a split ring. Flash a card and have children call out that number. To whole-group assess, flash a card and have children silently hold up that many fingers. You can tell at a glance who is having difficulty.
The packet includes a set of large teacher dice cards, a smaller set for students to sequence, + a mini set so you can play a whole-group game of "Show Me What I Need To Make __________." Teacher holds up her big card and asks children to show them what they need to make another number. i.e. I hold up the #2 dice, and ask children to show me what other dice they need to make the sum of 5. They would hopefully show me the #3 card.
I've also included math symbol cards, so students can make equations, a bookmark you can use as a whole-group assessment game, a roll & dot dice game, 2 trace-write and match worksheets, + a What's Missing? activity.
Laminate a set of bookmarks and use them for another math dice activity. Review the numbers orally and have children point to that number and count with you. You can count from a certain number up to 6 or even count backwards.
Make extra copies of the medium-sized cards so students can play a Memory Match game. They can match the dice to the number box, or the number word, or all three. I've also included a cover so students can sequence the cards and make an Itty Bitty booklet. There's a separate set of dice-number-number word cards to print, laminate and cut into puzzles too.
These are a wonderful whole-group assessment tool too. Give students one M&MM (mighty math marker) to move to whatever number is called out. After glancing around, jot down names of children and the numbers they are having problems identifying. I used sticky notes and a clipboard. After the game, students can eat their candy.
Children can also practice one-to-one correspondence, by having them place however many pony beads or other small items, onto the square that will match the number amount on the dice picture. Click on the link to view/download the Dice Math Packet
As far as dice are concerned, I really like the large foam dice that they sell at The Dollar store. They are easy for little ones to hold, don't fly on the floor as much, and are blessedly quiet! If your Dollar Store doesn't have them, you can also purchase them from Oriental Trading. They are only \$4 for a dozen. They come in an assortment of rainbow colors, so i also used them for patterning.
Another quiet way I had my students "roll dice" was to recycle those mini water bottles. I'd toss two dice inside, fill with water and a bit of glitter and glue the caps shut with Gorilla Glue.
Students enjoyed shaking up the dice and then peeking on the bottom to see what their numbers were. Use a drop of food coloring or a pinch of plastic seasonal confetti, for extra pizzazz or to make special ones for Halloween, Valentine's Day etc.
I wanted to include a photo here, so I Googled waterbottle dice and found a teacher who also uses them, over at Kids Count. Shari has some math FREEBIES using dice as well. Click on the link to check out her wonderful creativity.
As mentioned yesterday, some clever person has come up with a little dice INSIDE a larger dice. Woo hoo for creativity. I'm sure they'll be a hit with your kiddo's. You can get a pack of 8 for only \$2.28 from Pure Fun or \$2.69 from On The Fly Supply.
One of my favorite ways to review the numbers on a dice was with a "magic trick". I'd use a big foam dice and choose a child. They'd come up to the front of the class, look at the dice and choose a number they wanted to show the other children.
I reminded the class NOT to shout out the answer, or they'd ruin the trick. Carefully, so they didn't reveal the face of the dice and the number to me, they'd keep it facing the class and hold it above their head. I stood behind the child so I could see the number on the back of the dice. I'd pretend to be "reading" their minds and then ask: "Are you looking at the number 3?"
I also had a dice and would show them that number. To their utter amazement they were looking at that number! "Do it again! Do it again!" could be heard, as well as, "How did you do that?" I did not reveal the answer to the trick 'til I was done using this as a number review game. I told my students I'd let them know the answer, when everyone could recognize numbers 1 to 6, then they could practice and do the trick for their families.
One of the parents of my Y5's told me at conferences that her son Garret couldn't wait to find out. She asked about the trick, so I showed her and shared the secret. Karen taught high school math and wondered how she could do it with her students. I told her to use it as a math problem. Demonstrate the trick and then have students try and figure out how it was mathematically done. She reported back that it was a HUGE success, and has used it every year!
The secret? The front and back numbers of a dice, when added together, will always-equal 7, so if you are looking at the number 5, your students will be looking at the number 2. Cool huh? I hope you have as much fun with this as I do.
I found this photo of a tot with a jumbo dice and thought that would be a really fun size for this activity. Even after searching, I could not find a source to buy just one jumbo dice. I found really humongous "cheese" ones with green dots (Go Packers!), but nothing this size. Anyone out there know? You can leave a comment here, or shoot me an e-mail: This email address is being protected from spambots. You need JavaScript enabled to view it.
Thanks for visiting today. I design and blog daily, so I hope you can stop by again tomorrow for the newest FREEBIES. You can PIN anything from my site. Just think how much easier our lives would be, if more people made the time to share.
To ensure that "pinners" return to THIS blog article, click on the green title at the top; it will turn black, now click on the "Pin it" button on the burgundy menu bar. If you'd like to take a peek at all of the wonderfully educational items I pin, click on the heart button to the right of the blog.
"I am learning all of the time. My tombstone will be my diploma." -Eartha Kitt
## Spider Shape Activities
1-2-3 Come Do Some Spider Stuff With Me!
Even though I am absolutely creeped out by spiders, I LOVED teaching our spider unit to my Y5's. These spiders were cute and not creepy. The reason I hate real ones, is a huge pine spider dropped from the ceiling onto my shoulder, when I was lying on a cot at our cottage. I was only 5, but I still remember it. Yikes!
Anything I design with shapes seems to be downloaded quite often, so I decided to whip together some 2D flat shape activities, featuring some sweet spiders. These lessons are quite versatile. Use them for independent math centers, table top lessons, a Daily 5 option, review, game, or even a whole-group assessment!
Inky is a quick and easy "craftivity." Students trace, cut and glue their spider slider together. Add some wiggle eyes for extra pizzazz and have students trace and color the shapes. Cut slits and insert the shape slider.
Teacher calls out a shape and children slide their strip up and down 'til they locate Inky's "tongue." If you want to whole-group assess, have students show you their answer. Click on the link to view/download the spider slider shape craftivity.
Peek-A-Eek is another "craftivity" that you can simply make for yourself and share as a read-aloud to review the basic 2D-flat shapes. I used a file folder to make my easy-reader sturdier.
If you want your kiddo's to have their own, simply trim some folders and have them glue the cover (circle web page) to the front, and the hexagon web page to the inside.
Make a fluffy spider, by gluing a black pom pom to the center of the hexagon shape. This is the last page. Trim and assemble the rest of the pages. Cut the "web window" shapes out so that the spider will peek through all of the pages. Click on the link to view/download Peek-A-Eek the spider shape booklet.
Spin A Spider is also quick and easy. Your little ones will enjoy taking turns spinning. Whatever shape they spin, they color or bingo dot the matching spider on their web.
I've included spider cards with the shapes as well as the shape words on them. Laminate and trim into puzzles
Besides putting together a puzzle, use the cards for a Memory Match, or "I Have; Who Has?" game. There's also a "Match the spider shape to the shape word" activity. Students can use the spinner to fill in this worksheet as well. Click on the link to view/download the Spin A Spider game packet.
Finally, I made a Spider Shape game, that matches the other themed ones that have been so popular.
Run off the shape tiles on a variety of colors of construction paper; laminate and trim. Students place the tile onto the matching spider card. Click on the above link to view/download.
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## Math Activities For October
1-2 3 Come Do Some Skelton Activities With Me!
I was diddling around with the idea of making a math packet around the play on words "Numb Skulls." Since it's October it seemed fitting to plug in a few skeletons. If you don't do "Halloween" themed things, the skulls are perfect for a pirate theme too, or perhaps you can use them as centers when your kiddo's study about bones and the human body.
I think your students will enjoy rolling 2 dice to make additon or subtraction equations on their "Numb Skull" and then solving them. They write in their answer and color that many teeth. Students can play independently or with a partner. Once I started designing with the skulls, more ideas kept popping into mine, 'til I had a whopping 46-page Numb Skull packet, that covers a variety of Common Core State Standards
Lots of the items are versatile. The number cards with number words, can be cut into puzzles, or run off so students can make an Itty Bitty Counting booklet, which is a nice activity for your Daily 5 word work. You can also use them for a Memory Match game, or to play "I Have; Who Has?" Add the bomb cards to make things more exciting.
The packet includes:
• A Numb Skull slider. Students trace the numbers from 0-30, or insert a skip counting by 2's, 3's, 5's, or 10's number strip. There's also a slider for counting backwards from 10 to 0 and 20 to 0.
• A Numb Skull addition and subtraction game.
• A Count to 100 Numb Skull game. Students add the dice that they roll and X-off that many skulls 'til they have added their way to 100.
• Trace and Write the numbers from 0-120 worksheets
• What's Missing worksheets for numbers 0-120, and all of the skip counted numbers
• Skull number cards from 0-120 + matching math symbol cards so students can make equations.
• Blank skull cards to program with whatever, or use to make groups/sets.
• Odd Todd and Even Steven skeleton sorting mats.
• Numb Skull puzzle cards. Cut them into puzzles, and/or run them off so students can make an Itty Bitty Booklet.
• Bomb cards to make number games more exciting.
• 3 skull number strip puzzles: 1-10, 10 to 1, and skip counting by 10's to 100. Use them as puzzles, or run them off and have students cut and glue them to a sheet of black construction paper, leaving a space between for an interesting look.
• A certificate of praise for a job well done.
Since I get quite a few requests for telling time activities, I decided to whip together a Numb Skull clock and a few telling time to the hour and half hour games. The packet includes analog as well as digital time cards that you can use as flashcards, or to play games with. Click on the link to view/down load the It's Numb Skull Time packet.
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"One man who has a mind and knows it, can always beat ten men who haven't and don't." -George Bernard Shaw
## Leaf Games
1-2-3 Come Play Some Leaf Games With Me!
You can make all sorts of number games and math centers with these leaf cards. Print, laminate and trim. There are 2 sets of cards: the bear with a leaf, as well as the yellow maple leaf. Students can play independently or with a partner. Children can match number cards to number word cards, or mix and match the sets and match numbers to numbers etc.
Besides the Memory Match games, toss a set of cards in a basket and have children choose one to play I Have; Who Has? "I have the number one card; who has the matching number one word card?" Add the "Kaboom" bomb cards, to make the game even more fun. There are many more games and ideas listed in the 3-page tip-list that's included in the packet.
I've also included mini-leaf tiles. so students can choose a numbered leaf card and count out that many leaves. They can sort odd & even numbers onto a leaf math mat, (included) or use the leaf math symbol cards to make addition and subtraction equations, or show greater and less than.
If you'd like your students to sequence and collate the cards into their own itty bitty booklet, run off the cards plus the cover master. Click on the link to view/download the Leaf Math Game packet.
For more leaf game fun, you can prit off a set of alphabet leaf cards. There's a set of separate uppercase and lowercase letter cards too, as well as a blank set for you to program with whatever. A "What Else Can I Do With the Cards?" is a list of other ideas and games you can play with the alphabet leaf cards. Click on the link to view/download the Alphabet Leaf Cards
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"The universe is transformation. Our life is what our thoughts make it." -Marcus Aurelius
## Back To School Bubble Blowing Math Activities
1-2-3 Come Play Some Interesting and Fun Math Games With Me!
Are you looking for some quick and easy ideas to do for the 1st day of school? Then I think you'll enjoy these simple bubble activities.
Start things off, by leaving the "I'm bubbling with excitement that you are in my class." bookmarks, as a cute surprise left on your students' desks. I found this sweet saying on Pinterest as valentine cards with heart bubbles. Click on the link to check out this creative teacher's original post. Adding a small bottle of bubbles is an inexpensive way to help make children feel especially welcome. (The Dollar Store sells 3 to 6 in a pack. You can also buy a box of 20 mini wedding bubbles at most craft stores.) Let students know that they will be allowed to blow bubbles at recess or at the end of the day. Have them count how many bubbles they blew in 1 breath and then graph the results. (Template included.) What a simple icebreaker sure to get your kiddo's excited about being in school.
Print off the bubble picture cards from 1-20; laminate and cut out. Students use opalescent flat-backed glass "marbles" as manipulatives, to show how many bubbles. Use the larger pieces of glass for numbers less than 10 and the smaller ones for numbers 11-20. The "marbles" are not only inexpensive, but they have the appearance of being a flattened bubble!
Where did I get this idea? While in Hobby Lobby, I overheard a little girl ask her mom if she could buy a bag. When her mom asked her why she wanted them, "Kara" replied: "Because they are flat bubbles that won't pop!" I thought, "Wow! What can I make with 'flat bubbles'?" and the rest is history...
I've also included a set of number-word bubbles. Run the templates off on blue construction paper, laminate & trim. Older students can match the number word bubble, to the picture cards.
For more fun, run off the "bubble wand" on a variety of colors of construction paper and laminate. You can have these pre-cut by a room helper, or let your students trim off the excess, for fine-motor practice. This also enables you to see who might be struggling with scissor coordination. The center of the wand now looks like it's filled with bubble solution and can double as a magnifying glass.
For a "get the wiggles" out game, have students use their paper bubble wands, to find hidden bubbles around the room, or use them as an assessment tool for a whole group identification activity. i.e. you display a bubble card and students raise their wand if they know the answer, providing a quick way to whole group assess comprehension. Play "Swish." After the number is correctly identified, have students swish their wand that many times. (Swish left-right-left for a number 3 bubble card.)
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"Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world." -Einstein
## Fact Family Fun For October
A Spook-tacular Math Game:Haunted Fact Family Houses!
Since the Fact Family Schoolhouses were such a huge hit for back to school, I decided to make a Fact Family set for October too.
This "craftivity" and game help reinforce addition and subtraction math standards in a fun way.
The 17-page packet includes:
• Fact Family "craftivity" templates.
• Fact Family spinner game templates.
• Directions and link for a home-school math connection for fact family online homework skill building practice
• Directions and sample of how to make mini dry erase envelope boards.
• Fact Family T-bar "mad-minute" skill sheets +
• A certificate of praise.
If your students did the Schoolhouse Packet, they are already "empowered"! This one will be a real self-esteem builder for them.
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"Let us endeavor so to live, that when we die, even the undertaker will be sorry." - Mark Twain
## Math Dice Games Part 3
Let's Keep Things Rolling! More Math Games With A Dice Theme
I made Dice Game Stuff to go with the addition, subtraction, greater & less than dice games featured in the last 2 articles.
Whenever I taught a concept to my Y5’s I liked to stick with a theme.
It kept things simple, organized and less complicated for them.
I also had everything I needed handy and things just seem to flow from one transition into the next.
I could also overlap the various subjects too.
Here are some things you can do with these items:
The Make your own dice is a nice home-school connection where students can practice their cutting skills, something for a sub folder, or that extra activity students can do when they’ve finished everything else.
Run it off on cardstock. Give students a jingle bell to glue inside for added fun.
The large red dice make perfect flashcards when young students are learning to identify groups with a number.
Print them off, laminate, cut them out and keep them with your calendar or story time “stuff”.
You can also punch a hole in one corner and put them on a split ring.
Run off the smaller copies for students to make a split ring flipbook as well. You flash your large number and they flip through their little ones to see who can find it the fastest.
Run off the Smaller Red-Dot Dice, laminate and cut out and make Memory Match Concentration games. Students can match them dice to dice or dice to number.
Laminate the number and symbol cards as well. These too, can be used as Memory Match games or have students make equations with them.
Students can roll real dice, make an equation with the laminated paper dice, and then write down the equation on a sheet of scratch paper.
Set the timer to ring after 5 minutes. Students can play individually or with a partner.
The person with the most equations completed when the timer rings, is the winner.
The traceable number flashcards offer a nice way to review skip counting by 2’s, 3’s and 5’s.
I’m always looking for easy and interesting ways to plug that concept in, for a quick review my kiddo’s would think was fun, so they’d want to continue practicing.
I made covers for the traceable flashcards so they can be turned into Itty Bitty booklets.
Run off extra sets on different colors to make Memory Match Concentration games. You can also play I Have; Who Has? with them as well.
I hope you enjoy getting things rolling with your little ones and they have fun with these activities.
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Thanks for visiting!
"Life is a great big canvas and you should throw all the paint you can on it." -Danny Kaye
## Another Math Dice Game: Teaching the Concept of More and Less
Is it greater than, less than, or equal to? Whatever it is, it's in the bag!
Yesterday I posted the fun addition and subtraction dice games.
While I was making those example Baggies, I thought how perfect this idea would be for the greater than, less than concept, as the Baggies are clear, and when flipped over would reveal the opposite symbol!
All I had to do was include a small square that said = on it, for students to cover the < > signs, when they rolled doubles, and I was in business!
Students can either work independently or choose a partner and play against them, seeing who can solve the most equations before the timer rings.
Here’s how to play the game:
Children roll 2 dice and find that equation on their paper.
They rewrite it, and then show it in their manipulative Baggie, flipping the bag to whatever side they need to show greater than or less than, or covering the symbol with an equals sign if they roll doubles.
If they roll the same 2 dice that they already have an equation for, they lose their turn.
Baggie Manipulatives:
Put 10 buttons, or whatever manipulatives you have, in large Ziploc Baggies. Draw a greater than sign in the middle. Trace a black line above and below it.
Give it a few seconds to dry and then flip it over and retrace to make the less than symbol. Using index cards or old file folders, cut squares and label them with equal signs. Tuck one in each baggie. I’ve also made greater than and less than label templates if you want to stick those on the top of your Baggies to help your students associate the words with the symbols.
Simply put a 30-on-a-page Avery label sheet into your printer and print. Students move the manipulatives to the right and left to show the equation they rolled. For example: 5 < 6, 3 = 3, 4 > 1
My Baggie idea was inspired by Click on the link to see how this creative teacher uses her Baggies.
If you like this greater than less than game, you will also enjoy Alligobbler. It's a quick and easy "craftivity" where you make an aligator out of a long envelope.
His toothy grin is the symbol. Students have fun "feeding" him numbers.
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Do you have a greater than / less than concept that helps your students understand things? I'd enjoy hearing from you. This email address is being protected from spambots. You need JavaScript enabled to view it. You can also post a comment here as well.
Do drop in tomorrow for another teaching tip, until then, remember:
"Students don't care how much you know, until they know how much you care!"
## Playing Dice Games To Teach Addition
Whenever you can think of a way to teach a concept via a game your students will enthusiastically want to learn.
Dice are the perfect way to introduce simple addition for numbers 1-6 and then move students to subtraction.
To make the game even more fun, I’ve included clip art to guide them. Because I want students to practice writing their numbers, I have them not only solve the dice equation, but rewrite it in all numbers.
I also feel that student need to “see” counters to visualize the true concept of addition and subtraction.
I have a variety of ways for my students to do this, but stumbled across bead bracelets and manipulative Baggies via 2 creative teachers on Pinterest.
I decided to incorporate the “seeing-is-believing” and the “doing-is-understanding!” principal to this dice game by making it even more hands on. After students write the equation they SHOW it, using either the bracelet or Baggie.
Ta Da! Hopefully the light bulbs will be going on while the kiddo’s are having a fun time.
Students can either work independently or choose a partner and play against them, seeing who can solve the most equations before the timer rings.
Here’s how to play the game:
Children roll 2 dice and find that equation on their paper.
They rewrite it, solve the problem and work it out on either their bead bracelet or manipulative bag.
If they roll the same 2 dice that they already have an equation for, they lose their turn.
After students have played the addition version of the game, have them switch to subtraction.
To make a class set of bead bracelets for this game put 6 pony beads on 25 pipe cleaners. (Or however many students you usually have in class.) Twist the ends so they look like a bracelet. Students move the beads to show the various rolls of the dice. i.e. 3http://tunstalltimes.blogspot.com/2011/08/number-bracelets.htmlQ + 2 = 6
I got the bead bracelet idea from: Mrs. Tunstall’s Teaching Tidbits click on the link to check out her cute site and how else she uses her bracelets.
Baggie Manipulatives:
Put 6 buttons, or whatever manipulatives you have, in small Ziploc Baggies. I used poker chips becaus you can buy them at The Dollar Store. Draw a blue or red + sign in the middle of the bag with a black line above and below it so that the line runs down the center.
Make another set of Baggies for subtraction and put a minus sign in the middle. If you only want to make one set of Baggies, simply put a line down the middle.
I really believe that it is worth the few extra dollars to make separate addition and subtraction bags, because I think that the more students see thosee math symbols, the more the concept gets ingrained in their brains.
Students move the manipulatives to the right and left of the line to show what equation they rolled. i.e. 3 + 3 = 6
I got the Baggie idea from Mrs. T’s First Grade Blog click on the link to see her sweet site and how else she uses her Baggies.
Hopefully your students will enjoy this game and things really will start to add up in your class!
Thanks for visiting today. I hope you can drop in tomorrow for another teaching tip!
Feel free to PIN if so inspired.
## Teaching With Pattern Tanagram Blocks
Math Games That Teach
My Y5’s really enjoyed Tummy Time and hauling out a tub of manipulatives to “play” with.
I put that word in quotations because they thought they were playing, but in reality, they were reinforcing a huge number of math skills as well as strengthening their upper body, by being on their tummies and working on exercising those fine motor skills as well, which helped to strengthen their finger muscles, that is so important in developing their ability to write.
One of their favorite tubs was the tanagram pattern blocks. I had a variety of puzzle-sheets for them to fill in , by placing the colorful tanagram blocks on them, as well as a variety of patterning strips for them to complete.
I like making up monthly activities that follow the same format, because it empowers students, as once they’ve played a game, or learned the directions, they can get down to business to practice the skill.
This builds their self-esteem, they get better at the activity, and the teacher is freed up to work one-on-one with struggling students, or do assessing, because other children can work independently because they know what to do.
I decided to make some monthly tanagram pattern block sheets that involved dice, as number recognition is a skill that the Y5’s needed to acquire.
Students can play independently, with a partner, or in groups of 2 to 4.
Whatever number they roll, they trace that tanagram piece, the tanagram’s number, and then place the tanagram over that piece.
You can also have students color their papers to match the real tanagrams.
The first child to complete their seasonal tanagram picture, by rolling all of the numbers and covering them, is the winner.
I hope you enjoy adding this game to your bag of tricks.
Do you have one you can share with us? I’d enjoy hearing from you. This email address is being protected from spambots. You need JavaScript enabled to view it. or feel free to leave a comment here, especially if you use one of my ideas.
You may also PIN anything on my site that you think will help others. I truly feel life is all about sharing. Just think how easy our teaching would be, if everyone took the time to post just one thing on the internet that turned the light bulb on for kids!
Have a great day and may all your puzzles be fun to solve!
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posted by on .
Given θ = arcsin (tan45°) find the exact degree measure of θ without usiing calculator.
• Math - ,
tan 45° = 1
θ = arcsin(1) = 90° | 54 | 156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-26 | latest | en | 0.734339 |
https://www.doubtnut.com/qna/195231887 | 1,726,167,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00289.warc.gz | 683,864,310 | 37,872 | # A particle is moving in a straight line and passes through a point O with a velocity of 10ms−1. The particle moves with a constannt retardation of 5ms−2 for 3 s and there after moves with constant velocity. How long after leaving O does the particle return to O:-
A
3s
B
8s
C
Never
D
4.5s
Text Solution
Verified by Experts
## The correct Answer is:D
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## A particle is moving in a straight line and passes through a point O with a velocity of 6ms−1 The particle moves with a constant retardation of 2ms−2 for 4 s and there after moves with constant velocity. How long after leaving O does the particle return to O
A2t
B(2+2)t
Ct2
DCan not perdicted unless accelerationis given
• Question 2 - Select One or More
## A particle is moving on a straight line with constant retardation of 1m/s2. Its initial velocity is 10m/s.
AThe distance travelled by the particle till it comes to rest is 50 m.
BThe average velocity till it comes to rest is 5m/s.
CIt comes to rest in 5 sec.
DIts velocity after 5 sec is 5m/s
• Question 3 - Select One
## A particle of mass 1mg has the same wavelength as an electron moving with a velocity of 3×106ms−1. The velocity of the particle is
A2.7×1018ms1
B9×102ms1
C3×1031ms1
D2.7×1021ms1
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• info@doubtnut.com | 683 | 2,435 | {"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 | 0.879364 |
https://d2mvzyuse3lwjc.cloudfront.net/doc/X-Function/ref/interp3 | 1,712,993,407,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00708.warc.gz | 188,534,225 | 38,792 | # 2.8.15 interp3
Analysis: Mathematics: 3D Interpolation
## Brief Information
Perform 3D interpolation
## Command Line Usage
1. interp3 (Col(1), Col(2), Col(3), Col(4)) pts:=5;
## Variables
Display
Name
Variable
Name
I/O
and
Type
Default
Value
Description
Input irng
Input
Range
The input range.
Number of Points in Each Dimension pts
Input
int
10
Number of points in each dimension.
X Minimum xmin
Input
double
Specifies the X minimum value of this interpolation.
X Maximum xmax
Input
double
Specifies the X maximum value of this interpolation.
Y Minimum ymin
Input
double
Specifies the Y minimum value of this interpolation.
Y Maximum ymax
Input
double
Specifies the Y maximum value of this interpolation.
Z Minimum zmin
Input
double
Specifies the Z minimum value of this interpolation.
Z Maximum zmax
Input
double
Specifies the Z maximum value of this interpolation.
Output rd
Output
ReportData
[<input>]<new>
The output result.
## Examples
1. Import the data \Samples\Mathematics\3D Interpolation.dat.
2. Highlight column C and right-click. In the fly-out menu, select Set As : Z to set it as Z column. Then click Plot : 3D XYZ : 3D Scatter to create a 3D graph.
3. To give a visual impression on what the original data is, we can set column D as symbol size. Double click the 3D Scatter to bring up the Plot Details dialog. Expand all branches on the left panel and select Original. Then in the Symbol tab of right panel, select Col(F) from Size drop-down list and click OK. The 3D Scatter will looks like:
4. To interpolate these data, active the worksheet, and select Analysis : Mathematics : 3D Interpolation to bring up the X-Function dialog. Enter column X, Y, Z, F to Input branch respectively. Enter 5 on Number of Points in Each Dimension box, which means it will generate 5*5*5 = 125 interpolated points. Then click OK to do interpolation.
5. To verify the result, active the output result worksheet, and plot a 3D scatter plot as above, using Col(F) as symbol size. The interpolated graph will looks like: | 516 | 2,070 | {"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} | 2.640625 | 3 | CC-MAIN-2024-18 | longest | en | 0.595161 |
https://www.justfreetools.com/en/kev-to-ev-conversion | 1,642,842,009,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303779.65/warc/CC-MAIN-20220122073422-20220122103422-00599.warc.gz | 903,384,103 | 12,918 | keV to eV free online conversion
Kiloelectron-volts (keV) to electron-volts (eV) conversion calculator and how to convert.
keV to eV conversion calculator
keV to eV, energy conversion calculator.
Enter the energy in kiloelectron-volts and press the Convert button:
keV
eV
eV to keV conversion »
How to convert keV to eV
One kiloelectron-volt is equal to 1000 electron-volts:
1keV = 1000eV
The energy in electron-volts E(eV) is equal to the energy in kiloelectron-volts E(keV) times 1000:
E(eV) = E(keV) × 1000
Example
Convert 5keV to eV:
E(eV) = 5keV × 1000 = 5000eV
keV to eV conversion table
Energy (keV) Energy (eV)
0.001 keV 1 eV
0.002 keV 2 eV
0.003 keV 3 eV
0.004 keV 4 eV
0.005 keV 5 eV
0.006 keV 6 eV
0.007 keV 7 eV
0.008 keV 8 eV
0.009 keV 9 eV
0.01 keV 10 eV
0.02 keV 20 eV
0.03 keV 30 eV
0.04 keV 40 eV
0.05 keV 50 eV
0.06 keV 60 eV
0.07 keV 70 eV
0.08 keV 80 eV
0.09 keV 90 eV
0.1 keV 100 eV
0.2 keV 200 eV
0.3 keV 300 eV
0.4 keV 400 eV
0.5 keV 500 eV
0.6 keV 600 eV
0.7 keV 700 eV
0.8 keV 800 eV
0.9 keV 900 eV
1 keV 1000 eV
2 keV 2000 eV
3 keV 3000 eV
4 keV 4000 eV
5 keV 5000 eV
6 keV 6000 eV
7 keV 7000 eV
8 keV 8000 eV
9 keV 9000 eV
10 keV 10000 eV
100 keV 100000 eV
eV to keV conversion »
Currently, we have around 1976 calculators, conversion tables and usefull online tools and software features for students, teaching and teachers, designers and simply for everyone. | 627 | 1,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-05 | latest | en | 0.413784 |
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MasteringPhysics2a
# MasteringPhysics2a - MasteringPhysics Assignment Print View...
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MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmentI... 2 of 11 10/4/2007 3:29 PM Part C Select from the graphs pictured the one that most plausibly represents the graph of x velocity vs. time. ANSWER: a b c d e f Correct. First the ball moves with constant speed. Then, as it rolls down the ramp, its speed increases linearly. The ball encounters another flat section, and its speed is once again constant. Finally, as the ball rolls up the ramp, it decelerates. Part D Select from the graphs pictured the one that most plausibly represents the graph of x acceleration vs. time. (Note that the origin of the y axis of the graphs is arbitrary; zero acceleration is not necessarily at the bottom of the graphs.) ANSWER: a b c d e f Correct. Initially, the ball's acceleration is zero. When the ball rolls down the first ramp, its acceleration becomes positive and constant. When the ball encounters the second flat section, its acceleration is once again zero. Finally, as the ball rolls
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Ask a homework question - tutors are online | 386 | 1,671 | {"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-2018-13 | latest | en | 0.863415 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-10th-edition/chapter-1-section-1-4-radical-equations-equations-quadratic-in-form-factorable-equations-1-4-assess-your-understanding-page-117/22 | 1,527,394,177,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867995.55/warc/CC-MAIN-20180527024953-20180527044953-00086.warc.gz | 746,575,047 | 13,032 | ## College Algebra (10th Edition)
The solution set is $\left\{-2\right\}$.
Square both sides to obtain: $x^2=(2\sqrt{-x-1})^2$ Use the rule $(ab)^m = a^mb^m$ to obtain: $x^2=2^2(\sqrt{-x-1})^2 \\x^2=4(-x-1) \\x^2=-4x-4$ Add $4x$ and $4$ on both sides of the equation to obtain: $x^2+4x+4 = -4x-4+4x+4 \\x^2+4x+4=0$ Factor the trinomial to obtain: $(x+2)(x+2) = 0$ Equate each factor to zero, and then solve each equation to obtain: $\begin{array}{ccc} &x+2=0 &\text{ or } &x+2=0 \\&x=-2 &\text{ or } &x=-2 \end{array}$ Thus, the solution set is $\left\{-2\right\}$. | 247 | 566 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2018-22 | longest | en | 0.757617 |
http://web.cs.wpi.edu/~cs4445/b06/Projects/Project3/ | 1,679,640,847,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00401.warc.gz | 55,147,055 | 10,566 | ### CS 4445 Data Mining and Knowledge Discovery in Databases - B Term 2006 Homework and Project 3: Numeric Predictions
#### PROF. CAROLINA RUIZ
DUE DATES:
• Part I (the individual homework assignment) is due on Tuesday, Nov. 28 at 11:50 am, and
• Part II (the group project) is due on Friday, Dec. 1st at 11:50 am.
#### PROJECT DESCRIPTION
The purpose of this project is to construct accurate numeric prediction models for the two datasets under consideration using the following techniques:
• Linear Regression
• Regression Trees
• Model Trees
Also, to gain close understanding of how those methods work, this project also include following those methods by hand on a toy dataset.
#### PROJECT ASSIGNMENT
1. Part I. INDIVIDUAL Homework
Consider the dataset below. This dataset is an adaptation of the World Happiness Dataset.
```@relation world_happiness
% - Life Expectancy from UN Human Development Report (2003)
% - GDP per capita from figure published by the CIA (2006), figure in US\$.
% - SWL (satisfaction with life) index calculated from data published
% by New Economics Foundation (2006).
@attribute country string
@attribute continent {Americas,Africa,Asia,Europe}
@attribute life-expectancy numeric
@attribute GDP-per-capita numeric
@attribute access-to-education-score numeric
@attribute SWL-index numeric
@data
Switzerland, Europe, 80.5, 32.3, 99.9, 273.33
Canada, Americas, 80, 34, 102.6, 253.33
Usa, Americas, 77.4, 41.8, 94.6, 246.67
Germany, Europe, 78.7, 30.4, 99, 240
Mexico, Americas, 75.1, 10, 73.4, 230
France, Europe, 79.5, 29.9, 108.7, 220
Thailand, Asia, 70, 8.3, 79, 216.67
Brazil, Americas, 70.5, 8.4, 103.2, 210
Japan, Asia, 82, 31.5, 102.1, 206.67
India, Asia, 63.3, 3.3, 49.9, 180
Ethiopia, Africa, 47.6, 0.9, 5.2, 156.67
Russia, Asia, 65.3, 11.1, 81.9, 143.3
```
For this homework, we want to predict the SWL-index attribute (prediction target) from the other predicting attributes continent, life-expectancy, GDP-per-capita, access-to-education-score. Note that the attribute country identifies each data instance uniquely and as such will be disregarded in our analysis. It is provided just for context.
1. (15 points) Linear Regression
• (5 points) Describe the linear regression equation that would result from using linear regression to solve this numeric prediction problem.
• (5 points) Describe in detail the procedure that would be followed by linear regression to find appropriate parameters for this equation.
• (5 points) Run linear regression in the Weka system over this dataset and provide the precise linear regression equation output by Weka (you'll need it for the testing part below). Set the "eliminateColinearAttributes" parameter of linear regression to False so that your linear regression formula includes all the predicting attributes.
2. (45 points) Regression Trees and Model Trees
Follow the procedure described in the textbook to construct a model tree and a regression tree to solve this numeric prediction problem. Remember to:
1. (5 points) Start by translating the nominal attribute continent into boolean/numeric attributes. This is done by taking the average of the CLASS values associated with each of the continent values Americas,Africa,Asia,Europe. Sort them in decresing order by average. Now, create new boolean attributes, one for each possible split of these four nominal values in the order listed. After this translation, all the predicting attributes are numeric.
2. (5 points) Sort the values of each attribute in say increasing order. Define a "split point" of an attribute as the midpoint between two subsequent values of the attribute.
3. (5 points) Consider the set of split points of all attributes. Select as the condition for the root node on your tree, the split point that maximizes the value of the following formula:
``` SDR = sd(CLASS over all instances)
- ((k1/n)*sd(CLASS of instances with attribute value below split point)
+ (k2/n)*sd(CLASS of instances with attribute value above split point))
where sd stands for standard deviation.
k1 is the number of instances with attribute value below split point.
k2 is the number of instances with attribute value above split point.
n is the number of instances.
```
To reduce the number of calculations that you need to perform, your HW solutions can be limited to the following split points:
``` binary attributes life-expectancy GDP-per-capita access-to-education-score
0.5 (63.3+47.6)/2 (3.3+0.9)/2 (49.9+5.2)/2
(70.0+65.3)/2 (8.3+3.3)/2 (73.4+49.9)/2
(75.1+70.5)/2 (29.9+11.1)/2 (94.6+81.9)/2
(78.7+77.4)/2 (32.3+31.5)/2 (102.1+99.9)/2
(82.0+80.5)/2 (41.8+34.0)/2 (108.7+103.2)/2
```
If during the construction of the tree you encounter an attribute such that none of the split points listed above apply to the instances in the node, then use instead all the attribute's split points that apply to that collection of instances.
4. (20 points) Continue the construction of the tree following the same procedure recursively. Remember that for each internal node, the procedure above is applied only to the data instances that belong that that node. You can stop splitting a node when the node contains less than 4 data instances and/or when the standard deviation of the CLASS value of the node's instances is less than 0.05*sda, where sda is the standard deviation of the CLASS attribute over the entire input dataset. See Figure 6.15 on p. 248 of your textbook.
5. (10 points) For each leaf node in the tree:
• Compute the value that would be predicted by that leaf in the case of a Regression Tree.
• Compute the linear regression formula that would be used by that leaf to predict the CLASS value in the case of a Model Tree. In order to find the coefficients of the linear regression formula, run the linear regression method implemented in the Weka system for the appropriate data instances (those that belong to the leaf).
3. (15 points) Testing
Use each of the three numeric predicting models constructed (linear regression equation, model tree and regression tree), to predict the CLASS values (i.e., the SWL-index) for each of the test instances below. That is, complete the following table:
``` LINEAR MODEL TREE REGRESSION TREE
REGRESSION PREDICTION PREDICTION
Costa_Rica, Americas, 78.2, 11.1, 50.9, 250 __________ ___________ __________
United_Kingdom, Europe, 78.4, 30.3, 157.2, 236.67 __________ ___________ __________
South_Africa, Africa, 48.4, 12, 90.2, 190 __________ ___________ __________
Lithuania, Europe, 72.3, 13.7, 93.4, 156.67 __________ ___________ __________
LINEAR MODEL TREE REGRESSION TREE
REGRESSION ERROR ERROR
ERROR
root mean-square error (see p. 178) __________ ___________ __________
mean absolute error (see p. 178) __________ ___________ __________
```
SHOW IN DETAIL ALL THE STEPS OF THE PROCESS.
• Part II. Project
1. Datasets: Two datasets will be analyzed in this part:
1. The World Happiness Dataset. See also the World Happiness Dataset with Continents information added by Paul Sader and converted into arff. Thanks, Paul!!
Use the attribute SWL-index as the prediction target. After you run experiments predicting this attribute you may, if you wish, run additional experiments using a different predicting target of your choice. Since the SWL-ranking can be derived from SWL-index, remove SWL-ranking from consideration. Also, remove the attribute country as each of its values identifies an instance uniquely. Note that the access-to-education-score attribute contains missing values, marked with ".". Remember to replace them with "?" in your arff file.
2. Together with your project partner, choose one dataset from the following options:
• The Abalone Dataset available from the The University of California Irvine (UCI) Data Repository.
Use the attribute age as the prediction target. After you run experiments predicting this attribute you may, if you wish, run additional experiments using a different predicting target of your choice.
• The Automobile Dataset available from the The University of California Irvine (UCI) Data Repository.
Use the attribute price as the prediction target. After you run experiments predicting this attribute you may, if you wish, run additional experiments using a different predicting target of your choice.
• A dataset of your choice. This dataset can be one available on a public, online data repository (including but not limited to the datasets used on Projects 1 or 2) or any other valid source. The dataset should contain at least 500 data instances with at least 5 different attributes (ideally some numeric and some nominal). For ideas, see the Data Sets listed on the course webpage/syllabus.
• Textbook: Read in great detail the following Sections from your textbook:
• Numeric Predictions: Sections 4.6, 6.5, 5.8.
• Weka Code: Read the code of the relevant techiques implemented in the Weka system. Some of those techniques are enumerated below:
• Numeric Predictions:
• Linear Regression (under "functions")
• M5P (under "trees"): Regression Trees, Model Trees
#### Experiments:
For each of the datasets, use the Weka system to perform the following operations:
• A main part of the project is the PREPROCESSING of your dataset. You should apply relevant filters to your dataset before doing the mining and/or using the results of previous mining tasks. For instance, you may decide to remove apparently irrelevant attributes, replace missing values if any, etc. Missing values in arff files are represented with the character "?". See the weka.filter.ReplaceMissingValuesFilter in Weka. Play with the filter and read the Java code implementing it. Your report should contain a detailed description of the preprocessing of your dataset and justifications of the steps you followed. If Weka does not provide the functionality you need to preprocess your data as you need to obtain useful patterns, preprocess the data yourself either by writing the necessary filters (you can incorporate them in Weka if you wish).
• To the extent possible/necessary, modify the attribute names and the nominal value names so that the resulting models are easy to read.
• You may restrict your experiments to a subset of the instances in the input data IF Weka cannot handle your whole dataset (this is unlikely). But remember that the more accurate your models, the better.
• Ideas for making this project even more interesting:
• Start by choosing a dataset in a domain that excites you.
• Before you start running experiments, look at the raw data in detail. Figure out 5 or more specific, interesting questions that you want to answer with your experiments. These questions may be phrased as conjectures that you want to confirm/refute with your experimental results. Sample questions on the World Happiness are: what's the relative importance of access to education, per-capita income, and life expectancy in satisfaction with life? What factor is most important? Are there general trends by continent?
If you pursue this last question, add continent information to your dataset. The first group/individual to submit a correct dataset including continent information to the course mailing list will receive a 25 points bonus. To make it standard, let's assume there are 6 continents: Antarctica, Americas, Europe, Asia, Africa, Australia.
Paul Sader was the first to submit a correct dataset with continent information added to it. Thanks, Paul!
• Analyze your resulting models in the light of your 5 questions.
• WRITTEN REPORT.
Your project report should contain discussions of all the parts of the work you do for this project. In particular, it should elaborate on the the following topics:
1. Code Description: Describe algorithmicly the Weka code of the 3 methods covered by this project: LINEAR REGRESSION, MODEL TREES, AND REGRESSION TREES; and filters that you use in the project. More precisely, explain the ALGORITHM underlying the code in terms of the input it receives, the output it produces, and the main steps it follows to produce this output. PLEASE NOTE THAT WE EXPECT A DETAIL DESCRIPTION OF THE ALGORITHMS USED NOT A LIST OF OBJECTS AND METHODS IMPLEMENTED IN THE CODE.
2. Experiments: For EACH EXPERIMENT YOU RAN describe:
• 5 or more specific questions/conjectures about the dataset domain that you aim to answer/validate with your experiments. See the section on "Ideas for making this project even more interesting" above.
• Instances: What data did you use for the experiment? That is, did you use the entire dataset of just a subset of it? Why?
• Any pre-processing done to the data. That is, did you remove any attributes? Did you replace missing values? If so, what strategy did you use to select a replacement of the missing values?
• For each of the methods covered by this project (i.e. LINEAR REGRESSION, MODEL TREES, AND REGRESSION TREES):
• Results and detail ANALYSIS of results of the experiments you ran using different ways of testing (split ratio and N-fold cross-validation) the model.
• Error measures of the resulting models.
• Comparison of the results across the 3 numeric prediction methods used in this project.
• Detailed analysis of the models obtained. Examine the linear regression formulas and the trees obtained. Elaborate on interesting attribute weights in those formulas, as well as presence/absence of predicting attributes in the trees. Do these models answer any of your 5 questions? If so, what's the answer? If not, why not and what modifications to the experiments are needed to answer those questions?
• Investigate the pruning method used in M5P (i.e., when the option UNPRUNED=FALSE is chosen from the M5P GUI), which is explained in your textbook. Describe in detail how the pruning is done and run experiments to see how it modifies the resulting tree and how this affects the error values reported for the tree.
Investigate also the smoothing method used in M5PRIME (i.e. when the option useUNSMOOTHED=FALSE is chosen from the M5PRIME GUI). Describe in detail how the smoothing is done and run experiments to see how it modifies the resulting tree and/or how it affects the error values reported for the tree.
3. Summary of Results
• For each of the datasets, what were the lowest error values obtained in your project? Include this model in your report as well as interesting aspects of this model.
• Strengths and weaknesses of your project.
• ORAL REPORT. We will discuss the results from the individual projects during the class on Tuesday, Dec. 5. Your oral report should summarize the different sections of your written report as described above. Each group will have about 4 minutes to explain your results and to discuss your project in class. Be prepared!
#### PROJECT SUBMISSION AND DUE DATE
Part II is due Friday, Dec. 1st at 11:50 am. BRING A HARDCOPY OF THE WRITTEN REPORT WITH YOU TO CLASS. In addition, you must submit your report electronically as specified below. Submissions received on Friday, Dec. 1st between 11:51 am and 12:00 midnight will be penalized with 30% off the grade, submissions received on Saturday Dec. 2th between 12:01 am (early morning) and 8:00 am will be penalized with 60% off the grade; and submissions received after Saturday Dec. 2th at 8:00 am won't be accepted.
1. [lastname]_proj3_report.[ext] for those of you who are working alone on this project. If you are taking this course for grad. credit, state this fact at the beginning of your report. This file should be either a PDF file (ext=pdf), a Word file (ext=doc), or a PostScript file (ext=ps). For instance my file would be named (note the use of lower case letters only):
• ruiz_proj3_report.pdf
2. [lastname1_lastname2]_proj3_report.[ext] containing your group written reports. This file should be either a PDF file (ext=pdf), a Word file (ext=doc), or a PostScript file (ext=ps). For instance my file would be named (note the use of lower case letters only):
• ruiz_smith_proj3_report.pdf if I worked with Joe Smith on this project.
3. [lastname1_lastname2]_proj3_slides.[ext] (or [lastname]_proj3_slides.[ext] in the case of students taking this course for graduate credit) containing your slides for your oral reports. This file should be either a PDF file (ext=pdf) or a PowerPoint file (ext=ppt). Your group will have only 4 minutes in class to discuss the entire project.
```
(TOTAL: 75 points) FOR THE HOMEWORK (PART I) as stated in the Homework assignment above.
(TOTAL: 200 points) FOR THE PROJECT (PART II) as follows:
```
NOTES:
Along with scoring criteria, provided are lists of suggestions. Positive suggestions (marked with green "+") include aspects of a project that have positive impacts on the scoring criteria. Negative suggestions (marked with red "-") have negative impacts. Finally, suggestions marked with "!" result in extra points.
!this can earn a few extra points (0-8)
! !this can earn a more extra points (0-16)
! ! !this can earn even more extra points (0-24)
! ! ! !this can earn a large extra points (0-32)
! ! ! ! !this can earn a very large amount of extra points (0-40)
+this has a positive impact
+ +this has a greater positive impact
+ + +this has a significant positive impact
+ + + +this has very significant positive impact
+ + + + +this has huge positive impact
-this has a negative impact
- -this has a greater negative impact
- - -this has a significant negative impact
- - - -this has very significant negative impact
- - - - -this has huge negative impact
METHODS for this project are:
linear regression,
model tree, and
regression tree.
DATASETS for this project are:
dataset1: world happiness dataset
dataset2: abalone dataset, automobile dataset, OR a dataset of your choice
TOTAL: 200 points: Project 3 Report
TOTAL: 20 points: Algorithmic Description of the Code
(05 points) Description of the algorithms underlying the WEKA filters used. If you have already described some used filters in previous projects then copying and pasting your work would be a good idea. If issues were raised about the old descriptions, consider revising them.
(TOTAL: 15 points) Description of the ALGORITHMS underlying the data mining methods used in this project (see METHODS). This includes:
(10 points) Descriptions of model construction
(5 points) Descriptions of instance classification/prediction
+(very) high-level pseudo code with explanation and justification
- -raw pseudo code with no justification or reasoning behind the steps involved
- - - - -structural description of WEKA code, i.e. classes/members/methods
TOTAL: 5 points: Slides
(5 points) How well do they summarize concisely the results of the project? We suggest you summarize the setting of your experiments and their results in a tabular manner.
! excellent summary and presentation of results in the slides
+ (potential) unanswered questions presented to class
+ visual aids
+ +main ideas and observations summarized
- nothing but accuracy measures
- lack of visual aids
TOTAL: 5 points: Pre-Processing of the Datasets
(5 points) Preprocess attributes as needed and dealing with missing values appropriately
!Trying to do "fancier" things with attributes (i.e. combining two attributes highly correlated into one, using background knowledge, etc.)
TOTAL: 15 points: Class Presentation
(15 points) How well your oral presentation summarized concisely the results of the project and how focus your presentation was on the more creative/interesting/useful of your experiments and results. This grade is given individually to each team member.
TOTAL: 155 points: Experiments Goals
(30 points) Good description of the experiment setting and the results. This applies to all the experiments done for this project.
+ + + motivation for experiments described
+ specify testing method used (cross-validation / % split / etc.)
+ + overall summary of your experiments in one concise list that includes relevant parameters
- included are trivial experiment details that are not used in dicscussion (full algorithm parameters, full WEKA output, etc.)
- - ambiguous or unclear setting and/or results
(30 points) Good analysis of the results of the experiments INCLUDING discussion of evaluations statistics returned by the WEKA systems (accuracy and/or errors) and discussion of particularly interesting results. This applies to all the experiments done for this project.
! ! ! !excellent analysis of the results and comparisons
! ! ! !running additional interesting experiments
- - - - results rewritten in prose used in place of discussion and analysis
TOTAL: 35 points: Method-Oriented Goals
(15 points) ran a good number of experiments to get familiar with the data mining methods in this project (see METHODS) using a variety of datasets (see DATASETS)
+explore variation in non-trivial method parameters (debug/verbose mode IS a trivial parameter)
- - -use only one of the datasets
(10 points) Investigate the pruning method used in M5P (i.e., when the option UNPRUNED=FALSE is chosen from the M5P GUI), which is explained in your textbook. Describe in detail how the pruning is done and run experiments to see how it modifies the resulting tree and how this affects the error values reported for the tree.
- - -use only one of the datasets
(10 points) Investigate also the smoothing method used in M5PRIME (i.e. when the option useUNSMOOTHED=FALSE is chosen from the M5PRIME GUI). Describe in detail how the smoothing is done and run experiments to see how it modifies the resulting tree and/or how it affects the error values reported for the tree.
- - -use only one of the datasets
TOTAL: 40 points: Dataset-Oriented Goals
For each dataset (see DATASETS):
(10 points) 5 or more specific questions/conjectures about the dataset domain that you aim to answer/validate with your experiments
(10 points) Comparison of the results across all the methods used in this project (see METHODS). That is, compare the results of one method on this dataset to the results of the other methods on the same dataset.
! !comparison with methods from past projects
TOTAL: 20 points: Holistic Goals
For each dataset (see DATASETS):
(10 points) Argumentation of weaknesses and/or strengths of each of the methods on this dataset, and argumentation of which method should be preferred for this dataset and why. | 5,245 | 22,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-14 | latest | en | 0.751777 |
https://community.wolfram.com/groups/-/m/t/1835194 | 1,611,043,174,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518201.29/warc/CC-MAIN-20210119072933-20210119102933-00366.warc.gz | 284,054,201 | 29,232 | # Generate a network for the entropy values
Posted 1 year ago
2604 Views
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13 Replies
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14 Total Likes
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Dear all,I calculated the entropy values for the time series of 44 cities. Is it possible to define a simple or complex network for these values? I attached a related reference for getting more information. shannonEntropy[data_, binWidth_] := Module[{iter, n = Length[data], pi, pi1}, iter = {Min[data], Max[data] + binWidth, binWidth}; pi = N[BinCounts[data, iter]/n]; pi1 = DeleteCases[pi, 0.]; -pi1.Log[pi1]] entr = {1.21, 0.55, 0.84, 1.06, 1.33, 0.82, 1.51, 0.27, 1.21, 1.14, 0.92, 1.23, 1.37, 0.82, 0.69, 1.46, 0.1, 0.09, 1.63, 1.58, 0.89, 0.21, 0.93, 1.33, 1.31, 1.09, 0.46, 0.54, 0.94, 0.04, 0.88, 0.87, 1.24, 1.62, 0.96, 1.35, 1.43, 0.9, 0.72, 1.07, 1.02, 1.16, 1.38, 0.07}; lat = {36.37, 35.44, 35.44, 36.69, 37., 35.75, 36.69, 34.81, 36.06, 36.69, 35.13, 36.37, 36.37, 34.81, 33.56, 37., 33.25, 32.63, 36.37, 36.69, 35.44, 33.88, 35.44, 36.37, 36.37, 36.06, 35.75, 33.25, 35.75, 32., 35.73, 35.75, 34.81, 36.06, 35.13, 37., 36.37, 35.44, 35.13, 35.44, 36.69, 35.44, 36.69, 33.25}; lon = {1.875, 2.5, -0.9375, 3.125, 7.8125, 6.25, 5., 5.625, 4.6875, 3.4375, 4.0625, 0.9375, 6.5625, 3.125, 0.9375, 8.125, 6.875, 3.75, 7.5, 5.625, 7.1875, 2.8125, 0.3125, 2.8125, 6.25, 0., 4.6875, -0.3125, -0.3125, 5.625, 7.3693, 0.625, 0.3125, 5.3125, -0.625, 6.875, 7.8125, 7.8125, -1.875, 1.25, 2.5, 1.5625, 3.75, 5.9375}; Attachments:
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Posted 1 year ago
Hello Alex,Maybe I do not jet fully understand this seemingly interesting concept. Is it this you are having in mind? (I have no idea on how to use your position data ...) v4 = Partition[entr, 4, 1]; order4 = FromDigits@*Ordering /@ v4; rels = UndirectedEdge @@@ Partition[order4, 2, 1]; Graph[rels, VertexLabels -> Automatic] EDIT:I guess the partition of the data should be done differently and my code should rather read like so: v4 = Partition[entr, 4]; order4 = FromDigits@*Ordering /@ v4; rels = UndirectedEdge @@@ Partition[order4, 2, 1]; Graph[rels, VertexLabels -> Automatic]
Posted 1 year ago
Henrik you did an incredible job. Congratulations!Is there any concept behind the this network.? What does this network give us from the mathematical or application view?
Posted 1 year ago
Thank you Mohammad - but there is definitely no reason for any congratulations! I simply tried to follow the idea/concept in the above attached publication.
Posted 1 year ago
Dear Henrik,I plotted for different orders and got interesting results. Thank you. Attachments:
Posted 1 year ago
Hi Henrik,I found another paper. I thought it may be interesting to you, especially Fig3 and Fig4. Is it possible to investigate this approach to your city air temperature time series (for example).Thanks, Alex Attachments:
Posted 1 year ago
Hi Alex, thank you for this publication! I will try to find time looking into it. Regards -- Henrik
Posted 1 year ago
Hi Henrik, Did you have time to take a look at the paper that I sent you a few days ago? I got Oberpfaffenhofen Airport air temperature time series, may be it be useful for the method. Oberpfaffenhofen Airport ICAO code | EDMO tt0 = WeatherData["EDMO", "MeanTemperature", {{2000}, {2018}, "Day"}]; temp0 = Last[tt0\[Transpose]]; Normal[temp0]; DateListPlot[temp0] Attachments:
Posted 1 year ago
Hello Alex. sorry for the delay!I finally found some time to look into this problem. But as a word of caution: I am not sure whether I am understanding the respective publication correctly, and I do not have any experiences with graphs !!!Here is what I tried: ClearAll["Global*"] (* function for calculating the edge weight between two vertices v1,v2: *) getEdgeWeight[v1_List, v2_List] := Module[{trans1, trans2}, If[v1 == v2, Return[{Null, 0}]]; {trans1, trans2} = Partition[#, 2, 1] & /@ {v1, v2}; {UndirectedEdge[v1, v2], Total[Count[trans1, #] & /@ trans2]} ] tt0 = WeatherData[Entity["City", {"Munich", "Bavaria", "Germany"}], "MeanTemperature", {{2008}, {2018}, "Day"}]; temperatures = First@Normal@tt0["ValueList"]; temperatNums = Round@QuantityMagnitude[temperatures]; lenght = tt0["PathLength"]; vertexNameLength = Round[N@Sqrt[lenght]]; (* according to Ferreira et al *) vertexList = Partition[temperatNums, vertexNameLength]; combs = Select[Flatten[Outer[getEdgeWeight, vertexList, vertexList, 1], 1], Last[#] != 0 &]; {conns, weigth} = Transpose[combs]; gr = Graph[conns, EdgeWeight -> weigth, ImageSize -> Large] The resulting graph is highly connected, but the connections/edges are weighted; Mathematica can easily disentangle this: CommunityGraphPlot[gr, ImageSize -> Large] One still has to convince oneself of the fact that this has something to do with the periodicity of the data. So I plot the data with a colored background according to the graph communities: grcomms = FindGraphCommunities[gr]; indx = Flatten /@ Map[Position[vertexList, #] &, grcomms, {2}]; colr = {Red, Yellow, Magenta}; prolog = MapIndexed[{colr[[First[#2]]], Rectangle[{1 + (#1 - 1) vertexNameLength, -14}, {#1 vertexNameLength, 30}]} &, indx, {2}]; ListLinePlot[temperatNums, Prolog -> prolog, ImageSize -> Large] `Does that help? In any case: You should check carefully what I did! Regards -- Henrik
Posted 1 year ago
Hi Henrik,Please take a look at the attached manuscript. Studying periodicity using the complex network is a novel approach in water engineering, so I decided to apply this concept for 44 rain stations in the northern region of Algeria.If you remember, you have extracted a graph for one of the stations about two months ago (Figure 6). I need your help in providing graphs for all stations. We will write the section related to complex network and also the results and other details. May I send you data from all the stations?It's my pleasure and honor to work with you.Regards, Mohammad Attachments:
Posted 1 year ago
Dear Prof. Ghorbani,oops - I nearly missed that post, sorry! I will try to contact you via ResearchGate.netRegards -- Henrik
Posted 24 days ago
Dear Henrik,I hope your Christmas is filled with joy this year!I applied your program for the hourly river water temperature time series of the Skokomish river in the USA. If possible, please have a look at the notebook and let me know the output is correct or not?Have a nice weekend Attachments: | 2,082 | 6,302 | {"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} | 2.953125 | 3 | CC-MAIN-2021-04 | latest | en | 0.62466 |
https://www.physicsforums.com/threads/rotational-velocity.144527/ | 1,544,985,250,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827963.70/warc/CC-MAIN-20181216165437-20181216191437-00361.warc.gz | 992,002,372 | 14,979 | # Homework Help: Rotational velocity
1. Nov 19, 2006
### cd80187
I have been trying this question for so long and I still cannot figure out how to do it...
A solid brass ball of mass 8.2 g will roll smoothly along a loop-the-loop track when released from rest along the straight section. The circular loop has radius R = 4.9 m, and the ball has radius r << R. (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop? (b) If the ball is released at height h = 6.00R, what is the magnitude of the horizontal force component acting on the ball at point Q?
The picture looks like a straight line angling towards the ground, and when it gets to the ground, it shallows out and does a loop-the-loop. The radius of the loop is R, and the ball is started from height h. Point Q is a line drawn from the center of the loop, directly right to the edge of the loop.
So for this problem, you use the conservation of energy. I have been using M x g x h= M x g x h + 1/2m(v squared) + (1/2 Icom Omega squared). So I know that since the ball is on the verge of leaving at the top of the loop, the kinetic energy must be zero, but my guess is that it still has rotational energy, so how am I supposed to find that out? And I know that for part b, I must use F= m(v squared)/r, but once again, I don't know how to find v, since there is v and omega. But thank you for the help in advance
2. Nov 19, 2006
### Staff: Mentor
That's not true. The minimum speed at the top of the loop to just barely maintain contact is not zero. Hint: Analyze the forces acting on the ball at that point and apply Newton's 2nd law. What kind of acceleration is the ball undergoing?
Another hint: Since the ball "rolls smoothly", its translational and rotational speed are related by what simple formula?
3. Nov 19, 2006
### cd80187
Well, the ball is exerting falling under gravity, therefore, would the way to find v simply be F (m x a) = m x (v squared)/R in which V squared is the only unknown, and then once that is found, I can use the equation V = omega x R and substitute it to get rid of either v or omega, and just go from there?
4. Nov 19, 2006
### Staff: Mentor
Good, but what forces act on the ball when it's at the top of the loop?
5. Nov 19, 2006
### cd80187
You have the force of gravity acting down, the centripital force acting down, and the normal force acting down... So would it be F normal + (m x g) = m(v squared)/R?
6. Nov 19, 2006
### Staff: Mentor
Careful here. "Centripetal force" is not a kind of force, it's just the name given to the net force when the acceleration is centripetal. ("Centripetal" just means "toward the center".)
Only two forces act on the ball: gravity and the normal force. They add up to be the net force, which in this case is a centripetal force.
Good. And if the ball barely maintains contact with the track, what can you say about the normal force?
7. Nov 19, 2006
### cd80187
It would have to be 0 because there is nothing pushing back on it, right?
8. Nov 19, 2006
### Staff: Mentor
That's correct.
9. Nov 19, 2006
### cd80187
Thank you very much for the help
10. Nov 19, 2006
### cd80187
But just to ensure, I find v using the m x g = m(v squared)/ R, and then using that, I will use the equation m x g x h = (m g h) + .5 m (v squared) + (.5 (I)(v/r)squared? And I = m (rsquared), correct?
11. Nov 19, 2006
### OlderDan
You need to look up the moment of inertia of a sphere about its CM. The h values on the two sides of your equation are not the same. On is the starting height and the other is related to the size of the loop. | 989 | 3,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-51 | latest | en | 0.947999 |
https://numberworld.info/111001221210112 | 1,603,705,575,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891203.69/warc/CC-MAIN-20201026090458-20201026120458-00262.warc.gz | 460,521,579 | 4,181 | # Number 111001221210112
### Properties of number 111001221210112
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 13549953761
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
64f47c5c2000
Base 32:
34uhu5o800
sin(111001221210112)
0.99766537623909
cos(111001221210112)
-0.068291998460448
tan(111001221210112)
-14.608817998157
ln(111001221210112)
32.340562319073
lg(111001221210112)
14.045327756822
sqrt(111001221210112)
10535711.70876
Square(111001221210112)
1.2321271110136E+28
### Number Look Up
111001221210112 (one hundred eleven trillion one billion two hundred twenty-one million two hundred ten thousand one hundred twelve) is a very impressive figure. The cross sum of 111001221210112 is 16. If you factorisate the number 111001221210112 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 13549953761. The figure 111001221210112 has 28 divisors ( 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 13549953761, 27099907522, 54199815044, 108399630088, 216799260176, 433598520352, 867197040704, 1734394081408, 3468788162816, 6937576325632, 13875152651264, 27750305302528, 55500610605056, 111001221210112 ) whith a sum of 221988892482846. 111001221210112 is not a prime number. 111001221210112 is not a fibonacci number. 111001221210112 is not a Bell Number. The figure 111001221210112 is not a Catalan Number. The convertion of 111001221210112 to base 2 (Binary) is 11001001111010001111100010111000010000000000000. The convertion of 111001221210112 to base 3 (Ternary) is 112120000121112221220002221021. The convertion of 111001221210112 to base 4 (Quaternary) is 121033101330113002000000. The convertion of 111001221210112 to base 5 (Quintal) is 104022121000112210422. The convertion of 111001221210112 to base 8 (Octal) is 3117217427020000. The convertion of 111001221210112 to base 16 (Hexadecimal) is 64f47c5c2000. The convertion of 111001221210112 to base 32 is 34uhu5o800. The sine of the number 111001221210112 is 0.99766537623909. The cosine of the figure 111001221210112 is -0.068291998460448. The tangent of the number 111001221210112 is -14.608817998157. The root of 111001221210112 is 10535711.70876.
If you square 111001221210112 you will get the following result 1.2321271110136E+28. The natural logarithm of 111001221210112 is 32.340562319073 and the decimal logarithm is 14.045327756822. You should now know that 111001221210112 is very unique figure! | 960 | 2,587 | {"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-2020-45 | latest | en | 0.572976 |
https://www.1onepsilon.com/single-post/2017/07/15/Seeking-Patterns-with-Math | 1,585,463,980,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00369.warc.gz | 816,770,304 | 134,863 | © 2016 by One on Epsilon PTY LTD
# Seeking Patterns with Math
July 15, 2017
Is math important to you? Is it because you make use of basic math facts in your daily life, such as when you go shopping or walking down the street? Sure, but some people use it for technical work in fields such as finance, engineering, chemistry and architecture, just to name a few. Pretty much every field has some math applications. Math can be an incredibly useful tool. Is it a mistake, however, for us to think of math as only being a tool? We believe it is so much more.
In fact, math can be studied for its own sake. As Arthur Benjamin, author of The Magic of Math: Solving for x and Figuring Out Why, beautifully says, “Math is the science of finding patterns.” Those patterns could involve anything from numeric sequences to geometric relationships. For example, modular arithmetic, first developed by Carl Friedrich Gauss, is a type of arithmetic done on a number circle rather than a number line. Numbers reset to 0 upon reaching a certain fixed number. The classic example is arithmetic on a clock: when we get to 12, the count resets to 0. So, for example, on a clock, 10 o’clock plus 4 o’clock equals 2 o’clock. This is mod 12 arithmetic.
While modular arithmetic does have some wonderful properties, one of its major practical uses only came about in the 1970s when mathematicians used it in developing RSA encryption -- a method used to send and receive secure data like private love letters.
In a short ten minute Numberphile video James Grime wonderfully explains how the RSA key works:
If math is the science of finding patterns, then how do we actually find them? Well, there may be many avenues; here, let’s consider two popular themes:
1) Induction: we find a new or a unique rule that helps explain the pattern.
2) Deduction: we use other existing patterns to help uncover the new pattern.
In this blog, we will use the deductive method to help us solve the following puzzle:
In the above diagram, we have two primary shapes: a quarter circle with a radius, r, of length 6 inches, and a rectangle inscribed inside the quarter circle whose side lengths of x and y add up to 8 inches. The problem is to find the area of the white triangle.
To solve this problem, we can use our existing knowledge of these shapes (hint: deduction). Observe that the area of the white triangle is half the area of the rectangle with sides of length x and y as in the picture.
Now, since the area of the whole rectangle is xy (that is, x multiplied by y), the area of the white triangle is:
Identifying the area formula is a really important step. What we need now is a way to calculate xy. Well, don't forget that the problem teases us with a clue: x + y = 8. Unfortunately, this equation alone is not enough to find xy. We need something else.
Actually, there is a very special pattern hidden in this picture. We know that when we are attempting to measure a missing side length of a right-angled triangle, we can use Pythagoras' Theorem:
Clearly, we can tell that the perpendicular side lengths a and in the formula will be substituted by x and y, but what about - the hypotenuse? Before you read on, give it a try on your own.
Notice that the hypotenuse of the triangle we are interested in has the same length as the radius of the circle. Hence, c = 6. Using the Pythagoras' Theorem, we can then see:
This might just be the extra something that we need! We now make an unexpected move that combines the given algebraic information x + y = 8 and the deduced Pythagorean equation: We take the square of x + y = 8. Let’s see:
Notice that in the second to last step, we isolate xy. Also, in the last step, do you see how Pythagoras' Theorem played a role in the substitution of x^2+ y^2? Additionally, every time we write "=>", we read it as “which means that”, so it gives us a sequence of statements, each following on from the previous one.
Now that we have worked out the product of x and y, then the area of the white triangle is just:
Reflection time: What do you think was the most difficult part of our working?
Perhaps it was the point where we decided to take the square of the equation x + y = 8. What on earth made us decide to do that? It was because we had insight to deduce where that mathematical process would lead us. We knew that to complete this mathematical puzzle, we needed to find xy. We did it by using the information given, deducing new information (the Pythagorean equation) from the geometric patterns of the puzzle, and then putting it all together in a clever way.
Here's an open-ended question: Is there only one such triangle that will match the conditions given in the problem? In other words, what are all the values of x and y that would still give us an area of 7?
We appreciate feedback! You may have also tried solving the problem differently. Do you have any insight from that? Did you explore your solution with your loved ones? We would love you to share your working with us.
With a little inspiration, curiosity and persistence, people of all different math comfort levels can make their own discoveries. In the link below, Arthur Benjamin, in his TED talk, “The Magic of Fibonacci Numbers” shares his inspiration.
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September 29, 2018 | 1,287 | 5,426 | {"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-2020-16 | latest | en | 0.908931 |
https://tikz.net/electric_circuit_resistor/ | 1,723,122,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00458.warc.gz | 477,178,219 | 11,653 | # Circuit with resistors & Kirchhoff’s laws
Some basic circuit diagrams with one resistor, or more in parallel or series, as well as current branches to study Kirchhoff’s circuit laws.
For more related figures, please see the “circuits” tag.
Edit and compile if you like:
% Author: Izaak Neutelings (Februari, 2020)
% http://texample.net/tikz/examples/tag/circuitikz/
% http://texample.net/tikz/examples/circuitikz/
% https://www.overleaf.com/learn/latex/CircuiTikz_package
% http://texdoc.net/texmf-dist/doc/latex/circuitikz/circuitikzmanual.pdf
% http://repositorios.cpai.unb.br/ctan/graphics/pgf/contrib/circuitikz/circuitikzmanual.pdf
\documentclass[border=3pt,tikz]{standalone}
\usepackage{amsmath} % for \dfrac
\usepackage{physics}
\usepackage{tikz,pgfplots}
\usepackage[siunitx]{circuitikz} %[symbols]
\usepackage[outline]{contour} % glow around text
\usetikzlibrary{arrows}
\usetikzlibrary{decorations.markings}
\tikzset{>=latex} % for LaTeX arrow head
\usepackage{xcolor}
\colorlet{Icol}{blue!50!black}
\colorlet{Ccol}{orange!90!black}
\colorlet{Rcol}{green!50!black}
\colorlet{loopcol}{red!90!black!25}
\colorlet{pluscol}{red!60!black}
\colorlet{minuscol}{blue!60!black}
\newcommand\EMF{\mathcal{E}} %\varepsilon}
\contourlength{1.5pt}
\tikzstyle{EMF}=[battery1,l=$\EMF$]
\tikzstyle{internal R}=[R,color=Rcol,Rcol,l=$r$,/tikz/circuitikz/bipoles/length=30pt]
\tikzstyle{loop}=[->,red!90!black!25]
\tikzstyle{loop label}=[loopcol,fill=white,scale=0.8,inner sep=1]
\tikzstyle{thick R}=[R,color=Rcol,thick,Rcol,l=$R$]
%\tikzset{
% loop/.style={thick,red!80!black!30,decoration={markings,
% mark=at position #1 with {\arrow{latex}}},
% postaction={decorate}},
% loop/.default=0.6}
\newcommand{\myvoltmeter}[2]
{ % #1 = name , #2 = rotation angle
\begin{scope}[transform shape,rotate=#2]
\draw[thick] (#1)node(){$\mathbf V$} circle (11pt);
\draw[rotate=45,-latex] (#1) +(-17pt,0) --+(17pt,0);
\end{scope}
}
\begin{document}
% RESISTOR without battery
\begin{tikzpicture}
\draw (0,2) to [short,*-] (3,2) to[R,color=Rcol,thick,l=$R$] (3,0) to [short,-*] (0,0);
\node[below left] at (0,2) {$+$};
\node[above left] at (0,0) {$-$};
\node at (0,1) {$\Delta V$};
\end{tikzpicture}
% RESISTOR with battery and arrow
\begin{tikzpicture}
\draw (0,0) to[EMF] (0,2) -- (3,2)
to[thick R] (3,0) -- (0,0);
\node at (-0.35,0.7) {$-$};
\node at (-0.35,1.4) {$+$};
\draw[->,Icol] (0.5, 2.15) --++ (1.2,0) node[midway,above=1] {current $I$};
\draw[->,Icol] (0.5,-0.15) --++ (1.2,0) node[midway,below=1] {electron flow};
\end{tikzpicture}
% RESISTOR with battery and arc arrow
\begin{tikzpicture}
\def\ang{120}
\def\a{0.8}
\def\b{0.7}
\draw (0,0) to[EMF] (0,2) -- (3,2)
to[thick R] (3,0) -- (0,0);
\node at (-0.35,0.7) {$-$};
\node at (-0.35,1.4) {$+$};
\draw[->,Icol] ({1.5+\a*cos(\ang)},{1+\b*sin(\ang)}) arc (120:-100:{\a} and {\b});
\end{tikzpicture}
% RESISTOR with EMF + internal resistance
\begin{tikzpicture}
\draw (0,0) to[EMF] (0,1.4)
to[internal R] (0,2.5) -- (0,3) --++ (3,0)
to[thick R] ++(0,-3) -- (0,0);
%\draw (3,2.5) --++ (0.6,0) to[voltmeter,color=white,name=M] ++(0,-2) --++ (-0.6,0);
%\myvoltmeter{M}{0} % rotate
\fill[black] (3,2.5) circle (0.05) node[right] {$V_a$};
\fill[black] (3,0.5) circle (0.05) node[right] {$V_b$};
\node at (-0.35,0.44) {$-$};
\node at (-0.35,1.05) {$+$};
\end{tikzpicture}
% RESISTOR with EMF + internal resistance
\begin{tikzpicture}
\draw (0,0) to[EMF] (0,1.4)
to[internal R] (0,2.5) -- (0,3) --++ (3,0)
to[thick R] ++(0,-3) -- (0,0);
\draw (3,2.5) --++ (1.4,0) to[voltmeter,color=white,name=M] ++(0,-2) --++ (-1.4,0);
\myvoltmeter{M}{0} % rotate
\fill[black] (3,2.5) circle (0.05) node[left] {$V_a$};
\fill[black] (3,0.5) circle (0.05) node[left] {$V_b$};
\node at (-0.35,0.44) {$-$};
\node at (-0.35,1.05) {$+$};
\end{tikzpicture}
%% RESISTOR with EMF + internal resistance
%\begin{tikzpicture}
% \draw (0,0) to[EMF] (0,1.4)
% to[internal R] (0,2.5) -- (0,3) --++ (3,0)
% --++ (0,-0.5) coordinate (T) --++ (-0.9,0)
% to[thick R] ++(0,-2) --++ (0.9,0) |- (0,0)
% (T) --++ (0.9,0) to[voltmeter,color=white,name=M] ++(0,-2) --++ (-0.9,0); %,l=$R$
% \myvoltmeter{M}{0} % rotate
% \node at (-0.35,0.44) {$-$};
% \node at (-0.35,1.05) {$+$};
%\end{tikzpicture}
%
%
%% RESISTOR with EMF + internal resistance + boxes
%\begin{tikzpicture}
% \draw (0,0) to[EMF] (0,1.4)
% to[internal R] (0,2.5) -- (0,3) --++ (3,0)
% --++ (0,-0.5) coordinate (T) --++ (-0.9,0)
% to[thick R] ++(0,-2) --++ (0.9,0) |- (0,0)
% (T) --++ (0.9,0) to[voltmeter,color=white,name=M] ++(0,-2) --++ (-0.9,0); %,l=$R$
% \myvoltmeter{M}{0} % rotate
% \node at (-0.35,0.44) {$-$};
% \node at (-0.35,1.05) {$+$};
% \draw[dashed,very thin] (-0.9,0.3) rectangle (0.8,2.7);
% \draw[dashed,very thin] ( 1.6,0.3) rectangle (4.5,2.7);
%\end{tikzpicture}
% RESISTOR with EMF + AMPEREMETER + VOLTMETER
\begin{tikzpicture}
\draw (0,0) to[EMF] (0,1.4)
to[internal R] (0,2.5) -- (0,3) to[ammeter] ++(3,0)
to[thick R] ++(0,-3) -- (0,0);
\draw (3,2.5) --++ (1.5,0) to[voltmeter,color=white,name=M] ++(0,-2.0) --++ (-1.5,0);
\fill[black] (3,2.5) circle (0.05);
\fill[black] (3,0.5) circle (0.05);
\myvoltmeter{M}{0} % rotate
\node at (-0.35,0.44) {$-$};
\node at (-0.35,1.05) {$+$};
\node[] at (1.5,3.65) {$R_\mathrm{A}$};
\node[] at (5.25,1.5) {$R_\mathrm{V}$};
\end{tikzpicture}
% KIRCHHOFF's RULE 1
\begin{tikzpicture}
\def\a{0.2}
\def\b{0.4}
\def\H{2}
\def\W{2}
\draw[rounded corners=10,loop] (\a,2*\a) |- (2*\W-\a,\H-\a) |- (2*\a,\a);
\draw[rounded corners=10,loop] (\b,\b+\a) |- (\W-.6*\b,\H-\b) |- (\b+\a,\b);
\draw[rounded corners=10,loop] (\W+.6*\b,\b+\a) |- (2*\W-\b,\H-\b) |- (\W+.6*\b+\a,\b);
\fill[black] (0,\H) circle (0.05); %node[above] {$V_i$};
\fill[black] (0,0) circle (0.05); %node[below] {$V_f$};
\fill[black] (\W,\H) circle (0.05); %node[above] {$V_{1a}$};
\fill[black] (\W,0) circle (0.05); %node[below] {$V_{1b}$};
\fill[black] (2*\W,\H) circle (0.05); %node[above] {$V_{2a}$};
\fill[black] (2*\W,0) circle (0.05); %node[below] {$V_{2b}$};
\draw ( 0,0) |- ++(\W,\H) |- cycle;
\draw (\W,0) -| ++(\W,\H) --++ (-2,0);
\node[fill=white,rotate=90,inner sep=2] at (0,\H/2) {$.\,.\,.$};
\node[fill=white,rotate=90,inner sep=2] at (\W,\H/2) {$.\,.\,.$};
\node[fill=white,rotate=90,inner sep=2] at (2*\W,\H/2) {$.\,.\,.$};
\node[left] at (0,\H/2) {$\Delta V$};
\node[left=-1] at (\W,\H/2) {\contour{white}{$\Delta V_a$}};
\node[right] at (2*\W,\H/2) {$\Delta V_b$};
\node[loop label] at (0.40*\W,0.90*\H) {$c$};
\node[loop label] at (0.55*\W,0.80*\H) {$a$};
\node[loop label] at (1.50*\W,0.80*\H) {$b$};
\end{tikzpicture}
% KIRCHHOFF's RULE 2
\begin{tikzpicture}
\def\r{0.6}
\def\R{1}
\coordinate (O) at (0,0);
\draw (O) --++ (180:\R);
\draw (O) --++ (60:\R);
\draw (O) --++ (-60:\R);
\draw[<-,Icol] (O)++(140:0.4*\r) --++ (180:\r) node[midway,above] {$I_0$};
\draw[->,Icol] (O)++( 30:0.5*\r) --++ ( 60:\r) node[midway,right=1] {$I_1$};
\draw[->,Icol] (O)++(-30:0.5*\r) --++ (-60:\r) node[midway,right=1] {$I_2$};
\end{tikzpicture}
% RESISTOR in series
\begin{tikzpicture}
\draw (0,2) to [short,*-] (0.6,2)
to[thick R,l=$R_1$] ++(1.5,0)
to[thick R,l=$R_2$] ++(1.5,0)
to[thick R,l=$R_3$] ++(1.5,0)
-- ++(1.5,0) node[midway,fill=white,inner sep=5,scale=1.2] {$.\,.\,.$}
-- (7,2) -- (7,0) to[short,-*] (0,0);
\node at (0,1) {$\Delta V$};
\node[below left] at (0,2) {$+$};
\node[above left] at (0,0) {$-$};
\end{tikzpicture}
% RESISTOR in parallel
\begin{tikzpicture}
\node[fill=white,inner sep=5,scale=1.2] (ET) at (7.4,2) {$.\,.\,.$};
\node[fill=white,inner sep=5,scale=1.2] (EB) at (7.4,0) {$.\,.\,.$};
\node at (0,1) {$\Delta V$};
\draw (0,2) to[short,*-] (2,2) to[thick R,l=$R_1$] (2,0) to[short,-*] (0,0);
\draw (2,2) -- (4,2) to[thick R,l=$R_2$] (4,0) -- (2,0);
\draw (4,2) -- (6,2) to[thick R,l=$R_3$] (6,0) -- (4,0);
\draw (6,2) -- (ET.180);
\draw (6,0) -- (EB.180);
\node at (0,1) {$\Delta V$};
\node[below left] at (0,2) {$+$};
\node[above left] at (0,0) {$-$};
\end{tikzpicture}
% RESISTOR in series - zoomed in
\begin{tikzpicture}
\def\a{1.8}
\def\b{0.35}
\draw (-\b,0) -- (0,0) to[thick R,l=$R_1$] (\a,0) to[thick R,l=$R_2$] (2*\a,0) --++ (\b,0);
\fill (0,0) circle (0.05) node[above] {$V_a$};
\fill (\a,0) circle (0.05) node[above] {$V_b$};
\fill (2*\a,0) circle (0.05) node[above] {$V_c$};
\end{tikzpicture}
% RESISTOR in parallel - R1, R2
\begin{tikzpicture}
\draw[rounded corners=12,loop] (0.3,0.3) |- (3.7,2.8) |- (0.4,0.2);
\draw[rounded corners=10,loop] (0.5,0.6) |- (1.7,2.6) |- (0.6,0.4);
\draw[rounded corners=10,loop] (2.3,0.6) |- (3.5,2.6) |- (2.4,0.4);
\draw (0,0) to[EMF] (0,1.4)
to[internal R] (0,2.5)
-- (0,3) coordinate (I0) --++ (2,0) coordinate (I1)
to[thick R,l=$R_1$] (2,0) coordinate (IF) -- (0,0);
\draw (2,3) --++ (2,0) coordinate (I2)
to[thick R,l=$R_2$] (4,0) -- (0,0);
\node at (-0.35,0.44) {$-$};
\node at (-0.35,1.05) {$+$};
\draw[->,Icol] (I0)++(0.5, 0.1) --++ (1,0) node[midway,above] {$I_0$};
\draw[->,Icol] (I1)++(0.1,-0.1) --++ (0,-0.65) node[midway,right] {\contour{white}{$I_1$}};
\draw[->,Icol] (I1)++(0.5, 0.1) --++ (1,0) node[midway,above] {$I_2$};
\draw[->,Icol] (I2)++(0.1,-0.1) --++ (0,-0.65) node[midway,right] {$I_2$};
\draw[->,Icol] (IF)++(-0.5,-0.1) --++ (-1,0) node[midway,below] {$I_0$};
\node[loop label] at (0.90,2.78) {$c$};
\node[loop label] at (1.12,2.62) {$a$};
\node[loop label] at (2.92,2.62) {$b$};
\end{tikzpicture}
% RESISTOR in parallel - R12
\begin{tikzpicture}[scale=0.8]
\draw (0,0) to[EMF] (0,1.4)
to[internal R] (0,2.5) -- (0,3) coordinate (I0) --++ (3,0)
to[thick R,l=$R_\mathrm{eq}$] (3,0) -- (0,0); %R_{12}
\node at (-0.35,0.43) {$-$};
\node at (-0.35,1.08) {$+$};
\draw[->,Icol] (I0)++(1.0, 0.1) --++ (1,0) node[midway,above] {$I_0$};
\end{tikzpicture}
% RESISTOR in parallel - R12r
\begin{tikzpicture}[scale=0.7]
\draw (0,0) to[EMF] (0,3) --++ (3,0)
to[thick R,l=$R$] (3,0) -- (0,0); %_{12r}
\node at (-0.35,1.15) {$-$};
\node at (-0.35,1.90) {$+$};
\draw[->,Icol] (I0)++(1.0, 0.1) --++ (1,0) node[midway,above] {$I_0$};
\end{tikzpicture}
\end{document} | 4,735 | 10,084 | {"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} | 3.203125 | 3 | CC-MAIN-2024-33 | latest | en | 0.317049 |
https://physics-network.org/what-is-the-volume-of-1-0-kg-of-mercury/ | 1,695,312,218,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00820.warc.gz | 502,896,540 | 56,927 | # What is the volume of 1.0 kg of mercury?
One kilogram of mercury converted to liter equals to 0.074 L.
## What is the volume of mercury in?
Mercury has a mass of 3.3 x 1023kilograms. This mass is contained in a volume of 14.6 billion cubic miles (60.8 billion cubic km).
## How do you find volume using specific gravity?
1. Where Vsg is the Volume from Specific Gravity (m^3)
2. m is the mass of the substance (kg)
3. SG is the specific gravity.
## What is the density of mercury in kg m3?
The value for the density of mercury refers to an ambient pressure of 101325 Pa. The density of mercury at 20 degree C is 13545.848 kg/m3.
## How do I find the volume?
To find the volume of a box, simply multiply length, width, and height — and you’re good to go! For example, if a box is 5×7×2 cm, then the volume of a box is 70 cubic centimeters.
## What is the mass of 40.0 ml of mercury density 13.6 g ml )?
The mercury metal has a mass of 48.3g.
## What is the volume of mercury in cm3 if mass of Mercury is 1kg and density is 13.6 g cm3?
Hence, volume of the given mercury is 10 cm³.
## What is the formula for calculating volume of a sphere?
The formula for the volume of a sphere is V = 4/3 πr³. See the formula used in an example where we are given the diameter of the sphere.
## What is the specific volume calculator?
The specific volume is the reciprocal of the density. It is denoted by the symbol “v”. In short, specific is termed as volume per unit mass of a substance. This online Specific Volume Calculator is useful in calculating the specific volume of a mass unit of substance.
## How do you find volume from density?
Divide the mass by the density of the substance to determine the volume (mass/density = volume). Remember to keep the units of measure consistent. For example, if the density is given in grams per cubic centimeter, then measure the mass in grams and give the volume in cubic centimeters.
## What is the density of mercury 13.6 g cm3 in units of kg m3?
=13. 6×10−3×106=13. 6×103 kg/m3.
## Why is density of mercury is 13600 kg m3?
Density of mercury = 13600 Kgm-3. Density of water at 4oC = 1000 kg m-3. Relative density = density of substance /density of water at 40C. Relative density of mercury = 13600 Kgm-3/1000 kg m-3 = 13.6.
## What is the density of mercury in physics?
The density of mercury is 13.6 g/mL.
## What are 3 ways to calculate volume?
1. Solve for Volume by Space. All physical objects occupy space, and you can find the volume for some of them by measuring their physical dimensions.
2. Solve for Volume by Density and Mass. Density is defined as an object’s mass per a given unit of volume.
3. Solve for Volume by Displacement.
## How do you calculate volume of a liquid?
1. First, determine the total mass of the liquid. Pour the liquid into a container to measure the total mass.
2. Next, determine the mass of the liquid through a table lookup. For water density, this would be approx. 997 kg/m^3.
3. Finally, calculate the liquid volume using the formula LV = M/d.
## What volume of mercury density 13.6 g mL will 0.23 grams occupy?
Therefore, the volume of mercury is 0.017 mL.
## What is the volume of 35.0 g of ethyl alcohol which has a density of 0.79 g mL?
1 Answer. The volume would be 2.01 qt .
## What is the volume of mercury incm3 if the mass m of mercury is 136grams g and density R of mercury is 13.6g cm3 *?
What is the volume of mercury in cm3, if the mass (m) of mercury is 136 grams (g) and density (r)of mercury is 13.6 g/cm3? A:136 cm3.
## What is the block weight in kg if volume is 320 cm3 and density 8.9 g cm3?
| दिये गए ब्लॉक का भार कलोग्राम मे क्या होगा ,यदि इसका आयतन (V) 320 सेमी3. और घनत्व 8.9 ग्राम / सेमी3 है? ans:- 2.848
## What is the specific volume of mercury in CM³ g?
The answer is: The change of 1 cc – cm3 ( cubic centimeter ) unit of a mercury amount equals = to 13.53 g ( gram ) as the equivalent measure for the same mercury type.
## How do you find volume from area?
It gives the proportion of surface area per unit volume of the object (e.g., sphere, cylinder, etc.). Therefore, the formula to calculate surface area to volume ratio is: SA/VOL = surface area (x2) / volume (x3) SA/VOL = x-1 , where x is the unit of measurement. | 1,179 | 4,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-40 | latest | en | 0.860239 |
https://fred.stlouisfed.org/series/MHICILBNC37127A052NCEN | 1,603,948,093,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902745.75/warc/CC-MAIN-20201029040021-20201029070021-00393.warc.gz | 331,399,379 | 13,547 | # 90% Confidence Interval Lower Bound of Estimate of Median Household Income for Nash County, NC (MHICILBNC37127A052NCEN)
Observation:
2018: 46,046
Updated: Dec 13, 2019
Units:
Dollars,
Frequency:
Annual
1Y | 5Y | 10Y | Max
NOTES
Source: U.S. Census Bureau
Frequency: Annual
#### Notes:
The U.S. Census Bureau provides annual estimates of income and poverty statistics for all school districts, counties, and states through the Small Area Income and Poverty Estimates (SAIPE) program. The bureau's main objective with this program is to provide estimates of income and poverty for the administration of federal programs and the allocation of federal funds to local jurisdictions. In addition to these federal programs, state and local programs use the income and poverty estimates for distributing funds and managing programs.
Household income includes income of the householder and all other people 15 years and older in the household, whether or not they are related to the householder. Median is the point that divides the household income distributions into two halves: one-half with income above the median and the other with income below the median. The median is based on the income distribution of all households, including those with no income.
A confidence interval is a range of values, from the lower bound to the respective upper bound, that describes the uncertainty surrounding an estimate. A confidence interval is also itself an estimate. It is made using a model of how sampling, interviewing, measuring, and modeling contribute to uncertainty about the relation between the true value of the quantity we are estimating and our estimate of that value. The "90%" in the confidence interval listed above represents a level of certainty about our estimate. If we were to repeatedly make new estimates using exactly the same procedure (by drawing a new sample, conducting new interviews, calculating new estimates and new confidence intervals), the confidence intervals would contain the average of all the estimates 90% of the time. For more details about the confidence intervals and their interpretation, see this explanation.
#### Suggested Citation:
U.S. Census Bureau, 90% Confidence Interval Lower Bound of Estimate of Median Household Income for Nash County, NC [MHICILBNC37127A052NCEN], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/MHICILBNC37127A052NCEN, October 29, 2020.
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Updating graph. | 531 | 2,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-45 | latest | en | 0.905942 |
https://oeis.org/A137985 | 1,623,638,817,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00185.warc.gz | 408,197,244 | 5,712 | 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!)
A137985 Complementing any single bit in the binary representation of these primes produces a composite number. 8
127, 173, 191, 223, 233, 239, 251, 257, 277, 337, 349, 373, 431, 443, 491, 509, 557, 653, 683, 701, 733, 761, 787, 853, 877, 1019, 1193, 1201, 1259, 1381, 1451, 1453, 1553, 1597, 1709, 1753, 1759, 1777, 1973, 2027, 2063, 2333, 2371, 2447, 2633, 2879, 2917 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS If 2^m is the highest power of 2 in the binary representation of the prime p, there is no requirement that p+2^(m+1) be composite. Sequence A065092 imposes this extra requirement. The prime 223 is the first number in this sequence that is not in A065092. Mentioned Feb 25 2008 by Terence Tao in his blog http://terrytao.wordpress.com. Tao proves that there are an infinite number of these primes in every fixed base. Digitally delicate primes in base 2. - Marc Morgenegg, Apr 21 2021 REFERENCES Cohen, Fred; Selfridge, J. L., Not every number is the sum or difference of two prime powers. Collection of articles dedicated to Derrick Henry Lehmer on the occasion of his seventieth birthday. Math. Comp. 29 (1975), 79-81. MR0376583 (51 #12758). LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 Warren D. Smith et al., Primes such that every bit matters?, Yahoo group "primenumbers", April 2013. Warren D. Smith and others, Primes such that every bit matters?, digest of 14 messages in primenumbers Yahoo group, Apr 3 - Apr 9, 2013. [Cached copy] Terence Tao, A remark on primality testing and decimal expansions, arXiv:0802.3361 [math.NT], 2008-2010; Journal of the Australian Mathematical Society 91:3 (2011), pp. 405-413. EXAMPLE The numbers produced by complementing each of the 8 bits of 223 are 95, 159, 255, 207, 215, 219, 221 and 222, which are all composite. MAPLE P:=proc(n) local a, b, c, d, j, k, ok; a:=ithprime(n); b:=convert(a, base, 2); ok:=1; for k from 1 to nops(b) do c:=b; c[k]:=(c[k]+1) mod 2; d:=0; for j from 1 to nops(c) do d:=2*d+c[-j]; od; if isprime(d) then ok:=0; break; fi; od; if ok=1 then a; fi; end: seq(P(i), i=1..420); # Paolo P. Lava, Dec 24 2018 MATHEMATICA t={}; k=1; While[Length[t]<100, k++; p=Prime[k]; d=IntegerDigits[p, 2]; n=Length[d]; i=0; While[i z; Select[Prime /@ Range[3, PrimePi[10^6]], isWPbase2@# &] (* Terentyev Oleg, Jul 17 2011 *) PROG (PARI)f(p)={pow2=1; v=binary(p); L=#v; forstep(k=L, 1, -1, if(v[k], p-=pow2; if(isprime(p), return(0), p+=pow2), p+=pow2; if(isprime(p), return(0), p-=pow2)); pow2*=2); return(1)}; forprime(p=2, 2879, if(f(p), print1(p, ", "))) \\ Washington Bomfim, Jan 18 2011 (PARI) is_A137985(n)=!for(k=1, n, isprime(bitxor(n, k)) && return; k+=k-1) && isprime(n) \\ Note: A bug in early versions of PARI 2.6 (execute "for(i=0, 1, i>3 && error(buggy); i=9)" to check) makes that this is is_A065092 rather than is_A137985 as expected. For these versions, replace the upper limit n with n\2. \\ M. F. Hasler, Apr 05 2013 (Python) from sympy import isprime, primerange def ok(p): # p assumed prime return not any(isprime((1<
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Last modified June 13 22:46 EDT 2021. Contains 345016 sequences. (Running on oeis4.) | 1,232 | 3,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-25 | latest | en | 0.782495 |
https://www.meracalculator.com/physics/thermodynamics/suvat-calculator.php | 1,675,205,675,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00422.warc.gz | 879,850,494 | 18,374 | # SUVAT Calculator
Choose the unknown term and enter the required values in this SUVAT calculator.
Formula t = v - ua
## SUVAT Calculator
This SUVAT calculator solves for the values of quantities involved in the constantly accelerated motion. Using this uniformly accelerated motion calculator you can find the values of:
• Distance (S)
• Time (t)
• Initial velocity (u)
• Final velocity (v)
• Acceleration (a)
Let’s see what do we mean by SUVAT and what are the formulas used to calculate these values.
## What is SUVAT?
SUVAT is the acronym of five motion quantities: Distance, initial velocity, final velocity, acceleration, and time respectively. These formulae are only applicable if the acceleration is uniform throughout the motion.
## SUVAT formulas:
The five formulas used to calculate these values are:
s = ut + ½ * at2
### Initial velocity:
u = (s - ½ * at2) / t
### Final velocity:
v = (s + ½ * at2) / t
a = v2 - u2 / 2s
t = v - u / a
## How to calculate SUVAT quantities?
The most difficult equation by glance is of distance. So let’s see an example of calculating this quantity.
Example:
If a man has been running with an acceleration of 2ms-2 for two 5 seconds. The initial velocity is 0ms-1 and the final velocity is 10ms-1. Find the distance.
Solution:
s = (0)(5) + ½ * (2)(5)2
s = ½ * 50
s = 25 m | 348 | 1,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2023-06 | longest | en | 0.829189 |
https://cm2feet.com/converter/computer/data-storage/megabyte-to-gigabyte/ | 1,726,248,500,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00808.warc.gz | 154,002,861 | 6,434 | # =
Precision:
#### Megabyte to Gigabyte Conversion Formula:
gigabyte (GB) = megabyte (MB) / 1022.2022585963
#### How to Convert megabyte (MB) to gigabyte (GB)?
To get Gigabyte data storage, simply divide Megabyte by 1022.2022585963. With the help of this data storage converter, we can easily convert Megabyte to Gigabyte. Here you are provided with the converter, proper definitions,relations in detail along with the online tool to convert megabyte (MB) to gigabyte (GB).
#### How many Gigabyte in one Megabyte?
1 megabyte (MB) is 0.00097827997501514 gigabyte (GB).
megabyte (MB) to gigabyte (GB) converter is the data storage converter from one unit to another. It is required to convert the unit of data storage from Megabyte to Gigabyte, in data storage. This is the very basic unit conversion, which you will learn in primary classes. It is one of the most widely used operations in a variety of mathematical applications. In this article, let us discuss how to convert megabyte (MB) to gigabyte (GB), and the usage of a tool that will help to convert one unit from another unit, and the relation between Megabyte and Gigabyte with detailed explanation.
#### Megabyte Definition
A megabyte (MB or Mbyte) is a decimal multiple of the unit byte for digital information or computer storage. The prefix mega (symbol M) is defined in the International System of Units (SI) as a multiplier of 10⁶, therefore, 1 megabyte = 10⁶ bytes = 1,000,000 bytes. At the same time, traditionally this metric prefix is used to designate binary multiplier 2²⁰, so 1 MB = 1024 Kbytes (note the capital K). The correct prefix for 2²⁰ is a mebibyte (MiB), introduced by the International Electrotechnical Commission (IEC) in 1999. Binary prefixes are increasingly used in technical literature, open source software and cloud services.
#### Gigabyte Definition
A gigabyte (GB or Gbyte) is a decimal multiple of the unit byte for digital information or computer storage. The prefix giga (symbol G) is defined in the International System of Units (SI) as a multiplier of 10⁹, therefore, 1 gigabyte = 10⁹ bytes = 1,000,000,000 bytes. At the same time, traditionally this metric prefix is used to designate binary multiplier 2³⁰, so 1 GB = 1024 Mbytes. The correct prefix for 2³⁰ is a gibibyte (GiB), introduced by the International Electrotechnical Commission (IEC) in 1999. Binary prefixes are increasingly used in technical literature, open source software and cloud services.
#### megabyte (MB) to gigabyte (GB) Conversion table:
Computer Convertersions | 629 | 2,549 | {"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-2024-38 | latest | en | 0.75944 |
https://www.scribd.com/doc/91954327/Stiffness-10 | 1,498,432,296,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320593.91/warc/CC-MAIN-20170625221343-20170626001343-00545.warc.gz | 886,588,507 | 25,618 | # المحاضرة العاشرة
Lecture No. : 10
Remember
olution Steps of assembly method :
Drive the member local stiffness matrix
l
l
l
F = K D
Drive the member transformation matrix
T
Obtain the member global stiffness matrix
l
Km = T Km T
g
T
2
Remember
g
Km
Make assembly
g
Km
g
g
Km
F = K
Km
D
Make partition
Fu
Fr
=
Kuu Kur
Du
Kru Krr
Dr
3
Remember
Extract the stiffness equation
Fu
Fr
=
Kuu Kur
Du
Kru Krr
Dr
Fu = Kuu Du + Kur
Dr
Obtain the deformation
Du = Kuu
-1
{F
u
- Kur
Dr
}
4
Remember
Calculate the reactions
Fr = Kru
Du + Krr
Dr
Find internal forces in members
T
g
l
l
Fm = Km T Dm
5
d2
d3
d1
7
Drive the member local stiffness matrix
d2
d3
d1
d6
d5
d4
l
l
l
F = K D
8
Local
d2
d3
F1
F2
F3
d1
d6
F5
d5
F6
d4
F4
9
F1
k11
k12 k13 k14
k15 k16
D1
F2
k21
k22
k23 k24
k25
k26
D2
F3
k31
k32
k33 k34
k35
k36
D3
F4
k41
k42 k43 k44
k45 k46
D4
F5
k51
k52
k53 k54
k55
k56
D5
F6
k61
k62
k63 k64
k65
k66
D6
=
10
First column in
Local Stiffness matrix
d1 =1
EA
L
d2
d3
d1
∆=
1
EA
F1 =
L
EA
F4 = L
F2 = 0
F5 = 0
F3 = 0
F6 = 0
d6
d5
d4
EA
L
11
First column in Local Stiffness matrix
k11
E A
L
k21
0
k31
0
k41
=
E A
L
k51
0
k61
0
12
Second column in
Local Stiffness matrix
d2 =1
d2
d3
d1
d6
d5
d4
6 EI ∆
L2
6 EI ∆
L2
12 EI∆
L3
12 EI∆
L3
13
Second column in Local Stiffness matrix
F4 =
F1 = 0
F2 =
F3 =
12 EI
L
3
6 EI
L2
F5 =
F6 =
0
12 EI
L3
6 EI
L2
k12
0
12 EI
k22
L3
6 EI
k32
k42
k52
k62
=
L2
0
12 EI
L3
6 EI
L2
14
Third column in
Local Stiffness matrix
d3 =1
d2
d3
d1
θ
4 EI θ
L
d6
d5
d4
2 EI θ
L
6 EI θ
L2
6 EI θ
L2
15
Third column in Local Stiffness matrix
F4 =
F1 = 0
F2 =
F3 =
6 EI
L2
4 EI
L
F5 =
F6 =
0
k13
_ 6 EI
k23
L2
k33
2 EI
L
k43
k53
k63
0
6 EI
L2
4 EI
=
L
0
_ 6 EI
L2
2 EI
L
Fourth column in
Local Stiffness matrix
d4 =1
EA
L
d2
d3
d6
d1
∆=
1
EA
L
EA
F1 = L
EA
F4 =
L
F2 = 0
F5 = 0
F3 = 0
F6 = 0
d5
d4
Fourth column in Local Stiffness matrix
k14
E A
L
k24
0
k34
0
k44
=
E A
L
k54
0
k64
0
18
Fifth column in
Local Stiffness matrix d
d5 =1
d2
3
d1
d6
d5
d4
6 EI ∆
L2
6 EI ∆
L2
12 EI∆
L3
12 EI∆
L3
19
fifthcolumn in Local Stiffness matrix
F4 =
F1 = 0
F2 =
F3 =
12 EI
L
3
6 EI
L
2
F5 =
F6 =
0
12 EI
L3
6 EI
L2
k15
0
k25
-12 EI
k35
-6 EI
k45 =
k55
k65
L3
L2
0
12 EI
L3
-6 EI
L2
20
sixth column in
Local Stiffness matrix
d6 =1
2 EI θ
L
d2
d3
d6
d1
d5
4 EI θ
L
6 EI θ
L2
6 EI θ
L2
θ
d4
6 column in Local Stiffness matrix
F4 =
F1 = 0
F2 =
6 EI
F3 =
2 EI
L2
L
F5 =
F6 =
0
6 EI
L2
4 EI
L
k16
k26
k36
k46 =
k56
k66
0
6 EI
L2
4 EI
L
0
6 EI
L2
2 EI
L
22
E
AL
0
l
K
=
0
EA
L
0
0
0
12
L
EI
3
6
L
EI
2
-0
12
EIL
63
L
EI
2
0
6
L
EI
2
4
EI
L
0
6E
L
I2
2
EI
L
EA
L
0
0
E
AL
0
0
-0
12
EIL
-3
6E
L
I2
0
6
L
EI
2
2
EI
L
0
12
L
EI
-3
6E
L
I2
0
6E
L
I2
4
EI
L
23
Drive the member transformation matrix
F
g
F5
l
5
F4
F6
θ
g
F2
l2
l
g
F4
l1
F
F
F3
g
F1
24
F2l cosθ
F1l sinθ
g
F2
l2
F
l
F2 sinθ
l1
F
θ
θ
F3
F1l cosθ
g
F1
F1 = Fl1 cos θ – Fl2 sin θ
g
F2 = Fl1 sin θ + Fl2 cos θ
g
l
F3 = F3
g
25
F5l cosθ
F4l sinθ
g
F5
l5
F5 sinθ
F
θ
F
l
l4
l
F4 cosθ
θ
F6
g
F4
g
l
l
g
l
l
g
l
F4 = F4 cos θ – F4 sin θ
F5 = F5 sin θ + F5 cos θ
F 6 = F6
26
cos θ
F1g
0
0
0
0
0
0
Fl2
0
0
0
F3l
F2g
sin θ
cos θ
0
F3g
0
0
1
0
0
F4g
=
F1l
- sin θ
0
x
F4l
0
cos θ - sin θ
sin θ cos θ 0
F5
1
F6l
F5g
0
0
0
F6g
0
0
0
0
0
0
l
27
T=
cos θ
- sin θ
0
0
0
sin θ
cos θ
0
0
0
0
0
0
1
0
0
0
0
0
0
cos θ - sin θ
0
0
0
sin θ cos θ 0
0
0
0
0
0
0
0
1
28
Obtain the member global stiffness matrix
g
l
Km = T Km T
T
29
Example 1:
Draw N.F, S.F & B.M.Ds for the shown frame
where E = 106 kN/m2
12 kN/m
A
B
A=0.6 m
I = 0.02 m4
2
3
A=0.4 m2
I = 0.005 m4
8
50 kN
2
C
30
memb
L
θ
C
S
EA/L
12EI/L3
6EI/L2
4EI/L
2EI/L
AB
8
0
1
0
75000
469
1875
10000
5000
CB
5
90
0
1
80000
480
1200
4000
2000
First element : ( AB)
E
AL
0
l
K
=
0
EA
L
0
0
0
12
L
EI
3
6
L
EI
2
0
12EI
L
63
L
EI
2
0
6
L
EI
2
4
EI
L
0
6E
L
I2
2
EI
L
EA
L
0
0
E
AL
0
0
0
-12EI
L
-3
6E
L
I2
0
12
L
EI
-3
6E
L
I2
0
6
L
EI
2
2
EI
L
0
6E
L
I2
4
EI
L
l
K1
=
75000
0
0
469
1875
0
1875
-75000
-7500
0
0
0
-469
1875
10000
0
-1875
5000
0
0
75000
0
0
0
-469
-1875
0
469
-1875
0
1875
5000
0
-1875
10000
0
33
T
=
cos θ
- sin θ
0
0
0
0
sin θ
cos θ
0
0
0
0
0
0
1
0
0
0
0
0
0
cos θ - sin θ 0
0
0
0
sin θ cos θ 0
0
0
0
0
0
1
34
1
0
T =
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
1
0
1
35
Obtain the member global stiffness matrix
l
Km = Tm Km T
g
g
K1
g
K1
=
T
L
K1
75000
0
0
469
1875
0
1875
0
0
0
-469
1875
10000
0
-1875
5000
0
0
75000
0
0
-469
-1875
0
469
-1875
0
1875
5000
0
-1875
10000
= -75000
-7500
0
0
B
A
g
K1
75000
0
0
469
1875
0
1875
0
0
0
-469
1875
10000
0
-1875
5000
0
0
75000
0
0
-469
-1875
0
469
-1875
0
1875
5000
0
-1875
10000
= -75000
-7500
0
A
0
B
37
second element : ( CB )
E
AL
0
l
K2
=
0
EA
L
0
0
0
12
L
EI
3
6
L
EI
2
0
12EI
L
63
L
EI
2
0
6
L
EI
2
4
EI
L
0
6E
L
I2
2
EI
L
EA
L
0
0
E
AL
0
0
0
-12EI
L
-3
6E
L
I2
0
12
L
EI
-3
6E
L
I2
0
6
L
EI
2
2
EI
L
0
6E
L
I2
4
EI
L
l=
K2
80000
0
0
-80000
0
0
0
480
1200
0
-480
1200
0
1200
4000
0
-1200
2000
-80000
0
0
80000
0
0
0
-480
-1200
0
480
-1200
0
1200
2000
0
-1200
4000
39
l=
K2
40
T
=
cos θ
- sin θ
0
0
0
0
sin θ
cos θ
0
0
0
0
0
0
1
0
0
0
0
0
0
cos θ - sin θ 0
0
0
0
sin θ
0
0
0
0
cos θ
0
0
1
41
T =
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
42
l
Km = T Km T
g
0
1
0
0
0
0
-1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
-1
0
0
0
0
0
0
0
1
80000
0
0
480
0
1200
0
-80000
0
-480
0
1200
-80000
0
0
1200
0
4000
80000
0
0
-1200
2000
0
0
-480
-1200
0
480
-1200
0
-1
0
0
0
0
0
-480
80000
0
0
1200
0
480
0
-80000
0
1200
0
480
-1200
0
4000
-80000
0
0
-1200
0
2000
1200
0
2000
0
80000
0
-480
0
-1200
1200
0
4000
-1200
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
-1
0
T
0
1200
2000
0
-1200
4000
1
0
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
43
d1
75480
0
Ks
1200
= -480
1200
d2
0
d3
1200
80469 -1875
d7
-480
0
d9
1200
d1
0
d2
-1875
14000
-1200
2000
0
-1200
480
-1200
d3
d7
0
2000
-1200
4000
d9
44
Obtain the member global stiffness matrix
l
Km = T Km T
g
480
g
K2
=
T
0
-1200
-480
0
-1200
0
80000
0
0
-80000
0
-1200
0
4000
1200
0
2000
-480
0
1200
480
0
1200
0
-80000
0
0
80000
0
-1200
0
2000
1200
0
4000
45
B
C
480
g
K2
=
0
-1200
-480
0
-1200
0
80000
0
0
-80000
0
-1200
0
4000
1200
0
2000
-480
0
1200
480
0
1200
0
-80000
0
0
80000
0
-1200
0
2000
1200
0
4000
C
B
46
Assembly :
A
B
75480
0
0
80469
1200
-1875
0
-469 -1875
14000
0
1875
0
-80000
0
5000 -1200
0
0
0
0
1875 10000
0
0
0
0
0
-1875
5000
0
-480
0
-1200
0
0
-80000
0
0
0
0
0
75000 0
0
0
0
0
480
0
-1200
B
2000
469 1875
1875
2000
1200
0
-649
0
0
0
0
1200
-480
0
0
0
0
-1875
0
-75000
Ks =
1200 -75000 0
C
0
A
-1200
80000
0
0
4000
C
47
Modeling
d5
d6
A d4
d2
d3
d1
B
d9
d8
d7
C
48
48
B
d1
d2
75480
0
0
80469
1200
-1875
d3
d4
d5
1200 -75000 0
d9
0
-480
0
1200
0
-80000
0
d1
d2
2000
d3
14000
0
1875
0
0
0
469 1875
0
0
0
d4
d5
1875 10000
0
0
0
d6
-1200
d7
0
d8
1875
0
0
-1875
5000
0
-480
0
-1200
0
0
-80000
0
0
0
0
0
2000
5000 -1200
0
-649
0
d8
-469 -1875
0
1200
d7
0
0
0
d6
-1875
0
-75000
Ks =
C
A
75000 0
0
0
0
0
480
0
-1200
0
80000
0
4000
d9
B
A
C
49
Force vector
12 kN/m
A
B
A=0.6 m
I = 0.02 m4
2
3
A=0.4 m2
I = 0.005 m4
8
50 kN
2
C
50
Force vector
12x82
12
64 kNm
12 kN/m
A
64 kNm
B
48 kN
48 kN
8
51
Force vector
50x3x22
52
24 kNm
B
3
17.6 kN
50 kN
2
32.4 kN
C
36 kNm
50x2x32
52
52
Fixed End
Reaction
(FER)
Force vector
64 kNm
A
48 kN
64 kNm
24 kNm
B
B
48 kN
17.6 kN
32.4 kN
C
53
53
36 kNm
Fixed End
Action
(FEA)
Force vector
64 kNm
64 kNm
A
B
48 kN
d2
A
d3
d1
B
d9
C
F
d7
48 kN
F1
F2
-17.6
- 48
= F3 =
F7
40
-32.4
F9
36
24 kNm
B
17.6 kN
32.4 kN
C
54
54
F= K D
-17.6
- 48
40
75,480
0
= 1200
0
1200
80,469 -1875
-1875
- 480
1200
0
0
14,000 -1,200 2,000
-32.4
- 480
0
-1,200
36
1200
0
2,000 -1,200 4,000
480
-1,200
d1
d2
d3
d7
d9
55
D = K-1 F
-1
1200 - 480 1200
d1
75,480
0
80,469 -1875
0
0
d2
0
d3 = 1200 -1875 14,000 -1,200 2,000
d7
0
- 480
-1,200 480 -1,200
d9
0
2,000 -1,200 4,000
1200
-17.6
- 48
40
-32.4
36
56
D = K-1 F
d1
- 0.0007
- 0.0008
d2
d3 = - 0.0088
- 0.2244
d7
d9
- 0.0538
57
Internal forces
d1
d2
d3
d7
d9
A=0.6 m2
I = 0.02 m4
- 0.0007
- 0.0008
=
- 0.0088
- 0.2244
- 0.0538
A=0.4 m2
I = 0.005 m4
d2
E = 106 kN/m2
A
EIAB = 2x10
EIBC = 5x103
d3
d1
B
4
d9
d7
C
58
Fixed End
Reaction
(FER)
Force vector
64 kNm
A
48 kN
64 kNm
24 kNm
B
B
48 kN
17.6 kN
32.4 kN
C
59
64 kNm
64 kNm
d3
A
B
48 kN
d9
d2
d1
d7
48 kN
d1
d2
d3
d7
d9
- 0.0007
- 0.0008
=
- 0.0088
- 0.2244
- 0.0538
Find internal forces in members
[ ] = [ ] [ T] [ ] + [ ]
Fm
Km
T
g
Dm
Fm FER
60
Member (AB)
0 7500 0
0
1870
0 469
0 -469 1875
6
187 1000
-1876 500
0
0
0
5
0
0
0 7500 0
0
75000
0
-469
469
0
0
1875
1875
1000
0 1875 5000 0 187
0
5
7500
0
0
l
Fm =
0
0
0
- 0.0007
- 0.0008
- 0.0088
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
48
+
64
0
48
-64
61
l
Fm
52.5
-16. 125
= -42.5
-52.5 +
16. 25
-86.5
52.5
31.875
l
21.5
Fm =
-52.5
64.125
150.5
0
48
64
0
48
-64
62
B
d3
24 kNm
17.6 kN
d2
d7
d9
C
d1
d1
d2
d3
d7
d9
36 kNm
- 0.0007
- 0.0008
=
- 0.0088
- 0.2244
- 0.0538
Member (CB)
l
g
l
Fm = Km T Dm
T
63
Member (CB)
0 8000 0
0
1200
0 480
0 -480 1200
0
120
200
4000 0 -1200
0
0
0
0
0 8000 0
0
80000
0
-480
480
0
0
1200
1200
0 1200 2000 0 120 4000
0
8000
0
0
l
Fm =
- 0.2244
0
- 0.0538
- 0.0007
- 0.0008
- 0.0088
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
-1
0
0
0
0
1
0
0
0
0
0
0
0
1
0
+
0
-36
17.6
0
24
64
64
32.25
l
36
Fm =
-64
-32.25
126
+
0
0
-36
17.6
0
24
64
32.25
l
Fm =
0
46.4
-32.25
150
65
21.75 kNm
12 kN/m
150 kNm
150 kNm
A
B
B
3
32 kN
50 kN
50 kN
64 kN
8
2
0 kN
C
66
21.75 kNm
A
32 kN
150 kNm
12 kN/m
150 kNm
B
64 kN
50 kN
64 kN
50 kN
B
50 kN
C
67
21.75 kNm
150 kNm
12 kN/m
A
150 kNm
B
32 kN
64 kN
-
50 kN
64 kN
50 kN
B
50 kN
50
64 -
C
N.F.D
68
21.75 kNm
A
32 kN
32
150 kNm
12 kN/m
150 kNm
B
50 kN
64 kN
64 kN
50 kN
+
B
50
50 kN
64
S.F.D
+
C
50
69
21.75 kNm
150 kNm
12 kN/m
150 kNm
A
32 kN
21.75
B
85.875
50 kN
64 kN
64 kN
150
50 kN
96
-
+
10.125
-
B
150
-
50 kN
C
B.M.D
70 | 6,685 | 11,197 | {"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-2017-26 | latest | en | 0.573038 |
http://math.stackexchange.com/questions/119954/newton-raphsons-method-to-find-sqrt2012 | 1,469,815,623,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00258-ip-10-185-27-174.ec2.internal.warc.gz | 156,275,936 | 17,492 | # Newton-Raphson's Method to find $\sqrt{2012}$
I am asked to find $\sqrt{2012}$ using Newton-Raphson's Method with the following recursive method
$$x_{n+1} = \frac{1}{2} (x_n + \frac{a}{x_n})$$
I notied that give same answers as using
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
This is easy, but the next part asks to find a similar recursive method to find $\sqrt[3]{2012}$. How do I find such a method?
UPDATE
I did
$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 12.7551$$
$$x_2 = 12.6257$$
$$x_3 = 12.6244$$
$$x_4 = 12.6244$$
$$x_4 ^ 3 = 2012.02$$
Which seems correct. But I didn't use a similar recursive method like the question asked?
-
Hint: if you want to find $a$, then write it as a root of some function $f(x)$. In your case, write $\sqrt[3]{2012}$ as a root of a <fill_here> degree polynomial. Now, you can read Robert Israel's answer. – user2468 Mar 14 '12 at 5:44
If you use the idea described by Robert Israel, and fool around with the resulting formula a bit, you will end up with something that has a shape very similar to the one for square root. – André Nicolas Mar 14 '12 at 5:58
Hint: What does Newton's method say for $f(x) = x^3 - a$? | 398 | 1,165 | {"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} | 3.578125 | 4 | CC-MAIN-2016-30 | latest | en | 0.891477 |
https://expskill.com/question/a-projectile-has-both-change-in-kinetic-and-potential-energy-still-the-work-done-calculated-between-any-two-points-is-zero-this-is-due-to-_________/ | 1,679,488,927,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.76/warc/CC-MAIN-20230322114226-20230322144226-00320.warc.gz | 306,775,353 | 22,689 | # A projectile has both change in kinetic and potential energy. Still the work done calculated between any two points is zero. This is due to _________
Category: QuestionsA projectile has both change in kinetic and potential energy. Still the work done calculated between any two points is zero. This is due to _________
Editor">Editor Staff asked 1 year ago
A projectile has both change in kinetic and potential energy. Still the work done calculated between any two points is zero. This is due to _________
(a) Presence of no external force
(b) No displacement
(c) The displacement being in both vertical and horizontal directions
(d) Inter-conversion of kinetic and potential energies
I got this question in unit test.
The question is from Work-Energy Theorem topic in division Work, Energy and Power of Physics – Class 11
Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
Editor">Editor Staff answered 1 year ago
The correct choice is (d) Inter-conversion of kinetic and potential energies
Easiest explanation: When a projectile moves, there is an inter-conversion of kinetic and potential energies. This is what causes the body to cover a maximum vertical and horizontal distance. One might argue over the presence of gravity which is an external force. Well, the work done by gravity is nothing but the potential energy of the body which has already been taken care of in the expression for total energy.
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Articles: 40701 | 463 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-14 | longest | en | 0.891654 |
https://onlinejudge.org/board/viewtopic.php?f=22&t=11220 | 1,582,624,769,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146064.76/warc/CC-MAIN-20200225080028-20200225110028-00049.warc.gz | 495,137,291 | 6,966 | ## how to find Derangement of a multiset
Let's talk about algorithms!
Moderator: Board moderators
rushel
Learning poster
Posts: 67
Joined: Sat Jan 22, 2005 5:57 am
### how to find Derangement of a multiset
can anybody give me a link or any combinatorics formula or any algorithm of
number of derangement of a multiset.
misof
A great helper
Posts: 430
Joined: Wed Jun 09, 2004 1:31 pm
Just to clarify, what do you understand under "derangement"?
For permutations, derangement is a permutation without fixed points. The best definition I could come up with for multisets:
Suppose M is a multiset. Let sorted(M) be the vector containing all elements of M in sorted order.
For example, if M = {3, 4, 2, 3, 1}, then sorted(M) = (1,2,3,3,4).
Let isPermutation(V,M) be true if and only if the vector V contains exactly the elements of M, in any order.
For example:
isPermutation( (1,3,2), {1,2,3} ) = true
isPermutation( (1,3,2,3), {1,2,3} ) = false
Then, a derangement of M is a vector V such that isPermutation(V,M) && (V and sorted(M) differ on all places).
For example, for M = {3, 4, 2, 3, 1} the vector (3,3,1,4,2) is a derangement, and the vector (4,1,3,2,3) is not.
Is this what you want?
rushel
Learning poster
Posts: 67
Joined: Sat Jan 22, 2005 5:57 am
misof, yes thats is the thing i want . sorry for my late reply. | 427 | 1,334 | {"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-2020-10 | latest | en | 0.804298 |
https://dsp.stackexchange.com/questions/3026/picking-the-correct-filter-for-accelerometer-data%20%22%22 | 1,566,532,561,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317817.76/warc/CC-MAIN-20190823020039-20190823042039-00411.warc.gz | 431,140,832 | 35,960 | # Picking the correct filter for accelerometer data
I am fairly new to DSP, and have done some research on possible filters for smoothing accelerometer data in python. An example of the type of data Ill be experiencing can be seen in the following image:
Essentially, I am looking for advice as to smooth this data to eventually convert it into velocity and displacement. I understand that accelerometers from mobile phones are extremely noisy.
I dont think I can use a Kalman filter at the moment because I cant get hold of the device to reference the noise produced by the data (I read that its essential to place the device flat and find the amount of noise from those readings?)
FFT has produced some interesting results. One of my attempts was to FFT the acceleration signal, then render low frequencies to have a absolute FFT value of 0. Then I used omega arithmetic and inverse FFT to gain a plot for velocity. The results were as follows:
Is this a good way to go about things? I am trying to remove the overall noisy nature of the signal but obvious peaks such as at around 80 seconds need to be identified.
I have also tired using a low pass filter on the original accelerometer data, which has done a great job of smoothing it, but I'm not really sure where to go from here. Any guidance on where to go from here would be really helpful!
EDIT: A little bit of code:
for i in range(len(fz)):
testing = (abs(Sz[i]))/Nz
if fz[i] < 0.05:
Sz[i]=0
Velfreq = []
Velfreqa = array(Velfreq)
Velfreqa = Sz/(2*pi*fz*1j)
Veltimed = ifft(Velfreqa)
real = Veltimed.real
So essentially, ive performed a FFT on my accelerometer data, giving Sz, filtered high frequencies out using a simple brick wall filter (I know its not ideal). Then ive use omega arithmetic on the FFT of the data. Also thanks very much to datageist for adding my images into my post :)
• Welcome to DSP! Is the red curve in your second picture a "smoothed" version of the original (green) data? – Phonon Aug 2 '12 at 19:41
• The red curve is (hopefully!) a velocity curve generated from fft followed by filtering, followed by omega arithmetic (dividing by 2*pifj), following by inv. fft – Michael M Aug 2 '12 at 20:24
• Perhaps if you include a more precise mathematical expression or pseudocode for what you did would clear things up a bit. – Phonon Aug 2 '12 at 20:53
• Added some now, thats the general feel of the code.. – Michael M Aug 4 '12 at 10:55
• My question would be: what do you expect to see in the data? You won't know whether you have a good approach unless you know something about the underlying signal that you expect to see after filtering. In addition, the code that you showed is confusing. Although you don't show the initialization of the fz array, it appears that you're applying a highpass filter instead. – Jason R Aug 6 '12 at 14:13
As pointed out by @JohnRobertson in this post, Total Variaton (TV) denoising is another good alternative if your signal is piece-wise constant. This may be the case for the accelerometer data, if your signal keeps varying between different plateaux.
Below is a Matlab code that performs TV denoising in such a signal. The code is based on this paper. The parameters $\mu$ and $\rho$ have to be adjusted according to the noise level and signal characteristics.
If $y$ is the noisy signal and $x$ is the signal to be estimated, the function to be minimized is $\mu\|{x-y}\|^2+\|{Dx}\|_1$, where $D$ is the finite differences operator.
function denoise()
f = [-1*ones(1000,1);3*ones(100,1);1*ones(500,1);-2*ones(800,1);0*ones(900,1)];
plot(f);
axis([1 length(f) -4 4]);
title('Original');
g = f + .25*randn(length(f),1);
figure;
plot(g,'r');
title('Noisy');
axis([1 length(f) -4 4]);
fc = denoisetv(g,.5);
figure;
plot(fc,'g');
title('De-noised');
axis([1 length(f) -4 4]);
function f = denoisetv(g,mu)
I = length(g);
u = zeros(I,1);
y = zeros(I,1);
rho = 10;
eigD = abs(fftn([-1;1],[I 1])).^2;
for k=1:100
f = real(ifft(fft(mu*g+rho*Dt(u)-Dt(y))./(mu+rho*eigD)));
v = D(f)+(1/rho)*y;
u = max(abs(v)-1/rho,0).*sign(v);
y = y - rho*(u-D(f));
end
function y = D(x)
y = [diff(x);x(1)-x(end)];
function y = Dt(x)
y = [x(end)-x(1);-diff(x)];
Results:
• Really like this answer, gonna go ahead and try it. Sorry it took me so long to reply! – Michael M Aug 28 '12 at 16:51
• Excellent answer. Thanks for the details. I am looking for the C version of this code. Anyone here ported this matlab code to C they would like to share? Thanks. – pixbroker Jan 12 '16 at 18:24
• What does piece-wise constant mean? – tilaprimera Jun 1 '18 at 20:25
The problem is that your noise has a flat spectrum. If you assume white Gaussian noise (which turns out to be a good assumption) its power spectrum density is constant. Roughly speaking, it means that your noise contains all frequencies. That's why any frequency approach, e.g. DFT or low-pass filters, is not a good one. What would be your cut-off frequencies since your noise is all over the spectrum?
One answer to this question is the Wiener filter, which requires knowledge of the statistics of your noise and your desired signal. Basically, the noisy signal (signal + noise) is attenuated over the frequencies where the noise is expected to be grater than your signal, and it is amplified where your signal is expected be grater than your noise.
However, I would suggest more modern approaches that use non-linear processing, for example wavelet denoising. These methods provide excellent results. Basically, the noisy signal is first decomposed into wavelets and then small coefficients are zeroed. This approach works (and DFT doesn't) because of the multi-resolution nature of wavelets. That is, the signal is processed separately in frequency bands defined by the wavelet transform.
In MATLAB, type 'wavemenu' and then 'SWT denoising 1-D'. Then 'File', 'Example Analysis', 'Noisy signals', 'with Haar at level 5, Noisy blocks'. This example uses Haar wavelet, which should work fine for your problem.
I'm not good at Python, but I believe you can find some NumPy packages which perform Haar wavelet denoising.
• I would disagree with your first statement. You're assuming that the signal of interest covers the full bandwidth of the input sequence, which is unlikely. It is still possible to obtain improved signal-to-noise ratio using linear filtering in this case, eliminating the out-of-band noise. If the signal is highly oversampled, then you may obtain a large improvement with such a simple approach. – Jason R Aug 7 '12 at 1:37
• It's true, and this is achieved by the Wiener filter, when you know the statistics of your signal and your noise. – Daniel R. Pipa Aug 7 '12 at 10:57
• Although the theory behind wavelet denoising is complicated, the implementation is as simple as the approach you described. It involves only filter banks and thresholding. – Daniel R. Pipa Aug 7 '12 at 12:08
• Im researching into this now, will post my progress above, thanks to both of you and Phonon for all of your help so far! – Michael M Aug 7 '12 at 14:11
• @DanielPipa I don't have access to the matlab packages in question. Can you provide a paper or other reference that describes the method that corresponds to your matlab code. – John Robertson Aug 8 '12 at 22:44
As per suggestion of Daniel Pipa, I took a look at wavelet denoising and found this excellent article by Francisco Blanco-Silva.
Here I have modified his Python code for image processing to work with 2D (accelerometer) rather than 3D (image) data.
Note, the threshold is "made up" for the soft-thresholding in Francisco's example. Consider this and modify for your application.
def wavelet_denoise(data, wavelet, noise_sigma):
'''Filter accelerometer data using wavelet denoising
Modification of F. Blanco-Silva's code at: https://goo.gl/gOQwy5
'''
import numpy
import scipy
import pywt
wavelet = pywt.Wavelet(wavelet)
levels = min(15, (numpy.floor(numpy.log2(data.shape[0]))).astype(int))
# Francisco's code used wavedec2 for image data
wavelet_coeffs = pywt.wavedec(data, wavelet, level=levels)
threshold = noise_sigma*numpy.sqrt(2*numpy.log2(data.size))
new_wavelet_coeffs = map(lambda x: pywt.threshold(x, threshold, mode='soft'),
wavelet_coeffs)
return pywt.waverec(list(new_wavelet_coeffs), wavelet)
Where:
• wavelet - string name of wavelet form to be used (see pywt.wavelist(), e.g. 'haar')
• noise_sigma - standard deviation of noise from data
• data - array of values to filter (e.g. x, y, or z axis data) | 2,196 | 8,492 | {"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} | 3 | 3 | CC-MAIN-2019-35 | latest | en | 0.943145 |
https://stats.libretexts.org/Courses/Taft_College/PSYC_2200%3A_Elementary_Statistics_for_Behavioral_and_Social_Sciences_(Oja)/Unit_2%3A_Mean_Differences/08%3A_One_Sample_t-test/8.05%3A_Confidence_Intervals/8.05.01%3A_Practice_with_Confidence_Interval_Calculations | 1,669,718,872,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00633.warc.gz | 589,478,810 | 32,261 | # 8.5.1: Practice with Confidence Interval Calculations
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation $$s$$ as an estimate for $$\sigma$$ and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.
William S. Goset (1876–1937) of the Guinness brewery in Dublin, Ireland ran into this problem. His experiments with hops and barley produced very few samples. Just replacing $$\sigma$$ with $$s$$ did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to "discover" what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name "Student."
Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and only used the Student's $$t$$-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever $$s$$ is used as an estimate for $$\sigma$$. If you draw a simple random sample of size $$N$$ from a population that has an approximately a normal distribution with mean $$\mu$$ and unknown population standard deviation $$\sigma$$ and calculate the $$t$$-score, then the $$t$$-scores follow a Student's t-distribution with $$n – 1$$ degrees of freedom. The $$t$$-score has the same interpretation as the z-score. It measures how far $$\bar{x}$$ is from its mean $$\mu$$. For each sample size $$n$$, there is a different Student's t-distribution.
The degrees of freedom, $$n – 1$$, come from the calculation of the sample standard deviation $$s$$. Previously, we used $$n$$ deviations ($$x - \bar{x}$$ values) to calculate $$s$$. Because the sum of the deviations is zero, we can find the last deviation once we know the other $$n – 1$$ deviations. The other $$n – 1$$ deviations can change or vary freely. We call the number $$n – 1$$ the degrees of freedom (df).
For each sample size $$N$$, there is a different Student's t-distribution.
#### Properties of the Student's $$t$$-Distribution:
• The graph for the Student's $$t$$-distribution is similar to the standard normal curve.
• The mean for the Student's $$t$$-distribution is zero and the distribution is symmetric about zero.
• The Student's $$t$$-distribution has more probability in its tails than the standard normal distribution because the spread of the $$t$$-distribution is greater than the spread of the standard normal. So the graph of the Student's $$t$$-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution.
• The exact shape of the Student's $$t$$-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of Student's $$t$$-distribution becomes more like the graph of the standard normal distribution.
• The underlying population of individual observations is assumed to be normally distributed with unknown population mean $$\mu$$ and unknown population standard deviation $$\sigma$$. The size of the underlying population is generally not relevant unless it is very small. If it is bell shaped (normal) then the assumption is met and doesn't need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.
A probability table for the Student's $$t$$-distribution can be used. The table gives $$t$$-scores that correspond to the confidence level (column) and degrees of freedom (row).
A Student's $$t$$-table gives $$t$$-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's $$t$$-probabilities.
##### Example $$\PageIndex{1}$$
Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.
8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9
Solution
To find the confidence interval, you need the sample mean, $$\bar{x}$$, and the $$EBM$$.
$$\bar{x} = 8.2267$$
$$s = 1.6722$$
$$n = 15$$
$$df = 15 – 1 = 14$$
$$\frac{\alpha}{2} = 0.025$$, so $$t_{crit} = t_{0.025}$$
The area to the right of $$t_{0.025}$$ is 0.025, and the area to the left of $$t_{0.025}$$ is 1 – 0.025 = 0.975
$$t_{0.025} = 2.145$$ from the Table of Critical t-scores.
\begin{align*} EBM &= \left(t\right)\left(\frac{s}{\sqrt{n}}\right) \\[4pt] &= (2.14)\left(\frac{1.6722}{\sqrt{15}}\right) = 0.924 \end{align*}
Now it is just a direct application of Equation \ref{confint}:
\begin{align*} \bar{x} – EBM &= 8.2267 – 0.9240 = 7.3 \\[4pt] \bar{x} + EBM &= 8.2267 + 0.9240 = 9.15 \end{align*}
The 95% confidence interval is (7.30, 9.15).
We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.
You can try one on your own:
##### Exercise $$\PageIndex{1}$$
You do a study of hypnotherapy to determine how effective it is in increasing the number of hourse of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.
8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5
(8.1634, 9.8032)
##### Example $$\PageIndex{2}$$: The Human Toxome Project
The Human Toxome Project (HTP) is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists at HTP tested cord blood samples for 20 newborn infants in the United States. The cord blood of the "In utero/newborn" group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous system toxicity, immune system toxicity, and reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table $$\PageIndex{1}$$ shows how many of the targeted chemicals were found in each infant’s cord blood.
79 145 147 160 116 100 159 151 156 126 137 83 156 94 121 144 123 114 139 99
Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant’s blood.
Solution
From the sample, you can calculate $$\bar{x} = 127.45$$ and $$s = 25.965$$. There are 20 infants in the sample, so $$n = 20$$, and $$df = 20 – 1 = 19$$.
You are asked to calculate a 90% confidence interval: $$CL = 0.90$$, so
$\alpha = 1 – CL = 1 – 0.90 = 0.10 \frac{\alpha}{2} = 0.05, t_{\frac{\alpha}{2}} = t_{0.05}$
By definition, the area to the right of $$t_{0.05}$$ is 0.05 and so the area to the left of $$t_{0.05}$$ is $$1 – 0.05 = 0.95$$.
Use a table, calculator, or computer to find that $$t_{0.05} = 1.729$$.
$EBM = t \times \left(\frac{s}{\sqrt{n}}\right) = 1.729\times \left(\frac{25.965}{\sqrt{20}}\right) \approx 10.038 \nonumber$
$\bar{x} – EBM = 127.45 – 10.038 = 117.412\nonumber$
$\bar{x} + EBM = 127.45 + 10.038 = 137.488\nonumber$
We estimate with 90% confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.
##### Example $$\PageIndex{3}$$
A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table $$\PageIndex{2}$$. Use this sample data to construct a 98% confidence interval for the mean number of hours statistics students will spend watching television in one week.
0 3 1 20 9 5 10 1 10 4 14 2 4 4 5
Solution
• $$\bar{x} = 6.133$$,
• $$s = 5.514$$,
• $$n = 15$$, and
• $$df = 15 – 1 = 14$$.
$$\frac{\alpha}{2} = 0.01 t_{\frac{\alpha}{2}} = t_{0.01} = 2.624$$
$\bar{x} – EBM = 6.133 – 3.736 = 2.397\nonumber$
$\bar{x} + EBM = 6.133 + 3.736 = 9.869\nonumber$
We estimate with 98% confidence that the mean number of all hours that statistics students spend watching television in one week is between 2.397 and 9.869.
## Reference
1. “America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013).
2. Data from Microsoft Bookshelf. | 2,879 | 9,956 | {"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} | 4.71875 | 5 | CC-MAIN-2022-49 | latest | en | 0.683048 |
https://ch.mathworks.com/matlabcentral/cody/problems/22-remove-the-vowels/solutions/942091 | 1,590,969,749,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00464.warc.gz | 289,699,481 | 15,802 | Cody
# Problem 22. Remove the vowels
Solution 942091
Submitted on 25 Aug 2016 by Sienna Grace
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
s1 = 'Jack and Jill went up the hill'; s2 = 'Jck nd Jll wnt p th hll'; assert(isequal(s2,refcn(s1)))
expression = [aeiou] match = 'a' 'a' 'i' 'e' 'u' 'e' 'i' noMatch = 'J' 'ck ' 'nd J' 'll w' 'nt ' 'p th' ' h' 'll' s2 = Jck nd Jll wnt p th hll
2 Pass
s1 = 'I don''t want to work. I just want to bang on the drum all day.'; s2 = ' dn''t wnt t wrk. jst wnt t bng n th drm ll dy.'; assert(isequal(s2,refcn(s1)))
expression = [aeiou] match = Columns 1 through 14 'I' 'o' 'a' 'o' 'o' 'I' 'u' 'a' 'o' 'a' 'o' 'e' 'u' 'a' Column 15 'a' noMatch = Columns 1 through 11 '' ' d' 'n't w' 'nt t' ' w' 'rk. ' ' j' 'st w' 'nt t' ' b' 'ng ' Columns 12 through 16 'n th' ' dr' 'm ' 'll d' 'y.' s2 = dn't wnt t wrk. jst wnt t bng n th drm ll dy. | 405 | 997 | {"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-2020-24 | latest | en | 0.575396 |
http://reference.wolfram.com/legacy/v8/guide/GraphsAndNetworks.html | 1,510,980,078,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00384.warc.gz | 265,820,009 | 10,852 | This is documentation for Mathematica 8, which was
based on an earlier version of the Wolfram Language.
MATHEMATICA GUIDE Functions »| More About »
Graphs & Networks
Graphs and networks are all around us including technological networks (the internet, power grids, telephone networks, transportation networks, ...), social networks (social graphs, affiliation networks, ...), information networks (world wide web, citation graphs, patent networks, ...), biological networks (biochemical networks, neural networks, food webs, ...) and many more. Graphs provide a structural model that makes it possible to analyze and understand how many separate systems act together.
Mathematica provides state-of-the-art functionality for modeling, analyzing, synthesizing, and visualizing graphs and networks. Whether those graphs are small and diagrammatic or large and complex, Mathematica provides numerous high-level functions for creating or computing with graphs. Graphs are first-class citizens in Mathematica; they can be used as input and output and they are deeply integrated into the rest of the Mathematica system.
Graph construct a graph from vertices and edges
GraphData, ExampleData curated collection of theoretical and empirical graphs
CompleteGraph, GridGraph generate special parametric graphs
AdjacencyGraph, IncidenceGraph construct a graph from matrices
RandomGraph construct random graphs from symbolic graph distributions
Import import graphs from a variety of file formats
Graph graph object with vertex and edge properties
UndirectedEdge (,Esc ue Esc ▪ DirectedEdge (,Esc de Esc ▪ ...
HighlightGraph highlight vertices, edges, or whole subgraphs
NeighborhoodGraph graph neighborhood of some vertex, edge etc.
IsomorphicGraphQ test whether two graphs are the same after vertex renaming
FindShortestPath find the shortest path between two vertices
ConnectedComponents give groups of vertices that are strongly connected
FindClique find complete subgraphs
DepthFirstScan scan a graph in depth-first order | 391 | 2,018 | {"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-2017-47 | latest | en | 0.871111 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_TERM/TRS/Ste92/perfect.trs.Thm26:EMB:NO.html.lzma | 1,718,279,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00594.warc.gz | 83,205,413 | 1,839 | Term Rewriting System R:
[x, y, u, z]
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) -> F(x, u, z, u)
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Argument Filtering and Ordering`
Dependency Pairs:
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) -> F(x, u, z, u)
Rules:
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
The following dependency pairs can be strictly oriented:
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) -> F(x, u, z, u)
There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
F(x1, x2, x3, x4) -> x1
s(x1) -> s(x1)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳AFS`
` →DP Problem 2`
` ↳Dependency Graph`
Dependency Pair:
Rules:
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
Using the Dependency Graph resulted in no new DP problems.
Termination of R successfully shown.
Duration:
0:00 minutes | 777 | 1,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.489497 |
https://answers.opencv.org/answers/1105/revisions/ | 1,680,092,074,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00244.warc.gz | 128,077,501 | 6,792 | # Revision history [back]
I was able to decode the strange behavior of the Gabor functions by keeping the other parameters constant in a more favorable position; This way, as seen from the graphic, the lambda parameter controls the frequency of the sinusoidal function that composes the kernel.
Conclusion: it controls the harmonic's frequency.
Bonus: The other parameters explained for non-math students:
psi - shift, in radians of the sinusoidal from center.
• 0 means that the max value in in center (symetrical, max positive)
• CV_PI/2 means that the max and min crescents are at the sides of the center (antisymetrical) - as in the above example
• CV_PI - min (negative value) in center (symetrical, max negative)
gamma - how elongated the filter is in the lateral direction. Seem to be the complement for the gaussian sigma, in the lateral direction. Note that low values mean elongated filters. 1 seem to a good default value.
sigma the good old sigma from the gaussian distribution. Controls the spread (radius)of the kernel
theta The rotation angle of the kernel. This way, you can select vertical stripes, horizontal stripes, or any other angle. It is the parameter of choice if you want to select, by example, edges at a given angle in an image. If you look for edges at 45 degrees, your gabor kernel will have to have a theta of pi/4
I was able to decode the strange behavior of the Gabor functions by keeping the other parameters constant in a more favorable position; This way, as seen from the graphic, the lambda parameter controls the frequency of the sinusoidal function that composes the kernel.
Conclusion: it controls the harmonic's frequency.
Edit This gorgeous answer from two math scholars shed some light on the strange pattern seen in the second graph. It's all about small details... here, numerical stability. Setting lambda to bigger values ( > 2) solves the issue. The comments are as useful as the post.
Bonus: The other parameters explained for non-math students:
psi - shift, in radians of the sinusoidal from center.
• 0 means that the max value in in center (symetrical, max positive)
• CV_PI/2 means that the max and min crescents are at the sides of the center (antisymetrical) - as in the above example
• CV_PI - min (negative value) in center (symetrical, max negative)
gamma - how elongated the filter is in the lateral direction. Seem to be the complement for the gaussian sigma, in the lateral direction. Note that low values mean elongated filters. 1 seem to a good default value.
sigma the good old sigma from the gaussian distribution. Controls the spread (radius)of the kernel
theta The rotation angle of the kernel. This way, you can select vertical stripes, horizontal stripes, or any other angle. It is the parameter of choice if you want to select, by example, edges at a given angle in an image. If you look for edges at 45 degrees, your gabor kernel will have to have a theta of pi/4
I was able to decode the strange behavior of the Gabor functions by keeping the other parameters constant in a more favorable position; This way, as seen from the graphic, the lambda parameter controls the frequency of the sinusoidal function that composes the kernel.
Conclusion: it controls the harmonic's frequency.
Edit
This gorgeous answer from two math scholars shed some light on the strange pattern seen in the second graph. It's all about small details... here, numerical stability. Setting lambda to bigger values ( > 2) solves the issue. The comments are as useful as the post.
Bonus: The other parameters explained for non-math students:
psi - shift, in radians of the sinusoidal from center.
• 0 means that the max value in in center (symetrical, max positive)
• CV_PI/2 means that the max and min crescents are at the sides of the center (antisymetrical) - as in the above example
• CV_PI - min (negative value) in center (symetrical, max negative)
gamma - how elongated the filter is in the lateral direction. Seem to be the complement for the gaussian sigma, in the lateral direction. Note that low values mean elongated filters. 1 seem to a good default value.
sigma the good old sigma from the gaussian distribution. Controls the spread (radius)of the kernel
theta The rotation angle of the kernel. This way, you can select vertical stripes, horizontal stripes, or any other angle. It is the parameter of choice if you want to select, by example, edges at a given angle in an image. If you look for edges at 45 degrees, your gabor kernel will have to have a theta of pi/4 | 1,010 | 4,552 | {"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} | 3.03125 | 3 | CC-MAIN-2023-14 | latest | en | 0.855406 |
http://what-when-how.com/Tutorial/topic-497t64c3d5/HTML5-Mobile-Game-Development-45.html | 1,544,936,045,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827252.87/warc/CC-MAIN-20181216025802-20181216051802-00589.warc.gz | 317,306,869 | 4,431 | HTML and CSS Reference
In-Depth Information
The constructor first copies three sets of objects into the this object: the base parameters, the blueprint,
and the override. Because the enemy can have different sprites depending on the blueprint, the width and the
height are set afterward based on the sprite property of the object. Finally, a t parameter is initialized to
0 to keep track of how long this sprite has been alive.
If the repetition in this code bothers you, don't worry! You clean it up in the section “Refactoring the Sprite
Classes” later in this chapter.
Stepping and Drawing the Enemy Object
The step function (see Listing 2-3 ) for the enemy should update the velocity based on the aforementioned
equation. The this.t property needs to be incremented by dt to keep track of how long the sprite has been
alive. Next, the equation from earlier in this chapter can be plugged directly into the step function to calculate
the x and y velocity. From the x and y velocity, the x and y location are updated. Finally, the sprite needs to check
if it's gone off the board to the right or the left, in which case the enemy can remove itself from the page.
Listing 2-3: The Enemy Step and Draw Methods
Enemy.prototype.step = function(dt) {
this.t += dt;
this.vx = this.A + this.B * Math.sin(this.C * this.t + this.D);
this.vy = this.E + this.F * Math.sin(this.G * this.t + this.H);
this.x += this.vx * dt;
this.y += this.vy * dt;
if(this.y > Game.height ||
this.x < -this.w ||
this.x > Game.width) {
this.board.remove(this);
}
}
Enemy.prototype.draw = function(ctx) {
SpriteSheet.draw(ctx,this.sprite,this.x,this.y);
}
The draw function is a near duplicate of the PlayerMissile object; the only difference is that it must
look up which sprite to draw in a property called sprite .
Now you add some initial enemy sprites to the top of game.js along with a simple enemy blueprint for one
enemy that can fly down the page:
var sprites = {
ship: { sx: 0, sy: 0, w: 37, h: 42, frames: 1 },
missile: { sx: 0, sy: 30, w: 2, h: 10, frames: 1 } ,
enemy_purple: { sx: 37, sy: 0, w: 42, h: 43, frames: 1 },
enemy_bee: { sx: 79, sy: 0, w: 37, h: 43, frames: 1 },
enemy_ship: { sx: 116, sy: 0, w: 42, h: 43, frames: 1 },
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This collection from Teachers TV features videos that are intended to be shown to students in primary mathematics lessons. Many of the videos are ideal as lesson starters, often posing a problem for students to consider.
Algebra - What's the Pattern? - an number sequence investigation which involves tables and chairs.
Algebra in Action - solve the formula which links apples, scarves and coats.
Angles - Pirate's Lost Treasure - use a series of directions to draw a treasure map.
Decimals Forever - an interactive lesson which includes recurring decimals.
Division Volume and Other Topics - eight dramatised lesson starters highlighting different problems.
Fractions Out Shopping - prompts discussion of the fractions and decimals found on the packaging of every day products.
Pizza Fractions - a prompt for discussion about fractions.
Polygons - an interactive lesson which explores the interior angles of hexagons and tessellation of shapes.
Problem Solving - Football Problems - students follow clues to find out where each player should stand for a team photo.
Problem Solving - Is It a Bargain? - students have to decide whether Dodgy Dave's offers are good.
Problem Solving and Other Topics - a selection of dramatised situations for students to solve.
The Shape Show - properties of triangles, quadrilaterals and their use in the environment.
## Resources
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Showing 12 result(s)
### Algebra - What's the Pattern?
This is a Teachers TV Key Stage Two algebra lesson starter based around how many people can sit at a birthday party table. The birthday boy and girl must sit at opposite ends of the table with all their guests sitting down either side. A single table can seat six children, one at each end and two on either side....
### Problem-Solving and Other Topics
This classroom resource from Teachers TV is aimed at Key Stage Two students and presents eight dramatised problems for use with an interactive whiteboard. This resource addresses fractions; decimals and percentages; shape, space and measures; numbers and algebra.
Each problem contextualises a different...
### Division, Volume and Other Topics
This Teachers TV classroom resource for Key Stage Two mathematics presents eight dramatised situations highlighting a problem to be solved. The clips are intended to be used as starters for primary lessons on fractions, decimals and percentages, or numbers and algebra.
Bring maths lessons...
### Pizza Fractions
As part of the series Primary Lesson Starters, this short Teachers TV film clip is designed to be shown in class to prompt discussion about fractions. When a customer orders a pizza she does not expect her friends to arrive at the restaurant and want to share it with her. How should the waitress cut the pizza so... | 553 | 2,820 | {"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-2024-38 | latest | en | 0.879377 |
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# How can you make 24 using only 1 6 6 8?
Updated: 9/19/2023
Wiki User
13y ago
Best Answer
A way of doing this would be:
1 + 8 = 9;
sqrt(9)= 3
3 * 6 = 18
18 + 6 = 24
1+8=9; sqrt(9)=3; 3*6+6=24
Another solution can be:
(1*6-(8-6))!
(6-2)!=4!=4*3*2*1=24
Wiki User
13y ago
This answer is:
## Add your answer:
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Q: How can you make 24 using only 1 6 6 8?
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Related questions
### How do you make 24 by using 3188?
You cannot make 24 using only the digits 3, 1, 8, and 8.
### How do you make 24 using only 6 2 7 1?
6 * (7 - 1 - 2) = 6*4 = 24
### How can you make 9 7 5 1 equal 24 using number only once?
It is: (9-5)*(7-1) = 24
7 + 9 + 8*1
### How can you make 24 using only 2 3 5 12?
12 ÷ (3 - 5/2) = 12 ÷ 1/2 = 24
5x5=25 25-1=24
### Can you make 24 using 1 1 8 8?
8 x (1 + 1) + 8 = 24
9 +16 -1 = 24
6/(1-3/4)=24
(9-4-1)*6=24
### Can 24 be formed using each number of tiles?
No. You cannot form 24 using only 1 tile, for example.
5x5=25 25-1=24 | 462 | 1,050 | {"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} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.731383 |
https://www.intellectualmath.com/converse-of-the-pythagorean-theorem-for-classifying-triangle.html | 1,722,749,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640389685.8/warc/CC-MAIN-20240804041019-20240804071019-00551.warc.gz | 664,419,815 | 6,357 | # CONVERSE OF THE PYTHAGOREAN THEOREM FOR CLASSIFYING TRIANGLE
Let a, b and c be the sides of the triangle.
• If a2 + b2 > c2, the triangle is acute triangle.
• If a2 + b2 = c2, the triangle is right triangle.
• If a2 + b2 < c2, the triangle is obtuse triangle.
Where a and b are the lengths of the two shorter sides and c be the length of the longest side.
Classify the triangle as acute, right, or obtuse, explain.
Problem 1 :
Solution:
Let c represent the length of the longest side of the triangle.
c2 ? a2 + b2
62 ? 52 + 22
36 ? 25 + 4
36 > 29
Because c2 is greater than a2 + b2, the triangle is obtuse.
Problem 2 :
Solution:
Let c represent the length of the longest side of the triangle.
c? a2 + b2
172 ? 82 + 152
289 ? 64 + 225
289 = 289
Because c2 is equal to a2 + b2, the triangle is right.
Problem 3 :
Solution:
Let c represent the length of the longest side of the triangle.
c? a2 + b2
72 ? 72 + 72
49 ? 49 + 49
49 < 98
Because c2 is less than a2 + b2, the triangle is acute.
Use the side lengths to classify the triangle as acute, right, or obtuse.
Problem 4 :
7, 24, 24
Solution:
By Triangle Inequality Theorem, the above set of numbers can represent the side lengths of a triangle.
Compare the square of the length of the longest side with the sum of the squares of the lengths of the two shorter sides.
c2 ? a2 + b2
24272 + 242
576 49 + 576
576 < 625
Because c2 is less than a2 + b2, the triangle is acute.
Problem 6 :
7, 24, 25
Solution:
By Triangle Inequality Theorem, the above set of numbers can represent the side lengths of a triangle.
Compare the square of the length of the longest side with the sum of the squares of the lengths of the two shorter sides.
c2 ? a2 + b2
252 72 + 242
625 ? 49 + 576
625 = 625
Because c2 is equal to a2 + b2, the triangle is right.
Problem 7 :
7, 24, 26
Solution:
By Triangle Inequality Theorem, the above set of numbers can represent the side lengths of a triangle.
Compare the square of the length of the longest side with the sum of the squares of the lengths of the two shorter sides.
c2 ? a2 + b2
262 72 + 242
676 ? 49 + 576
676 > 625
Because c2 is greater than a2 + b2, the triangle is obtuse.
Determine whether the triangle is acute, right, or obtuse.
Problem 8 :
Solution:
Let c represent the length of the longest side of the triangle.
c2 ? a2 + b2
52 ? 42 + 42
25 ? 16 + 16
25 < 32
Because c2 is less than a2 + b2, the triangle is acute.
Problem 9 :
Solution:
Let c represent the length of the longest side of the triangle.
c2 ? a2 + b2
142 ? 122 + 62
196 144 + 36
196 > 180
Because c2 is greater than a2 + b2, the triangle is obtuse.
Problem 10 :
Solution :
Let c represent the length of the longest side of the triangle.
c2 ? a2 + b2
152 ? 122 + 92
225 144 + 81
225 = 225
Because c2 is equal to a2 + b2, the triangle is right.
Problem 11 :
Match the side lengths of a triangle with the best description.
1) 2, 10, 112) 8, 5, 73) 5, 5, 54) 6, 8, 10 A. rightB. acuteC. obtuseD. equiangular
Solution:
1) By Pythagorean Theorem,
112 = 22 + 10
121 = 4 + 100
121 > 104
It is obtuse triangle.
2) By Pythagorean Theorem,
72 = 82 + 52
49 = 64 + 25
49 < 89
It is acute triangle.
3)
5, 5, 5
It is equiangular triangle.
4) By Pythagorean Theorem,
102 = 62 + 82
100 = 36 + 64
100 = 100
It is right triangle.
1) 2, 10, 112) 8, 5, 73) 5, 5, 54) 6, 8, 10 C. obtuseB. acuteD. equiangularA. right
## Recent Articles
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May 21, 24 08:51 PM
Finding Range of Values Inequality Problems
2. ### Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
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Periodic Motion Problems - Assignment Example
Summary
1. For a massive block oscillating up and down on a spring like in Fig 9.1.5, label how the following changes would affect the oscillation period. Label it as making it shorter (S), longer (L), or unchanged (U). Explain your response…
Extract of sample"Periodic Motion Problems"
Download file to see previous pages The force applied on the spring reduces hence the tension. The spring therefore accelerate mass faster, therefore the period would be shorter. b. Taking it to the moon where gravity is weaker. (U) The gravity change would have no effect on the period taken since the mass and spring are still the same, therefore no change is expected.
c. Weakening the spring (reducing the spring constant). (L) For a weakened spring, the force the spring exerts is decreased. The oscillations period would be therefore lengthened, would be longer d. Making the amplitude of the oscillation larger.(U) The amplitude does not affect frequency since the distance from relaxation position would increase the restoring force. Therefore, the frequency remains unchanged.
2. For a pendulum as in Fig 9.1.2: label how the following changes would affect the oscillation period. Label each as making the period shorter (S), longer (L), or unchanged (U). Explain your response. a. Taking it to a planet where gravity is larger.(S) Gravity affects the oscillation period from the formula of finding period using length and gravity. Therefore, as the gravity increases, the period decreases as they are inversely proportional b. Increase the mass hanging on the pendulum.(U) Period is mass independent. Therefore, at gravity all masses accelerate equally, hence the period is unchanged. c. Making the pendulum shorter. (S) The length is directly proportional to the period. Therefore a decrease in the pendulum length decreases the period d. Reducing the amplitude of the oscillation (assuming that it was not very big to start with). (U) The oscillations period remains constant due to the lack of relation to the amplitude. 3. The frequency of the tone produced by a violin string is higher (H), lower (L) or unchanged (U) if we make the following changes (note that here we are asking about the frequency, whereas on the earlier problems we were asking about the period of the oscillation, which is just the inverse of the frequency) : a. Making the string shorter. (H) The frequency of the tone is high. The shorter the string the higher the pitch, therefore the high frequency experienced. b. Making the string thicker. (L) The increased thickness increases the mass per unit length. Therefore the string moves slower which decreases the pitch, hence the frequency. c. Pressing the string down on the fingerboard.(H) The vibration reduces when the spring is pressed to the fingerboard; the active part is shortened. Therefore the pitch and frequency rose. d. Reducing the tension of the string. (L) The reduced tension of string causes slow movement of the string therefore the pitch and frequency reduced Explain your response. 4. I take a violin and make an exact copy of it, except that it is bigger. The strings are identical except for the length; they have the same material and the same tension. If the new violin is 2.30 times the size of the original, at what frequency would the string that was previously the A4 string (that is 440 Hz on a regular violin) oscillate? Use units of "Hz." Explain your response. When the size increases the pitch decreases, therefore 440Hz divided by 2.3 440Hz / 2.3= 191.30Hz 5. If your hearing cuts off at 17440 Hz, what is the highest harmonic of E5 string you can hear? The answer is an integer without units. Hint: The E5 string vibrates at 660 Hz. Explain your response. The highest harmonic is 17440Hz divided by 660Hz 17440/660=26.42 Rounding off, the highest harmonic to be heard is the 26th Harmonic 6. The frequency of the sound coming from the organ pipe is higher (H), lower (L) or the same (S) if we make the following changes to the organ: a. Moving the organ to a higher elevation. (H) The air is less dense at higher elevation, therefore the molecules move more ...Download file to see next pagesRead More
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Let us find you another Assignment on topic Periodic Motion Problems for FREE!
+16312120006 | 1,777 | 8,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-43 | longest | en | 0.907961 |
https://mathematica.stackexchange.com/questions/221408/how-can-i-use-lighting-for-specific-surface-of-a-3d-object | 1,723,769,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00846.warc.gz | 293,558,500 | 41,658 | # How can I use Lighting for specific surface of a 3D object?
Graphics3D[{Cylinder[{{(-0.1 + 10.128)/2, (13 + 1.3)/2,
0}, {(-0.1 + 10.128)/2, (13 + 1.3)/2, 0.001}}, 3.2]},
Lighting -> {{"Spot", Red}}]
I would like to use Lighting on the top surface ofCylinder above to produce the same lighting profile as this one
plane = DiscretizeRegion[
InfinitePlane[{0, 0, 0}, {{1, 0, 0}, {0, 1, 0}}], {{-1, 1}, {-1,
1}, {0, 1}}, MaxCellMeasure -> {"Length" -> 0.01},
BaseStyle -> {EdgeForm[], White}];
gty = Show[{plane},
Lighting -> {{"Spot", Red, {{0, 0, 1}, {0, 0, 0}}, {Pi/2, 4}}}]
You have a very thin cylinder. Ellipsoid seems to perform better if you are trying to emulate a disk.
cyl = Cylinder[{{(-0.1 + 10.128)/2, (13 + 1.3)/2,
0}, {(-0.1 + 10.128)/2, (13 + 1.3)/2, 0.001}}, 3.2];
centroid = RegionCentroid@cyl;
bnds = First@Differences@Transpose@RegionBounds@cyl;
ell = Ellipsoid[centroid, bnds];
Manipulate[
Graphics3D[{Specularity[White, 5], Style[ell, c]},
Lighting -> {{"Spot",
Red, {centroid + {0, 0, 3}, centroid}, θ}},
ImageSize -> Small], {{θ, π/2.5}, π/40, π/2}, {c,
White, ColorSlider}]
The following shows how the image changes at discrete surface colors with constant lighting.
Grid[{Graphics3D[{Style[ell, #]},
Lighting -> {{"Spot", Red, {centroid + {0, 0, 3}, centroid},
Pi/2.5}}] & /@ {White, Yellow, LightBlue, Blue}}]
The following shows how the image changes with underlying surface color under a White, Red, Green, and Blue spotlight.
mean = First@Mean@Transpose@RegionBounds@cyl;
tt = TranslationTransform /@ (DeleteCases[
Tuples[{0, #}, 3], {_, _, #}] &@bnds[[1]]*2.5);
ctt = Transpose[{{"White", "Red", "Blue", "Green"}, tt}];
Manipulate[
Graphics3D[{Specularity[White, 5],
Flatten@({Lighting -> {{"Spot",
ToExpression@#[[1]], #[[2]] /@ {centroid + {0, 0, 3},
centroid}, s}},
GeometricTransformation[
Style[{ell, Text[#[[1]], {mean, bnds[[1]]*2.3, 0}]}, c], #[[
2]]]} & /@ ctt)},
ViewPoint -> Top], {{s, Pi/5, "Spotlight Angle"}, Pi/20,
Pi/2.5}, {{c, Yellow, "Underlying Object Color"}, ColorSlider},
ControlPlacement -> Top]
• Thanks a lot @Tim Laska! Is it possible to assign a specific color for the desk instead of Black? Commented May 9, 2020 at 2:21
• @HD2006 The disk is black due to the lack of lighting. I added the ability to change the surface color through the Style command and specularity if desired. Commented May 9, 2020 at 2:54
• it is not working with me, it gives black disk without lighting once I change the color inside style to blue. May you please let the disk blue while lighting is red as it is and add a picture of the new shape? Commented May 9, 2020 at 3:09
• @HD2006 I added a ColorSlider to the Manipulate function to show that the image changes with changes in the surface color. Commented May 9, 2020 at 3:43
• @HD2006 I am pretty sure the black you are seeing is a function of the lighting. If evaluate Graphics3D[{Style[ell, Yellow]}], then you should see a Yellow disk. If you add lighting by evaluating Graphics3D[{Style[ell, Yellow]}, Lighting -> {{"Spot", Red, {centroid + {0, 0, 3}, centroid}, Pi/2.5}}], then the disk appears darker because less light is reflecting back. Commented May 9, 2020 at 4:24 | 1,049 | 3,180 | {"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} | 2.875 | 3 | CC-MAIN-2024-33 | latest | en | 0.654876 |
http://www.gtoal.com/compilers101/gtlanguage/coco/samples/pascal.atg | 1,545,218,045,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-51/segments/1544376832259.90/warc/CC-MAIN-20181219110427-20181219132427-00383.warc.gz | 369,949,592 | 2,537 | COMPILER Pascal /* J & W Pascal - not Turbo Pascal */ /* This grammar is not LL(1) */ CHARACTERS eol = CHR(13) . letter = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz" . digit = "0123456789". noQuote1 = ANY - "'" - eol . IGNORE CHR(9) .. CHR(13) IGNORE CASE COMMENTS FROM "(*" TO "*)" COMMENTS FROM "{" TO "}" TOKENS identifier = letter { letter | digit } . integer = digit { digit } | digit { digit } CONTEXT ("..") . real = digit { digit } "." digit { digit } [ "E" ["+" | "-"] digit { digit } ] | digit { digit } "E" ["+" | "-"] digit { digit } . string = "'" { noQuote1 | "''" } "'" . PRODUCTIONS Pascal = "program" NewIdent [ ExternalFiles ] ";" Block "." . ExternalFiles = "(" NewIdentList ")" . Block = DeclarationPart StatementPart . DeclarationPart = LabelDeclarations ConstDefinitions TypeDefinitions VarDeclarations { ProcDeclarations } . /* -------------------------------------------------------------------- */ LabelDeclarations = [ "label" Labels ";" ] . Labels = Label { "," Label } . Label = UnsignedInt . /* -------------------------------------------------------------------- */ ConstDefinitions = [ "const" ConstDef { ConstDef } ] . ConstDef = NewIdent "=" Constant ";" . Constant = [ "+" | "-" ] ( UnsignedNumber | ConstIdent ) | String . UnsignedNumber = UnsignedInt | UnsignedReal . ConstIdent = identifier . /* -------------------------------------------------------------------- */ TypeDefinitions = [ "type" TypeDef { TypeDef } ] . TypeDef = NewIdent "=" Type ";" . Type = SimpleType | [ "packed" ] StructType | "^" TypeIdent . SimpleType = TypeIdent | EnumerationType | SubrangeType . TypeIdent = identifier . EnumerationType = "(" NewIdentList ")" . SubrangeType = Constant ".." Constant . StructType = ArrayType | RecordType | SetType | FileType . ArrayType = "array" "[" IndexList "]" "of" Type . IndexList = SimpleType { "," SimpleType } . RecordType = "record" FieldList "end" . SetType = "set" "of" SimpleType . FileType = "file" "of" Type . FieldList = [ ( fixedPart [ ";" VariantPart ] | VariantPart ) [ ";" ] ] . fixedPart = RecordSection { ";" RecordSection } . RecordSection = NewIdentList ":" Type . VariantPart = "case" VariantSelector "of" Variant { ";" Variant } . VariantSelector = [ NewIdent ":" ] TypeIdent . Variant = CaseLabelList ":" "(" FieldList ")" . /* -------------------------------------------------------------------- */ VarDeclarations = [ "var" VarDecl { VarDecl } ] . VarDecl = NewIdentList ":" Type ";" . /* -------------------------------------------------------------------- */ ProcDeclarations = ( ProcHeading | FuncHeading ) ";" Body ";" . ProcHeading = "procedure" NewIdent [ FormalParams ] . FuncHeading = "function" NewIdent [ FormalParams ] ReturnType . ReturnType = [ /* empty if forward referenced */ ":" TypeIdent ] . Body = Block | "forward" . FormalParams = "(" FormalSection { ";" FormalSection } ")" . FormalSection = [ "var" ] ParamGroup | ProcHeading | FuncHeading . ParamGroup = NewIdentList ":" ParamType . ParamType = TypeIdent | "array" "[" IndexSpecList "]" "of" ParamType | "packed" "array" "[" IndexSpec "]" "of" TypeIdent . IndexSpecList = IndexSpec { ";" IndexSpec } . IndexSpec = NewIdent ".." NewIdent ":" TypeIdent . /* -------------------------------------------------------------------- */ StatementPart = CompoundStatement . CompoundStatement = "begin" StatementSequence "end" . StatementSequence = Statement { ";" Statement } . Statement = [ Label ":" ] [ AssignmentOrCall | CompoundStatement | GotoStatement | WhileStatement | RepeatStatement | IfStatement | CaseStatement | ForStatement | WithStatement ] . AssignmentOrCall = Designator ( ":=" Expression | [ ActualParams ] ) . ActualParams = "(" ActualParameter { "," ActualParameter } ")" . ActualParameter = Expression [ FieldWidth /* only in i/o */ ] . FieldWidth = ":" IntegerExpression [ ":" IntegerExpression ] . GotoStatement = "goto" Label . WhileStatement = "while" BooleanExpression "do" Statement . RepeatStatement = "repeat" StatementSequence "until" BooleanExpression . IfStatement = "if" BooleanExpression "then" Statement [ "else" Statement ] . CaseStatement = "case" OrdinalExpression "of" CaseList "end" . CaseList = OneCase { ";" OneCase } [ ";" ] . OneCase = CaseLabelList ":" Statement . CaseLabelList = CaseLabel { "," CaseLabel } . CaseLabel = Constant . ForStatement = "for" ControlVariable ":=" OrdinalExpression ( "to" | "downto" ) OrdinalExpression "do" Statement . ControlVariable = identifier . WithStatement = "with" RecVarList "do" Statement . RecVarList = Designator { "," Designator } . /* -------------------------------------------------------------------- */ IntegerExpression = Expression . BooleanExpression = Expression . OrdinalExpression = Expression . Expression = SimpleExpression [ RelOp SimpleExpression ] . RelOp = "=" | "<" | ">" | "<=" | ">=" | "<>" | "in" . SimpleExpression = ( "+" Term | "-" Term | Term ) { AddOp Term } . AddOp = "+" | "-" | "or" . Term = Factor { MulOp Factor } . MulOp = "*" | "/" | "div" | "mod" | "and" . Factor = Designator [ ActualParams ] | UnsignedLiteral | SetConstructor | "(" Expression ")" | "not" Factor . Designator = identifier { "." identifier | "[" ExpList "]" | "^" } . ExpList = Expression { "," Expression } . UnsignedLiteral = UnsignedNumber | "nil" | String . SetConstructor = "[" Member { "," Member } "]" . Member = Expression [ ".." Expression ] . /* -------------------------------------------------------------------- */ NewIdentList = NewIdent { "," NewIdent } . NewIdent = identifier . UnsignedInt = integer . UnsignedReal = real . String = string . END Pascal. | 1,278 | 5,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-51 | latest | en | 0.338854 |
https://www.experts-exchange.com/questions/10317533/Which-letter-or-digit.html | 1,529,323,345,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859766.6/warc/CC-MAIN-20180618105733-20180618125733-00483.warc.gz | 812,035,340 | 15,966 | # Which letter or digit ??
How can i determine the letter or digit from its ASCII code easily/efficiently. I currently use the switch statement and i have 26 cases for letter and 10 for digit.
i.e.
switch (letr)
{
case 97:
{
strcpy(ltr,"a");
ltr[1] = '\0';
strcat(expression_buffer,ltr);
ltr[0] = '\0';
};break;//end
.
.
.
case 122:
{
strcpy(ltr,"z");
ltr[1] = '\0';
strcat(expression_buffer,ltr);
ltr[0] = '\0';
};break;//end
and the same thing for digits!!!!
This makes my code too big
Pls help me find a better way of doing this.
I'm using turbo C++
All suggestions are welcome.
###### Who is Participating?
Commented:
You could just place the ASCII character in your character array. For example:
if (isalpha(letr) || isdigit(letr))
{
ltr[0] = letr;
ltr[1] = '\0';
}
0
Commented:
Also (and this is a step in the wrong direction, because Steve's suggestion is clearly better, but for future reference) you can always use a single character in place a number representing the character's aSCII value, for example you could have done
switch (letr)
{
case 'a':
{
strcpy(ltr,"a");
ltr[1] = '\0';
strcat(expression_buffer,ltr);
ltr[0] = '\0';
};break;//end
case 'b':
etc.
That should be much clearer to understand. Obviously, don't do it now!
0
Question has a verified solution.
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Have a better answer? Share it in a comment. | 396 | 1,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-26 | latest | en | 0.802506 |
https://oeis.org/A335870/internal | 1,620,387,811,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.80/warc/CC-MAIN-20210507090724-20210507120724-00024.warc.gz | 449,648,805 | 3,074 | 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!)
A335870 a(n) is the least k > 0 such that T^k(n) = T^(2*k)(n) (where T^k denotes the k-th iterate of A006370, the Collatz map); a(n) = -1 if no such k exists. 1
%I
%S 1,3,3,6,3,3,6,15,3,18,6,12,9,9,15,15,3,12,18,18,6,6,15,15,9,21,9,111,
%T 18,18,18,105,3,24,12,12,21,21,21,33,6,108,6,27,15,15,15,102,9,24,24,
%U 24,9,9,111,111,18,30,18,30,18,18,105,105,6,27,27,27,12
%N a(n) is the least k > 0 such that T^k(n) = T^(2*k)(n) (where T^k denotes the k-th iterate of A006370, the Collatz map); a(n) = -1 if no such k exists.
%C If the Collatz conjecture is true, then a(n) > 0 for all n >= 0.
%H Rémy Sigrist, <a href="/A335870/b335870.txt">Table of n, a(n) for n = 0..10000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cycle_detection#Floyd's_Tortoise_and_Hare">Floyd's Tortoise and Hare</a>
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%e For n = 3 we have:
%e k T^k(3) T^(2*k)(3)
%e - ------ ----------
%e 1 10 5
%e 2 5 8
%e 3 16 2
%e 4 8 4
%e 5 4 1
%e 6 2 2
%e so a(3) = 6.
%o (PARI) a(n, T=x->if (x%2, 3*x+1, x/2)) = my (x1=n, x2=n); for (k=1, oo, x1=T(x1); x2=T(T(x2)); if (x1==x2, return (k)))
%Y Cf. A006370, A139399.
%K nonn
%O 0,2
%A _Rémy Sigrist_, Jun 28 2020
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Last modified May 7 07:37 EDT 2021. Contains 343636 sequences. (Running on oeis4.) | 739 | 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} | 3.359375 | 3 | CC-MAIN-2021-21 | latest | en | 0.555808 |
http://dml.cmnh.org/2002Mar/msg00238.html | 1,553,314,855,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202723.74/warc/CC-MAIN-20190323040640-20190323062640-00217.warc.gz | 57,921,169 | 2,235 | # Re: Re[1] Froude numbers, GRF
```I understand what the speed would be, but I'd like to know what the GRF is
for FR < 16. The GRF is necessary to figure out how much of the mass would
have to be extensors and therefore whether the speed is possible or not.
----- Original Message -----
From: "dexter dexter" <dexter1647@caramail.com>
To: <dinosaur@usc.edu>
Sent: Tuesday, March 05, 2002 9:12 PM
Subject: Re[1] Froude numbers, GRF
<< After reading the new rex paper, I've beenconsidering
calculating the T value for lower Froude values (such as 4).
What would the GRF be for that? >>
Scale from that : It says Fr.: 1 at 5 m/sec-1, Tom Holtz
says Fr.: 3.8 at 9.1 m/sec-1 and Hutchinson says Fr.: 5 at
11 m/sec-1. That would give about fr.:4 = 9.5-10.0 m/sec-1.
(35-36 km/h or 21.5-22 mph).
Thomas Miller
_________________________________________________________
Le journal des abonnés Caramail - http://www.carazine.com
``` | 290 | 929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-13 | latest | en | 0.873497 |
http://oeis.org/A213053 | 1,545,066,349,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828697.80/warc/CC-MAIN-20181217161704-20181217183704-00339.warc.gz | 226,352,830 | 4,374 | This site is supported by donations to The OEIS Foundation.
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A213053 Decimal expansion of the absolute minimum of sinc(x) = sin(x)/x (negated). 1
2, 1, 7, 2, 3, 3, 6, 2, 8, 2, 1, 1, 2, 2, 1, 6, 5, 7, 4, 0, 8, 2, 7, 9, 3, 2, 5, 5, 6, 2, 4, 7, 0, 7, 3, 4, 2, 2, 3, 0, 4, 4, 9, 1, 5, 4, 3, 5, 5, 8, 7, 4, 8, 2, 3, 6, 5, 4, 4, 9, 0, 2, 7, 7, 1, 4, 5, 0, 5, 3, 4, 3, 5, 8, 9, 0, 6, 3, 2, 2, 9, 1, 8, 5, 5, 6, 8, 0, 5, 0, 6, 5, 3, 9, 2, 3, 5, 4, 9, 5, 1, 5, 2, 0, 1 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Minimum value of the first negative lobe of sinc(x), attained for abs(x) = A115365. The involute of the unit circle which starts at (1,0) crosses the x-axis for the first time at x = 1/a. - Álvar Ibeas, Jul 28 2017 LINKS FORMULA Equals -1 / sqrt(1 + A115365^2) = cos(A115365). - Álvar Ibeas, Jul 28 2017 EXAMPLE min[real x](sinc(x)) = -0.2172336282112216574082... MATHEMATICA digits = 105; NMinimize[ Sinc[x], x, WorkingPrecision -> digits+5] // First // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Mar 05 2013 *) RealDigits[Sinc[BesselJZero[3/2, 1]], 10, 100][[1]] (* Vladimir Reshetnikov, May 13 2016 *) PROG (PARI) y=solve(x=4, 4.5, tan(x)-x); -sin(y)/y \\ Charles R Greathouse IV, Jun 10 2012 CROSSREFS Cf. A115365. Sequence in context: A121416 A317547 A089329 * A200236 A292191 A239155 Adjacent sequences: A213050 A213051 A213052 * A213054 A213055 A213056 KEYWORD nonn,cons AUTHOR Stanislav Sykora, Jun 09 2012 STATUS approved
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Last modified December 17 12:05 EST 2018. Contains 318201 sequences. (Running on oeis4.) | 877 | 2,147 | {"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-2018-51 | latest | en | 0.612897 |
https://www.esaral.com/q/is-zero-a-rational-number-78276 | 1,713,049,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00564.warc.gz | 705,381,086 | 10,925 | # Is zero a rational number?
Question:
Is zero a rational number? Can you write it in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$ ?
Solution:
Yes, write $\frac{0}{1}$ (where 0 and 1 are integers and $q=1$ which is not equal to zero). | 85 | 263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-18 | latest | en | 0.847177 |
https://www.got-it.ai/solutions/excel-chat/excel-help/row-function?page=11 | 1,713,471,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817239.30/warc/CC-MAIN-20240418191007-20240418221007-00318.warc.gz | 715,921,853 | 59,558 | # Get instant live Excel expert help with ROW Functions
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Question :: 211
Write a program to generate the checksum for given multiple byte data, you can use any CRC algorithm. And verify the same given byte of data with generated checksum.
Input : NA
Output : NA
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Question :: 212
Write a program to find the no of Palindrome can be generated form the given string "abayacbbcayxdb" Ex: abayacbbcayxdb Palindrome: aba, acbbca, aya, cbbc
Input : NA
Output : NA
| Basics | | | | Experience
No Discussion on this question yet!
Question :: 213
Write your own strtok program in c.
Input : NA
Output : NA
| Basics | | | | Experience
#include <stdio.h>
#include <string.h>
int main()
{
char str[] = "qqq44in-tttterv-vviieew-www...-...com-mm";
strtoken(str,'-');
//printf("%s",str);
return 0;
}
void strtoken(char *x, char c)
{
char *y = x;
int i = 0, n = 0;
while (*x != '\0')
{
if (*x == c)
{
print(y,x);
y = x+1;
}
x++;
}
}
print(char *x, char *y)
{
while (x<y)
{
printf("%c",*x);
x++;
}
printf("\n");
}
Question :: 214
Write a program to delete a node in a Linked list.
Input : NA
Output : NA
| Basics | | | | Experience
No Discussion on this question yet!
Question :: 215
Write a message queue program. Read
Input : NA
Output : NA
| Basics | | | | Experience
No Discussion on this question yet!
Question :: 216
Write a program to find the no of word in given sentence, and count the repeating word. Ex: "abc helloabc mu multiple test onlinetest" There are 6 word and "abc" 2 time, "mu" 2 time "test" 2 time.
Input : NA
Output : NA
| Basics | | | | Experience
No Discussion on this question yet!
Question :: 217
Write a program to find the no. of occurrence of "bc" in given sting "acbcadgixbccbbcx"
Input : NA
Output : NA
| Basics | | | | Experience
No Discussion on this question yet!
Question :: 218
Write stack overflow program in c.
Input : NA
Output : NA
| Basics | | | | Experience
No Discussion on this question yet!
Question :: 219
Write a program in c,where memory leak happen.
Input : NA
Output : NA
| Basics | | | | Experience
No Discussion on this question yet!
Question :: 220
Write your own strncpy program in c.
Input : NA
Output : NA
| Basics | | | | Experience | 820 | 2,746 | {"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-2018-43 | latest | en | 0.506487 |
https://fastracblog.wordpress.com/2018/03/14/why-are-fat-people-fat/ | 1,561,006,985,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999141.54/warc/CC-MAIN-20190620044948-20190620070948-00335.warc.gz | 450,263,637 | 17,001 | # Why are Fat People Fat?
Yesterday, I traveled in a public transport bus. There were a few seats available. Three vacant seats, one in a two-seat and one each in the other two three-seat, were available to me. They were not bucket seats. The two persons in the three-seat were sitting at the extreme seats comfortably. They never bothered me. When I approached, they offered me the middle seat. But it was quite difficult to get into the middle seat. I had to physically disturb them to get into the middle seat. They were not ready to get up and ease my going to the middle seat.
They were ready to get disturbed. However, I decided not to disturb them. Instead, I opted to take the one in the two-seat. In that case, I don’t need to disturb anybody. But the person in the two-seat was really a big fat boy. He was keeping his two legs so wide that only 20% of seating area was available to me. I decided to take that seat. The boy did not bother to adjust for my comfort. He was playing with his cell-phone. He bought only one ticket. I also bought one ticket. However, he boarded the bus before I.
Bus fare is not charged according to a person’s weight. Why? I was thinking to find a solution for this problem while traveling the rest of my journey. Finally, I realized that this problem is a complexity. If a rule comes into force that bus fare should be collected according to the weight of a person, then that rule will benefit only the bus owners because the thin and lean persons will be charged the same fare while the fat persons will be charged more and that extra amount goes straight into bus owners’ pocket. Therefore, people don’t raise this issue. Instead, the thin and lean continue to suffer. It’s at the sacrifice of the thin and lean that the fat is enjoying.
Why is the problem a complexity? There are practical problems. How do you weigh a person before they board the bus? Even if a weighing machine is available, it will take extra time to weigh a person and a calculator to calculate a fare proportion to their weight. Busy buses cannot do this. Assuming that this is done, how do you seat them so that all passengers are seated proportionate to their fare and weight? Ultimately, the bus needs AI to allocate seats to optimize maximum number of persons in order to get maximum profit. Execution of this rule involves a cost. That will be again collected from the passengers. One simple solution, though not perfect, is to have bucket seats for all buses. But again you have to pay more for this comfort and this will be collected equally from all passengers fat or thin. In any case the thin and lean is not benefited financially.
A similar situation exists in middle class restaurants. A complete meal is priced at a fixed cost. Whatever quantity you eat you pay the same price. The fat eats more but pays the same price. The fat eats more at the cost of the thin and lean. Why does it happen perpetually? Why is the check not proportionate to quantity you eat? The lean becomes leaner and the fat becomes fatter. It seems it’s good to be fat.
In upper class restaurants you pay according to what you order but you pay more in such restaurants. The thin and lean have no choice but pay more. The restaurants are deliberately doing this in order to get more fat customers. If they increase the price the fat customers are not happy and the restaurants will lose many customers. In order to satisfy the fat customers they take something from the thin and lean and give it to fat. There is no other choice for the thin. The restaurants don’t lose the thin customers. The restaurants optimize their profit at the cost of the thin and lean. It’s taxing the thin and lean. | 784 | 3,701 | {"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} | 3.828125 | 4 | CC-MAIN-2019-26 | latest | en | 0.991428 |
https://www.manhattanprep.com/lsat/blog/category/lsat-strategies/ | 1,555,884,417,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532929.54/warc/CC-MAIN-20190421215917-20190422001917-00446.warc.gz | 751,252,601 | 19,291 | ## Articles published in LSAT Strategies
### What’s Tested on LSAT Reading Comprehension
LSAT Reading Comprehension can be both a blessing and a curse for LSAT takers. Read more
### How to Study for the LSAT
If you’re reading this blog post, you probably already know how hard it can be to study for the LSAT. The three different sections cover vastly different subject matter (I’m looking at you, Logic Games), the test is about how you think, not what you know, and on top of all that, the stakes are incredibly high! Because of all this, when you’re studying for the LSAT, you need to be strategic. This article will explore how to study for the LSAT to get the most out of your practice. Read more
### What’s Tested on LSAT Logical Reasoning
More than any other section of the test, the LSAT Logical Reasoning section has a clear mandate that directly pertains to your future as a law student: to make sure you can understand the ins and outs of argumentation. For that reason, one of my favorite LSAT Logical Reasoning tips—indeed, one of the first LSAT Logical Reasoning tips I share with all of my students—is to think of the Logical Reasoning section not as a hurdle you have to jump to get to law school, but as part of your essential preparation for law school. Read more
### Why and How LSAT Conditional Logic Wrecks Test-Takers
This post is inspired by some recent in-class interactions with students, with an inspirational assist from Ally Bell’s post “Conditional Logic Doppelgangers.” I hope you enjoy! Read more
### Introducing The 5 lb. Book of LSAT Practice Drills: An Innovative New Book to Supplement Any Study Plan
5 years ago, when we released LSAT Interact, my colleague Noah announced the project on our blog by saying, “Have you ever given birth to a baby? I have. And I did it along with some fellow LSAT geeks here at Manhattan Prep.”
Well, if LSAT Interact was the firstborn child of the Manhattan Prep LSAT team, The 5 lb. Book of LSAT Practice Drills is the second. Two years, 1,100 pages, and 5,000 LSAT practice problems later, we are so proud to present our new baby to the world. Read more
### How Your Science Fair Project Prepared You for LSAT Logical Reasoning
When I was in fourth grade, I designed a bizarre, painful, and deeply flawed experiment for the school science fair. My goal was to test the relative effectiveness of garlic and bug spray for repelling mosquitoes. I sacrificed myself for science and covered one-third of my arm in garlic, one-third in bug spray, and one-third in nothing, then stood outside next to the swampy forest at dusk to tally the bug bites. Even with fake arms hanging off of my project board, dotted with permanent marker “mosquito bites,” I still only took home an honorable mention. Read more
### The Spookiest Parts of the LSAT
If you’re a new trick-or-treater to the neighborhood, you have no strategy but to try every house. However, once you’ve lived there a few years, you’ve been around the block (literally). You know your different neighbors’ tendencies. You know what kind of candy they’re likely to give out. You know which houses to avoid: | 715 | 3,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-18 | longest | en | 0.909249 |
http://clay6.com/qa/1637/let-a-1-1-then-dicuss-whether-the-following-functions-defined-on-a-are-one- | 1,480,934,185,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541692.55/warc/CC-MAIN-20161202170901-00217-ip-10-31-129-80.ec2.internal.warc.gz | 57,030,460 | 27,469 | Browse Questions
# Let $A=[-1,1].$ Then,dicuss whether the following functions defined on $A$ are one-one,onto or bijective:$\; f(x)\;=\;\frac{x}{2}$
Note:This is the 1st part of the 4 part question
Toolbox:
• A function $f: A \rightarrow B$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
• A function is bijective if it is both one-one and onto
Given $f(x)=\frac{x}{2}\qquad x \in [-1,1]$
Let $f(x)=f(y)$
Step1: Injective or One-One function:
=>$\frac{x}{2}=\frac{y}{2}$
=>$x=y$
f is one one
Step 2: Surjective or On-to function:
$f(x)=y=1$
$=>\frac{x}{2}=1$
$x=2$
$2 \in [-1,1]$
There does not exists an element n in A
Such that $f(x)=y\qquad \;for\;y=1$
Therefore f is not onto
Solution:Hence f is one-one but not onto
edited Mar 27, 2013 by meena.p | 364 | 1,030 | {"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": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2016-50 | longest | en | 0.769419 |
https://sengkang-grand-residencescondo.com/qa/quick-answer-how-many-car-lengths-is-50-feet.html | 1,611,698,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803737.78/warc/CC-MAIN-20210126202017-20210126232017-00738.warc.gz | 555,078,949 | 8,719 | # Quick Answer: How Many Car Lengths Is 50 Feet?
## How much feet is a car length?
The average length of a typical current model intermediate sized automobile, such as a Toyota Corolla, is 14 to 15 feet.
A mid-sized vehicle, such as an Accord or Camry, 15 to 16 feet.
The standards state any vehicles more than 15.4 ft long & all passenger vehicles cannot exceed 16.8 ft length..
## How many car lengths should you be?
Figure one car length for every ten miles an hour,” Barndt said. “So if you’re doing 55 miles an hour you should have six car lengths between you so that if something happens to the car in front of you, you have time to stop or react.” The number two item Barndt says drivers are all guilty of is being distracted.
## How many car lengths is 2 seconds?
The two-second rule is useful as it works at most speeds. It is equivalent to one vehicle- length for every 5 mph of the current speed, but drivers can find it difficult to estimate the correct distance from the car in front, let alone to remember the stopping distances that are required for a given speed.
## How many car lengths does it take to stop?
Car: 243 feet (about 16 car lengths) – This gives you the necessary space to stop safely. Semi-Truck: 300 feet (about 20 car lengths) – Semis carry heavy loads, so more than slamming on the brakes, something can fall off or out of the truck, and you need time to react and avoid the debris.
## How long is a SUV in feet?
The average length of an SUV is around 175 to 190 inches or between 14 ½ feet to 16 feet. This changes if you take into consideration Mini SUVs who are within 9 to 12 feet in length and Full Size and Extended Length SUVs that can be as long as 18 feet or more.
## What should your following distance be at night?
Dim your lights before they cause glare for other drivers. Dim your lights within 500 feet of an oncoming vehicle and when following another vehicle within 500 feet. With low-beam headlights, approximately how far ahead can you see?
## How tall is a normal car?
An average large sedan is about 6 1/4 feet wide, while the average height is slightly less than 5 feet. A sedan’s length averages just more than 17 feet. The typical weight is more than 4,300 pounds.
## What is the 3 second plus rule?
The three-second rule is recommended for passenger vehicles during ideal road and weather conditions. Slow down and increase your following distance even more during adverse weather conditions or when visibility is reduced. Also increase your following distance if you are driving a larger vehicle or towing a trailer.
## How many feet does it take to stop at 35 mph?
136 feetAt 30mph the stopping distance is much greater—109 feet. At 35 mph it goes up to 136 feet, and you’re not really speeding yet. Switch up the numbers to freeway speeds—60 mph has a stopping distance of around 305 feet. That’s the length of an entire football field to stop.
## When should you use the 4 second rule?
You should apply the four-second rule when it’s wet, frosty or when you are towing a trailer. The four-second rule means that you leave four seconds between you and the vehicle in front. It gives you more time to react and more time to stop.
## How far should a car be behind another car?
two secondsThe two-second rule is a rule of thumb by which a driver may maintain a safe trailing distance at any speed. The rule is that a driver should ideally stay at least two seconds behind any vehicle that is directly in front of his or her vehicle.
## How long is a full size car?
The sales of full-size vehicles in the United States declined after the early 1970s fuel crisis. By that time, full-size cars had grown to wheelbases of 121–127 inches (3.1–3.2 m) and overall lengths of around 225 in (5,715 mm).
## How far do you travel when you look at your phone?
Keeping your eyes on the road is essential for safe driving. But when you look at your phone, you’re oblivious to what’s around you. In fact, at 50km per hour, even a 2 second glance at your phone means you’ll travel up to 28 metres blind.
## When should you use the 2 second rule?
The 2 second rule is used regardless of speed because the distance between your vehicle and the one in front will extend the faster you travel. Using the 2 second rule helps to significantly reduce accidents or reduce collision damage if one occurs.
## What is the safest way to drive up to intersections?
When driving up to the intersection signal your intention for at least three seconds if you are turning. If you are passing straight through an intersection check left and right in case another vehicle runs a red light, or is likely to pull out in front of you.
## How many car lengths is 50 mph?
Driver Care – Know Your Stopping DistanceSpeedPerception/Reaction DistanceEqual to Approx Number of Car Lengths (@15 feet)40 mph59 feet950 mph73 feet1460 mph88 feet1870 mph103 feet232 more rows
## When should you approach another car at night?
When following another vehicle, keep your low-beams on to avoid blinding the driver ahead of you. If you have car trouble at night, pull off the road as far as possible and turn on your hazard lights. Use your high-beam lights when driving in rural areas and on open highways away from urban and metropolitan areas.
## How do you calculate total stopping distance?
All you need to do is multiply the speed by intervals of 0.5, starting with 2. That’ll give you the stopping distance in feet, which is acceptable for the theory test. For example… There are 3.3 feet in a metre – so divide the distance in feet by 3.3 to get the stopping distance in metres. | 1,280 | 5,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-04 | latest | en | 0.956529 |
https://mathoverflow.net/questions/24280/converse-of-principal-ideal-theorem/24285 | 1,469,606,409,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826736.89/warc/CC-MAIN-20160723071026-00158-ip-10-185-27-174.ec2.internal.warc.gz | 850,971,084 | 15,084 | # Converse of Principal Ideal Theorem
$(A, \mathfrak{m})$ a Noetherian local ring, $a\in\mathfrak{m}$ a zero divisor. Then is it true that $\mbox{dim}\ A/(a) = \mbox{dim}\ A$ ?
-
No. Let A be the ring k[x,y,z]/(xz,yz) localized at (x,y,z). Then the dimension of A is 2, but the dimension of A/(x) is one. Note that xz = 0 in A.
The geometric picture is this: In 3-space, take the union of the horizontal z=0 plane with the vertical line x = y = 0, and look at the local ring at the origin. Then restricting to x=0, we get the union of the y-axis and the z-axis.
-
Such pictures are good for the memory. – Martin Brandenburg May 12 '10 at 0:05
@Charles Staats: What is the most easiest way to see that dimA=2? Thanks a lot! – TmobiusX May 13 '10 at 7:02
Necessary and sufficient conditions for your question to have a positive answer are that $A$ be $\textit{unmixed}$, that is, every associated prime is a minimal prime, and $\textit{equidimensional}$, that is, $\mathrm{dim} (A/\mathfrak{p}) = \mathrm{dim}(A)$ for every minimal prime $\mathfrak p$. Charles Staats' example is not equidimensional. The one-dimensional ring $A=k[x,y]/(x^2,xy)$ is not unmixed, and $A/(y)$ has dimension zero. Sufficiency follows from the fact that the set of zerodivisors of $A$ is equal to the union of the associated primes.
- | 404 | 1,318 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2016-30 | latest | en | 0.840574 |
https://community.anaplan.com/t5/Best-Practices/Workarounds-To-Sum-Data-While-Using-NONE-as-Summary-Method/ta-p/94700 | 1,611,712,743,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00716.warc.gz | 273,305,833 | 23,576 | Workarounds To Sum Data While Using NONE as Summary Method Within Source Module
As the Anaplan model size capacity has recently grown from roughly 18 billion cells to 100 billion cells with the introduction of HyperModel, it is still important to structure models efficiently, while creating as little 'wasted space' as possible.
The Planual states that the Summary Method for most line items should be set to None. This is a great rule to follow—unless:
• The totals need to be displayed on a dashboard/page.
• The data needs to be aggregated.
Let's assume we have several levels in a cost center hierarchy as shown below:
- L1 Region
- L2 Sub Region
- L4 Cost Center
Let's assume we have a source module dimensioned by Accts, L4 Cost Centers, Months, and a single line item where NONE is the Summary Method.
Now, let's say we have a target module dimensioned by Accts, L2 Sub Region, Months, and some line items.
If we simply reference the source module without any functions then this will not roll the data up (see above line item on the left). A quick way around this is to use the SUM formula and reference an SYS module that has the mapping of the L4 Cost Centers to the L3, L2, and L1.
In this example, the formula would be Source Data.'Amt (No Summary)'[SUM: 'SYS L4 Cost Center'.Sub Region]. Simple enough, right?
But what if the target module does not contain a higher level dimension of the L4 Cost Center structure? A workaround is to create a one-item dimension...
...and map it to the L4 Cost Centers list within the L4 Cost Center SYS module.
Whereas the L4 Cost Center mapping to L3, L2, etc. is based on the composite structure and uses the PARENT formula, the mapping to this one-item list is just OneItemList.MemberOfOneItemList.
Then, in order to sum the data within a target module, the one-item dimension must be included in the structure and reference the L4 Cost Center SYS module for the one-item mapping via a SUM formula.
If it is not desired to have the one-item dimension in the final output module, then the above module can be used as the source in conjunction with a line item to select the 'Dummy Total'...
...and then feed into a target model that references the staging module using a LOOKUP statement to point to the above line item.
Depending on the number of levels and size of the dimensions for the source and target modules this can save a lot of space, roughly 30-70%. For small modules, using this approach is likely not worth the effort, but this can provide tremendous space savings on very large modules.
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• Modeling
Version history
Revision #:
4 of 4
Last update:
3 weeks ago
Updated by:
The content in this article has not been evaluated for all Anaplan implementations and may not be recommended for your specific situation. | 629 | 2,817 | {"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-2021-04 | latest | en | 0.875354 |
https://www.cnblogs.com/xiandnc/p/10506440.html | 1,642,461,112,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300624.10/warc/CC-MAIN-20220117212242-20220118002242-00597.warc.gz | 784,831,454 | 9,746 | # 在Xunit中使用FsCheck
## 目录
Install-Package FsCheck.Xunit -Version 2.13.0
private int Add(int x, int y)
{
return x + y;
}
[Property]
public bool Commutative(int x, int y)
{
}
F#
[<Property>]
let Commutative x y =
#### 1. 不同的执行顺序，同样的执行结果
[Property]
{
var result1 = list.OrderBy(x => x).Select(Add1);
var result2 = list.Select(Add1).OrderBy(x => x);
return result1.SequenceEqual(result2);
}
F#
[<Property(Verbose=true)>]
let +1 then sort should be same as sort then +1 aList =
let add1 x = x + 1
let result1 = aList |> List.sort |> List.map add1
let result2 = aList |> List.map add1 |> List.sort
result1 = result2
#### 2.连续执行操作，结果跟之前一致
[Property]
public bool ReverseThenReverseShouldSameAsOriginal(int[] list)
{
var result= list.Reverse().Reverse();
return result.SequenceEqual(list);
}
F#
[<Property>]
let reverse then reverse should be same as original
(aList:int list) =
let reverseThenReverse = aList |> List.rev |> List.rev
reverseThenReverse = aList
#### 3. 有一些属性是永远不会改变的
public bool SomethingNeverChanged(List<int> list)
{
var result = list.OrderBy(x => x);
return result.Count() == list.Count;
}
F#
let sort should have same length as original (aList:int list) =
let sorted = aList |> List.sort
List.length sorted = List.length aList
## 为OO代码编写Property-based测试
public class Dollar
{
private int _amount;
public Dollar(int amount)
{
_amount = amount;
}
public int Amount => _amount;
{
}
public void Multiplier(int multiplier)
{
_amount = _amount * multiplier;
}
public static Dollar Create(int amount)
{
return new Dollar(amount);
}
}
F#
type Dollar(amount : int) =
let mutable privateAmount = amount;
member this.Amount = privateAmount
member this.Times multiplier =
privateAmount <- this.Amount * multiplier
static member Create amount =
Dollar amount
[Property]
public bool SetAndGetShouldGiveSameResult(int amount)
{
var dollar = Dollar.Create(0);
return dollar.Amount == amount;
}
F#
[<Property>]
let set then get should give same result value =
let obj = Dollar.Create 0
let newValue = obj.Amount
value = newValue
[Property]
public bool AddThenMultiplierSameAsCreate(int start, int times)
{
var dollar = Dollar.Create(0);
dollar.Multiplier(times);
var dollar2 = Dollar.Create(start * times);
return dollar.Amount == dollar2.Amount;
}
F#
[<Property>]
let add then multiplier same as create value times =
let dollar = Dollar.Create 0
dollar.Times times
let dollar2 = Dollar.Create(value*times);
dollar.Amount = dollar2.Amount
## 编写自定义Generator
public class User
{
public string Name { get; set; }
public int Age { get; set; }
}
public class UserArbitrary: Arbitrary<User>
{
public override Gen<User> Generator =>
from x in Arb.Generate<string>()
from int y in Gen.Choose(20, 30)
where x != string.Empty
select new User {Name = x, Age = y};
}
public class MyGenerators {
public static Arbitrary<User> User() {
return new UserArbitrary();
}
}
Arb.Register<MyGenerators>();
[Property]
public bool GenerateUsers(User user)
{
return user.Name != string.Empty;
}
posted @ 2019-03-10 19:01 .NET西安社区 阅读(376) 评论(0编辑 收藏 举报 | 833 | 3,097 | {"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} | 3.046875 | 3 | CC-MAIN-2022-05 | latest | en | 0.492189 |
http://de.metamath.org/mpeuni/issstrmgm.html | 1,723,139,634,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00267.warc.gz | 7,089,547 | 7,577 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > issstrmgm Structured version Visualization version GIF version
Theorem issstrmgm 17075
Description: Characterize a substructure as submagma by closure properties. (Contributed by AV, 30-Aug-2021.)
Hypotheses
Ref Expression
issstrmgm.b 𝐵 = (Base‘𝐺)
issstrmgm.p + = (+g𝐺)
issstrmgm.h 𝐻 = (𝐺s 𝑆)
Assertion
Ref Expression
issstrmgm ((𝐻𝑉𝑆𝐵) → (𝐻 ∈ Mgm ↔ ∀𝑥𝑆𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆))
Distinct variable groups: 𝑥,𝐵,𝑦 𝑥,𝐻,𝑦 𝑥,𝑆,𝑦 𝑥,𝑉,𝑦
Allowed substitution hints: + (𝑥,𝑦) 𝐺(𝑥,𝑦)
Proof of Theorem issstrmgm
StepHypRef Expression
1 simplr 788 . . . . 5 ((((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) ∧ (𝑥𝑆𝑦𝑆)) → 𝐻 ∈ Mgm)
2 simplr 788 . . . . . . . . . 10 (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → 𝑆𝐵)
3 issstrmgm.h . . . . . . . . . . 11 𝐻 = (𝐺s 𝑆)
4 issstrmgm.b . . . . . . . . . . 11 𝐵 = (Base‘𝐺)
53, 4ressbas2 15758 . . . . . . . . . 10 (𝑆𝐵𝑆 = (Base‘𝐻))
62, 5syl 17 . . . . . . . . 9 (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → 𝑆 = (Base‘𝐻))
76eleq2d 2673 . . . . . . . 8 (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → (𝑥𝑆𝑥 ∈ (Base‘𝐻)))
87biimpcd 238 . . . . . . 7 (𝑥𝑆 → (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → 𝑥 ∈ (Base‘𝐻)))
98adantr 480 . . . . . 6 ((𝑥𝑆𝑦𝑆) → (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → 𝑥 ∈ (Base‘𝐻)))
109impcom 445 . . . . 5 ((((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) ∧ (𝑥𝑆𝑦𝑆)) → 𝑥 ∈ (Base‘𝐻))
116eleq2d 2673 . . . . . . . 8 (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → (𝑦𝑆𝑦 ∈ (Base‘𝐻)))
1211biimpcd 238 . . . . . . 7 (𝑦𝑆 → (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → 𝑦 ∈ (Base‘𝐻)))
1312adantl 481 . . . . . 6 ((𝑥𝑆𝑦𝑆) → (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → 𝑦 ∈ (Base‘𝐻)))
1413impcom 445 . . . . 5 ((((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) ∧ (𝑥𝑆𝑦𝑆)) → 𝑦 ∈ (Base‘𝐻))
15 eqid 2610 . . . . . 6 (Base‘𝐻) = (Base‘𝐻)
16 eqid 2610 . . . . . 6 (+g𝐻) = (+g𝐻)
1715, 16mgmcl 17068 . . . . 5 ((𝐻 ∈ Mgm ∧ 𝑥 ∈ (Base‘𝐻) ∧ 𝑦 ∈ (Base‘𝐻)) → (𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻))
181, 10, 14, 17syl3anc 1318 . . . 4 ((((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) ∧ (𝑥𝑆𝑦𝑆)) → (𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻))
19 fvex 6113 . . . . . . . . . 10 (Base‘𝐺) ∈ V
204, 19eqeltri 2684 . . . . . . . . 9 𝐵 ∈ V
2120ssex 4730 . . . . . . . 8 (𝑆𝐵𝑆 ∈ V)
2221adantl 481 . . . . . . 7 ((𝐻𝑉𝑆𝐵) → 𝑆 ∈ V)
23 issstrmgm.p . . . . . . . 8 + = (+g𝐺)
243, 23ressplusg 15818 . . . . . . 7 (𝑆 ∈ V → + = (+g𝐻))
2522, 24syl 17 . . . . . 6 ((𝐻𝑉𝑆𝐵) → + = (+g𝐻))
2625adantr 480 . . . . 5 (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → + = (+g𝐻))
2726oveqdr 6573 . . . 4 ((((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) = (𝑥(+g𝐻)𝑦))
286adantr 480 . . . 4 ((((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) ∧ (𝑥𝑆𝑦𝑆)) → 𝑆 = (Base‘𝐻))
2918, 27, 283eltr4d 2703 . . 3 ((((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) ∧ (𝑥𝑆𝑦𝑆)) → (𝑥 + 𝑦) ∈ 𝑆)
3029ralrimivva 2954 . 2 (((𝐻𝑉𝑆𝐵) ∧ 𝐻 ∈ Mgm) → ∀𝑥𝑆𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆)
315adantl 481 . . . . 5 ((𝐻𝑉𝑆𝐵) → 𝑆 = (Base‘𝐻))
3225oveqd 6566 . . . . . . 7 ((𝐻𝑉𝑆𝐵) → (𝑥 + 𝑦) = (𝑥(+g𝐻)𝑦))
3332, 31eleq12d 2682 . . . . . 6 ((𝐻𝑉𝑆𝐵) → ((𝑥 + 𝑦) ∈ 𝑆 ↔ (𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻)))
3431, 33raleqbidv 3129 . . . . 5 ((𝐻𝑉𝑆𝐵) → (∀𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆 ↔ ∀𝑦 ∈ (Base‘𝐻)(𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻)))
3531, 34raleqbidv 3129 . . . 4 ((𝐻𝑉𝑆𝐵) → (∀𝑥𝑆𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆 ↔ ∀𝑥 ∈ (Base‘𝐻)∀𝑦 ∈ (Base‘𝐻)(𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻)))
3635biimpa 500 . . 3 (((𝐻𝑉𝑆𝐵) ∧ ∀𝑥𝑆𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆) → ∀𝑥 ∈ (Base‘𝐻)∀𝑦 ∈ (Base‘𝐻)(𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻))
3715, 16ismgm 17066 . . . 4 (𝐻𝑉 → (𝐻 ∈ Mgm ↔ ∀𝑥 ∈ (Base‘𝐻)∀𝑦 ∈ (Base‘𝐻)(𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻)))
3837ad2antrr 758 . . 3 (((𝐻𝑉𝑆𝐵) ∧ ∀𝑥𝑆𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆) → (𝐻 ∈ Mgm ↔ ∀𝑥 ∈ (Base‘𝐻)∀𝑦 ∈ (Base‘𝐻)(𝑥(+g𝐻)𝑦) ∈ (Base‘𝐻)))
3936, 38mpbird 246 . 2 (((𝐻𝑉𝑆𝐵) ∧ ∀𝑥𝑆𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆) → 𝐻 ∈ Mgm)
4030, 39impbida 873 1 ((𝐻𝑉𝑆𝐵) → (𝐻 ∈ Mgm ↔ ∀𝑥𝑆𝑦𝑆 (𝑥 + 𝑦) ∈ 𝑆))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 195 ∧ wa 383 = wceq 1475 ∈ wcel 1977 ∀wral 2896 Vcvv 3173 ⊆ wss 3540 ‘cfv 5804 (class class class)co 6549 Basecbs 15695 ↾s cress 15696 +gcplusg 15768 Mgmcmgm 17063 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-ndx 15698 df-slot 15699 df-base 15700 df-sets 15701 df-ress 15702 df-plusg 15781 df-mgm 17065 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 3,800 | 5,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.124814 |
https://www.definitions.net/definition/countability | 1,679,556,785,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00464.warc.gz | 832,322,380 | 17,338 | # Definitions for countabilitycount·abil·i·ty
### WiktionaryRate this definition:0.0 / 0 votes
1. countabilitynoun
The quality of being countable.
### WikipediaRate this definition:0.0 / 0 votes
1. countability
In mathematics, a set is countable if either it is finite or it can be made in one to one correspondence with the set of natural numbers. Equivalently, a set is countable if there exists an injective function from it into the natural numbers; this means that each element in the set may be associated to a unique natural number, or that the elements of the set can be counted one at a time, although the counting may never finish due to an infinite number of elements. In more technical terms, assuming the axiom of countable choice, a set is countable if its cardinality (its number of elements) is not greater than that of the natural numbers. A countable set that is not finite is said countably infinite. The concept is attributed to Georg Cantor, who proved the existence of uncountable sets, that is, sets that are not countable; for example the set of the real numbers.
### Numerology
1. Chaldean Numerology
The numerical value of countability in Chaldean Numerology is: 2
2. Pythagorean Numerology
The numerical value of countability in Pythagorean Numerology is: 7
### Popularity rank by frequency of use
countability#100000#259402#333333
## Translations for countability
• 가산 성
• 可数
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##### an almost pleasurable sensation of fright
• A. recital
• B. ditch
• C. reciprocal
• D. tingle | 859 | 3,032 | {"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-2023-14 | longest | en | 0.803436 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?p=89823 | 1,585,510,705,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370495413.19/warc/CC-MAIN-20200329171027-20200329201027-00546.warc.gz | 560,236,676 | 11,458 | ## 15.17
$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$
Moderators: Chem_Mod, Chem_Admin
Katelyn 2E
Posts: 35
Joined: Sat Jul 22, 2017 3:01 am
### 15.17
I'm a little confused as to how [C] is independent of the rate, as stated in the solutions manual.
Alvin Tran 2E
Posts: 39
Joined: Fri Sep 29, 2017 7:06 am
### Re: 15.17
The rate doesn't depend on the concentration of C, which means it is independent, because the reaction is zero order with respect to C. If you compare experiments 1 and 4 (where the concentrations of A and B stay the same), the initial concentration of C changes but the initial rate doesn't change.
Eli Aminpour 2K
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am
### Re: 15.17
compare the reaction rates when all other reactants remain constant except for C. The reaction rate doesn't change even though C does, meaning it isn't in the rate law because it is a zero order reactant.
Ammarah 2H
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am
### Re: 15.17
If you write the rate law, [C] will be to the zero power, meaning that [C] will always be equal to one, meaning it does not change the rate (so it is independent of the rate).
RohanGupta1G
Posts: 34
Joined: Sat Jul 22, 2017 3:00 am
### Re: 15.17
As the others said, C is zero order because looking at experiments 1 and 4, even though the concentration of C changes, the initial rate remains the same. This means C has no impact and therefore its exponent is 0.
Return to “General Rate Laws”
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https://forum.alphasoftware.com/showthread.php?100815-Programming-Puzzle-29-Another-math-treat&s=bcd971f59564f77538d1952051322566 | 1,569,138,446,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575402.81/warc/CC-MAIN-20190922073800-20190922095800-00512.warc.gz | 465,165,089 | 12,819 | # Thread: Programming Puzzle 29 - Another math treat
1. ## Programming Puzzle 29 - Another math treat
Puzzle 29
If you prepend a 1 to a five digit number, the result
is exactly 3 times smaller than if you appended a 1 to
the same five digit number. What's the five digit number?
i.e. Assume a five digit number NNNNN
1NNNNN is exactly 3 times smaller than
NNNNN1. What is NNNNN?
The "professor" thinks this is too easy, but I insisted you needed a break after Puzzle 28, so
have fun. Don't let conversion from string to numeric to string to numeric to string, and so on, drive
you crazy!
Your task boys and girls is to write an xbasic script that solves this puzzle. Display your
answer in the trace window. For extra credit don't post your script in the body of a reply,
but figure out how to "export" your script to a text file, and then "attach" your text file
Enjoy!
Credit for this puzzle: http://www.mathsisfun.com/
2. ## Re: Programming Puzzle 29 - Another math treat
Professor,
I hope my comments are clear. Five lines of code did it.
3. ## Re: Programming Puzzle 29 - Another math treat
hello
here is mine
answer in three places 1> message box, 2> trace window 3> status bar - overkill?
Attachment 32339
4. ## Re: Programming Puzzle 29 - Another math treat
end to the for next loop
Attachment 32343
5. ## Re: Programming Puzzle 29 - Another math treat
gandhi,
Very nice! When I saw your solution I woke the "professor" up to show it to him. He was his usual grumpy sellf, and muttered something about EXIT FOR before falling back asleep. You might check it out in the helps, since that can be a better way to jump out of a FOR ... NEXT loop.
Good work.
6. ## Re: Programming Puzzle 29 - Another math treat
Stephen,
Very nice!
You did a great job of narrowing the range of values to be checked. Your decision to step through the values 30 at a time was inspired. In my own solution I was looking for a way to "spin" the 10's digit ten at a time, without noticing that the script could be speeded up by spinning 30 at a time. Have you studied with the "professor" before?
7. ## Re: Programming Puzzle 29 - Another math treat
Of course, just for fun, if I was a smart-alic mathematician, I would do this:
justforfun.txt
8. ## Re: Programming Puzzle 29 - Another math treat
Andy, your solution illustrates another approach altogether. The key to all of these puzzles is figuring out how to express the problem in terms the machine can understand. Thanks for showing us a different way.
10. ## Re: Programming Puzzle 29 - Another math treat
Very good, Tony.
You should read up on "exporting" and "importing" scripts. The professor awards "extra credit" if solutions are pre-formatted for easy import. The same technique can be a big help in maintaining customer databases, as well. Much better than copying and pasting, or using simple text files.
-- tom
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• | 750 | 3,039 | {"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-2019-39 | latest | en | 0.894662 |
https://stats.stackexchange.com/questions/499788/can-you-write-a-geometric-random-variable-as-some-combination-of-bernoulli-rando | 1,713,276,848,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817095.3/warc/CC-MAIN-20240416124708-20240416154708-00325.warc.gz | 500,244,320 | 41,485 | Can you write a Geometric random variable as some combination of Bernoulli random variables?
Background
Given $$Y \sim \text{Binomial(n,p)}$$, we can write $$Y = \sum_{i=1}^{n} X_i$$ where $$X_1,X_2,...,X_n$$ are iid $$\text{Bernoulli}(p)$$. This is useful in, for example, determining the mean of a binomial random variable: $$E(Y)=E\left(\sum X_i\right) = \sum E(X_i) = np$$
Question
If we are given $$Y \sim \text{Geometric(p)}$$, can we similarly write $$Y$$ as some combination of Bernoulli random variables?
• It would have to be an infinite combination, wouldn't it? By "combination" do you mean a sum of independent variables or would you include more general operations? If so, what would they be?
– whuber
Dec 7, 2020 at 20:14
• I was trying to think about how to do it with a sum, but I couldn't make it work. So I decided to put "combination" instead. Yes, I would include more general operations. I'm not sure what they would be... I was thinking we need some way to mathematically "keep track" of whether we have gotten a success, because once we get a success, we stop the Bernoulli trials. But I'm not sure what that would look like. I think if we did have some kind of infinite sequence of operations and the "keep track" operation automatically becomes 0 once we get a success, it might work? Dec 7, 2020 at 20:19
• Trivially, every discrete distribution is a mixture of (shifted or scaled) Bernoulli distributions. A less trivial result is that the geometric distribution cannot be expressed as the sum of independent Bernoulli distributions, as you have already figured out.
– whuber
Dec 7, 2020 at 20:25
For clarity, I am going to look at the version of the geometric distribution with support on the non-negative integers, with expected value $$\mathbb{E}(Y) = (1-p)/p$$. Now, suppose we have a sequence of Bernoulli random variables $$X_1,X_2,X_3,... \sim \text{IID Bern}(p)$$ to use for the construction.
The geometric random variable $$Y$$ can be interpreted as the number of "failures" that occur before the first "success", so it can be written as:
\begin{align} Y &\equiv \max \ \{ y = 0,1,2,... | X_1 = \cdots = X_{y} = 0 \} \\[12pt] &= \max \Bigg\{ y = 0,1,2,... \Bigg| \prod_{\ell = 1}^{y} (1-X_\ell) = 1 \Bigg\} \\[6pt] &= \sum_{i=1}^\infty \prod_{\ell = 1}^{i} (1-X_\ell). \\[6pt] \end{align}
This is probably the "simplest" you can write the expression, since it accords with the descriptive intuition of what the random variable represents. As whuber notes in the comments, every discrete distribution can be written as a mixture of a countable number of shifted or scaled Bernoulli random variables (and indeed, there are an infinite number of ways to do this).
(Note: For the other version of the geometric distribution, with support on the positive integers, the random variable can be interpreted as the number of trials that occur by the time of the first "success", which is one more than the number of "failures". In this case you just add one to the above expression to get construct the random variable.)
Confirming the result: To see that this expression is adequate, note that:
\begin{align} F_Y(y) \equiv \mathbb{P}(Y \leqslant y) &= 1 - \mathbb{P}(Y \geqslant y+1) \\[12pt] &= 1 - \mathbb{P} \Bigg( \prod_{\ell = 1}^{y+1} (1-X_\ell) = 1 \Bigg) \\[6pt] &= 1 - \prod_{\ell = 1}^{y+1} \mathbb{P}(1-X_\ell = 1) \\[6pt] &= 1 - \prod_{\ell = 1}^{y+1} \mathbb{P}(X_\ell = 0) \\[6pt] &= 1 - \prod_{\ell = 1}^{y+1} (1-p) \\[6pt] &= 1 - (1-p)^{y+1}, \\[6pt] \end{align}
which is the CDF of the (chosen version of the) geometric distribution.
• This is cool! It looks like we can compute $E(Y)$ using your result, although I'm not sure about one step. $$E(Y) = E\left(\sum_{i=0}^{\infty} \prod_{l=0}^{i} (1-X_i)\right)=\sum_{i=0}^{\infty} \prod_{l=0}^{i}E(1-X_i)=\sum_{i=0}^{\infty}(1-p)^i=\frac{1-p}{p}$$ We can move the expectation inside the product because of independence. The only thing I'm not sure about is moving the expectation inside an infinite sum. I think it's justified because (with probability 1) the tail terms of the sum are 0. Dec 7, 2020 at 22:56
• You can move an expectation inside the infinite sum via the linearity property. You can then move it inside the product via independence. There are some errors in your working (e.g., wrong subscripts, etc). It should be: $$\mathbb{E}(Y) = \mathbb{E} \Bigg( \sum_{i=1}^\infty \prod_{\ell = 1}^{i} (1-X_\ell) \Bigg) = \sum_{i=1}^\infty \prod_{\ell = 1}^{i} \mathbb{E} (1-X_\ell) = \sum_{i=1}^\infty (1-p)^i = \frac{1-p}{1-(1-p)} = \frac{1-p}{p}.$$
– Ben
Dec 7, 2020 at 23:01
• Yes, I see that now. Thanks! Dec 7, 2020 at 23:06 | 1,482 | 4,645 | {"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": 12, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-18 | latest | en | 0.941111 |
https://chemistry.stackexchange.com/questions/97099/preparing-1m-acetic-acid-buffered-with-1-m-ca-acetate-to-specific-ph/97102 | 1,560,957,611,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00017.warc.gz | 400,035,203 | 34,781 | # Preparing 1M acetic acid buffered with 1 M Ca-acetate to specific pH
I am prepping a buffer solution of 1M acetic acid and 1M calcium acetate with a pH of 5, but it has been a long time since chem 101. I have some notes from an old lab mate that suggested that 600mL of acetic acid and 150mL Ca-acetate (with water added to make 1L) makes about a pH of 4.5. However when try using Henderson-Hasselbalch to get amount for a pH of 5 I get vastly different results - 640mL Ca-acetate and 360 mL of acetic acid. Where am I going wrong? Any tips on how to calculate the amounts needed correctly?
Thank you
This is the way I would approach the problem, for what it's worth.
Henderson Hasselbalch equation is pH = pKa + log [base]/[acid]
If we take the pKa of acetic acid (HAc) to be 4.74, then we have
5 = 4.74 + log [B]/[A] where B is the base and A is the acid (B = acetate)
Solving for B/A we get
log [B]/[A] = 5 - 4.74 = 0.26
[B]/[A] = 1.82 and ...
[B] = 1.82[A]
But since you are using calcium acetate as the salt, there are two acetates for each calcium acetate, i.e. CaAc_2 ==>2Ac^- + Ca^2+
Thus, [B] = 0.91[A] and
A + B = 1000 ml (to make 1 liter of buffer)
Simultaneous equations and solve for A
0.91A + A = 1000 ml
1.91A = 1000 ml
A = 524 ml
B = 476 ml
• Sorry for poor formatting. I'm new here. – Dr. J. May 17 '18 at 17:16
• Ok, thanks. I was definitely missing the two acetate ions part. – RNocked May 18 '18 at 18:30 | 486 | 1,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2019-26 | latest | en | 0.932318 |
http://mathoverflow.net/questions/106114/lie-subgroups-of-so2son | 1,397,667,575,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00177-ip-10-147-4-33.ec2.internal.warc.gz | 156,423,082 | 15,218 | # Lie Subgroups of SO(2)×So(n)
Hello, I need to know (connected closed) Lie subgroups of SO(2)×So(n), indeed these are compact Lie subgroups of SO(2,n) which I am looking for. But I don't know what we can say about Lie subgroups of product Lie groups. I'll be thankful if someone help me in this subject. Thanks in advanced.
-
Have you look in Hammermesh's book? Another useful reference could be Robert Gilmore's book. Cheers – Dox Sep 1 '12 at 16:39
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## 2 Answers
It is not clear to me what kind of answer is expected. Generally speaking, subgroups of Lie groups can be classified by Lie correspondence combined with combinatorial analysis resulting from structure theory of semisimple Lie groups. Below I address two particular cases.
If $K$ is a compact connected semisimple subgroup of $SO(2)\times SO(n)$ then its projection onto the first factor is trivial and the question is reduced to the $SO(n)$ case. (Closed) Lie subgroups of $SO(n)$ are precisely (compact) Lie groups with a faithful $n$-dimensional real orthogonal representation, so there are quite a few of them (the maximal connected ones were classified long time ago by Dynkin). If you need a complete description for small values of $n$, the Atlas of Lie groups is very handy.
In the other extreme case where $K=SO(2)$ you are, in effect, asking about the maps
$$f: SO(2)\to SO(2)\times SO(n).$$
They can be classified by passing to the Lie algebras. More precisely, the differential of $f$ is a linear map $so(2)\to so(2)\oplus so(n).$ Identifying $so(2)$ with $\mathbb{R}$ and $so(n)$ with the skew-symmetric matrices, it may be viewed as a pair $(d,A),$ where $d$ is an integer and $A$ is a skew-symmetric matrix whose eigenvalues are integral multiples of $i.$ Explicitly,
$$f:R(\varphi)\mapsto (R(d\varphi), \exp(\varphi A)),$$
where
$$R(\varphi)=\begin{bmatrix} \phantom{-}\cos(\varphi) & \sin(\varphi) \cr -\sin(\varphi) & \cos(\varphi) \end{bmatrix}$$
is the counterclockwise rotation by $\varphi$ and $\exp$ is the matrix exponential function.
The maps $f$ and $f'$ associated with non-zero pairs $(d,A)$ and $(d',A')$ have the same image if and only if the pairs are proportional. The case $d=0$ corresponds to an $SO(2)$ subgroup of the second factor $SO(n).$ In the case $d=1$, the subgroup $f(K)$ is the graph of a map $SO(2)\to SO(n).$
Note that for the original question about subgroups of $SO(2,n)$ one must impose further equivalences in the case $n=2$, because different subgroups of $SO(2)\times SO(2)$ can be conjugate in $SO(2,2)$.
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I wasn't able to make bmatrix to work correctly. Feel free to fix it if you can. – Victor Protsak Sep 2 '12 at 0:45
I fixed your rotation matrix. You need to triple up the backslashes (use \\\ instead of \\ for reasons of how the website and the LaTeX interact here) or else use \cr instead of \\. – Theo Buehler Sep 2 '12 at 2:33
thanks for your useful comments. I need to know the compact subgroups up to conjugacy, but if it is not possible to do, we may add some nice hypothesis like as semisimle , etc. It is also helpful to know any special classes of such subgroups. – nerd-math Sep 2 '12 at 14:27
I don't understand, why in the first case (second paragraph of the answer) when $K$ is semisimple it's projection on the first factor is trivial? – nerd-math Sep 20 '12 at 9:41
The first factor is abelian and semisimple groups do not have non-trivial characters. – Victor Protsak Sep 20 '12 at 19:49
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Let me answer this question for the Lie algebras, which already goes part way to answering the original question. The question is then what are the Lie subalgebras of $\mathfrak{so}(2)\oplus\mathfrak{so}(n)$. The general question of which are the Lie subalgebras of the direct sum of Lie algebras is solved by (the Lie algebraic version of) the Goursat Lemma.
It does not hurt to work in more generality. So Let $\mathfrak{g}_L$ and $\mathfrak{g}_R$ be two real Lie algebras and let $\mathfrak{g} = \mathfrak{g}_L \oplus \mathfrak{g}_R$ be their product. Elements of $\mathfrak{g}$ are pairs $(X_L,X_R)$ with $X_L \in \mathfrak{g}_L$ and $X_R \in \mathfrak{g}_R$. The Lie bracket in $\mathfrak{g}$ of two such elements $(X_L,X_R)$ and $(Y_L,Y_R)$ is given by the pair $([X_L,Y_L], [X_R,Y_R])$.
We are interested in Lie subalgebras $\mathfrak{h}$ of $\mathfrak{g}$.
Let $\pi_L : \mathfrak{g} \to \mathfrak{g}_L$ and $\pi_R : \mathfrak{g} \to \mathfrak{g}_R$ denote the projections onto each factor: they are Lie algebra homomorphisms. Let $\mathfrak{h}_L$ and $\mathfrak{h}_R$ denote, respectively, the image of the subalgebra $\mathfrak{h}$ under $\pi_L$ and $\pi_R$. They are Lie subalgebras of $\mathfrak{g}_L$ and $\mathfrak{g}_R$, respectively. Let us define $\mathfrak{h}^0_L := \pi_L(\ker \pi_R \cap \mathfrak{h})$ and $\mathfrak{h}^0_R := \pi_R(\ker \pi_L \cap \mathfrak{h})$. One checks that they are ideals of $\mathfrak{h}_L$ and $\mathfrak{h}_R$, respectively. This means that on $\mathfrak{h}_L/\mathfrak{h}^0_L$ and $\mathfrak{h}_R/\mathfrak{h}^0_R$ we can define Lie algebra structures. Goursat's Lemma says that these two Lie algebras are isomorphic.
Goursat's Lemma suggests a systematic approach to the determination of the Lie subalgebras of $\mathfrak{g}_L \oplus \mathfrak{g}_R$, which is particularly feasible when $\mathfrak{g}_L$ and $\mathfrak{g}_R$ have low dimension.
Namely, we look for Lie subalgebras $\mathfrak{h}_L \subset \mathfrak{g}_L$ and $\mathfrak{h}_R \subset \mathfrak{g}_R$ which have quotients isomorphic to $\mathfrak{q}$, say. Let $f_L:\mathfrak{h}_L \to \mathfrak{q}$ and $f_R:\mathfrak{h}_R \to \mathfrak{q}$ be the corresponding surjections. Let $\varphi$ denote an automorphism of $\mathfrak{q}$. Then we may define a Lie subalgebra $\mathfrak{h}$ of $\mathfrak{h}_L \oplus \mathfrak{h}_R$ by
$$\mathfrak{h} := \lbrace(X_L,X_R) \in \mathfrak{h}_L \oplus \mathfrak{h}_R | f_L(X_L) = \varphi(f_R(X_R))\rbrace$$ Of course, we need only consider automorphisms $\varphi$ which are not induced by automorphisms of $\mathfrak{h}_L$ or $\mathfrak{h}_R$. We record here the following useful dimension formula: $$\dim \mathfrak{h} = \dim \mathfrak{h}_L + \dim \mathfrak{h}_R - \dim \mathfrak{q}.$$
A commonly occurring special case is when one of $\mathfrak{h}_L \to \mathfrak{q}$ or $\mathfrak{h}_R \to \mathfrak{q}$ is an isomorphism. For definiteness let us assume that it is $\mathfrak{h}_R \to \mathfrak{q}$ which is an isomorphism. Then we get a Lie algebra homomorphism $\mathfrak{h}_L \to \mathfrak{h}_R$ obtained by composing $\mathfrak{h}_L \to \mathfrak{q}$ with the inverse of $\mathfrak{h}_R \to \mathfrak{q}$. In fact, we get a family of such homomorphisms labelled by the automorphisms of $\mathfrak{q}$ or, equivalently, of $\mathfrak{h}_R$. The fibred product which Goursat's Lemma describes is now the graph in $\mathfrak{h}_L \oplus \mathfrak{h}_R$ of such a homomorphism $\mathfrak{h}_L \to \mathfrak{h}_R$. The resulting Lie algebra is abstractly isomorphic to $\mathfrak{h}_L$.
In your case, since $\mathfrak{h}_R$, say, is one-dimensional, $\mathfrak{q}$ is either trivial or else isomorphic to $\mathfrak{h}_R$. In the trivial case, you have direct products of subalgebras, and in the latter case, the previous paragraph applies.
The Lie subalgebras of $\mathfrak{so}(n)$ are tabulated at least for small $n$ in several places. If you are a physicist, then perhaps Slansky's Physics Report Group theory for unified model building might be the most readable.
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add comment | 2,350 | 7,518 | {"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} | 3.015625 | 3 | CC-MAIN-2014-15 | latest | en | 0.881481 |
https://www.love2d.org/forums/viewtopic.php?p=232839 | 1,597,382,828,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739177.25/warc/CC-MAIN-20200814040920-20200814070920-00324.warc.gz | 726,956,781 | 10,152 | Template for using Löve as if it were not event-oriented
Showcase your libraries, tools and other projects that help your fellow love users.
pgimeno
Party member
Posts: 2228
Joined: Sun Oct 18, 2015 2:58 pm
Re: Template for using Löve as if it were not event-oriented
Yeah, the maze is a "perfect maze" (a spanning tree of a grid graph, in graph theory terms). A consequence is that from any point to any other point, there's only one path that does not retrace any steps. In other words, the solution is always unique.
monolifed
Party member
Posts: 127
Joined: Sat Feb 06, 2016 9:42 pm
Re: Template for using Löve as if it were not event-oriented
I see that is a mathematical maze solver I was thinking about something like this:
https://www.youtube.com/watch?v=LAYdXIREK2I
neku
Prole
Posts: 12
Joined: Thu May 17, 2018 3:07 pm
Re: Template for using Löve as if it were not event-oriented
Subsequent to the post of (Sep 21, 2019 12:11 pm) pgimeno and his reference to my post by: May 17, 2018 5:17 pm.
The program that was in question at that time I have finally finished with the love-template (published here).
it went well after all. you just have to forget about the love-mechanics.
the main code:
Code: Select all
-------------------------------------------------------------------------
-- main.lua (KT01 template: pgimeno)
-------------------------------------------------------------------------
--[[
concept of:
http://code.activestate.com/recipes/578382-knights-tour-using-warnsdorff-algorithm/
various considerations:
http://www.cs.utsa.edu/~wagner/python/knight/knight_tour.html
]]
-- possible moves
local pm = {{-2,1},{-1,2},{1,2},{2,1},{-2,-1},{-1,-2},{1,-2},{2,-1}}
local n = 8 -- number rowscollons
local cbx = n -- rows
local cby = n -- colls
-- startfeld - randomized (for testing)
math.randomseed (os.time ()) -- random start
local kx =math.random(1, 8)
local ky =math.random(1, 8)
print("Startfeld kx/ky: " ..kx.."/"..ky)
-- make chessboard nn
local cb = {} -- cb definition with zeros
for i = 1, n do
cb[i] = {}
for j = 1, n do
cb[i][j] = 0
end
end
math.randomseed (os.time ()) -- random (coin toss)
-- schachbrett sollte im zentrum sein (wie grid layout)
local gw =320 -- grid width: 8 x 40 = 320
local n =8 -- no of fields
local bs =gw/n -- feldbreite 40 (boxsize)
local count =0 -- Zaehler ( in drawLine() )
-- cb zeichnen (img geht auch)
function drawCB()
for r=0, n-1 do --row
for c=0, n-1 do --col
if (r+c)%2 == 0 then
love.graphics.setColor(255,0,0) --red
else
love.graphics.setColor(255,255,255) --white
end
love.graphics.rectangle("fill", r*bs, c*bs, bs, bs)
end -- for c
end -- for r
end --func
function printCB() -- show cb
for y = 1, n do
io.write()
for x = 1, n do
if cb[y][x] == 0 then
io.write("..")
elseif cb[y][x] <10 then
io.write( " "..cb[y][x].."|" )
else
io.write( cb[y][x].."|" )
end
end
print()
end
return
end
-- between inclusive
--if nx = 1 and nx = N and ny = 1 and ny = N then -- cb border
function between(x, min, max)
return (x>=min) and (x<=max)
end
k = 0 -- jumps-counter
function doJump()
while(k <65) do
cb[ky][kx] = k+1 -- counter set & save in cb
local start_x = kx * bs - (bs/2) -- make start x, y (f. line)
local start_y = ky * bs - (bs/2)
pq = {} -- make priority queue
for i = 1, n do -- row
nx = kx + pm[i][1]; ny = ky + pm[i][2]
if between(nx, 1, cbx) and between(ny, 1, cby) then
if cb[ny][nx] == 0 then ctr = 0 -- lua is 1-based!
for j = 1, n do -- col
ex = nx + pm[j][1]; ey = ny + pm[j][2]
if between(ex, 1, cbx) and between(ey, 1, cby) then
if cb[ey][ex] == 0 then ctr=ctr+1 end
end
end
table.insert(pq, (ctr*100)+i) -- looking for a better solution
end
end
end
--[[ Warnsdorff's algorithmus; extended
move to the neighbor that has min number of available neighbors
randomization we could take it - or not (coin toss)
]]
if #pq > 0 then
table.sort (pq) -- ascending (min to supreme)
minVal = 11 -- max loop nr
minD = 0 -- min value
for dd = 1, #pq do -- provisionally:
x = table.remove(pq,1) -- delete head-element
p = math.floor(x / 100) -- p wert extrahieren
m = x % 10 -- m (row) wert extrahieren
if p == minVal and math.random(100) <55 then
minVal = p
minD = m
end
if p < minVal then
minVal = p
minD = m
end
end --dd
m = minD
kx = kx + pm[m][1]
ky = ky + pm[m][2]
local end_x = kx * bs - (bs/2) -- make new end x, y
local end_y = ky * bs - (bs/2)
--print("start: ",start_x, start_y)
--print("end: ",end_x, end_y)
love.graphics.setColor(0,0,255) --blue
--love.graphics.setColor(255,255,0) --yellow
love.graphics.line(start_x, start_y, end_x, end_y)
sleep(0.1)
else
if k < 63 then
--print("Fehler im Feld-Nr.: " ..k)
print("Error in Field-No.: " ..k)
break
else
--print("Erfolg.")
print("Success.")
break
end
end
k = k+1
end -- while
end --func doJump()
-- end main pgm
-- **************************** main ****************************
drawCB()
doJump()
printCB()
-- **************************** end pgm *********************
Attachments
Knigt's Tour (finished)
KT01-love-PG - 30_09.png (44.63 KiB) Viewed 3630 times
neku
Prole
Posts: 12
Joined: Thu May 17, 2018 3:07 pm
Re: Template for using Löve as if it were not event-oriented
and the complete code (KT01.zip):
Attachments
Nestor-KT01.zip
(5.59 KiB) Downloaded 85 times
neku
Prole
Posts: 12
Joined: Thu May 17, 2018 3:07 pm
Re: Template for using Löve as if it were not event-oriented
Addendum to the topic: "the nested "for" loops are atrocious."
a nQueens variant with a variable chessboard, flexible window (with info) and wait for the next output (key or mouse button).
a screen or field for parameter value for largnes to chessboard would be nice .. but not yet realized.
Nesstor-nQueens-PG.zip
(32.51 KiB) Downloaded 83 times
neku
Prole
Posts: 12
Joined: Thu May 17, 2018 3:07 pm
Re: Template for using Löve as if it were not event-oriented
aiii .. the screenshot is stuck somewhere ...
;-)
Nestor-nQueens-PGcont - 02_11.png (36.45 KiB) Viewed 3068 times
pgimeno
Party member
Posts: 2228
Joined: Sun Oct 18, 2015 2:58 pm
Re: Template for using Löve as if it were not event-oriented
Update: Basically rewritten to accommodate all events. Now in https://notabug.org/pgimeno/alg-visualizer with full docs and examples. OP updated. Many thanks to babulous for the inspiration when writing this version.
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https://ccssmathanswers.com/180-days-of-math-for-fifth-grade-day-13-answers-key/ | 1,709,599,981,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00764.warc.gz | 154,880,070 | 56,005 | # 180 Days of Math for Fifth Grade Day 13 Answers Key
By accessing our 180 Days of Math for Fifth Grade Answers Key Day 13 regularly, students can get better problem-solving skills.
## 180 Days of Math for Fifth Grade Answers Key Day 13
Directions Solve each problem.
Question 1.
49 + 5 = ______
Perform an addition operation to find the sum of the two given numbers.
49 + 5 = 54
Question 2.
By performing the multiplication operation for the two given numbers we can find the product of 6 and 6.
We get 36
Question 3.
32 ÷ 8 = ____
Answer: 32 ÷ 8 = 4
By performing the division operation for the two given numbers we can find the quotient of 32 and 8
32 ÷ 8 = 4
Question 4.
Is 16,563 less than 16,653?
___________
Answer: 16,563 is greater than 16,653
Question 5.
Write 0.25 as a fraction.
___________
0.25 as a fraction is 25/100.
Question 6.
(3 × 5) + 6 = ___
(3 × 5) + 6
15 + 6 = 21
Question 7.
× 4 = 20
Let the unknown value be x.
x × 4 = 20
x = 20/4
x = 5
Question 8.
How many mL are in 7 L?
_____________
Convert from liters to milliliters.
1 Liter = 1000 milliliters
7 liters = 7 × 1000 milliliters = 7000 milliliters.
Question 9.
Does a pentagon have five right angles?
___________
Answer: A pentagon has a maximum of three right angles.
Question 10.
Favorite Foods
Tacos Spaghetti Pizza Hot Dogs 17 18 26 11
How many children were surveyed?
_____________
Answer: Children who love pizza were most surveyed.
Question 11.
What is the probability that you toss a coin and it lands with tails up?
_________________
There are two chances of getting tails of heads TH or HT
So, the probability that you toss a coin and it lands with tails up is 1 out of 2.
Question 12.
Sam had ten dollars to spend.
He buys 3 milkshakes. How many cheeseburgers can he buy with the rest of his money?
The cost of 1 milkshake is $1 3 milkshakes = 3 ×$1 = $3.00 The cost of cheeseburger is$2
$10 –$3 = \$7.00 | 580 | 1,922 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-10 | latest | en | 0.914482 |
https://electrowiring.herokuapp.com/post/states-of-matter-study-guide-answers | 1,581,976,214,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143373.18/warc/CC-MAIN-20200217205657-20200217235657-00479.warc.gz | 367,386,343 | 18,974 | 9 out of 10 based on 491 ratings. 2,476 user reviews.
# STATES OF MATTER STUDY GUIDE ANSWERS
[PDF]
10 States of Matter - Website
States of Matter SECTION 2 SHORT ANSWER Answer the following questions in the space provided. 1. a Liquids possess all the following properties except (a) relatively low density. (c) relative incompressibility. (b) the ability to diffuse. (d) the ability to change to a gas. 2. a. Chemists distinguish between intermolecular and intramolecular forces.
States of Matter Study Guide Flashcards | Quizlet
Start studying States of Matter Study Guide. Learn vocabulary, terms, and more with flashcards, games, and other study tools.
States of Matter Study Guide | Chemistry Flashcards | Quizlet
The 4th state of matter is a super-heated gas. The particles in this have less of an attraction than a solid. The particles in this overcome their attraction to each other. They move quickly and take the
Quiz & Worksheet - Different States of Matter | Study
A state of matter that has a definite volume but not a definite shape. A state of matter that does not have a definite volume or shape. A state of matter that has both a fixed volume and shape. A state of matter that conducts electricity well. A state of matter similar to gas that has free electrons.
States of Matter (Study Guide) - Mr Brennan's Science Page
matter. states of matter. solid, liquid, gas, plasma (difference between gas and plasma and an example of plasma) viscous. melting, freezing, boiling points. Vaporization, evaporation, condensation, sublimation. absolute zero. The boiling point and freezing point of water in Celsius, Fahrenheit and Kelvin. (100, 0 - 212, 32 - 373 - 273)
Unit 2: States of Matter - SMHS Physical Science
Unit 2: States of Matter. Unit 3: Classification of Matter. Unit 4: Atoms & the Periodic Table States of Matter; Answer Key: Study Guide: States of Matter; Subpages (5): Lesson 1: States of Matter Lesson 2: State Changes Lesson 3: Thermal Expansion Lesson 4: State [PDF]
Matter Unit Test
Matter Unit Study Guide [2] KEY. TCSS 1. 1. Samantha cools 100g of gaseous Nitrogen (N. 2. ) until it becomes liquid and then cools it even more until. it becomes a solid. Explain what happens to the energy of the molecules as the temperature is lowered.[PDF]
ch 12 Study guide TE - Mr. McKnight Clawson High School
TEACHER GUIDE AND ANSWERS Chemistry: Matter and Change Teacher Guide and Answers 7 Study Guide - Chapter 12 – States of Matter Section 12.1 Gases 1. motion 2. a. small b. forces c. random d. elastic; kinetic 3. KE 1/2 mv2 4. Temperature 5. true 6. true 7. false 8. true 9. true 10. false 11. true 12. false 13. a 14. a 15. d 16. d 17. b 18. b 19. b 20. barometer 21.
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matter study guide answersstates of matter study guidematter unit study guide 2 answer keymatter test study guideall about matter study guidematter study guide pdfmatter unit study guide 2study guide unit test matter | 758 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-10 | latest | en | 0.873373 |
https://knowledge.adaptiveplanning.com/Model_Administration/Managing_the_Model/Formulas/0020_Create_Formulas | 1,571,817,814,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829507.97/warc/CC-MAIN-20191023071040-20191023094540-00530.warc.gz | 525,975,739 | 12,391 | # Create Formulas
Includes some of the basic steps you take to create and work with formulas.
Formulas allow the calculation of numeric values in cells. In most cases, you can add a formula to any editable, numeric cell.
Formulas cannot be used in cells containing dates, text, and list boxes. Formulas cannot be created in read-only cells or in cells that show calculated totals.
### Enter a Formula
You can enter a formula by typing directly in the cell, or using the formula bar.
When creating formulas, click Apply to apply changes. To delete a formula, select the cell with the formula and click Cancel . After you enter a formula, an f(x)? symbol appears. This means that the formula hasn't been calculated, yet. Save the sheet and the formula will display its evaluated result.
Item Description
A blue f(x)? indicates the formula hasn't been calculated, yet. Save the sheet to display the calculated results.
A small triangle in the lower right corner of the cell indicates that the cell value was generated by a formula. Use the Cell Explorer to view formula details.
Cells with formulas that contain errors appear in red.
### Using the Capture Method to Reference Other Cells
A formula typed into a cell can reference other cells on the sheet. Start by typing a equal sign (`=`) into a cell. The cell formula contains an equal sign and nothing else. Now, click another cell in the sheet to include it in the formula. The cell now contains the code of the referenced account, as well as any other pertinent information about the selected cell.
The formula can be further modified by typing in simple operators symbols such as plus (`+`), minus (`-`), or multiply (`*`). When the formula is complete, click Apply
The Capture method only works when you are entering the information directly in a cell. It does not work if you edit or update the formula in the formula bar.
### Formula Errors
If a formula contains improper syntax (for example, a circular reference) the cell appears in red when the formula is saved. Hovering over the cell displays a tooltip to explain the cause of the error. For information on formula errors, refer to Troubleshooting Formulas.
### Use Formula Assistant
Anywhere a formula can be written, the Formula Assistant is available. The Formula Assistant helps users construct syntactically-correct formulas. You can use it to insert valid accounts, assumptions, qualifiers, and functions into a formula.
To open the formula assistant:
• Select a cell to contain a formula, then click the Formula Assistant button .
• Click the Formula Assistant link in Account Details, etc.
### Explore a Formula in a Cell
A small triangle in the lower right corner of the cell indicates that the cell value was generated by a formula.
To view the formula associated with a cell:
1. Select the cell that contains the formula.
2. Right-click the cell and select Explore Cell.
### Recommended Development Process
Adaptive Insights recommends the following process for developing formulas:
1. Plan what areas of your model will be impacted by the formula you’re creating.
You can create simple formulas that are used locally on a sheet or in a report. Or, you can create shared formulas that are used globally across your model. Formulas can be used in:
• Accounts – metric, modeled, cube
• Standard sheet data entry
• Cube data entry
• Formulas page (for level-based/shared formulas)
• Report calculations
• Allocation Rules
For more information, refer to Locations for Creating Formulas and Shared Formulas.
1. Get familiar with formula functions and syntax.
Adaptive Insights provides a broad range of formula functions (mathematical, logical, date, and string) and account references you can use as building blocks. Refer to Reference: Formula Syntax for Account References and Modifiers and Reference: Formula Functions.
Don't forget that a formula will often include both a calculation and some conditions that must be met to apply the calculation. For example, a pay raise may be defined as a simple 10% pay increase, but you may also need to include other conditions in the formula. For example: Is the person a full-time employee?; Have they reached their anniversary date?; and so on.
1. Get familiar with formula examples.
We've provided some common examples of formulas you can use for reference to see how formulas are typically constructed. In some cases, you can use these formulas as a starting point for your own models. Refer to Analyzing a Pay Rate Formula - A Detailed Walkthrough and Formula Examples.
1. Pick your formula tools. You can:
• Create simple formulas by typing directly into a cell or formula bar on a sheet. Or, by typing into a Formula text field.
• Use the Formula Assistant. The Formula Assistant can help you construct formulas by guiding you in the use of formula syntax including terms, modifiers, and functions.
• Define shared formulas by navigating to Formulas.
• Import or paste a formula from an outside source (Excel, Word)
Adaptive Insights generally recommends using the Formula Assistant to prevent syntax and naming errors. Refer to Using the Formula Assistant.
###### Formula Assistant
1. Start debugging your formula and correct errors.
As a formula developer, you may be inclined to write the complete formula and then begin debugging and checking for errors. This approach can save time for smaller, simpler formulas. However, more complicated formulas can include many variables, functions, and dependencies that need more attention. Therefore, it’s a good idea to test and debug each function separately to make sure it’s working as expected before combining the functions into a single formula. This saves time in the long run and helps you to quickly pinpoint the cause of issues if they occur.
Consider using a sandbox instance (if you have one) or a test account to try out your new formulas before rolling out the formula to a wider audience. Alternatively, have a rollback strategy (or backup) so you can easily restore an earlier version if things aren’t working as expected. | 1,251 | 6,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-43 | latest | en | 0.872003 |
https://www.geeksforgeeks.org/quizzes/top-mcqs-on-puzzles-with-answers/ | 1,713,850,649,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00037.warc.gz | 710,604,898 | 62,141 | # Top MCQs on Puzzles with Answers
Question 1
There are 25 horses among which you need to find out the fastest 3 horses. You can conduct race among at most 5 to find out their relative speed. At no point you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 horses.
• 5
• 6
• 7
• 8
Question 2
A Lady (L) bought an item of Rs 100 from the Shopkeeper (C). She paid him with a 500 Rs Note. Realizing that he did not have a change, the shopkeeper C got change for that note from another shopkeeper (S) and paid Rs 400 to the Lady.
After a few days, S realized that the note is fake, And he railed at C and took 500 Rs back from him.
So in this whole process, how much money did C lose in the end?
• 100
• 400
• 600
• 500
Question 3
A car has 4 tyres and 1 spare tyre. Each tyre can travel a maximum distance of 20000 miles before wearing off. What is the maximum distance the car can travel before you are forced to buy a new tyre? You are allowed to change tyres (using the spare tyre) an unlimited number of times.
• 20000
• 25000
• 15000
• 40000
Question 4
Your friend said, “If yesterday was tomorrow, today would be Friday.”
On which day did your friend make this statement ?
• Sunday
• Wednesday
• Friday
• Thursday
Question 5
There are two trains(Train A and Train B) running on the same track towards each other at a speed of 100 km/hr. They enter a tunnel 200 km long at the same time. As soon as they enter, a supersonic bee flying at a rate of 1000 km/hr also enters the tunnel from one side (say Train A side). The bee flies towards the other Train B and as soon as it reaches the train B, it turns back and flies back to the Train A. This way it keeps flying to and fro between the Trains A and B. The trains collide after a certain point of time leading to a massive explosion. The task is to find the total distance travelled by the bee until the collision occurred.
• 100
• 200
• 1000
• 500
Question 6
A farmer bought chickens for 4 unique clients on a selected day. Each customer buys half the amount of chicken left till his turn and half a chicken (i.e., if x chicken were left he buys x/2 + 1/2). The fourth customer buys a single chicken and after his turn, no chicken was left. Can you find the number of chickens the farmer bought on that day?
• 15
• 17
• 14
• 16
Question 7
Suppose there are four cards labeled with the letters A, B, C, and D and the numerals 3, 4, 5, and 6. It is known that every card has a letter on one side and a number on the other. The rule of the game is that a card with a vowel on it always has an even number on the other side. How many and which cards should be turned over to prove this rule to be true?
• 1
• 2
• 4
• 3
Question 8
There are 3 jars, namely, A, B, C. All of them are mislabeled. Following are the labels of each of the jars:
• A: Candies
• B: Sweets
• C: Candies and Sweets (mixed in a random proportion)
You can put your hand in a jar and pick only one eatable at a time. Tell the minimum number of eatable(s) that has/have to be picked in order to label the jars correctly. Assume that the shape of the candies and sweets are identical and there is no way to differentiate them by touching alone.
• 1
• 3
• 4
• 2
Question 9
Given a 8×8 chessboard, figure out the maximum number of kings that can be placed on the chessboard so that no two kings attack each other, i.e., none of the kings is under check. A king can move only one step at a time any direction on the chessboard (horizontally, vertically and diagonally).
• 32
• 18
• 8
• 16
Question 10
Ram used to arrive at the railway station every day at 6 pm from work. As soon as Ram arrived at the station, his wife Sita too arrives at the station by car to pick him up. Both drive back home. One day Ram arrived at the station one hour early and thought of taking a walk back to home. He started walking when after covering a specific distance, he meets with Sita. Both drive back home and this time they reached 30 minutes earlier than the usual time. How long has Ram been walking?
It is known that Sita drove every day at a uniform speed.
• 15
• 30
• 45
• 60
There are 10 questions to complete.
Last Updated :
Take a part in the ongoing discussion | 1,138 | 4,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | latest | en | 0.956941 |
http://robertsilvernail.com/gauss/ | 1,586,482,287,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371883359.91/warc/CC-MAIN-20200410012405-20200410042905-00426.warc.gz | 140,389,384 | 8,041 | Gauss
Gauss
Carl Friedrich Gauss. Have you ever heard of him? He was a German mathematician from the early 1800’s. Considered to be one of the greatest mathematicians ever. Belongs on the Mount Rushmore of math geniuses, up there with Euclid, Newton, Einstein, and maybe your high school trigonometry teacher. Personal friends with Ludwig von Beethoven (maybe not, but they were Germans of right about the same age, although Beethoven had moved to Austria).
Why should an investor care about Gauss? Because he came up with The Bell Curve, also known as the Normal Distribution Curve. All of us have seen it:
There are several bodies of academic study at play here. You have Probability Theory, which addresses the likelihood of a random event actually occurring. You have Randomness, which, in the investment world, is the notion that investment returns are Random, and not a result of some reason or design. Burton Malkiel, an American economist and author who remains active today, wrote “A Random Walk Down Wall Street”, which made famous the notion that investment returns are random. Closely related is the Efficient Market Hypothesis, which posits that stock prices immediately reflect all available information, meaning that technical analysis and possibly fundamental analysis of individual stocks is futile. All of these may be future blog topics. Perhaps I can find an artist who can draw Gauss and Malkiel randomly walking hand-in-hand.
If investment returns are indeed Random, then they can be measured and graphed, and their graph would resemble the above Normal Distribution Curve. Then, taking the next step, one can use that curve (and its inherent mathematics) to predict (with perhaps 68% or 95% probability) that future returns will fall within a plus-or-minus range of percentages.
All of today’s algorithmic stock trading has its roots in Gauss’s work. You could say C.F. Gauss is the gross-gross-grossvater of algorithms. Gazillions of dollars, Euro, Yen, and other currencies are now at work under the assumption that Gauss was onto something.
But what if Gauss was wrong? What if future event outcomes do not look like past event outcomes? Gauss himself offered answers to this, with the concepts known as Skewness and Kurtosis. The following chart represents a Normal Distribution (red line), a Skewed Distribution (green line) and two Kurtotic Distributions (flat, or Platykurtic, which is the yellow line, and steep, or Leptokurtic, which is the blue line).
The green Skewed line shows that the bulk of the returns (or random events) are not concentrated around the Mean. The Platykurtic yellow line shows returns that are not as concentrated around the mean as in the Normal distribution, and the Leptokurtic blue line shows returns highly concentrated around the mean. So, Gauss had alternatives to the Normal state of things.
However, what none of these graphs show is a situation where many returns or events are far away from the Mean. In other words, none of these show “Fat Tail” or “Black Swan” events. The “tails” on these graphs are the extremes, at -5 or +5 standard deviations or even farther out on each extreme. Imagine investment returns which are 5 or more standard deviations from the Mean – the Black Monday 1987 crash comes to mind, as does the 2008 correction that occurred over a longer (but still short) period of time.
Notice that these “Fat Tail” events I cited are to the downside. There are numerous other examples in the world of individual stocks (i.e., Enron, and employees who lost everything there because all they owned was Enron stock). An old saying is that stocks take the stairs up to the top and the elevator down. So maybe the Distributions should have their right (or positive) leg close to the X axis but the left (negative) leg raised up a bit, maybe about to kill a bug on the ground?
IMO
I believe the Normal Distribution is a mostly but not fully accurate way of graphing investment returns and therefore a somewhat flawed way of predicting future returns. If it was fully accurate, Long Term Capital Management would still be around (if you don’t know about its collapse in 1998, Google it). I believe trading algorithms are very useful and have indeed made billionaires out of their best practitioners. However, I also believe in Fat Tail/Black Swan risk, and that an investor needs to have some form of protection in place to guard against these risks. Being diversified among asset classes, being liquid (i.e., able to sell quickly if Armageddon happens), and having a strong portfolio manager are all good forms of portfolio protection. | 1,034 | 4,654 | {"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-2020-16 | latest | en | 0.952461 |
http://stats.stackexchange.com/questions/87037/which-variables-explain-which-pca-components-and-vice-versa | 1,464,076,693,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049270513.22/warc/CC-MAIN-20160524002110-00059-ip-10-185-217-139.ec2.internal.warc.gz | 270,088,032 | 21,902 | # Which variables explain which PCA components, and vice versa?
Using this data:
head(USArrests)
nrow(USArrests)
I can do a PCA as thus:
plot(USArrests)
otherPCA <- princomp(USArrests)
I can get the new components in
otherPCA$scores and the proportion of variance explained by components with summary(otherPCA) But what if I want to know which variables are mostly explained by which principal components? And vice versa: is e.g. PC1 or PC2 mostly explained by murder? How can I do this? Can I say for instance that PC1 is 80% explained by murder or assault? I think the loadings help me here, but they show the directionality not the variance explained as i understand it, e.g. otherPCA$loadings
Comp.1 Comp.2 Comp.3 Comp.4
Murder 0.995
Assault -0.995
UrbanPop -0.977 -0.201
Rape -0.201 0.974
-
Note that the signs of the loadings are arbitrary. The three crime variables are all positively correlated with each other, but you would be smart to work that from the signs of the loadings above. – Nick Cox Dec 3 '14 at 19:53
Unfortunately, I believe that the accepted answer to this question is incorrect. I posted my own answer below. – amoeba Jan 15 '15 at 22:59
The problem with princomp is, it only shows the "very high" loadings. But since the loadings are just the eigenvectors of the covariance matrix, one can get all loadings using the eigen command in R:
loadings <- eigen(cov(USArrests))$vectors explvar <- loadings^2 Now, you have the desired information in the matrix explvar. - thank you random guy, could you possibly show me for example assault or urban pop we could do this? partly confused because there is only one correlation present in the matrix for assault – user1320502 Feb 18 '14 at 16:10 Sorry, I improved my answer and did not notice you commented my post already. assault loads with -0.995 on PC1. Thus, one can conclude after squaring this value PC1 explains 99% of the variance of the variable assault. After squaring the values of urban pop, you can conclude PC3 explains 4% and PC2 95.5% of the variance of urban pop. – random_guy Feb 18 '14 at 16:24 Doesn't OP ask about how much of the PCA can be attributed to a variable? Your answer is about how much of a variable can be explained by a CPA – Heisenberg Dec 3 '14 at 18:40 @amoeba Of course, they are not the correlations. Thanks for pointing that out. – random_guy Dec 3 '14 at 18:51 @Heisenberg Yes, however, I think the first part of the question goes in the direction I answered it and in the other part he got it the other way around. Nevertheless, what I wrote in response to his comment makes clear from what perspective I looked at it and I think that's okay for him because the answer was accepted. – random_guy Dec 3 '14 at 19:00 You can do a backwards or forwards stepwise variable selection predicting a component or a linear combination of components from their constituent variables. The$R^2$will be 1.0 at the first step if you use backwards stepdown. Even though stepwise regression is pretty much of a disaster when predicting$Y$it can work well when the prediction is mechanistic as is the case here. You can add or remove variables until you explain 0.8 or 0.9 (for example) of the information in the principal components. - Neat idea--thank you. – rolando2 Dec 4 '14 at 0:21 The US arrests data bundled with R are just an example here, but I note that the loadings calculations in the question come from a PCA of the covariance matrix. That's somewhere between arbitrary and nonsensical, as the variables are measured on different scales. Urban population looks like a percent. California is 91% and highest. The three crime variables appear to be number of arrests for crimes expressed relative to population size (presumably for some time period). Presumably it's documented somewhere whether it's arrests per 1000 or 10000 or whatever. The mean of the assault variable in the given units is about 171 and the mean murder is about 8. So, the explanation of your loadings is that in large part the pattern is an artefact: it depends on the very different variability of the variables. So, although there is sense in the data in that there are many more arrests for assaults than for murders, etc., that known (or unsurprising) fact dominates the analysis. This shows that, as any where else in statistics, you have to think about what you are doing in a PCA. If you take this further: 1. I'd argue that percent urban is better left out of the analysis. It's not a crime to be urban; it might of course serve proxy for variables influencing crime. 2. A PCA based on a correlation matrix would make more sense in my view. Another possibility is to work with logarithms of arrest rates, not arrest rates (all values are positive; see below). Note: @random_guy's answer deliberately uses the covariance matrix. Here are some summary statistics. I used Stata, but that's quite immaterial. Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- urban_pop | 50 65.54 14.47476 32 91 murder | 50 7.788 4.35551 .8 17.4 rape | 50 21.232 9.366384 7.3 46 assault | 50 170.76 83.33766 45 337 - I think that the accepted answer can be dangerously misleading (-1). There are at least four different questions mixed together in the OP. I will consider them one after another. • Q1. How much of the variance of a given PC is explained by a given original variable? How much of the variance of a given original variable is explained by a given PC? These two questions are equivalent and the answer is given by the square$r^2$of the correlation coefficient between the variable and the PC. If PCA is done on the correlations, then the correlation coefficient$r$is given (see here) by the corresponding element of the loadings. PC$i$is associated with an eigenvector$\mathbf V_i$of the correlation matrix and the corresponding eigenvalue$s_i$. A loadings vector$\mathbf L_i$is given by$\mathbf L_i = (s_i)^{1/2} \mathbf V_i$. Its elements are correlations of this PC with the respective original variables. Note that eigenvectors$\mathbf V_i$and loadings$\mathbf L_i$are two different things! In R, eigenvectors are confusingly called "loadings"; one should be careful: their elements are not the desired correlations. [The currently accepted answer in this thread confuses the two.] In addition, if PCA is done on covariances (and not on correlations), then loadings will also give you covariances, not correlations. To obtain correlations, one needs to compute them manually, following PCA. [The currently accepted answer is unclear about that.] • Q2. How much of the variance of a given original variable is explained by a given subset of PCs? How to select this subset to explain e.g.$80\%$of the variance? Because PCs are orthogonal (i.e. uncorrelated), one can simply add up individual$r^2$values (see Q1) to get the global$R^2$value. To select a subset, one can add PCs with the highest correlations ($r^2$) with a given original variable until the desired amount of explained variance ($R^2$) is reached. • Q3. How much of the variance of a given PC is explained by a given subset of original variables? How to select this subset to explain e.g.$80\%$of the variance? An answer to this question is not automatically given by PCA! E.g. if all original variables are very strongly inter-correlated with pairwise$r=0.9$, then correlations between the first PC and all the variables will be around$r=0.9$. One cannot add these$r^2$numbers to compute the proportion of variance of this PC explained by, say, five original variables (this would result in a nonsensical result$R^2 = 0.9\cdot0.9\cdot5>1$). Instead, one would need to regress this PC on these variables and obtain the multiple$R^2\$ value. | 1,894 | 7,805 | {"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} | 2.921875 | 3 | CC-MAIN-2016-22 | latest | en | 0.905821 |
https://studydaddy.com/question/what-is-the-x-coordinate-of-the-point-of-inflection-on-the-graph-of-y-1-10x-5-1 | 1,527,368,951,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867904.94/warc/CC-MAIN-20180526210057-20180526230057-00505.warc.gz | 649,618,585 | 8,991 | Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.
QUESTION
# What is the x-coordinate of the point of inflection on the graph of y=1/10x^(5)+1/2X^(4)-3/10?
We find the Inflection Points of y by finding the second derivative of the function (y''), and the x-values at which y'' equals 0.
We look for the zeroes because at those points the concavity (or the direction in which the slope of the function f(x) is trending) has leveled off; it is at these points that the concavity is most likely to turn from positive to negative, or vice-versa.
Just as background, Math is Fun offers notes on inflection points.
y= 1/10x^5 +1/2x^4 -3/10 y'=1/2x^4+ 2x^3 y''=2x^3+6x^2
We set our second derivative to 0: y''=x^2(2x+6)=0
And we can find our inflection points through these equations:
• x^2=0
• 2x+6=0
Our Inflection Points are then at x= 0, and x=-3 | 278 | 916 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.919313 |
https://convertoctopus.com/965-grams-to-ounces | 1,680,435,136,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00789.warc.gz | 222,380,911 | 7,396 | ## Conversion formula
The conversion factor from grams to ounces is 0.03527396194958, which means that 1 gram is equal to 0.03527396194958 ounces:
1 g = 0.03527396194958 oz
To convert 965 grams into ounces we have to multiply 965 by the conversion factor in order to get the mass amount from grams to ounces. We can also form a simple proportion to calculate the result:
1 g → 0.03527396194958 oz
965 g → M(oz)
Solve the above proportion to obtain the mass M in ounces:
M(oz) = 965 g × 0.03527396194958 oz
M(oz) = 34.039373281345 oz
The final result is:
965 g → 34.039373281345 oz
We conclude that 965 grams is equivalent to 34.039373281345 ounces:
965 grams = 34.039373281345 ounces
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 ounce is equal to 0.029377744170984 × 965 grams.
Another way is saying that 965 grams is equal to 1 ÷ 0.029377744170984 ounces.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that nine hundred sixty-five grams is approximately thirty-four point zero three nine ounces:
965 g ≅ 34.039 oz
An alternative is also that one ounce is approximately zero point zero two nine times nine hundred sixty-five grams.
## Conversion table
### grams to ounces chart
For quick reference purposes, below is the conversion table you can use to convert from grams to ounces
grams (g) ounces (oz)
966 grams 34.075 ounces
967 grams 34.11 ounces
968 grams 34.145 ounces
969 grams 34.18 ounces
970 grams 34.216 ounces
971 grams 34.251 ounces
972 grams 34.286 ounces
973 grams 34.322 ounces
974 grams 34.357 ounces
975 grams 34.392 ounces | 460 | 1,712 | {"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} | 3.96875 | 4 | CC-MAIN-2023-14 | latest | en | 0.789998 |
https://debunkingrelativity.com/2012/12/21/a-brief-journey-into-the-weird-sciences/ | 1,500,906,781,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424884.51/warc/CC-MAIN-20170724142232-20170724162232-00367.warc.gz | 649,257,077 | 64,704 | ## Relativity stupidity
In our everyday world we know that different observers measure the speed of a moving object differently depending upon their own speed. For example an observer standing on a platform may measure the speed of a motor bike as 100kmph. Another observer travelling in a bus at 40kmph in the same direction will measure the speed of the same motor bike as 60kmph. And the motor cyclist himself will measure the speed of his bike as zero with reference to him. So the speed of any object is relative and depends upon the reference frame of the observer. This is what commonsense tells us. But apparently this commonsense can’t be applied to Light. Relativity preaches us that light always travels with the speed ‘c’ (3×108m/sec) irrespective of the reference frame of the observer.
If we ask why, some relativists put that down to Maxwell. It is true that Maxwell had deduced the value of ‘c’ (speed of light) mathematically after experimenting on electromagnetism but he didn’t know to which reference frame this speed of light applies. While scientists were pondering on this reference frame issue, Einstein mesmerised the scientific folk with his weird maths and said that the SOL (3x108m/sec) deduced by Maxwell must be applicable to every observer irrespective of their reference frame and made the crowd to believe in the absurd law he proposed i.e. the law of constant speed of light.
Having lost the commonsense, the mesmerized scientific folk then interpreted every experiment as proof of relativity. As discussed elsewhere no experiment straight away supports any notion, rather we the humans apply our commonsense, interpret the data and decide whether the experimental data supports a notion or not. So we need commonsense and reasoning to interpret any experiment. But the mesmerized scientific folk had abandoned them in favour of weird maths. Great physicists like Stephen Hawking believe that our commonsense and logic may get affected by our earthly ‘illusions’ but not our mathematics. Scientists argue that what we see and how we experience the world depends upon how our brain processes and interprets the data it receives from the sense organs. So, what we see and experience i.e. our perceived reality may not be the actual reality, and another creature’s brain may interpret the same in a different way depending upon its neuronal anatomy and physiology. So our ‘picture’ of the universe could just be an illusion created by our brain. Hence the physicists argue that our logic and commonsense can’t be sworn upon to explain Nature and its actual behaviour.
But then, how come mathematics which is also the result of our brain’s activity can be relied upon any better? How come only Logic gets affected by our earthly illusions but not mathematics? I believe that Logic is the basis of all our knowledge and understanding of the Nature. And logic is the basis of mathematics. If some mathematical model predicts something that is against logic, there is no reason to discard our logic and uphold the mathematical prediction. Every mathematical model, however complex it may be, is ultimately built upon bits of simple reasoning and logic. Then how can mathematics contradict logic? How can anything contradict its own basic pillars of foundation and still be valid?
Most physics students do agree that the theory of relativity is weird, but they put that down to their ignorance and inability to grasp the ‘complex’ mathematics behind the theory (like the crowd in the Emperor story who believe that it must be their ignorance that is stopping them from appreciating the Emperor’s magical costume!). And to progress in their career, students have to believe in the weird theory and live up to the expectations of their professors (who themselves have also gone through the same indoctrination process as students). After years of chanting and studying the same physics, some ‘bright’ students at some point of time in their career get ‘enlightened’ and they ‘realize’ that relativity is not at all weird but actually represents the ultimate reality or truth. Having studied and chanted the weird theory for years, now they don’t see anything weird in relativity. And having suppressed common sense during all these years of study, now it is the commonsense that appears weird to them. At this stage they get opportunities to join and interact with the top class physicists of the world (who had also gone through the same phases of ‘transformation’) and keep spreading the weird science. This is how science students ascend in their career and become physicists. And the process is no different from someone becoming a priest.
But most science students aren’t ‘bright enough’ to reach to that celebrity stage and hence settle somewhere much down in the social hierarchy of the ‘science religion’. And they continue to believe that it is their ignorance that stops them from fully understanding the weird theory and from experiencing the truth. “Because the theory has been endorsed by all the top class physicists, and accepted and taught all over the world, though the theory sounds weird and its predictions absurd, it must probably be true” an average student is right to think this way. But a logician doesn’t blindly believe in what the majority think or what some celebrity professors and scientists teach. Every scientific theory is amenable to logical deduction unless it is based upon some weird magical assumption. As I said earlier, Logic is the basis of all our knowledge including science and it can’t be defeated by weird theories masquerading as science. In this work I have argued why the theory of relativity and its predictions are absurd and illogical by all means of reasoning. I have also exposed the distorted interpretation of many experiments which the mesmerized physicists claim as proof of the weird theory.
Overthrowing someone’s theory doesn’t automatically make that someone stupid. For example Ptolemy’s geocentric model of the Universe was thrown away later by better reasoning in the wake of newer information gained as part of the mankind’s ongoing quest to understand Nature. But that shouldn’t make Ptolemy and his followers any stupid, because the model was true and very much logical up to that point of time. But that’s not the case with the theory of relativity. We don’t need any newer information or more sophisticated experiments to disprove the absurd theory which the modern physicists hail as the greatest scientific theory and whose principles they chant every day. Disproving relativity just involves exposing the relativists’ weird thinking and their stupid interpretation of the various experiments. So unlike the case with the Ptolemy’s Geocentric theory, disproving relativity also proves relativists as stupid.
The stupid thinkers claim that their weird theory has been proved beyond doubt by many experiments. Obviously no experiment straight away supports any theory but the data needs logical interpretation to arrive at correct conclusions. If some folk strongly believes that our world is fundamentally weird and hence declares that logic isn’t the best way of understanding nature, how can we expect such weird folk to draw logically valid conclusions out of any experimental data? No doubt that, physicists are the most intelligent crowd amongst the humans and I agree that we all need to respect them for advancing our knowledge and technology. But what if they get affected by a mania and that mania masquerades as science? It will be a big shame not only for them but to all the humans. It will also be a shame to our Planet Earth if some aliens realise how stupid the most intelligent race on earth thinks! So to save science from weird theories and to save ourselves from the embarrassment, our physicists must be rescued from the relativity mania.
### Quantum ignorance
Quantum physicists are not as stupid as relativists – While relativity starts with the weird assumption of constant SOL and is supported by false interpretation of experimental observations (which could have been easily explained by classical science unlike what the relativity maniacs claim), quantum physics is ‘woven’ to explain some ‘really’ weird observations to which classical physics couldn’t offer logical explanation. For example the results of double slit experiment suggest that an electron travels via both the slits simultaneously. This observation and others forced the physicists to propose the weird laws of the quantum world.
While I don’t call quantum physicists as stupid, I blame them for one reason- Rather than trying to find out the missing logical ‘link’ connecting the classical and quantum worlds, the ‘tired’ physicists have taken the easy path of ‘blaming’ the Nature for being weird at the quantum scale. They teach that events in the quantum world (e.g. radioactive decay) occur ‘by chance’ or at random and hence what we can expect to know is only the probability of such events. For example we can only know how many atoms in a given radioactive substance may decay in a certain period but can’t exactly predict which individual atom decays and when. Apparently even Nature doesn’t ‘know’ when each individual atom decays. The overconfident physicists claim that they know as much as the Nature knows and the reason why they are unable to accurately predict any individual event at the quantum level is because Nature itself doesn’t know! In other words, we are ignorant because the Nature is ignorant. This is where I feel the quantum physicists are wrong. We can accept that quantum world is weird and hence we are unable to accurately predict individual events in the microcosm but that weirdness and unpredictability of quantum world must be to do with our ignorance and inability.
Of course, its again the Relativity religion which distorted the face of science altogether and there by necessitated establishment of the quantum religion. If only physicists hadn’t misinterpreted Michelson’s experiment and thus forced the scientific community abandon the Ether theory, quantum physics with all its absurd notions wouldn’t have come into existence. Because, then physicists would have realized that Ether model would provide a very simple and straight forward explanation for the double slit experiment. But unfortunately, as Ether was ‘disproved’, the later physics pastors had no choice but to establish the quantum religion with all its mythical and illogical notions in order to explain the results of double slit experiment.
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• Galacar On April 14, 2014 at 2:38 am
“we all need to respect them for advancing our knowledge and technology”
Well, really?
There is nothing good that ‘modern science (physics) has brought us.
Nothing at all!
Not the atomic bomb (is also not good), not the transistor, not the MRI, not the microwave oven, not remote control and the list goes on and on and on…
Warm greetings
Galacar (Holland)
Liked by 1 person
• pimikepi On April 30, 2014 at 8:37 pm
There is nothing illogical about the predictions of mathematics. Unfamiliar, yes, but not illogical. Take the Monty Hall problem. Are you going to say that because the prediction seems illogical that the mathematics is false? Put your money where your logic is, then. I will play you, I will follow the mathematics and beat you using mathematics.
QM was the “easy path?” Do you REALLY think they didn’t agonize about it for years? Einstein never accepted the conclusions of QM. It was only with Bell’s Theorem, and its experimental verification, that Einstein was proved wrong. Bell showed that if QM predictions were only due to ignorance, you would get one result, but if QM was fundamentally how nature was, you would get another. The latter case was true.
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• drgsrinivas On May 7, 2014 at 3:23 pm
Nobody here is arguing against the value of correct mathematics built upon correct logic. Just like there exists stupid people like relativists, there also exists stupid mathematics built upon stupid notions. I have only been arguing against that stupid maths.
Mathematics just represents a symbolic version of a logical argument. Something that can be explained in mathematical terms can also be expressed in terms of logical statements. And there isn’t any puzzle that will only ‘yield’ to mathematics but not to logic. Either the puzzle is a stupid one ‘woven’ around a delusion (like your twins’ paradox) or it will yield to both maths and logic.
When someone fails to predict things correctly using calculations, it either shows the inability of the person to use the correct method of calculation or it may be that the correct way of calculation is not yet known or devised. (When someone doesn’t know how to add numbers, one would obviously get at a wrong sum. It amounts to stupidity if one blames mathematics for the wrong result). Similarly when people fail to solve a puzzle using logic, the problem is not with the discipline of Logic, but is due to the ignorance and inability of the people to apply the correct logic.
Are you arguing that Monty Hall paradox can’t be explained by Logic? Unfortunately for your religious crowd, this oft-repeated paradox can be easily solved by simple logic. You want to play with me? No problem, you may come with all your crowd. If I lose I will pay you. If I win I don’t want your money. Only ignorant people crave for money. My only expectation is to set free a poor mind from the clutches of a stupid religion. And let me correct you that Logic can never be unfamiliar; one just has to follow the correct logical sequence.
Of course, even if I were to fail to solve that by Logic, that doesn’t in any case underestimate the value of Logic. That would just indicate my inability. And if you fail to solve a puzzle by Maths; that would just indicate your inability and not that of the Maths per se. It would be stupid to portray a competition between individuals as that between Maths and Logic.
Einstein is surely the most intelligent person amongst your stupid folk. While he managed to ‘convert’ all the ‘scientific’ folk into his ‘religion’ and made them to religiously chant his stupid theory, he remained highly sensible when it came to accepting your other great religious theory – he didn’t fall prey to the quantum religion unlike the rest of your stupid scientific religious folk.
Bell’s theorem only proves the ignorance of your stupid crowd. How can any ignorant and stupid crowd prove or disprove anything? If you presuppose the shape of an egg as a pyramid, then you could prove that the oval white thing laid by a hen is not an egg. That is, you can prove and disprove things as per your religious beliefs by making some wrong stupid presuppositions.
I am not exaggerating your pastors’ stupidity at all. This is what they usually do. For example let me tell you how your pastors disproved Ether drag. They presupposed that if Ether drag was true, Earth would drag a large blob of Ether around it. But only a stupid mind can accept that presupposition- when a ball moves in air, it wouldn’t drag a blob of air. And similarly when a ball moves inside a stationary pool of water, it wouldn’t drag a fixed blob of water around it. (https://debunkingrelativity.com/ether-wind-and-ether-drag/)
And even if we imagine that we drag a ‘blob’ of air around us as we walk in rain, the rain drops wouldn’t fall straight down unlike what your pastors preach. They would still fall at an angle. https://debunkingrelativity.com/2014/09/07/aberration-of-star-light/
The way your pastors think reminds me of a saying in Telugu- when a stupid person was asked how many are pancha pandavas (five pandavas), he apparently said “they are like the three legs of a cot” and showed two fingers. Not only that he didn’t know how many are pandavas, he also didn’t know how many legs a cot will have. And when someone shows two fingers to represent three, we can imagine how stupid is that someone. Your pastors are not too different.
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• Angela Stahlfest-Moller On December 13, 2014 at 6:07 am
Hello you wonderful man (or woman).
I thought I was the only one thinking, if you can say gravity affects TIME then I can say temperature affects DISTANCE!
Do you know how I found you? Not by ‘gravity does NOT affect time’ which just gave me the usual crap. What finally worked was something like ‘crackpot gravity & time’. I think that says a lot. My 13yr old already thinks gravity affects time.
Thankyou again and if you are interested I have a theory.
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• drgsrinivas On December 14, 2014 at 1:06 pm
Thank you for your interest. I am glad to see more and more people like you successfully ‘navigating’ through the vast ocean of ignorance and reaching here. Yes, people who are truly rational and not blinded by the superstitious religion of science are welcome here to share their views and interact.
And thank you Galacar for keeping the site active!
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• Galacar On December 13, 2014 at 12:14 pm
Angela Stahlfest-Moller .
I am not the owner of this site, but Welcome!
Your reaction moved me! So good to see that people are waking up to all the bullshit in ‘science’ . I love that!!
And see how the young are being indoctrinated!
I don’t know where all your people are coming from, but I am really thinking of lectures of a few ours about how we are being lied to on all fronts!
We are sold so much bullshit, it is unbelievable.
Anyway, yes, I am very much interested in your theory!
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• Angela Stahlfest-Moller On December 13, 2014 at 2:35 pm
Hello Galacar. It is wonderful that you are interested, I really need feedback.
I do not know if this is allowed, but my email is ladygreen69@outlook.com.
It is a work in progress, as I really need collaborates. Trying to go to my local university was a bad move.
Thankyou Angela
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• Galacar On December 14, 2014 at 3:10 pm
Angela Stahlfest-Moller,
I have just send you an e-mail. I am very curious.
You also wrote
“Trying to go to my local university was a bad move.”
For me too. However I never finished. I did physics./ math and clinical psychology.But I have seen enough to see what a madhouse it really is!
Furthermore I am glad I find out what I know now.
But , were I live, I see students going to university every day and
I feel for them.
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• Angela Stahlfest-Moller On December 15, 2014 at 3:35 pm
Hello drgsrinivas and Galacar. And other reading this.
Is it possible to find how many different types of clock have been used in gravity vs time experiments? ie pendulum, water, spring without batteries etc. Or just atomic.
Thankyou
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• Angela Stahlfest-Moller On December 15, 2014 at 6:51 pm
Hello again.
How do you feel about massless anything and energy mass equivalence?
I think it is more mania. E=mc2 contains an equals sign not an equivalance sign.
Ice=frozen water, when water freezes into ice neither the frozen nor the water disappears, the equals sign means they are different ways of saying the same thing.
I will say it another way. kg.m2/s2= Joules= kg.m2/s2= Energy= kg.m2/s2= mc2= kg.m2/s2. If kg or m=0 then E=0. If there is E then there HAS to be mass.
Liked by 1 person
• drgsrinivas On December 16, 2014 at 10:05 am
I agree, if there is E, there has to be mass. I did present my views about energy (and the silly equation E=MC2) at few places.
But for every sensible question that we pose to the relativists, they will have a stupid answer ready. For example, a photon apparently can have momentum (mc) despite its zero mass (m=0). And for them, it is the momentum multiplied by the velocity that gives the Energy. So energy could exist without mass!
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• Galacar On December 15, 2014 at 11:46 pm
to Angela Stahlfest-Moller
About those clocks, I only know of atomic clocks. But I may be wrong of course,
About mass, in my opinion mass is non-existent
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• Angela Stahlfest-Moller On December 16, 2014 at 5:03 am
Hello again.
As you can tell I am finding the chance to talk irresistable. And my phone limits post length.
I am going to use my favourite estimation and guess that 50% to 90% of the people that find this site, and the few others like it, are not only thinkers rather than believers but also have an idea of how things actually work.
I am giving a thumbs up to this site as the simplest and most rational I’ve seen just trying to say NO, & although not the most polite it is true.
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• Angela Stahlfest-Moller On December 16, 2014 at 5:58 am
I have always thougt that Einstein was spot on, Everything IS Relative.
Including numbers.
2 is only 2 relatiue to; base 10, divided by 1, + not -, to power 1.
Should include; multiplied by ‘the absolute of the square root of’ (or ‘§’) +1 or -1.
Now ‘§’ +1 is half + & half -, a binary system. ‘§’ -1 is also a + or – binary. Together it is a dual binary system from which complex numbers emerge.
DNA is a dual binary of AT or TA & CG or GC from which protein production emerges.
Brain too. A
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• Angela Stahlfest-Moller On December 16, 2014 at 6:42 am
Hate;
Forget constant SOL, it is at least debatable, half of relativity is based on ‘if gravity did’t exist’!
Forget logic, I refuse to base my view of the univese on a theory that includes a false statement on moral & mental grounds. That is almost the very definition of ‘a mental problem’.
Relativity explains the universe from the viewpoint of a particle. Explain flight with ‘the bird remains still, all the rest moves, etc’.
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• drgsrinivas On December 16, 2014 at 10:12 am
Very nicely put. I couldn’t control my laughter!
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• Angela Stahlfest-Moller On December 16, 2014 at 7:06 am
Particle theory explains using particles only, but they can point to stars for ‘why particles?’.
String theory uses only strings & no answer to ‘why strings?’.
Colliders are following a logic train like this; ‘I am going to smash apart this uncooked pasta, & analyze the fragments with respect only to each other, to find the flour & water it is made from’. Not possible.
Strong Nuclear ignores: Still repels!! Any explaination of atoms must include a change in rules at very close so + & + attract.
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• Angela Stahlfest-Moller On December 16, 2014 at 12:08 pm
Hi Guys.
Yeah I found the E=mc2 comments after that post, but limited post space made me not correct my self.
Reading ‘A Grand Design’ gets me cranky, because of the ‘Philosophy’ thing. Although ‘renormalization’ does not help; ‘oh dear we got the wrong answer, to prove we know what we are doing we need this answer, so we will renormalize our answer to give the answer we need to prove us right’. Arrgh!!
Also thanks for response.
Angela
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• LadyGreen On December 17, 2014 at 3:49 am
Hi Galacar.
I am guessing you do not like my take on the universe.
What is your idea to explain what we call mass?
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• Galacar On December 17, 2014 at 3:06 pm
to Angela Stahlfest-Molle
Do not like it? Well I have recieved it , and thanks for that,
but still needs to read it and I will.
My take is that this ‘universe’ is a hologram. a holographic universe.
There is only light(waves) and what we see is there because we decode it with our brains.For a good example look up:
Jill Bolte Taylor: My stroke of insight,
Because of her stroke she doesn’t decode the ‘input’ anymore!
Furthermore, most scientist are left brain thinkers, an If you are going to do research into rigt brain, you can find out that the left brain is extremely short sighted and stupid (Hence the stupidity of ‘scientists, see also my remarks on
the education system, which shuts out the right brain as far as possible)
Anyway, if here is only light, hence I think there is no ‘mass’
Because there is no matter at all!
The interesting thing is that if you start research from this viewpoint, one is able
to calculate a lot of interesting things, because if this world is indeed only light,
and human have been adjusted to this. lots of things you can calculate, e.g. the optimoum length of pregnancy, the form and length of dna, the optimum temperature for health, the number of accupuncture points, radioactive decay and so on and so forth,
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• LadyGreen On December 18, 2014 at 5:00 am
Hi.
Come join me at PhysForum.com for some frustrating fun with ‘belivers’.
Thankyou
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• drgsrinivas On December 27, 2014 at 2:54 pm
You have really been making rational arguments to prove why slowing of an atomic clock isn’t same as time dilation. But I don’t think the religious scientific community will be ready to accept ideas that challenge their religious superstitions and beliefs. We can discuss with religious believers who know that they are religious believers. But we can never discuss with and convey anything to those religious people who don’t realise that they are religious believers, but call themselves as ‘scientific’ ‘rational’ ‘sceptic’ etc and adamantly claim their stupid beliefs as proven facts!
May be we should all come together and launch a ‘crusade’ against the stupid religion of modern physics. Just thinking how we should do that!
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• Galacar On December 18, 2014 at 2:09 pm
Nope, I was banned there because , while I was very very serious, they thought I was ‘trolling’. You see, they like new ideas, but to too new. 😉
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• LadyGreen On December 18, 2014 at 5:37 pm
Hi Galacar.
My first knee jerk reaction was no. I thought when you break a hologram each piece has all the info of the original.
But then went, yes and when you pull us apart you get cells, each with all the info of the original.
Also agree that real is just our brain interpreting data.
So in the end I find no fault with the idea.
Do you consider us as a copy of another nonholographic universe? Or is this it, not a hologram of something, just a hologram?
Last post for this week.
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• Galacar On December 19, 2014 at 3:04 pm
I don’t know yet about yours being non-holographic yet. I only stated my viewpoint.
And you are on the right track about the cells, however, I really think it goes deeper than that and that all the information in the universe is at every point and there are ways to access that information. I also use energy psychology techniques in whicht the premise is that this world is a hologram.(TAT).And there is so much more, pointing at this.And I find this extremely effective!
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• LadyGreen On December 25, 2014 at 4:02 am
A very MERRY CHRISTMAS to everyone!
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• Blue Heffnir On January 1, 2015 at 4:20 am
And a very HAPPY NEW YEAR to everyone too 🙂
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• Galacar On January 1, 2015 at 4:35 pm
HAPPY NEW YEAR!!! to you all
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• skycentrism On January 24, 2015 at 10:58 pm
Hello. Check out the Concave Earth theory. I make videos myself and here’s a great source of information as well: http://www.wildheretic.com/concave-earth-theory/
Regards.
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• David P On February 11, 2015 at 9:28 pm
Excellent site — I’ve been think much the same things here for the past 20 years. SR goes against common sense and engineering. “Nothing can go faster than light” is just the new flat earth philosophy — humans in general always seem to need an “edge” to their universe to feel comfortable. Two things i’d like to see more on:
1. Black holes. Singularities can’t exist. All particles in the collapsar would have to collapse uniformly inwards at the same moment. In reality their entropy would give them an ever increasing acceleration resulting in an extremely fast orbit around one another — the closest you could get to a black hole is a “dark doughnut”. All of this would have to happen outside any theoretical “event horizon” so light would always escape.
2. Expanding universe. More ether nonsense. Perhaps some environmental phenomenon (e.g. gravity) accounts for Doppler shift other than motion towards/away. Perhaps if we were at the center of the galaxy we would see less red shift or maybe even our physicists would say we live in a contracting universe?
Keep up the good work!
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• Scott On February 14, 2015 at 9:14 am
If you seriously believe relativists are stupid…then why start this “article” with – “In our everyday world we know that different observers measure the speed of a moving object differently depending upon their own speed. For example an observer standing on a platform may measure the speed of a motor bike as 100kmph. Another observer travelling in a bus at 40kmph in the same direction will measure the speed of the same motor bike as 60kmph…” Kinda paradoxical in nature isn’t it? Seeing as you share, at least, some relativistic views.
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• drgsrinivas On February 14, 2015 at 5:51 pm
That just shows the plight of science students and science believers: most of them confuse relativity of motion of classical mechanics(“Galilean- Newtonian relativity”) with Einstein’s theory of relativity! Despite the fact that they don’t really know what Einstein had preached, they chant and adorn Einstein everyday.
Anyway, let me clarify your doubt: By the word ‘relativists’ I only mean people who believe in Einstein’s theory of relativity (and not “classical” relativity of motion)
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• woodside On March 13, 2015 at 8:47 pm
In accordance to classical physics, the Global Positioning System gives an example where the speed of light measured by an observer varies with their own velocity.
A Global Positioning System receiver records that a transmission from a satelite to the east of the receiver approaches faster, that is it arrives earlier, than that of a signal from a satelite that is west of the receiver. The earth rotates towards the east. Radio wave and light are both electromagnetic propogation.
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• truerelativist On March 27, 2015 at 4:09 am
Hi, Dr. Srinivasa
I love, that You have taken the time and created this interesting and necessary website. There is not much material in the web about this, so thank you!
I only recently took interest in relativity and quantum theory, thinking before that it is too complicated and I am incapable of understanding any of it. But since I have always been interested about how stuff works and I started to run out of interesting things to learn about, I took a shot at it. I watched a video in youtube, explaining the whole thing. Then it hit me – THIS HAS GOT TO BE WRONG!
After a few weeks intensely reading and thinking about this, I finally understood, where the erroneous thinking came from. I started searching for similar explanations, but couldn´t find any, so I am posting this here for the first time.
Here is the reasoning witch should put an end to the „relativity“ mania once and for all:
I put relativity in braces for a reason – I 100% agree – all movement in space is relative. I am sure, there is no grid in space and no center of the universe relative to which all else moves. In that light we have to define, what a relative motion really is.
But first:
Due to our 2D perception of the world we identify very easily movement in that plane, but poorly in depth. We measure the angle between points seen by the eye and knowing how far they are, can estimate the approximate distance between these points and vice versa – knowing the size of the object, can estimate its distance from us. Our brains build a 3D model of the world with experience from interacting with the world. When we try to estimate the size of an unfamiliar object at a distance, we can compare familiar objects (which size we know) close to the object and give a pretty accurate measure.
Due to the way our vision works, it is very easy to detect movement of an object orbiting us or in a tangential motion relative to us, when there is a reference frame (backround). It is much harder to tell the motion of an object (especially when it is far away) moving towards or away from us, due to little change in the field of vision. This is a reason why most people, when thinking about motion, visualize an object moving in a frame of reference, tangentially relative to them. And that is the fatal flaw in thinking that Einstein and all the relativists after him made.
Back to relative motion.
Wikipedia: Motion is observed by attaching a frame of reference to a body and measuring its change in position relative to that frame.
In absolute space, there are three basic types of motion:
Radial motion – a body is moving towards or away from the other in a straight line.
Circular motion – a body is circulating another body, moving in a curved tradjectory.
Tangential motion – one body passing the other, moving in a straight line in the frame of reference.
In relative space the only frame of reference is a second body, relative to which the motion happens. I would define relative motion as change of distance between two objects.
Let´s visualize an empty space, with only two bodies in it. In this situation there can be only one type of motion – two bodies moving towards or away from each other (radial motion). There is no change in distance when one body is circulating the other. Without a grid of reference there is no way to verify, whether the first body is circulating the other or the latter is simply rotating. Anyway, there is no change in distance between them, hence no motion.
Tangential motion in absolute space is a composite of the two – an object, in a uniform motion, approaching stationary object from a distance is initially almost in a radial motion relative to the other, with a very little circulating element. When the object gets closer, the radial element (relative speed) decreases and the circulating element increases, the change in distance decelerates. When the object closing, reaches the closest point to the other, its radial (relative) motion drops to 0. Next the opposite happens – the object rapidly accelerates away from the second, with increasing radial movement and decreasing circulating element, until reaching almost uniform radial motion. This is how we perceive tangential motion in a frame of reference (absolute space). In a relative space, the motion would look like this – two bodies approach each other at almost constant speed. When they get close, their approaching speed starts to drop and at the same time they rotate around their axis in opposite directions. At the point closest between them, the bodies have rotated 90 degrees and for a very brief moment there relative motion has dropped to 0. Next, the bodies accelerate away from each other until they reach approaching speed and at the same time perform another 90 degree rotation.
Now analyzing Einstein´s thought experiment, where one person is standing on the platform and three on a train passing by:
The train in this experiment is performing classical tangential motion. When the man standing in the middle of the coach lights his lighter at the moment he is closest to the man on the platform, there is no relative motion between these men. Hence, the man on the platform will observe light reaching men at the opposite ends of the coach exactly at the same time and this violates no logic, since the men are not in relative motion at that moment. No need for time dilation, length contraction and other nonsense.
In case the man in the coach lights the lighter, when the train is approaching the man on the platform, the latter would observe light reaching first the man at the front of the car and then the one at the rear. Also, the light would have a blueshift. No logic violated here and again, no need for time dilation,…
If the man on the coach lights his lighter, when the train has passed the one on the platform, the latter would observe light reaching man at the rear first and then the one at the front. The light would have a redshift. Once again, no need for time dilation, etc.
What Einstein did – he took motion in an absolute space and used it in the frame of relativity. In fact all these absurd properties, that have been assigned to space, time and mass with this weird theory, goes against true relativity. Therefore all experiments that have allegedly proven the theory are either misinterpretations of a different phenomenon, malfunctions of equipment or outright forgeries.
For instance – The twin flight experiment with planes flying in opposite directions is absurd from the beginning. The experiment is again based on motion in absolute frame, not relative. Relatively, there is no difference, whether the earth is rotating, or the entire universe is circulating around it (that is the preconception that STR is based on). Relatively, the earth (together with the rest of the world) could travel at the speed of light in any direction if we look at things from the perspective of a photons emitted from the earth. So expecting to find any, let alone different time dilation on these planes (also applies to the GPS satelites) is directly against relativity. Any time difference measured is either fabrication or some other phenomena interacting with the equipment (aether wind?).
Finally,
The maximum relative speed in space is at least double the speed of light. When a body in space emits simultaneously two photons (for arguments sake, let´s assume there exist this mysterious particle), which travel in opposite directions, relative to either photon, the other travels double the speed of light. Eat this, Einstein!
Best regards,
True Relativist
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• drgsrinivas On March 30, 2015 at 10:12 pm
True Relativist,
Glad to see you not succumbed to the weird teachings of relativity, but I have to tell you that I disagree with your description of relative motion and your argument overall. And I request to have a critical rethink on that.
When we describe motion from the perspective of an observer, we imagine that there exists a grid, fixed to the observer, spanning the entire universe. And displacement of an object within that grid constitutes motion according to the observer, whether it is radial, tangential or circular. And that motion can be objectively quantified from the perspective of the observer including total displacement, rate of displacement and direction.
You seem to be defining motion as change in the distance between the observer and the body. But truly, any change in the position of an object with respect to the grid/observer constitutes motion from the perspective of the observer. So, circular motion does constitute motion.
Coming to your train explanation, what if the relativists argue that their observer is not actually on the platform but is in the same straight line as the people on the train, so that all motion becomes ‘radial’? Does it make relativists’ distortion of space/time true?
But I agree with what you said in the last two paragraphs.
Luckily, for the truth seekers, it isn’t difficult at all to understand why relativity and its teachings are utterly wrong.
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• woodside On March 27, 2015 at 7:56 pm
True Relavatist
In all three cases of the train and the man on the station the man would see the light arriving at man at the rear of the carriage first, because that man has moved closer to the position that the lighter was when it was lit and so the light from the lighter has travelled less distance when it reaches the man at the rear of the carriage than the distance it travels to rach the man at the front of the carriage, who has moved away from the position that the lighter was when it was lit.
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• truerelativist On April 1, 2015 at 12:19 pm
Dr. Srinivasa,
That’s exactly it – we IMAGINE the grid, which does not mean it really exists. This is what my argument and true relativity is all about – there cannot be an objective/absolute grid in space. We are so accustomed to a reference grid backround /foreground objects, that it is very hard to imagine an empty space without it and this is why we automatically attribute a grid to a reference point when there in fact is none. It is simply one perspective among unaccountable others and everyone of them is just as valid. We could look at things from the perspective of a photon, and say that the whole universe is in extreme motion and there are other hpotons travelling twice as fast. Or linking the grid to an object rotating, we could attribute enormous speeds to an object very far away from the first. Imagine what would be the speed of a galaxy 1000 light years away, if we tie the grid to a spinning rotor of an electric motor, turning at 100 000 rpm. My point is, that there is no RELATIVE MOVEMENT between the rotor and the distant galaxy, regardless the rotation of the rotor.
Imagine we have a solar system in an empty space, where nothing else exists. This solar system would have planets orbiting the star at fixed positions relative to each other. In this case, it is impossible to tell, whether the planets are orbiting the star or the star is simply rotating. There is no grid – other reference points beyond the planets, compared to which to determine the motion. It is only when we fix the grid to the star at the center, we may say, the planets are orbiting. This would not make it absolute. Just as well we could take the perspective of a planet and say that the star is rotating. It would be just as valid.
To make things more interesting, let´s imagine, the planet is also rotating.Fixing the grid to this planet, we could rightfully say that the whole solar system is orbiting the planet. In this case – which is right – the perspective of the sun or the perspective of the planet? Anyway, there is no change in distance between the star and the planets, hence no relative motion, regardless the rotation.
Before Copernicus/Galilei, the scientific paradigm was that the sun, the moon and all the stars are orbiting the earth. And they were right, from their perspective. Only after the accustomization to the new thinking and indoctrination from an early childhood, we think that the sun is in stationary and the earth is circling it. This is also the reason, why they expected to find relativistic results from the twin flight experiment – they looked at earths rotation from the suns perspective and attributed absoluteness/objectiveness to it (forgetting the earth orbiting the sun). If the experiment would have been conducted fixing the grid to the moon, I´m sure the results would have been different 😉
If in the train example, the poor suicidal observer would stand on the tracks in front of the train, he would definitely perceive light reaching the man at the front first and then the man at the rear of the coach, because the light reaching the man at the rear has to travel extra distance to reach us. Light would reach the two men at the ends simultaneously. But since the man at the front is closer to the observer, the later would see light reaching him sooner because it has the length of the carriage to travel extra from the man at the rear. Also, since the train is in motion towards the observer, the light has to have blueshift (Doppler effect).
In case the observer is on the tracks behind the train and the latter is moving away, he would see light reaching the man at the rear first (same principle). And since the light source is moving away, the observer would perceive redshift.
In both cases there is no violation to logic and hence no need to attribute strange behavior to time and length.
The observers perception of light reaching either man first is just a PERCEPTION not a true representation of the reality. Just as if there where two explosions far away from each other, at exactly the same moment, and we would be at equal (safe) distant from the two. In this case we would hear the bang from the two at the same time – just one bang. But if we would be positioned so that the explosions were one behind the other, we would hear two bangs in sequence, in spite of visual perception of a simultaneous event.
Let´s make things interesting and put an explosion (smaller) on a moving train. There are two sound detectors at each ends of a carriage and the exploding device in the center at equal distance from the detectors. The detectors give off a light pulse the moment they register the explosion. The carriage is closed and there is no air movement in it.
Just as the carriage is passing and is closest to us (standing on the platform), the explosion device goes off. In this case, would we see the light turning on at the rear first, and then at the front?
I wonder what would Einstein say. Time dialation? ;D
Best regards,
True Relativist
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• drgsrinivas On April 3, 2015 at 3:51 pm
The fact that your platform observer and the train are not real but are just imaginary doesn’t mean that we can’t draw valid conclusions from that imaginary scenario.
Whether there really is a grid or not doesn’t change the things. It’s just a way of explaining relativity of motion. In stead of grid, one could use the coordinates to specify the position a body in space with respect to the observer. A change in the position of the body constitutes motion in the reference frame of the observer.
Relativity of motion doesn’t mean what the human observer just sees or feels. For example a blind observer may not see a ball moving in front of him but it doesn’t mean that there doesn’t exist a ball near by him or there is no motion of ball.
An observer may not see a distant star moving. But that doesn’t mean that the star is stationary relative to him. And an observer may only see light signals coming from the front but not those coming from behind. Another observer who has got eyes both in front and back of his head will see both the signals. A dead observer or a wooden block may not appreciate any of them. These differences occur because of the different capabilities of perception of different observers. You may call that as ‘subjective relativity’ if you want.
But what we discuss in physics is ‘objective relativity’. And you seem to be confusing between the two. Of course even relativists do the same and mess up things when they ‘explain’ twins paradox and relativity of simultaneity etc. The fact that there exists some time delay for the signal to reach the observer, doesn’t mean that the actual event occurred late even in the perspective of the observer. An intelligent observer would know that the event had actually occurred earlier than he saw it. And we must design the experiments in such a way that we eliminate the bias caused by the signal delays and also the perception defects of the blind/ deaf observers.
Of course what you said at other places is correct. For example you are talking about objective relativity here “It is simply one perspective among unaccountable others and everyone of them is just as valid. We could look at things from the perspective of a photon, and say that the whole universe is in extreme motion and there are other hpotons travelling twice as fast. Or linking the grid to an object rotating, we could attribute enormous speeds to an object very far away from the first. Imagine what would be the speed of a galaxy 1000 light years away, if we tie the grid to a spinning rotor of an electric motor, turning at 100 000 rpm”
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• woodside On April 1, 2015 at 5:32 pm
the air that the sound is moving in is moving with the train and so the sound would arrive at the front and back simultaneously. But the light would not arrive simutaneously. Its the same when you watch a distant thunderstorm the wind effects the sound not the light.
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• truerelativist On April 3, 2015 at 3:10 pm
Woodside,
That’s exactly it! The medium, in which the sound is traveling is motionless RELATIVE to the men in the train, regardless the motion of the train. So why should the “medium” in which the light is traveling be in motion? The only case, the light would reach the man at the rear sooner, would be if it travels in ether and the train is in motion relative to it. In this case all observers would register it exactly the same and there would be no reason to implement time dilation etc. But this would contradict the fundamental precondition of Einstein´s “relativity” and the reason it was conceived – to eradicate ether from physics. If the relativists say – “light would have to reach man at the rear first”, means it travels at different speeds in two directions relative to the emitter. This is also a gross violation of “relativity”. There is a fundamental contradiction in this experiment – if light travels at the speed c from the (imaginary) position in space it was emitted from, it cannot travel at the same speed relative to the emitter, which is (allegedly) in motion. There is just no two ways of looking at this.
To make things clear and easier to imagine, let´s put the men in spaceships in an empty space. In this case, there is no backround to confuse us. The man in one spaceship lights a LED bulb (no open flames in a spaceship!) at the moment the other spaceship with the observer is passing by and is closest. As relativity says – there is no objective way to say, which ship is in motion, since there is no grid in space or either, relative to which they move. We can say, that both ships are in motion, the one carrying the observer or the one with the three men. All cases are correct! Only the relative motion BETWEEN these ships is of significance and there is NONE at the moment the light is turned on – the only RELATIVE motion is radial motion (change of distance between objects under observation). So, which man at the ends of the spaceship the light reaches sooner?
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• Alex On December 16, 2015 at 4:49 pm
I haven’t finished reading all of your posts but still, I need a comment on Lorentz’s factor (γ). When one points out to a physicist that the astronaut moving away from Earth can’t be treated differently from an Earth moving away from the astronaut, he ‘ll always answer that only the astronaut is accelerating and that’s why relativistic time and length appear on him. This implies that acceleration is paramount to dilation and γ factor. Yet, Lorentz’s factor has no acceleration factor/parameter. I has (v) for velocity and not (a), which implies that acceleration can’t be the answer. And one can’t sidestep the problem by substituting (v) with (a). Looking at the formula again, if you substitute, say, (v)=10m/sec with (a)=10m/sec^2, you end up with a result that doesn’t say [T=xT’] but something like [T=xT’sec], which is meaningless. I don’t know if I made myself clear, since English are not my native language but I don’t see any way around this. If the Lorenz Factor is meaningfull, then one can’t explain dilation according to acceleration. If, on the other hand, acceleration is important (when it comes to time dilation), then the Lorenz Factor is wrong or at least incomplete. And if the Lorenz Factor is wrong, so is most of Einstein’s work.
If my scepticism is grounded, the next question is “how on Earth don’t people see something that obvious when they give you the acceleration answer”?
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• Galacar On December 17, 2015 at 2:15 am
Alex wrote:
“If my scepticism is grounded, the next question is “how on Earth don’t people see something that obvious when they give you the acceleration answer”?”
I really don’t ‘get’ your whole post, but to your question why people don’t see the obvious the answer is ‘education’ better called ‘indoctrination’
The thing is like a magician playing his tricks with smoke and mirrors.
When you know how it is done, the trick, you are no longer distracted
by the smoke and mirrors and can , very very simply. see throgh the whole thing.
A real life example.
Did you know banks create money out of nothing, out of thin air?
Then they lend this money to you. Then you later have to pay back.
But the bank actually never gave you real money. No coins were moved
no gold was used., because there is no connection to gold and money anymore.
So, it is all an illusion! albeit a persisten one, to quote einstein! lol
It nearly always goes something like this: “well, it can’t be true.There are controls in banks, there are controls in government. Government would never allow thar.
besides the whole system is wayyyyyy too complex to let it work this way.
You have derivatives, put options,…”
and so on and so forth.
If you do the research it really works they way I wrote.
The rest is nearly all, smoke and mirrors!
So, to come back to your question why people don’t see the obvious.
It is their ‘education’ telling them that the smoke and mirrors are real!
My two cents.
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• Nona Maus On December 17, 2015 at 10:29 am
@Alex very well put. I have similar questions. Care to help us out, @drgsrivanis ?
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• drgsrinivas On December 17, 2015 at 11:27 pm
Alex’s criticism is absolutely right. The problem is obviously with relativity and not with his rationality or lack of understanding of the weird theory. Relativists brought the issue of acceleration into twin’s paradox just to confuse people and to save their stupid theory. If they didn’t, they knew that their religious theory would crumble and vanish from this world.
They do a similar kind of mess up when they talk about circular motion. At one time, relativists argue circular motion as accelerated motion and use that to illustrate the gravity-acceleration equivalence, and at other times, they take circular motion as uniform motion and use special relativity while calculating the time dilation for particles moving in circles in ring accelerators. https://debunkingrelativity.com/muons-time-dilation/
And there exist so many absurdities and self contradictory explanations in relativity religion. I have talked about some of them here- https://debunkingrelativity.com/photon-clock-and-the-maya-of-time-dilation/
Basically, whenever relativists feel that their theory is in trouble, they introduce some weird argument or propose some absurd phenomenon by default and the story goes on.
Actually it doesn’t require great IQ to realize the absurdities and self contradictions in relativity. And most people do realize them while studying relativity. But people have so much faith in scientists that they blame upon their own ignorance and limited IQ for the perceived absurdness. People think that they are not intelligent enough to question scientists however absurd the latter sound. “If scientists are wrong, we wouldn’t be having all these gadgets” they argue. So they always blame upon their ignorance whenever they feel that some scientific theory is absurd. It is upon that blind faith, the physicists of our time survive.
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• K Sean Proudler On December 27, 2015 at 3:51 am
Something will only seem weird if the absolute truth is not being exposed. Special Relativity(SR) says that being at absolute spatial rest can not be detected, that absolute motion can not be detected, that an absolute length can not be measured, etc. If this is accepted, then it becomes impossible to understand SR absolutely, since SR excludes many absolutes in the first place.
Meanwhile, if you start with the acceptance of the existence of an absolute 4 dimensional environment, an environment known as Space-Time, and you analyze the concept of absolute motion ongoing within that environment, you end up independently discovering with SR and independently deriving all of its equations. Due to having started such an analysis with just these bare bone simple basics, no prior knowledge regarding physics is required to achieve such an outcome. Thus, SR does exist within an absolute foundation, the foundation of which physicists ignore.
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• Galacar On December 28, 2015 at 12:38 pm
to K Sean Proudler
Ok, I am still watching the video’s. So haven’t finished them, yet.
Anyway, First I congratulate you that
“Due to my unique but unappreciated way of thinking, my parents had pulled me out of school, and had done so before I had a chance to acquire any education in the field of physics.”
http://blogs.scienceforums.net/IME/2014/04/27/einstein-relatively-simple/
Very good! You are blessed in this regard.Maybe more than you know.
However, It looks like you are still ‘defending’ relativity and I feel it is wrong on all sides.I feel it is here to disguise any real physics. And , according to me, real physics needs the aether in place. ALL, and I mean ALL inventions are done with
the aether in place. Tesla rejected Einsteins relativity. Guess which one has the most patents?
And you wrote:
“Something will only seem weird if the absolute truth is not being exposed”
Yes, this is true.But then the problem becomes what the ‘absolute truth’ is!
That might be that the whole of science and therefore ‘education’ is wrong, or that somethings in education are discarded so we don’t go into sensitive areas
like anti-gravitation.That math can be wrong, and so on and so forth,.
See what I mean? because you don’t mention what ‘absolute truth’ is,
it is rather vague too me.
That is not to say that I don’t value your independent thinking, but if I watch the videos it feels like you don’t want to reject einsteins relativity totally, which is what I do.Not that I am right by definition of course. I might be wrong as well.
But so far I haven’t seen anything to defend the totally wrong theory of special and, for that matter, general relativty.
Might it be true, that special relativity is defended by you, because somewhere in the back of your mind it must be true because it is taught at schools and universities?
I am kind of a radical, I admit. When something is pushed hard at schools and universities I take it as not true at all!. I immeditaley assume there is someone with an agenda to push it, As I have written here before, if something is true, it will never be taught at universities and schools.
Truth might have some people at the top of this world lose their power.
My two cents
Namaste
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• Galacar On December 29, 2015 at 6:03 pm
As my research progress, it comes more and more clear to me that the whole of ‘education” is here to put us in a box, so to speak.
I mean by this that our consciousness is closed off and mentally we are living in an extremely small box.(left brain anyone?)
But once you become aware of this.You can start looking outside this box.
(actually, while you are in the box, you are enslaved. But you don’t see the bars!).
Looking outside this box is what I am doing now for years and that way you can find out how extremely powerfulll our consciousness is!
There are still things I have a hard time to believe, but know them to be true.
If what I have seen, and of course others with me. then I am very positive about our future. IF we get our head of the box!!
Anyway, the more I see what our consciounsess can do, the more I understand that schooling, culture conditioning and what have you, is to here to clamp down
our consciousness.(put it in a box).
To keep all this in the context of this site, once you are aware what consciousness can do , it is more easy to understand we don’t need schooling and theories like relativity and quantum bogus etc.
The ONLY thing these absurd theories will l do is ‘teaching us that there are limits.
However, in the consciousness there are NO LIMITS!
Literally.
I know it sounds preposterous to some and I understand. My left brain still has some difficulty in believing this kind of stuff. But I do accept more of it then when I started on this journey.
I can put things here that will turn a lot of people immediately off.
It will be too far out for them.I went there gradually. So I understand.
The following might look off topic. However I am trying to illustrate my point with this.
So, something simple.
Put a glass of water (simple tap water) in your room and send your intent that it will do whatever you want.)
Some examples of intent
Heal an ailment
get some money
a job
flush out poison and that kind of a thing.
etc
Now a lot of people will make this very complex , but don’t.
Then just very simply just drink this water, and see what happens.
I have seen this work by a lot of people.
Why not try it? It is free and see it as an interesting experiment.
One of the things here is that you need to get rid or don’t take your limits seriously,
If you this and it works you have prove to yourself that you don’t need any theory and that theories limit your use of your consciousness.
Just sayin 😉
Like
• Galacar On January 1, 2016 at 4:32 am
HAPPY NEW YEAR TO YOU ALL!
Like
• geocentric101 On January 10, 2016 at 4:22 am
I have recently come across the work of Walter Russel and his theories of light and how everything stems from electricity etc.
I wonder if anyone on this site/thread knows of his work and theories?
I think he may well have more of a grasp of reality than our current understanding of life, the Universe and everything as espoused by today’s scientist.
Like
• drgsrinivas On January 10, 2016 at 7:29 pm
I have just googled Walter Russell.
Found his teachings truly enlightening:
-The Universe we believe to be solid and real is an illusion
-Our senses trick us into believing motion is matter
-The Universe is a projection of Mind, and is generated in the same way you imagine a thought in your own mind
(http://walter-russell.com/documents/novice.html)
And what surprised even more is that, these are exactly the same conclusions that I have also arrived at https://debunkingrelativity.com/2014/03/05/double-slit-experiment-electrons/
But I must confess that I didn’t really get the explanation provided for the double slit experiment on that site. I think Photon ether model is the only rational explanation for DSE. And photon Ether provides the framework to explain everything in this universe in terms of waves.
Thank you geocentric101 for pointing us to such an enlightening personality.
Like
• Galacar On January 10, 2016 at 12:00 pm
@geocentric101
Yes I have and I love his work! And oh yes, he has for sure more grasp of realty then today’s ‘scientists’
(Well, everybody has per definition a better grasp than today ‘scientists’ by definition. lol)
And if you are interested in his work you probably are also interested in the works of Bruce Cathie and Buckminster Fuller.
If you understand these works you will laugh and scoff and want to mock that cheap religion called ‘science’ today!
You will see (‘science’ that is) it for what it is. A bag full of bollocks and shite.
Just my two regular cents 😉
Like
• Galacar On January 10, 2016 at 8:42 pm
drgsrinivas,
“-The Universe we believe to be solid and real is an illusion”
Yes! Enter Bill Hicks, a comedian:
““Today a young man on acid realized that all matter is merely energy condensed to a slow vibration, that we are all one consciousness experiencing itself subjectively, there is no such thing as death, life is only a dream, and we are the imagination of ourselves. Heres Tom with the Weather.”
― Bill Hicks”
““The world is like a ride in an amusement park, and when you choose to go on it you think it’s real because that’s how powerful our minds are. The ride goes up and down, around and around, it has thrills and chills, and it’s very brightly colored, and it’s very loud, and it’s fun for a while. Many people have been on the ride a long time, and they begin to wonder, “Hey, is this real, or is this just a ride?” And other people have remembered, and they come back to us and say, “Hey, don’t worry; don’t be afraid, ever, because this is just a ride.” And we … kill those people. “Shut him up! I’ve got a lot invested in this ride, shut him up! Look at my furrows of worry, look at my big bank account, and my family. This has to be real.” It’s just a ride. But we always kill the good guys who try and tell us that, you ever notice that? And let the demons run amok … But it doesn’t matter, because it’s just a ride. And we can change it any time we want. It’s only a choice. No effort, no work, no job, no savings of money. Just a simple choice, right now, between fear and love. The eyes of fear want you to put bigger locks on your doors, buy guns, close yourself off. The eyes of love instead see all of us as one. Here’s what we can do to change the world, right now, to a better ride. Take all that money we spend on weapons and defenses each year and instead spend it feeding and clothing and educating the poor of the world, which it would pay for many times over, not one human being excluded, and we could explore space, together, both inner and outer, forever, in peace.”
― Bill Hicks”
Don’t you love these quotes!
And now, you know one of the main reasons, why drugs are FORBOTTEN!!
They let some people see through the illusions in which we are caughtt!
lol, there is more wisdom in a comedian, then all scientists combined!
Namaste!
Galacar
Like
• Galacar On February 10, 2016 at 1:00 pm
And, of course, once again, things are not coming from “Modern Science” as
I state all the time.It is impossible with “modern science’ to create new technology.
Here is another example, the TV and even the “SMART”-phone.
(The reason I put ‘smart’ apart is a whole theory by itself, I won’t go into
that here and now.)
“Years Before The First TV, Tesla Predicted And Helped To Develop The Smartphone And FaceTime
Five years prior to the world seeing the first practical demonstration of the television, and scores before the first instance of the smartphone, Nikola Tesla, not only imagined it but had completed crucial steps for making it possible.
Television was only a small portion of Tesla’s predictions, however. In Volume 100 of Popular Science Monthly, January-June, 1922, Tesla would predict FaceTime.
A real good “educated” scientist would say:
“Yeah, the smartphone is here because of ‘modern physics”
IT IS BOLLOCKS!
When do people en masse wake up to the all pervading lies surrounding them?!
Well, ithings are changing.
Namaste!
Galacar!
Like
• Galacar On February 11, 2016 at 12:19 pm
Now for some real controversial info: 😉
In reality we are all infinite consciousness having an experience here.
So we are not Ethel, Galacar, Jim or what have you.
(We call this “The Phantom self”)
Now , nearly everything in mainstream is here because it is trying to convince us
we are ‘little people’ (NOT infinite consciousness)
Hence all the lies everywhere.
If we buy into this, and from the moment we are born this is the message., just
look around you, we can easily be controlled!
Once we understand we are all infinite consciousness, it will be impossible
to control ‘us’.
Do people here now understand whty I write all the time that ‘science’ is one of
the tools to control us? It is a “surpressing/control ” tool.So it MUST be made of
lies, bollocks and shite! Of course all this by design.
There is no other way!
Really hope this might open some doors for some people.
I am pretty sure it will resonate with some here.
Others will rather die clinging to their belief systems. so be it.
Nuff said.
Galacar
Like
• Robert Borer On January 7, 2017 at 7:34 pm
I found your site by googling “time dilation fallacy.” Love what I’m seeing.
Like
• Rajesh Swarnkar On May 20, 2017 at 7:50 pm
Haven’t Quantum Mechanics became just an obscure pop science topic than a real scientific challenge?
I dislike the fuzziness going around the modern day QM or QFT etc etc. I feel the obscure and far-fetched mathematical complexity is keeping the young minds from re-discovering it. I feel Science is facing an elitism and I hate it.
I frigging want some “REAL” science with “meat” in it. I feel like PHYSICS NEEDS A REBOOT.
Liked by 1 person | 15,073 | 67,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-30 | longest | en | 0.938571 |
http://mathonline.wikidot.com/continuity-of-linear-functions-on-lctvs | 1,716,905,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00752.warc.gz | 20,668,604 | 5,762 | Continuity of Linear Functions on LCTVS
# Continuity of Linear Functions on LCTVS
Proposition 1: Let $X$ be a locally convex topological vector space. If $f : X \to \mathbb{R}$ is linear then $f$ is continuous on $X$ if and only if there is an open neighbourhood $N_0$ of the origin such that $f$ is bounded on $N_0$.
If $X$ is a topological space and $f : X \to \mathbb{R}$ then $f$ is continuous at a point $x_0$ if for all $\epsilon > 0$ there exists an open neighbourhood $N_{x_0}$ of $x_0$ such that if $x \in N_{x_0}$ then $|f(x) - f(x_0)| < \epsilon$. We use this in the proof below.
• Proof: $\Rightarrow$ Suppose that $f$ is continuous on $X$. Then in particular, $f$ is continuous at the origin $0$. So if $\epsilon = 1 > 0$ there exists a neighbourhood $N_0$ of $0$ such that if $x \in N_0$ then $|f(x) - f(0)| < \epsilon = 1$. But observe that $f(0) = 0$ since $f$ is linear. So $|f(x)| < 1$ for all $x \in N_0$, that is, $f$ is bounded on $N_0$.
• $\Leftarrow$ Suppose that there exists an open neighbourhood $N_0$ of the origin such that $|f(x)| \leq M$ for all $x \in N_0$. Since $N_0$ is open, so is $tN_0$ for all $t > 0$. Furthermore, since $f$ is linear, then $|f(x)| \leq tM$ for all $x \in tN_0$.
• Let $x_0 \in X$ and let $\epsilon > 0$ be given. Choose $t > 0$ such that $tM < \epsilon$. Since $X$ is a locally convex topological vector space, $x_0 + tN_0$ is an open neighbourhood of $x_0$. Note that if $x \in x_0 + tN_0$ then $x - x_0 \in tN_0$ and so:
(1)
\begin{align} \quad |f(x) - f(x_0) = |f(x - x_0)| < tM < \epsilon \end{align}
• Therefore $f$ is continuous at $x_0$, and since $x_0 \in X$ was arbitrary, $f$ is continuous on $X$. $\blacksquare$
Proposition 2: Let $X$ be a locally convex topological vector space. If $f : X \to \mathbb{R}$ is linear then $f$ is continuous on $X$ if and only if $f$ is continuous at the origin.
Recall that if $X$ and $Y$ are topological spaces and $f : X \to Y$ then $f$ is continuous at $x \in X$ if for every open neighbourhood $V$ of $f(x)$ there exists an open neighbourhood $U$ of $x$ such that $f(U) \subseteq V$.
• Proof: $\Rightarrow$ Suppose that $f$ is continuous on $X$. Then trivially since $0 \in X$ we have that $f$ is continuous at $0$.
• $\Leftarrow$ Suppose that $f$ is continuous at the origin $0$. Then for every open neighbourhood $V_{f(0)}$ of $f(0)$ there exists an open neighbourhood $U_0$ of $0$ such that $f(U_0) \subseteq V_{f(0)}$.
• Let $x \in X$ and let $V_{f(x)}$ be an open neighbourhood of $f(x)$. Consider the set $V_{f(x)} - f(x)$. Note that since $\mathbb{R}$ itself a topological vector space we have that $V_{f(x)} - f(x)$ is an open set. Furthermore, since $f(x) \in V_{f(x)}$ we see that $0 \in V_{f(x)} - f(x)$ and so $V_{f(x)} - f(x)$ is an open neighbourhood of $0 = f(0)$.
• Since $f$ is continuous at the origin $0$ there exists an open neighbourhood $U_0$] of [[$0$ such that:
(2)
\begin{align} \quad f(U_0) \subseteq V_{f(x)} - f(x) \end{align}
• Consider the set $U_0 + x \subseteq X$. Since $U_0$ is open and $X$ is a topological vector space, $U_0 + x$ is open. Furthermore, since $f$ is linear:
(3)
\begin{align} \quad f(U_0 + x) = f(U_0) + f(x) \subseteq V_{f(x)} \end{align}
• Therefore $f$ is continuous at $x$. Since $x \in X$ is arbitrary, $f$ is continuous on all of $X$.
Proposition 3: Let $X$ be a locally convex topological vector space. If $f : X \to \mathbb{R}$ is linear then $f$ is continuous on $X$ if and only if there exists exists a convex open neighbourhood $O$ of the origin $0$ such that $f(O) \neq \mathbb{R}$.
• Proof: $\Rightarrow$ Suppose that $f$ is continuous on $X$. Let $V \neq \mathbb{R}$ be an open neighbourhood of $0 = f(0) \in \mathbb{R}$. Then by definition there exists an open neighbourhood $O$ of the origin $0 \in X$ such that $f(O) \subseteq V \neq \mathbb{R}$.
• $\Leftarrow$ Suppose that there exists an open neighbourhood $O$ of the origin $0$ such that $f(O) \neq \mathbb{R}$. We may assume that $O$ is symmetric, i.e., if $x \in O$ then $-x \in O$ (if not, take a convex subset of $O$ that is symmetric about $0$). Let $\alpha \in \mathbb{R} \setminus f(O)$. Note that $\alpha \neq 0$ since $0 = f(0) \in f(O)$.
• Suppose instead that $f(O)$ is unbounded. Since $f(O)$ is symmetric convex and contains $0$ and unbounded there exists an $o_n \in O$ such that $|f(o_n)| > |\alpha| > 0$. But then $f(O)$ cannot be convex since $f(O)$ does not contain the line segment joining $0$ and $f(o_n)$. Thus $f(O)$ must be bounded. By proposition 2 we have that $f$ is continuous. $\blacksquare$ | 1,577 | 4,554 | {"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": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-22 | latest | en | 0.834422 |
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P2_1030
# P2_1030 - Math 415.01 Dr Huseyin Coskun Exam 2 Practice...
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Unformatted text preview: Math 415.01 - Dr. Huseyin Coskun - Exam 2 Practice Problems - 10230 Class (Vutha) February 14, 2011 The following list of problems is meant to serve as an extra resource for your preparations for Exam 1. This list is in by no means exhaustive, nor is it meant to predict the questions in the exam. Instead it can be used as a tool to assess how well prepared you may be, and to highlight problems that you might face. Pay careful attention to your timing and clarity of ideas — as this is something that can greatly afiect the outcome of your exam. (3.1.22) Solve the initial value problem 41/” - y = 0, 31(0) = 1, y'(0) = -5 Find B for which all solutions approach zero as t -—> 00 (3.2.8) Find the largest interval in which the given initial value problem will certainly have a unique solution. Do NOT find the solution. (t - 1)?!” ~ 3ty’ + 4y = sin“), y(-2) = 2, y’(-2) = 1 (3.3.11) If the functions yl and y2 are linearly independent solutions to y” + p(t)y’ + q(t)y = 0, prove that clyl and c2y2 are also linearly independent solutions provided c1 and c2 are not zero. (3.4.20) Find the general solution and sketch a graph of the solution and describe its behavior for increasing t. 7r 77 3/"+y=°’ 9(3) =2, y'(§)=-4 1 (3.5.18) Consider the initial value problem 9y” + 12y’+4y = 0, y(0) = a > 0, y’(0) = —1 (a) Solve the initial value problem. (b) Find the critical value of a that separates solutions that become negative than those that are always positive. (3.6.15) Find the solution of the given initial value problem 9" - 231’ + y = te‘ +4. 11(0) = 1. y’(0) = 1 (3.7.10) Find the general solution to the given differential equation 61 1+t2 y”-2y’+y= (3.8.7) ( Refer Text ) (3.9.9) If an undamped spring-mass system with a mass that weights 6 lb and a spring constant 1 lb/z'n is suddenly set in motion at t = 0 by an external force of 4 cos(7t) lb, determine the position of the mass at any time and draw a graph of the displacement versus t. (10.1.15) Find the eigenvalues and eigenvectors of the given boundary value problem. 14” + M) = 0, y’(0) = 0, 9(a) = 0 ( Note: There are no problems from sections 10.2 and 10.3 ) ® Scuba“; m yCe):<:i?)€/Vg+(w)ézyb 2- L m Mano; 44 Aetmmga 17 e“. :6, he. =) 2132:) =>- 2 PCE : it ) 4;" Oqu—onlmm <11: (2 l L4 CE):- 53E 9 L-‘l @ Q) and slums: CH. 00‘ 0‘72 0“,:ng NH 4 C403) LHS (9“) + QM) P“) + CQMQUC Q C!” 4 decoy.) : O esul\ak:a Sch—Judd WVHW C‘ONHLOJ wahL‘JL A; ‘erfgz m4 W Pew/Dd 24" MA [MK 4erqu 00%.?) Q3) @ no“ Efflkéo: qwz'i'D—Y4—410 “(1 ——I2 : J \44 —4(9)(1+) 1C9) _ “i?“ 43-2% © E‘ASOVKW‘ ~10); QézéLJr geejlfl CMaOfiCp/w age)- néQelli’fing-gt )efi': Sag State, 0.70, AQLLHOJc 523me flow/- 01: a‘ljxhfi’afiy -I 40/ SOlu‘l—O-«fil J’LQL'L W38 QbLSY‘WOWC‘S O—Q-zét + (£3): glib: O 3 3:) Q4 2g :0 3 =J e = \$2: Zoré ‘ 1,; Lake, ObfyofiLv-e vWGAWmROM LC 8 Quake, 9C6); aefzéewkid. 4;, “(”07“ FOJiLCL- Pm‘mm‘ow SdHJ-Lory @xsjr Gvé55> f CAB/Qet +C WM Mw; @mkd,‘° W’Ff)(7 1:7! E;- VPCE) a. CABS-g— BLL) e‘: + C A’Uo W VP' Ce) .- e} [BM—fl 2%) + (M3+BL}] g ___ (\$ka + (3M3) (5’- + (2.5) 4;] VFHCQ): Q}: [<3Atz+ C3A+6)G)L +243 ) +[ 1453+ (3 4+3) *Qz)‘: E? 1 O N :_ 1‘- J‘tA ‘ VCE)‘ C4 (LO-5 +AL 495‘“ (\fié‘ E) v m GA) [gamma 954M L1 QC): ‘hFA[ [—OC‘San/AL )4 QwJCfiAf—J] ”ICN) )—O 9) {TIA W) WA”) I 7(0):.(3 3) (ls/O VCWFO 2) C. £04 (£3 rt) 20 =) @A 11.)” Q”) ”A .1) JiA: é:“)IA:“" @421 Lb ‘—-— ...
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(Resting Energy Requirements or RER), which can be calculated by multiplying the animal’s body weight in kilograms raised to the ¾ power by 70, for example, a 10kg (22lb) adult neutered dog of healthy weight needs RER = 70(10kg)3/4 ≈ 400 Calories/day.
How many calories should I feed my dog chart?
Dog and Cat Daily Caloric Needs
CALORIE CHART
Typical Total Daily Calories to achieve Weight Loss in Adult Spayed or Neutered Dogs Lightly Active Indoor Spayed or Neutered Dog Maintenance Diet (kcals per day)
10 206 248
11 220 264
12 234 280
How many calories does a 60 lb dog need? A 60 pound dog will need to eat at least 1337 calories per day if they are not very active. If they are moderately active then they will need 2505 calories per day.
How many calories should a 80 pound dog eat per day? Here’s the scoop: In general, most dogs need to eat between 25 and 30 calories per pound of body weight to keep from gaining or losing weight.
How many calories are in 1 cup of dry dog food? The average 1-cup serving of dry dog food has anywhere between 325 and 600 calories. Also, keep in mind all of the treats that you feed your dog. A simple dog bone has about 35-40 calories.
How many calories does my dog need? – Related Asked Question
How many calories do dogs burn walking?
Calorie Burning in Dogs
8 calories per pound per mile. By extrapolation, you can estimate that a small to average-sized dog will burn about 64 calories in a single 1-hour walk.
How many calories should a 45 lb dog eat?
Daily Calorie Requirements for Dogs
Body weight in pounds Pupppy up to 4 mos. Intact adult
35 1674 1004
40 1848 1109
45 2019 1211
50 1312
How do you calculate calories in dog treats?
First needed is the weight of a single treat or a cup of product in grams. Dividing the kcal/kg value as determined above by 1000 converts it to kcal per gram. Then, multiplying by the number of grams per treat or cup gives you the calories per treat or cup.
How many calories do dogs burn running?
If your dog is healthy enough to run for a full half hour, he can burn up about 3.75 calories per pound of body weight, one study on heart rates in dogs had them walking through water tanks on a treadmill. Although few people will want to set that up at home, that exercise was good enough to stimulate a full run.
How many calories should a 90 pound dog eat?
dog needs about 366 calories, a 40 lb. dog 616, a 60 lb. dog 835 and 100 lb. giant dogs need about 1225 calories each day.
Are dog calories the same as human calories?
It’s commonplace to think that overfeeding isn’t an issue for exceedingly large dogs (those that weigh 32kg or more), but a growing dog this size still only needs 1688 calories per day. That’s nearly 900 calories less than the average man, and almost 400 calories less than the average woman.
How many calories are in moist and meaty dog food?
Calorie Content (calculated) (ME): 2794 kcal/kg, 474 kcal/pouch. Purina Moist &, Meaty Burger with Cheddar Cheese Flavor is formulated to meet the nutritional levels established by the AAFCO Dog Food Nutrient Profiles for maintenance of adult dogs.
How many calories should a dog eat per day to lose weight?
Here’s another approach: 3,500 calories = one pound of weight. If your goal is for your dog to lose one pound every two weeks (appropriate for a dog who should weigh 50 pounds), you must reduce calories by 3,500 per two weeks, or 250 calories per day.
Is canned dog food high in calories?
Some types of canned dog food, like Purina Pro Plan Sport, Energy &, Vitality Support, are high in calories.
Is a half hour walk enough for a dog?
As a general rule of thumb, most healthy, large-breed canines need a minimum of 30 minutes to two hours of walking every day. If your pooch has high energy levels, however, it may need more than two hours.
Is a 2 mile walk good for a dog?
The average adult dog benefits from at least 30 minutes of exercise daily, which can be broken up into two or three walks. Harper, now 10 1/2 , still enjoys a 2-mile walk or several short walks daily. It’s good for her health and mine, and best of all, it makes both of us happy.
How many calories does a dog burn on a 30 minute walk?
However, there is another study that shows that a 22-pound dog walking on a treadmill can burn around 64 calories in 30 minutes while being submerged in about 10 inches of water and maintaining a 6.8-mile-per-hour pace. It’s good to note that this pace is 2 miles an hour less than that of a marathon runner.
How do I calculate calories in homemade dog food?
This is the amount of energy (also called calories) that your dog’s body burns in a typical day. Before you can find the DER, you’ll need to find Fido’s Resting Energy Requirement (RER).
70(body weight in kg. ^.75)
Inactive/obese prone =1.2-1.4 x RER
Weight loss =1.0 x RER for ideal weight
How much should my dog eat by weight?
What are Dog Feeding Charts?
Adult Dog Size (lbs) Dry Food Feeding Amount (Cups)
26 to 50 2 to 2-2/3
51 to 75 2-2/3 to 3-1/3
76 to 100 3-1/3 to 4-1/4
100+ 4-1/4 plus 1/4 cup for each 10 lbs of body weight over 100 lbs
Is kcal the same as calories?
The “calorie” we refer to in food is actually kilocalorie. One (1) kilocalorie is the same as one (1) Calorie (uppercase C). A kilocalorie is the amount of heat required to raise the temperature of one kilogram of water one degree Celsius.
What is the fastest way for a dog to lose weight?
One simple solution to jump-start your dog’s weight loss is to feed your dog his normal food in the morning but replacing his second meal with mostly green beans (low sodium), a bit of kibble, and a doggie multi-vitamin in the evening. Switching your dog’s treats to healthier options will help him lose weight, too.
How far should I walk my overweight dog?
This should be about a 12-15 minute per mile pace (7-9 minute per kilometer). It should feel like a brisk walk and you should break into a light sweat. The key is to keep it up!
How much should you feed an old dog?
In terms of dog food for senior dogs, you should start by feeding about 2% of their body weight, and adjust to suit their needs and activity. A 50lbs senior dog of moderate activity level would be eating roughly one pound of food per day.
How do I know how much to feed my dog?
Calculate the daily caloric value for your dog’s weight and life stage. Locate the caloric value of the food in calories per kilo. Divide the first number by the second and multiply by 1,000 to get the daily serving in grams. Use the kitchen scale to measure out the right amount of food.
Is it okay to feed a dog cheese?
While cheese can be safe to feed to your dog, there are some things to remember. Cheese is high in fat, and feeding too much to your dog regularly can cause weight gain and lead to obesity. Even more problematic, it could lead to pancreatitis, a serious and potentially fatal illness in dogs.
Do dogs live longer eating human food?
More time with your best friend. Switching your dog to a fresh food diet helps maintain a healthy body weight, which has been linked to a 20% longer lifespan. Think about it: a human diet consisting primarily of processed foods would leave you sluggish, overweight, and at risk for host of health issues.
Why is dog food in kcal?
Though it sounds complicated, the requirement that all pet food labels express calorie statements in terms of kilocalories per kilogram of product as fed makes it easier to compare similar products. A kilocalorie is the same as a calorie (aka a big calorie or food calorie).
How much moist and meaty should I feed my dog?
Adult Dog Size: Over 100 lbs, Feeding Amount (pouches): 4 pouches plus 1/4 pouch for each 10 lbs of body weight over 100 lbs. Amounts are recommended for an average adult dog with normal activity. Remember food intake requirements vary depending on age, activity and environment, and should be adjusted accordingly.
Does Purina moist and meaty expire?
It’s recommended to use this food before the expiration date stamped on the box. To extend the freshness, it’s best to put the dry food in a sealed container.
Does Purina moist and meaty contain grain?
Purina Moist and Meaty is a grain-inclusive semi-moist dog food using a moderate amount of named by-products or chicken as its dominant source of animal protein, thus earning the brand 1 star.
What high calorie foods can I feed my dog?
Eggs – Raw, scrambled, over easy. Just make sure they are plain. Cottage Cheese – Full fat. Lean meat – Raw or cooked to match their diet.
Here are some tasty toppers that will make your dog’s diet more calorie-dense:
• Stella &, Chewy’s Meal Mixers.
• Honest Kitchen Goat’s Milk.
• Nature’s Logic Bone Broth.
• Tripett Canned Tripe.
What can I feed my dog to put weight on?
High protein and fat foods are great for dogs who need to gain weight. Feeding your dog a diet higher in fat and protein, which will help your dog gain weight steadily over time. Foods high in protein and fat will help your dog healthily put on weight, and you should notice a weight increase in just a couple of weeks.
Should I mix wet and dry dog food?
Mixing wet and dry pet food is fine as long as both options are of high-quality and meet your pup’s nutritional and health needs. Quality wet and dry dog foods are formulated to provide all the protein, vitamins, fat, minerals, and other vital nutrients your pup needs to thrive.
Is it OK not to walk your dog everyday?
Walking: Walking should be part of every dog’s daily routine to keep them physically and mentally healthy. Most dogs need at least 1-2 walks per day (unless otherwise specified by your vet).
Which dog breeds need the most exercise?
Which dogs need the most exercise?
• Labrador Retriever. Britain’s best-loved dog, there are more labrador retrievers registered in the UK than any other breed. …
• Dalmatian. …
• Border Collie. …
• Boxer. …
• English Springer Spaniel. …
• German Shepherd. …
• Golden Retriever.
How do I know if my dog is tired of walking?
1. Wear-and-Tear on Paw Pads. For some dogs, playing is more important than painful feet, says Dr. …
2. Sore Muscles. Muscular pain and stiffness is another sign your dog may be getting too much exercise, Downing says. …
3. Heat Sickness. …
4. Joint Injury. …
5. Behavioral Changes.
Can you over walk a dog?
However, we often get asked ‘Can I over exercise my dog? ‘. It’s an important question because the answer is yes, you can. Just like humans, dog’s have their limits in terms of exercise, and this varies wildly depending on the age, breed, health and the fitness level of your dog.
What time should I walk my dog?
Morning. Morning walks are recommended for puppies and senior dogs since they need to potty first thing in the morning to prevent accidents. Morning walks also allow dogs to get their energy out early in the day, this often equates to better behavior and more receptivity to training.
Is one walk a day enough for a dog?
The government recommends that we get at least 30 minutes of exercise each day and this is something that everybody can achieve – and go beyond – on a daily dog walk. The amount of exercise your dog needs will vary according to its breed, but every dog should have at least one walk a day, often two.
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- - markov query (http://www.velocityreviews.com/forums/t355451-markov-query.html)
kpp9c 03-14-2006 04:19 AM
markov query
markov query
I have noticed a couple markov implementations in python, but none
quite seem to do what i would like. Most seem to do an analysis of some
text and create a new text based on that... I think, (sorry i just
don't know terminology well) a markov table (or is it called a
transition table) to describe how to get from one event to another. So
if i have 3 events, say, A, B, and C which can come in any order, a
Markov chain describes the probability of what the next event will be
using a table. The following table demonstrates a first-order Markov
chain. There are three possible states. Either the current event is A,
B, or C. For each possible current state, there are three possible next
letters. Each row in the table indicates the relative probability of
going to the next letter. For example, if you are currently on letter
A, there is a 20% chance of repeating letter A, a 50% chance of going
to B, and a 30% chance of going to C. Note that the sum of changes for
each row is 100% (20 + 50 + 30 = 100).
Current -- next -- next -- next
A ----- B ----- C
A: 20% -- 50% -- 30%
B: 35% -- 25% -- 40%
C: 70% -- 14% -- 16%
Here the sequence C B and C C would be rare and the sequence C A
common.
This is a first-order Markov chain, which means that only the current
state affects the choice of the next event. A second-order Markov chain
would mean that the current state and the last state affect the choice
of the next event. A third-order Markov chain would indicate that the
current state and the last two states in the sequence will affect the
choice of the next state, and so on. Here is an example transition
table for a 2nd order Markov chain.
Current -- next -- next -- next
A ----- B ----- C
A A 15% 55% 30%
A B 20% 45% 35%
A C 60% 30% 10%
B A 35% 25% 40%
B B 49% 48% 3%
B C 60% 20% 20%
C A 5% 75% 20%
C B 0% 90% 10%
C C 70% 14% 16%
For example, if the current event is B and the last one was A, then the
probability of the next event being A is 20%, B is 45% and C is 35%.
The pattern C B A will never occur, and the pattern B B C will occur
rarely.
Does anyone know of any modules or packages that include markov tables
for creating patterns like shown above.
P.S. Does any one know first of all whether these are called markov
tables, transition tables or probability tables? I am not sure i am
referring to this correctly and what the differences would be if any
cheers,
kp
Stefan Behnel 03-14-2006 07:43 AM
Re: markov query
Don't know of a Python module (although this doesn't look complex enough for a
package anyway...), but
kpp9c wrote:
> P.S. Does any one know first of all whether these are called markov
> tables, transition tables or probability tables? I am not sure i am
> referring to this correctly and what the differences would be if any
as for terminology:
http://www.csse.monash.edu.au/~lloyd...tured/HMM.html
http://en.wikipedia.org/wiki/Hidden_Markov_model
Hope it helps,
Stefan
Erik Max Francis 03-14-2006 08:15 AM
Re: markov query
kpp9c wrote:
> I have noticed a couple markov implementations in python, but none
> quite seem to do what i would like. Most seem to do an analysis of some
> text and create a new text based on that... I think, (sorry i just
> don't know terminology well) a markov table (or is it called a
> transition table) to describe how to get from one event to another.
Yes, a system which does this has to build a Markov chain from a data
set and then traverse it.
> Does anyone know of any modules or packages that include markov tables
> for creating patterns like shown above.
Any program that actually uses Markov chains to generate new text based
on existing input as you've described will necessarily create a Markov
chain. So if what you want is the Markov chain itself, then it's
> P.S. Does any one know first of all whether these are called markov
> tables, transition tables or probability tables? I am not sure i am
> referring to this correctly and what the differences would be if any
They're called Markov chains.
--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
Nothing spoils a confession like repentence.
-- Anatole France
kpp9c 03-15-2006 12:14 AM
Re: markov query
> Yes, a system which does this has to build a Markov chain from a data
> set and then traverse it.
>Any program that actually uses Markov chains to generate new text based
>on existing input as you've described will necessarily create a Markov
>chain.
I think you misunderstood. If you see my original post my whole point
was that this was exactly what i don't want. There are several
algorithms out that that already do that. I want to use a table that i
define from scratch to shape a stochastic process. In this case there
is no original input to analyze and i don't want a chain built by
analysis and i am not using necessarily texts.
kpp9c 03-15-2006 12:36 AM
Re: markov query
>Yes, a system which does this has to build a Markov
>chain from a data set and then traverse it.
>>Any program that actually uses Markov chains to generate
>> new text based on existing input as you've described
Hi. That isn't really what i have described. If i did i could use
exsisting algorithms. What you describe is exactly what i don't want in
this particular case. I am actually not wanting it to build a chain
from some input that it analyzes. There is no input. If you see, i am
trying to define a table that tells it how to proceed forward from
scratch as a stochastic process.
cheers,
-kp--
Robert Kern 03-15-2006 01:12 AM
Re: markov query
kpp9c wrote:
>>Yes, a system which does this has to build a Markov
>>chain from a data set and then traverse it.
>
>>>Any program that actually uses Markov chains to generate
>>>new text based on existing input as you've described
>
> Hi. That isn't really what i have described. If i did i could use
> exsisting algorithms. What you describe is exactly what i don't want in
> this particular case. I am actually not wanting it to build a chain
> from some input that it analyzes. There is no input. If you see, i am
> trying to define a table that tells it how to proceed forward from
> scratch as a stochastic process.
Any such program has two main parts, one that trains the transition matrix from
a data set, and one that generates output from that transition matrix. You
should be able to take such a program and only use the latter component with
your transition matrix, however you choose to create it. Googling for 'python
--
Robert Kern
robert.kern@gmail.com
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# Functions & Differentiation eBook
This functions and differentiation eBook contains a selection of A4 printable worksheets for GCSE mathematics . Solutions are provided to enable quick and easy feedback.
### Topics Included
• Working with functions
• Composite functions
• Inverse functions
• Translating functions
• Stretching functions
• Iteration
• Differentiation
Usual price: £4.99
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### Mr Mathematics Blog
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There are three common ways to organise data that fall into multiple sets: two-way tables, frequency diagrams and Venn diagrams. Having blogged about frequency diagrams before I thought I would write about how to draw a Venn Diagram to calculate probabilities. Recapping Two-Way Tables This activity works well to review two-way tables from the previous […]
#### Calculations with Percentages
Students learn how to find a percentage of an amount using calculator and non-calculator methods. As learning progresses they use decimal multipliers to find a percentage change and calculate a simple interest in financial mathematics. This topic follows on from Fractions, Decimals and Percentages and takes place in Year 8 Term 5. Calculations with Percentages […]
#### Proving Geometrical Relationships using Algebra
Back in May 2017 maths teachers around the country eagerly awaited the first exam for the new GCSE Mathematics syllabus. Proving geometrical relationships using algebra featured at grade 9. In Paper 1 of Edexcel’s test paper the last question of the higher tier looked like this. Edexcel wrote about student’s performance on this question in […] | 347 | 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} | 2.640625 | 3 | CC-MAIN-2018-13 | latest | en | 0.843886 |
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# BUSN 379 Week 2 Homework Chapter4 (8, 17, 18) and Chapter 5 (1, 4, 12) LATEST
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## This Tutorial contains following Attachments:
• BUSN 379 Week 2 Homework Chapter 4 (8, 17, 18) and Chapter 5 (1, 4, 12).docx
BUSN 379 Week 2 Homework CHAPTER 4 (8, 17, 18) and CHAPTER 5 (1, 4, 12) NEW
CHAPTER 4 (8, 17, 18)
8. Calculating the Number of Periods. Calculating Rates of Return. In 2011, an 1880-O Morgan silver dollar sold for \$13,113. What was the rate of return on this investment?
17. Calculating Present Values. Suppose you are still committed to owning a \$150,000 Ferrari (see Question 9). If you believe your mutual fund can achieve a 10.25 percent annual rate of return, and you want to buy the car in 10 years on the day you turn 30, how much must you invest today?
18. Calculating Future Values. You have just made your first \$5,000 contribution to your individual retirement account. Assuming you earn a 10.1 percent rate of return and make no additional contributions, what will your account be worth when you retire in 45 years? What if you wait 10 years before contributing? (Does this suggest an investment strategy?)
CHAPTER 5 (1, 4, 12)
1.Present Value and Multiple Cash Flows. Rooster Co. has identified an investment project with the following cash flows. If the discount rate is 10 percent, what is the present value of these cash flows? What is the present value at 18 percent? At 24 percent?
4. Calculating Annuity Present Values. An investment offers \$6,700 per year for 15 years, with the first payment occurring 1 year from now. If the required return is 8 percent, what is the value of the investment? What would the value be if the payments occurred for 40 years? For 75 years? Forever?
12. Calculating EAR. Find the EAR in each of the following cases: | 516 | 1,894 | {"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} | 3.796875 | 4 | CC-MAIN-2019-39 | latest | en | 0.910118 |
https://socratic.org/questions/which-state-of-matter-has-the-highest-entropy-1 | 1,669,808,421,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00664.warc.gz | 574,876,335 | 5,969 | # Which state of matter has the highest entropy?
Nov 21, 2015
${S}_{g a s} \text{>>} {S}_{l i q u i d} > {S}_{s o l i d}$
#### Explanation:
Entropy by definition is the degree of randomness in a system.
If we look at the three states of matter: Solid, Liquid and Gas, we can see that the gas particles move freely and therefore, the degree of randomness is the highest.
For liquid state, still the particles are moving but less freely than the gas particles and more than the solid particles.
Therefore, ${S}_{g a s} \text{>>} {S}_{l i q u i d} > {S}_{s o l i d}$
Here is a video that explains more this topic and how it is related to spontaneity.
Thermodynamics | Spontaneous Process & Entropy. | 192 | 704 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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} | 2.84375 | 3 | CC-MAIN-2022-49 | latest | en | 0.887369 |
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# Math Worksheet Numbers Printable Trace And Write Dotted To
By Felix Glockner at December 19 2018 19:02:30
There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign "%." Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom's delicious apple pie is gone.
Remember to select worksheets that are the right level difficulty for your child. Get something too hard, and your child will become discouraged. Make it too easy, and they won't learn much.
When a child learns to relate math to everyday questions, he will be great at it from the simplest addition all the way to trigonometry. To convert percentages, decimals and fractions is thus one essential skill.
They're also available for nearly all grade levels. There are printable middle school, high school, elementary school, and even pre-school worksheets. | 288 | 1,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-13 | latest | en | 0.946835 |
https://www.physicsforums.com/threads/what-is-the-probability-of-their-child-having-sickle-cell-anemia.219148/ | 1,721,932,638,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00488.warc.gz | 801,301,781 | 15,859 | # What Is the Probability of Their Child Having Sickle Cell Anemia?
• User Name
In summary, the frequency of sickle cell anemia among black Americans is about .0025. When two black Americans marry, the probability that both will be heterozygotes is .25. If both are heterozygotes, the probability that one of their children will have sickle cell anemia is also .25. However, it is important to note that this question is looking at an individual cross, rather than population genetics, so the original frequency of .0025 may not apply.
User Name
Among black Americans, the frequency of sickle cell anemia is about .0025. What is the frequency of heterozygotes? When one black American marries another, what is the probability that both will be heterozygotes? If both are heterozygotes, what is the probability that one of their children will have sickle cell anemia?
I've already solved the first two questions, and now I'm just focusing on the last question. My question is, would the probability of one of their children having sickle cell anemia be the original .0025 that was given to me? Or would it be .25 (the probability of two recessive alleles when you cross two heterozygotes)?
Thanks, in advance. =]
For the last question, you're switching from population genetics to an individual cross. I suspect that's the reason for including that question, to remind you to think about whether you're working at the population level or individual level. That should tell you which of your answers to use.
Think of this question as three separate questions. Could your teacher still ask you the last question if you did not have the first two (assuming you still know you are working with sickle cell anemia)?
Yes, you could.
So now using ONLY the information in the 3rd question, can you figure out the answer?
## 1. What is the Hardy-Weinberg Equilibrium?
The Hardy-Weinberg Equilibrium (HWE) is a mathematical model that describes the genetic equilibrium of a population over generations. It predicts the frequencies of alleles and genotypes in a population that is not evolving.
## 2. What are the assumptions of the Hardy-Weinberg Equilibrium?
The assumptions of HWE include a large population size, random mating, no mutations, no migration, and no natural selection. These assumptions allow for the prediction of allele and genotype frequencies in a stable population.
## 3. How is the Hardy-Weinberg Equilibrium calculated?
The equation for HWE is p2 + 2pq + q2 = 1, where p and q represent the frequencies of the two alleles in a population. p2 represents the frequency of homozygous dominant individuals, 2pq represents the frequency of heterozygous individuals, and q2 represents the frequency of homozygous recessive individuals.
## 4. What is the significance of the Hardy-Weinberg Equilibrium?
The HWE model is important in population genetics as it allows scientists to compare observed allele and genotype frequencies to those predicted by the model. Deviations from HWE can indicate evolutionary forces such as natural selection, genetic drift, or gene flow.
## 5. How is the Hardy-Weinberg Equilibrium used in research?
The HWE model is used in research to study the genetic structure of populations and to determine if a population is evolving. It can also be used to estimate the probability of inheriting genetic diseases and to track changes in allele frequencies over time.
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## Break a number into array of digits
hi all,
I am trying to break a number into an array of digits using split function but it keeps giving the whole number as the first item of the array.
How can this be broken down into array items for each diigt?
Split function used, EmployeeID contains the number 789545
split(split(items('Apply_to_each_2')?['EmployeeID'],'')[0],'')
thanks
5 REPLIES 5
Multi Super User
Hey @aarnav
I have made a flow to get all digits of a number in an array. Its a little long, I hope you understand it.
Heres how the input looks like:
Here is the Output:
Variable A to store the number, variable C to store the digits, Variable D for index number, Variable E just stored variable A's temporary value. And also the condition for do until is given.
Inside Do Until: I am taking last digit of integer. And putting it in array. Its a reverse array.
Append to Array Value: mod(variables('E'),10)
Compose Value: div(variables('E'),10)
Set Variable: outputs('Compose')
Variables F to store reverse array(The result you want).
Variable G: sub(length(variables('C')),1)
Apply to each loop, to get the final values:
Compose 2: variables('C')[variables('G')]
Append to Array 2: outputs('Compose_2')
Last compose to show the output:
I hope you like the solution. If you have any doubts just comment.
Community Champion
Try to use this:
``split(trim(replace(replace(replace(replace(replace(replace(replace(replace(replace(replace(variables('strDigits'),'0',' 0'),'1',' 1'),'2',' 2'),'3',' 3'),'4',' 4'),'5',' 5'),'6',' 6'),'7',' 7'),'8',' 8'),'9',' 9')),' ')``
Super User
This should do what you want:
The output generated from the select action is:
``````[
"7",
"8",
"9",
"5",
"4",
"5"
]``````
If you want to understand better how this works, this blog post I wrote should help:
The subject of the post isn't what you want to do, but it uses the same technique.
Blog: tachytelic.net
Community Champion
Really elegant approach, @Paulie78. Bravo!
Super User
Thanks @VictorIvanidze I love the range function, useful in so many situations.
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http://www.thefullwiki.org/Autocorrelation | 1,539,633,071,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509690.35/warc/CC-MAIN-20181015184452-20181015205952-00274.warc.gz | 560,898,183 | 16,462 | # Autocorrelation: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
A plot showing 100 random numbers with a "hidden" sine function, and an autocorrelation (correlogram) of the series on the bottom.
Autocorrelation is the cross-correlation of a signal with itself. Informally, it is the similarity between observations as a function of the time separation between them. It is a mathematical tool for finding repeating patterns, such as the presence of a periodic signal which has been buried under noise, or identifying the missing fundamental frequency in a signal implied by its harmonic frequencies. It is often used in signal processing for analyzing functions or series of values, such as time domain signals.
## Definitions
Different fields of study define autocorrelation differently, and not all of these definitions are equivalent. In some fields, the term is used interchangeably with autocovariance.
### Statistics
In statistics, the autocorrelation of a random process describes the correlation between values of the process at different points in time, as a function of the two times or of the time difference. Let X be some repeatable process, and i be some point in time after the start of that process. (i may be an integer for a discrete-time process or a real number for a continuous-time process.) Then Xi is the value (or realization) produced by a given run of the process at time i. Suppose that the process is further known to have defined values for mean μi and variance σi2 for all times i. Then the definition of the autocorrelation between any two time s and t is
$R(s,t) = \frac{\operatorname{E}[(X_t - \mu_t)(X_s - \mu_s)]}{\sigma_t\sigma_s}\, ,$
where "E" is the expected value operator. Note that this expression is not well-defined for all time series or processes, because the variance may be zero (for a constant process) or infinite. If the function R is well-defined, its value must lie in the range [−1, 1], with 1 indicating perfect correlation and −1 indicating perfect anti-correlation.
If Xt is a second-order stationary process then the mean μ and the variance σ2 are time-independent, and further the autocorrelation depends only on the difference between t and s: the correlation depends only on the time-distance between the pair of values but not on their position in time. This further implies that the autocorrelation can be expressed as a function of the time-lag, and that this would be an even function of the lag τ = t − s. This gives the more familiar form
$R(\tau) = \frac{\operatorname{E}[(X_t - \mu)(X_{t+\tau} - \mu)]}{\sigma^2}\, ,$
and the fact that this is an even function can be stated as
$R(\tau) = R(-\tau).\,$
It is common practice in some disciplines, other than statistics and time series analysis, to drop the normalization by σ2 and use the term "autocorrelation" interchangeably with "autocovariance". However, the normalisation is important both because the interpretation of the autocorrelation as a correlation provides a scale-free measure of the strength of statistical dependence, and because the normalisation has an effect on the statistical properties of the estimated autocorrelations.
### Signal processing
In signal processing, the above definition is often used without the normalization, that is, without subtracting the mean and dividing by the variance. When the autocorrelation function is normalized by mean and variance, it is sometimes referred to as the autocorrelation coefficient.[1]
Given a signal f(t), the continuous autocorrelation Rff(τ) is most often defined as the continuous cross-correlation integral of f(t) with itself, at lag τ.
$R_{ff}(\tau) = \overline{f}(-\tau) * f(\tau) = \int_{-\infty}^{\infty} f(t+\tau)\overline{f}(t)\, dt = \int_{-\infty}^{\infty} f(t)\overline{f}(t-\tau)\, dt$
where $\bar f$ represents the complex conjugate and * represents convolution. For a real function, $\bar f = f$.
The discrete autocorrelation R at lag j for a discrete signal xn is
$R_{xx}(j) = \sum_n x_n \overline{x}_{n-j} \ .$
The above definitions work for signals that are square integrable, or square summable, that is, of finite energy. Signals that "last forever" are treated instead as random processes, in which case different definitions are needed, based on expected values. For wide-sense-stationary random processes, the autocorrelations are defined as
$R_{ff}(\tau) = \operatorname{E}\left[f(t)\overline{f}(t-\tau)\right]$
$R_{xx}(j) = \operatorname{E}\left[x_n \overline{x}_{n-j}\right].$
For processes that are not stationary, these will also be functions of t, or n.
For processes that are also ergodic, the expectation can be replaced by the limit of a time average. The autocorrelation of an ergodic process is sometimes defined as or equated to[1]
$R_{ff}(\tau) = \lim_{T \rightarrow \infty} {1 \over T} \int_{0}^{T} f(t+\tau)\overline{f}(t)\, dt$
$R_{xx}(j) = \lim_{N \rightarrow \infty} {1 \over N} \sum_{n=0}^{N-1}x_n \overline{x}_{n-j}.$
These definitions have the advantage that they give sensible well-defined single-parameter results for periodic functions, even when those functions are not the output of stationary ergodic processes.
Alternatively, signals that last forever can be treated by a short-time autocorrelation function analysis, using finite time integrals. (See short-time Fourier transform for a related process.)
Multi-dimensional autocorrelation is defined similarly. For example, in three dimensions the autocorrelation of a square-summable discrete signal would be
$R(j,k,\ell) = \sum_{n,q,r} (x_{n,q,r})(x_{n-j,q-k,r-\ell}).$
When mean values are subtracted from signals before computing an autocorrelation function, the resulting function is usually called an auto-covariance function.
## Properties
In the following, we will describe properties of one-dimensional autocorrelations only, since most properties are easily transferred from the one-dimensional case to the multi-dimensional cases.
• A fundamental property of the autocorrelation is symmetry, R(i) = R(−i), which is easy to prove from the definition. In the continuous case, the autocorrelation is an even function
$R_f(-\tau) = R_f(\tau)\,$
when f is a real function and the autocorrelation is a Hermitian function
$R_f(-\tau) = R_f^*(\tau)\,$
when f is a complex function.
• The continuous autocorrelation function reaches its peak at the origin, where it takes a real value, i.e. for any delay τ, $|R_f(\tau)| \leq R_f(0)$. This is a consequence of the Cauchy–Schwarz inequality. The same result holds in the discrete case.
• The autocorrelation of a periodic function is, itself, periodic with the same period.
• The autocorrelation of the sum of two completely uncorrelated functions (the cross-correlation is zero for all τ) is the sum of the autocorrelations of each function separately.
• Since autocorrelation is a specific type of cross-correlation, it maintains all the properties of cross-correlation.
• The autocorrelation of a continuous-time white noise signal will have a strong peak (represented by a Dirac delta function) at τ = 0 and will be absolutely 0 for all other τ.
$R(\tau) = \int_{-\infty}^\infty S(f) e^{j 2 \pi f \tau} \, df$
$S(f) = \int_{-\infty}^\infty R(\tau) e^{- j 2 \pi f \tau} \, d\tau.$
• For real-valued functions, the symmetric autocorrelation function has a real symmetric transform, so the Wiener–Khinchin theorem can be re-expressed in terms of real cosines only:
$R(\tau) = \int_{-\infty}^\infty S(f) \cos(2 \pi f \tau) \, df$
$S(f) = \int_{-\infty}^\infty R(\tau) \cos(2 \pi f \tau) \, d\tau.$
## Estimation
For a discrete process of length n {X1X2, … Xn} with known mean and variance, an estimate of the autocorrelation may be obtained as
$\hat{R}(k)=\frac{1}{(n-k) \sigma^2} \sum_{t=1}^{n-k} [X_t-\mu][X_{t+k}-\mu]$
for any positive integer k < n. When the true mean μ and variance σ are known, this estimate is unbiased. If the true mean and variance of the process are not known there are a several possibilities:
• If μ and σ2 are replaced by the standard formulae for sample mean and sample variance, then this is a biased estimate.
• A periodogram-based estimate replaces n-k in the above formula with n. This estimate is always biased; however, it usually has a smaller mean square error.[2][3]
• Other possibilities derive from treating the two portions of data {X1X2, … Xn-k} and {Xk+1X2, … Xn} separately and calculating separate sample means and/or sample variances for use in defining the estimate.
The advantage of estimates of the last type is that the set of estimated autocorrelations, as a function of k, then form a function which is a valid autocorrelation in the sense that it is possible to define a theoretical process having exactly that autocorrelation. Other estimates can suffer from the problem that, if they are used to calculate the variance of a linear combination of the X's, the variance calculated may turn out to be negative.
## Regression analysis
In regression analysis using time series data, autocorrelation of the residuals ("error terms", in econometrics) is a problem.
Autocorrelation violates the ordinary least squares (OLS) assumption that the error terms are uncorrelated. While it does not bias the OLS coefficient estimates, the standard errors tend to be underestimated (and the t-scores overestimated) when the autocorrelations of the errors at low lags are positive.
The traditional test for the presence of first-order autocorrelation is the Durbin–Watson statistic or, if the explanatory variables include a lagged dependent variable, Durbin's h statistic. A more flexible test, covering autocorrelation of higher orders and applicable whether or not the regressors include lags of the dependent variable, is the Breusch–Godfrey test. This involves an auxiliary regression, wherein the residuals obtained from estimating the model of interest are regressed on (a) the original regressors and (b) k lags of the residuals, where k is the order of the test. The simplest version of the test statistic from this auxiliary regression is TR2, where T is the sample size and R2 is the coefficient of determination. Under the null hypothesis of no autocorrelation, this statistic is asymptotically distributed as χ2 with k degrees of freedom.
Responses to nonzero autocorrelation include generalized least squares and the Newey–West HAC estimator (Heteroskedasticity and Autocorrelation Consistent).[4]
## Applications
• For measuring particle size distributions of very fine particles or micelles suspended in a fluid. A laser shining into the mixture produces flicker, which correlates with the motion of the particles. Autocorrelation of the signal gives a picture of the diffusion speeds of the particles. From this, knowing the viscosity of the fluid, the sizes of the particles can be calculated.
• In optics, normalized autocorrelations and cross-correlations give the degree of coherence of an electromagnetic field.
• Autocorrelation in space rather than time, via the Patterson function, is used by X-ray diffractionists to help recover the "Fourier phase information" on atom positions not available through diffraction alone.
• In statistics, spatial autocorrelation between sample locations also helps one estimate mean value uncertainties when sampling a heterogeneous population.
• The SEQUEST algorithm for analyzing mass spectra makes use of autocorrelation in conjunction with cross-correlation to score the similarity of an observed spectrum to an idealized spectrum representing a peptide.
• In Astrophysics, auto-correlation is used to study and characterize the spatial distribution of galaxies in the Universe and in multi-wavelength observations of Low Mass X-ray Binaries.
## References
1. ^ a b Patrick F. Dunn, Measurement and Data Analysis for Engineering and Science, New York: McGraw–Hill, 2005 ISBN 0-07-282538-3
2. ^ Spectral analysis and time series, M.B. Priestley (London, New York : Academic Press, 1982)
3. ^ Percival, Donald B.; Andrew T. Walden (1993). Spectral Analysis for Physical Applications: Multitaper and Conventional Univariate Techniques. Cambridge University Press. pp. pp190--195. ISBN 0-521-43541-2.
4. ^
5. ^ Tyrangiel, Josh (2009-02-05), "Auto-Tune: Why Pop Music Sounds Perfect", Time Magazine | 3,002 | 12,466 | {"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": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-43 | longest | en | 0.921189 |
https://usethinkscript.com/threads/simple-color-change-on-watchlist-for-value-above-x.10522/ | 1,723,651,697,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641118845.90/warc/CC-MAIN-20240814155004-20240814185004-00173.warc.gz | 464,911,342 | 18,370 | # Simple Color Change on Watchlist for value Above "X"
#### Luke91-DT
##### New member
Hey everyone,
I'm looking for something pretty simple, I am pretty inexperienced to writing script for ThinkScript. I am using a simple script for showing volume spikes on my WL tickers, what I'd like to be able to do is have it turn the background color green if it is above X value (5 or 10 for example). Below is the script and a screen shot of how it presents on my watchlist currently.
Any help is appreciated, thanks!
(volume / Average(volume, 20))
Solution
I'm looking for something pretty simple, I am pretty inexperienced to writing script for ThinkScript. I am using a simple script for showing volume spikes on my WL tickers, what I'd like to be able to do is have it turn the background color green if it is above X value (5 or 10 for example). Below is the script and a screen shot of how it presents on my watchlist currently.
this is a column study.
it has 4 number levels to compare to, so there are 5 color ranges.
the levels are set to, 2,4,6,8
lower numbers are dark colors.
higher numbers are lighter. above 8 will be green.
column study , 1 minute
zvolrng1 , http://tos.mx/Rhjxbj2
Ruby:
``````# zvolrng1
# r = (volume / Average(volume, 20))
def v = volume;
def len = 20;
def...``````
I'm looking for something pretty simple, I am pretty inexperienced to writing script for ThinkScript. I am using a simple script for showing volume spikes on my WL tickers, what I'd like to be able to do is have it turn the background color green if it is above X value (5 or 10 for example). Below is the script and a screen shot of how it presents on my watchlist currently.
this is a column study.
it has 4 number levels to compare to, so there are 5 color ranges.
the levels are set to, 2,4,6,8
lower numbers are dark colors.
higher numbers are lighter. above 8 will be green.
column study , 1 minute
zvolrng1 , http://tos.mx/Rhjxbj2
Ruby:
``````# zvolrng1
# r = (volume / Average(volume, 20))
def v = volume;
def len = 20;
def vavg = average(v, len);
def r = round(v/vavg,2);
plot z = r;
# font color
z.setdefaultcolor(color.black);
#z.setdefaultcolor(color.white);
def level1 = 2;
def level2 = 4;
def level3 = 6;
def level4 = 8;
assignbackgroundcolor(
if r < level1 then color.dark_gray
else if ( r >= level1 and r < level2) then color.gray
else if ( r >= level2 and r < level3) then color.magenta
else if ( r >= level3 and r < level4) then color.yellow
else color.green);``````
this is a column study.
it has 4 number levels to compare to, so there are 5 color ranges.
the levels are set to, 2,4,6,8
lower numbers are dark colors.
higher numbers are lighter. above 8 will be green.
column study , 1 minute
zvolrng1 , http://tos.mx/Rhjxbj2
Ruby:
``````# zvolrng1
# r = (volume / Average(volume, 20))
def v = volume;
def len = 20;
def vavg = average(v, len);
def r = round(v/vavg,2);
plot z = r;
# font color
z.setdefaultcolor(color.black);
#z.setdefaultcolor(color.white);
def level1 = 2;
def level2 = 4;
def level3 = 6;
def level4 = 8;
assignbackgroundcolor(
if r < level1 then color.dark_gray
else if ( r >= level1 and r < level2) then color.gray
else if ( r >= level2 and r < level3) then color.magenta
else if ( r >= level3 and r < level4) then color.yellow
else color.green);``````
Thanks a ton! I'll be able to learn from this as well.
Easy to adjust and might help with my future scripts!
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What are the benefits of VIP Membership? | 1,160 | 4,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.894761 |
https://encyclopediaofmath.org/index.php?title=Permanent&oldid=35182 | 1,591,152,985,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347428990.62/warc/CC-MAIN-20200603015534-20200603045534-00396.warc.gz | 330,748,244 | 8,196 | Permanent
of an -matrix
The function
where are elements from a commutative ring and summation is over all one-to-one mappings from into . If , then represents all possible permutations, and the permanent is a particular case of the Schur matrix function (cf. Immanant)
for , where is a character of degree 1 on the subgroup (cf. Character of a group) of the symmetric group (one obtains the determinant for , , in accordance with the parity of ).
The permanent is used in linear algebra, probability theory and combinatorics. In combinatorics, a permanent can be interpreted as follows: The number of systems of distinct repesentatives for a given family of subsets of a finite set is the permanent of the incidence matrix for the incidence system related to this family.
The main interest is in the permanent of a matrix consisting of zeros and ones (a -matrix), of a matrix containing non-negative real numbers, in particular doubly-stochastic matrices (in which the sum of the elements in any row and any column is 1), and of a complex Hermitian matrix. The basic properties of the permanent include a theorem on expansion (the analogue of Laplace's theorem for determinants) and the Binet–Cauchy theorem, which gives a representation of the permanent of the product of two matrices as the sum of the products of the permanents formed from the cofactors. For the permanents of complex matrices it is convenient to use representations as scalar products in the symmetry classes of completely-symmetric tensors (see, e.g., [3]). One of the most effective methods for calculating permanents is provided by Ryser's formula:
where is the set of submatrices of dimension for the square matrix , is the sum of the elements of the -th row of and . As it is complicated to calculate permanents, estimating them is important. Some lower bounds are given below.
a) If is a -matrix with , , then
for , and
if and .
b) If is a -matrix of order , then
where are the sums of the elements in the rows of arranged in non-increasing order and .
c) If is a positive semi-definite Hermitian matrix of order , then
where if .
Upper bounds for permanents:
1) For a -matrix of order ,
2) For a completely-indecomposable matrix of order with non-negative integer elements,
3) For a complex normal matrix with eigen values ,
The most familiar problem in the theory of permanents was van der Waerden's conjecture: The permanent of a doubly-stochastic matrix of order is bounded from below by , and this value is attained only for the matrix composed of fractions . A positive solution to this problem was obtained in [4].
Among the applications of permanents one may mention relationships to certain combinatorial problems (cf. Combinatorial analysis), such as the "problème des rencontresproblème de rencontres" and the "problème d'attachementproblème d'attachement" (or "hook problemhook problem" ), and also to the Fibonacci numbers, the enumeration of Latin squares (cf. Latin square) and Steiner triple systems (cf. Steiner system), and to the derivation of the number of -factors and linear subgraphs of a graph, while doubly-stochastic matrices are related to certain probability models. There are interesting physical applications of permanents, of which the most important is the dimer problem, which arises in research on the adsorption of di-atomic molecules in surface layers: The permanent of a -matrix of a simple structure expresses the number of ways of combining the atoms in the substance into di-atomic molecules. There are also applications of permanents in statistical physics, the theory of crystals and physical chemistry.
References
[1] H.J. Ryser, "Combinatorial mathematics" , Wiley & Math. Assoc. Amer. (1963) [2] V.N. Sachkov, "Combinatorial methods in discrete mathematics" , Moscow (1977) (In Russian) [3] H. Minc, "Permanents" , Addison-Wesley (1978) [4] G.P. Egorichev, "The solution of van der Waerden's problem on permanents" , Krasnoyarsk (1980) (In Russian) [5] D.I. Falikman, "Proof of the van der Waerden conjecture regarding the permanent of a doubly stochastic matrix" Math. Notes , 29 : 6 (1981) pp. 475–479 Mat. Zametki , 29 : 6 (1981) pp. 931–938 | 971 | 4,189 | {"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-2020-24 | latest | en | 0.920569 |
http://utter.chaos.org.uk/~eddy/math/linear/ramble.xhtml | 1,713,456,828,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00430.warc.gz | 33,475,303 | 6,271 | ]> Thoughts on linearity
# Thoughts on linearity
Where orthodoxy addresses a collection of values which may be added, I'm identifying each of those values, v, with the associated translation through v, (: v+u ←u :). Composing two such translations gives the translation through the sum of the values added by the two translations, (:a+u←u:)&on;(b+u←u) = (:a+b+u←u:). Once a and b are replaced by x = (:a+u←u:) and y = (:b+u←u:), we can write x+y for x&on;y and, indeed, recover the ability to write all orthodox statements about vectors as statements about translations. Because translations are, however, relations we can repeat them, to obtain scaling by the positive natural numbers; in so far as these scalings are monic (on some given collection of translations) we can introduce their reverses to obtain rational scalings – scale up by one positive integer and down by another.
We can define respects addition as a property on relations (from a set of translations, or from a set embeddable in such) by: r respects addition iff r relates u to v and r relates x to y implies r relates u+x to v+y. Any such relation then trivially commutes with every rational scaling. Within the collection of relations that respect addition, one naturally has the positive rational numbers as an abelian multiplicative group with an associated (cancellable abelian associative) addition over which the multiplication (compsition) distributes; in so far as we can find some field – or, at least, some abelian multiplicative group (the positive members of the field) with associated addition over which the multiplication distributes – which subsumes the positive rationals, we can allow its members as further scalings, extending the rationals. I thus chose to leave it to context to select and specify which other relations to deem scalings, provided {scalings} supports abelian multiplication distributing over abelian addition, with the addition cancellable and, aside from the additive identity (if any), the multiplication also cancellable.
So, consider a collection V of relations (e.g. our translations) which is closed under composition and among which composition is abelian - i.e. a&on;b = b&on;a is in V for all a, b in V - and write + for composition's restriction to V, i.e. a+b = a&on;b for a, b in V. For any equivalence I, extend + to act on {relations (V::I)} according to:
given (V:f:I) and (V:g:I)
f+g = unite({ (V: f(i)+g(i)←i :I), (V: i not in (:g|); f(i)←i :I), (V: i not in (:f|); g(i)←i :I) })
which adds f and g pointwise throughout I, implicitly extending each, wherever only the other is defined, with an additive identity; but does so without needing to insist that an additive identity be present.
Describe a relation (V:f:V) as real linear on V iff: f relates u to x and v to y implies f relates u+v to x+y, i.e. iff f respects addition. Describe a real linear (V:f:V) as a real scaling of V iff f commutes, i.e. f&on;g = g&on;f, with every real linear (V:g:V). Because every real linear respects addition, it also respects repeated addition; so, for each positive natural n, (V:repeat(n):V) is a real scaling of V; hereinafter written as n.v←v using . as a binary operator (to be construed as multiplication of a vector by a scalar).
For example, in a vector space (real or complex), the real scalings are multiplication by real scalars; in the complex case, antilinear and linear maps respect addition but don't commute with one another, aside from the real scalings (which are linear; though the zero map, if V has an additive identity, is also antilinear). I need some way to identify the remaining scalings, when there are any; they commute with all (complex) linear (V:|V) but conjugate-commute with antilinear (V:|V), which seems to depend on having the notion of linearity sewn up already. And they're only present in complex contexts.
The identity on V is manifestly (:repeat(1):) hence real linear and, indeed, a real scaling. If V subsumes U, we can read U as the identity on U, which is a monic mapping (V::V), and ask whether it is real linear: since the identity is (: u←u :U), we can reduce U relates u to x and v to y implies U relates x+y to u+v to u in U and v in U implies u+v in U, which may be read as U is closed under addition. So a collection is real linear iff it is closed under addition.
For any real linear (V:r:V) we have r relates x to u and y to v implies r relates x+y to u+v whence, in particular, any sum of r's left values is a left value of r, likewise for right values; i.e. (:v←v:r) and (r:v←v:) are real linear whenever (V:r:V) is.
So now, given real linear (V:r:V), consider (|r:); it relates u to v iff ({u}&on;r: x←x :) = ({v}&on;r: x←x :) is non-empty; is (|r:) also real linear ? Its collection of values is just (:v←v:r), which we know to be real linear. If (|r:) relates u to v, we have non-empty ({v}&on;r: x←x :) = ({u,v}&on;r: x←x :) = ({u}&on;r: x←x :). If r relates w to z and x is in ({v,u}&on;r: x←x :), then r relates w+v to z+x and w+u to z+x. If r relates w+v to g, not necessarily given to be a sum of members of V (e.g. the collection of positive integers is an additive domain in which 1 is not a sum), can I show that it must also relate w+u to g ? RTP: ({v+w}&on;r: x←x :) = ({u+w}&on;r: x←x :) technical hitch: are there a in ({u+w}&on;r: x←x :) which are not s+t for any s, t ? will it matter if there are ?
In so far as we have an addition on V, we induce one on {(V::A)} for any A via: (V:f:A)+(V:g:A) relates z to a iff: f relates u to a, g relates v to a and u+v = z or one of f, g relates z to a, but a is not a right value of the other. This works regardless of whether we have an addition on A; it is symmetric, associative and cancellable because the addition on V has these properties, so {(V::A)} is a linear context.
Given a linear context, V, describe U as closed under composition iff u, v in U implies u&on;v in U (which we can construe as U respects composition).
## Linearity
I'll entertain the possibility that context may wish to deal with some further (V:f|V), which respect addition, as scalings, though I'll shortly impose some restrictions on how far context can push this. The first restriction is simply that context must provide a self-inverse conjugation on scalings, ({scalings}: z* ← z |{scalings}), whose restriction to real scalings is the identity and for which: for any scaling z, (z* + z) and z*&on;z are real scalings. In the absence of non-real scalings, one may use the identity as conjugation.
Given scalings on two linear contexts, U and V, we can use a
Describe a relation (V:g|U) which respects addition as: linear precisely if it commutes with all scalings; and as antilinear precisely if, for every scaling z, z*&on;g = g&on;z. If all scalings are real, all relations which respect addition are both linear and antilinear.
The scalings of a linear context form a linear context [I still need to verify that its addition is cancellable] isomorphic to its own {scalings}; I need to introduce a representation of the isomorphism classes of the relevant kind of linear context, via which to abstract scalars with {scalars for V} being a canonical representation of the isomorphism class of which {scalings of V} is a member, thereby establishing a natural (multiplicative) action of scalars on linear contexts, which I need to show will ensure {scalars for V} subsumes {scalars for U} whenever there is some linear (V:|U) of which at least one output is not an additive identity.
I'll describe {linear ({scalars for V}: |V)} as the dual of any given linear context, V, and write it dual(V).
Since several flavours of multiplication are going to come along, I'll distinguish three standard kinds by the binary operator I use to represent them:
.
e.g. n.v
where (at least) one of the values combined (n or v) is a scalar: this is the most mundane (and fundamental) variety. Formally, given the way I've defined scalars, when n is a scalar for V with v in V, n.v is synonymous with n(v).
·
e.g. f·v
where the binary operator is contraction (which I'd better explain later): this generalises the notion of an inner product. Each operand is in a linear context and there is a linear action of one on the other; e.g. when f is a linear map (U:|V) and v is in V, f·v is synonymous with f(v), but were v a linear (V:|W), f·v would be synonymous with f&on;v = (U: f(v(w)) ←w |W).
×
e.g. u×v
denoting the tensor product (which I'd better explain later): this generalises the notion of an outer product. Again, each operand is in a linear context, but there is no contraction involved.
Note that the first – and only the first – of these is always symmetric: n.v = v.n. When one operand is a scalar, all three are synonyms: and I'll usually use the first. The last is seldom symmetric unless synonymous with the first (the exception is when the operands are members of the same one-dimensional space). The second only gets to be symmetric by virtue of special circumstances: when an operand is a scalar; when one operand's linear context is the collection of symmetric linear maps from the other's linear context to this last's dual; there may be further cases, but I trust you get the idea - don't presume symmetry of multiplication except when I write it in the first form.
Note that, despite my notation generally borrowing from programming languages, I use * to denote generally any binary operator, not necessarily presumed to be in any sense multiplicative.
## Ramifications
For each v in any linear context V, define 1.v = v and, for every positive integer, n, (1+n).v = v+(n.v). Infer, in particular, that: for every positive integers n, m; (n+m).v = (n.v)+(m.v).
Notice that the identity (V:v←v|V) on V is exactly the relation I use to encode the collection V itself: this saves me having to have a mapping, Identity, which maps each linear context to its identity linear map. The identity is trivially linear in every linear context: indeed, it is trivially a scaling.
Given addition on U, for any A whatsoever, U's addition delivers an addition on {(V:|A)} defined by (V:f|A)+(V:g|A) = (V: f(a)+g(a) ←a |A); for any scaling n of V, there is an associated scaling (: n&on;f ←f :) of {(V:|A)}.
Given linear (W:g|V) and (V:f|U),
• g(f(u+v)) = g(f(u)+f(v)) = g(f(u)) + g(f(v)) so g&on;f is linear
Given any linear (V:g|U) and (V:f|U),
• (f+g)(u+v) = f(u+v) + g(u+v) = f(u) + f(v) + g(u) + g(v) = (f+g)(u) + (f+g)(v) so f+g is linear
In particular, using g=V (which we already know to be linear) and f=n.V for any positive integer n, we find that (1+n).V is linear whenever n.V is; hence, inductively, for all positive integer n. Further, 1.V = V = (V: v=1.v ←v |V) and, whenever n.V = (V: n.v ←v |V),
• (1+n).V = V + n.V = (V: v+n.v ←v |V) = (V: (1+n).v ←v |V)
so, inductively, n.V means what we might have hoped it would mean, namely (V: n.v ←v |V), for every positive natural n.
Given scalings (V:n|V) and (V:m|V), consider any linear (V:f|V); we're given that it commutes with n and m, so f&on;(n+m) = (V: f(n(v)+m(v)) ←v :V) = f&on;n + f&on;m = n&on;f + m&on;f = (V: n(f(v)) + m(f(v)) ←v :V) = (n+m)&on;f, so n+m commutes with f; likewise n&on;m&on;f = n&on;f&on;m = f&on;n&on;m so n&on;m commutes with f; this being for arbitrary f, we can infer that n+m and n&on;m are scalings. In particular, V being a scaling, n.V is a scaling for every positive integer n.
Given linear (V|f|U), its reverse is (U| u ← f(u) |V) which is trivially linear (though it may be a linear relation rather than a linear mapping - you'll notice the definitions above don't mention mappings). If f is a mapping, f&on;(:u←f(u):) is the identity on V; if its reverse is a mapping (a.k.a. f is monic) then (:u←f(u):)&on;f is the identity on V.
If (V:f|W) is linear and we're given (U:g|V) with g&on;f linear, can we infer that g is linear ? Only on (|f:), but for x, y in (|f:) we have u, v with x=f(u), y=f(v) and can apply g(x+y) = g(f(u)+f(v)) = g(f(u+v)) = g(f(u)) + g(f(v)) = g(x) + g(y); so g is linear if f is epic (i.e. (:f|) = V). Likewise, if g and g&on;f were linear, could we infer that f is linear ? No, but g must be unable to distinguish f(u+v) and f(u)+f(v), so f is linear if g is monic.
Given a scaling (V:n|V)
Written by Eddy. | 3,407 | 12,282 | {"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-2024-18 | latest | en | 0.941569 |
https://www.physicsforums.com/threads/vector-and-components-problem.130665/ | 1,606,262,777,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00610.warc.gz | 814,641,580 | 16,060 | # Vector and components problem
A radar station, located at the origin of xz plane, as shown in the figure, detects an airplane coming straight at the station from the east. At first observation (point A), the position of the airplane relative to the origin is R_vec_A. The position vector R_vec_A has a magnitude of 360m and is located at exactly 40 degrees above the horizon. The airplane is tracked for another 123 degrees in the vertical east-west plane for 5.0s, until it has passed directly over the station and reached point B. The position of point B relative to the origin is R_vec_B (the magnitude of R_vec_B is 880 m).
I'm suppose to find the ordered pair (x,z) for components of the vector R(AB), which I am suppose to be able to find by R(AB) = R(B) - R(A).
http://server6.theimagehosting.com/image.php?img=phytest.jpg
So far, I solved for the components of the vector of B, and the vector of A.
Vector A:
cos 40 = x/360; x = 276
sin 40 degrees = y/360; y = 231
= (276, 231)
Vector B:
I guess to use sin, cos, tan I need a right angle. So I do the bottom of B to do it. (123+40=163; 180-163=17 degrees)
cos 17 = x/880; x = 842
sin 17 = y/880; y = 257
= (842, 257)
Vector B - Vector A = (842, 257) - (276, 231) = (566, 26)
Which is wrong. What am I doing wrong? Should I be doing Vector B a different way? Or did I do the entire thing wrong?
Related Introductory Physics Homework Help News on Phys.org
Kurdt
Staff Emeritus
Gold Member
What you've forgotten to do is to take into account that the B x coordinate should be negative because you have crossed the perpendicular to the origin.
B = (-842,257)
easy mistake to make.
Well, that makes it:
(-842, 257) - (276, 231) = (-1118, 26)
Unfortunately it's still wrong, I guess I'll have to think of something else.
Kurdt
Staff Emeritus | 508 | 1,808 | {"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} | 3.890625 | 4 | CC-MAIN-2020-50 | latest | en | 0.927976 |
https://discuss.leetcode.com/topic/56349/o-n-java-solution-with-213-ms-runtime | 1,513,054,206,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515165.6/warc/CC-MAIN-20171212041010-20171212061010-00234.warc.gz | 556,146,482 | 9,297 | # O(N) Java Solution with 213 ms runtime
• ``````public class Solution {
class Point{
int x;
int y;
int index;
Point (int x, int y, int index){
this.x =x;
this.y = y;
this.index = index;
}
@Override
public boolean equals(Object o){
if(o instanceof Point){
Point b = (Point)o;
return this.x == b.x && this.y == b.y;
}
return false;
}
@Override
public int hashCode(){
return (new Integer(x).hashCode() + new Integer(y).hashCode());
}
}
public boolean isRectangleCover(int[][] rectangles) {
if(rectangles.length <=1){
return true;
}
int sum = 0;
int minX = Integer.MAX_VALUE;
int minY = Integer.MAX_VALUE;
int maxX = Integer.MIN_VALUE;
int maxY = Integer.MIN_VALUE;
HashMap<Point,Integer> hm = new HashMap<>();
for (int[] rect : rectangles){
Point p1 = new Point(rect[0],rect[1],1);
Point p2 = new Point(rect[2],rect[3],2);
Point p3 = new Point(rect[0],rect[3],3);
Point p4 = new Point(rect[2],rect[1],4);
sum += (rect[2] - rect[0])*(rect[3]-rect[1]);
minX = Math.min(minX,rect[0]);
minY = Math.min(minY,rect[1]);
maxX = Math.max(maxX,rect[2]);
maxY = Math.max(maxY,rect[3]);
Point[] points = {p1,p2,p3,p4};
for(Point p : points){
if(hm.containsKey(p) && hm.get(p) != p.index){
hm.remove(p);
}
else{
hm.put(p,p.index);
}
}
}
return (hm.size() ==4) && (sum == (maxX-minX)*(maxY-minY));
}
}
``````
• I really like your idea! WIth this method, we do not need to worry about how to arrange those rectangles.
The runtime can be significantly improved (from 200ms+ to around 80ms) by changing the hashCode() to:
return (new Integer(x).hashCode() * 31 + new Integer(y).hashCode())
Other improvements include:
• if(hm.containsKey(p) && hm.get(p) == p.index){ return false;}
• Directly contructs 4 points in an array.
With all those improvements, the runtime is 61ms.
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 541 | 1,865 | {"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-2017-51 | latest | en | 0.529915 |
https://www.painphr.com/q407_squared_four_over_one | 1,537,901,379,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162385.83/warc/CC-MAIN-20180925182856-20180925203256-00534.warc.gz | 830,503,619 | 7,031 | what is one over four squared
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Guide :
# what is one over four squared
1/4 squared
## Research, Knowledge and Information :
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### Isolate seven circles in three squares
The picture shows one solution. The squares don't have to be tilted: However, this gives the appearance of more than four squares the same size.
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### If a square based pyramind has sides of 20 what is the the height.
Take one of the equilateral triangles and drop a perpendicular from the top to the base. The perpendicular bisects the base and forms two back-to-back right-angled triangles. The length of the perpendicular is sqrt(20^2-10^2) by Pythagoras. That's sqrt(300)=10sqrt(3). Now view the pyramid side on an
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### what are the properties of the quadrilateral whose vertices are the center of the squares?
The quadrilateral so formed is a rhombus (a parallelogram in which all the sides are of equal length). The interior angles of the rhombus are 90+x and 90-x where x is the angle of the parallelogram and where 90+x are the measures of opposite angles of the rhombus. The adjacent angles add up to
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### cost of white washing on its four walls
tu get area av 4 walls av a room: leng=8, wide=6, hite=4 total leng=2*(8+6)=2*14=28 area=length*hite=28*4=112 (square meters)
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• Start your question with What, Why, How, When, etc. and end with a "?"
• Be clear and specific
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# AWA Prep
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Intern
Joined: 17 Sep 2007
Posts: 16
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### Show Tags
09 Oct 2007, 00:42
I have completely ignored the AWA section of the GMAT in my prep as well as when I give practice CATs. Any suggestions on how to ace this section in the actual GMAT exam. I've been told its good practice to give atleast 1 or 2 CATs with the AWA. But are there any good books with sample essays etc? Any other resources?
Intern
Joined: 10 Aug 2006
Posts: 5
Location: NYC
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### Show Tags
09 Oct 2007, 03:04
The Official Guide has sample essays near the back of the book.
Also in the OG, there's a complete list of all the topics in the question pool. It's a waste of time to go through them all very carefully,but you can get a very good idea of what you're up against.
I would strongly recommend writing an essay or two under the 30-minute time limit just for practice.
I wrote more about the AWA here:
http://www.gmathacks.com/gmat-awa/all-a ... t-awa.html
There's a section called "How to Prepare" near the bottom of the article.
--
http://www.gmathacks.com
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### Show Tags
09 Oct 2007, 11:11
Thanks. I read your post. So 250 words essay is not too short? What's the ideal length?
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### Show Tags
10 Oct 2007, 18:07
What practice materials are you using? Most of them should cover the AWA and provide templates and topics to practice with. I have Cracking the GMAT and it's got a really great chapter on the AWA, with key transitional words to use, a template, and tips.
10 Oct 2007, 18:07
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# AWA Prep
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# Find a recursive formula for the arithmetic sequence 8, - 2, - 12, - 22
+1
Answers (2)
1. 11 March, 19:05
0
xₙ₊₁ = xₙ - 10 with x₀=8
Step-by-step explanation:
Next element is the previous element minus 10.
2. 11 March, 19:48
0
Step-by-step explanation:
hello:
Un+1 = Un - 10
Know the Answer?
Not Sure About the Answer?
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Question: 71 of 74
# How to find the 2 minimum salaries in a table?
Manoj Thomas
Answered On : Aug 18th, 2006
set rowcount 2
select salary from employee order by salary
Gorilla Killa
Answered On : Sep 1st, 2006
The posted answer in wrong.Correct answer would beset rowcount 2select distinct salary from employee order by salaryThis will take care of any repeats.
visweswar
Answered On : Sep 13th, 2006
May be you only mistaken.
If the Query is for 2 minimum salary rows. Then even the Salary amount can be equal. In that case the 1st answer is correct.
sa10
Answered On : Sep 26th, 2006
You guys forgot to add "ORDER BY salary ASC"
vimal
Answered On : Sep 29th, 2006
Hi,
select min(salary) from employee where salary >(select min(salary) from employee)
Thanks,
Vimal.
gurveen singh rekhi
Answered On : Oct 11th, 2006
Vimal told us the second part .. ( the second minimum salary ) . For the first min its simple :
select salary, min(salary) from employee where salary >(select min(salary) from employee) group by salary.
Answered On : Oct 28th, 2006
The following article can clarify your doubt:
http://searchoracle.techtarget.com/ateQuestionNResponse/0,289625,sid41_cid504195_tax301455,00.html
Hrushikesh Thite
Answered On : Oct 30th, 2006
select TOP 2 * from emp_salary
order by salary desc....
this is quite simple.....
ashish gag
Answered On : Feb 19th, 2007
select top 2 salary from sal order by salary asc;
Shailendra
Answered On : Mar 13th, 2007
hi,
Since you are using the order by clause, please be informed that asc should not be mentioned, as Order by cluase by default sorts the required column in the descending order....
amy
Answered On : Mar 20th, 2007
In oracle, you should do this to get the 2 minimun salaries:
select salary from (select salary from employee
order by salary asc)
where rownum <= 2
Hi All,
This should work by a simple self join as:
select b.sal
from
emp a, emp b
where
b.sal >= a.sal
group by b.sal
having count(b.sal) = X
Where X = ( if you put 1 : It will be the minimum salary
sly if you put 2 : It will be the second minimum salary
Sybase_Guru
Answered On : Apr 10th, 2007
To get the 2 minimum in a single statement
select b.salary from test a,test b where b.salary>=a.salary group by b.salary
having count(b.salary) in(1,2)
Lakshmi Mandava
Answered On : May 15th, 2007
Will this query works if the table field contains duplicate values?
Abhijeet
Answered On : May 28th, 2007
Vimal told us about finding the 2nd min salary...how about 3rd min will that query work? for finding 3rd,4rth and so on..you need self join?
since we are doing group by this should obviously remove duplicates. So the following query is still correct irrespective of duplicates.
select b.sal
from emp a, emp b
where b.sal > a.sal
group by b.sal
having count(b.sal) in (1,2,3...)
Abhijit S
Answered On : Aug 21st, 2007
The last query works, but only if the values in the operated column are distinct.
For this purpose one needs to make sure that the resultset on which the self join is to be done contains unique values:
Example:
select a.salary, count(a.salary) from (select salary from emp) a ,
(select salary from emp) b
where a.salary > b.salary
group by a.salary
having count(a.salary) in (1,2)
Abhijit S
Answered On : Aug 21st, 2007
Sorry, forgot to mention the 'distinct' keyword in the earlier query.
The query is:
select a.salary, count(a.salary) from (select distinct salary from emp) a , (select distinct salary from emp) b
where a.salary > b.salary
group by a.salary
having count(a.salary) in (1,2)
nivas4u
Answered On : Sep 28th, 2007
Above query returns 2nd min and 3rd min as forget to put "=" in query. Below query returns correct result event it have duplicate salary records.
select a.salary
from (select distinct salary from test) a ,
(select distinct salary from test) b
where a.salary >= b.salary
group by a.salary
having count(a.salary) in (1,2)
gomzy007
Answered On : Jun 16th, 2008
I think you can use below one..
select * from (select * from table_name order by asce)
where rownum >=2
rso2003
Answered On : Sep 14th, 2008
This is a more elegant solution:
select ee.name, ee.salary from employee ee
where 2<(select count(*) from employee e
where ee.salary< e.salary)
This will give you exactly the two minimum salaries in the table.
Thanks
rso2003
silambazhagan
Answered On : Jan 2nd, 2009
Top function is only working in sql server, please ignore it.
if you include any clolumn with max() function it will not give proper output.
guys, if you want to need minimum 2salary from emp table in syabse pleasefollowthe below query.
selectsalary fromemployee wheresalary < (
select min(salary) from employee a where 2 =
(select count(*) fromemployee b where a.salary > b.salary)
)
gcvpgeek
Answered On : Mar 4th, 2009
The query is
select max(sal) from emp where sal < (select max(sal) from emp where rownum <3) group by sal | 1,388 | 4,974 | {"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-2013-20 | latest | en | 0.897267 |
https://stats.stackexchange.com/questions/117442/factor-analysis-using-outliers-only-time-series | 1,571,106,230,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00058.warc.gz | 662,922,112 | 30,147 | # Factor analysis using “outliers-only” time series
Some background
I run a factor analysis of a time series $Y$ using a standard OLS model with n+1 independent variables $(F,X_1...X_n)$, where $F$ is the main factor (from an explanatory power perspective).
$X_i$ and $F$ are highly correlated ($\rho>0.8$) but on certain days $t$, $X_i(t)$ will diverge significantly from its estimated value using the coefficient of its regression against $F$ : $X_i=\beta_i F$. In other words the distribution of $\epsilon_i(t)=X_i(t)-\beta_i F(t)$ has fat tails.
The reason why those highly collinear factors are included in the regression is precisely the presence of outliers which have a significant impact on the dependent variable $Y$.
My approach
To get rid of the multicollinearity problem yet keep the outliers, I create new factors $X_i'$ which are constructed like this:
• $X_i'(t)=\epsilon_i(t)$ if $\epsilon_i(t)$ is in the top or bottom $n$ percentiles of $\epsilon_i$
• $X_i'(t)=0$ otherwise
so if $n=5$ for example, $X_i'$ will be zero 90% of the time.
I then regress $Y$ against $(F, X_1'...X_n')$
My questions
• Is there a flaw in the method?
• Are there alternatives that would help me capture that effect?
(I know that I could orthogonalise each $X_i$ vs $(F,X_0...X_{i-1})$ but the results are the same as and as unstable as regressing against the original factors).
Note: in practice, $Y$ represents the returns of a stock, $F$ represents the returns of the market and the $X_i...X_n$ represents other factors such as industry indices. | 410 | 1,556 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-43 | latest | en | 0.904791 |
https://maanumberaday.blogspot.com/2011/07/447.html | 1,726,051,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00188.warc.gz | 352,093,309 | 12,371 | ## Friday, July 22, 2011
### 447
447 = 3 x 149.
447 is a divisor of 444 - 1.
447 is the smallest number of convex quadrilaterals formed by 15 points in general position.
447 is 677 in base 8 and 377 in base 11. It is 313 in base 12.
447 is a number that cannot be written as a sum of three squares.
Air France Flight 447 from Rio de Janeiro to Paris crashed into the Atlantic Ocean on June 1, 2009. | 119 | 405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.94157 |
https://alexaanswers.amazon.com/question/2v8qteqG26OZsZAIxK3olw | 1,680,255,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00601.warc.gz | 111,216,690 | 14,054 | Food
# How many tablespoons is three and a half teaspoons?
There are 3 teaspoons in a tablespoon. So 3 and a half is slightly more than 1 tablespoon.
LIVE
Points 54
Rating | 45 | 174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.958821 |
https://nptel.ac.in/courses/Webcourse-contents/IIT-KANPUR/wasteWater/pbose%20quiz/Wastewater%20Treatment%20(5)%20(Solution).htm | 1,539,822,649,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511365.50/warc/CC-MAIN-20181018001031-20181018022531-00523.warc.gz | 744,330,736 | 6,745 | WASTEWATER TREATMENT-V
SOLUTION
1. Describe the differences in the nature of recycling between activated sludge process and trickling filter. (3)
Solution:
Activated Sludge Process · The sludge or biomass is recycled in the activated sludge process · Purpose of recycling is to maintain a high biomass concentration in the aeration tank Trickling Filter · Treated wastewater is recycled in the trickling filter · Purpose of recycling is to maintain adequate hydraulic loading rate, without changing organic loading rate, so that all portions of the filter may be wetted adequately all the time.
2. A tricking filter with the following dimensions is available. Depth: 2 m, Surface area: 150 m2. The media consists of stones of 7-10 cm diameter. This filter will be used to treat 0.6 MLD wastewater with BOD5 = 300 mg/L. The trickling filter will be operated in the high-rate mode, i.e., OLR: 0.48 – 0.96 Kg/m3/d, HLR: 10 – 40 m3/m2/d, re-circulation ratio: 1-2. Based on this information, calculate the expected BOD5 removal efficiency.
Hint:
, ,
Where, So = BOD5 in Raw Wastewater, mg/L
Se = Total BOD5 of settled effluent from the filter, mg/L
Sa = Total BOD5 of wastewater applied to the filter, mg/L
k = Treatability constant, 2.36
D = Depth of the Trickling Filter, m
Q = Total Flow rate applied to the filter without recirculation, m3/d
A = Surface Area of the Trickling Filter, m2
n = 0.5
V = Volume of the Trickling Filter, m3 (7)
Solution:
Volume of trickling filter = D.A = (2).150 = 300 m3
(The above value is inadequate and hence must be increased). So, let R = 2
Hence, HLR = . This value is adequate.
Now, to calculate BOD5 removal efficiency.
It is known that, , Where,
Also,
or, or,
or, ;
Therefore, BOD5 removal efficiency: or, 96.64% | 523 | 2,053 | {"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-2018-43 | latest | en | 0.778809 |
http://www.algebra.com/algebra/homework/Human-and-algebraic-language/Human-and-algebraic-language.faq.question.115282.html | 1,369,380,473,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704288823/warc/CC-MAIN-20130516113808-00005-ip-10-60-113-184.ec2.internal.warc.gz | 308,670,955 | 6,064 | # SOLUTION: 3x+4y=5 x-2x=-5
Algebra -> Algebra -> Human-and-algebraic-language -> SOLUTION: 3x+4y=5 x-2x=-5 Log On
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Word Problems: Translating English into Algebreze Solvers Lessons Answers archive Quiz In Depth
Question 115282: 3x+4y=5
x-2x=-5
Answer by jim_thompson5910(28595) (Show Source):
You can put this solution on YOUR website! | 161 | 532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2013-20 | latest | en | 0.709841 |
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1. 5(2x+4)2. -2(x-3)3. -8(4x+5)4. 3(-x+4)5. 7(4x-2)6. -2(-6x-1)7. 3(2x+1)8. -(x-12)9. -5(6x-2)10. 12(3x+5)
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A202424 Primes of the form n!*n!! - 1. 1
3, 17, 191, 13934591999, 414935135999, 841488455807999, 12256784251917004799999, 91886617089132974573617151999999, 20572604964026488636856632501862399999999, 624332713268595066448813603451600045741761894966886399999999999 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS For n > 4, the last digits of the prime numbers are of the form 999, 999, 99999, 999999,...,...99999. LINKS EXAMPLE 191 is in the sequence because, for n = 4, 4!*4!! - 1 = 24*8 - 1 = 191. MATHEMATICA a={}; Do[p=n!*n!!-1; If[PrimeQ[p], AppendTo[a, p]], {n, 10^3}]; Print[a]; PROG (MAGMA) a:=func< n | Factorial(n)*(&*[n..2 by -2])-1 >; [ a(n): n in [0..78] | IsPrime(a(n)) ]; // Bruno Berselli, Dec 19 2011 CROSSREFS Cf. A000142, A006882, A202426. Sequence in context: A088678 A195067 A158885 * A335343 A133991 A210898 Adjacent sequences: A202421 A202422 A202423 * A202425 A202426 A202427 KEYWORD nonn AUTHOR Michel Lagneau, Dec 19 2011 STATUS approved
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Last modified May 27 12:58 EDT 2022. Contains 354097 sequences. (Running on oeis4.) | 505 | 1,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-21 | latest | en | 0.57834 |
https://www.physicsforums.com/threads/the-mysterious-phenomenon-of-knotting-a-string-around-a-rod.903338/ | 1,723,128,257,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00596.warc.gz | 720,579,130 | 20,987 | The Mysterious Phenomenon of Knotting a String Around a Rod
• john101
In summary: It seems that as the string is twisted, it forms loops. These loops reduce the tension on the string, and when the loop is finally released, the tension is still present in the string.
john101
Hi.
I've been searching for some description of the following. Something that will explain it and perhaps even some maths about it.
Take a rod or a stick and tie a string around its middle.
Hang it so the string is unwound, still.
Start turning the rod so the string twists around.
Keep twisting.
If at any time you let the rod go the string unwinds again to rest.
Keep twisting and you can see the string twists to a point where it suddenly creates a knot.
If you at that point let the rod go it only unwinds to the knot but no further.
If you keep twisting it forms knot after knot until the whole string is a row of knots that won't unwind.
If you keep twisting a third set of knots pop into existence.
It seems the thickness of the string, the length of the string, the weight of the rod are factors that determing things.
Is this something that anyone anywhere has looked at and have words to describe it and perhaps even something that can predict and generally explain it.
Last edited:
It seems it's only possible to get a third set of knots. Probably depends on thickness of string.
Reminds me of the trick re folding a sheet of paper more than nine times. Mythbusters tried that one in a hangar using a roller and forklift. From memory they set a record.
Would it be possible to post a video or even a diagram of what you trying to describe? I don't understand what you mean.
Yes.
The string is fixed at the top and the rod is spun around and the string twists and at some point creates a knot that is difficult to unwind...etc.
Okay, I think I understand now.
The "knots" you describe are not really knots, so to speak. They are just loops that form on the string.
This is not my field of expertise, but I'm sure somebody else here might be able to help better. The mechanical engineering sub-forum might be the best place to look.
In the mean time, I believe this might be explained in terms of stresses on the string. As the string is twisted it gains torsional stresses. Each loop that forms in the string effectively reduces the torsional stress. If the string was rigid, these loops would introduce bending stress, and in that case, it would be the result of the string attempting to balance torsional and bending stresses. In the case of a normal string though, bending stresses are negligible, and the loops merely shorten the effective length of the string (increasing the potential energy of the rod by raising it up) -- still, however, each loop reduces the torsional stress of the string.
So why don't the loops always vanish when the rod is released? I'm guessing friction.
Again though, this isn't my area of expertise and someone in the mechanical engineering subforum that knows more about stresses could probably help better.
Hmmm.. makes sense. It seems the 'knotting' momentarily reduces tension in the string. That it doesn't unwind becuse of friction makes sense.
Then it looks to me that that tension is stored in the unwound loop.
When watching the formation of a loop they all seem form in the same way. As the knots form the 'string' becomes stiffer. And the rod rises and becomes harder to turn until the next knot forms.
Here's something related to this phenomena
http://www.livescience.com/43536-yarn-muscles-100x-stronger-human-muscles.html
A researcher twists and coils fish line to make a muscle that reacts to heat.
It also may be related to supercoiling but I couldn't find an example of a rubber band or string doing this although as a kid we would wind up model plane powered by rubberbands and watch the coil up when they couldn't twist anymore and that was essential to adding more flight time.
1. What is the mysterious phenomenon of knotting a string around a rod?
The mysterious phenomenon of knotting a string around a rod is a physics experiment where a string is wrapped around a rod and pulled tight in a way that creates a series of knots. The knots are formed due to the friction and tension between the string and the rod.
2. How is the knotting of a string around a rod related to physics?
The knotting of a string around a rod is related to physics because it involves the principles of friction and tension. These forces play a crucial role in the formation of the knots and can be studied and analyzed using mathematical equations and scientific theories.
3. What are the real-world applications of studying the knotting of a string around a rod?
The study of the knotting of a string around a rod has several real-world applications. It can be used to understand and improve the design of various mechanical systems that involve ropes and pulleys. It can also be used in the fields of engineering and architecture to design structures that can withstand strong forces and stresses.
4. Can the knotting of a string around a rod be used to create stronger knots?
Yes, the knotting of a string around a rod can be used to create stronger knots. By experimenting with different wrapping techniques and tensions, scientists have found that certain knot formations can increase the strength and stability of the knot. This knowledge can be applied in various industries, such as sailing and rock climbing, to create safer and more secure knots.
5. Are there any other factors that can affect the knotting of a string around a rod?
Yes, there are other factors that can affect the knotting of a string around a rod. The thickness and material of the string, the shape and texture of the rod, and the environment in which the experiment is conducted can all have an impact on the formation of the knots. This makes the phenomenon a complex and fascinating subject for scientific research.
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7K | 1,403 | 6,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.961412 |
https://www.physicsforums.com/threads/non-equilibrium.735809/ | 1,531,897,786,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590069.15/warc/CC-MAIN-20180718060927-20180718080927-00381.warc.gz | 956,504,875 | 18,570 | # Homework Help: Non equilibrium
1. Jan 30, 2014
### courtney101ann
A 15-g bullet is fired from a rifle. It takes 2.5 × 10^-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 715 m/s . Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet
I need step by step explains please
2. Jan 30, 2014
Welcome to PF :)
Please Show what you have done so far.
3. Jan 31, 2014
### klimatos
Neither the acceleration nor the speed can remain constant. The acceleration ceases as soon as the propellant stops acting on the bullet. The bullet speed will diminish steadily due to friction with the surrounding air.
4. Jan 31, 2014
As this question does not mention friction,it should be neglected.
5. Jan 31, 2014
### collinsmark
Hello courtney101ann,
Welcome to Physics Forums!
In the mean time, let me throw out a couple of questions for consideration:
• What is the change in the bullet's momentum?
• What is an impulse, and how is it defined? (Defined in the context of force, time, momentum, etc.)
[Edit: klimatos and adjacent: as it turns out, the acceleration doesn't need to constant (uniform) to determine the average, net force. The average, net force can be determined quite easily, in fact, even if the acceleration varies all over the place. The problem statement's assumption about the acceleration being constant is superfluous; it's an unnecessary assumption.
In addition, friction can be present too. It doesn't matter. Friction can be part of the "net" force. Take it or leave it. Whether friction is present or not doesn't change the final answer.]
Last edited: Jan 31, 2014
6. Jan 31, 2014
### haruspex
I presume you are suggesting using ΔE/Δs = ∫F.ds/∫ds. I'm afraid that would be an error. I see it several times a year in book questions posed on this forum.
Consider whether average force should be defined as ∫F.dt/∫dt or ∫F.ds/∫ds. I would argue that the appropriate definition would have to match that of average acceleration, namely Δv/Δt = ∫a.dt/∫dt. If acceleration is not constant then this will generally not be equal to ∫a.ds/∫ds.
7. Jan 31, 2014
### collinsmark
It's simpler than that. I'm just saying that the impulse equals the change in momentum.
[Edit: but yes, if you wish, I'm also saying that the average, net force is $\vec F_{ave} = \frac{\int_{t_1}^{t_2} \vec F_{net}(t) \ dt}{t_2 - t_1}$ But that's a little more complicated than I wanted to get into. So I'll summarize that by saying the impulse is also equal to the average net force times the time interval in question; $\vec J = \vec p_2 - \vec p_1 = \Delta \vec p = \vec F_{ave} \ \Delta t$ ]
Last edited: Jan 31, 2014
8. Jan 31, 2014
### haruspex
My mistake - I didn't read the information provided properly. I thought it gave the length of the barrel, not the time.
9. Jan 31, 2014
### Staff: Mentor
What is the speed at time zero?
What is the speed at time 2.5 × 10^-3 s?
What is the acceleration?
What is the relationship between force, mass, and acceleration?
What is the force?
10. Jan 31, 2014
It seems the OP is inactive.
However,only after answering @Chestermiller's questions,should any help be provided.
11. Feb 1, 2014
### courtney101ann
I can't answer @Chestermiller's question I only have the information I posted and as for work I don't have any I'm completely clueless we only got half way through the notes when we had an emergency release due to the snow and we havent returned to school yet and my teacher still wants it done by Monday
12. Feb 1, 2014
### courtney101ann
I do have my free body diagram and summations
13. Feb 1, 2014
### courtney101ann
This is the work
#### Attached Files:
• ###### 1391312586537.jpg
File size:
24.9 KB
Views:
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14. Feb 1, 2014
### Staff: Mentor
Do you have any idea what the answers to my first two questions are? The answers to these are mentioned right in your problem statement.
Chet
15. Feb 2, 2014 | 1,080 | 3,981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-30 | latest | en | 0.926994 |
https://freevideolectures.com/course/2561/algebra/32 | 1,721,020,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00627.warc.gz | 236,115,741 | 10,105 | x
# Algebra
Khan Academy, , Prof. Salman Khan
Updated On 02 Feb, 19
##### Overview
Contents:
Simple Equations - Equations - Linear Equations - Solving Inequalities - graphing lines - Slope and Y - intercept intuition - Slope - Equation of a line - Slope and Y-intercept Intuition - Averages - Integer sums -Taking percentages - Growing by a percentage - Another Percent Word Problem - More percent problems - systems of equations - Introduction to Ratios - Ratio problem with basic algebra - More advanced ratio problem - with Algebra
Alternate Solution to Ratio Problem - Introduction to Ratios - Advanced ratio problems - Age word problems - multiplying expressions - Solving a quadratic by factoring - i and Imaginary numbers - Complex Numbers - Introduction to the quadratic equation - Quadratic Equation - Completing the square - Quadratic Formula - Quadratic Inequalities - Introduction to functions - Functions - Domain of a function - Proof: log a + log b = log ab-Proof: A(log B) = log (B^A), log A - log B = log (A/B) - Proof: log_a (B) = (log_x (B))/(log_x (A))-Algebraic Long Division-Introduction to Conic Sections-Conic Sections: Intro to Circles
## Lecture 32: Complex Numbers (part 1)
4.1 ( 11 )
###### Lecture Details
Introduction to complex numbers. Adding, subtracting and multiplying complex numbers.
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