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https://www.superprof.co.uk/resources/academic/maths/algebra/polynomials/polynomial-formulas.html | 1,660,434,206,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571989.67/warc/CC-MAIN-20220813232744-20220814022744-00052.warc.gz | 891,927,988 | 20,839 | Chapters
What is Polynomial
The word poly means more than 1 or in easy words "many". The word nominal means mathematical expression. If we combine both definitions then polynomial means an algebraic expression or equation which contains more than one term. Almost all equations in algebra are polynomial equations. Whether it is a quadratic equation or straight-line equation, you can call it a polynomial equation.
However, polynomial equations have some limitations. These limitations are discussed in the previous post but let's just revise it here. There are two conditions for a polynomial expression or equation.
1. The exponent of the variable should be a positive integer
2. The coefficient should be a real number
3. There should be more than one mathematical expression
This means that if any condition is void then you can't call it a polynomial equation or expression. Let's create some expressions and you try to think either the expressions are a polynomial equation or not?
The first three and the sixth option are polynomial expressions and the rest are not. If you look carefully, the first two expressions satisfy all the conditions and therefore we can declare it a polynomial equation. However, in the third expression, there is a square root on a variable with a square. If we simplify it then it will be which means that it is a polynomial expression.
In the case of the fourth expression, you can see a negative sign with the exponent. The exponent is an integer, there is no doubt but remember that the integer should be a positive number otherwise it won't be a polynomial expression. In the fifth one, the same reason but a different approach. The variable is in fraction form. It might look like a polynomial but don't let the question deceive you. If you reciprocate it, the exponent will be converted into a negative term and hence negative exponent voids the condition of the polynomial.
The sixth one is also a polynomial. The reason is that it fulfills all the conditions. The name of this polynomial is monomial. Last but not least, the seventh expression has a problem. The variable and the exponent satisfy their conditions but the coefficient has a problem. The coefficient of this expression is a negative number inside a square root. In simple words, it is an imaginary number (a number which doesn't exist in real life). In conclusion, it is not a polynomial expression.
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What is the difference between an expression and an equation?
To easily understand, let's examine the below equation and expression:
The first one is the algebraic expression and the next one is the algebraic equation. How can we differentiate between an equation and an expression? The word equation means to equate to something. If you look at the equation, the left-hand side is equal to the right-hand side in this case, it is equal to zero. It is not necessary that the left-hand side is always zero, it can be any number or variable. The keyword is the "is equal to" symbol. If you see this symbol that means it is an equation.
The word expression means something which is expressed. It doesn't mean to express it in words but it just an expression telling you something. In the first option, you see an expression that tells you the variables, coefficients, and constants. In the mathematical world, we deal with the equations mostly because you can't do anything to an expression (unless you have been asked to factorize it).
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Formulas
There are many formulas for different polynomial expressions or equations. Before we start with the formulas, you should know the categories of polynomials. There are three types of polynomial, below are the types of polynomials.
1. Monomials
2. Binomials
3. Trinomials
Monomials
This means if the exponent and the variables are the same then you can take out as common and add or subtract their coefficients.
If the bases (which is the variable in this case) are the same and there is a multiplication sign then it means that the exponent will be the sum of both and the coefficient will be the product of both.
If the bases (which is the variable in this case) are the same and there is a division or ratio sign then it means that the exponent will be the difference of the second term from the first term and the coefficient will also be divided in that order.
If there is an exponent on a term that also has an exponent then the resulting exponent will be the product of both exponents.
Binomials
We can split the term binomial into two which are bi and nomial. The word bi means 2 and nomial mean mathematical expression. In other words, if there are two mathematical terms then that expression or equation will be binomial. Below are the formulas for binomials.
Trinomials
Like binomial, we can also break trinomial into two terms which are tri and nomial. Tri means "3" and nomial means mathematical term. In conclusion, we will declare an equation or expression as trinomial if the equation or expression contains "3" terms. Below are all the formulas for trinomials equation or expression. | 1,361 | 5,912 | {"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} | 4.75 | 5 | CC-MAIN-2022-33 | latest | en | 0.927817 |
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# It’s scarily easy to see how much money you’ve spent at Amazon
Amazon has completely changed the way people shop in every market where the company operates. No, it certainly wasn’t the first e-commerce company. No, it wasn’t the first web retailer to offer deep discounts. And no, it wasn’t the first company to provide a single online marketplace where thousands of third-party sellers can offer products directly to consumers. But it was the first company in many markets to check all those boxes and really resonate with shoppers.
How long have you been buying products on Amazon? Five years? Fifteen years? Maybe even 20 years? If you had to guess, how much money do you think you’ve spent in all that time? Well, you don’t have to guess because we’re going to show you exactly how to figure it out in just a few easy steps.
We’ve seen this tip on a bunch of different websites over the years, but it was posted most recently on Business Insider. In a nutshell, Amazon allows you to download spreadsheets that show all of your orders over a certain period of time. Of course, you can easily set that time span so that it covers your entire order history, and then use a simple formula to add up the grand total.
Here’s what you need to do:
1. First, visit this page on the Amazon website.
2. In the Orders section, click on Download Order Reports (you’ll be prompted to log in if you haven’t already).
3. In the Start Date box, put January 1 of the oldest available year. Then click the Use today link next to the End Date box.
4. Click Request Report. After a few moments, you’ll be prompted to download a “.CSV” file.
5. Save the file on your PC and then open it using Microsoft Excel. If you don’t have Excel, you can use Google Docs for free by visiting http://drive.google.com.
6. In the cell at the bottom of the column showing your transaction totals, input the formula =SUM(X1:X1000) and then hit enter. In place of “X1” and “X1000,” be sure to enter the actual cell names for the first and last numbers in that column.
So what’s your total? Remember, it might seem like an insanely high number if you’ve been an Amazon customer for a long time, but just think about how much money you’ve saved on all those orders by purchasing from Amazon instead of pricier retailers. | 517 | 2,296 | {"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-2023-06 | latest | en | 0.930145 |
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# John has 20 ounces of a 20% of salt solution. How much salt
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John has 20 ounces of a 20% of salt solution. How much salt [#permalink]
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10 Oct 2006, 07:03
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
John has 20 ounces of a 20% of salt solution. How much salt should he add to make it a 25% solution?
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10 Oct 2006, 07:22
Add a ounces more of salt, we got
(a+4)/(20+a)=1/4 => a=4/3
So 4/3 ounces more of salt pliz!
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10 Oct 2006, 08:14
quangviet512 wrote:
Add a ounces more of salt, we got
(a+4)/(20+a)=1/4 => a=4/3
So 4/3 ounces more of salt pliz!
I have a doubt: is the volume supposed to increase or is it constant? If I had not seen your work, I'd have assumed that volume would remain at 20ounces.
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10 Oct 2006, 08:53
Obviously the volume will increase.
It must be 4/3 ounces.
I have explained a technique called alligation so that questions on mixtures like these can be answered fast...........
http://www.gmatclub.com/phpbb/viewtopic ... 676#251676
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12 Oct 2006, 05:38
cicerone wrote:
Obviously the volume will increase.
It must be 4/3 ounces.
I have explained a technique called alligation so that questions on mixtures like these can be answered fast...........
http://www.gmatclub.com/phpbb/viewtopic ... 676#251676
Cicerone,
Can you please explain how you would apply the alligation technique for this problem? Thanks!
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12 Oct 2006, 05:47
Just saw your explanation for this problem in the other thread. Thanks much, Cicerone!
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13 Oct 2006, 02:00
so (20+x)*25/100=20*20/100 + x
solving for x = 4/3
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13 Oct 2006, 12:21
Using the alligation technique cicerone so kindly informed us of :
20----------------------------100
--------------25-----------------
75------------------------------5
(20/75) * 5 = 20/15 = 4/3
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14 Oct 2006, 06:30
this is the best technique ever!!
thanks Ciceron
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# John has 20 ounces of a 20% of salt solution. How much salt
Moderator: chetan2u
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,303 | 4,082 | {"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} | 3.984375 | 4 | CC-MAIN-2017-51 | latest | en | 0.891749 |
https://blog.csdn.net/gongweixin2018/article/details/113532813 | 1,618,884,845,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038921860.72/warc/CC-MAIN-20210419235235-20210420025235-00374.warc.gz | 242,341,969 | 22,718 | # 每日算法 -- 最长公共子串
package 牛客;
/**
* 最长公共子串
* 题目描述
* 给定两个字符串str1和str2,输出两个字符串的最长公共子串，如果最长公共子串为空，输出-1。
* 示例1:
* 输入 "1AB2345CD","12345EF"
* 返回值 "2345"
* 核心思路：（动态规划）
*
*/
public class NC127 {
/**
* 自己动态规划求解法超时
* 重复比较，耗时严重
*/
public static String LCS (String str1, String str2) {
//零界值判断
if (str1.isEmpty()||str2.isEmpty()) {
return "-1";
}
//定义一个数组确定哪些是比较过的
int len1 = str1.length(), len2 = str2.length();
int[][] dp = new int[len1][len2];
//找到相同字符
String result = "";
for (int i =0;i<str1.length();i++){
String temp = "";
for (int j =0;j<str2.length();j++){
if(dp[i][j]==0){
if(str1.charAt(i)==str2.charAt(j)){
//判断连续相等字符
String s1 = str1.substring(i,len1);
String s2 = str2.substring(j,len2);
if(!s1.isEmpty()&&!s2.isEmpty()){
temp = getMaxlen(s1,s2,dp,i,j);
}
}
if(temp.length()>result.length()){
result = temp;
temp = "";
}
}
if(str2.length()-j<result.length()){
break;
}
}
}
return result;
}
private static String getMaxlen(String str1,String str2,int[][] dp,int indexi ,int indexj){
String result = "";
int len = str1.length()<str2.length()? str1.length():str2.length();
for (int i = 0;i<len;i++){
if(str1.charAt(i)!=str2.charAt(i)){
return result;
}else{
dp[indexi++][indexj++]+=1;
result+=str1.charAt(i);
}
}
return result;
}
/**
* 大神的动态规划
* @param str1
* @param str2
* @return
*/
public static String LCS2(String str1, String str2) {
if (str1.isEmpty()||str2.isEmpty()) {
return "-1";
}
int n1 = str1.length(), n2 = str2.length();
int[][] dp = new int[n1 + 1][n2 + 1];
int maxLen = 0;
int x = 0;
for (int i = 1; i <= n1; i++) {
char ch1 = str1.charAt(i - 1);
for (int j = 1; j <= n2; j++) {
char ch2 = str2.charAt(j - 1);
if (ch1 == ch2) {
dp[i][j] = dp[i - 1][j - 1] + 1;
if (dp[i][j] > maxLen) {
maxLen = dp[i][j];
x = i;
}
}
}
}
return maxLen == 0 ? "-1" : str1.substring(x - maxLen, x);
}
/**
* 滑动窗口算法
* @param str1 string字符串 the string
* @param str2 string字符串 the string
* @return string字符串
*/
public static String LCS1(String str1, String str2) {
// write code here
StringBuilder sb = new StringBuilder();
int start = 0, end = 1;
while (end < str1.length() + 1) {
if (str2.contains(str1.substring(start, end))) {
if (sb.length() < end - start) {
sb.delete(0, sb.length());
sb.append(str1, start, end);
}
} else {
start++;
}
end++;
}
if (sb.length() == 0) {
return "-1";
}
return sb.toString();
}
public static void main(String[] args) {
System.out.println(LCS("1AB2345CD","12345EF"));
System.out.println(LCS1("1AB2345CD","12345EF"));
}
}
03-28 2150
08-21 723 | 908 | 2,480 | {"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} | 3.484375 | 3 | CC-MAIN-2021-17 | latest | en | 0.148029 |
https://smart-answers.com/mathematics/question12713465 | 1,603,788,313,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893845.76/warc/CC-MAIN-20201027082056-20201027112056-00229.warc.gz | 535,536,650 | 35,745 | , 12.11.2019 08:31, calmicaela12s
# What is the quotient (6x4 − 15x3 + 10x2 − 10x + 4) ÷ (3x2 + 2)? 2x + 6 2x − 6 2x + 4 2x − 4, r = 1
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# Q2) Determine which of the following subsets of R^n are in fact subspaces of R^n (n>2)?
a) {X│x1 x2=0}
b) {X│AX=b,A_(m×n )≠0 and b_(m×n )≠0}
c) {X│x_i≥0}
d) { X ∑_(j=1)^n▒x_j =1}
Relevance
• kb
Lv 7
7 years ago
a) Not a subspace.
Counterexample (n = 2): (1, 0) and (0, 1) are in the subset, but their sum (1, 1) is not.
b) Not a subspace, since 0 is not in the space.
c) Not a subspace, since it is not closed under scalar multiplication.
For instance, although (1, ..., 1) is in the subset, -1 * (1, ..., 1) = (-1, ..., -1) is not.
d) Not a subspace, since it is not closed under scalar multiplication.
For instance, although (0, ..., 0, 1) is in the subset, 2(0, ..., 0, 1) = (0, ..., 0, 2) is not.
I hope this helps! | 301 | 795 | {"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-2020-45 | latest | en | 0.912934 |
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Help
0
211
1
Work out the area of a rectangle with base, b = 18mm and perimeter, P = 50mm.
Oct 3, 2018
#1
+4
+1
18+18=36
50-36=14
14/2=7
7*18=126
Area=126
Oct 3, 2018
#1
+4
+1
18+18=36
50-36=14
14/2=7
7*18=126
Area=126
moffitt110242 Oct 3, 2018 | 141 | 265 | {"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} | 3.6875 | 4 | CC-MAIN-2019-30 | latest | en | 0.691247 |
https://www.heizung-ts.com/dvvjsr/0e8181-exponential-growth-equation | 1,652,799,280,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517485.8/warc/CC-MAIN-20220517130706-20220517160706-00197.warc.gz | 952,591,804 | 13,130 | # exponential growth equation
of Half lives, y=16 Need help with a homework or test question? Do NOT follow this link or you will be banned from the site. A population of French children are offered a riddle, which shows an aspect of exponential growth: "the apparent suddenness with which an exponentially growing quantity approaches a fixed limit". {\displaystyle x(t)=x(0)e^{kt}} ) T Initially, the small population (3 in the above graph) is growing at a relatively slow rate. t. Suppose a culture of This bias can have financial implications as well. t Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. Exponential growth is a specific way that a quantity may increase over time. t Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function. DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. The function’s initial value at t=0 is A=3. Log in to rate this practice problem and to see it's current rating. Do you have a practice problem number but do not know on which page it is found? a. Since the population is said to be growing, the growth factor is b = 1 + r. y = ? Years. Amazingly, the original handful of bacteria will blossom into a colony of nearly a thousand in one day’s time. Half-life is the time it takes for half the substance to decay. For any fixed b not equal to 1 (e.g. where b is a positive real number not equal to 1, and the argument x occurs as an exponent. r The king readily agreed and asked for the rice to be brought. However, as the population grows, the growth rate increases rapidly. Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC. What is the half-life of the substance when the initial amount is 100g? \[\begin{array}{rcl} 100 Solution: The initial size of the account is \$100, so A=100. Find an expression for the number of bacteria after t hours. That would be $$10/2=5$$ grams at time $$t=20$$ days. Exponential growth occurs when the instantaneous rate of change of a quantity with respect to time is proportional to the quantity itself. When $$b > 1$$, we call the equation an exponential growth equation. Half-Life . The rate of increase keeps increasing because it is proportional to the ever-increasing number of bacteria. For a nonlinear variation of this growth model see logistic function. The larger the value of k, the faster the growth will occur. The variable k is the growth constant. is the growth rate (for example, a When a population becomes larger, it’ll start to approach its carrying capacity, which is the largest population that can be sustained by the surrounding environment. Carbon-14 has a half-life of 5730 years. Exponential growth models apply to any situation where the growth is proportional to the current size of the quantity of interest. A half-life, the amount of time it takes to deplete half the original amount, infers decay. In the case of a discrete domain of definition with equal intervals, it is also called geometric growth or geometric decay since the function values form a geometric progression. If τ < 0 and b > 1, or τ > 0 and 0 < b < 1, then x has exponential decay. Suppose a material decays at a rate proportional to the quantity of the material and there were 2500 grams 10 years ago. . y = ? How much remains after 75 days? The growth of a bacterial colony is often used to illustrate it. For example, comparing $$f(t)=t^2$$ and $$g(t)=2^t$$, notice that $$t$$ is in the exponent of the $$g(t)$$, so $$g(t)$$ is considered an example of exponential growth but $$f(t)$$ is not (since $$t$$ is not in the exponent).
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https://audiologysource.com/fourier-transform/ | 1,607,093,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00518.warc.gz | 197,430,881 | 6,569 | # Fourier Transform
Written by Travis M. Moore
Last edited 26-Sep-2019
### Time and Frequency Domains
The concept of time and frequency "domains" might sound intimidating, but it's really just describing two different ways to look at the same sound. In fact, we have already been viewing signals in both domains already. The time domain representation is simply a plot with amplitude on the y axis and time on the x axis. The frequency domain plots amplitude on the y axis and frequency on the x axis. Take a look at Figure 1 to refresh your memory.
The time domain view of a sine wave is the increases and decreases in amplitude over time: the typical shape of a "sine wave" we have come to know and love. What we don't know from the time domain plot, at least not from just a quick glance, is the frequency of the wave. (For the smarty-pants out there, yes, we could use the time axis to calculate the period of a cycle, then the frequency, but that's hardly information "at a glance.") The frequency view of a sine wave explicitly tells us what frequency (or frequencies) we're dealing with, but nothing about the duration of the sound.
So far we have been dealing with known frequencies, so we have just plotted in the time or frequency domain as necessary. However, what if we didn't know the frequency ahead of time? Or, given just the frequency, how would we be able to view the morphology of the wave over time? Well, a mathematician by the name of Joseph Fourier figured out how to do this, and he did so in the early 1800s, which means by hand. The concept is fairly simple, but extremely complex to carry out.
Moving from the time domain to the frequency domain is admittedly tricky. Luckily we do not need to concern ourselves with the math. All you need to know is that given any periodic (repeating) signal plotted as amplitude over time, the Fourier transform can figure out the sine waves (i.e., frequencies) needed to build (compose) that time waveform. This is called the Fourier Transform.
The reverse ("inverse") condition is moving from the frequency domain to the time domain. Appropriately, this is called the inverse Fourier transform. We have already seen sine waves combine to make a wave shaped like a square and a triangle. For example, dumping a bunch of odd harmonics (along with amplitudes!) into a Fourier transform is all that's needed to compose the square wave in time. You may recall learning about this process in the module on additive synthesis. | 533 | 2,491 | {"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-2020-50 | latest | en | 0.954642 |
https://www.coursehero.com/file/6630763/688-Physics-ProblemsTechnical-Physics/ | 1,529,660,440,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864387.54/warc/CC-MAIN-20180622084714-20180622104714-00340.warc.gz | 792,827,738 | 137,287 | 688_Physics ProblemsTechnical Physics
# 688_Physics ProblemsTechnical Physics - 28 Electric Fields...
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28 Electric Fields ANSWERS TO EVEN PROBLEMS P23.2 (a) 2 62 10 24 . × ; (b) 2.38 electrons for every 10 9 present P23.36 (a) 200 pC; (b) 141 pC; (c) 58.9 pC P23.38 see the solution P23.4 57 5 . N P23.40 (a) 1 3 ; (b) q 1 is negative and q 2 is positive P23.6 2 51 10 9 . × P23.42 electron: 4 39 . Mm s ; proton: 2 39 . km s P23.8 514 kN P23.10 x d = 0 634 . . The equilibrium is stable if the third bead has positive charge. P23.44 (a) 57 6 . ± i Tm s 2 ; (b) 2 84 . ± i Mm s; (c) 49.3 ns P23.46 (a) down; (b) 3 43 . C µ P23.12 (a) period = π 2 3 md k qQ e where m is the mass of the object with charge Q ; (b) 4 3 a k qQ md e P23.48 The particle strikes the negative plate after moving in a parabola 0.181 mm high and 0.961 mm. P23.50 Possible only with + 51 3 . C µ at x = − 16 0 . cm P23.14 1 49 . g P23.16 720 kN C down P23.52 (a) 24 2 . N C at 0 ° ; (b) 9 42 . N C at 117 ° P23.18 (a) 18 0 218 . ± ± i j kN C; P23.54 5 25 . C µ (b) 36 0 436 . ± ± i j e j mN P23.56 (a) mg A B cot θ + ; (b) mgA A B cos sin θ θ + P23.20 (a)
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22nd November 2007, 10:38 AM #1 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 30,810 WTC1 - A bit of simple geometry This is just something fairly obvious that occurred to me concerning that well-known conspiracist question: "Why didn't the tops of the WTC towers fall off?" Let's just look at WTC1 for a moment. Tower height = 1368 feet 1368/110 = 12.4 feet per storey Collapse initiated at the 97th floor, so the upper block had 13 floors. 13x12.4 = 161 feet high. Assume the centre of gravity is half this height, so it's about 80 feet above the base of the falling part of the tower. For the top part to fall off, we need it to tip over far enough that this centre of gravity is outside the footprint of the tower. Now the sides of the WTC towers were 208 feet long, so the edge was 104 feet from the centre. Assuming an axis of rotation right on the centre of the building - the most favourable place for sideways movement of the centre of gravity - if the top part somehow rotated all the way over on to its side, its centre of gravity would therefore still be 24 feet inside the footprint of the tower. Move the axis closer to the edge and the tower can't tip that way, because there would have to be some force lifting it upwards. Place the axis towards the opposite edge and the centre of gravity ends up even further inside the footprint. Therefore, there's no conceivable way the top of WTC1 could have fallen off - it's a simple geometrical impossibility. WTC2 requires some thought and analysis - it requires about a 42 degree rotation about the most favourable axis to get the centre of gravity outside the footprint, which requires that the lower corner has crushed its way through about 140 feet more of building than the upper corner - but for WTC1 it's fairly trivial to prove that it's a stupid question. Dave __________________ Inspiring discussion of Sharknado is not a good sign for the audience expectations of your new high-concept SF movie sequel. - Myriad | 487 | 2,132 | {"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} | 3.671875 | 4 | CC-MAIN-2020-24 | latest | en | 0.953351 |
https://scicomp.stackexchange.com/questions/31149/mass-matrix-and-how-to-handle-it-odes-references | 1,713,730,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00037.warc.gz | 443,920,916 | 42,434 | # Mass Matrix and how to handle it (ODEs) - References
I'm interested in a good reference/paper about how to handle numerically a mass-matrix system as
\begin{align} \mathbf{M}(t,y)\dot{y} =F(y,t) \end{align}
I know that such a problem can be solved by using classical ODEsolvers in MatLab such as ode15s, and others. Of course, due to copiright issues, it's not possible to read the source code and, IMHO, there's a lack of reference.
I've found something interesting in Hairer & Wanner's books (both volumes), but honestly there's not so much stuff. In such problems, the non-singularity of $$M(t,y)$$ plays a fundamental role, and I'm quite interested in some techniques.
Suppose to have a not-time-dependent and non-singular mass matrix $$M(t,y)=M(y)$$ and a system \begin{align} M(y)\dot{y} =F(y,t) \end{align}
and say I want to compute the numerical solution with Backward Euler (or any implicit method, in order to avoid numerical instabilities)
So, one can write $$\dot{y} = M^{-1}(y) f(y,t):= \tilde{F}(y,t)$$ and formally we have
\begin{align} y_{n+1}=y_n + \Delta t \tilde{F}(t_{n+1},y_{n+1})=y_n+ M(y_{n+1})^{-1} \Delta t F(t,y_{n+1}) \end{align}
and hence
\begin{align} M(y_{n+1}) [y_{n+1}-y_n]=\Delta t F(t_{n+1},y_{n+1}) \end{align}
Another interesting problem is how to solve with Newton's method this non-linear system of equations. As far as I know, a simplified Newton's method is used, but I can't find any reference about it.
I'd be very grateful if someone has references/hints or something else !
EDIT (after Bill's comment) [REMARK: I'd like not to copy explicitely MatLab's approach] For the simplfied Newton's method, a possible approach could be the following.
Starting from the functional we have to set to zer written above, I'd like to differentiate w.r.t. $$y_{n+1}$$, namely
\begin{align} J_n=\frac{\partial (M(y_{n+1}) \cdot y_{n+1})}{\partial y_{n+1}}- \Delta t \frac{\partial F(t_{n+1},y_{n+1})}{\partial y_{n+1}} \end{align}
and consider the constant matrix $$M(y_{n+1})$$ at each step, so the jacobian would become
\begin{align} J_n=M(y_{n+1})-\Delta t \frac{\partial F(t_{n+1},y_{n+1})}{\partial y_{n+1}} \end{align}
Could it be a good starting point?
EDIT$$^2$$ [02/03/19']
I tried to use the previous approach to solve simple autonomous and non-singular problems like
\begin{align} \begin{bmatrix} y(1) & 0 \\ 0 & y(2) \end{bmatrix} \dot{y} = [y(1),\sin(y(2))] \end{align}
with $$y(1)(0)=y(2)(0)=1$$ and it worked in the right way.
Also with other non-linear and more complicated problems, the numerical solution was ok.
I also tried to generalize this approach to the trapezoidal rule: startinf from $$M(y) \dot{y}=f(y)$$, in the hypotesis that $$M$$ is non-singular, I have \begin{align} \dot{y}= M^{-1}(y)f(y)=\tilde{f}(y) \end{align}
So, the standard trapezoidal rule leads to : \begin{align} y_{n+1}=y_{n} +\frac{\Delta t}{2}(M^{-1}(y_{n+1})f(y_{n+1})+M^{-1}(y_n)f(y_n)) \end{align} and multiplying first for one and then for the other inverse, one gets
\begin{align} M(y_n)M(y_{n+1})[y_{n+1}-y_n]=\frac{\Delta t}{2} (M(y_n)f(y_{n+1})+M(y_{n+1})f(y_n)) \end{align}
The jacobian has been computed with the same logic as before, "freezing" the $$M(y_{n+1})$$ term.
A rapid numerical experiment with the same problem above shows the correct order of convergence.
Any other hints, suggestions, comments, are really appreciated.
• On the contrary, the matlab ode solvers including ode15s are well documented (IMHO). I suggest starting with this reference: mathworks.com/help/pdf_doc/otherdocs/ode_suite.pdf Feb 28, 2019 at 16:47
• @BillGreene you're right! Honestly, I read it some time ago, and probably too much quickly! At page 8 there's an approach which is quite familiar with mine
– VoB
Feb 28, 2019 at 16:53
• Edited my post, adding a possible way to handle Newton's method
– VoB
Feb 28, 2019 at 17:23
• A quick question, don’t you need commutativity between $M(y_n)$ and $M(y_{n+1}$ in your trapezoidal rule derivation (last equation in Edit 2)? Is there a reason to assume that such commutativity holds? Mar 2, 2019 at 20:11
• With the test problem I used above, the order was right with both the expressions, but your question makes sense: there's nothing that ensures commutativity for that product
– VoB
Mar 3, 2019 at 9:52
Ignoring Newton's method here is the wrong approach! The fact that you're using Newton's method is what makes this cheap to add, and is what makes singular mass matrices possible. Essentially look at Implicit Euler. You'll notice that the only matrix you need to use in your quasi-Newton steps is
$$W = (I - \gamma J)$$
and if you have a mass matrix, this becomes
$$W = (M - \gamma J)$$
This is true for any BDF/SDIRK/Rosenbrock method, since you can simplify any of their rootfinding equations to
$$M u_{n+1} - \gamma f(u_{n+1},p,t) - C = 0$$
where $$C$$ is all of the stuff that doesn't depend on $$u_{n+1}$$. So with that small change to your Newton method, you now converge to the solution with the mass matrix for any of these (semi-)implicit schemes. Newton's method just uses this for a linear solve, so you only ever have to LU it, never inverse. If $$M$$ is singular, you can still have that $$W$$ isn't, and also just use something robust like QR-factorization, and then this will still converge. A singular mass matrix is another representation of a DAE (use the terms without a derivative to define a constraint relation).
(Note that for higher order methods, you may have to handle FSAL separately with an $$M$$ linear solve.)
• The equations you wrote are OK, (I assume $\gamma$ is the step-size $\Delta t$). Looking at mine, I see that the third your equation (the one I have to set to zero) is the same I have written in the first edit, i.e. : $M(y_{n+1}) [y_{n+1}-y_n]=\Delta t F(t_{n+1},y_{n+1})$. So actually the matrix I need in my quasi-Newton's method is $M(y_{n})- \Delta t J$, where $J=\frac{\partial f(t_{n+1},y_{n+1})}{\partial y_{n+1}}$. This seems to be, basically, the same thing you wrote. As you said, I could LU it at each time step in order to save computational cost. Am I wright?
– VoB
Mar 4, 2019 at 8:47
• Well, actual ODE solvers are much more complicated than that. $\gamma$ is proportional to $\Delta t$ and is dependent on the diagonal of the tableau (which happens to be 1 for Implicit Euler), and you don't necessarily have to recompute $W$ every step, but yes the general idea is there that there's no reason to invert $M$ and then $W$ since it can be done simultaneously, and doing it simultaneously as numerically sound and handles the singular $M$ case. Mar 4, 2019 at 10:48 | 1,995 | 6,636 | {"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": 28, "wp-katex-eq": 0, "align": 10, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-18 | latest | en | 0.904196 |
http://dic.academic.ru/dic.nsf/eng_rus_technic/56/ratio | 1,493,116,478,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120338.97/warc/CC-MAIN-20170423031200-00504-ip-10-145-167-34.ec2.internal.warc.gz | 95,095,065 | 15,103 | # 4:1:1 ratio это:
4:1:1 ratio
отношение частот дискретизации 4:1:1
Англо-русский словарь технических терминов. 2005.
### Смотреть что такое "4:1:1 ratio" в других словарях:
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• 1 + 2 + 4 + 8 + · · · — In mathematics, 1 + 2 + 4 + 8 + hellip; is the infinite series whose terms are the successive powers of two. As a geometric series, it is characterized by its first term, 1, and its common ratio, 2. :sum {i=0}^{n} 2^i.As a series of real numbers… … Wikipedia
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• 1.85:1 — Format de projection Cinéma … Wikipédia en Français
• Ratio — ist ein lateinischer Begriff und wird häufig mit Vernunft oder Verstand übersetzt. In der Mathematik dagegen bedeutet das Wort meist „Verhältnis“ beziehungsweise „Quotient“ und weist darauf hin, dass eine Größe zu einer anderen in Beziehung… … Deutsch Wikipedia
• Ratio — This article is about the mathematical concept. For the Swedish institute, see Ratio Institute. For the academic journal, see Ratio (journal). For the philosophical concept, see Reason. For the legal concept, see Ratio decidendi. The ratio of… … Wikipedia
• Ratio distribution — A ratio distribution (or quotient distribution ) is a statistical distribution constructed as the distribution of the ratio of random variables having two other distributions.Given two stochastic variables X and Y , the distribution of the… … Wikipedia
• 4:3 — Unter Seitenverhältnis im weiteren Sinne versteht man das Verhältnis von mindestens zwei unterschiedlich langen Seiten eines Polygons. Meistens wird damit das Verhältnis von der Breite eines Rechtecks, eines Bildschirms oder einer Leinwand zu… … Deutsch Wikipedia
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Нажмите правой клавишей мыши и выберите «Копировать ссылку» | 786 | 2,654 | {"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} | 2.875 | 3 | CC-MAIN-2017-17 | longest | en | 0.318506 |
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1. using integral test for sum of a series
the question is :
Determine whether the integral test can be used to determine the convergence or divergence of the following series. If it can be used, use it to determine the convergence or divergence. If the integral test can not be used, explain why.
$\sum_{n = 0}^{\infty} \frac{\cos(n)}{n}$
what i see is that
$-1\leq\cos(x)\leq1$
so
$\frac{-1}{x}\leq\frac{\cos(x)}{x}\leq\frac{1}{x}$
so because it is not always positive we can not use the integral test.... is this correct?
2. Maybe the amount from one to infinity?
$\sum\limits_{n = {\color{red}{1}}}^\infty {\frac{{\cos n}}{n}}$
3. I don't see how changing the lower index from 0 to 1 helps at all.
cos(2)= -0.41614 which is negative and the integral tests, as acosta0809 suggests, does not apply here.
4. sorry! yes initial n=1 | 263 | 893 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2016-50 | longest | en | 0.793966 |
https://www.varsitytutors.com/intermediate_geometry-help/how-to-find-an-angle-in-a-pentagon | 1,716,161,787,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00522.warc.gz | 961,955,298 | 43,519 | # Intermediate Geometry : How to find an angle in a pentagon
## Example Questions
### Example Question #1 : How To Find An Angle In A Pentagon
The angles at 3 verticies of a pentagon are 60, 80 and 100. Which of the following could NOT be the measures of the other 2 angles?
Possible Answers:
Correct answer:
Explanation:
The sum of the angles in a polygon is
For a pentagon, this equals 540. Since the first 3 angles add up to 240, the remaining 2 angles must add up to
### Example Question #1 : How To Find An Angle In A Pentagon
What is the sum of three angles in a hexagon if the perimeter of the hexagon is ?
Possible Answers:
Correct answer:
Explanation:
The perimeter in this question is irrelevant. Use the interior angle formula to determine the total sum of the angles in a hexagon.
There are six interior angles in a hexagon.
Each angle will be a sixth of the total angle.
Therefore, the sum of three angles in a hexagon is:
### Example Question #1 : How To Find An Angle In A Pentagon
Add four interior angles in a regular pentagon. What is the result?
Possible Answers:
Correct answer:
Explanation:
Use the interior angle formula to find the total sum of angles in a pentagon.
for a pentagon, so substitute this value into the equation and solve:
Divide this number by 5, since there are five interior angles.
The sum of four interior angles in a regular pentagon is:
### Example Question #4 : How To Find An Angle In A Pentagon
What is the sum of two interior angles of a regular pentagon if the perimeter is 6?
Possible Answers:
Correct answer:
Explanation:
The perimeter of a regular pentagon has no effect on the interior angles of the pentagon.
Use the following formula to solve for the sum of all interior angles in the pentagon.
Since there are 5 sides in a pentagon, substitute the side length .
Divide this by 5 to determine the value of each angle, and then multiply by 2 to determine the sum of 2 interior angles.
The sum of 2 interior angles of a pentagon is .
### Example Question #5 : How To Find An Angle In A Pentagon
Suppose an interior angle of a regular pentagon is . What is ?
Possible Answers:
Correct answer:
Explanation:
The pentagon has 5 sides. To find the value of the interior angle of a pentagon, use the following formula to find the sum of all interior angles.
Substitute .
Divide this number by 5 to determine the value of each interior angle.
Every interior angle is 108 degrees. The problem states that an interior angle is . Set these two values equal to each other and solve for .
### Example Question #1 : How To Find An Angle In A Pentagon
Let the area of a regular pentagon be . What is the value of an interior angle?
Possible Answers:
Correct answer:
Explanation:
Area has no effect on the value of the interior angles of a pentagon. To find the sum of all angles of a pentagon, use the following formula, where is the number of sides:
There are 5 sides in a pentagon.
Divide this number by 5 to determine the value of each angle.
### Example Question #1 : Pentagons
True or false: Each of the five angles of a regular pentagon measures .
Possible Answers:
True
False
Correct answer:
False
Explanation:
A regular polygon with sides has congruent angles, each of which measures
Setting , the common angle measure can be calculated to be
The statement is therefore false.
### Example Question #2 : How To Find An Angle In A Pentagon
True or false: Each of the exterior angles of a regular pentagon measures .
Possible Answers:
False
True
Correct answer:
True
Explanation:
If one exterior angle is taken at each vertex of any polygon, and their measures are added, the sum is . Each exterior angle of a regular pentagon has the same measure, so if we let be that common measure, then
Solve for :
The statement is true.
### Example Question #9 : How To Find An Angle In A Pentagon
Given: Pentagon .
True, false, or undetermined: Pentagon is regular.
Possible Answers:
False
True
Undetermined
Correct answer:
Undetermined
Explanation:
Suppose Pentagon is regular. Each angle of a regular polygon of sides has measure
A pentagon has 5 sides, so set ; each angle of the regular hexagon has measure
Since one angle is given to be of measure , the pentagon might be regular - but without knowing more, it cannot be determined for certain. Therefore, the correct choice is "undetermined". | 1,029 | 4,431 | {"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} | 4.3125 | 4 | CC-MAIN-2024-22 | latest | en | 0.8281 |
https://jhlchan.wordpress.com/ | 1,500,669,615,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423809.62/warc/CC-MAIN-20170721202430-20170721222430-00305.warc.gz | 636,645,160 | 35,547 | My research and investigations
Fermat’s Little Theorem
I have a joint paper with my supervisor, on combinatoric and group-theoretic proofs of Fermat’s Little Theorem in HOL4.
It all started when I came across the Necklace proof of Fermat’s Little Theorem, which is different from the textbook number-theoretic proof. The number-theoretic proof is also standard in automatic theorem provers, e.g. it is a worked example in HOL Light Tutorial. I was wondering how to put a combinatoric proof in HOL4.
It turns out that it is just a matter of counting with sets. There are tricks to set up bijections to establish the sizes of sets. Digging deeper, I find that there is a kind of symmetry underlying the bijections, and I am surprised to find group action is involved. Eventually I use the Orbit-Stabilizer Theorem to prove the Necklace Theorem, thereby proving Fermat’s Little Theorem.
At the end, we did several proofs of the theorem in HOL4, made a comparison, and gave the paper this title: “A String of Pearls: Proofs of Fermat’s Little Theorem”. The paper was accepted by CPP 2012, and I gave a talk at the conference held in Kyoto this December.
During this process, I have build up a library of group theory in HOL4. This should help my further work in finite fields.
Math – Finding Patterns
Everyone learn math at school, but it’s a subject most would rather forget. Simple calculations are fine. Solving problems is a headache. Proving theorem is like climbing mountains.
However, mathematician study math with pleasure. What are they studying, and what pleasure do they get?
If you’re looking for a short answer to the question “What’s math?”, here is one: “Math is the study of patterns.”
Yes, to mathematicians, there are patterns — visible or hidden — in their imaginary math world. They seek pleasure in finding patterns, show them off, perhaps apply them to dig out deeper patterns.
You don’t believe this? I’ll let YOU find a pattern.
Math deals with many things: numbers, shapes, changes, structures, patterns, patterns of patterns, … Let’s start with the basic stuff: numbers.
What patterns are hidden in numbers? Nothing much for 1, 2, 3, 4, 5, ….
How about the squares? 1, 4, 9, 16, 25, 36, 49, ….
A pattern is something that repeats. To find a pattern, we may need to see a long list. In this case, compute as many squares as possible — the first 100 squares, 1000 squares, even more. Perhaps you are willing to compute the first 10 squares. Beyond that, you know you have the skills (just multiply a number with itself), but you’d rather not to waste your time doing so. Nowadays, there are computers (most likely you’re using one right now) which can compute at almost lightning speed. I’ll help you to use the computer for this task.
However, it’s rather tricky to teach you how to program (give instructions to) a computer to perform calculations.
Or is it? You’re using a browser to read this blog. You can instruct the browser to ask the computer to do calculations!
Modern browser knows something called Javascript, and you can do wonderful things with it. Also, it helps if your browser allows new tabs. If not, you’ll need to open a new browser instance, in order to keep reading this.
So click for a new tab in your current browser, copy the following line (highlight it, right-click, copy), and paste it in the new tab’s address box (right-click the long blank rectangle near the top, paste). Press ENTER (or click “->”).
`javascript:m=100;for(n=1;n<=m;n++)document.write(n+' sq = '+n*n+'<br>')`
On iPad/iPhone, just select the above line, copy, switch to Safari, tab address box, clear by (x), then paste, “Go”.
You should see the first 100 squares printed in your browser page, one square per line. I won’t teach you Javascript here. I’ll just say that m=100 above means “set maximum to 100”, and “for…” will repeat the computation (in this case printing the squaring line) for the maximum number of times.
If you close the tab, start an new one, and copy/paste again: this time change m=100 to m=500 before “->”. The browser will print the first 500 squares, faster than you can scroll!
Try this in your browser now, and see if you can spot any pattern in the squares.
Steve Jobs and Apple
We’ve all shared the news. Steve Jobs had left a legacy that perhaps no one can match. I still remember my Apple ][ — with high-resolution color graphics and 64KB memory, no hard disk, but boot from 5.25″ floppy — very “modern” in 1980!
To appreciate the achievements of Steve Jobs, you have to understand his philosophy. We can get a lot of inspiration from Wikipedia:
I am really inspired to read how Steve talks about himself, in his 2005 Stanford Commencement Address, with video (also this Chinese translation: 喬布斯三個故事啟發人心).
Last year, Steve Jobs was named “Person of the Year for 2010” by the Financial Times. The article ends with this story from the Apple CEO when Steve left the company in 1985:
“In his autobiography, John Sculley, the former PepsiCo executive who once ran Apple, said this of the ambitions of the man he had pushed out: “Apple was supposed to become a wonderful consumer products company. This was a lunatic plan. High-tech could not be designed and sold as a consumer product.” How wrong can you be.”
However, this was quoted from John Sculley’s autobiography in 1987. Since then, John understands Steve better, and this interview in 2010 is really informative.
Did you ever wonder why the Apple Logo has a bite? Here is the story.
Oh yes, Steve is a Beatles fan. Here is an interesting episode.
The Proof Path
When doing a math proof, it’s like walking along the math landscape: your proof is a path (a logical path) from the hypotheses (starting point) to the conclusion (finishing point).
The HOL theorem-prover provides an interactive proof manager. When it shows “all goals proved”, it is indeed a revelation: suddenly the logical path is complete.
Because the logical path is taken by the machine, you’re sure that there are no logical gaps — no missing steps. This is the advantage of using the theorem-prover: it won’t allow logical gaps.
However, you can be driven into logical blind alleys — the theorm-prover won’t tell you. This is because along the proof path, there are many decision points: which symbol to induct on? should cases of a symbol be taken?
Ideally all paths will lead to the same goal, but some paths are easier to walk, while other paths go through mountains or deserts. Some even lead to a door — you just can’t open it without a key. The key may be nearby, or far away; but you need to find the key to open the door if you want to continue on with that particular path. That’s the key idea to overcome the stumbling block.
All in all, a theorem prover is a good guide for the proof path, as you can see the effect of tactics, and where you’re going. Most goals will branch off into subgoals, and it’s tedious to keep track of subgoals by hand — better let the machine do the book-keeping.
Key idea in a proof
To spell out all the logical details of a proof, as required by a theorem-prover, can be routine — not very illuminating, even boring. The HOL theorem-prover provides various rewrite tactics and simplification sets to ease the task, but picking out the appropriate tools to do the job still needs experience.
However, sometimes you’ll hit a stumbling block: something that looks simple and must be true, but you just can’t see any simple proof of it. That’s when the proof becomes interesting.
This is not related to a particular theorem prover. Ask any mathematician and he or she will tell you stories.
The solution of the “stumbling block” usually involves a key idea.
For example, given a number N, how to show that there is a prime greater than N? There is no known formula to generate primes. It was Euclid, around 300 BC, who recorded this key idea: use N to generate a number M such that M cannot be divided by any primes up to N. By case-analysis of this M, the infinitude of primes can be proved.
Of course, during case-analysis of M, you’ll need to show that any number must be a product of primes. If this is a stumbling block, you’ll need another key idea. And so on.
How to find the key idea? Using examples or pictures can help visualizing the stumbling block. After some deep thoughts, the key idea will dawn upon. Sometimes you need to look outside the box. And you’ll need pencil-and-paper to try out the key idea, as the theorem prover will not allow unproven stuff.
To find the key idea in a proof is almost an art — there is no algorithm for it.
Theorem Proving by Induction
The HOL theorem-prover includes tactics on induction, a standard technique for proofs involving a recursive data structure.
For example, the whole numbers can be constructed recursively as follows: starting from 0, successive numbers can be obtained by adding 1 to the predecessor: n to n+1. So if a property P can be shown such that:
1. P is valid for 0, and
2. If P is valid for n, it is also valid for n+1.
Then we can claim that the property P is valid for all whole number n.
Similarly, the list data structure can be constructed recursively as: starting from the empty list [], elements are inserted into the list (h::t), with head h and tail t. If tail t = [], the list has one element [h]. If tail = [k], the list has two elements [h; k], and so on.
Corresponding to the induction for proofs involving whole number properties, there is Structural Induction for proofs involving lists:
1. Property P is valid for [], and
2. If P is valid for list t, it is also valid for list (h::t).
The we can claim that the property P is valid for all lists.
Usually the base case (1) is trivial, easily proved by some fundamental results from definition.
The trick to prove the step case (2) is to match the pattern of the induction hypothesis: P is valid for t, to the pattern in the goal: P is valid for (h::t).
Sometimes the definition provides decomposition of the pattern (h::t) to t, then the step case is easy to match.
Otherwise, the best strategy is:
• figure out what’s the stumbling block,
• formulate a lemma that can solve the stumbling block,
• prove the lemma.
It is the solving of stumbling blocks that make theorem-proving endeavours interesting.
Theorem Proving in HOL
Recently I’ve been doing actual theorem proving using HOL. This is a real eye-opener. Here is a summary of my first experience:
• Simple theorems can be proved just by using definitions.
• It’s not good to prove everything from definition — this makes the proof tedious.
• Some fundamental results should first be established by definition — they become theorems for proving more complex results.
This corresponds to the working mode of mathematicians. To build a theory one starts from definitions, properties, lemma, then theorems. There are many proof techniques, called tactics in HOL, and choosing the most efficient one needs a lot of experience.
Meanwhile, HOL upon loading includes some basic theories, and there is mechanism to load in more specific theories if required. Users of HOL can write up the proofs in scripts, and the Holmake utility can compile the script and generate the theory file and binary files, ready for loading so that the theorems will be available in other HOL sessions.
HOL is a very useful too, and I’m learning the basics of the tool. The HOL documentations are a big help here. I alway learn a lot by following the theory scripts provided in the source and examples. | 2,643 | 11,542 | {"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} | 3.5625 | 4 | CC-MAIN-2017-30 | longest | en | 0.955726 |
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Re: SMU Cox 60K scholarship vs McCombs Austin UT [#permalink]
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28 May 2013, 06:33
Guys, I couldn't decide. I ended up putting my deposit down for both schools. I am going to take the next month to talk to international students to help me make a decision.
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Re: SMU Cox 60K scholarship vs McCombs Austin UT [#permalink]
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12 Jan 2015, 12:20
Hi,
So where did you finally matriculate ?
And the pros and cons of your decision that you feel now ?
_________________
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Re: SMU Cox 60K scholarship vs McCombs Austin UT [#permalink] 12 Jan 2015, 12:20
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 812 | 2,867 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-09 | longest | en | 0.887281 |
https://wikimho.com/us/q/3dprinting/8482 | 1,670,651,003,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711712.26/warc/CC-MAIN-20221210042021-20221210072021-00136.warc.gz | 682,576,493 | 13,464 | ### After some time stepper motor is hot
• I have three stepper motors.
One Nema 17 - 2.4 ohm, the second smaller noname from color printer - 9.5 ohm and third the smallest noname from cdrom - 10.5 ohm.
I have connected them to arduino mega 2560 with ramps 1.4(set to 1/32 micro stepping) and drivers drv8825. See my previous question.
After some time (less than one minute) the first is cold. The second motor is hot. And the third is very hot. I can not even touch it.
What can I do to fix it.
there should be labels what they are rated for on the smaller motors.
Note that you should unlock your steppers when not in use. When an "idle" stepper gets hot, it's using energy to keep rigidly locked at a precise rotation (often for no good reason).
@Davo, they work constantly.
• The second motor is hot. And the third is very hot. I can not even touch it.
This is to some degree, completely normal and expected. From the datasheet for a typical NEMA 17 stepper, the rated temperature rise is 80 °C above ambient and the maximum operating temperature is 130 °C (implying an ambient temperature of 50 °C). It is normal that stepper motors (in general) get a bit hot.
"Too hot to touch" is still relatively cold. 60 °C is already too hot to touch, and that's only a 40 °C rise above a 20 °C ambient temperature.
You can reduce the temperature rise of the motors by reducing the current they receive. The stepper driver has a small potentiometer that can be turned to adjust the current, but keep in mind that doing so will also reduce the torque of the motors and thus they might skip steps if you reduce the current too much.
Technical details: Note that stepper motor drivers used in 3D printers are constant current drivers, and the little potentiometer controls the current. If you had not paid much attention to this potentiometer, the drivers might all have been set for the same constant current of $$1.0\ \text A$$. The stepper driver would (to achieve the same constant current) send a higher voltage to the higher resistance motors. This would imply a power dissipation of $$2.4\ \text W$$ in the Nema 17, and a power dissipation of $$10.5\ \text W$$ in the small stepper. $$2.4\ \text W$$ in the Nema 17 would only heat it up by about $$20\ °\text C$$ above ambient. A dissipation of $$10\ \text W$$ in the small stepper, which also has much less surface area to dissipate the power, would heat it up by a lot (and probably, given that you didn't fry it, the current was set lower -- or a technical peculiarity limited the current given that the motor likely also has very low inductance).
2 of his motors are *not* NEMA 17 but salvaged from other machines.
Stepper motors, in general, may be rated to operate at higher temperatures. "Too hot to touch" could still be completely normal. I am just using that datasheet (for a NEMA17) as one example of how hot steppers can normally get.
true. note though that a typical CD-drive Stepper with 5V/10 Ohm is like *20* (or 24) steps per turn...
Thank you! I need to check it.
I have reduced current by reduced voltage from 0.65 V to 0.15(cold)-0.2(warm) V for cd motor and 0.2(cold)-0.4(warm) V for color printer motor.
@burtsevyg can you measure the current instead? but reducing the current to about a third means that about that is for the current too, which means about 1/9th of the power dissipated.
@Trish A reference voltage of 0.65V translates to a current of 1.2A, so the current has been reduced to around 300-500mA for the small motor and 500-700mA for the large one.
@Trish I will try. Should I measure the current on single сoil? | 922 | 3,616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.945652 |
https://www.unitconverters.net/speed/foot-minute-to-centimeter-minute.htm | 1,721,029,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00758.warc.gz | 897,762,555 | 3,377 | Home / Speed Conversion / Convert Foot/minute to Centimeter/minute
Convert Foot/minute to Centimeter/minute
Please provide values below to convert foot/minute [ft/min] to centimeter/minute [cm/min], or vice versa.
From: foot/minute To: centimeter/minute
Foot/minute to Centimeter/minute Conversion Table
Foot/minute [ft/min]Centimeter/minute [cm/min]
0.01 ft/min0.3048 cm/min
0.1 ft/min3.048 cm/min
1 ft/min30.48 cm/min
2 ft/min60.96 cm/min
3 ft/min91.44 cm/min
5 ft/min152.4 cm/min
10 ft/min304.8 cm/min
20 ft/min609.6 cm/min
50 ft/min1524 cm/min
100 ft/min3048 cm/min
1000 ft/min30480 cm/min
How to Convert Foot/minute to Centimeter/minute
1 ft/min = 30.48 cm/min
1 cm/min = 0.032808399 ft/min
Example: convert 15 ft/min to cm/min:
15 ft/min = 15 × 30.48 cm/min = 457.2 cm/min | 259 | 787 | {"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-2024-30 | latest | en | 0.33613 |
https://studylib.net/doc/10595938/20-3---- | 1,603,275,568,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876307.21/warc/CC-MAIN-20201021093214-20201021123214-00252.warc.gz | 555,235,783 | 13,515 | # 20 3 ) (
```The Chain Rule , finding the derivative of a composition.
A composition of functions is a function , u(x) inside another function, f(u), making a
function of x, f(u(x)).
For example
f (x)
The inside function is
x 3 x 20
3
u ( x ) x 3 x 20 .
3
The outside function is f ( u ) u
If x changes , then u changes and then f changes.
Consider two linear functions,
u( x)
1
f ( u ) 3 u The composition is
x
2
1
f ( u ( x )) 3
x
2
3
x
2
The slopes multiplied. The chain rule says the same is true for nonlinear functions.
The Chain Rule:
df
df du
dx
Examples in which
f ' (u )u ' ( x )
du dx
u( x) x 3 :
2
a ) h ( x ) ( x 3)
2
5
h ' ( x ) 5( x 3) ( 2 x )
2
4
Here h ( x ) f ( u ( x )) where f ( u ) u
5
f ' ( u ) u ' ( x ) 5u ( 2 x ) 5( x 3) ( 2 x )
4
b) f ( x)
1
x 3
2
2
( x 3)
2
2
1
2x
2
f ' ( x ) ( x 3) 2 x
4
x
2
3
2
f(x) in part b could also be done using the quotient rule.
More examples : Find the derivative of each function.
1)
f ( x ) ( x 1)
2)
g ( x ) ( 2 x 5)
3)
h ( x ) ( x 1)
4)
f (x)
3
3
2
x
( x 1)
2
2
4
5)
g(x)
2x
3
5x 7
2
6) A snowball is a perfect sphere with volume given by V
(r)
4
r
3
cubic cm.
3
Find the rate of change of the volume with respect to the radius.
If the radius is decreasing by 2 cm per min. due to melting, how fast is the volume
decreasing when the radius is 30 cm?
7) The revenue as a function of x=number of units sold in the month is given by
R ( x ) 0 . 2 x 40 x . If the quantity sold is decreasing by 12 units per month,
2
at what rate is revenue changing with respect to t=time in months if x=150?
``` | 757 | 1,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-45 | latest | en | 0.704174 |
http://mathsfunplaynlearn.com/2015/10/30/mathematical-symbols-and-its-representation/ | 1,516,729,562,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00018.warc.gz | 212,595,422 | 23,083 | # Mathematical symbols and their representation
This post is a response to the article, ‘Why 5+5+5 doesn’t always make 15: Maths exam question divides the internet’.
Mathematical symbols convey information for the reader to interpret. In the example of question 1 shown above, is it fair for the child to have one mark deduced for ‘wrong’ representation?
When does ‘5 x 3 = 15’ be presented in form of 5 + 5 + 5 = 15 or 3 + 3 + 3 + 3 + 3 = 15?
5 x 3 = 15 in form of repeated addition; 3 + 3 + 3 + 3 + 3 = 15
• 5 times of 3, when ‘times of’ is represented by the multiplication sign ‘x’
• 5 groups of 3, when ‘groups of’ is represented by the multiplication sign ‘x’
5 x 3 = 15 in form of repeated addition; 5 + 5 + 5 = 15
• 5 multiply 3 times, though this is usually not used in communication but the statement stands true.
Hence, both 3 + 3 + 3 + 3 + 3 = 15 and 5 + 5 + 5 = 15 are correct, depending on the interpretation of the reader. In the case of the paper setter, he/she interpret it as ‘5 times of 3’ / ‘5 groups of 3’. The child interpret it as ‘5 multiply 3 times’. Though the syllabus focuses on the former interpretation, that does not mean a child should be penalised for another logical interpretation.
If the required answer is 3 + 3 + 3 + 3 + 3 = 15, this question can be modified in the following suggestions:
1. Solve 5 x 3 by completing the repeated addition statement.
5 x 3 = 3 + 3 + ____________________ = __________. (Answer: 3 + 3 + 3 = 15)
2. Based on the diagram below, solve 5 x 3 by using repeated addition.
(Answer: 3 + 3 + 3 + 3 + 3 = 15)
This kind of questions reminds me of a comic I posted before in my post ‘Do schools kill creativity?’ | 503 | 1,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-05 | longest | en | 0.886641 |
https://www.jiskha.com/display.cgi?id=1360722134 | 1,503,319,064,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108268.39/warc/CC-MAIN-20170821114342-20170821134342-00190.warc.gz | 922,609,571 | 3,795 | # cherokee
posted by .
If the regular price is \$
12.00 and the discount is 25% what is the sale price?
• math -
0.75 * 12 = ?
• cherokee -
12.00*.25=3 so 12.00-3.00= sale price of \$9.00
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More Similar Questions | 648 | 2,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-34 | latest | en | 0.87234 |
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WanderingWinder
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The Key to Big Money Part I: Money Density
« on: January 19, 2012, 08:53:14 am »
+4
In a big-money kind of deck, there's really two concepts you need to be aware of; the first is money density, the second is opportunity cost.
Money density is the average value in coin production of cards in your deck, i.e. copper produces one, silvers two, estates and such 0. It's important to keep in mind that, based on 5-card hands, you need a money density of 1.6 to buy a province and 2.2 for a colony. You need only 1 for duchies or dukes, and less for things like gardens, islands, tunnels, whatever.
Calculating your money density is very simple if you know what's in your deck: add up all the production values of the money, divide by the total cards in your decks. So for your initial deck, you have 7*1 for the coppers +3*0 for the estates, all divided by the 10 total cards for a money density of 0.7.
Branching out slightly, you probably want to buy at least one card that's not a silver or gold or province or duchy, right? How do other cards fit in to money density? Well, the simplest are cards like Woodcutter. Woodcutter (at least, the first one) provides an obvious benefit over silver in that it gives you a buy. But, for all intents and purposes, it still counts as \$2 in your money density.
There's another very simple, very common kind of card to deal with when making your money density calculations: cantrips. (I'm using 'cantrip' here to define any kind of card that always draws at least one card and gives at least one action back to you). Cantrips are what I call, for the purposes of money density calculations, 'virtual cards'. What I mean by that is, because they replace themselves totally in your hand, they don't count toward the total count of cards which you're using as the denominator for your money density calculations. So, if you buy a village and a militia with your two starting buys (not, by the way, a good strategy), you have 7 coppers, 3 estates, 1 village, 1 militia, producing 7, 0, 0, and 2 money respectively and with a total of 7, 3, 0, and 1 cards to count against your deck total. Your total money density is therefore 9/11 = .818181.....
Further expanding on that, if you get a slightly more interesting (in this respect anyway) card, the peddler, into your deck, you've increased your effective deck size by 0 (because it's a cantrip), but as it produces \$1 extra, you've increased your buying power by one. Add peddler to your starting deck, and you have \$8 total money in 10 effective cards for a density of \$0.8.
Okay, once you get that down, you need to think about terminal collision. I think that most of you know that buying only treasures and VP won't get you very far in terms of success (or fun). So you probably want to buy some terminals, and by the end of the game, you probably want to buy more than one. This creates some chance that your terminal actions will collide. The big key to playing big money decks is weighing out the benefits that actions provide you with versus the chances that they collide. Of course, with non-terminals, you don't have to worry about that, but very often, you're better served by taking the risk at some point. Fortunately, calculating the chances for terminals collision isn't too hard in general, you just have to remember to use your effective deck size rather than the actual number of cards in your deck. As for figuring out which benefits are worth it... well, I'll let you guys work that out for yourselves. Just keep in mind that you aren't optimizing your results in a vacuum, you have to beat another player. Which means, generally, that you have to count on yourself getting a little luckier than you should expect to on average, because in those really unlucky cases, you've probably already lost anyway. And the amount you have to count on yourself getting lucky, i.e. the amount of risks you have to take, increases more with the more players you add to the game. Villages will help to ease these wrinkles, but you have to get the village together in the hand that the terminals collide in, which doesn't happen so often as people think.
Of course, this leads us to the very important subject of terminal card-draw. In general, by the time you're mixing multiple terminal draws... you're probably engine building*. And for engine building, things like getting your engine to be able to fire consistently and having a sufficient payload are far more important than the money density concept. But for one to two terminal drawers, this money density look at things is still quite effective. For your first terminal drawer, it's a virtual card to your deck, once more, and then you have a percentage (based on the size of your deck) of having a larger handsize. After all, the reason why average card value is important is so that you can calculate the average value of your hand. So take smithy; if I have 2 silvers, a gold, a smithy, and my starting cards in the deck, that's 13 effective cards, 14 total money, and you've got your chance of getting a 7 card hand rather than a 5. Calculating the exact probability is not as easy as you might think, given how reshuffles work. But you can come up with ways to approximate it. And then for your second (and every subsequent) terminal card draw, they do count as cards in your deck, with 0 money value. So adding a second smithy to your deck, you have a higher chance of getting your 7 card hand, but your money density has dropped from 14/13 to 14/14 (or \$1).
Understanding money density is also helpful in understanding how much your deck will stall out. A deck with 3 gold, 7 silver, 7 copper and 3 estates has a money density of \$1.5. A deck with 1 gold, 3 silver, 2 copper, and a chapel has a money density of 11/7, or just over \$1.57. But if we add two provinces to both decks... the first deck drops to an average money density of ~\$1.364. The second drops to ~\$1.222. So we can see that thinner decks generally require more padding, and/or choke more on green cards. Whereas decks rich with big money are much more resilient.
In actuality, things are a little bit more complicated than this model would have you look at, because you don't actually draw average hands. Dominion isn't a game that's continuous; it's discrete. So there's a difference between having two silvers and having a gold and a copper. Sometimes you want more variance, sometimes you want less.
*Actually, this isn't really true for most terminal card draw - envoy being the huge exception; you probably want two smithies even before the end. And lots of terminal card draw have ways of mitigating the collision; vault, embassy, courtyard...
Smartie
• Coppersmith
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Re: The Key to Big Money Part I: Money Density
« Reply #1 on: January 19, 2012, 09:11:50 am »
0
I must say that this is an excellent article! Especially when one is going big money, one often wonder, should I buy a gold to increase money density or should I just province 1st? Of course like any statistics models, it can be difficult to calculate totally, especially with effects like discard effects, curses, attacks etc. Often, one also relies on card draws to boost chances for getting provinces or colonies which can be difficult to predict too! Though simple model, but creative and covers most aspects! Lucky that this game is not entirely based on calculations
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Geronimoo
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Re: The Key to Big Money Part I: Money Density
« Reply #2 on: January 19, 2012, 09:30:19 am »
+1
If this is published, add some graphs from simulations to clarify things. For instance, the explanation how exactly Smithy increases your average \$density is not so easy to follow, but the graph clearly shows the turbo injection effect of the Smithy on the average \$-production in comparison to a pure treasure deck.
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timchen
• Minion
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Re: The Key to Big Money Part I: Money Density
« Reply #3 on: January 19, 2012, 10:47:06 am »
0
Personally I've never found the money density concept useful. Maybe I am horribly wrong, but usually my VP buying rule (for BM-ish deck, anyway) is always something like, ok, I am getting the two golds. Then I am heading for Provinces. If I were to guess my effective money density when I start buying Provinces, I would say it is probably lower than 1.6. This is due to the distribution. The point is, what you want actually is to maximize the chance you have above \$8 in a Province game (given the limited time frame), not to have your average money per turn to be \$8. Given the starting deck without trashing, I suspect the chance will be quite a bit higher than 50% when you reach the average of \$1.6...
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WanderingWinder
• Adventurer
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Re: The Key to Big Money Part I: Money Density
« Reply #4 on: January 19, 2012, 10:53:53 am »
0
Personally I've never found the money density concept useful. Maybe I am horribly wrong, but usually my VP buying rule (for BM-ish deck, anyway) is always something like, ok, I am getting the two golds. Then I am heading for Provinces. If I were to guess my effective money density when I start buying Provinces, I would say it is probably lower than 1.6. This is due to the distribution. The point is, what you want actually is to maximize the chance you have above \$8 in a Province game (given the limited time frame), not to have your average money per turn to be \$8. Given the starting deck without trashing, I suspect the chance will be quite a bit higher than 50% when you reach the average of \$1.6...
Oh, well you're probably actually waiting longer to green than I am then. I probably ought to make it clear that you don't need the money density of your whole deck to be 1.6 to start greening - that would mean you'd be able to buy province like every turn. Or close to it. Somewhere between 50% and 100%. Anyway, yes, it's not meant to be 'oh my average money per hand is \$1.6, it's now safe for me to green'; it's more keeping an eye on how each purchase affects that density.
DG
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Re: The Key to Big Money Part I: Money Density
« Reply #5 on: January 19, 2012, 11:08:44 am »
0
I (predictably) apply some slightly different maths to look at drawing effects. When adding a treasure card with value t to a deck with T total treasure and C cards in deck, the average card value moves from T/C to (T+t)/C+1. When adding a drawing card to a deck however I would approximate this by subtracting the number of cards drawn from the denominator. So adding a single smithy would change the average card value from T/C to T/(C-2). The smithy itself adds a card to the deck but then draws three, and one minus three equals minus two.
This method is a very broad and inaccurate approximation that I'm not going to justify beyond it's simplicity. Nevertheless this approach does allow some conclusions to be drawn. The first thing you can do is set up an (inaccurate) equation for when buying one smithy increases your hand values more than buying a silver. This is (T+2)/(C+1) < T/(C-2). A little bit of algebra gives you a result of buying the smithy when T/C > 2/3 - 4/3C. This suggest that the drawing card has less impact the more cards you have in your deck (true) and that the first smithy is almost always better than a silver, unless you are drawing a very poor deck (true).
« Last Edit: January 19, 2012, 11:18:10 am by DG »
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WanderingWinder
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Re: The Key to Big Money Part I: Money Density
« Reply #6 on: January 19, 2012, 11:17:44 am »
0
I (predictably) apply some slightly different maths to look at drawing effects. When adding a treasure card with value t to a deck with T total treasure and C cards in deck, the average card value moves from T/C to (T+t)/C+1. When adding a drawing card to a deck however I would approximate this by subtracting the number of cards drawn from the denominator. So adding a single smithy would change the average card value from T/C to T/(C-2). The smithy itself adds a card to the deck but then draws three, and one minus three equals minus two.
This method is a very broad and inaccurate approximation that I'm not going to justify beyond it's simplicity. Nevertheless this approach does allow some conclusions to be drawn. The first thing you can do is set up an (inaccurate) equation for when buying one smithy increases your hand values more than buying a silver. This is (T+2)/(C+1) < T/(C-2). A little bit of algebra gives you a result of buying the smithy when T/C > 1/2 - 2/C. This suggest that the drawing card has less impact the more cards you have in your deck (true) and that the first smithy is almost always better than a silver, even if you are drawing a very poor deck (true).
The reason I don't like doing things this way, apart from the nonsensical answers it gives you when you have really few cards in your deck, has to do with the point I make at the end, that dominion is discrete rather than continuous. Smithy doesn't at all help hands without smithy in it, and in those hands WITH smithy, the bump is significantly larger. So (to oversimplify) it's half the hands 40% better and half the hands 10% worse (as compared to silver), rather than an across-the-board kind of thing. Of course that's also true of the silver itself, but it's felt more when the power is concentrated in one hand, like smithy does for you.
But yeah, you can do it that way.
ecq
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Re: The Key to Big Money Part I: Money Density
« Reply #7 on: January 19, 2012, 11:21:06 am »
0
In actuality, things are a little bit more complicated than this model would have you look at, because you don't actually draw average hands. Dominion isn't a game that's continuous; it's discrete. So there's a difference between having two silvers and having a gold and a copper. Sometimes you want more variance, sometimes you want less.
I've been thinking about this recently (and really wishing I had paid more attention in stat class). If your average money per hand is \$7 with no variance, you'll never buy a province. If it's \$7 with moderate variance, you may be able to afford a province every couple of hands. With high variance, you may have several hands with too little money followed by a hand with far too much.
I'd be really interested to hear someone with better math skills than me shed some light onto how variance impacts things, and how to assess the variance of a deck.
« Last Edit: January 19, 2012, 11:23:41 am by ecq »
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DG
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Re: The Key to Big Money Part I: Money Density
« Reply #8 on: January 19, 2012, 11:31:45 am »
0
I initially got the maths wrong. I don't do much algebra these days.
The maths can give some useful answers sometimes. You generally get a mix of the obvious - a smithy is worthy more than a silver if your average card value is more than 2/3 - plus some extra factor that shows how card power changes with the size of the deck or treasure. If you put reasonable limits on the maths, such as you're using your smithy with at least 7 treasures in your deck, then it no longer gives nonsensical answers.
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DStu
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Re: The Key to Big Money Part I: Money Density
« Reply #9 on: January 19, 2012, 11:36:11 am »
0
I'd be really interested to hear someone with better math skills shed some light onto how variance impacts things, and how to assess the variance of a deck.
First is that variance alone does not tell you much. You can have mean \$7 with high variance because you have \$6 with probability 1/2 and \$8 with probability 1/2, or you can have \$7 in 90% of the cases, but \$0 in say 5% and \$14 in 5%, that will also give you a high variance, whereas the possibility to buy a Province with it is quite low.
So what you really would need to say something is the distribution. But this in general is a too complicated thing, and most decks locks kind of the same (decks with 90% \$7 hands, but 5% \$0 or \$14 hands is not something you usually see), so you take the variance as an indicator of how the distribution will look like.
And then it tells you basically: The larger the variance is, the more likely it is to get hands that are significantly better (and worse) than the average hand. But I don't think there is a good way to really quantify this (beside simulations), as first you have to few cards in your deck compared to your hand that you could say you just apply some Gaussian limit, and also the deck (and it's variance) of course changes all the time. Also your ability to influence it is quite limited, without harming your mean to much (buying Coppers/Curses instead of Silvers of course would increase the variance of most deck, but you don't really want to do this) There are some strategies where you can say you rely on variance (Treasure map), but that is more to rely on the variance of the strategy (sometimes it is really good, sometimes it is really bad), than to rely on a deck whose hand have high variance.
Hmm, written a lot of stuff, but basically said nothing...
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timchen
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Re: The Key to Big Money Part I: Money Density
« Reply #10 on: January 19, 2012, 12:12:39 pm »
0
Personally I've never found the money density concept useful. Maybe I am horribly wrong, but usually my VP buying rule (for BM-ish deck, anyway) is always something like, ok, I am getting the two golds. Then I am heading for Provinces. If I were to guess my effective money density when I start buying Provinces, I would say it is probably lower than 1.6. This is due to the distribution. The point is, what you want actually is to maximize the chance you have above \$8 in a Province game (given the limited time frame), not to have your average money per turn to be \$8. Given the starting deck without trashing, I suspect the chance will be quite a bit higher than 50% when you reach the average of \$1.6...
Oh, well you're probably actually waiting longer to green than I am then. I probably ought to make it clear that you don't need the money density of your whole deck to be 1.6 to start greening - that would mean you'd be able to buy province like every turn. Or close to it. Somewhere between 50% and 100%. Anyway, yes, it's not meant to be 'oh my average money per hand is \$1.6, it's now safe for me to green'; it's more keeping an eye on how each purchase affects that density.
Precisely my point! So, what should one do with his eyes on the money density? For me it is just a rough feeling; do I have bought a few silvers that I can afford Duchy-dancing? something like that. It would be much more useful, to have some quantitative criteria.
And it is really important not to mislead people to think that they should somehow get their money density to 1.6 in a Province game...
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ecq
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Re: The Key to Big Money Part I: Money Density
« Reply #11 on: January 19, 2012, 03:02:00 pm »
+1
Maybe variance is a loaded word, because it has such a specific meaning in the stat world. The question is how the quality of your money affects your expected outcome, all other things being equal.
I just ran a couple of sims. Let's say it's turn 9, and we're both just buying money. I've bought 2 Golds, 4 Silvers, 2 Coppers (yeah, hypothetical). You've bought 8 Silvers. We both have the same total money, the same total cards, and therefore the same average money. We both start buying Provinces and only Provinces. I win 54%/25%.
Things change when we add more cards to the mix. At 16 cards purchased (me with 6 Golds, 4 Silvers, 6 Coppers and you with 16 Silvers), you're winning. It seems the reason you win is that you can grab the first couple of Provinces far more consistently than I can. Once things get greener, the higher-variance buys relatively more Provinces, but not enough to catch up.
The conclusion seems to be:
1. When things get green, it's better to have a few Golds, even with the same total money in-deck.
2. If things don't get very green, consistency (lots of Silver) wins. (See: Double-Jack)
Can anyone speak to this from a stat perspective? (I'd love to know the math)
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HiveMindEmulator
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Re: The Key to Big Money Part I: Money Density
« Reply #12 on: January 19, 2012, 03:10:46 pm »
+1
This a very nice idea for an article with some good information in it, but it could definitely use some cleaning up. As it is currently formatted, it seems it would be very easy for one to get lost in the math and not get the big picture out of it. It's nice to have some big bullet points or conclusions for the reader to take home. How is this money density important? Do I really need to know the difference between \$1.6 and \$1.57? Or is "relatively high" vs "relatively low" enough? Or is actually important to keep track of total money vs total cards so you can think about the derivative of money density with purchases (in some sense)? There are basically 2 major decisions you have to make in big money decks: when to add more actions, and when to buy duchies. Is there some sort of rule of thumb for how to make these decisions?
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Asklepios
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Re: The Key to Big Money Part I: Money Density
« Reply #13 on: January 21, 2012, 07:17:50 pm »
0
There's another very simple, very common kind of card to deal with when making your money density calculations: cantrips. (I'm using 'cantrip' here to define any kind of card that always draws at least one card and gives at least one action back to you). Cantrips are what I call, for the purposes of money density calculations, 'virtual cards'. What I mean by that is, because they replace themselves totally in your hand, they don't count toward the total count of cards which you're using as the denominator for your money density calculations. So, if you buy a village and a militia with your two starting buys (not, by the way, a good strategy), you have 7 coppers, 3 estates, 1 village, 1 militia, producing 7, 0, 0, and 2 money respectively and with a total of 7, 3, 0, and 1 cards to count against your deck total. Your total money density is therefore 9/11 = .818181.....
Not sure if it was mentioned somewhere within this article (if so, I may have missed it), but I think its worth noting that cantrips DO effect your money density calculations on the final draw.
So, say you play a deck that has four golds, four silvers, two coppers, ten Schemes and one Smithy. Before the Smithy is played, the Schemes are "virtual cards" as you say. But when you play the Smithy, suddenly they come into play again, as if you draw them on the Smithy draw, they're \$0 cards, not virtual cards, so you need to factor in the number of Schemes left in the deck.
I guess the way to put it is that if you are using your last action to draw, then the virtual cards need to be counted again.
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WanderingWinder
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Re: The Key to Big Money Part I: Money Density
« Reply #14 on: January 21, 2012, 11:10:00 pm »
0
There's another very simple, very common kind of card to deal with when making your money density calculations: cantrips. (I'm using 'cantrip' here to define any kind of card that always draws at least one card and gives at least one action back to you). Cantrips are what I call, for the purposes of money density calculations, 'virtual cards'. What I mean by that is, because they replace themselves totally in your hand, they don't count toward the total count of cards which you're using as the denominator for your money density calculations. So, if you buy a village and a militia with your two starting buys (not, by the way, a good strategy), you have 7 coppers, 3 estates, 1 village, 1 militia, producing 7, 0, 0, and 2 money respectively and with a total of 7, 3, 0, and 1 cards to count against your deck total. Your total money density is therefore 9/11 = .818181.....
Not sure if it was mentioned somewhere within this article (if so, I may have missed it), but I think its worth noting that cantrips DO effect your money density calculations on the final draw.
So, say you play a deck that has four golds, four silvers, two coppers, ten Schemes and one Smithy. Before the Smithy is played, the Schemes are "virtual cards" as you say. But when you play the Smithy, suddenly they come into play again, as if you draw them on the Smithy draw, they're \$0 cards, not virtual cards, so you need to factor in the number of Schemes left in the deck.
I guess the way to put it is that if you are using your last action to draw, then the virtual cards need to be counted again.
Actually, all that matters is terminal card draw. That's the only way a non-terminal's gonna hit you. And I imply it I think, but probably need to make it more explicit.
I plan on making some revisions either tomorrow or Monday.
dondon151
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Re: The Key to Big Money Part I: Money Density
« Reply #15 on: January 22, 2012, 05:44:52 am »
0
Can anyone speak to this from a stat perspective? (I'd love to know the math)
This is not going to be extremely rigorous math-wise, but let's take your provided example of a deck (player A) with 2 Golds, 4 Silvers, 9 Coppers, 3 Estates against a deck (player B) with 8 Silvers, 7 Coppers, and 3 Estates. Then let's add 3 Provinces to each deck (21 cards total).
Let it be given that player A draws a Gold as 1 of the 5 cards in his hand. He can get to \$8 in the following ways:
No Victory cards in hand - 14C4 / 20C4 = 20.66%
All combinations get a Province - 100%
Cumulative: 20.66% * 100% = 20.66%
1 Victory card in hand - 6C1 * 14C3 / 20C4 = 45.08%
All combinations with exactly 0 Copper get a Province - 5C3 / 14C3 = 2.75%
All combinations with exactly 1 Copper get a Province - 5C2 * 9C1 / 14C3 = 24.73%
Only 1 Gold with 2 Copper get a Province - 1C1 * 9C2 / 14C3 = 9.89%
Cumulative: 45.08% * 37.36% = 16.84%
2 Victory cards in hand - 6C2 * 14C2 / 20C4 = 28.17%
Must draw 1 Gold and 1 Silver - 1C1 * 4C1 / 14C2 = 4.40%
Cumulative: 28.17% * 4.40% = 1.24%
Chance to buy Province given that Gold is in hand: 38.74%
Player B's treasure is more homogeneous, so we don't have to do conditional analysis to figure out how he can get to \$8:
4 Silvers in hand - 8C4 * 13C1 / 21C4 = 15.20%
3 Silvers, 2 Coppers in hand - 8C3 * 7C2 / 21C4 = 19.65%
Chance to buy Province: 34.85%
How about we add 4 Provinces apiece instead in order to figure out the chances of each deck closing out the game with a 5/3 Province split?
Player A's chance to buy Province given that Gold is in hand: 37.08%
Player B's chance to buy Province: 29.47%
Now, OK, it's not quite fair to always assume that player A has a Gold in hand and directly compare the probabilities of buying a Province. But player A will draw a Gold into his hand at some point. In a deck bloated with Victory cards, player B has trouble closing out a win when he has a 4/3 Province advantage, but player A doesn't have nearly as much trouble as long as he finds his Golds. That can probably explain your lopsided win ratios.
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Asklepios
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Re: The Key to Big Money Part I: Money Density
« Reply #16 on: January 22, 2012, 12:01:24 pm »
0
On the topic of Money Density, what do people think of Cache?
Obviously if there are good \$5 actions out there its pretty weak, but say you're playing 2 Smithys and Big Money towards Provinces, and you hit \$5. Is Cache a good call or is Silver better?
To me, I think it may well be but I find it hard to explain why in statistical terms. Gold and two coppers is only \$1.666 average, which makes it worse than buying single silver, but in experience Cache seems to have worked out for me more often than not. Has anyone simulated 2 Smithy + Big Money + Caches on \$5 vs 2 Smithy + Big Money?
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Thisisnotasmile
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Re: The Key to Big Money Part I: Money Density
« Reply #17 on: January 22, 2012, 12:07:15 pm »
0
Completely unoptimized it's about 59-33 to the Cache player.
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HiveMindEmulator
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Re: The Key to Big Money Part I: Money Density
« Reply #18 on: January 22, 2012, 12:14:44 pm »
0
Cache is better than silver in basically any non-trimmed deck that relies on gold to buy provinces. When money density is even reasonably close, gold-heavy is generally better than silver-heavy.
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DG
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Re: The Key to Big Money Part I: Money Density
« Reply #19 on: January 22, 2012, 12:47:26 pm »
0
Another simple situation is woodcutter + treasure, where you might get the option to buy an gold+copper or silver+silver instead of a gold. In both cases you should buy the one gold if you're trying to buy provinces. If you are not trying to raise your average hand value much above 5, perhaps to buy duchies and dukes, then more/lower treasures can become superior.
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Asklepios
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Re: The Key to Big Money Part I: Money Density
« Reply #20 on: January 23, 2012, 03:48:19 am »
0
Completely unoptimized it's about 59-33 to the Cache player.
That was what I expected. Can someone explain why that should be the case statistically though, given that Cache has a lower average money per card than silver? I suspect its to do with distributions of probabilities rather than averages.
That is, in statistical terms I think it must be more important to know p(\$8+) than the \$mean when you are analysing money density.
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DStu
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Re: The Key to Big Money Part I: Money Density
« Reply #21 on: January 23, 2012, 04:42:22 am »
0
Completely unoptimized it's about 59-33 to the Cache player.
That was what I expected. Can someone explain why that should be the case statistically though, given that Cache has a lower average money per card than silver? I suspect its to do with distributions of probabilities rather than averages.
I think you can also argue for Cache without even going to the probabilities, and I think the article even gives all you need. It's not a clear call, I think I could argue in the other direction also if the facts where different, but anyway, here is what is in Caches favour in BM-decks:
1) \$5/3 is more than you need on average to get to Provinces, even without Smithies. It's also more than your average hand usually is in BM-decks when you go for Provinces.
2) Silver is even higher, but it's only one card. Cache is three cards. Three cards have an higher influence on the deck than one. And you are more resilent to clogging with a larger deck. Extreme case: You have a deck consistent of only 2 victory cards, and the alternative of gaining a Cache or a Silver. Cache brings you deck to average \$1, Silver to average 2/3.
I don't believe there is really a situation where Cache overtakes Silver in this way, as the effect gets smaller with both more money in the deck and more cards in the deck, just to illustrate that gaining 3 cards for 5/3 must be valued stronger in many cases than gaining just one of it.
3) Two Smithies are already pretty much for BM, might be optimal but I think it was close, as they tend to collide. Adding 3 cards to your deck reduced the risk of collisions more than adding one.
4) Variance should also help the Cache.
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Asklepios
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Re: The Key to Big Money Part I: Money Density
« Reply #22 on: January 23, 2012, 08:16:45 am »
0
Yes, I agree.
I think as another extreme circumstance the following example might indicate why mean\$ is less important than p(\$8).
First deck: \$1 x 8, \$2 x2 has got a mean \$ of \$1.2, but zero chance of making \$8.
Second deck: \$1 x 10, \$2 x1, \$3 x1 \$ of \$1.15, but has a fair chance of making \$8.
By the same count, I think Cache likely increases your chance of reaching \$6 faster than Silver does, so Cache is a quicker than silver as a route to gold.
Does this make sense? I mention this only because I think its a trap to think of money density purely in terms of mean \$ per card.
That is, the second from last paragraph in the article is probably the most important one. Gold + Copper =/= Silver + Silver
Or to put it in Cache terms:
Gold + Copper + Copper > Silver.
« Last Edit: January 23, 2012, 08:19:19 am by Asklepios »
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HiveMindEmulator
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Re: The Key to Big Money Part I: Money Density
« Reply #23 on: January 23, 2012, 03:30:37 pm »
0
To add to the cache vs silver thing, think about what hands that buy provinces look like. Unless you're going for a really heavy silver strat like trader or double jack, you basically need either a gold (or cache) or a +cards card to hit \$8. And if you have 2, you're basically guaranteed a province. Given reasonable money density, gold (+ cache) density actually becomes more important than money density. Similarly with buying golds. Pre-greening, a gold (or cache) in hand basically allows you to buy another gold almost regardless of what you draw it with.
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GendoIkari
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Re: The Key to Big Money Part I: Money Density
« Reply #24 on: January 23, 2012, 04:00:35 pm »
0
To add to the cache vs silver thing, think about what hands that buy provinces look like. Unless you're going for a really heavy silver strat like trader or double jack, you basically need either a gold (or cache) or a +cards card to hit \$8. And if you have 2, you're basically guaranteed a province. Given reasonable money density, gold (+ cache) density actually becomes more important than money density. Similarly with buying golds. Pre-greening, a gold (or cache) in hand basically allows you to buy another gold almost regardless of what you draw it with.
So I assume that this means also that in a hand of Mine + Copper + Silver, Silver -> Gold is always better than Copper -> Silver?
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posted by .
12. In a calorimeter, 10 g of ice absorbs heat with an enthalpy of fusion of 334 J/g. What is the heat absorbed?
q = mCHf
I am having a lot of trouble on this type of problem in chem. Help and an explanation would be greatly appreciated. Thank you.
• Chemistry -
When you are moving within a phase (for example, liquid water at one T to liquid water at another T OR ice at one T to ice at another T OR steam at one T to steam at another T), the equation to use is
q = mass x specific heat of the substance in that phase x (Tfinal-Tintial).
When you are at the phase change point (ice or liquid water at zero C) the formula is
q = mass x heat fusion. + if ice to liquid or - if liquid to solid.
When you are at the phase change point at the other end (liquid water or steam at 100 C) the formula is
q = mass x heat vaporization. + if liquid to steam or - if steam to liquid.
In the problem above it is
q = mass x heat fusion
q = 10 g x 334 J/g = 3340 J. Since the ice is absorbing heat it is + 334 J.
• Chemistry -
Thanks so much DrBob222. This helped a bunch!
## Similar Questions
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Conclusion question(s) from a lab we did to find the heat of fusion of ice: Does the value obtained for the molar heat of fusion depend on the volume of water used?
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When 1 kg of ice at 0 degrees Celsius absorbs 80 kilocalories of heat, the ice undergoes a) a change of state b) a loss of energy c) a rise in temperature d) an increase in volulme The heat of fusion of water is 80 kcal/kg. That means …
3. ### chemistry
When ice at 0C melts to liquid water at 0C, it absorbs 0.334 kj of heat per gram. Suppose the heat needed to melt 31.5 g of ice is absorbed from the water contained in a glass. If this water has a mass of .210 kg and a temperature …
4. ### Chemistry
When ice at 0°C melts to liquid water at 0°C, it absorbs 0.334 kJ of heat per gram. Suppose the heat needed to melt 30.7 g of ice is absorbed from the water contained in a glass. If this water has a mass of 0.189 kg and a temperature …
5. ### Chemistry
In a calorimeter, 100 g of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g. How much heat was absorbed?
How would i find the following? I just need an explanation. I'm stuck. I have many other questions like this and i need to know how to do them. The specfic heat of ice is 2.05 J/g. In a calorimeter, 10.0 g of ice melts at 0oC. The
7. ### Will Someone please respond! Chem
How would i find the following? I just need an explanation. I'm stuck. I have many other questions like this and i need to know how to do them. The specfic heat of ice is 2.05 J/g. In a calorimeter, 10.0 g of ice melts at 0oC. The
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Calculate the amount of heat absorbed when 58 g of steam at 117◦C are completely converted to ice at −32.5◦C. Specific Heat: H2O(s) = 2.09 J/g◦C H2O(l) = 4.18 J/g◦C H2O(g) = 2.03 J/g◦C Heat of Fusion …
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Published on Sunday, 18th May 2014, 01:00 am; Solved by 219;
Difficulty rating: 60%
### Problem 472
There are N seats in a row. N people come one after another to fill the seats according to the following rules:
1. No person sits beside another.
2. The first person chooses any seat.
3. Each subsequent person chooses the seat furthest from anyone else already seated, as long as it does not violate rule 1. If there is more than one choice satisfying this condition, then the person chooses the leftmost choice.
Note that due to rule 1, some seats will surely be left unoccupied, and the maximum number of people that can be seated is less than N (for N > 1).
Here are the possible seating arrangements for N = 15:
We see that if the first person chooses correctly, the 15 seats can seat up to 7 people.
We can also see that the first person has 9 choices to maximize the number of people that may be seated.
Let f(N) be the number of choices the first person has to maximize the number of occupants for N seats in a row. Thus, f(1) = 1, f(15) = 9, f(20) = 6, and f(500) = 16.
Also, ∑f(N) = 83 for 1 ≤ N ≤ 20 and ∑f(N) = 13343 for 1 ≤ N ≤ 500.
Find ∑f(N) for 1 ≤ N ≤ 1012. Give the last 8 digits of your answer. | 359 | 1,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-35 | longest | en | 0.931201 |
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Comparing and Ordering Fractions Problem Solving: Comparing and Ordering Fractions Problem Solving includes 6 problem solving situations, and each situation has 3 - 4 questions to be answered. The questions incorporate comparing and ordering
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Problem solving can be much more fun and engaging for students when they work together! This resource includes 28 problem solving pages. Each page gives a situation and then has a set of 3-4 questions to be answered. Can be used for: * cooperative
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Decimal Operations Task Cards (Footloose Activity) and Problem Solving: Using the task cards in Decimal Operations Footloose is a fantastic way to get students moving while they are reviewing concepts! This product offers: * high level of student
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Fraction Operations Task Cards (Footloose Activity) and Problem Solving: This Fraction Operations Footloose is a fantastic way to get students moving while they are reviewing concepts. Students enjoy the activity; teachers enjoy the high level of
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Fractions, Decimals, and Percents Problem Solving: Problem solving can be challenging, but when students discuss problems, they can learn so much from each other! This product was created with cooperative problem solving in mind, but can also be
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Students love Footloose! Students enjoy the movement while they are reviewing concepts; teachers enjoy the high level of student engagement and the quick/easy preparation. There are 13 files in this bundle, with a total of 14 sets of cards
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Bundle
This bundle has 16 files, with a total of 17 sets of cards (Factors and GCF has 2 sets). Footloose activities offer: * high level of student engagement * easy prep for teachers * 30 question cards for each activity Students solve each of the 30
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Use these Problem Solving Strategies Doodle Notes to help your students get organized for problem solving, use symbols and color coding to help remember the strategies, and practice writing more in math class. The notes can be kept in students’
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You can use the wheel to introduce different math conversation starters, and it can then be kept in students’ notebooks as a reference for the year. The sections include: 1) I agree because… 2) I disagree because… 3) I solved by… 4) To solve, I
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Fraction Resources, Grades 4-6: This is a bundle of great resources that provide fraction practice in fun, active ways! Each item in the bundle is listed below and is linked to its individual product page, in case you'd like to look at any of them
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This is the last third of the book, 6th Grade Math Warm Ups, 180 Days of Spiral Review. It can be copied for each student and kept in a binder for them to work on as soon as they get to class. I started out using these as warm ups for students to
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This grade 6 math bundle includes: daily math warm-ups, Footloose task cards activities, interactive notebook fold it ups, problem solving and color by numbers. This combination of resources will take you through your entire school year with a great
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Mixed Math Practice Task Cards - Fall Footloose: Using the task cards in this Fall/Halloween - themed Footloose is a fantastic way to get students moving while they are reviewing concepts! This product offers: * high level of student engagement *
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Mixed Math Practice Task Cards - Winter Footloose: Using the task cards in Winter Mixed Math Practice Footloose is a fantastic way to get students moving while they are reviewing concepts! This product offers: * high level of student engagement *
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Give students this handy bookmark to help them remember steps to take to solve those word problems! They can use it to keep their places in their notebooks or textbooks, or any books they are currently reading! Print and laminate as many as you
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TEACHING EXPERIENCE
I've been teaching for more than 20 years:-) I taught elementary school for 12 years, in 2nd, 4th, and 5th grades. Now I teach math and language arts in middle school.
MY TEACHING STYLE
I like to give students as much opportunity to move and discuss their learning as possible.
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
B.S. in Elementary Education Masters in Computers in Education
Join my Facebook page for access to special offers and more! Middle School Math Moments - and more! or check out my blog, Cognitive Cardio with Middle School Math Moments | 1,695 | 6,820 | {"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} | 2.765625 | 3 | CC-MAIN-2018-43 | latest | en | 0.890276 |
https://www.scriptol.com/programming/algorithm-definition.php | 1,719,164,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00186.warc.gz | 841,602,523 | 9,923 | Definition of algorithm
An algorithm is deterministic automaton for accomplishing a goal which, given an initial state, will terminate in a defined end-state. The efficiency of implementation of the algorithm depends upon speed, size, resources consumption. We will discuss definitions, classifications and the history.
What is an algorithm?
No agreed-to definition of "algorithm" exists.
A simple definition: A set of instructions for solving a problem.
The algorithm is either implemented by a program or simulated by a program. Algorithms often have steps that iterate (repeat ) or require decisions such as logic or comparison.
An very simple example of an algorithm is multiplying two numbers: on first computers with limited processors, this was accomplished by a routine that in a number of loop based on the first number adds the second number. The algorithm translates a method into computer commands.
Algorithms are essential to the way computers process information, because a computer program is essentially an algorithm that tells the computer what specific steps to perform (in what specific order) in order to carry out a specified task, such as calculating employees' paychecks or printing students' report cards. Thus, an algorithm can be considered to be any sequence of operations which can be performed by a Turing-complete system. Authors who assert this thesis include Savage (1987) and Gurevich (2000):
...Turing's informal argument in favor of his thesis justifies a stronger thesis: every algorithm can be simulated by a Turing machine" ...according to Savage [1987], an algorithm is a computational process defined by a Turing machine.
Typically, when an algorithm is associated with processing information, data is read from an input source or device, written to an output sink or device, and/or stored for further processing. Stored data is regarded as part of the internal state of the entity performing the algorithm. In practice, the state is stored in a data structure.
For any such computational process, the algorithm must be rigorously defined: specified in the way it applies in all possible circumstances that could arise. That is, any conditional steps must be systematically dealt with, case-by-case; the criteria for each case must be clear (and computable).
Because an algorithm is a precise list of precise steps, the order of computation will almost always be critical to the functioning of the algorithm. Instructions are usually assumed to be listed explicitly, and are described as starting 'from the top' and going 'down to the bottom', an idea that is described more formally by flow of control.
So far, this discussion of the formalization of an algorithm has assumed the premises of imperative programming. This is the most common conception, and it attempts to describe a task in discrete, 'mechanical' means. Unique to this conception of formalized algorithms is the assignment operation, setting the value of a variable. It derives from the intuition of 'memory' as a scratchpad.
For some alternate conceptions of what constitutes an algorithm see functional programming and logic programming.
Definitions of algorithms
Blass and Gurevich
Blass and Gurevich describe their work as evolved from consideration of Turing machines, Kolmogorov-Uspensky machines (KU machines), Schönhage machine (storage modification machine SMM), and pointer machines (linking automata) as defined by Knuth. The work of Gandy and Markov are also described as influential precursors.
Gurevich offers a 'strong' definition of an algorithm (that is summarized here):
The Turing's informal argument in favor of his thesis justifies a stronger thesis: every algorithm can be simulated by a Turing machine. In practice, it would be ridiculous. Can one generalize Turing machines so that any algorithm, never mind how abstract, can be modeled by a generalized machine? But suppose such generalized Turing machines exist. What would their states be?A first-order structure. A particular small instruction set suffices in all cases. Computation could be an evolution of the state, could be nondeterministic, interact with its environment, could be parallel and multi-agent, could have dynamic semantics. The two underpinings of thier work are: Turing's thesis andthe notion of first order structure of Tarski.
The above phrase computation as an evolution of the state differs markedly from the definition of Knuth and Stone, the "algorithm" as a Turing machine program. Rather, it corresponds to what Turing called the complete configuration, and includes both the current instruction (state) and the status of the tape. Kleene (1952) shows an example of a tape with 6 symbols on it, all other squares are blank, and how to "Gödelize" its combined table-tape status.
Boolos and Jeffrey
Their definition is:
"Explicit instructions for determining the nth member of a set, for arbitrary finite n. Such instructions are to be given quite explicitly, in a form in which they could be followed by a computing machine, or by a human who is capable of carrying out only very elementary operations on symbols."
Knuth
Knuth (1968, 1973) has given a list of five properties that are widely accepted as requirements for an algorithm:
1. Finiteness: "An algorithm must always terminate after a finite number of steps"
2. Definiteness: "Each step of an algorithm must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case"
3. Input: "...quantities which are given to it initially before the algorithm begins. These inputs are taken from specified sets of objects"
4. Output: "...quantities which have a specified relation to the inputs"
5. Effectiveness: "... all of the operations to be performed in the algorithm must be sufficiently basic that they can in principle be done exactly and in a finite length of time by a man using paper and pencil"
Knuth offers as an example the Euclidean algorithm for determining the greatest common divisor of two natural numbers.
Knuth admits that, while his description of an algorithm may be intuitively clear, it lacks formal rigor, since it is not exactly clear what "precisely defined" means, or "rigorously and unambiguously specified" means, or "sufficiently basic", and so forth. He makes an effort in this direction in his first volume where he defines in detail what he calls the "machine language" for his "mythical MIX... the world's first polyunsaturated computer".
Many of the algorithms in his books are written in the MIX language. He also uses tree diagrams, flow diagrams and state diagrams.
Markov
A. A. Markov (1954) provided the following definition of algorithm:
1. In mathematics, "algorithm" is commonly understood to be an exact prescription, defining a computational process, leading from various initial data to the desired result....
The following three features are characteristic of algorithms and determine their role in mathematics:
a) the precision of the prescription, leaving no place to arbitrariness, and its universal comprehensibility -- the definiteness of the algorithm;
b) the possibility of starting out with initial data, which may vary within given limits -- the generality of the algorithm;
c) the orientation of the algorithm toward obtaining some desired result, which is indeed obtained in the end with proper initial data -- the conclusiveness of the algorithm.
He admitted that this definition "does not pretend to mathematical precision". His 1954 monograph was his attempt to define algorithm more accurately; he saw his resulting definition -- his "normal" algorithm -- as "equivalent to the concept of a recursive function". His definition included four major components:
1. Separate elementary steps, each of which will be performed according to one of the given substitution rules.
2. Steps of local nature (the algorithm won't change more than a certain number of symbols).
3. The scheme of the algorithm is a list of rules for the substitution formulas.
4. A means to distinguish a concluding substitution (a final state).
In his Introduction Markov observed that the entire significance for mathematics of efforts to define algorithm more precisely would be in connection with the problem of a constructive foundation for mathematics.
Minsky
Minsky (1967) asserts that an algorithm is an effective procedure and replaces further in his text algorithm by effective procedure. The term is also used by Knuth. Here is its definition of effective procedure:
A set of rules which tell us, from moment to moment, precisely how to behave.
But he recognizes that this is subject to a criticism:
The interpretation of the rules is left to depend on some person or agent.
He made a refinement: to specify, along with the statement of the rules, the details of the mechanism that is to interpret them. To avoid the cumbersome process of having to do this over again for each individual procedure he hopes to identify a reasonably uniform family of rule-obeying mechanisms. Here is his formulation:
(1) a language in which sets of behavioral rules are to be expressed, and
(2) a single machine which can interpret statements in the language and thus carry out the steps of each specified process.
In the end, though, he still worries that "there remains a subjective aspect to the matter. Different people may not agree on whether a certain procedure should be called effective".
But Minsky is undeterred. He immediately introduces "Turing's Analysis of Computation Process". He quotes what he calls Turing's thesis:
Any process which could naturally be called an effective procedure can be realized by a Turing machine.
(This is also called Church's thesis).
After an analysis of "Turing's Argument" he observes that equivalence of many intuitive formulations of Turing, Church, Kleene, Post, and Smullyan "leads us to suppose that there is really here an objective or absolute notion".
Stone
Stone (1972) and Knuth (1968, 1973) were professors at Stanford University at the same time so it is not surprising if there are similarities in their definitions:
To summarize ... we define an algorithm to be a set of rules that precisely defines a sequence of operations such that each rule is effective and definite and such that the sequence terminates in a finite time.
Stone is noteworthy because of his detailed discussion of what constitutes an effective rule for his robot, or person-acting-as-robot, must have some information and abilities within them, and if not the information and the ability must be provided in "the algorithm":
For people to follow the rules of an algorithm, the rules must be formulated so that they can be followed in a robot-like manner, that is, without the need for thought... however, if the instructions [to solve the quadratic equation, his example] are to be obeyed by someone who knows how to perform arithmetic operations but does not know how to extract a square root, then we must also provide a set of rules for extracting a square root in order to satisfy the definition of algorithm.
(...) not all instructions are acceptable, because they may require the robot to have abilities beyond those that we consider reasonable.
He gives the example of a robot confronted with the question: "Is Henry VIII is a King of England", print 1 if yes and 0 if no, but the robot has not been previously provided with this information. And worse, if the robot is asked if Aristotle was a King of England and the robot only had been provided with five names, it would not know how to answer. Thus:
An intuitive definition of an acceptable sequence of instructions is one in which each instruction is precisely defined so that the robot is guaranteed to be able to obey it.
After providing us with his definition, Stone introduces the Turing machine model and states that the set of five-tupes that are the machine's instructions is an algorithm... known as a Turing machine program. Puis he says that a computation of a Turing machine is described by stating:
1. The tape alphabet
2. The form in which the parameters are presented on the tape
3. The initial state of the Turing machine
4. The form in which answers will be represented on the tape when the Turing machine halts
5. The machine program.
This is in the spirit of Blass and Gurevich.
Some issues
Expressing algorithms
Algorithms can be expressed in many kinds of notations:
- Natural language expressions of algorithms tend to be verbose and ambiguous, and are rarely used for complex or technical algorithms.
- Pseudocode and flowcharts are structured ways to express algorithms that avoid the ambiguities, while remaining independent of a particular implementation language.
- Programming languages are primarily intended for expressing algorithms in a form that can be executed by a computer, but are often used as a way to define or document algorithms.
Must an algorithm halt?
Some writers restrict the definition of algorithm to procedures that eventually finish. Others, as Kleene, include procedures that could run forever without stopping. Such a procedure has been called a "computational method" by Knuth or "calculation procedure or algorithm" by Kleene. However, Kleene notes that such a method must eventually exhibit "some object".
Minksy (1967) makes the observation that, if an algorithm hasn't "terminated" then how can we answer the question: "Will it terminate with the correct answer?'"
Thus the answer is: undecidable. We can never know, nor can we do an analysis beforehand to find out. The analysis of algorithms for their likelihood of termination is called "Termination analysis".
Algorithm analysis
The terms "analysis of algorithms" was coined by Knuth. Most people who implement algorithms want to know how much of a particular resource, such as time or storage, is required for the execution. Methods have been developed for the analysis of algorithms to obtain such quantitative answers.
The analysis and study of algorithms is one discipline of computer science, and is often practiced abstractly (without the use of a specific programming language or hardware). But the Scriptol code is portable, simple and abstract enough for such analysis.
Simple example: multiplication
```int multiply(int x, int y)
int sum = 0
while y > 0
sum + x
let y - 1
return sum
int a = 5
int b = 7
print a,"x", b, "=", multiply(a, b)```
References
• Martin Davis. The Undecidable: Basic Papers On Undecidable Propostions, Unsolvable Problems and Computable Functions. New York: Raven Press, 1965.
Davis gives commentary before each article.
• Yuri Gurevich. Sequential Abstract State Machines Capture Sequential Algorithms, ACM Transactions on Computational Logic, Vol 1, no 1 (July 2000), pages 77-111.
Includes bibliography of 33 sources.
• A. A. Markov. Theory of algorithms. Imprint Moscow, Academy of Sciences of the USSR, 1954 . Original title: Teoriya algerifmov.
• Marvin Minsky. Computation: Finite and Infinite Machines, First, Prentice-Hall, Englewood Cliffs, NJ, 1967.
• Harold S. Stone. Introduction to Computer Organization and Data Structures, 1972, McGraw-Hill, New York.
Cf in particular the first chapter titled: Algorithms, Turing Machines, and Programs. | 3,118 | 15,296 | {"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 | 3 | CC-MAIN-2024-26 | latest | en | 0.935814 |
https://www.time4learning.com/blog/methods-tools-resources/effective-strategies-to-help-students-struggling-with-math/ | 1,726,879,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00436.warc.gz | 941,097,615 | 21,528 | If you have a student struggling with math, it can be a challenge to get them the help they need. The first step is to determine where the issue lies. Is he or she lacking foundational skills? Is it a particular concept like decimals, fractions or graphing? Once you know where the problem lies, you can better tailor the kind of supplemental material that you can use.
Whether you homeschool or your child attends traditional school, there are plenty of effective strategies for teaching students with difficulties in math. This page will explore a number of options for helping students with math in elementary, middle and high school, as well as provide information on how Time4Learning’s award-winning curriculum provides all the tools your child needs to practice and improve their math skills.
## Helping Elementary School Students With Math
Play Math Games: Although there are tons of online games, many of them aren’t educational, and are free at the cost of having inappropriate ads, so be careful when choosing which games to play. One easy and safe way is Time4MathFacts, an online math fact fluency program that uses games and rewards to help students master math facts. With engaging and accessible gamified lessons, practice is a breeze!
Reinforce reading comprehension: Believe it or not, reading comprehension plays a big role in math, especially when students get into word problems. If a student doesn’t understand the question being asked, there is no way they will be able to work out the problem and provide a correct answer. Ensuring that your child has solid comprehension skills can help improve their math skills. Additionally, it’s always a good idea to have students read the question several times, and even break it down into parts for multi-step problems.
Use visual aids: This strategy can be especially helpful for visual learners and kinesthetic learners who learn by observing and doing. Posters, videos, or manipulatives like toys, candy or cereal pieces, can help bring concepts to life. In addition, having students color and draw when learning geometry and fractions can be very helpful references.
## Helping Middle School Students With Math
Thorough, step-by-step instruction makes a huge difference in how well a student understands a math concept. Sometimes the key to helping students struggling with math can be as simple as teaching a topic in a different way. Whether you homeschool or just want to provide your student some after school math help, learning at home under your own terms makes a big difference.
Emphasize understanding: Building a strong middle school math foundation requires students to reinforce their elementary school math concepts and expand their understanding into more complex concepts. If your child isn’t grasping a concept, reviewing the lesson can help further their understanding, and increase their confidence. Our math curriculum for middle schoolers allows students to review material at their own pace and retake quizzes to make sure understand the concepts they learned.
For example, mastering order of operations is critical for students to thrive at higher grade levels. In chapter 2 of our 6th grade math curriculum, you’ll find activities to ensure students master this key skill before moving on to more complex concepts.
Provide and support multiple paths: “It’s not the destination, it’s the journey”. Sometimes there are multiple ways to arrive at the same answer. Whenever possible, show students the various ways they can solve a problem. Support them when they find a new path, and explore their understanding of that pathway. Challenge them to see how many different methods they can use to solve it.
Take breaks: If a concept isn’t sinking in, frustration can get the best of students. When this happens, it’s important to take a break and step back to clear one’s mind. A few minutes can make a big difference and help students tackle the material with a fresh set of eyes.
## Helping High School Students With Math
Use engaging materials: Whether it’s interactive math labs like those in our high school trigonometry and pre-calculus courses, positive reinforcement, or engaging video lessons taught by funny characters, it’s important to make math fun for students of all ages.
Reinforce through practice: Students struggling with math benefit the most with this strategy. This is true at any grade level, not just high school. Once a student is finally able to grasp a difficult concept, it is important to keep at it to ensure those hard-earned skills don’t just fade away quickly. Whether it’s worksheets, flash cards, whiteboarding, or even online activities, the more a student is able to practice, the better their math skills will be.
Connect with real life applications: How many times have you heard your child say, ‘How will I use this in real life’? You can answer that question by making real-world connections which is a major feature of our high school math curriculum where teachers step into real-world settings to engage students and help apply what they are learning to the world around them.
• For example, if your child is an athlete, you can explain how knowledge of physics, geometry, and trigonometry can help improve their performance. Additionally, there are careers in sports that use math to analyze data during sporting events for a wide variety of sports.
• If your student is interested in arts and crafts, fashion design or other hands-on projects, improving math skills can help them make more accurate designs and creations.
Present students with choices: As students get older, they crave independence. Help them feel empowered by giving them the option to choose which problems to work on from a “menu” or “bank” of assignments. This can help them gain the motivation they need to improve their math skills and give them a sense of freedom and control over their studies.
Making sure students who struggle with math get the help they need is not just about improving their grades or making sure they get into a good school. Children with solid math skills become adept at the process of problem-solving. They’re able to identify the root of a problem by taking it apart and exploring possible solutions. This ultimately leads to enhanced critical thinking skills, increased confidence, and an appreciation for teamwork and effort.
Have any strategies or tips you’d like to share with other parents? Share them in the comments below!
Learn tricks and strategies to help support students struggling in math! | 1,256 | 6,557 | {"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-2024-38 | latest | en | 0.949437 |
https://infinitylearn.com/surge/question/mathematics/a-stone-is-dropped-into-a-quiet-lake-if-the-waves-moves-in/ | 1,675,220,042,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499899.9/warc/CC-MAIN-20230201013650-20230201043650-00807.warc.gz | 307,952,103 | 12,308 | A stone is dropped into a quiet lake. If the waves moves in circles at the rate of 30 cm/sec when the radius is 50 m, the rate of increase of enclosed area, is
# A stone is dropped into a quiet lake. If the waves moves in circles at the rate of 30 cm/sec when the radius is 50 m, the rate of increase of enclosed area, is
1. A
$30\pi {\mathrm{m}}^{2}/\mathrm{sec}$
2. B
$30{\mathrm{m}}^{2}/\mathrm{sec}$
3. C
$3\pi {\mathrm{m}}^{2}/\mathrm{sec}$
4. D
none of these
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### Solution:
Let the radius of the circular wave ring be r cm at any time t.
Then $\frac{dr}{dt}=30\mathrm{cm}/\mathrm{sec}$ (Given)
Let A be the area of the enclosed ring. Then,
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GATE CE 2017 Set 2
MCQ (Single Correct Answer)
+2
-0.6
Consider the following definite integral $${\rm I} = \int\limits_0^1 {{{{{\left( {{{\sin }^{ - 1}}x} \right)}^2}} \over {\sqrt {1 - {x^2}} }}dx}$$\$
The value of the integral is
A
$${{{\pi ^3}} \over {24}}$$
B
$${{{\pi ^3}} \over {12}}$$
C
$${{{\pi ^3}} \over {48}}$$
D
$${{{\pi ^3}} \over {64}}$$
2
GATE CE 2017 Set 2
Numerical
+1
-0
The divergence of the vector field $$\,V = {x^2}i + 2{y^3}j + {z^4}k\,\,$$ at $$x=1, y=2, z=3$$ is ________.
3
GATE CE 2017 Set 2
Numerical
+1
-0
A two-faced fair coin has its faces designated as head $$(H)$$ and tail $$(T)$$. This coin is tossed three times in succession to record the following outcomes: $$H, H,H.$$ If the coin is tossed one more time, the probability (up to one decimal place ) of obtaining $$H$$ again, given the previous realizations of $$H, H$$ and $$H$$, would be _____.
4
GATE CE 2017 Set 2
Numerical
+2
-0
The analysis of a water sample produces the following results: The total hardness (in mg/L as $${\mathrm{CaCO}}_3$$) of the water sample is ______.
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Class 12 | 552 | 1,394 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-33 | latest | en | 0.749671 |
https://web2.0calc.com/questions/triangles-abc-and-dfg-are-given-find-the-lengths | 1,596,667,189,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735989.10/warc/CC-MAIN-20200805212258-20200806002258-00386.warc.gz | 570,734,337 | 6,801 | +0
# Triangles ABC and DFG are given. Find the lengths of all other sides of these triangles if:
0
804
1
+57
Triangles ABC and DFG are given. Find the lengths of all other sides of these triangles if:
Chapter Reference
b
∠A≅∠D, AB·DG=AC·DF, AC=7 cm, BC=15 cm, FG=20 cm, and DF-AB=3 cm
I found that AB is 9, and that DC is 12, but can't find AC or DG because there aren't that many clues that lead to the answer.
Feb 2, 2019
#1
+111435
+1
AB / BC = DF / FG
AB / 15 = (AB + 3)/20 (cross-multiply)
20AB = 15(AB + 3)
20AB = 15AB + 45
5AB = 45
AB = 9
DF - AB = 3 which implies that
Then DF = AB + 3 = 12
And we know that
AB * DG = AC * DF
9 * DG = 7 * 12
9* DG = 84
DG = 84/9 = 28/ 3
So AB = 9 AC = 7 BC = 15
And
DF = 12 DG = 28/3 FG = 20
Note that this is true.....the scale factor from ABC to DFG = 4/3
Feb 2, 2019 | 349 | 870 | {"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-2020-34 | latest | en | 0.877681 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/798/2/k/i/ | 1,701,651,573,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00573.warc.gz | 957,104,658 | 58,317 | # Properties
Label 798.2.k.i Level $798$ Weight $2$ Character orbit 798.k Analytic conductor $6.372$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [798,2,Mod(463,798)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(798, base_ring=CyclotomicField(6))
chi = DirichletCharacter(H, H._module([0, 0, 2]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("798.463");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$798 = 2 \cdot 3 \cdot 7 \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 798.k (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$6.37206208130$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{2} - x + 1$$ x^2 - x + 1 Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - \zeta_{6} + 1) q^{2} + ( - \zeta_{6} + 1) q^{3} - \zeta_{6} q^{4} + ( - 3 \zeta_{6} + 3) q^{5} - \zeta_{6} q^{6} - q^{7} - q^{8} - \zeta_{6} q^{9} +O(q^{10})$$ q + (-z + 1) * q^2 + (-z + 1) * q^3 - z * q^4 + (-3*z + 3) * q^5 - z * q^6 - q^7 - q^8 - z * q^9 $$q + ( - \zeta_{6} + 1) q^{2} + ( - \zeta_{6} + 1) q^{3} - \zeta_{6} q^{4} + ( - 3 \zeta_{6} + 3) q^{5} - \zeta_{6} q^{6} - q^{7} - q^{8} - \zeta_{6} q^{9} - 3 \zeta_{6} q^{10} + 4 q^{11} - q^{12} - 3 \zeta_{6} q^{13} + (\zeta_{6} - 1) q^{14} - 3 \zeta_{6} q^{15} + (\zeta_{6} - 1) q^{16} + ( - 2 \zeta_{6} + 2) q^{17} - q^{18} + (5 \zeta_{6} - 3) q^{19} - 3 q^{20} + (\zeta_{6} - 1) q^{21} + ( - 4 \zeta_{6} + 4) q^{22} - \zeta_{6} q^{23} + (\zeta_{6} - 1) q^{24} - 4 \zeta_{6} q^{25} - 3 q^{26} - q^{27} + \zeta_{6} q^{28} - 3 q^{30} - 6 q^{31} + \zeta_{6} q^{32} + ( - 4 \zeta_{6} + 4) q^{33} - 2 \zeta_{6} q^{34} + (3 \zeta_{6} - 3) q^{35} + (\zeta_{6} - 1) q^{36} + 4 q^{37} + (3 \zeta_{6} + 2) q^{38} - 3 q^{39} + (3 \zeta_{6} - 3) q^{40} + (4 \zeta_{6} - 4) q^{41} + \zeta_{6} q^{42} + ( - 6 \zeta_{6} + 6) q^{43} - 4 \zeta_{6} q^{44} - 3 q^{45} - q^{46} + 2 \zeta_{6} q^{47} + \zeta_{6} q^{48} + q^{49} - 4 q^{50} - 2 \zeta_{6} q^{51} + (3 \zeta_{6} - 3) q^{52} + 4 \zeta_{6} q^{53} + (\zeta_{6} - 1) q^{54} + ( - 12 \zeta_{6} + 12) q^{55} + q^{56} + (3 \zeta_{6} + 2) q^{57} + (13 \zeta_{6} - 13) q^{59} + (3 \zeta_{6} - 3) q^{60} - 5 \zeta_{6} q^{61} + (6 \zeta_{6} - 6) q^{62} + \zeta_{6} q^{63} + q^{64} - 9 q^{65} - 4 \zeta_{6} q^{66} + 12 \zeta_{6} q^{67} - 2 q^{68} - q^{69} + 3 \zeta_{6} q^{70} + ( - 7 \zeta_{6} + 7) q^{71} + \zeta_{6} q^{72} + ( - 2 \zeta_{6} + 2) q^{73} + ( - 4 \zeta_{6} + 4) q^{74} - 4 q^{75} + ( - 2 \zeta_{6} + 5) q^{76} - 4 q^{77} + (3 \zeta_{6} - 3) q^{78} + ( - 8 \zeta_{6} + 8) q^{79} + 3 \zeta_{6} q^{80} + (\zeta_{6} - 1) q^{81} + 4 \zeta_{6} q^{82} + 17 q^{83} + q^{84} - 6 \zeta_{6} q^{85} - 6 \zeta_{6} q^{86} - 4 q^{88} - 4 \zeta_{6} q^{89} + (3 \zeta_{6} - 3) q^{90} + 3 \zeta_{6} q^{91} + (\zeta_{6} - 1) q^{92} + (6 \zeta_{6} - 6) q^{93} + 2 q^{94} + (9 \zeta_{6} + 6) q^{95} + q^{96} + ( - 10 \zeta_{6} + 10) q^{97} + ( - \zeta_{6} + 1) q^{98} - 4 \zeta_{6} q^{99} +O(q^{100})$$ q + (-z + 1) * q^2 + (-z + 1) * q^3 - z * q^4 + (-3*z + 3) * q^5 - z * q^6 - q^7 - q^8 - z * q^9 - 3*z * q^10 + 4 * q^11 - q^12 - 3*z * q^13 + (z - 1) * q^14 - 3*z * q^15 + (z - 1) * q^16 + (-2*z + 2) * q^17 - q^18 + (5*z - 3) * q^19 - 3 * q^20 + (z - 1) * q^21 + (-4*z + 4) * q^22 - z * q^23 + (z - 1) * q^24 - 4*z * q^25 - 3 * q^26 - q^27 + z * q^28 - 3 * q^30 - 6 * q^31 + z * q^32 + (-4*z + 4) * q^33 - 2*z * q^34 + (3*z - 3) * q^35 + (z - 1) * q^36 + 4 * q^37 + (3*z + 2) * q^38 - 3 * q^39 + (3*z - 3) * q^40 + (4*z - 4) * q^41 + z * q^42 + (-6*z + 6) * q^43 - 4*z * q^44 - 3 * q^45 - q^46 + 2*z * q^47 + z * q^48 + q^49 - 4 * q^50 - 2*z * q^51 + (3*z - 3) * q^52 + 4*z * q^53 + (z - 1) * q^54 + (-12*z + 12) * q^55 + q^56 + (3*z + 2) * q^57 + (13*z - 13) * q^59 + (3*z - 3) * q^60 - 5*z * q^61 + (6*z - 6) * q^62 + z * q^63 + q^64 - 9 * q^65 - 4*z * q^66 + 12*z * q^67 - 2 * q^68 - q^69 + 3*z * q^70 + (-7*z + 7) * q^71 + z * q^72 + (-2*z + 2) * q^73 + (-4*z + 4) * q^74 - 4 * q^75 + (-2*z + 5) * q^76 - 4 * q^77 + (3*z - 3) * q^78 + (-8*z + 8) * q^79 + 3*z * q^80 + (z - 1) * q^81 + 4*z * q^82 + 17 * q^83 + q^84 - 6*z * q^85 - 6*z * q^86 - 4 * q^88 - 4*z * q^89 + (3*z - 3) * q^90 + 3*z * q^91 + (z - 1) * q^92 + (6*z - 6) * q^93 + 2 * q^94 + (9*z + 6) * q^95 + q^96 + (-10*z + 10) * q^97 + (-z + 1) * q^98 - 4*z * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + q^{2} + q^{3} - q^{4} + 3 q^{5} - q^{6} - 2 q^{7} - 2 q^{8} - q^{9}+O(q^{10})$$ 2 * q + q^2 + q^3 - q^4 + 3 * q^5 - q^6 - 2 * q^7 - 2 * q^8 - q^9 $$2 q + q^{2} + q^{3} - q^{4} + 3 q^{5} - q^{6} - 2 q^{7} - 2 q^{8} - q^{9} - 3 q^{10} + 8 q^{11} - 2 q^{12} - 3 q^{13} - q^{14} - 3 q^{15} - q^{16} + 2 q^{17} - 2 q^{18} - q^{19} - 6 q^{20} - q^{21} + 4 q^{22} - q^{23} - q^{24} - 4 q^{25} - 6 q^{26} - 2 q^{27} + q^{28} - 6 q^{30} - 12 q^{31} + q^{32} + 4 q^{33} - 2 q^{34} - 3 q^{35} - q^{36} + 8 q^{37} + 7 q^{38} - 6 q^{39} - 3 q^{40} - 4 q^{41} + q^{42} + 6 q^{43} - 4 q^{44} - 6 q^{45} - 2 q^{46} + 2 q^{47} + q^{48} + 2 q^{49} - 8 q^{50} - 2 q^{51} - 3 q^{52} + 4 q^{53} - q^{54} + 12 q^{55} + 2 q^{56} + 7 q^{57} - 13 q^{59} - 3 q^{60} - 5 q^{61} - 6 q^{62} + q^{63} + 2 q^{64} - 18 q^{65} - 4 q^{66} + 12 q^{67} - 4 q^{68} - 2 q^{69} + 3 q^{70} + 7 q^{71} + q^{72} + 2 q^{73} + 4 q^{74} - 8 q^{75} + 8 q^{76} - 8 q^{77} - 3 q^{78} + 8 q^{79} + 3 q^{80} - q^{81} + 4 q^{82} + 34 q^{83} + 2 q^{84} - 6 q^{85} - 6 q^{86} - 8 q^{88} - 4 q^{89} - 3 q^{90} + 3 q^{91} - q^{92} - 6 q^{93} + 4 q^{94} + 21 q^{95} + 2 q^{96} + 10 q^{97} + q^{98} - 4 q^{99}+O(q^{100})$$ 2 * q + q^2 + q^3 - q^4 + 3 * q^5 - q^6 - 2 * q^7 - 2 * q^8 - q^9 - 3 * q^10 + 8 * q^11 - 2 * q^12 - 3 * q^13 - q^14 - 3 * q^15 - q^16 + 2 * q^17 - 2 * q^18 - q^19 - 6 * q^20 - q^21 + 4 * q^22 - q^23 - q^24 - 4 * q^25 - 6 * q^26 - 2 * q^27 + q^28 - 6 * q^30 - 12 * q^31 + q^32 + 4 * q^33 - 2 * q^34 - 3 * q^35 - q^36 + 8 * q^37 + 7 * q^38 - 6 * q^39 - 3 * q^40 - 4 * q^41 + q^42 + 6 * q^43 - 4 * q^44 - 6 * q^45 - 2 * q^46 + 2 * q^47 + q^48 + 2 * q^49 - 8 * q^50 - 2 * q^51 - 3 * q^52 + 4 * q^53 - q^54 + 12 * q^55 + 2 * q^56 + 7 * q^57 - 13 * q^59 - 3 * q^60 - 5 * q^61 - 6 * q^62 + q^63 + 2 * q^64 - 18 * q^65 - 4 * q^66 + 12 * q^67 - 4 * q^68 - 2 * q^69 + 3 * q^70 + 7 * q^71 + q^72 + 2 * q^73 + 4 * q^74 - 8 * q^75 + 8 * q^76 - 8 * q^77 - 3 * q^78 + 8 * q^79 + 3 * q^80 - q^81 + 4 * q^82 + 34 * q^83 + 2 * q^84 - 6 * q^85 - 6 * q^86 - 8 * q^88 - 4 * q^89 - 3 * q^90 + 3 * q^91 - q^92 - 6 * q^93 + 4 * q^94 + 21 * q^95 + 2 * q^96 + 10 * q^97 + q^98 - 4 * q^99
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/798\mathbb{Z}\right)^\times$$.
$$n$$ $$115$$ $$211$$ $$533$$ $$\chi(n)$$ $$1$$ $$-\zeta_{6}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
463.1
0.5 − 0.866025i 0.5 + 0.866025i
0.500000 + 0.866025i 0.500000 + 0.866025i −0.500000 + 0.866025i 1.50000 + 2.59808i −0.500000 + 0.866025i −1.00000 −1.00000 −0.500000 + 0.866025i −1.50000 + 2.59808i
505.1 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i 1.50000 2.59808i −0.500000 0.866025i −1.00000 −1.00000 −0.500000 0.866025i −1.50000 2.59808i
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
19.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 798.2.k.i 2
3.b odd 2 1 2394.2.o.a 2
19.c even 3 1 inner 798.2.k.i 2
57.h odd 6 1 2394.2.o.a 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
798.2.k.i 2 1.a even 1 1 trivial
798.2.k.i 2 19.c even 3 1 inner
2394.2.o.a 2 3.b odd 2 1
2394.2.o.a 2 57.h odd 6 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(798, [\chi])$$:
$$T_{5}^{2} - 3T_{5} + 9$$ T5^2 - 3*T5 + 9 $$T_{11} - 4$$ T11 - 4 $$T_{13}^{2} + 3T_{13} + 9$$ T13^2 + 3*T13 + 9 $$T_{17}^{2} - 2T_{17} + 4$$ T17^2 - 2*T17 + 4
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2} - T + 1$$
$3$ $$T^{2} - T + 1$$
$5$ $$T^{2} - 3T + 9$$
$7$ $$(T + 1)^{2}$$
$11$ $$(T - 4)^{2}$$
$13$ $$T^{2} + 3T + 9$$
$17$ $$T^{2} - 2T + 4$$
$19$ $$T^{2} + T + 19$$
$23$ $$T^{2} + T + 1$$
$29$ $$T^{2}$$
$31$ $$(T + 6)^{2}$$
$37$ $$(T - 4)^{2}$$
$41$ $$T^{2} + 4T + 16$$
$43$ $$T^{2} - 6T + 36$$
$47$ $$T^{2} - 2T + 4$$
$53$ $$T^{2} - 4T + 16$$
$59$ $$T^{2} + 13T + 169$$
$61$ $$T^{2} + 5T + 25$$
$67$ $$T^{2} - 12T + 144$$
$71$ $$T^{2} - 7T + 49$$
$73$ $$T^{2} - 2T + 4$$
$79$ $$T^{2} - 8T + 64$$
$83$ $$(T - 17)^{2}$$
$89$ $$T^{2} + 4T + 16$$
$97$ $$T^{2} - 10T + 100$$ | 5,318 | 9,958 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.526659 |
https://www.wyzant.com/resources/answers/407623/solve_word_problem | 1,610,737,010,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00627.warc.gz | 1,208,435,125 | 14,246 | Amber R.
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ORDER NOW! | 973 | 4,337 | {"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-2018-17 | latest | en | 0.921582 |
https://minuteshours.com/33-88-hours-in-hours-and-minutes | 1,606,890,035,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00333.warc.gz | 384,152,398 | 4,816 | 33.88 hours in hours and minutes
Result
33.88 hours equals 33 hours and 52.8 minutes
You can also convert 33.88 hours to minutes.
Converter
Thirty-three point eight eight hours is equal to thirty-three hours and fifty-two point eight minutes. | 60 | 247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-50 | latest | en | 0.889784 |
https://gre-test-prep.com/iowa-algebra-test/complex-fractions/printable-order-of-operations.html | 1,621,352,062,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989637.86/warc/CC-MAIN-20210518125638-20210518155638-00169.warc.gz | 296,341,852 | 12,219 | Algebra Tutorials!
Tuesday 18th of May
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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• example of a factor/math | 918 | 3,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-21 | latest | en | 0.896559 |
https://www.studypool.com/discuss/516343/find-the-value-of-x-round-your-answer-to-the-nearest-tenth-1?free | 1,508,621,612,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00402.warc.gz | 992,218,236 | 13,895 | Find the value of x. Round your answer to the nearest tenth.
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Find the value of x. Round your answer to the nearest tenth.
sin 33° = 7/x
a. 8.3
b. 5.9
c. 10.8
d. 12.9
May 5th, 2015
sin 33 degrees= 0.544639035
= 0.544639035=7/x
= 0.544639035x=7
= x=7/0.544639035
= x= 12.8525... or a 12.9 option D.
May 5th, 2015
...
May 5th, 2015
...
May 5th, 2015
Oct 21st, 2017
check_circle | 201 | 476 | {"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 | 3 | CC-MAIN-2017-43 | latest | en | 0.820825 |
http://lotterywinnerstories.com/Lottery/picking-lotto-numbers-formula-powerball/ | 1,510,946,520,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803906.12/warc/CC-MAIN-20171117185611-20171117205611-00776.warc.gz | 192,600,733 | 25,226 | ## How To Crack The Lottery Code? PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## State Lotteries - Which State Has the Best Odds of Winning?
Have you been searching for Pick 3 lottery workout systems that actually work? Are you tired of guessing and relying on luck? Well here's one system that you should try. It only involves simple mathematics and a bit of concentration. This system is one of the oldest. Although it doesn't guarantee a hundred percent, it has a high percentage hit. It is called the 123 workout. Here is how it works:
o You begin by using the preceding numbers drawn from the Pick 3 lottery. Just to provide a concrete example, let us use the numbers 468. We will call it the stack number.
o Important Rule of the 123 workout: Do not carry over when adding or subtracting the numbers. Here is an example: 5 + 9 = 4 (instead of 14)
o Monitor all the numbers that come out in the following days. You will realize that one out of the 22 combinations will be the next hit. Let's face it; it is better than having to choose from a million other combinations. Your chance of winning the next lottery is now greater having to choose from a smaller set.
Others have tried other Pick 3 lottery workout systems. Some of them are just too complicated. They would require the work of your personal computer, plus it costs much and takes time to download them. This system only requires a pen and a piece of paper. Try this method and see if the numbers really come out. What have you got to lose?
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
How To Crack The Lottery Code? Massachusetts Lottery
## How To Win The Lottery Using The Secret? PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The PowerBall lottery?
Ok what are you selling? That is the first question that may come to your mind if you are reading a title like win the lotto. But if you can get pass the title and read through the entire article I promise to give you some key secrets that will put you ahead of 95% of the average lotto players. Like you I play the lotto and I hate for people to waste my time with easy cut answers that lead to no where. I want to be told what it really takes to win the lotto regularly. I know you are the same since you are reading this.
So let's get down to business shall we. The key to winning the lotto boils down to 4 closely held secrets they are:
• Research
• Strategy
• Patience
• System
System: Now here is the kicker since everything boils down to what system are you using. Without the correct system patience, strategy and research goes out the window. In fact it makes no sense to do if you do not have a system to apply your strategies or research to. Patience has no bearing if you don't have a system. Okay you got the point. Remember the system you choose determines everything so choose wisely.
## Winning Numbers For Mega Millions - How to Choose Mega Millions Numbers
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
Learn How To Win The Lottery California
## Best Lottery Secrets Revealed PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## 5 Tips How To Win The Lottery
Because you are sitting in front of your computer viewing this page, chances are, you're an eager and regular lotto player who has yet to win lotto prizes. Lottery cheats are some of the most researched items or articles here on the Internet, and it's easy to understand why: lottery games from all states can be addictive, attracting countless players from all across the country to play the lotto on a regular basis. What you need to understand is that while playing the lottery can be fun and exciting, the real treat lies in winning the lotto prizes - including the jackpot. And unlike what some lotto players believe, you can win the lotto using lottery cheats.
Lottery cheats are not cheats in the real sense of the word. They are not illegal and won't put you into some kind of federal trouble. In fact, lottery cheats have long been recognized by serious lottery players as the reason why they have higher chances of winning. Cheats at winning the lottery are actually guides on how to win any type of lotto game you choose to play. They offer sensible pieces of advice to increase the likelihood of holding a winning ticket.
As far as lottery cheats are concerned, statistics experts highly advise against forming lottery combinations in a mathematical sequence and playing patterns on lottery tickets. These acts are sure to reduce your chances of winning because mathematical sequences and patterns are almost never taken into consideration in lottery games. Past winning results show a tendency towards random combinations. If you use a proven system that can efficiently analyze lottery data, such as past winning numbers, trends, and angles, you are more likely to bring home the jackpot prize not just once but as many times as you please.
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
How Can You Win A Lottery? Rhode Island Lottery
## Best Lottery Numbers To Pick PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The PowerBall lottery?
We have 5 tips on how to win the lottery. We know you'll be interested - everybody dreams of winning the lottery one day. The lottery brings out some kind of instinct in people; it allows ordinary people to become rich simply over-night. This kind of thing doesn't happen often, but the lottery is one thing that makes these kinds of special events possible.
Good, helpful tips on how to win the lottery are always hard to find, especially for free. This is because most people simply want to cash in on their secrets, although to be honest I don't quite understand how people can pay for lottery winning tips. Surely if someone knows the secret to winning the lottery, they aren't going to give away their secret for a few dollars? We know we would much rather win the lottery using our own knowledge than share the secrets.
Be sure to follow all of these notes on how to win the lottery, but also be sure to remember that it is a completely random draw. Try and choose numbers at random, and be sure to join a syndicate if you can find one to join.
## Lottery Cheats - How to Cheat the System and Win!
Are you constantly thinking and saying, "I want to win the lottery?" Do you ask yourself, "Will I win the lottery?" If you think about winning the lottery all the time, you need to know this secret to winning the secret to winning the lottery.If you think winning the big bucks is just about picking the right lottery numbers, you're wrong. Yes, you do need to pick the right numbers, and it's best to have a system to win the lottery. Doing things like reading a lottery book can be useful too. But none of this will work for you if you don't do this one essential thing:The universe rewards those who are prepared. Did you know that?When you act as if you expect something, it's going to come to you. Did you know that?4. Make a list of people to whom you plan to give money and decide on how much you'll give. Take into account gift taxes when you do this. Be careful about who you tell, "I'll give you money when I win." You don't want to give away all your winnings.Combine this lottery win preparation method with a lottery winning system, and you will be able to answer "Yes," to "Will I win the lottery?" You'll be able to turn "I want to win the lottery," to "I WON the lottery."
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
Winning Lottery Strategy Arizona
## Winning Lottery Numbers PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The PowerBall lottery?
People who play Megamillions have one question in common and that would be which the best numbers to play Megamillions are? The thing is, there isn't a standard or a formula that one can follow in order to come to a conclusion regarding the best numbers to play Megamillions. There are, however, things that you can keep in mind when you are in the process of picking numbers. You can consider them strategies or systems or you can think of them, simply, as tips. They may not give you a concrete answer towards the best numbers to play Megamillions, but they can help take you a step closer.1. When choosing numbers, do not try and form patterns. Remember to be as random as you can. Think about it, if you decide to go with a pattern such as choosing all the numbers that go a certain way, whether upwards or downwards the sheets then you are already putting yourself at a disadvantage.8. When you ask people who have won jackpots for themselves, it is likely that they'd tell you that the numbers they chose were chosen mostly because of gut feel. A kind of intuition. You may not think much of it but this intuition could potentially win you millions. So go with what you feel and not what you believe to be the best numbers to play Megamillions. Remember, trust your gut and go with your intuition.
## Powerball: How To Use Easy Pick To Win The Powerball
When it comes to megamillions lottery, there are a number of secrets that are unveiled. Sometimes, these secrets are facts that only players have ideas about. In this article, there are secrets and facts that will help a player that aims to win the jackpot prize.One secret of megamillions lottery is that it usually draws many odd numbers than even numbers. Of course, this is based on studies and analysis done by lottery experts. Then knowing this secret, one can favor odd numbers to increase the chance of winning.Another secret in megamillions lottery are the strategies that can be used to be able to have a greater chance of winning. The odd that make this lottery possible is 1 in 176 million. There are two strategies that can be used. First, play in groups. The chance of winning is greater since there many tickets that are purchased. Also, one of the reasons behind to not playing on own is due to the too long odds. It is best to let all other players go for the jackpot and play in groups. The second strategy is to play once in group in a year but invest the most. It is said that the opportunity comes only once that is why it is important to be careful too.Further, it has been found out that almost all of the players of megamillions lottery choose the quick pick method in choosing their numbers. This is the method when the machine chooses the numbers in random and gives them to the player. The rest choose to personally choose their numbers. Maybe, they use their birthdates and anniversary dates which is a common method in choosing the numbers.In addition, it has been found out that the odds of a set of numbers repeating themselves have a very nil probability. Then, knowing this secret might tell the gamer in advance not to use repeating numbers. So it is important to know if the numbers in hand have been played already or not. Megamillions lottery's result in the internet might help the player to determine which numbers have been played before or remains not picked.On the other hand, there are also myths that must be known so as to decide whether to play lottery or not. One myth is that the lotteries take advantage of the poorer people. This is said to be a myth since the people who usually plays the lottery are those who have extra cash in their pockets for leisure. In fact, a great percentage of players are rich people. Another myth is that the lottery is a form of tax. It is said that no one is required to join the lottery. And when the player wins, this is the only times that he/she is asked to pay taxes. The reason behind this is that the jackpot prize is also a form of income and any income is taxed by the government. Therefore, given these secrets mentioned, the player will have the right information in order to increase the chances of winning.
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
How Do I Calculate Lotto Numbers? Delaware Lottery
## Richard Lustig Lottery Winning Formula Powerball
Playing the Powerball lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Powerball lottery?
Ok what are you selling? That is the first question that may come to your mind if you are reading a title like win the lotto. But if you can get pass the title and read through the entire article I promise to give you some key secrets that will put you ahead of 95% of the average lotto players. Like you I play the lotto and I hate for people to waste my time with easy cut answers that lead to no where. I want to be told what it really takes to win the lotto regularly. I know you are the same since you are reading this.
So let's get down to business shall we. The key to winning the lotto boils down to 4 closely held secrets they are:
• Research
• Strategy
• Patience
• System
System: Now here is the kicker since everything boils down to what system are you using. Without the correct system patience, strategy and research goes out the window. In fact it makes no sense to do if you do not have a system to apply your strategies or research to. Patience has no bearing if you don't have a system. Okay you got the point. Remember the system you choose determines everything so choose wisely.
## Powerball: How To Use Easy Pick To Win The Powerball
Winning the Powerball lottery is very rewarding. Give this system a spin and watch your winnings grow.
Learn How To Win The Lottery New York Lottery
## Lotto Dominator Winning Software PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## How To Answer "Yes," To "Will I Win The Lottery?" And Turn "I Want To Win The Lottery" Into "I Won!"
Lottery predictions is quite popular these days. People used to be skeptical with the predictions as they thought that the winning numbers are a matter of luck and fortunes. Not many people believe that lottery can be won by using some kind of a sophisticated science based predictions. It was not until the late 90s when lottery players began using lottery predictions to help them to win lottery or at least get closer to the winning numbers. When Gonzalo Garcia-Pelayo, a Spanish man who managed to study and analyze many games in two different countries, Spain and the US and win a lot of money by using different strategies. After him people started to believe that lottery results can be predicted.
Lottery players start thinking about how to win the lotteries using predictions. They use many kinds of predictions: from mechanical predictions on mechanical lotteries to technological predictions using computer software. A lot of people use algorithm to analyze and predict lottery results.
Analysis of Groups
There are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.
Analysis of Hot-Cold Trend
This algorithm analysis is one of the most favourite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings.
Analysis of Repetition Pattern
A lot of lottery players share the same opinion that repetition is quite important to predict the winning numbers as most of jackpots will appear again in the future.
The analysis mentioned above represents only a part of the strategies that lottery players can use. There are still many other algorithm analysis that can be done by predictors to help them win.
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
Lottery Dominator Secret Formula West Virginia Lottery
## Best Lottery Winning System PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## How To Answer "Yes," To "Will I Win The Lottery?" And Turn "I Want To Win The Lottery" Into "I Won!"
Are you constantly thinking and saying, "I want to win the lottery?" Do you ask yourself, "Will I win the lottery?" If you think about winning the lottery all the time, you need to know this secret to winning the secret to winning the lottery.If you think winning the big bucks is just about picking the right lottery numbers, you're wrong. Yes, you do need to pick the right numbers, and it's best to have a system to win the lottery. Doing things like reading a lottery book can be useful too. But none of this will work for you if you don't do this one essential thing:The universe rewards those who are prepared. Did you know that?When you act as if you expect something, it's going to come to you. Did you know that?4. Make a list of people to whom you plan to give money and decide on how much you'll give. Take into account gift taxes when you do this. Be careful about who you tell, "I'll give you money when I win." You don't want to give away all your winnings.Combine this lottery win preparation method with a lottery winning system, and you will be able to answer "Yes," to "Will I win the lottery?" You'll be able to turn "I want to win the lottery," to "I WON the lottery."
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
How To Win The Lottery Using The Secret? Georgia Lottery
## The Key To Winning The Lottery PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The PowerBall lottery?
We have 5 tips on how to win the lottery. We know you'll be interested - everybody dreams of winning the lottery one day. The lottery brings out some kind of instinct in people; it allows ordinary people to become rich simply over-night. This kind of thing doesn't happen often, but the lottery is one thing that makes these kinds of special events possible.
Good, helpful tips on how to win the lottery are always hard to find, especially for free. This is because most people simply want to cash in on their secrets, although to be honest I don't quite understand how people can pay for lottery winning tips. Surely if someone knows the secret to winning the lottery, they aren't going to give away their secret for a few dollars? We know we would much rather win the lottery using our own knowledge than share the secrets.
Be sure to follow all of these notes on how to win the lottery, but also be sure to remember that it is a completely random draw. Try and choose numbers at random, and be sure to join a syndicate if you can find one to join.
## Five Secrets To Winning The Lottery: How To Pick Megamillions Numbers
A lottery is a popular form of gambling which involves the drawing of lots for a prize. There are some states where lottery is forbidden but others endorse it to the extend of organizing a national lottery. Lotteries date back to the period of Romans. And it was common for emperors to give their dinner party guests gifts like slaves to lavish villas after drawings.
Two common lottery myths
Lottery is tax
Lottery take advantage of the poor economic strata of our society
Lottery facts
In China lottery was first played to fund the Great Wall
U.S. lotteries helped fund the Colonial Army in the Revolutionary War
At one point of time many churches and universities such as Harvard, Yale and Princeton were funded in part by lotteries
Where does the Lottery revenue expenditure go?
A good amount of lottery earnings is spend on economic development of the country
Steps are taken to improve job opportunities
Educational purposes
Human resources
Natural resources[environmental protection]
Transportation
Public health finance
Keeping apart money to improve the economic development of a country, lottery gains also go to cancer organizations, child care canters and places where it is needed
Today, many religious and social organizations condemn lotteries saying that it’s a one way ticket to accumulating lots of wealth. Since lotteries lotteries tempt people to play with money and gain huge amounts back if they got lucky, it can easily become an addiction. Many individuals waste much money over lotteries and other gambling games without taking care of their families properly. This is unethical to a population of today’s society. Of course, it is not right to spend all the money you earn on lotteries. But by taking part in lotteries you contribute a little to the development of your own country. That’s why we say lotteries are not just about gambling. It’s a lot about winning and giving.
ZZZZZZ
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
How To Beat The Lottery Using Math? Maryland Lottery
## Lotto Dominator Winning Software PowerBall
Playing the PowerBall lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The PowerBall lottery?
Here's 5 top tips to bear in mind next time you complete your lottery playslip. They won't increase your CHANCES of winning - because ONLY more entries can do that.
BUT they can significantly increase the AMOUNT you win when your numbers do come up.
How?
Because all 5 lottery tips are based on avoiding the way a lot of other people pick their numbers. If you pick lottery numbers the same way as most people do, then when you hit the jackpot, you share that prize with everybody else who picked the same numbers.
That can turn a jackpot pool of millions into a prize of just thousands! It happens all too often - so please don't let it happen to you.
The easiest, fastest way to pick better lottery numbers, is to pick them totally at random. So pull scraps of paper out of a bag. It won't guarantee NOT picking a 'bad' set of numbers, but at least there's a good chance you won't be sharing your lottery millions with a hundred other 'lucky' winners.
Are you ready for more tips on how to win more often when playing the lottery?
If so then read below for a free report that uncovers the biggest mistakes most lottery players are making, and how you can avoid them.
## 5 Top Tips To Win More With Any Lottery
Why bother going to a gas station and buy some Pick 3 tickets when you can go for Pick 3 Online? Yes, you read that right. Pick 3 Online, the same Pick 3 lottery game that everyone is crazed upon can now be played without leaving your own home. If you have been playing the famous lottery game for quite sometime now then going through the ABC's on how to play it would just be a waste of time. Perhaps some guidelines on how to cling closer to the chances of winning might interest you?
If you've been playing the lottery game by betting your favorite numbers, you probably noticed that it doesn't pay out the way you expected it. Using favorite numbers are already a thing of the past when it comes to betting. Upon getting to try the Pick 3 Online, try a new strategy other than guessing; that would have to be not guessing at all. Pick 3 Online lottery is not a game of chance but a somewhat mathematical and logistic inclined game for the itching winner.
Be careful of downloading systems that assure you of winning online betting games. Those that offer cheats with a price seldom work and would just make a run for your money. Be mindful of the fact that online lottery games like Pick 3 Online are more of taking good strategy than taking a chance.
Winning the PowerBall lottery is very rewarding. Give this system a spin and watch your winnings grow.
Learn How To Win The Lottery Pennsylvania Lottery | 6,623 | 32,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-47 | longest | en | 0.9607 |
https://www.scribd.com/presentation/326021419/Image-Enhancement-in-the-fd-ppt | 1,566,658,495,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321140.82/warc/CC-MAIN-20190824130424-20190824152424-00402.warc.gz | 982,341,018 | 59,838 | You are on page 1of 22
# Image Enhancement in the
Frequency Domain
Filtering in
the Frequency Domain
Compute Fourier transform of image
Multiply the result by a filter transfer
function (or simply filter).
Take the inverse transform to produce the
enhanced image.
Summary:
G(u,v) = H(u,v) F(u,v)
1
Filtered Image = G(u,v)
## Spatial & Frequency Domain
f(x,y) * h(x,y) F(u,v) H(u,v)
(x,y) * h(x,y) [(x,y)] H(u,v)
h(x,y) H(u,v)
Filters in the spatial and frequency domain
form a FT pair, i.e. given a filter in the
frequency domain we can get the
corresponding one in the spatial domain by
taking its inverse FT
Frequency Domain
## Image Enhancement in the
Frequency Domain
Enhancement in
the Frequency Domain
Types of enhancement that can be done:
Lowpass filtering: reduce the high-frequency
content -- blurring or smoothing
Highpass filtering: increase the magnitude of
high-frequency components relative to lowfrequency components -- sharpening.
## Image Enhancement in the
Frequency Domain
Lowpass Filtering
in the Frequency Domain
Edges, noise contribute significantly to the highfrequency content of the FT of an image.
Blurring/smoothing is achieved by reducing a
specified range of high-frequency components:
G (u , v) H (u , v) F (u, v)
Smoothing
in the Frequency Domain
G(u,v) = H(u,v) F(u,v)
Ideal
Butterworth (parameter: filter order)
Gaussian
These three filters cover the range from very sharp (ideal) to very
smooth (Gaussian) filter functions. The Butterworth filter has a
parameter , called the filter order. For high values of this
parameter the butterworth filter approaches the ideal filter.
For lower order values, the butterworth filter similar to the
Gaussian filter. Thus the butterworth filter may be viewed as a
transition between two extremes.
## Ideal Filter (Lowpass)
A 2-D ideal low-pass filter:
1 if D (u, v) D0
H (u, v)
0 if D (u , v) D0
where D0 is a specified nonnegative quantity and
D(u,v) is the distance from point (u,v) to the center of
the frequency rectangle.
Center of frequency rectangle: (u,v)=(M/2,N/2)
Distance from any point to the center (origin) of the FT:
D (u, v) (u 2 v 2 )1/ 2
Frequency Domain
## Ideal Filter (Lowpass)
Ideal:
all frequencies inside a circle of radius D0 are
passed with no attenuation
all frequencies outside this circle are completely
attenuated.
Cutoff-frequency: the point of transition between
H(u,v)=1 and H(u,v)=0 (D0)
To establish cutoff frequency loci, we typically
compute circles that enclose specified amounts of
total image power PT.
## Ideal Filter (cont.)
PT is obtain by summing the components of power
spectrum P(u,v) at each point for u up to M-1 and v
up to N-1.
A circle with radius r, origin at the center of the
frequency rectangle encloses a percentage of the
power which is given by the expression
u
Frequency Domain
## Butterworth Filter (Lowpass)
This filter does not have a sharp
discontinuity establishing a clear cutoff
between passed and filtered frequencies.
1
H (u , v)
1 [ D(u, v) / D0 ]2 n
## Butterworth Filter (Lowpass)
To define a cutoff frequency locus: at points
for which H(u,v) is down to a certain fraction
of its maximum value.
When D(u,v) = D0, H(u,v) = 0.5
i.e. down 50% from its maximum value of 1.
Frequency Domain
## Gaussian Lowpass Filter
GLPF in two dimensions is given by
H (u , v) e
D 2 ( u ,v ) / 2 2
## D(u,v) is the distance from the origin of the fourier transform
and is a measure of the spread of the Gaussian curve. =D
0
We can express the GLPF
H (u , v) e
D 2 ( u ,v ) / 2 D0 2
## Where D0 is the cutoff frequency . When D(u,v)=D0 , the filter down to
0.607 of its maximum value. | 980 | 3,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-35 | latest | en | 0.78005 |
https://justaaa.com/statistics-and-probability/33942-he-table-provides-summary-statistics-on-highway | 1,716,995,653,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00865.warc.gz | 281,136,487 | 10,175 | Question
# he table provides summary statistics on highway fuel economy of cars manufactured in 2012 (from Exercise...
he table provides summary statistics on highway fuel economy of cars manufactured in 2012 (from Exercise 5.32). Use these statistics to calculate a 98% confidence interval for the difference between average highway mileage of manual and automatic cars, and interpret this interval in the context of the data.
Hwy MPG, Automatic Hwy MPG, Manual
Mean 22.92 27.88
SD 5.29 5.01
n 26 26
lower bound: mpg (please round to two decimal places)
upper bound: mpg (please round to two decimal places)
Interpret your confidence interval in the context of the problem:
• We can be 98% confident that our confidence interval contains the difference in average highway mileage of the cars in our sample is contained within our confidence interval
• We can be 98% confident that the difference in average highway mileage of manual and automatic cars is contained within our confidence interval
• 98% of manual and automatic cars will have a difference in highway mileage that falls within our confidence interval
Does your confidence interval provide significant evidence for a difference in the highway fuel efficiency of automatic versus manual cars? Explain.
• no, since 0 is not contained within our confidence interval
• no, since negative highway mileage doesn't make sense
• yes, since highway mileage cannot be negative
• yes, since 0 is not contained within our confidence interval
This is attempt 1 of 4.
For Manual : x̅1 = 27.88, s1 = 5.01, n1 = 26
For Automatic : x̅2 = 22.92, s2 = 5.29, n2 = 26
df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 49.8529 = 50
98% Confidence interval for the difference :
At α = 0.02 and df = 50, two tailed critical value, t_c = T.INV.2T(0.02, 50) = 2.403
Lower Bound = (x̅1 - x̅2) - t_c*√(s1²/n1 +s2²/n2)
= (27.88 - 22.92) - 2.403*√(5.01²/26 + 5.29²/26) = 1.53
Upper Bound = (x̅1 - x̅2) + t_c*√(s1²/n1 +s2²/n2)
= (27.88 - 22.92) + 2.403*√(5.01²/26 + 5.29²/26) = 8.39
--
We can be 98% confident that the difference in average highway mileage of manual and automatic cars is contained within our confidence interval.
--
no, since 0 is not contained within our confidence interval.
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 684 | 2,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-22 | latest | en | 0.834007 |
http://betterlesson.com/next_gen_science/browse/2166/ngss-5-ps2-1-support-an-argument-that-the-gravitational-force-exerted-by-earth-on-objects-is-directed-down | 1,487,740,299,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170884.63/warc/CC-MAIN-20170219104610-00422-ip-10-171-10-108.ec2.internal.warc.gz | 27,321,208 | 27,913 | Support an argument that the gravitational force exerted by Earth on objects is directed down.
55 Lesson(s)
Weight, Mass, & Gravity
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Types of Forces
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The Sun, Earth, and Moon Relationship
5th Grade Science » Unit: Out Of This World- A Journey Through Our Solar System
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Day 2: Roller Coaster Engineering Challenge
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The Weight of a Can of Soda Day 1
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Measuring the Weight & Mass of Pennies
5th Grade Science » Unit: Gravity
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In this lesson, students will be measuring the weight and mass of pennies using a spring scale and a balance scale. Then, students will graph their findings and record observations.
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Planning an Investigation
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The Weight of a Can of Soda Day 2
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Day 7: Roller Coaster Funnels & Half-Pipes
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Day 6: Building Up!
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Day One of Plaid Pete is Going Down!
5th Grade Science » Unit: Plaid Pete is Finding Earth's Place in the Universe
5th Grade Science » Unit: Plaid Pete is Finding Earth's Place in the Universe
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Next Gen Science | 856 | 3,398 | {"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-2017-09 | latest | en | 0.749355 |
https://flash-x.github.io/Flash-X-docs/Gravity.html | 1,721,130,807,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00584.warc.gz | 212,268,010 | 12,740 | # Gravity Unit
## Introduction
The Gravity unit supplied with Flash-X computes gravitational source terms for the code. These source terms can take the form of the gravitational potential $$\phi({\bf x})$$ or the gravitational acceleration $${\bf g}({\bf x})$$,
${\bf g}({\bf x}) = -\nabla \phi({\bf x})\ .$
The gravitational field can be externally imposed or self-consistently computed from the gas density via the Poisson equation,
$\label{Eqn:Poisson} \nabla^2\phi({\bf x}) = 4\pi G \rho({\bf x})\ ,$
where $$G$$ is Newton’s gravitational constant. In the latter case, either periodic or isolated boundary conditions can be applied.
## Externally Applied Fields
The Flash-X distribution includes three externally applied gravitational fields, along with a placeholder module for you to create your own. Each provides the acceleration vector $${\bf g}({\bf x})$$ directly, without using the gravitational potential $$\phi({\bf x})$$ (with the exception of UserDefined, see below).
When building an application that uses an external, time-independent Gravity implementation, no additional storage in unk for holding gravitational potential or accelerations is needed or defined.
### Constant Gravitational Field
This implementation creates a spatially and temporally constant field parallel to one of the coordinate axes. The magnitude and direction of the field can be set at runtime. This unit is called Gravity/GravityMain/Constant.
### Plane-parallel Gravitational field
This PlanePar version implements a time-constant gravitational field that is parallel to one of the coordinate axes and falls off with the square of the distance from a fixed location. The field is assumed to be generated by a point mass or by a spherically symmetric mass distribution. A finite softening length may optionally be applied.
This type of gravitational field is useful when the computational domain is large enough in the direction radial to the field source that the field is not approximately constant, but the domain’s dimension perpendicular to the radial direction is small compared to the distance to the source. In this case the angular variation of the field direction may be ignored. The PlanePar field is cheaper to compute than the PointMass field described below, since no fractional powers of the distance are required. The acceleration vector is parallel to one of the coordinate axes, and its magnitude drops off with distance along that axis as the inverse distance squared. Its magnitude and direction are independent of the other two coordinates.
### Gravitational Field of a Point Mass
This PointMass implementation describes the gravitational field due to a point mass at a fixed location. A finite softening length may optionally be applied. The acceleration falls off with the square of the distance from a given point. The acceleration vector is everywhere directed toward this point.
### User-Defined Gravitational Field
The UserDefined implementation is a placeholder module for the user to create their own external gravitational field. All of the subroutines in this module are stubs, and the user may copy these stubs to their setup directory to write their own implementation, either by specifying the gravitational acceleration directly or by specifying the gravitational potential and taking its gradient. If your user-defined gravitational field is time-varying, you may also want to set PPDEFINE |flashx|_GRAVITY_TIMEDEP in your setup’s Config file.
## Self-gravity
The self-consistent gravity algorithm supplied with Flash-X computes the Newtonian gravitational field produced by the matter. The produced potential function satisfies Poisson’s equation [Eqn:Poisson]. This unit’s implementation can also return the acceleration field $${\bf g}({\bf x})$$ computed by finite-differencing the potential using the expressions
\begin{split}\begin{aligned} \nonumber g_{x;ijk} &= {1\over2\Delta x}\left(\phi_{i-1,j,k} - \phi_{i+1,j,k}\right) + {\cal O}(\Delta x^2) \\ g_{y;ijk} &= {1\over2\Delta y}\left(\phi_{i,j-1,k} - \phi_{i,j+1,k}\right) + {\cal O}(\Delta y^2) \\ \nonumber g_{z;ijk} &= {1\over2\Delta z}\left(\phi_{i,j,k-1} - \phi_{i,j,k+1}\right) + {\cal O}(\Delta z^2) \ .\end{aligned}\end{split}
In order to preserve the second-order accuracy of these expressions at jumps in grid refinement, it is important to use quadratic interpolants when filling guard cells at such locations. Otherwise, the truncation error of the interpolants will produce unphysical forces at these block boundaries.
Two algorithms are available for solving the Poisson equations: Gravity/GravityMain/Multipole and Gravity/GravityMain/Multigrid. The initialization routines for these algorithms are contained in the Gravity unit, but the actual implementations are contained below the Grid unit due to code architecture constraints.
The multipole-based solver described in for self gravity is appropriate for spherical or nearly-spherical mass distributions with isolated boundary conditions. For non-spherical mass distributions higher order moments of the solver must be used. Note that storage and CPU costs scale roughly as the square of number of moments used, so it is best to use this solver only for nearly spherical matter distributions.
The multigrid solver described in is appropriate for general mass distributions and can solve problems with more general boundary conditions.
The tree solver described in [Sec:GridSolversBHTree] is appropriate for general mass distributions and can solve problems with both isolated and periodic boundary conditions set independently in individual directions.
### Coupling Gravity with Hydrodynamics
The gravitational field couples to the Euler equations only through the momentum and energy equations. If we define the total energy density as
$\rho E \equiv {1\over 2}\rho v^2 + \rho\epsilon\ ,$
where $$\epsilon$$ is the specific internal energy, then the gravitational source terms for the momentum and energy equations are $$\rho{\bf g}$$ and $$\rho{\bf v}\cdot{\bf g}$$, respectively. Because of the variety of ways in which different hydrodynamics schemes treat these source terms, the gravity module only supplies the potential $$\phi$$ and acceleration $${\bf g}$$, leaving the implementation of the fluid coupling to the hydrodynamics module. Finite-difference and finite-volume hydrodynamic schemes apply the source terms in their advection steps, sometimes at multiple intermediate timesteps and sometimes using staggered meshes for vector quantities like $${\bf v}$$ and $${\bf g}$$.
For example, the PPM algorithm supplied with Flash-X uses the following update steps to obtain the momentum and energy in cell $$i$$ at timestep $$n+1$$
\begin{split}\begin{aligned} \nonumber (\rho v)_i^{n+1} & = (\rho v)_i^n + {\Delta t\over 2} g_i^{n+1} \left(\rho_i^n + \rho_i^{n+1}\right)\\ (\rho E)_i^{n+1} & = (\rho E)_i^n + {\Delta t\over 4} g_i^{n+1} \left(\rho_i^n + \rho_i^{n+1}\right)\left(v_i^n + v_i^{n+1}\right)\ .\end{aligned}\end{split}
Here $$g_i^{n+1}$$ is obtained by extrapolation from $$\phi_i^{n-1}$$ and $$\phi_i^n$$. The $${\tt Poisson}$$ gravity implementation supplies a mesh variable to contain the potential from the previous timestep; future releases of Flash-X may permit the storage of several time levels of this quantity for hydrodynamics algorithms that require more steps. Currently, $${\bf g}$$ is computed at cell centers.
Note that finite-volume schemes do not retain explicit conservation of momentum and energy when gravity source terms are added. Godunov schemes such as PPM, require an additional step in order to preserve second-order time accuracy. The gravitational acceleration component $$g_i$$ is fitted by interpolants along with the other state variables, and these interpolants are used to construct characteristic-averaged values of $${\bf g}$$ in each cell. The velocity states $$v_{L,i+1/2}$$ and $$v_{R,i+1/2}$$, which are used as inputs to the Riemann problem solver, are then corrected to account for the acceleration using the following expressions
\begin{split}\begin{aligned} \nonumber v_{L,i+1/2} &\rightarrow& v_{L,i+1/2} + {\Delta t\over 4}\left(g^+_{L,i+1/2} + g^-_{L,i+1/2}\right)\\ v_{R,i+1/2} &\rightarrow& v_{R,i+1/2} + {\Delta t\over 4}\left(g^+_{R,i+1/2} + g^-_{R,i+1/2}\right)~.\end{aligned}\end{split}
Here $$g^\pm_{X,i+1/2}$$ is the acceleration averaged using the interpolant on the $$X$$ side of the interface ($$X=L,R$$) for $$v\pm c$$ characteristics, which bring material to the interface between cells $$i$$ and $$i+1$$ during the timestep.
### Tree Gravity
The Tree implementation of the gravity unit in physics/Gravity/GravityMain/Poisson/BHTree is meant to be used together with the tree solver implementation Grid/GridSolvers/BHTree/Wunsch. It either calculates the gravitational potential field which is subsequently differentiated in subroutine physics/Gravity/Gravity_accelOneRow to obtain the gravitational acceleration, or it calculates the gravitational acceleration directly. The latter approach is more accurate, because the error due to numerical differentiation is avoided, however, it consumes more memory for storing three components of the gravitational acceleration. The direct acceleration calculation can be switched on by specifying bhtreeAcc=1 as a command line argument of the setup script.
The gravity unit provides subroutines for building and walking the tree called by the tree solver. In this version, only monopole moments (node masses) are used for the potential/acceleration calculation. It also defines new multipole acceptance criteria (MACs) that estimate the error in gravitational acceleration of a contribution of a single node to the potential (hereafter partial error) much better than purely geometrical MAC defined in the tree solver. They are: (1) the approximate partial error (APE), and (2) the maximum partial error (MPE). The first one is based on an assumption that the partial error is proportional to the multipole moment of the node. The node is accepted for calculation of the potential if
$D^{m+2} > \frac{GMS_\mathrm{node}^m}{\Delta a_\mathrm{p,APE}}$
where $$D$$ is distance between the point-of-calculation and the node, $$M$$ is the node mass, $$S_\mathrm{node}$$ is the node size, $$m$$ is a degree of the multipole approximation and $$\Delta a_\mathrm{p,APE}$$ is the requested maximum error in acceleration (controlled by runtime parameter Gravity/grv_bhAccErr). Since only monopole moments are used for the potential calculation, the most reasonable choice of $$m$$ seems to be $$m=2$$. This MAC is similar to the one used in Gadget2 (see Springel, 2005, MNRAS, 364, 1105).
The second MAC (maximum partial error, MPE) calculates the error in acceleration of a single node contribution $$\Delta a_\mathrm{p,MPE}$$ according to formula 9 from Salmon&Warren (1994; see this paper for details):
$\Delta a_\mathrm{p,MPE} \le \frac{1}{D^2}\frac{1}{(1-S_\mathrm{node}/D)^2}\left( \frac{3\lceil B_2\rceil}{D^2} - \frac{2\lfloor B_3 \rfloor}{D^3}\right)$
where $$B_n = \Sigma_i m_i |\mathbf{r}_i-\mathbf{r}_0|^n$$ where $$m_i$$ and $$\mathbf{r}_i$$ are masses and positions of individual grid cells within the node and $$\mathbf{r}_0$$ is the node mass center position. Moment $$B_2$$ can be easily determined during the tree build, moment $$B_3$$ can be estimated as $$B_3^2 \ge B_2^3/M$$. The maximum allowed partial error in gravitational acceleration is controlled by runtime parameters Gravity/grv_bhAccErr and Gravity/grv_bhUseRelAccErr (see 1.4.1).
During the tree walk, subroutine physics/Gravity/Gravity_bhNodeContrib adds contributions of tree nodes to the gravitational potential or acceleration fields. In case of the potential, the contribution is
$\Phi = -\frac{GM}{|\vec{r}|}$
if isolated boundary conditions are used, or
$\Phi = -GMf_\mathrm{EF,\Phi}(\vec{r})$
if periodic periodic or mixed boundary conditions are used. In case of the acceleration, the contributions are
$\vec{a}_g = \frac{GM\vec{r}}{|\vec{r}|^3}$
for isolated boundary conditions, or
$\vec{a}_g = GMf_\mathrm{EF,a}(\vec{r})$
for periodic or mixed boundary conditions. In In the above formulae, $$G$$ is the constant of gravity, $$M$$ is the node mass, $$\vec{r}$$ is the position vector between point-of-calculation and the node mass center and $$f_\mathrm{EF,\Phi}$$ and $$f_\mathrm{EF,a}$$ are the Ewald fields for the potential and the acceleration (see below).
Boundary conditions can be isolated or periodic, set independently for each direction. If they are periodic at least in one direction, the Ewald method is used for the potential calculation (Ewald, P. P., 1921, Ann. Phys. 64, 253). The original Ewald method is an efficient method for computing gravitational field for problems with periodic boundary conditions in three directions. Ewald speeded up evaluation of the gravitational potential by splitting it into two parts, $$Gm/r=Gm \, \mathrm{erf}(\alpha r)/r + Gm \, \mathrm{erfc}(\alpha r)/r$$ ($$\alpha$$ is an arbitrary constant) and then by applying Poisson summation formula on erfc terms, gravitational field at position $$\vec{r}$$ can be written in the form
$\phi (\vec{r}) = -G \sum_{a=1}^N m_a \left( \sum_{i_1,i_2,i_3} A_S(\vec{r},\vec{r}_a,\vec{l}_{i_1,i_2,i_3}) + A_L(\vec{r},\vec{r}_a,\vec{l}_{i_1,i_2,i_3}) \right) = -G \sum_{a=1}^N m_a f_\mathrm{EF,\Phi}(\vec{r}_a - \vec{r}) \ , \label{ewald_sum}$
the first sum runs over whole computational domain, where at position $$\vec{r}_a$$ is mass $$m_a$$. Second sum runs over all neigbouring computational domains, which are at positions $$\vec{l}_{i_1,i_2,i_3}$$ and $$A_S(\vec{r},\vec{r}_a,\vec{l}_{i_1,i_2,i_3})$$ and $$A_L(\vec{r},\vec{r}_a,\vec{l}_{i_1,i_2,i_3})$$ are short and long–range contributions, respectively. It is sufficient to take into account only few terms in eq. [ewald_sum]. The Ewald field for the acceleration, $$f_\mathrm{EF,a}$$, is obtained using a similar decomposition. We modified Ewald method for problems with periodic boundary conditions in two directions and isolated boundary conditions in the third direction and for problems with periodic boundary conditions in one direction and isolated in two directions.
The gravity unit allows also to use a static external gravitational field read from file. In this version, the field can be either spherically symmetric or planar being only a function of the z-coordinate. The external field file is a text file containing three columns of numbers representing the coordinate, the potential and the acceleration. The coordinate is the radial distance or z-distance from the center of the external field given by runtime parameters. The external field if mapped to a grid using a linear interpolation each time the gravitational acceleration is calculated (in subroutine physics/Gravity/Gravity_accelOneRow).
## Usage
To include the effects of gravity in your Flash-X executable, include the option
-with-unit=physics/Gravity
on your command line when you configure the code with setup. The default implementation is Constant, which can be overridden by including the entire path to the specific implementation in the command line or Config file. The other available implementations are Gravity/GravityMain/Planepar, Gravity/GravityMain/Pointmass and Gravity/GravityMain/Poisson. The Gravity unit provides accessor functions to get gravitational acceleration and potential. However, none of the external field implementations of Section explicitly compute the potential, hence they inherit the null implementation from the API for accessing potential. The gravitation acceleration can be obtained either on the whole domain, a single block or a single row at a time.
When building an application that solves the Possion equation for the gravitational potential, additional storage is needed in unk for holding the last, as well as (usually) the previous, gravitational potential field; and, depending on the Poisson solver used, additional variables may be needed. The variables GPOT_VAR and GPOT_VAR, and others as needed, will be automatically defined in Simulation.h in those cases. See physics/Gravity/Gravity_potentialListOfBlocks for more information.
### Tree Gravity Unit Usage
Calculation of gravitational potential can be enabled by compiling in this unit and setting the runtime parameter Gravity/useGravity true. The constant of gravity can be set independently by runtime parameter Gravity/grv_bhNewton; if it is not positive, the constant Newton from the Flash-X PhysicalConstants database is used. If parameters Grid/gr_bhPhysMACTW or Grid/gr_bhPhysMACComm are set, the gravity unit MAC is used and it can be chosen by setting Gravity/grv_bhMAC to either ApproxPartialErr or MaxPartialErr. If the first one is used, the order of the multipole approximation is given by Gravity/grv_bhMPDegree.
The maximum allowed partial error in gravitational acceleration is set with the runtime parameter Gravity/grv_bhAccErr. It has either the meaning of an error in absolute acceleration or in relative acceleration normalized by the acceleration from the previous time-step. The latter is used if Gravity/grv_bhUseRelAccErr is set to True, and in this case the first call of the tree solver calculates the potential using purely geometrical MAC (because the acceleration from the previous time-step does not exist).
Boundary conditions are set by the runtime parameter Gravity/grav_boundary_type and they can be isolated, periodic or mixed. In the case of mixed boundary conditions, runtime parameters Gravity/grav_boundary_type_x, Gravity/grav_boundary_type_y and Gravity/grav_boundary_type_z specify along which coordinate boundary conditions are periodic and isolated (possible values are periodic or isolated). Arbitrary combination of these values is permitted, thus suitable for problems with planar resp. linear symmetry. It should work for computational domain with arbitrary dimensions. The highest accuracy is reached with blocks of cubic physical dimensions.
If runtime parameter Gravity/grav_boundary_type is periodic or mixed, then the Ewald field for appropriate symmetry is calculated at the beginning of the simulation. Parameter Gravity/grv_bhEwaldSeriesN controls the range of indices $$i_1,i_2,i_3$$ in (eq. [ewald_sum]). There are two implementations of the Ewald method: the new one (default) requires less memory and it should be faster and of comparable accuracy as the old one. The default implementation computes Ewald field minus the singular $$1/r$$ term and its partial derivatives on a single cubic grid, and the Ewald field is then approximated by the first order Taylor formula. Parameter Gravity/grv_bhEwaldNPer controls number of grid points in the $$x$$ direction in the case of periodic or in periodic direction(s) in the case of mixed boundary conditions. Since an elongated computational domain is often desired when Gravity/grav_boundary_type is mixed, the cubic grid would lead to a huge field of data. In this case, the amount of necessary grid points is reduced by using an analytical estimate to the Ewald field sufficiently far away of the symmetry plane or axis.
The old implementation (from Flash4.2) is still present and is enabled by adding bhtreeEwaldV42=1 on the setup command line. The Ewald field is then stored in a nested set of grids, the first of them corresponds in size to full computational domain, and each following grid is half the size (in each direction) of the previous grid. Number of nested grids is controlled by runtime parameter Gravity/grv_bhEwaldNRefV42. If Gravity/grv_bhEwaldNRefV42 is too low to cover origin (where is the Ewald field discontinuous), then the run is terminated. Each grid is composed of Gravity/grv_bhEwaldFieldNxV42 $$\times$$ Gravity/grv_bhEwaldFieldNyV42 $$\times$$ Gravity/grv_bhEwaldFieldNzV42 points. When evaluation of the Ewald Field at particular point is needed at any time during a run, the field value is found by interpolation in a suitable level of the grid. Linear or semi-quadratic interpolation can be chosen by runtime parameter Gravity/grv_bhLinearInterpolOnlyV42 (option true corresponds to linear interpolation). Semi-quadratic interpolation is recommended only in the case when there are periodic boundary conditions in two directions.
The external gravitational field can be switched on by setting Gravity/grv_useExternalPotential true. The parameter Gravity/grv_bhExtrnPotFile gives the name of the file with the external potential and Gravity/grv_bhExtrnPotType specifies the field symmetry: spherical for the spherical symmetry and planez for the planar symmetry with field being a function of the z-coordinate. Parameters Gravity/grv_bhExtrnPotCenterY, Gravity/grv_bhExtrnPotCenterX and Gravity/grv_bhExtrnPotCenterZ specify the position (in the simulation coordinate system) of the external field origin (the point where the radial or z-coordinate is zero).
calculation.
Variable
Type
Default
Description
Gravit y/grv_bhNewton
real
-1.0
constant of gravity; if $$<$$ 0, it is obtained
from internal physical constants database
Gra vity/grv_bhMAC
string
“A pproxPartialErr”
MAC, other option: “MaxPartialErr”
Gravity/ grv_bhMPDegree
integer
2
degree of multipole in error estimate in APE MAC
Gravity/grv_ bhUseRelAccErr
logical
.false.
if .true., grv_bhAccErr has meaning of
relative error, otherwise absolute
Gravit y/grv_bhAccErr
real
0.1
maximum allowed error in gravitational
acceleration
conditions.
Variable
Type
Default
Description
Gravity/grav _boundary_type
string
“isolated”
or “periodic” or “mixed”
Gravity/grav_b oundary_type_x
string
“isolated”
or “periodic”
Gravity/grav_b oundary_type_y
string
“isolated”
or “periodic”
Gravity/grav_b oundary_type_z
string
“isolated”
or “periodic”
Gra vity/grv_bhEwald AlwaysGenerate
boolean
true
whether Ewald field should be regenerated
Gravity/grv_ bhEwaldSeriesN
integer
$$10$$
number of terms in the Ewald series
Gravity/g rv_bhEwaldNPer
integer
32
number of points+1 of the Taylor expansion
Gravity/gr v_bhEwaldFName
string
“ewald_coeffs”
file with coefficients of the Ewald field Taylor expansion
Gravity/grv_bhE waldFieldNxV42
integer
$$32$$
size of the Ewald field grid in x-direction
Gravity/grv_bhE waldFieldNyV42
integer
$$32$$
size of the Ewald field grid in y-direction
Gravity/grv_bhE waldFieldNzV42
integer
$$32$$
size of the Ewald field grid in z-direction
Gravity/grv_ bhEwaldNRefV42
integer
-1
number of refinement levels (nested grids) for the Ewald
field; if $$<$$ 0, determined automatically
Gravi ty/grv_bhLinearI nterpolOnlyV42
logical
.true.
if .false., semi-quadratic interpolation is used for
interpolation in the Ewald field
Gravity/grv_bhEw aldFNameAccV42
string
” ewald_field_acc”
file with the Ewald field for acceleration
Gravity/grv_bhEw aldFNamePotV42
string
” ewald_field_pot”
file with coefficients of the Ewald field for potential
Tree gravity unit parameters controlling the external gravitational field.
Variable
Type
Default
Description
Grav ity/grv_bhUseExt ernalPotential
logical
.false.
whether to use external field
Gra vity/grv_bhUsePo issonPotential
logical
.true.
whether to use gravitational field calculated by the
tree solver
Gravity/grv_ bhExtrnPotFile
string
“externa l_potential.dat”
file containing the external gravitational field
Gravity/grv_ bhExtrnPotType
string
“planez”
type of the external field: planar or spherical symmetry
Gravity/grv_bhE xtrnPotCenterX
real
0.0
x-coordinate of the center of the external field
Gravity/grv_bhE xtrnPotCenterY
real
0.0
y-coordinate of the center of the external field
Gravity/grv_bhE xtrnPotCenterZ
real
0.0
z-coordinate of the center of the external field
## Unit Tests
There are two unit tests for the gravity unit. Poisson3 is essentially the Maclaurin spheroid problem described in . Because an analytical solution exists, the accuracy of the gravitational solver can be quantified. The second test, Poisson3_active is a modification of Poisson3 to test the mapping of particles in Grid/Grid_mapParticlesToMesh. Some of the mesh density is redistributed onto particles, and the particles are then mapped back to the mesh, using the analytical solution to verify completeness. This test is similar to the simulation PoisParticles discussed in . PoisParticles is based on the Huang-Greengard Poisson gravity test described in . | 5,843 | 24,537 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.840773 |
https://hg.ucc.asn.au/dropbear/file/8bba51a55704/libtommath/bn_mp_mul_2d.c | 1,638,353,424,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00237.warc.gz | 381,897,242 | 4,051 | ### view libtommath/bn_mp_mul_2d.c @ 1470:8bba51a55704
Update to libtommath v1.0.1
author Matt Johnston Thu, 08 Feb 2018 23:11:40 +0800 60fc6476e044 f52919ffd3b1
line wrap: on
line source
```#include <tommath_private.h>
#ifdef BN_MP_MUL_2D_C
/* LibTomMath, multiple-precision integer library -- Tom St Denis
*
* LibTomMath is a library that provides multiple-precision
* integer arithmetic as well as number theoretic functionality.
*
* The library was designed directly after the MPI library by
* Michael Fromberger but has been written from scratch with
*
* The library is free for all purposes without any express
* guarantee it works.
*
* Tom St Denis, [email protected], http://libtom.org
*/
/* shift left by a certain bit count */
int mp_mul_2d (mp_int * a, int b, mp_int * c)
{
mp_digit d;
int res;
/* copy */
if (a != c) {
if ((res = mp_copy (a, c)) != MP_OKAY) {
return res;
}
}
if (c->alloc < (int)(c->used + (b / DIGIT_BIT) + 1)) {
if ((res = mp_grow (c, c->used + (b / DIGIT_BIT) + 1)) != MP_OKAY) {
return res;
}
}
/* shift by as many digits in the bit count */
if (b >= (int)DIGIT_BIT) {
if ((res = mp_lshd (c, b / DIGIT_BIT)) != MP_OKAY) {
return res;
}
}
/* shift any bit count < DIGIT_BIT */
d = (mp_digit) (b % DIGIT_BIT);
if (d != 0) {
mp_digit *tmpc, shift, mask, r, rr;
int x;
mask = (((mp_digit)1) << d) - 1;
/* shift for msbs */
shift = DIGIT_BIT - d;
/* alias */
tmpc = c->dp;
/* carry */
r = 0;
for (x = 0; x < c->used; x++) {
/* get the higher bits of the current word */
rr = (*tmpc >> shift) & mask;
/* shift the current word and OR in the carry */
*tmpc = ((*tmpc << d) | r) & MP_MASK;
++tmpc;
/* set the carry to the carry bits of the current word */
r = rr;
}
/* set final carry */
if (r != 0) {
c->dp[(c->used)++] = r;
}
}
mp_clamp (c);
return MP_OKAY;
}
#endif
/* ref: \$Format:%D\$ */
/* git commit: \$Format:%H\$ */
/* commit time: \$Format:%ai\$ */
``` | 639 | 1,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-49 | longest | en | 0.585778 |
https://stackoverflow.com/questions/26081118/iterative-and-conditional-deleting-of-lines-in-a-file | 1,576,358,841,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541294513.54/warc/CC-MAIN-20191214202754-20191214230754-00236.warc.gz | 554,623,294 | 32,588 | Intro
I have a file named `data.dat` with the following structure:
`````` 1: 67: 1 :s
1: 315: 1 :s
1: 648: 1 :ns
1: 799: 1 :s
1: 809: 1 :s
1: 997: 1 :ns
2: 32: 1 :s
``````
Algorithm
The algorithm that I'm looking for is:
1. Generate a random number between 1 and number of lines in this file.
2. Delete that line if the fourth column is "s".
3. Otherwise generate another random number and repeat this until the number of lines reaches to a certain value.
Technical Concepts
Though technical concepts are irrelevant to this algorithm, but I try to explain the problem. The data shows connectivity table of a network. This algorithm allows us to run it over different initial conditions and study general properties of these networks. Especially, because of randomness property of deleting bonds, any common behavior among these networks can be interpreted as a fundamental law.
Update: Another good reason to produce a random number in each step is that after removing each line, it's possible that property of being `s`/`ns` of remaining lines can be changed.
Code
Here is the code I have until now:
``````#!/bin/bash
# bash in OSX
While ((#there is at least 1 s in the fourth column)); do
LEN=\$(grep -c "." data.dat) # number of lines
RAND=\$((RANDOM%\${LEN}+1)) # generating random number
if [[awk -F, "NR==\$RAND" 'data.dat' | cut -d ':' -f 4- == "s"]]; then
sed '\$RANDd' data.txt
else
#go back and produce another random
done
exit
``````
I try to find the fourth column with `awk -F, "NR==\$RAND" 'data.dat' | cut -d ':' -f 4-` and deleting the line by `sed '\$RANDd' data.txt`.
Questions
1. How should I check that there is `s` pairs in my file?
2. I am not sure if the condition in `if` is correct.
3. Also, I don't know how to force loop after `else` to go back to generate another random number.
Thank you,
I really appreciate your help.
• You're deleting every line that ends with ":s". Why bother with random numbers and iteration? – Beta Sep 28 '14 at 4:27
• It might sound technical but this is a network that I am interested in percolation and its flexibility under random removing of bonds. – Mahdi Sep 28 '14 at 4:31
• @JohnB: I know that the fastest way is to remove all lined with `s` but this is only one part of the study. As in future, I am intended to study these networks before reaching to the specific threshold (with no s). Basically, if I can run this, I have a program to run for any final number of `s` rows. Also it's important to remove the lines randomly to avoid any biased result. – Mahdi Sep 28 '14 at 4:44
• Perhaps it would be better to update your question with details to justify exactly how `RANDOM` is needed for a network study. Also, there are numerous syntax errors in your code. shellcheck.net might help. – John B Sep 28 '14 at 5:00
• I was wondering, have you tried any of the answers? – Tom Fenech Oct 1 '14 at 17:11
Personally, I would recommend against doing this in bash unless you have absolutely no choice.
Here's another way you could do it in Perl (quite similar in functionality to Alex's answer but a bit simpler):
``````use strict;
use warnings;
my \$filename = shift;
open my \$fh, "<", \$filename or die "could not open \$filename: \$!";
chomp (my @lines = <\$fh>);
my \$sample = 0;
my \$max_samples = 10;
while (\$sample++ < \$max_samples) {
my \$line_no = int rand @lines;
my \$line = \$lines[\$line_no];
if (\$line =~ /:s\s*\$/) {
splice @lines, \$line_no, 1;
}
}
print "\$_\n" for @lines;
``````
Usage: `perl script.pl data.dat`
Read the file into the array `@lines`. Pick a random line from the array and if it ends with :s (followed by any number of spaces), remove it. Print the remaining lines at the end.
This does what you want but I should warn you that relying on built-in random number generators in any language is not a good way to arrive at statistically significant conclusions. If you need high-quality random numbers, you should consider using a module such as Math::Random::MT::Perl to generate them, rather than the built-in `rand`.
``````#!/usr/bin/env perl
# usage: \$ excise.pl < data.dat > smaller_data.dat
my \$sampleLimit = 10; # sample up to ten lines before printing output
my \$dataRef;
my \$flagRef;
while (<>) {
chomp;
push (@{\$dataRef}, \$_);
push (@{\$flagRef}, 1);
}
my \$lineCount = scalar @elems;
my \$sampleIndex = 0;
while (\$sampleIndex < \$sampleLimit) {
my \$sampleLineIndex = int(rand(\$lineCount));
my @sampleElems = split("\t", \$dataRef->[\$sampleLineIndex];
if (\$sampleElems[3] == "s") {
\$flagRef->[\$sampleLineIndex] = 0;
}
\$sampleIndex++;
}
# print data.dat to standard output, minus any sampled lines that had an 's' in them
foreach my \$lineIndex (0..(scalar @{\$dataRef} - 1)) {
if (\$flagRef->[\$lineIndex] == 1) {
print STDOUT \$dataRef->[\$lineIndex]."\n";
}
}
``````
``````NumLine=\$( grep -c "" data.dat )
while [ \${NumLine} -gt \${TargetLine} ]
do
# echo "Line at start: \${NumLine}"
RndLine=\$(( ( \${RANDOM} % \${NumLine} ) + 1 ))
RndValue="\$( echo " \${RANDOM}" | sed 's/.*\(.\{6\}\)\$/\1/' )"
sed "\${RndLine} {
s/^\([^:]*:\)[^:]*\(:.*:ns\$\)/\1\${RndValue}\2/
t
d
}" data.dat > /tmp/data.dat
mv /tmp/data.dat data.dat
NumLine=\$( grep -c "" data.dat )
#cat data.dat
#echo "- Next Iteration -------"
done
``````
tested on AIX (so not a GNU sed). Under Linux, use `--posix` for sed option and you can use a `-i` in place of temporary file + redirection + move in this case
Dont't forget that `RANDOM` is NOT a real RANDOM so study on network behavior based on not random value could not reflect a reality bu a specific case | 1,606 | 5,647 | {"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-2019-51 | latest | en | 0.867295 |
https://se.mathworks.com/help/optim/ug/optimconstr.html | 1,670,578,579,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00520.warc.gz | 531,671,840 | 21,916 | # optimconstr
Create empty optimization constraint array
## Description
Use optimconstr to initialize a set of constraint expressions.
Tip
For the full workflow, see Problem-Based Optimization Workflow.
example
constr = optimconstr(N) creates an N-by-1 array of empty optimization constraints. Use constr to initialize a loop that creates constraint expressions.
example
constr = optimconstr(cstr) creates an array of empty optimization constraints that are indexed by cstr, a cell array of character vectors or string vectors.
If cstr is 1-by-ncstr, where ncstr is the number of elements of cstr, then constr is also 1-by-ncstr. Otherwise, constr is ncstr-by-1.
constr = optimconstr(cstr1,N2,...,cstrk) or constr = optimconstr({cstr1,cstr2,...,cstrk}) or constr = optimconstr([N1,N2,...,Nk]), for any combination of cstr and N arguments, creates an ncstr1-by-N2-by-...-by-ncstrk array of empty optimization constraints, where ncstr is the number of elements in cstr.
## Examples
collapse all
Create constraints for an inventory model. The stock of goods at the start of each period is equal to the stock at the end of the previous period. During each period, the stock increases by buy and decreases by sell. The variable stock is the stock at the end of the period.
N = 12;
stock = optimvar('stock',N,1,'Type','integer','LowerBound',0);
sell = optimvar('sell',N,1,'Type','integer','LowerBound',0);
initialstock = 100;
stockbalance = optimconstr(N,1);
for t = 1:N
if t == 1
enterstock = initialstock;
else
enterstock = stock(t-1);
end
stockbalance(t) = stock(t) == enterstock + buy(t) - sell(t);
end
show(stockbalance)
(1, 1)
-buy(1) + sell(1) + stock(1) == 100
(2, 1)
-buy(2) + sell(2) - stock(1) + stock(2) == 0
(3, 1)
-buy(3) + sell(3) - stock(2) + stock(3) == 0
(4, 1)
-buy(4) + sell(4) - stock(3) + stock(4) == 0
(5, 1)
-buy(5) + sell(5) - stock(4) + stock(5) == 0
(6, 1)
-buy(6) + sell(6) - stock(5) + stock(6) == 0
(7, 1)
-buy(7) + sell(7) - stock(6) + stock(7) == 0
(8, 1)
-buy(8) + sell(8) - stock(7) + stock(8) == 0
(9, 1)
-buy(9) + sell(9) - stock(8) + stock(9) == 0
(10, 1)
-buy(10) + sell(10) - stock(9) + stock(10) == 0
(11, 1)
-buy(11) + sell(11) - stock(10) + stock(11) == 0
(12, 1)
-buy(12) + sell(12) - stock(11) + stock(12) == 0
Include the constraints in a problem.
prob = optimproblem;
prob.Constraints.stockbalance = stockbalance;
Instead of using a loop, you can create the same constraints by using matrix operations on the variables.
tt = ones(N-1,1);
d = diag(tt,-1); % shift index by -1
stockbalance2 = stock == d*stock + buy - sell;
stockbalance2(1) = stock(1) == initialstock + buy(1) - sell(1);
Show the new constraints to verify that they are the same as the constraints in stockbalance.
show(stockbalance2)
(1, 1)
-buy(1) + sell(1) + stock(1) == 100
(2, 1)
-buy(2) + sell(2) - stock(1) + stock(2) == 0
(3, 1)
-buy(3) + sell(3) - stock(2) + stock(3) == 0
(4, 1)
-buy(4) + sell(4) - stock(3) + stock(4) == 0
(5, 1)
-buy(5) + sell(5) - stock(4) + stock(5) == 0
(6, 1)
-buy(6) + sell(6) - stock(5) + stock(6) == 0
(7, 1)
-buy(7) + sell(7) - stock(6) + stock(7) == 0
(8, 1)
-buy(8) + sell(8) - stock(7) + stock(8) == 0
(9, 1)
-buy(9) + sell(9) - stock(8) + stock(9) == 0
(10, 1)
-buy(10) + sell(10) - stock(9) + stock(10) == 0
(11, 1)
-buy(11) + sell(11) - stock(10) + stock(11) == 0
(12, 1)
-buy(12) + sell(12) - stock(11) + stock(12) == 0
Creating constraints in a loop can be more time-consuming than creating constraints by matrix operations. However, you are less likely to create an erroneous constraint by using loops.
Create indexed constraints and variables to represent the calories consumed in a diet. Each meal has a different calorie limit.
meals = ["breakfast","lunch","dinner"];
constr = optimconstr(meals);
foods = ["cereal","oatmeal","yogurt","peanut butter sandwich","pizza","hamburger",...
diet = optimvar('diet',foods,meals,'LowerBound',0);
calories = [200,175,150,450,350,800,150,650,350,300]';
for i = 1:3
constr(i) = diet(:,i)'*calories <= 250*i;
end
Check the constraint for dinner.
show(constr("dinner"))
200*diet('cereal', 'dinner') + 175*diet('oatmeal', 'dinner')
+ 150*diet('yogurt', 'dinner')
+ 450*diet('peanut butter sandwich', 'dinner') + 350*diet('pizza', 'dinner')
+ 800*diet('hamburger', 'dinner') + 150*diet('salad', 'dinner')
+ 650*diet('steak', 'dinner') + 350*diet('casserole', 'dinner')
+ 300*diet('ice cream', 'dinner') <= 750
## Input Arguments
collapse all
Size of the constraint dimension, specified as a positive integer.
• The size of constr = optimconstr(N) is N-by-1.
• The size of constr = optimconstr(N1,N2) is N1-by-N2.
• The size of constr = optimconstr(N1,N2,...,Nk) is N1-by-N2-by-...-by-Nk.
Example: 5
Data Types: double
Names for indexing, specified as a cell array of character vectors or a string vector.
Example: {'red','orange','green','blue'}
Example: ["red";"orange";"green";"blue"]
Data Types: string | cell
## Output Arguments
collapse all
Constraints, returned as an empty OptimizationConstraint array. Use constr to initialize a loop that creates constraint expressions.
For example:
x = optimvar('x',8);
constr = optimconstr(4);
for k = 1:4
constr(k) = 5*k*(x(2*k) - x(2*k-1)) <= 10 - 2*k;
end
## Limitations
• Each constraint expression in a problem must use the same comparison. For example, the following code leads to an error, because cons1 uses the <= comparison, cons2 uses the >= comparison, and cons1 and cons2 are in the same expression.
prob = optimproblem;
x = optimvar('x',2,'LowerBound',0);
cons1 = x(1) + x(2) <= 10;
cons2 = 3*x(1) + 4*x(2) >= 2;
prob.Constraints = [cons1;cons2]; % This line throws an error
You can avoid this error by using separate expressions for the constraints.
prob.Constraints.cons1 = cons1;
prob.Constraints.cons2 = cons2;
## Tips
• It is generally more efficient to create constraints by vectorized expressions rather than loops. See Create Efficient Optimization Problems.
• You can use optimineq instead of optimconstr to create inequality expressions. Similarly, you can use optimeq instead of optimconstr to create equality expressions.
## Version History
Introduced in R2017b | 2,022 | 6,219 | {"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-2022-49 | longest | en | 0.702691 |
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# exam 4 fall 2007 - Engineering Mathematics(E35 317 Exam 4...
• Test Prep
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Engineering Mathematics (E35 317) Exam 4 December 5, 2007 This exam contains seven multiple-choice problems worth two points each, eleven true-false problems worth one point each, and two free-response problems worth 15 points altogether, for an exam total of 40 points. There is a formula list on the last page of the exam. Part I. Multiple-Choice Clearly circle the only correct response. Each is worth two points. 1. Consider the differential equation .B C BC Ð*B %ÑC œ !#www# 2. Consider the following Sturm-Liouville problem. C C œ ! CÐ!Ñ œ ! C Ð Ñ œ ! ww w - 1 One eigenfunction is . (You do not need to verify this.) Find the corresponding C œ sin B # eigenvalue. (You do not need to solve the Sturm-Liouville problem in order to find the eigenvalue.) (A) (B) " " % # (C) (D) 1 1 % # (E) (F) 1 1 # # % # (G) (H) B B % # (I) (J) 1 1 B B % #
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3. Let on . Find . 0ÐBÑ œ B " Ÿ B Ÿ " ll B ll (A) !
• Fall '10
• Sin, Cos, Boundary value problem, Partial differential equation, initial boundary value
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I'm having a difficult time understanding this statement. Can someone please explain with a concrete example? | 452 | 1,734 | {"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} | 3.359375 | 3 | CC-MAIN-2016-30 | latest | en | 0.927496 |
http://mathhelpforum.com/pre-calculus/153996-conjecturing-sinusoids-using-cas.html | 1,508,788,109,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826283.88/warc/CC-MAIN-20171023183146-20171023203146-00837.warc.gz | 211,189,166 | 11,980 | # Thread: Conjecturing sinusoids using a CAS?
1. ## Conjecturing sinusoids using a CAS?
I really have no idea what any of this means, and I think I just need a translation...
Let $y=sin(ax) + cos(ax)$.
Use the symbolic manipulator of a computer algebra system (CAS) to conjecture the following:
a) Express y as a sinusoid for a=2,3,4 and 5
b) Conjecture another formula for y for a equal to any positive integer n.
c) Check your conjecture with a CAS.
d) Use the formula for the sine of the sum of two angles to confirm your conjecture.
Now, I have no idea what a symbolic manipulator, computer algebra system, or sinusoid is. I have a TI-84 and I really have no idea what just happened... help?
2. Do you know what the functions sin and cos are? You probably should but if you don't, here is something that may help Trigonometric functions - Wikipedia, the free encyclopedia. The stuff inside the functions is just a times x, where a is a constant, usually a real number.
Symbolic manipulator is a program like Mathematica, which lets you plug in equations without number (such as the above equation) and you can literally manipulate the expressions. In Mathematica for example, there are powerful algorithms that let you factor and simplify expressions and also transform expression using known identities. A CAS is a computer program with which you can do numerical math calculations. Nowadays a CAS and symbolic manipulator come hand in hand. Mathematica and Maple are the one I use.
In part a), you need to use trig identities for the double angles and for the sums of angles to express y as a sum of products of sine and cosine functions that involve only a single x. Once you do this for the first couple of values of a, you should see a clear pattern. Then you need to conjecture a formula.
In part c) you need to use a CAS to manipulate either the conjectured formula or simplify y.
3. I understand it a little more, but I still can't figure anything out.
I do understand the trig functions, but I can't figure out how to get the x alone using them. This is what I have, and it's only a few lines:
a) $y=sin(2x)+cos(2x)$
$y=2sin(x)cos(x)+(cos(x))^2-(sin(x))^2$
$y=2sin(x)cos(x)+1-2(sin(x))^2$
And yeah, I hope that part's right. I downloaded that Mathematica thingy, but holy cow is it complicated. It keeps telling me I'm wrong, and I can't help but to agree with it. Where should I go from here...?
4. You were on the right track. Just leave it as it is in the 2nd line. Do the 3x case and you'll see the pattern.
5. This is what I've collected so far, with help from the website: Summary of Trigonometric Identities.
(I omitted the x's.)
$1a) y=sin(2cos)-1+2cos^2$
$2a) y=sin(-1+4cos^2)-3cos+4cos^3$
$3a) y=sin(-4cos+8cos^3)+1-8cos^2+8cos^4$
$4a) y=sin(1-12cos^2+16cos^4)+5cos-20cos^3+16cos^5$
$b) y=2sin((n-1)x)cos(x)-sin((n-2)x)+2cos((n-1)x)cos(x)-cos((n-2)x)$
Now, my mind is like blown because I tried to get the triple and quadruple angle ones by like putting 2x into the previous one in place of x, but everything got like ten million times more confusing so I just trusted the site's triple and quadruple angle formulas. And I can see the rising exponents thing, but everything else just looks like random gibberish to me and I cannot see how anybody can draw a formula from that.
Anyways, I have no way of checking this with a CAS and since the answer in the back of my book for part (c) is "It works," I think I'm okay. The only problem is that the answer in my book is this:
$sqrt(2)sin(ax+(pi/4)$
I don't even understand where any of that comes from... and the proof they show for part (d) is just as overwhelming.
Can someone explain where they got their answer wrong, or why my answer is so long and possibly not right...?
6. Here is something to think about.
Let $\displaystyle \cos(x) = a$ and $\sin(x) = b$. Then
$\displaystyle \cos(2x)+\sin(2x) = a^2+ 2ab - b^2$
$\displaystyle (a-b)^2 = a^2 -2ab +b^2$
Next
$\displaystyle \cos(3x)+\sin(3x) = a^3+3a^2b-3ab^2+b^3$
$\displaystyle (a-b)^3 = a^3 - 3a^2b + 3ab^2 -b^3$
This is a definite hint on the general form of the identity. I believe the general identity is not far from this. | 1,174 | 4,176 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.908423 |
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December 3, 2020 2:40 pm 30 pts
51. Number of maximal ideals of Z36 is (A) 3 (B) 2 C) 4 (D) None of these 52. Let R be a commutative ring with unity and let A be an ideal of R. ' Then cho0se the correct statement: (A) R/A is an integral domain iff A is prime. (B) R/A is an integral domain iff A is maximal. (C) R/A is a field iff A is prime. (D) None of these 33. Let R be a ring with unity such that the only left ideals of R are (0) and R. Then R 1s a: (A) field (B) division ring (C)principal ideal domain (D) None of these 54. Let R be a commutative ring with unity. Then choose the correct statement: (A) Every maximal ideal is a prime ideal. (B) Every prime ideal is a maximal ideal. (C) Both (A) and (B) (D) None of these 55. The characteristic of an integral domain is either (A) 0 or 1 (B) 0 or prime (C) 1 or prime (D) None of these XE(H-3)-MAT(6) (12) | 295 | 878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-04 | latest | en | 0.877832 |
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Assignment help-CASE QUESTIONS: Warren E Buffett 2005
CASE QUESTIONS: Warren E Buffett 2005
1. Market reaction: What does the stock market seem to say about the acquisition of PacifiCorp by Berkshire Hathaway? Specifically, does the deal create value? If so, for which party? What does the \$2.55 billion increase in Berkshire Hathaway’s market value represent?
2. Choice of valuation methods: What do you think PacifiCorp is worth on its own before its acquisition by Berkshire? Which valuation method should you use to value PacifiCorp and why? Show clearly the steps to arrive at the following estimates in Exhibit 10:
Enterprise Value as Multiple of: MV Equity as Multiple of: Revenue EBIT EBITDA Net Income EPS Book Value Median 6,252 8,775 9,023 7,596 4,277 5,904 Mean 6,584 9,289 9,076 7,553 4,308 5,678 If you need to use a discount rate to discount cash flows then an appropriate discount rate estimate for PacifiCorp is approximately 9%.
3. Bid assessment: How do you assess the bid for PacifiCorp by Berkshire Hathaway? How much does Buffett pay for PacifiCorp for its equity and as a whole? How do these values compare with the firm’s intrinsic values estimated in Exhibit 10?
4. Investment valuation: Evaluate Berkshire Hathaway’s investment in MidAmerican Energy Holdings in 2000. Using the information from Exhibit 6, calculate the net gain to Berkshire Hathaway in 2000 dollars. First, calculate the free cash flow accruing to Berkshire from MidAmerican each year from 2001 to 2004. Second, discount the cash flows to year 2000 and compare with Berkshire’s investment in MidAmerican. Assume tax = 40% on EBIT, discount rate r = 9%, growth rate for the steady period starting 2004 g = 2%. Correction: Exhibit 6, Balance sheets, Assets: “Properties, plants, and equipment” and “Goodwill” figures are “gross” amount, not “net”. Clearly state your assumptions if any.
5. Past performance: Discuss Berkshire Hathaway’s business strategy during 1965-2004. How well did Berkshire Hathaway perform during that period? Using financial databases (such as Bloomberg, Datastream, Yahoo Finance, etc.), assess Berkshire Hathaway’s recent performance during the period of 10 years up the end of last year. Comment on your findings.
6. Investment assessment: Assess Berkshire’s investment in Buffett’s Big Four: American Express, Coca-Cola, Gillette, and Wells Fargo in terms of strategy and return performance (each individual company and portfolio as a whole).
7. Investment philosophy: Critically assess Buffett’s investment philosophy. Identify the points where you agree and disagree with him.
Assignment help-CASE QUESTIONS: Warren E Buffett 2005
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http://www.numbersaplenty.com/104260 | 1,596,509,348,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735851.15/warc/CC-MAIN-20200804014340-20200804044340-00050.warc.gz | 159,754,164 | 3,764 | Search a number
104260 = 22513401
BaseRepresentation
bin11001011101000100
312022000111
4121131010
511314020
62122404
7612652
oct313504
9168014
10104260
1171372
1250404
13385c0
1429dd2
1520d5a
hex19744
104260 has 24 divisors (see below), whose sum is σ = 236376. Its totient is φ = 38400.
The previous prime is 104243. The next prime is 104281. The reversal of 104260 is 62401.
Adding to 104260 its reverse (62401), we get a palindrome (166661).
It can be written as a sum of positive squares in 4 ways, for example, as 21316 + 82944 = 146^2 + 288^2 .
It is a Harshad number since it is a multiple of its sum of digits (13).
It is a self number, because there is not a number n which added to its sum of digits gives 104260.
It is an unprimeable number.
104260 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 60 + ... + 460.
It is an arithmetic number, because the mean of its divisors is an integer number (9849).
104260 is a Friedman number, since it can be written as 4010*26, using all its digits and the basic arithmetic operations.
It is a vampire number, since it can be written as 260401 (with its own digits).
2104260 is an apocalyptic number.
104260 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
It is an amenable number.
It is a practical number, because each smaller number is the sum of distinct divisors of 104260, and also a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (118188).
104260 is an abundant number, since it is smaller than the sum of its proper divisors (132116).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
104260 is a wasteful number, since it uses less digits than its factorization.
104260 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 423 (or 421 counting only the distinct ones).
The product of its (nonzero) digits is 48, while the sum is 13.
The square root of 104260 is about 322.8931711882. The cubic root of 104260 is about 47.0658500537.
The spelling of 104260 in words is "one hundred four thousand, two hundred sixty". | 644 | 2,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-34 | latest | en | 0.919566 |
https://electronics.stackexchange.com/questions/363502/lm317-equation-help | 1,716,343,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00639.warc.gz | 190,743,015 | 38,912 | LM317 equation help
So I have this voltage regulator circuit. Applying the voltage divider,
then
Vout = Vadj * (R1+R2)/R2 = Vadj * (1 + R1/R2).
Why every site shows Vout = Vadj * (1 + R2/R1)?
Am I doing something wrong?
• You are confusing Vadj with Vref Mar 19, 2018 at 20:01
You are confusing Vadj with Vref.
The device setps up Vref between the right output pin and the adjustment pin.
As such the current through R1 is
$I_{R1} = V_{REF}/R_1$
So, ignoring the small adjustment current through the device for now..
$V_{OUT} = V_{REF} + I_{R_1} * R_2$
$= V_{REF} + V_{REF} * R_2 /R_1$
$= V_{REF}(1 + R_2 /R_1)$
Of course in reality you need to add in $I_{ADJ} * R_2$ to that.
• Thanks! I found where I started to confuse things, but I didn't fully understand. Is the current through R1 passing R2 also? Isn't there a current node where it splits throug adj and R2? Mar 20, 2018 at 18:57
• So I did a little bit more research after your answer gave me something new to search. After one hour of researching I opened the LM317 datasheet, and at basic circuit operation everything was explained. I wrote down Kirchhoff's law on the output and the result is exactly the equation that I didn't understand in the first place. Thank you very much for your help! Mar 20, 2018 at 19:31 | 387 | 1,293 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.937773 |
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# sentence correction
Author Message
Intern
Joined: 20 Mar 2005
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### Show Tags
20 Mar 2005, 21:11
hi everyone : )
Why this is wrong?
The department defines a private passenger vehicle to be one that is registered to an individual with a gross weight of less than 8000 pounds.
Why this is correct?
The department defines a private passenger vehicle as one that is registered to an individual and that has a gross weight of less than 8000 pounds.
If you have any questions
New!
VP
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### Show Tags
20 Mar 2005, 21:44
the first sentence has two problems. one is idiom and second is parllel structure.
the correct idiom is define ........ as
parrallel strucuture is "that is registered ............... and that has a gross weight ....................... "
GMAT Club Legend
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### Show Tags
20 Mar 2005, 21:56
same reasoning.
'to be one' suggests that it isn't one yet
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### Show Tags
27 Mar 2005, 21:34
The department defines a private passenger vehicle to be one that is registered to an individual with a gross weight of less than 8000 pounds.
this first sentence also means that the weight of individual is 8000 pounds and not the truck......[/u]
_________________
i hate when people do'nt post the OA, it leaves in guessing!!!!
Manager
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### Show Tags
27 Mar 2005, 21:34
The department defines a private passenger vehicle to be one that is registered to an individual with a gross weight of less than 8000 pounds.
this first sentence also means that the weight of individual is 8000 pounds and not the truck......[/u]
_________________
i hate when people do'nt post the OA, it leaves in guessing!!!!
27 Mar 2005, 21:34
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#### Hawaii00000
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• Posts: 347
##### Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« on: January 28, 2009, 12:03:09 AM »
I replaced the potentiometer with two 2.2k resistor (I soldered the common ends to the red wire and the other ends to the yellow and green wires) ,but when I run the program the motor just keeps on rotating at the same speed. The program has different values so the servo should rotate at different speeds. What am I doing Wrong!?!?!?
"God chose to make the world according to very beautiful mathematics."
-Paul Dirac
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#### cosminprund
• Robot Overlord
• Posts: 284
##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #1 on: January 28, 2009, 12:37:18 AM »
I'd first try sending the servo an 1 ms and then an 2 ms pulse. The servo should rotate in opposite directions. If it does then start with the 1ms pulse and increase it until it reverses direction. Before you get to the point where it reverses direction you'll find the spot where the servo is standing still (the new "center" position)! You'll need to do this to compensate for the differences in the two resistors values (they're both 2k2 but they're likely of 5% precision). IF the two resistors would have been perfect (and there's no such thing) your servo center would be at exactly 1.5 ms.
The best way to do what I've described here is to use an "Divide and Conquer" algorithm.
Here's an example:
_delay_ms(1) => makes the servo turn clockwise
_delay_ms(2) => makes the servo turn counter clockwise
(1+2)/2 = 1.5 => I'll next try the 1.5 ms pulse
_delay_ms(1.5) => makes the servo trun clockwise (just an example) => 1.5 takes the place of "1" so now I have:
_delay_ms(1.5) => makes the servo turn clockwise
_delay_ms(2) => makes the servo turn counter clockwise
(1.5 + 2)/2 = 1.75 => I'll next try the 1.75 ms pulse
_delay_ms(1.75) => makes the servo turn counter clockwise => 1.75 takes the place of "2" sonow I have:
_delay_ms(1.5) => makes the servo turn clockwise
_delay_ms(1.75) => makes the servo turn counter clockwise
(1.5 + 1.75)/2 = 1.625; I'll next try the 1.625 ms pulse
As you can see the interval you're looking at gets divided in half with every single test! With a maximum of 7 tests you should find your servo's center point. Once you've got the servo's center point, sending an pulse that's shorter makes the servo turn one way, sanding an longer pulse makes the servo go the other way!
#### Hawaii00000
• Contest Winner
• Supreme Robot
• Posts: 347
##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #2 on: January 28, 2009, 01:35:59 AM »
The code I'm using sends a signal for 0, 180, and 360 degrees and I nknow It works because I tried on the another servo. Another site did say that you need 2.4k resistors and I haven't put the gears back in yet, but It should still work without the gears.
"God chose to make the world according to very beautiful mathematics."
-Paul Dirac
**************************************************************
Its Hawaii Five-O. Get it?
#### cosminprund
• Robot Overlord
• Posts: 284
##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #3 on: January 28, 2009, 05:46:41 AM »
You're saying the servo is rotating at the same speed and in the same direction for all 3 signals: 0, 180, 360? I care more about direction then speed: for my servos speed varies very little in a very narrow band around the midpoint: the motor is stopped at 180, moves slowly a 183, faster at 185, full speed at 190 (just an example, I'm using milliseconds, not degrees to set up my servos)
If 0 and 360 make the motor turn in the same direction and you have checked with an other servo and with the other servo it works as expected (makes the servo turn all the way to one side and then all the way to the other side) then it must be related to your soldering.
#### Hawaii00000
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##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #4 on: January 28, 2009, 10:30:51 AM »
I have checked it and there's nothing wrong with the program that I know of. I can't see how I could have gone wrong when soldering its not that hard. Maybe I soldered the resistors to the wrong wires.
"God chose to make the world according to very beautiful mathematics."
-Paul Dirac
**************************************************************
Its Hawaii Five-O. Get it?
#### Hawaii00000
• Contest Winner
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• Posts: 347
##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #5 on: January 28, 2009, 08:18:46 PM »
Heres a picture if that helps the resistors are 2.2k
« Last Edit: January 28, 2009, 08:25:07 PM by Hawaii00000 »
"God chose to make the world according to very beautiful mathematics."
-Paul Dirac
**************************************************************
Its Hawaii Five-O. Get it?
#### Hawaii00000
• Contest Winner
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• Posts: 347
##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #6 on: January 28, 2009, 08:26:27 PM »
The common side goes to the red and the individual sides go to the yellow and green.
"God chose to make the world according to very beautiful mathematics."
-Paul Dirac
**************************************************************
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#### airman00
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##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #7 on: January 28, 2009, 08:39:10 PM »
you didnt use precision ( 1% to 2% tolerance) resistors .
You used resistors with a precision band of gold - 5% tolerance
Check out the Roboduino, Arduino-compatible board!
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#### Hawaii00000
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##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #8 on: January 28, 2009, 08:44:31 PM »
Is that the only problem? Are the 2.2k resistors and all the rest right, because I don't wanna keep going to radio shack! This project is getting kind of expensive.
"God chose to make the world according to very beautiful mathematics."
-Paul Dirac
**************************************************************
Its Hawaii Five-O. Get it?
#### Razor Concepts
• Supreme Robot
• Posts: 1,856
##### Re: Still can't get this servo modification thing!!!!!!!!!!!!!!!!
« Reply #9 on: January 28, 2009, 09:23:56 PM »
Just adjust the values in the code until the servo stops spinning. It's not the best way but it works if you don't want to spend more money.
• Supreme Robot
• Posts: 11,658 | 1,767 | 6,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2014-35 | longest | en | 0.898724 |
http://www.instructables.com/id/LinkIT-ONE-Battery-Tester-1/ | 1,498,749,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128329344.98/warc/CC-MAIN-20170629135715-20170629155715-00156.warc.gz | 559,145,608 | 14,005 | It's an age-old question: Is this battery dead? Sometimes it can be a real drag, and you would hate to throw out a perfectly good battery. Alas, you can worry no more! Today we'll create a simple project that will be able to tell us if a battery is properly charged or not!
## Step 1: Supplies
We'll be creating a fairly basic circuit here, nothing too complicated. Hopefully these are all things most introductory kits would have on hand.
• 560 ohm resistors (3x)
• Green LED
• Blue LED (OK, this should really be Yellow, but my last yellow one burnt out!)
• Red LED
• Some Jumper Wires
## Step 2: Create the Circuit
Relatively basic circuit here, nothing a beginner shouldn't be able to create. The key here is to leave two jumper cables hanging off. We will do this so that we can put whatever size battery we want in between. Simply touching the ends of the jumper wires to our battery will complete our circuit, and give us a battery reading.
Follow the diagram above to create the circuit, then move on to the next step.
## Step 3: Deploying the Code
Our code mostly is composed of 3 if statements. We read the voltage from the battery and...
If it is greater than 1.6 volts, we say the battery is fully charged (GREEN LED)
If it is less then 1.6 but greater than 1.4 volts, we say it still has some juice left in it (BLUE LED)
And if we see less than 1.4 volts, we say it is dead (RED LED)
## Step 4: Test Those Batteries!
And there you have it! A fully functional battery tester! Now you don't have to worry and stress about throwing away perfectly good Batteries! Just throw them in your brand new battery charger and you're all good to go!
<p>Cool battery tester</p> | 408 | 1,690 | {"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-2017-26 | longest | en | 0.929609 |
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Author Message
deegajvit
Registered: 08.12.2005
From: Derry, Northern Ireland
Posted: Tuesday 12th of Apr 10:19 I have problems with algebra formulas for parabolas. I tried hard to find someone who can help me out with this. I also searched for a coach to tutor me and explain my problems on radical inequalities, solving inequalities and difference of cubes. Though I found some who could perhaps explain my problem, I realized that I cannot find the money for them. I do not have a great deal of time too. My assignment is coming up in a little while. I am distressed. Can anyone help me out of this situation? I would very much welcome any assistance or any advice .
oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Thursday 14th of Apr 07:42 I have been in your place some time agowhen I was studying algebra formulas for parabolas. What part of exponent rules and simplifying fractions poses more problems ? Because I think that what you really need is a good program to help you understand the basic concepts and methods of solving the exercises. Did you ever use a program like that? I have tried a few but I have to say that Algebra Helper is the best and the easiest to use. It's not like those other programs because it teaches you how to solve, it doesn't just give you the solutions.
Svizes
Registered: 10.03.2003
From: Slovenia
Posted: Thursday 14th of Apr 10:05 Even I’ve been through times when I was trying to figure out a way to solve certain type of questions pertaining to distance of points and syntehtic division. But then I found this piece of software and it was almost like I found a magic wand. In a flash it would solve even the most difficult problems for you. And the fact that it gives a detailed and elaborate explanation makes it even more handy. It’s a must buy for every math student.
guctvishop
Registered: 15.09.2003
From: Greenville, SC
Posted: Friday 15th of Apr 18:15 You guys have really caught my attention with what you just said. Can someone please provide the website URL where I can purchase this program? And what are the various payment options available?
cmithy_dnl
Registered: 08.01.2002
From: Australia
Posted: Sunday 17th of Apr 11:09 Don’t worry pal. Just visit http://www.algebra-answer.com/faq.shtml and read all the details about it. Good luck with your studies!
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ORDER NOW! | 1,067 | 4,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-22 | latest | en | 0.943962 |
https://playtaptales.com/10-to-the-power-of-7545/ | 1,675,592,705,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00448.warc.gz | 481,698,132 | 3,987 | ## 10 to the power of 7,545
What is 10 to the power of 7545 (10^7545)? The answer is a Billiquattuordeciquingentillion. This is based on the Conway-Wechsler system for naming numbers.
## 10 to the power of 7,545 written out
10^7545 is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 | 5,114 | 10,316 | {"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-2023-06 | latest | en | 0.59682 |
https://rdrr.io/cran/Rfast/man/dista.html | 1,529,323,225,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859766.6/warc/CC-MAIN-20180618105733-20180618125733-00234.warc.gz | 688,174,324 | 15,174 | # dista: Distance between vectors and a matrix In Rfast: A Collection of Efficient and Extremely Fast R Functions
## Description
Distance between vectors and a matrix.
## Usage
`1` ```dista(xnew,x,type = "euclidean",k=0,index=FALSE,trans = TRUE,square = FALSE) ```
## Arguments
`xnew` A matrix with some data or a vector. `x` A matrix with the data, where rows denotes observations (vectors) and the columns contain the variables. `type` This can be either "euclidean" or "manhattan". `k` Should the k smaller distances or their indices be returned? If k > 0 this will happen. `index` In case k is greater than 0, you have the option to get the indices of the k smallest distances. `trans` Do you want the returned matrix to be transposed? TRUE or FALSE. `square` If you choose "euclidean" as the method, then you can have the optino to return the squared Euclidean distances by setting this argument to TRUE.
## Details
The target of this function is to calculate the distances between xnew and x without having to calculate the whole distance matrix of xnew and x. The latter does extra calculaitons, which can be avoided.
## Value
A matrix with the distances of each xnew from each vector of x. The number of rows of the xnew and and the number of columns of xnew are the dimensions of this matrix.
## Author(s)
Michail Tsagris
R implementation and documentation: Michail Tsagris <[email protected]> and Manos Papadakis <[email protected]>.
```mahala, Dist, total.dist, total.dista ```
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13``` ```xnew <- as.matrix( iris[1:10, 1:4] ) x <- as.matrix( iris[-c(1:10), 1:4] ) a <- dista(xnew, x) b <- as.matrix( dist( rbind(xnew, x) ) ) b <- b[ 1:10, -c(1:10) ] sum( abs(a - b) ) ## see the time x <- matrix( rnorm(1000 * 4), ncol = 4 ) system.time( dista(xnew, x) ) system.time( as.matrix( dist( rbind(xnew, x) ) ) ) x<-b<-a<-xnew<-NULL ```
### Example output
```Loading required package: Rcpp | 562 | 1,955 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-26 | longest | en | 0.752656 |
https://www.ic.sunysb.edu/Class/phy141md/doku.php?id=phy131studiof16:lectures:chapter17 | 1,601,393,503,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401643509.96/warc/CC-MAIN-20200929123413-20200929153413-00441.warc.gz | 900,323,573 | 9,277 | # Chapter 17 - Temperature, Thermal Expansion and Gas Laws
## Atomic Theory of Matter
To introduce the concept of temperature we need to start from the atomic theory, which considers matter to be composed of atoms.
A useful unit when discussing the mass of atoms is the atomic mass unit [$u$]
$1\mathrm{u}=1.6605\times10^{-27}\mathrm{kg}$
Atoms and molecules are in continual motion, the amount of motion is proportional to the temperature, which we can think of as a measure of energy.
Brownian motion, named after Robert Brown who observed it as random movements of pollen or dust suspended in water was explained by Albert Einstein in 1905 as a product of the thermal motion of molecules or atoms. In 2010 the instantaneous velocity of a particle in Brownian motion was measured by a group at University of Texas using optical trapping techniques.
We can consider temperature to be a measure of the energy contained in the motion of the microscopic constituents of a material.
## Temperature Scales
To be able to measure temperature we need a scale. The temperature scale in common usage in the United States, Fahrenheit, $\mathrm{^{o}F}$ is based on what may now seem to be fairly arbitrary reference points.
Most of the world uses the Celsius scale, $\mathrm{^{o}C}$, for everyday measurements. This is based on dividing the difference between the freezing point and boiling point of water in to 100 degrees and fixing $0\mathrm{^{o}C}$ as the freezing point.
For thermodynamics an absolute temperature scale, in which 0 is the complete absence of thermal energy, is appropriate. This scale is the Kelvin Scale. In this scale water freezes at $273.15\mathrm{K}$. Note that we do not use a degree symbol for temperatures in Kelvin.
To convert between Farenheit and Celsius you can use the fact that water freezes at $32\mathrm{^{o}F}$ and boils at $212\mathrm{^{o}}F$ to deduce that:
$T(\mathrm{^{o}C})=\frac{5}{9}(T(\mathrm{^{o}F})-32)$
or
$T(\mathrm{^{o}F})=\frac{9}{5}T(\mathrm{^{o}C})+32$
## Thermal Equilibrium
If two objects with different temperatures are brought in to contact with one another thermal energy will flow from one to another until the temperatures are the same, and we then say that the objects are in thermal equilibrium.
The zeroth law of thermodynamics states that:
“If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.”
This law may seem obvious but this is only because we already have some expectation from our everyday experience that temperature is a general quantity that we can measure with a thermometer. This law allows to declare this formally, and turns out to be important, in conjunction with the First and Second Laws (we'll talk about these later!), for establishing the true definition of thermal equilibrium.
## Thermal Expansion
Most, but not all, materials expand when heated. The change in length of material due to linear thermal expansion is
$\Delta l=\alpha l_{0}\Delta T$
$\alpha$ is the coefficient of linear expansion of the material, measured in $\mathrm{(^{o}C)^{-1}}$
The length of the object after it's temperature has been changed by $\Delta T$ is
$l=l_{0}(1+\alpha\Delta T)$
A material expands in all directions, and if we are interested in the volume changes of a rectangular object, that is isotropic, meaning it expands in the same way in all directions, then
$\Delta V = \beta V_{0}\Delta T$
$V_{0}=l_{0}w_{0}h_{0}$ → $V=l_{0}(1+\alpha\Delta T)w_{0}(1+\alpha\Delta T)h_{0}(1+\alpha\Delta T)$
$\Delta V=V-V_{0}=V_{0}(1+\alpha\Delta T)^{3}-V_{0}=V_{0}[3(\alpha\Delta T)+3(\alpha\Delta T)^{2}+(\alpha\Delta T)^{3}]$
If $\alpha\Delta T << 1$ then $\beta \approx 3\alpha$
Coefficients of thermal expansion can be found here or your textbook.
## Problem 17.8
To solve this problem you need to consider linear thermal expansion. You can use the $l$ given at $20^{o}C$ even though the question asks for $\Delta l$ at $15^{o}C$ ($\Delta l << l$). You will need the linear coefficient of thermal expansion of concrete which can be found here, or in your textbook.
## Problem 17.16
In this problem you need to calculate the volume expansion of the radiator, engine and coolant. The difference between the volume of the coolant and the sum of the volumes of the two containers will be the volume that overflows. You will need the volume coefficients of thermal expansion which you can find either here, or in your textbook.
## Some Thermal Expansion demos
A Bimetallic Strip combines invar with either copper or brass (ours is brass). The very different coefficients of thermal expansion of the two materials means it bends when heated. Invar is very useful, because of it's low coefficient of thermal expansion. Another material with a very low coefficient of thermal expansion is fused quartz.
Does a hole expand or contract on heating?
## Negative Thermal Expansion
Some materials in certain temperature ranges do not expand with temperature. An example is water, which actually decreases in volume as it's temperature is increased from $0\mathrm{^{o}C}$ to $4\mathrm{^{o}C}$. This has an important effect, it explains why the surface of lakes freeze, which helps prevent the flow of heat out of the lake. If the thermal expansion was normal between $0\mathrm{^{o}C}$ and $4\mathrm{^{o}C}$ the circulation effect would continue below $4\mathrm{^{o}C}$ and the whole lake would freeze, starting from the bottom.
## Thermal Stress
What happens when an object wants to expand, but can't because it is fixed?
If an object is trying to expand by a length $\Delta l$ due to thermal expansion, but doesn't because something is holding it in place, that something must be exerting a force.
The force can be found by considering the relationship between stress and material deformation, which we discussed when we introduced elastic moduli.
$\frac{F}{A}=E \frac{\Delta l}{l_{0}}$
And as
$\Delta l=\alpha l_{0}\Delta T$
$\frac{F}{A}=\alpha E \Delta T$
These stresses can be reduced by the inclusion of expansion joints in bridges, roads and pipes.
## What makes a gas ideal?
There are a number of conditions which must be satisfied for a gas to be considered ideal
1. There must be a large number of molecules and they should move in random directions with a range of different speeds.
2. The spacing between molecules should be much greater than the size of the molecules.
3. Molecules are assumed to interact only through collisions.
4. The collisions are assumed to be elastic.
## Boyle's Law
At constant temperature, it is found that the product of the pressure and volume of an ideal gas are constant
$PV=\mathrm{constant}$
This is named Boyle's Law, after Robert Boyle who formulated it in 1662.
## Charles' Laws
Joesph Louis Gay-Lussac published Charles' Law in 1802, attributing it to unpublished work of Jacques Charles in the 1780's (Gay-Lussac has his own law..though it's not clear he should!).
Charles' Law states that at constant pressure the volume of a gas is proportional to the temperature.
$V\propto T$
## Gay-Lussac's law
Gay Lussac's Law states that for a fixed volume the pressure is proportional to the temperature
$P\propto T$
## Ideal Gas Law
The combination of the previous 3 laws implies that
$PV\propto T$
Our previous laws were for systems of constant mass, but we can see that the amount of mass should effect the volume (at a given pressure) or the pressure (at a given volume).
$PV\propto mT$
Measuring the amount of mass in moles will allow us to write the ideal gas law in terms of a universal constant. A mole of gas is a given number of molecules, Avagadro's number, $N_{A}=6.02\times 10^{23}$. If we have a certain mass $m$ of a gas which has a certain molecular mass (measured in atomic mass units, $\mathrm{u}$, which are also the number of grams per mole.), the the number of moles $n$ is given by
$n=\frac{m[\mathrm{g}]}{\textrm{molecular mass}[\mathrm{g/mol}]}$
and
$PV=nRT$ where $R=8.314\mathrm{J/(mol.K)}$
This equation is the ideal gas law
## Problem 17.42
This problem requires the use of the ideal gas law. Watch out for the correct units of temperature and the difference between gauge and absolute pressure.
## Problem 17.50
Another ideal gas law problem. The temperature is different at the bottom and surface of the lake, but you should also pay attention to the pressure as a function of depth.
## Problem 17.65
Another ideal gas law problem. Is the pressure in the ideal gas law absolute or gauge pressure?
## Ideal Gas Law for a number of molecules
The ideal gas law can also be written in terms of the number of molecules $N$
$PV=nRT=\frac{N}{N_{A}}RT=NkT$
where $k=\frac{R}{N_{A}}=\frac{8.314\mathrm{J/(mol.K)}}{6.02\times 10^{23}}=1.38\times 10^{-23}\mathrm{J/K}$ is the Boltzmann Constant.
## Using the Ideal Gas Law to determine Absolute Zero
If $PV=nRT$ the absolute zero temperature occurs when $P=0$. In practice most gases will liquefy before this point, but we can measure the pressure of a fixed volume of gas at a couple of reference points and extrapolate down to zero pressure to get an estimate for absolute zero.
Through laser cooling and molecular trapping techniques it is now possible (but difficult!) for temperatures on the order of a $\mathrm{nK}$ to be achieved. Prof. Dominik Schneble produces ultra-cold ($\mu K$) Bose-Einstein condensates in the basement of this building! Prof. Hal Metcalf was one of the key players in the original development of laser cooling.
## Ideal gas law from a molecular perspective
We can relate the pressure exerted by an ideal gas on it's container to the change in momentum of a molecule when it strikes the container wall
The average force due to one molecule is then
$F_{molecule}=\frac{\Delta(mv)}{\Delta t}=\frac{2mv_{x}}{2l/v_{x}}=\frac{mv_{x}^{2}}{l}$
The net force on the wall will be the sum of the forces from all $N$ molecules
$F_{net}=\frac{m}{l}\Sigma_{i=1..N} v_{xi}^{2}$
$\frac{\Sigma_{i=1..N} v_{xi}^{2}}{N}=\bar{v_{x}^{2}}$ → $F_{net}=\frac{m}{l}N\bar{v_{x}^{2}}$
$v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}$ → $\bar{v^{2}}=\bar{v_{x}^{2}}+\bar{v_{y}^{2}}+\bar{v_{z}^{2}}=\bar{3v_{x}^{2}}$
$F_{net}=\frac{m}{l}N\frac{\bar{v^{2}}}{3}$
$P=\frac{F}{A}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{Al}=\frac{1}{3}\frac{Nm\bar{v^{2}}}{V}$
$PV=\frac{2}{3}N(\frac{1}{2}m\bar{v^{2}})=NkT$
$\bar{KE}=\frac{1}{2}m\bar{v^{2}}=\frac{3}{2}kT$
phy131studiof16/lectures/chapter17.txt · Last modified: 2016/11/16 09:27 by mdawber | 2,760 | 10,531 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-40 | latest | en | 0.877339 |
https://www.sawaal.com/aptitude-reasoning/verbal-reasoning-mental-ability-questions-and-answers.htm?page=2&sort= | 1,624,527,091,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00293.warc.gz | 891,923,555 | 15,225 | # Verbal Reasoning - Mental Ability Questions
## What is Verbal Reasoning: Mental Ability ?
Verbal Reasoning - Mental ability is one of the common topics found in most of the entrance exams. Mental ability tests our level at which we learn things, understands instructions and solve problems. Verbal Reasoning - Mental Ability Questions encloses wide range of topics: verbal reasoning (vocabulary and missing letters), Arithmetic reasoning, Blood relations, Analogy, classifications, coding and decoding, Data sufficiency, Missing characters, logical sequences, series, numbers, spontaneous reaction tests, verification of truth statements, Venn diagrams, mathematical operations, direction sense test, statements and arguments, assumption tests and more.
## What Topics Are Included In Mental Ability Test ?
Mental ability questions are found commonly in competitive exams and entrance tests of APPSC Group 1 and Group2, UPSC ( Civil services, CPF(CA) ), TNPC(VAO, Group1, Group2), HPAS, WBCS, AP SI Selection process, NDA, IBPS and other Bank Exams, CAT, MAT, GMAT and other MBA entrance exams, GRE, GATE, TOEFL, IT and Non IT Placement papers.
Mental Ability Questions should be solved by cracking logic behind them. One can gain that logical thinking by thorough practice of number of questions in different patterns, good command in vocabulary, good command in maths numbers and formulas and good understanding abilities.
We have a large database for you to practice on verbal reasoning - mental ability questions and answers. You can expect these questions in most of the competitive exams.
• #### Verification of Truth of the statement
Q:
What should come in place of question mark (?) in the following number series?
132 156 ? 210 240 272
A) 196 B) 182 C) 199 D) 204
Answer & Explanation Answer: B) 182
Explanation:
The given series follows a logic that
11 x 12, 12 x 13, 13 x 14, 14 x 15, 15 x 16,...
So the missing number is 13 x 14 = 182
Report Error
511 106573
Q:
Arrange these words in alphabetical order and tick the one that comes last
1. Abandon 2. Actuate 3. Accumulate 4. Acquit 5. Achieve
A) Actuate B) Accumulate C) Acquit D) Achieve
Answer & Explanation Answer: A) Actuate
Explanation:
First letters are common. Second letters are:b, c, c, c, c. One of the four words having c is the last word. Let us see the third letters now ,there are: t, c, q, h. Clearly t is the last. Hence Actuate is the last word.
Report Error
531 101700
Q:
Find the wrong number in the series.
1, 2, 6, 15, 31, 56, 91
A) 31 B) 15 C) 56 D) 91
Answer & Explanation Answer: D) 91
Explanation:
The difference of the two consecutive numbers are 1, 4, 9, 16, 25, 35.
This is clearly ,
Here 35 is wrong, So our wrong number is 91
Report Error
576 101601
Q:
If South-East becomes North, North-East becomes West and so on. What will West become?
A) North-East B) North-West C) South-East D) South-West
Answer & Explanation Answer: C) South-East
Explanation:
It is clear from the diagrams that new name of West will become South-East.
Report Error
1513 99795
Q:
What should come next in the following letter sequence?
A A B A B C A B C D A B C D E A B C D
A) A B) E C) C D) B
Answer & Explanation Answer: B) E
Explanation:
Report Error
557 96751
Q:
In a row of boys, If A who is 10th from the left and B who is 9th from the right interchange their positions, A becomes 15th from the left. How many boys are there in the row ?
A) 23 B) 31 C) 27 D) 28
Answer & Explanation Answer: A) 23
Explanation:
Clearly, A’s new position is 15th from the left. But this is the same as B’s earlier position which is 9th from the right.
Report Error
722 95987
Q:
Four of the following five are alike in a certain way and so form a group. which one does not belong to that group?
1. Tooth 2. Chin 3. Nose 4. Ear 5. Eye
Answer
Answer : (2)
All others have two vowels each
Report Error
93259
Q:
In a family, there are six members A, B, C, D, E and F.
A and B are a married couple, A being the male member. D is the only son of C, who is the brother of A. E is the sister of D. B is the daughter-in-law of F, whose husband has died. How is E related to C ?
A) Sister B) Daughter C) Cousin D) Mother
Answer & Explanation Answer: B) Daughter
Explanation:
A is a male and married to B. So, A is the husband and B is the wife. C is the brother of A. D is the son of C. E. who is the sister of D will be the daughter of C. B is the daughter-in-law of F whose husband has died means F is the mother of A.
Clearly. E is the daughter of C.
Report Error
556 92582 | 1,237 | 4,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-25 | latest | en | 0.854673 |
https://www.ajjacobson.us/financial-statements-2/what-do-the-numbers-mean-wew.html | 1,550,323,653,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480472.38/warc/CC-MAIN-20190216125709-20190216151709-00196.warc.gz | 750,998,254 | 7,219 | ## What do the numbers mean
For most companies, taking advantage of these discounts makes sense as long as the annual interest rate calculated using this formula is higher than the one they must pay if they borrow money to pay the bill early. This becomes a big issue for companies because unless their inventory turns over very rapidly, 10 days probably isn't enough time to sell all the inventory purchased before they must pay the bill early. Their cash wouldn't come from sales but, more likely, from borrowing.
If cash flow is tight, a company has to borrow funds using its credit line to take advantage of the discount. For example, if the company buys \$100,000 in goods to be sold at terms of 2/10 net 30, it can save \$2,000 by paying within 10 days. If the company hasn't sold all the goods, it has to borrow the \$100,000 for 20 days, which wouldn't be necessary if it didn't try to take advantage of the discount. I assume that the annual interest on the company's credit line is 9 percent. Does it make sense to borrow the money?
The company would need to pay the additional interest on the amount borrowed only for 20 additional days (because that's the number of days the company must pay the bill early). Calculating the annual interest of 9 percent of \$100,000 equals \$9,000, or \$25 per day. Borrowing that money would cost an additional \$500 (\$25 times 20 days). So even though the company must borrow the money to pay the bill early, the \$2,000 discount would still save it \$1,500 more than the \$500 interest cost involved in borrowing the money.
0 0 | 356 | 1,581 | {"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-2019-09 | longest | en | 0.975583 |
https://www.esaral.com/q/the-perimeter-of-triangle-is-50-cm-one-side-of-the-triangle-is-4-cm-longer-than-the-smallest-side-and-the-third-side-is-6-cm-less-than-twice-the-smallest-side-40958 | 1,721,344,086,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00749.warc.gz | 645,513,264 | 11,884 | # The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side.
Question:
The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.
Solution:
Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x − 6.
Now,
x + x + 4 + 2x − 6 = 50 (∵ perimeter is 50 cm)
⇒ 4x − 2 = 50
⇒ 4x = 50 + 2
⇒ 4x = 52
⇒ x = 13
∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{13+17+20}{2}=\frac{50}{2}=25 \mathrm{~cm}$
∴ By Heron's formula,
Area of $\Delta A B C=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{25(25-13)(25-17)(25-20)}$
$=\sqrt{25(12)(8)(5)}$
$=20 \sqrt{30} \mathrm{~cm}^{2}$
Hence, the area of the triangle is $20 \sqrt{30} \mathrm{~cm}^{2}$. | 373 | 1,006 | {"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} | 4.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.666743 |
http://aimpl.org/crouzeix/9/ | 1,553,058,392,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202299.16/warc/CC-MAIN-20190320044358-20190320070358-00221.warc.gz | 9,977,109 | 6,669 | $\newcommand{\Cat}{{\rm Cat}}$ $\newcommand{\A}{\mathcal A}$ $\newcommand{\freestar}{ \framebox[7pt]{\star} }$
## 9. Inverse numerical range problems
1. ### A possible counterexample via FOM
The following inequality holds for FOM polynomials $p^F$ when applying the method to a linear system $Ax=b$ with $b$ a unit vector such that FOM does not break down at or before the $k$th iteration: $0=2\max_{\rho\in{\cal H}_k}|p_k^F(\rho)| < \| r_k^F\|\leq \|p_k^F(A)\|,$ where ${\cal H}_k$ are the Ritz values for the $k$th FOM iteration, $p_k^F$ is the corresponding FOM polynomial of degree at most $k$ with the value one in the origin and $r_k^F$ the residual vector. It is possible to construct matrices yielding any complex Ritz values ${\cal H}_k$ while the corresponding residual norm takes any positive real value (in fact, infinity might be included).
#### Problem 9.1.
[Jurjen Duintjer Tebbens] To extend the above inequality to a counterexample of Crouzeix’s conjecture, one may try among other ideas to (1) find matrices $A$ where (most of) the prescribed Ritz values lie on the boundary of the field of values ; (2) numerically follow the influence of the chosen prescribed residual norm $\| r_k\|$ on the max of $p_k$ for the field of values, with prescribed Ritz values.
Cite this as: AimPL: Crouzeix's conjecture, available at http://aimpl.org/crouzeix. | 394 | 1,368 | {"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} | 2.859375 | 3 | CC-MAIN-2019-13 | longest | en | 0.67022 |
https://justaaa.com/accounting/238413-on-january-1-2015-piper-co-issued-ten-year-bonds | 1,675,333,668,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00403.warc.gz | 358,906,365 | 10,721 | Question
# On January 1, 2015, Piper Co. Issued ten-year bonds with a face valueof \$3,000,000 and a...
On January 1, 2015, Piper Co. Issued ten-year bonds with a face valueof \$3,000,000 and a stated interest rate of 10%, payable semiannually on June 30 and December 31. The bonds were sold to yield 12%.
Questions:
1) Calculate the issue of price of the bonds
2) Prepare all journal entries for 2016.
USE THIS EFFECTIVE INTEREST METHOD CHART, THAT CONSIST OF THE FOLLOWING HEADINGS
1- DATE
2- CASH INTEREST
3- INTEREST EXPENSE
6- BOOK VALUE
Bond Issue Price = \$936,000 + \$1,720,500 = 2,656,500 Principal \$3,000,000 x 0.312 = \$936,000 Interest \$150,000 x 11.470 = \$1,720,500 Interest amount = \$3,000,000 x 10% x 6/12 = \$150,000 Date Interest Paid Interest Exp. Prem. Amortization Carrying Amount 01.01.2015 \$2,656,500 06.30.2015 \$150,000 \$132,825 \$17,175 \$2,639,325 12.31.2015 \$150,000 \$131,966 \$18,034 \$2,621,291 06.30.2016 \$150,000 \$131,065 \$18,935 \$2,602,356 12.31.2016 \$150,000 \$130,118 \$19,882 \$2,582,474 06.30.2016 Bond Interest Exp. \$131,065 Bonds Payable \$18,935 Cash \$150,000 12.31.2016 Bond Interest Exp. \$130,118 Bonds Payable \$19,882 Cash \$150,000
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 434 | 1,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-06 | latest | en | 0.794602 |
https://web2.0calc.com/questions/help-quick-please_13 | 1,601,085,127,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00068.warc.gz | 675,252,158 | 6,831 | +0
0
197
1
Jeremy is writing down a sequence of integers. He writes 1 as his first number. Then, he squares it and adds 3, writing down 4 next. He squares 4 and adds 3, writing down 19 as his third number. If Jeremy continues his square-and-add pattern, what will the units digit of the 20th number in his list be?
Jun 1, 2020
#1
0
a=1;p=0;b=a^2+3;printb;c=b;cycle:c=c^2+3;printc;p++;if(p<=18, goto cycle,0)
These are the 20 terms of your sequence. Notice that the last and 20th term is 335,690 digits long! The last 10 digits of that number are =1,569,412,499
4
19
364
132499
1 7555985004
3 0821260946 0672880019
9.499501263 E+40
9.024052425 E+81
8.143352216 E+163
6.631418532 E+327
4.397571175 E+655
1.933863224 E+1311
3.739826967 E+2622
1.398630575 E+5245
1.956167484 E+10490
3.826591226 E+20980
1.464280041 E+41961
2.144116039 E+83922
4.597233587 E+167844
2.113455665 E+335689
Note: There appears to be pattern of alternating between 4 for odd numbers and 9 for even numbers.
Jun 1, 2020
edited by Guest Jun 1, 2020 | 390 | 1,028 | {"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-2020-40 | latest | en | 0.827737 |
https://numbermatics.com/n/25820165069487125/ | 1,721,938,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00619.warc.gz | 369,486,510 | 6,969 | # 25820165069487125
## 25,820,165,069,487,125 is an odd composite number composed of two prime numbers multiplied together.
What does the number 25820165069487125 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 12 divisors.
25820165069487125 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of twelve divisors.
## Prime factorization of 25820165069487125:
### 53 × 591132
(5 × 5 × 5 × 59113 × 59113)
See below for interesting mathematical facts about the number 25820165069487125 from the Numbermatics database.
### Names of 25820165069487125
• Cardinal: 25820165069487125 can be written as Twenty-five quadrillion, eight hundred twenty trillion, one hundred sixty-five billion, sixty-nine million, four hundred eighty-seven thousand, one hundred twenty-five.
### Scientific notation
• Scientific notation: 2.5820165069487125 × 1016
### Factors of 25820165069487125
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 59118
### Divisors of 25820165069487125
• Number of divisors d(n): 12
• Complete list of divisors:
• Sum of all divisors σ(n): 25820273403458748
• Sum of proper divisors (its aliquot sum) s(n): 108333971623
• 25820165069487125 is a deficient number, because the sum of its proper divisors (108333971623) is less than itself. Its deficiency is 25820056735515502
### Bases of 25820165069487125
• Binary: 10110111011101101001101101001110000001000111100000101012
• Base-36: 728HG2XGS9H
### Squares and roots of 25820165069487125
• 25820165069487125 squared (258201650694871252) is 666680924215563070579710540765625
• 25820165069487125 cubed (258201650694871253) is 17213811511924074766357129845563832638123580078125
• The square root of 25820165069487125 is 160686542.9010379979
• 25820165069487125 is a perfect cube number. Its cube root is 295565
### Scales and comparisons
How big is 25820165069487125?
• 25,820,165,069,487,125 seconds is equal to 821,001,382 years, 10 weeks, 1 day, 1 hour, 32 minutes, 5 seconds.
• To count from 1 to 25,820,165,069,487,125 would take you about two billion, four hundred sixty-three million, four thousand, one hundred forty-six years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 25820165069487125 cubic inches would be around 24630.4 feet tall.
### Recreational maths with 25820165069487125
• 25820165069487125 backwards is 52178496056102852
• The number of decimal digits it has is: 17
• The sum of 25820165069487125's digits is 71
• More coming soon!
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Keywords: Divisors of 25820165069487125, math, Factors of 25820165069487125, curriculum, school, college, exams, university, Prime factorization of 25820165069487125, STEM, science, technology, engineering, physics, economics, calculator, twenty-five quadrillion, eight hundred twenty trillion, one hundred sixty-five billion, sixty-nine million, four hundred eighty-seven thousand, one hundred twenty-five.
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# When a charged particle enters a magnetic field in a direction perpendicular to the field, which one of the following does not change? (A) Kinetic energy (B) Velocity (C) Force (D) Momentum.
Last updated date: 23rd May 2024
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Hint: We are going to first explain what happens to the charged particle when it enters the magnetic field and the formula for the magnetic force that the particle experiences. Then based on the force value in the perpendicular magnetic field, the unchanged parameter is chosen.
Complete step-by-step solution:
When a charged particle enters a magnetic field, there is a magnetic force that is experienced by the charged particle. Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field. The magnetic force is given by the formula
$\vec F = q\left( {\vec v \times \vec B} \right)$
The magnitude of the force will be
$F = qvB\sin \theta$
Now, since the charged particle is moving in a direction perpendicular to the field, then the magnitude of the force becomes
$F = qvB\sin {90^ \circ } = qvB$
Now, also, the direction of the force is perpendicular to the direction of the velocity of the particle or the displacement, hence, the work that is done on the particle by the magnetic field is zero which means that by the Work-Energy theorem that the change in the kinetic energy is also zero which gives us the result that the kinetic energy of the particle remains the same.
The other factors like velocity, force and momentum change.
Note: It is important to note that moves in circular path direction of momentum will change but the magnitude is unchanged. Thus, the change in direction only also means a change in the velocity also. The force is applied which is the centripetal force that causes the particle to move in a circular path. | 444 | 1,978 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-22 | latest | en | 0.904533 |
http://stackoverflow.com/questions/18921302/how-to-incrementally-sample-without-replacement | 1,441,301,534,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645322940.71/warc/CC-MAIN-20150827031522-00077-ip-10-171-96-226.ec2.internal.warc.gz | 224,692,680 | 37,026 | # How to incrementally sample without replacement?
Python has `my_sample = random.sample(range(100), 10)` to randomly sample without replacement from `[0, 100)`.
Suppose I have sampled `n` such numbers and now I want to sample one more without replacement (without including any of the previously sampled `n`), how to do so super efficiently?
update: changed from "reasonably efficiently" to "super efficiently" (but ignoring constant factors)
-
Do you only want to sample integers in a `[0, x)` range? What is your expected `x`? – Chronial Sep 20 '13 at 16:54
[0, n) works for me. i can make any problem fit into it – necromancer Sep 20 '13 at 17:13
Is that what you need or not? Making another problem fit into that costs a severe amount of time and is very relevant considering the tight bounds you are asking for. – Chronial Sep 20 '13 at 17:21
You might want to look at the source for `random.sample` – Eric Sep 23 '13 at 13:44
What a thread! A simple question end up needing a 300 pt bounty as sheer gratitude for amazing answers. 4 answers from 1 person. 3 answers from another. 1 answer from OP that served as basis for correct answer. 1 amazing thesis-like answer which actually includes multiple sub-answers. Happy conclusions I hope. Thank you everybody. :-) – necromancer Sep 24 '13 at 23:00
Note to readers from OP: Please consider looking at the originally accepted answer to understand the logic, and then understand this answer.
Aaaaaand for completeness sake: This is the concept of no_answer_not_upvoted’s answer, but adapted so it takes a list of forbidden numbers as input. This is just the same code as in my previous answer, but we build a state from `forbid`, before we generate numbers.
• This is time `O(f+k)` and memory `O(f+k)`. Obviously this is the fastest thing possible without requirements towards the format of `forbid` (sorted/set). I think this makes this a winner in some way ^^.
• If `forbid` is a set, the repeated guessing method is faster with `O(k⋅n/(n-(f+k)))`, which is very close to `O(k)` for `f+k` not very close to `n`.
• If `forbid` is sorted, my ridiculous algorithm is faster with:
``````import random
def sample_gen(n, forbid):
state = dict()
track = dict()
for (i, o) in enumerate(forbid):
x = track.get(o, o)
t = state.get(n-i-1, n-i-1)
state[x] = t
track[t] = x
state.pop(n-i-1, None)
track.pop(o, None)
del track
for remaining in xrange(n-len(forbid), 0, -1):
i = random.randrange(remaining)
yield state.get(i, i)
state[i] = state.get(remaining - 1, remaining - 1)
state.pop(remaining - 1, None)
``````
usage:
``````gen = sample_gen(10, [1, 2, 4, 8])
print gen.next()
print gen.next()
print gen.next()
print gen.next()
``````
-
If you know in advance that you're going to want to multiple samples without overlaps, easiest is to do `random.shuffle()` on `list(range(100))` (Python 3 - can skip the `list()` in Python 2), then peel off slices as needed.
``````s = list(range(100))
random.shuffle(s)
first_sample = s[-10:]
del s[-10:]
second_sample = s[-10:]
del s[-10:]
# etc
``````
Else @Chronial's answer is reasonably efficient.
-
+1 thanks, still looking for perfect solution where i can pass in the list of previously sampled objects and it `smartly` (@Chronial's answer uses brute force) to sample the next one. – necromancer Sep 20 '13 at 16:35
@no_answer_not_upvoted I think you’ve seen all possibilities. If your range() is small, use my answer, if your sample size is small, use Veedrac’s answer. If both of them are massive, state that in your question and hope someone gives you a more complicated algorithm. But note that that would be slower in the first two cases. – Chronial Sep 20 '13 at 16:40
Nice solution, but I would recommend storing the current index instead of deleting, as that is quite slow. Or use queue. – Chronial Sep 20 '13 at 16:44
@no_answer_not_upvoted, good luck on that ;-) It will need to look at every "forbidden" value you pass in, and will need to look at every value in the base list to ensure that each is not in the forbidden list. Without smarter data structures to start with, it has to take time at least proportional to the sum of the sizes of both lists (base list and forbidden list). – Tim Peters Sep 20 '13 at 16:45
@Chronial, no, deleting is very fast here: that's why it deletes from the end of the list. No items need to be moved. CPython just decrements the length of the list, and decref's the pointers in the tail. – Tim Peters Sep 20 '13 at 16:47
Ok, here we go. This should be the fastest possible non-probabilistic algorithm. It has runtime of `O(k⋅log²(s) + f⋅log(f)) ⊂ O(k⋅log²(f+k) + f⋅log(f)))` and space `O(k+f)`. `f` is the amount of forbidden numbers, `s` is the length of the longest streak of forbidden numbers. The expectation for that is more complicated, but obviously bound by `f`. If you assume that `s^log₂(s)` is bigger than `f` or are just unhappy about the fact that `s` is once again probabilistic, you can change the log part to a bisection search in `forbidden[pos:]` to get `O(k⋅log(f+k) + f⋅log(f))`.
The actual implementation here is `O(k⋅(k+f)+f⋅log(f))`, as insertion in the list `forbid` is `O(n)`. This is easy to fix by replacing that list with a blist sortedlist.
I also added some comments, because this algorithm is ridiculously complex. The `lin` part does the same as the `log` part, but needs `s` instead of `log²(s)` time.
``````import bisect
import random
def sample(k, end, forbid):
forbidden = sorted(forbid)
out = []
# remove the last block from forbidden if it touches end
for end in reversed(xrange(end+1)):
if len(forbidden) > 0 and forbidden[-1] == end:
del forbidden[-1]
else:
break
for i in xrange(k):
v = random.randrange(end - len(forbidden) + 1)
# increase v by the number of values < v
pos = bisect.bisect(forbidden, v)
v += pos
# this number might also be already taken, find the
# first free spot
##### linear
#while pos < len(forbidden) and forbidden[pos] <=v:
# pos += 1
# v += 1
##### log
while pos < len(forbidden) and forbidden[pos] <= v:
step = 2
# when this is finished, we know that:
# • forbidden[pos + step/2] <= v + step/2
# • forbidden[pos + step] > v + step
# so repeat until (checked by outer loop):
# forbidden[pos + step/2] == v + step/2
while (pos + step <= len(forbidden)) and \
(forbidden[pos + step - 1] <= v + step - 1):
step = step << 1
pos += step >> 1
v += step >> 1
if v == end:
end -= 1
else:
bisect.insort(forbidden, v)
out.append(v)
return out
``````
Now to compare that to the “hack” (and the default implementation in python) that Veedrac proposed, which has space `O(f+k)` and (`n/(n-(f+k))` is the expected number of “guesses”) time:
I just plotted this for `k=10` and a reasonably big `n=10000` (it only gets more extreme for bigger `n`). And I have to say: I only implemented this because it seemed like a fun challenge, but even I am surprised by how extreme this is:
Let’s zoom in to see what’s going on:
Yes – the guesses are even faster for the 9998th number you generate. Note that, as you can see in the first plot, even my one-liner is probably faster for bigger `f/n` (but still has rather horrible space requirements for big `n`).
To drive the point home: The only thing you are spending time on here is generating the set, as that’s the `f` factor in Veedrac’s method.
So I hope my time here was not wasted and I managed to convince you that Veedrac’s method is simply the way to go. I can kind of understand why that probabilistic part troubles you, but maybe think of the fact that hashmaps (= python `dict`s) and tons of other algorithms work with similar methods and they seem to be doing just fine.
You might be afraid of the variance in the number of repetitions. As noted above, this follows a geometric distribution with `p=n-f/n`. So the standard deviation (=the amount you “should expect” the result to deviate from the expected average) is
Which is basically the same as the mean (`√f⋅n < √n² = n`).
**edit:
I just realized that `s` is actually also `n/(n-(f+k))`. So a more exact runtime for my algorithm is `O(k⋅log²(n/(n-(f+k))) + f⋅log(f))`. Which is nice since given the graphs above, it proves my intuition that that is quite a bit faster than `O(k⋅log(f+k) + f⋅log(f))`. But rest assured that that also does not change anything about the results above, as the `f⋅log(f)` is the absolutely dominant part in the runtime.
-
wow... +1 for now .. i promise to fully understand this ASAP!! thank you so much! – necromancer Sep 21 '13 at 8:20
Feel free to ask questions if you any problems. – Chronial Sep 21 '13 at 15:56
@Chronial, I don't know whether to laugh, cry, or applaud. So some of each: LOL, mwah, bravo! ;-) It is indeed an heroic solution to a problem nobody has ;-) – Tim Peters Sep 21 '13 at 21:41
@TimPeters yep, obviously I don’t have any useful algorithms to write ^^. – Chronial Sep 22 '13 at 1:49
i don't yet understand it, but hey, it is worth pushing you over to 10k whenever SO lets me do it (or if somebody else gives an even better answer, :D) – necromancer Sep 23 '13 at 0:13
# The short way
If the number sampled is much less than the population, just sample, check if it's been chosen and repeat while so. This might sound silly, but you've got an exponentially decaying possibility of choosing the same number, so it's much faster than `O(n)` if you've got even a small percentage unchosen.
# The long way
Python uses a Mersenne Twister as its PRNG, which is good. We can use something else entirely to be able to generate non-overlapping numbers in a predictable manner.
Here's the secret:
• Quadratic residues, `x² mod p`, are unique when `2x < p` and `p` is a prime.
• If you "flip" the residue, `p - (x² % p)`, given this time also that `p = 3 mod 4`, the results will be the remaining spaces.
• This isn't a very convincing numeric spread, so you can increase the power, add some fudge constants and then the distribution is pretty good.
First we need to generate primes:
``````from itertools import count
from math import ceil
from random import randrange
def modprime_at_least(number):
if number <= 2:
return 2
number = (number // 4 * 4) + 3
for number in count(number, 4):
if all(number % factor for factor in range(3, ceil(number ** 0.5)+1, 2)):
return number
``````
You might worry about the cost of generating the primes. For 10⁶ elements this takes a tenth of a millisecond. Running `[None] * 10**6` takes longer than that, and since it's only calculated once, this isn't a real problem.
Further, the algorithm doesn't need an exact value for the prime; is only needs something that is at most a constant factor larger than the input number. This is possible by saving a list of values and searching them. If you do a linear scan, that is `O(log number)` and if you do a binary search it is `O(log number of cached primes)`. In fact, if you use galloping you can bring this down to `O(log log number)`, which is basically constant (`log log googol = 2`).
Then we implement the generator
``````def sample_generator(up_to):
prime = modprime_at_least(up_to+1)
# Fudge to make it less predictable
fudge_power = 2**randrange(7, 11)
fudge_constant = randrange(prime//2, prime)
fudge_factor = randrange(prime//2, prime)
def permute(x):
permuted = pow(x, fudge_power, prime)
return permuted if 2*x <= prime else prime - permuted
for x in range(prime):
res = (permute(x) + fudge_constant) % prime
res = permute((res * fudge_factor) % prime)
if res < up_to:
yield res
``````
And check that it works:
``````set(sample_generator(10000)) ^ set(range(10000))
#>>> set()
``````
Now, the lovely thing about this is that if you ignore the primacy test, which is approximately `O(√n)` where `n` is the number of elements, this algorithm has time complexity `O(k)`, where `k` is the sample sizeit's and `O(1)` memory usage! Technically this is `O(√n + k)`, but practically it is `O(k)`.
### Requirements:
1. You do not require a proven PRNG. This PRNG is far better then linear congruential generator (which is popular; Java uses it) but it's not as proven as a Mersenne Twister.
2. You do not first generate any items with a different function. This avoids duplicates through mathematics, not checks. Next section I show how to remove this restriction.
3. The short method must be insufficient (`k` must approach `n`). If `k` is only half `n`, just go with my original suggestion.
1. Extreme memory savings. This takes constant memory... not even `O(k)`!
2. Constant time to generate the next item. This is actually rather fast in constant terms, too: it's not as fast as the built-in Mersenne Twister but it's within a factor of 2.
3. Coolness.
To remomve this requirement:
You do not first generate any items with a different function. This avoids duplicates through mathematics, not checks.
I have made the best possible algorithm in time and space complexity, which is a simple extension of my previous generator.
Here's the rundown (`n` is the length of the pool of numbers, `k` is the number of "foreign" keys):
### Initialisation time `O(√n)`; `O(log log n)` for all reasonable inputs
This is the only factor of my algorithm that technically isn't perfect with regards to algorithmic complexity, thanks to the `O(√n)` cost. In reality this won't be problematic because precalculation brings it down to `O(log log n)` which is immeasurably close to constant time.
The cost is amortized free if you exhaust the iterable by any fixed percentage.
This is not a practical problem.
### Amortized `O(1)` key generation time
Obviously this cannot be improved upon.
### Worst-case `O(k)` key generation time
If you have keys generated from the outside, with only the requirement that it must not be a key that this generator has already produced, these are to be called "foreign keys". Foreign keys are assumed to be totally random. As such, any function that is able to select items from the pool can do so.
Because there can be any number of foreign keys and they can be totally random, the worst case for a perfect algorithm is `O(k)`.
### Worst-case space complexity `O(k)`
If the foreign keys are assumed totally independent, each represents a distinct item of information. Hence all keys must be stored. The algorithm happens to discard keys whenever it sees one, so the memory cost will clear over the lifetime of the generator.
## The algorithm
Well, it's both of my algorithms. It's actually quite simple:
``````def sample_generator(up_to, previously_chosen=set(), *, prune=True):
prime = modprime_at_least(up_to+1)
# Fudge to make it less predictable
fudge_power = 2**randrange(7, 11)
fudge_constant = randrange(prime//2, prime)
fudge_factor = randrange(prime//2, prime)
def permute(x):
permuted = pow(x, fudge_power, prime)
return permuted if 2*x <= prime else prime - permuted
for x in range(prime):
res = (permute(x) + fudge_constant) % prime
res = permute((res * fudge_factor) % prime)
if res in previously_chosen:
if prune:
previously_chosen.remove(res)
elif res < up_to:
yield res
``````
The change is as simple as adding:
``````if res in previously_chosen:
previously_chosen.remove(res)
``````
You can add to `previously_chosen` at any time by adding to the `set` that you passed in. In fact, you can also remove from the set in order to add back to the potential pool, although this will only work if `sample_generator` has not yet yielded it or skipped it with `prune=False`.
So there is is. It's easy to see that it fulfils all of the requirements, and it's easy to see that the requirements are absolute. Note that if you don't have a set, it still meets its worst cases by converting the input to a set, although it increases overhead.
# Testing the RNG's quality
I became curious how good this PRNG actually is, statistically speaking.
Some quick searches lead me to create these three tests, which all seem to show good results!
Firstly, some random numbers:
``````N = 1000000
my_gen = list(sample_generator(N))
target = list(range(N))
random.shuffle(target)
control = list(range(N))
random.shuffle(control)
``````
These are "shuffled" lists of 10⁶ numbers from `0` to `10⁶-1`, one using our fun fudged PRNG, the other using a Mersenne Twister as a baseline. The third is the control.
Here's a test which looks at the average distance between two random numbers along the line. The differences are compared with the control:
``````from collections import Counter
def birthdat_calc(randoms):
return Counter(abs(r1-r2)//10000 for r1, r2 in zip(randoms, randoms[1:]))
def birthday_compare(randoms_1, randoms_2):
birthday_1 = sorted(birthdat_calc(randoms_1).items())
birthday_2 = sorted(birthdat_calc(randoms_2).items())
return sum(abs(n1 - n2) for (i1, n1), (i2, n2) in zip(birthday_1, birthday_2))
print(birthday_compare(my_gen, target), birthday_compare(control, target))
#>>> 9514 10136
``````
This is less than the variance of each.
Here's a test which takes 5 numbers in turn and sees what order the elements are in. They should be evenly distributed between all 120 possible orders.
``````def permutations_calc(randoms):
permutations = Counter()
for items in zip(*[iter(randoms)]*5):
sorteditems = sorted(items)
permutations[tuple(sorteditems.index(item) for item in items)] += 1
return permutations
def permutations_compare(randoms_1, randoms_2):
permutations_1 = permutations_calc(randoms_1)
permutations_2 = permutations_calc(randoms_2)
keys = sorted(permutations_1.keys() | permutations_2.keys())
return sum(abs(permutations_1[key] - permutations_2[key]) for key in keys)
print(permutations_compare(my_gen, target), permutations_compare(control, target))
#>>> 5324 5368
``````
This is again less than the variance of each.
Here's a test that sees how long "runs" are, aka. sections of consecutive increases or decreases.
``````def runs_calc(randoms):
runs = Counter()
run = 0
for item in randoms:
if run == 0:
run = 1
elif run == 1:
run = 2
increasing = item > last
else:
if (item > last) == increasing:
run += 1
else:
runs[run] += 1
run = 0
last = item
return runs
def runs_compare(randoms_1, randoms_2):
runs_1 = runs_calc(randoms_1)
runs_2 = runs_calc(randoms_2)
keys = sorted(runs_1.keys() | runs_2.keys())
return sum(abs(runs_1[key] - runs_2[key]) for key in keys)
print(runs_compare(my_gen, target), runs_compare(control, target))
#>>> 1270 975
``````
The variance here is very large, and over several executions I have seems an even-ish spread of both. As such, this test is passed.
A Linear Congruential Generator was mentioned to me, as possibly "more fruitful". I have made a badly implemented LCG of my own, to see whether this is an accurate statement.
LCGs, AFAICT, are like normal generators in that they're not made to be cyclic. Therefore most references I looked at, aka. Wikipedia, covered only what defines the period, not how to make a strong LCG of a specific period. This may have affected results.
Here goes:
``````from operator import mul
from functools import reduce
# Credit http://stackoverflow.com/a/16996439/1763356
# Meta: Also Tobias Kienzler seems to have credit for my
# edit to the post, what's up with that?
def factors(n):
d = 2
while d**2 <= n:
while not n % d:
yield d
n //= d
d += 1
if n > 1:
yield n
def sample_generator3(up_to):
for modulier in count(up_to):
modulier_factors = set(factors(modulier))
multiplier = reduce(mul, modulier_factors)
if not modulier % 4:
multiplier *= 2
if multiplier < modulier - 1:
multiplier += 1
break
x = randrange(0, up_to)
fudge_constant = random.randrange(0, modulier)
for modfact in modulier_factors:
while not fudge_constant % modfact:
fudge_constant //= modfact
for _ in range(modulier):
if x < up_to:
yield x
x = (x * multiplier + fudge_constant) % modulier
``````
We no longer check for primes, but we do need to do some odd things with factors.
• `modulier ≥ up_to > multiplier, fudge_constant > 0`
• `a - 1` must be divisible by every factor in `modulier`...
• ...whereas `fudge_constant` must be coprime with `modulier`
Note that these aren't rules for a LCG but a LCG with full period, which is obviously equal to the `mod`ulier.
I did it as such:
• Try every `modulier` at least `up_to`, stopping when the conditions are satisfied
• Make a set of its factors, `𝐅`
• Let `multiplier` be the product of `𝐅` with duplicates removed
• If `multiplier` is not less than `modulier`, continue with the next `modulier`
• Let `fudge_constant` be a number less that `modulier`, chosen randomly
• Remove the factors from `fudge_constant` that are in `𝐅`
This is not a very good way of generating it, but I don't see why it would ever impinge the quality of the numbers, aside from the fact that low `fudge_constant`s and `multiplier` are more common than a perfect generator for these might make.
Anyhow, the results are appalling:
``````print(birthday_compare(lcg, target), birthday_compare(control, target))
#>>> 22532 10650
print(permutations_compare(lcg, target), permutations_compare(control, target))
#>>> 17968 5820
print(runs_compare(lcg, target), runs_compare(control, target))
#>>> 8320 662
``````
In summary, my RNG is good and a linear congruential generator is not. Considering that Java gets away with a linear congruential generator (although it only uses the lower bits), I would expect my version to be more than sufficient.
-
Here's a way that doesn't build the difference set explicitly. But it does use a form of @Veedrac's "accept/reject" logic. If you're not willing to mutate the base sequence as you go along, I'm afraid that's unavoidable:
``````def sample(n, base, forbidden):
# base is iterable, forbidden is a set.
# Every element of forbidden must be in base.
# forbidden is updated.
from random import random
nusable = len(base) - len(forbidden)
assert nusable >= n
result = []
if n == 0:
return result
for elt in base:
if elt in forbidden:
continue
if nusable * random() < n:
result.append(elt)
n -= 1
if n == 0:
return result
nusable -= 1
assert False, "oops!"
``````
Here's a little driver:
``````base = list(range(100))
forbidden = set()
for i in range(10):
print sample(10, base, forbidden)
``````
-
+1 thank you, if i understand this correctly, this is `O(size(base))`? comparing with @Chronialis's answer, your space complexity is lower but the time complexity is on average the same? (ps: sampling just 1 would be ok to simplify the logic). – necromancer Sep 20 '13 at 17:38
Yes, `O(len(base))` time per call. But if you exhaust the base set (as my sample driver did), `len(forbidden) == len(base)` at the end of all the calls - you still end up with a set of the same size as `base`. Sorry, I don't understand what "sampling just 1" means. If you only want a sample of size 1, pass `1` for `n` to my `sample()` function. If you hard-coded `1` into it, the code would get a little simpler, but not a lot. The "hard part" remains skipping over all the elements returned on previous calls. – Tim Peters Sep 20 '13 at 17:46
BTW, you should think about my `shuffle` answer again. It's by far the most efficient way to do this if you're going to take multiple samples: `O(len(base))` time on the first call, and `O(n)` time for each subsequent call asking for a sample of size `n`. And it takes no extra space if you're willing to let `base` get destroyed. – Tim Peters Sep 20 '13 at 18:27
no doubt, but if the range is huge and I am selecting a small number of samples it is overkill. It is also an `O(range)` solution which I am trying to avoid. – necromancer Sep 20 '13 at 18:42
@no_answer_not_upvoted: If you detest that “hack”, I would strongly advice against using the python `sample()` function – Chronial Sep 20 '13 at 22:28
OK, one last try ;-) At the cost of mutating the base sequence, this takes no additional space, and requires time proportional to `n` for each `sample(n)` call:
``````class Sampler(object):
def __init__(self, base):
self.base = base
self.navail = len(base)
def sample(self, n):
from random import randrange
if n < 0:
raise ValueError("n must be >= 0")
if n > self.navail:
raise ValueError("fewer than %s unused remain" % n)
base = self.base
for _ in range(n):
i = randrange(self.navail)
self.navail -= 1
base[i], base[self.navail] = base[self.navail], base[i]
return base[self.navail : self.navail + n]
``````
Little driver:
``````s = Sampler(list(range(100)))
for i in range(9):
print s.sample(10)
print s.sample(1)
print s.sample(1)
``````
In effect, this implements a resumable `random.shuffle()`, pausing after `n` elements have been selected. `base` is not destroyed, but is permuted.
-
thanks so much :-) at the very least a +1 until i understand it. i do see that it uses `O(range)` memory but still if i don't write a better one myself i will accept it for effort. thank you again! – necromancer Sep 20 '13 at 22:23
LOL - you still haven't defined your problem ;-) If the size of your range is much larger than the total sizes of the samples you expect to extract, then @Veedrac's is an excellent approach. At least in Python 3, it will use a small amount of memory even if the `n` in `range(n)` is huge (in Python 2, much the same if you use `xrange(n)` instead). – Tim Peters Sep 21 '13 at 0:13
you guys have me caught between a rock and a hard place. that's how i feel about the two solutions. sigh.. – necromancer Sep 21 '13 at 0:44
woops did i forget the +1 originally? well +1-ed. feel free to triple dip too ;-) – necromancer Sep 21 '13 at 0:44
I have added my own answer. A critique would be appreciated. Thanks! – necromancer Sep 24 '13 at 15:32
You can implement a shuffling generator, based off Wikipedia's "Fisher--Yates shuffle#Modern method"
``````def shuffle_gen(src):
""" yields random items from base without repetition. Clobbers `src`. """
for remaining in xrange(len(src), 0, -1):
i = random.randrange(remaining)
yield src[i]
src[i] = src[remaining - 1]
``````
Which can then be sliced using `itertools.islice`:
``````>>> import itertools
>>> sampler = shuffle_gen(range(100))
>>> sample1 = list(itertools.islice(sampler, 10))
>>> sample1
[37, 1, 51, 82, 83, 12, 31, 56, 15, 92]
>>> sample2 = list(itertools.islice(sampler, 80))
>>> sample2
[79, 66, 65, 23, 63, 14, 30, 38, 41, 3, 47, 42, 22, 11, 91, 16, 58, 20, 96, 32, 76, 55, 59, 53, 94, 88, 21, 9, 90, 75, 74, 29, 48, 28, 0, 89, 46, 70, 60, 73, 71, 72, 93, 24, 34, 26, 99, 97, 39, 17, 86, 52, 44, 40, 49, 77, 8, 61, 18, 87, 13, 78, 62, 25, 36, 7, 84, 2, 6, 81, 10, 80, 45, 57, 5, 64, 33, 95, 43, 68]
>>> sample3 = list(itertools.islice(sampler, 20))
>>> sample3
[85, 19, 54, 27, 35, 4, 98, 50, 67, 69]
``````
-
Eric, this is basically the same as one of my earlier answers. Note that second argument to `xrange()` here should be 0, not 1 (e.g. `list(xrange(4, 1, -1)` is `[4, 3, 2]` - `range/xrange` always quit before the `stop` argument. – Tim Peters Sep 23 '13 at 19:33
@TimPeters: Had that before and decided to change it for some reason... Fixed now. Yes, this is algorithmically the same, but I think it's more cleanly implemented as an iterator. – Eric Sep 23 '13 at 19:51
+1 thanks for the answer @Eric – necromancer Sep 23 '13 at 22:13
@Eric I just added my own answer. Any thoughts? – necromancer Sep 24 '13 at 15:29
This is a rewritten version of @no_answer_not_upvoted's cool solution. Wraps it in a class to make it much easier to use correctly, and uses more dict methods to cut the lines of code.
``````from random import randrange
class Sampler:
def __init__(self, n):
self.n = n # number remaining from original range(n)
# i is a key iff i < n and i already returned;
# in that case, state[i] is a value to return
# instead of i.
self.state = dict()
def get(self):
n = self.n
if n <= 0:
raise ValueError("range exhausted")
result = i = randrange(n)
state = self.state
# Most of the fiddling here is just to get
# rid of state[n-1] (if it exists). It's a
# space optimization.
if i == n - 1:
if i in state:
result = state.pop(i)
elif i in state:
result = state[i]
if n - 1 in state:
state[i] = state.pop(n - 1)
else:
state[i] = n - 1
elif n - 1 in state:
state[i] = state.pop(n - 1)
else:
state[i] = n - 1
self.n = n-1
return result
``````
Here's a basic driver:
``````s = Sampler(100)
allx = [s.get() for _ in range(100)]
assert sorted(allx) == list(range(100))
from collections import Counter
c = Counter()
for i in range(6000):
s = Sampler(3)
one = tuple(s.get() for _ in range(3))
c[one] += 1
for k, v in sorted(c.items()):
print(k, v)
``````
and sample output:
``````(0, 1, 2) 1001
(0, 2, 1) 991
(1, 0, 2) 995
(1, 2, 0) 1044
(2, 0, 1) 950
(2, 1, 0) 1019
``````
By eyeball, that distribution is fine (run a chi-squared test if you're skeptical). Some of the solutions here don't give each permutation with equal probability (even though they return each k-subset of n with equal probability), so are unlike `random.sample()` in that respect.
-
+1 Thank you for the elegant rewrite. I am new to Python so this helps me learn too. – necromancer Sep 24 '13 at 19:49
TimPeters you get to pick the correct answer (hopefully between this one and Chronial's rewrite which I almost prefer but since he already has 300 rep and you have contributed hugely). Thanks so much for your contributions -- what started out as a simple question turned out to be a somewhat legendary thread. Multiple single-author answers, self-answer, and @Veedrac's semi-thesis which ended up in the community wiki. – necromancer Sep 24 '13 at 20:46
No contest - @Chronial's is prettiest! This was a lot of fun, but the cleanest code wins :-) Having said that, I wouldn't use his version - or mine. I want an interface that lets me specify how many samples to take "in one gulp". Messing with (e.g.) `itertools.slice()` to get that complicates life for the user, and obfuscates the intent for the code reader. Code like this would remain the core of it, though. Thanks to all for playing! :-) – Tim Peters Sep 24 '13 at 21:04
@TimPeters is there any reason why you did not use a generator here? It seems to just perfectly fit the situation. Or are there some drawbacks I’m missing? – Chronial Sep 24 '13 at 22:20
@TimPeters I think Chronial's "if-less" version is masterful indeed so I accepted it. I wanted to add another bounty for your contributions but the way StackOverflow bounties work is that the minimum amount I can give in a second bounty for this question is 500 points, and it would be too big a hit to my rep. Good to have your answers, esp. from THE Tim Peters!! It is a privilege :-) Will look forward to following your other answers. PS: Somebody made almost a 1000 points of rep just by asking a question based on something you wrote: stackoverflow.com/questions/228181/zen-of-python :) – necromancer Sep 24 '13 at 22:50
Edit: see cleaner versions below by @TimPeters and @Chronial. A minor edit pushed this to the top.
Here is what I believe is the most efficient solution for incremental sampling. Instead of a list of previously sampled numbers, the state to be maintained by the caller comprises a dictionary that is ready for use by the incremental sampler, and a count of numbers remaining in the range.
The following is a demonstrative implementation. Compared to other solutions:
• no loops (no Standard Python/Veedrac hack; shared credit between Python impl and Veedrac)
• time complexity is `O(log(number_previously_sampled))`
• space complexity is `O(number_previously_sampled)`
Code:
``````import random
def remove (i, n, state):
if i == n - 1:
if i in state:
t = state[i]
del state[i]
return t
else:
return i
else:
if i in state:
t = state[i]
if n - 1 in state:
state[i] = state[n - 1]
del state[n - 1]
else:
state[i] = n - 1
return t
else:
if n - 1 in state:
state[i] = state[n - 1]
del state[n - 1]
else:
state[i] = n - 1
return i
s = dict()
for n in range(100, 0, -1):
print remove(random.randrange(n), n, s)
``````
-
What's the advantage over the Veedrac Hack? Keeping track of a few auxiliary variables is less than keeping a whole dictionary so I don't see where you could ever use this but not mine. – Veedrac Sep 24 '13 at 16:33
Seriously cool! I'm going to post a rewrite that's easier to use, and perhaps to understand - I don't want to "enter" it, I just want to post it for posterity ;-) About the time complexity claim, doesn't make sense: this is `O(1)` per element extraction, so `O(k)` for getting a sample of size `k`. – Tim Peters Sep 24 '13 at 18:31
No, dict access is `O(1)` time in CPython. That's expected time. Worst-case time is `O(len(dict))` time, but that's never seen. But, to believe the `O(1)` claim, you need to have faith in probability ;-) – Tim Peters Sep 24 '13 at 18:52
@TimPeters The integers here are not from a continuous range, so it is definitely possible to hit the worst case somewhere during the loop. But if you assume that, then the algorithm is `O(number_previously_sampled)` and not `O(log(number_previously_sampled))`. – Chronial Sep 24 '13 at 19:14
@TimPeters Astronomically unlikely – yes, but that also doesn’t stop no_answer_not_upvoted from disliking Veedrac’s answer ;). Btw: I think I win the pretty contest :P. – Chronial Sep 24 '13 at 19:37
This is my version of the Knuth shuffle, that was first posted by Tim Peters, prettified by Eric and then nicely space-optimized by no_answer_not_upvoted.
This is based on Eric’s version, since I indeed found his code very pretty :).
``````import random
def shuffle_gen(n):
# this is used like a range(n) list, but we don’t store
# those entries where state[i] = i.
state = dict()
for remaining in xrange(n, 0, -1):
i = random.randrange(remaining)
yield state.get(i,i)
state[i] = state.get(remaining - 1,remaining - 1)
# Cleanup – we don’t need this information anymore
state.pop(remaining - 1, None)
``````
usage:
``````out = []
gen = shuffle_gen(100)
for n in range(100):
out.append(gen.next())
print out, len(set(out))
``````
-
+1 nice clean code! – necromancer Sep 24 '13 at 20:01
By the way, this code is waaaay too wordy ;-) You can get rid of the final line (`state.pop(...)`), and in the penultimate line replace `get` with `pop`. Then it's a good answer - LOL ;-) – Tim Peters Sep 25 '13 at 2:10
@TimPeters hehe, no. The reason why there is a `get` and and a `pop` is that it is possible that `i=remaining-1`. With just a `pop` we remove the item and just re-add it again. I would say that `remaining` is either large and this event very unlikely or `remaining` is small and we will be finished soon anyways, so the “leak” is not too problematic. But I wanted to be thorough :). – Chronial Sep 25 '13 at 2:25
Ah! Got it. My apologies for attempting to gild the lily ;-) – Tim Peters Sep 25 '13 at 2:28
Reasonably fast one-liner (`O(n + m)`, n=range,m=old samplesize):
``````next_sample = random.sample(set(range(100)).difference(my_sample), 10)
``````
-
+1 the problem is that if the n is large, say 10,000,000 then explicitly constructing a set will blow up – necromancer Sep 20 '13 at 16:33
That’s right, but know that constructing a set is still `O(n)` so even for 10,000,00 it takes less then a second. – Chronial Sep 20 '13 at 16:36 | 9,821 | 35,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2015-35 | latest | en | 0.941129 |
http://setiweb.org/percent-error/percentual-error.php | 1,537,325,807,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155817.29/warc/CC-MAIN-20180919024323-20180919044323-00156.warc.gz | 223,821,528 | 4,754 | Home > Percent Error > Percentual Error
# Percentual Error
## Contents
Tips Some teachers like the percent error to be rounded to a certain point; most people will be satisfied with the percent error rounded to three significant digits. Answer this question Flag as... The absolute value of the error is divided by an accepted value and given as a percent.|accepted value - experimental value| \ accepted value x 100%Note for chemistry and other sciences, Ex: 10 - 9 = 1 3 Divide the result by the real number. http://astro.physics.uiowa.edu/ITU/glossary/percent-error-formula/
## Percent Error Chemistry
PhysicsPreceptors 33.590 προβολές 14:52 Relative Error and Percent Error - Διάρκεια: 5:21. Place the fraction in decimal form. By using this site, you agree to the Terms of Use and Privacy Policy.
ISBN0-8247-0888-1. so divide by the exact value and make it a percentage: 65/325 = 0.2 = 20% Percentage Error is all about comparing a guess or estimate to an exact value. You can change this preference below. Κλείσιμο Ναι, θέλω να τη κρατήσω Αναίρεση Κλείσιμο Αυτό το βίντεο δεν είναι διαθέσιμο. Ουρά παρακολούθησηςΟυράΟυρά παρακολούθησηςΟυρά Κατάργηση όλωνΑποσύνδεση Φόρτωση... Ουρά παρακολούθησης Ουρά __count__/__total__ Error Negative Percent Error Learn more You're viewing YouTube in Greek.
The absolute value of a number is the value of the positive value of the number, whether it's positive or negative. Percent Error Calculator Becomean Author! archived preprint ^ Jorrit Vander Mynsbrugge (2010). "Bidding Strategies Using Price Based Unit Commitment in a Deregulated Power Market", K.U.Leuven ^ Hyndman, Rob J., and Anne B. http://astro.physics.uiowa.edu/ITU/glossary/percent-error-formula/ This value is your 'error'. continue reading below our video 4 Tips for Improving Test Performance Divide the error by the exact or ideal value (i.e., not your experimental or measured
The difference between At and Ft is divided by the Actual value At again. What Is A Good Percent Error Kick Images, Getty Images By Anne Marie Helmenstine, Ph.D. It usually expresses accuracy as a percentage, and is defined by the formula: M = 100 n ∑ t = 1 n | A t − F t A t | Also from About.com: Verywell & The Balance Υπενθύμιση αργότερα Έλεγχος Υπενθύμιση απορρήτου από το YouTube, εταιρεία της Google Παράβλεψη περιήγησης GRΜεταφόρτωσηΣύνδεσηΑναζήτηση Φόρτωση... Επιλέξτε τη γλώσσα σας. Κλείσιμο Μάθετε περισσότερα View this
## Percent Error Calculator
Ways of Expressing Error in Measurement: 1. http://chemistry.about.com/od/workedchemistryproblems/a/percenterror.htm Calculate the percent error of your measurement.Subtract one value from the other:2.68 - 2.70 = -0.02 Depending on what you need, you may discard any negative sign (take the absolute value): 0.02This Percent Error Chemistry Lalit Mohan Sharma 1.380 προβολές 15:38 Scientific Notation and Significant Figures (1.7) - Διάρκεια: 7:58. Percent Error Definition This little-known but serious issue can be overcome by using an accuracy measure based on the ratio of the predicted to actual value (called the Accuracy Ratio), this approach leads to
Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. Inputs: measured valueactual, accepted or true value Conversions: measured value= 0 = 0 actual, accepted or true value= 0 = 0 Solution: percent error= NOT CALCULATED Change Equation Variable Select to The difference between two measurements is called a variation in the measurements. While both situations show an absolute error of 1 cm., the relevance of the error is very different. Can Percent Error Be Negative
Operations Management: A Supply Chain Approach. Step 2: Divide the error by the exact value (we get a decimal number) Step 3: Convert that to a percentage (by multiplying by 100 and adding a "%" sign) As If you are measuring a football field and the absolute error is 1 cm, the error is virtually irrelevant. Once you find the absolute value of the difference between the approximate value and exact value, all you need to do is to divide it by the exact value and multiply
Percentage Difference Percentage Index Search :: Index :: About :: Contact :: Contribute :: Cite This Page :: Privacy Copyright © 2014 MathsIsFun.com Imaging the Universe A lab manual developed by Percent Error Worksheet Any measurements within this range are "tolerated" or perceived as correct. b.) the relative error in the measured length of the field.
## For example, you would not expect to have positive percent error comparing actual to theoretical yield in a chemical reaction.[experimental value - theoretical value] / theoretical value x 100%Percent Error Calculation
The actual length of this field is 500 feet. Please enter a valid email address. Accuracy is a measure of how close the result of the measurement comes to the "true", "actual", or "accepted" value. (How close is your answer to the accepted value?) Tolerance is Percent Error Definition Chemistry Another word for this variation - or uncertainty in measurement - is "error." This "error" is not the same as a "mistake." It does not mean that you got the wrong
Tyler DeWitt 102.551 προβολές 9:29 Class 10+1, Chapter 1E, Question 6, Absolute error, Relative error and percentage error - Διάρκεια: 15:38. Thanks for letting us know. A stopwatch has a circular dial divided into 120 divisions.time interval of 10 oscillation of a simple pendulum is measure as 25 sec.by using the watch Max. % error in the Percent of Error: Error in measurement may also be expressed as a percent of error.
The order does not matter if you are dropping the sign, but you subtract the theoretical value from the experimental value if you are keeping negative signs. In plain English: 4. The precision of a measuring instrument is determined by the smallest unit to which it can measure. 2. We can also use a theoretical value (when it is well known) instead of an exact value.
Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. | 1,442 | 6,053 | {"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} | 2.859375 | 3 | CC-MAIN-2018-39 | latest | en | 0.558267 |
https://reference.wolframcloud.com/language/ref/JacobiZeta.html | 1,716,413,348,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058568.59/warc/CC-MAIN-20240522194007-20240522224007-00276.warc.gz | 419,920,348 | 17,692 | # JacobiZeta
JacobiZeta[ϕ,m]
gives the Jacobi zeta function .
# Examples
open allclose all
## Basic Examples(4)
Evaluate numerically:
Plot over a subset of the reals:
Plot over a subset of the complexes:
Series expansion about the origin:
## Scope(30)
### Numerical Evaluation(5)
Evaluate to high precision:
The precision of the output tracks the precision of the input:
Evaluate for complex arguments and parameters:
Evaluate JacobiZeta efficiently at high precision:
JacobiZeta threads elementwise over lists:
JacobiZeta can be used with Interval and CenteredInterval objects:
### Specific Values(5)
Simple exact results are generated automatically:
Exact values after FunctionExpand is applied:
Value at infinity:
Find a local maximum as a root of :
JacobiZeta is an odd function with respect to the first argument:
### Visualization(3)
Plot JacobiZeta as a function of its first parameter :
Plot JacobiZeta as a function of its second parameter :
Plot the real part of :
Plot the imaginary part of :
### Function Properties(6)
JacobiZeta is not an analytic function:
However, for fixed , is an analytic function of :
Thus, for example, has no singularities or discontinuities:
is neither nondecreasing nor nonincreasing:
is not injective:
is not surjective:
is neither non-negative nor non-positive:
is neither convex nor concave:
### Differentiation and Integration(4)
First derivative:
Higher derivatives:
Plot higher derivatives for :
Differentiate with respect to its second parameter :
Definite integral of an odd function over an interval centered at the origin:
### Series Expansions(4)
Taylor expansion for JacobiZeta:
Plot the first three approximations for around :
Taylor expansion at the origin in the parameter :
Plot the first three approximations for around :
Find series expansions at a branch point:
JacobiZeta can be applied to a power series:
### Function Representations(3)
Primary definition:
Relation to other elliptictype functions:
## Applications(3)
Plot of the real part of JacobiZeta over the complex plane:
Supersymmetric zeroenergy solution of the Schrödinger equation in a periodic potential:
Check the Schrödinger equation:
Plot the superpotential, the potential and the wave function:
Define a conformal map:
## Properties & Relations(5)
Use FunctionExpand to express JacobiZeta in terms of incomplete elliptic integrals:
Expand special cases:
Some special cases require argument restrictions:
Numerically find a root of a transcendental equation:
For real arguments, if , then JacobiZN[u,m]JacobiZeta[ϕ,m] for :
JacobiZeta[ϕ,m] is real valued for real arguments subject to :
## Possible Issues(4)
Machine-precision input may be insufficient to give a correct answer:
A larger setting for \$MaxExtraPrecision may be needed:
JacobiZeta, function of amplitude , is not to be confused with JacobiZN, sometimes denoted as and a function of elliptic argument :
The Wolfram Language JacobiZeta[ϕ,m] is a function of amplitude and uses the following definition:
JacobiZN[u,m] is a function of elliptic argument and uses the definition , where is JacobiEpsilon[u,m]:
To avoid confusion, JacobiZN uses a different TraditionalForm:
In traditional form, the vertical separator must be used:
Wolfram Research (1991), JacobiZeta, Wolfram Language function, https://reference.wolfram.com/language/ref/JacobiZeta.html (updated 2020).
#### Text
Wolfram Research (1991), JacobiZeta, Wolfram Language function, https://reference.wolfram.com/language/ref/JacobiZeta.html (updated 2020).
#### CMS
Wolfram Language. 1991. "JacobiZeta." Wolfram Language & System Documentation Center. Wolfram Research. Last Modified 2020. https://reference.wolfram.com/language/ref/JacobiZeta.html.
#### APA
Wolfram Language. (1991). JacobiZeta. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/JacobiZeta.html
#### BibTeX
@misc{reference.wolfram_2024_jacobizeta, author="Wolfram Research", title="{JacobiZeta}", year="2020", howpublished="\url{https://reference.wolfram.com/language/ref/JacobiZeta.html}", note=[Accessed: 22-May-2024 ]}
#### BibLaTeX
@online{reference.wolfram_2024_jacobizeta, organization={Wolfram Research}, title={JacobiZeta}, year={2020}, url={https://reference.wolfram.com/language/ref/JacobiZeta.html}, note=[Accessed: 22-May-2024 ]} | 1,039 | 4,397 | {"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-2024-22 | latest | en | 0.756797 |
http://limayans.us/qoka/geometry-homework-help-circles-287.php | 1,503,283,563,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107487.10/warc/CC-MAIN-20170821022354-20170821042354-00368.warc.gz | 254,243,803 | 4,610 | # Geometry homework help circles
### Circles to Tangents Homework
Students, teachers, parents, and everyone can find solutions to their math.Free math problem solver answers your algebra homework questions with step-by-step explanations. Mathway. Visit Mathway on the web.She needs circles help homework geometry to take a dictionary or choose another toy and let her leave when he is mature and careful.
Geometry Math Help. Square. A square is a four sided regular polygon. The circle is a shape where all points along the shape are equal distance from a specific point.Video lessons for Geometry, Circle Basics, Arcs, Chords, and Angles and more - Hotmath.com.The Math Forum is the comprehensive resource for math education on the Internet.Help With Your Math Homework. Find Out Things About Circles.
### Circle Geometry Math
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An understanding of the attributes and relationships of geometric objects can be applied in diverse contexts.Core Connections Geometry is the second course in a five-year sequence of college preparatory mathematics courses that starts with Algebra I and continues through.Making an Email Message After you enter the content on the wishes of the new parking circles homework geometry help policy unfair.This page will show you how to calculate various things that pertain to a circle.Each week, all students are invited to an open Honors Geometry Math Help Room run by a rotating staff of instructors. | 781 | 3,912 | {"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-2017-34 | latest | en | 0.866078 |
https://prezi.com/d6oabf0_kmyz/decimal-place-rounding/ | 1,545,076,206,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829115.83/warc/CC-MAIN-20181217183905-20181217205905-00094.warc.gz | 703,665,852 | 23,632 | ### Present Remotely
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# Decimal Place Rounding
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by
## Mr Mattock
on 1 September 2018
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#### Transcript of Decimal Place Rounding
Decimal Place Rounding
Starter
What digits can fill this space: 27.2...64 ?
L.O. - Round a number to a given number of decimal places.
The calculation on the
1551.806452.
Round the answer to the nearest:
(a) 1000
(b) 100
(c) 10
Decimal places
1551
1552
1551.8
1551.9
Main Activity
Complete either the red, amber or green activity on the sheet.
Red
Amber
Green
(a) 7.638, 8.2, 7.749, 8.463, 8.34, 7.86, 8.45, 7.5.
(b) 7.5, 7.638, 7.749, 7.86, 8.2, 8.43, 8.45, 8.463.
(c) 7.5 - No. (d) 8.463 - Yes (8.499999....)
(1) 5, 5, 6, 7, 3, 5, 7, 4, 4, 3 (2) 3.5, 6.5, 4.0, 4.5, 3.2
(3) 4.50, 3.18
(4) The maximum number of decimal places
any of the numbers have is 3 decimal places.
(a) 3.450 and anything up to 3.549.
(b) 5.750 and anything up to 5.849.
(c) 5.750 - 3.549 = 2.201. (d) 5.849 - 3.450 = 2.399
What digits can fill this space: 27.2...64 ?
0, 1, 2, 3, 4 - Round down
5, 6, 7, 8, 9 - Round up
Decimal places
1551
1552
1551.8
1551.9
1551.80
1551.81
1551.806
1551.807
Starter
The calculation on the
1551.806452.
Round the answer to the nearest:
(a) 1000 -
2000
(b) 100 -
1600
(c) 10 -
1550
Activities
Activity
Key
Examples
Main Activity
On your whiteboard show me the number lines that the given numbers are between (to the given accuracy).
Decimal places
Whole numbers
Tenths
Hundredths
Thousandths
1551.80
1551.81
1551.806
1551.807
Worked
Example
Part 1
Worked
Example
Part 2
Worked
Example
Full transcript | 732 | 1,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-51 | latest | en | 0.76754 |
https://trac.sagemath.org/attachment/ticket/6962/trac_6962.patch | 1,611,196,819,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522150.18/warc/CC-MAIN-20210121004224-20210121034224-00486.warc.gz | 632,817,057 | 6,813 | # Ticket #6962: trac_6962.patch
File trac_6962.patch, 9.0 KB (added by ncohen, 11 years ago)
• ## sage/graphs/graph.py
# HG changeset patch
# User Nathann Cohen <nathann.cohen@gmail.com>
# Date 1253385913 -7200
# Node ID b2af0a8f0e5b97688407e37af24ed355592c8221
# Parent 021c67fcbec41f970d52b920b30ab91a28dd5cd8
diff -r 021c67fcbec4 -r b2af0a8f0e5b sage/graphs/graph.py
a """ return sorted(self.out_degree_iterator(), reverse=True) def feedback_edge_set(self,value_only=False): r""" Computes the minimum feedback edge set of a digraph ( also called feedback arc set ). The minimum feedback edge set of a digraph is a set of edges that intersect all the circuits of the digraph. Equivalently, a minimum feedback arc set of a DiGraph is a set S of arcs such that the digraph G-S is acyclic. For more informations, see ( http://en.wikipedia.org/wiki/Feedback_arc_set ) INPUT : - value_only (boolean) -- - When set to True, only the minimum cardinal of a minimum edge set is returned. - When set to False, the Set of edges of a minimal edge set is returned. This problem is solved using Linear Programming, which certainly is not the best way and will have to be updated. The program solved is the following : .. MATH: \mbox{Minimize : }&\sum_{(u,v)\in G} b_{(u,v)}\\ \mbox{Such that : }&\\ &\forall v\in G, \sum_{i\in [0,\dots,n-1]}x_{v,i}=1\\ &\forall i\in [0,\dots,n-1], \sum_{v\in G}x_{v,i}=1\\ &\forall v\in G,\sum_{i\in [0,\dots,n-1]} ix_{v,i}=d_v\\ &\forall (u,v)\in G, d_u-d_v+nb_{(u,v)}\geq 0\\ An explanation : An acyclic digraph can be seen as a poset, and every poset has a linear extension. This means that in any acyclic digraph the vertices can be ordered with a total order < in such a way that if (u,v)\in G, then `u | 563 | 1,733 | {"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} | 3.46875 | 3 | CC-MAIN-2021-04 | latest | en | 0.728542 |
http://en.wikipedia.org/wiki/Rigid_line_inclusion | 1,386,584,552,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163956743/warc/CC-MAIN-20131204133236-00074-ip-10-33-133-15.ec2.internal.warc.gz | 61,848,751 | 10,470 | # Rigid line inclusion
A rigid line inclusion, also called stiffener, is a mathematical model used in solid mechanics to describe a narrow hard phase, dispersed within a matrix material. This inclusion is idealised as an infinitely rigid and thin reinforcement, so that it represents a sort of ‘inverse’ crack, from which the nomenclature ‘anticrack’ derives.
From the mechanical point of view, a stiffener introduces a kinematical constraint, imposing that it may only suffer a rigid body motion along its line.
## Theoretical model
The stiffener model has been used to investigate different mechanical problems in classical elasticity (load diffusion,[1] inclusion at bi material interface [2]).
Sketch of a stiffener embedded in a matrix loaded at its boundary.
The main characteristics of the theoretical solutions are basically the following.
1. Similarly to a fracture, a square-root singularity in the stress/strain fields is present at the tip of the inclusion.
2. In a homogeneous matrix subject to uniform stress at infinity, such singularity only arises when a normal stress acts parallel or orthogonal to the inclusion line, while a stiffener parallel to a simple shear does not disturb the ambient field.
## Experimental validation
Dog-bone shaped sample of two-component epoxy resin containing a lamellar (aluminum) inclusion.
Photoelastic experiment to validate the rigid line inclusion model. Isochromatic fringe patterns around a steel platelet in a photo-elastic two-part epoxy resin compared to analytical solution obtained in plane-strain classical elasticity. Normal stress parallel to the inclusion line is applied.
The characteristics of the elastic solution have been experimentally confirmed through photoelastic transmission experiments.[3]
## Shear bands emerging at the stiffener tip
Analytical solutions obtained in prestressed elasticity show the possibility of the emergence of shear bands at the tip of the stiffener.[4][5][6][7]
## References
1. ^ Koiter, W.T., On the diffusion of load from a stiffener into a sheet. Q. J. Mech. Appl. Math. 1955, VIII, 164–178.
2. ^ Ballarini, R., A rigid line inclusion at a bimaterial interface. Eng. Fract. Mech., 1990, 37, 1–5.
3. ^ G. Noselli, F. Dal Corso and D. Bigoni, The stress intensity near a stiffener disclosed by photoelasticity. International Journal of Fracture, 2010, 166, 91–103.
4. ^ Bigoni, D. Nonlinear Solid Mechanics: Bifurcation Theory and Material Instability. Cambridge University Press, 2012 . ISBN 9781107025417.
5. ^ F. Dal Corso, D. Bigoni and M. Gei, The stress concentration near a rigid line inclusion in a prestressed, elastic material. Part I Full-field solution and asymptotics. Journal of the Mechanics and Physics of Solids, 2008, 56, 815–838.
6. ^ D. Bigoni, F. Dal Corso and M. Gei, The stress concentration near a rigid line inclusion in a prestressed, elastic material. Part II Implications on shear band nucleation, growth and energy release rate. Journal of the Mechanics and Physics of Solids, 2008, 56, 839–857.
7. ^ F. Dal Corso and D. Bigoni, The interactions between shear bands and rigid lamellar inclusions in a ductile metal matrix. Proceedings of the Royal Society A, 2009, 465, 143-163. | 745 | 3,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2013-48 | latest | en | 0.878557 |
https://www.physicsforums.com/threads/control-theory-state-space-method-with-derivative-input.805811/ | 1,518,946,923,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811795.10/warc/CC-MAIN-20180218081112-20180218101112-00688.warc.gz | 922,738,345 | 14,530 | # Control Theory State-Space method with derivative input
1. Mar 29, 2015
### Chacabucogod
Hi,
I'm reading Ogata's Modern Control Engineering, and when he talks about the representation of a differential equation in state space he divides the method in two. The first one is when the input of the differential equation involves no derivative term, for example:
x'(t)+x(t)=u(t)
The next step is doing it with a differential equation that has inputs that have derivatives. For example:
x'(t)+x(t)=u(t)+u'(t)
He then mention that the state varibles will be
x1=y-β0u
x2=y'-β1u-β0u' and so on...
I've tried finding a reason for this and the nearest I've come is the following PDF, which has errors:
http://www.ece.rutgers.edu/~gajic/psfiles/canonicalforms.pdf
Anybody got an idea how that can be derived?
2. Mar 30, 2015
### donpacino
note: this is not mine...
http://lpsa.swarthmore.edu/Representations/SysRepTransformations/TF2SS.html
Consider the third order differential transfer function:
We start by multiplying by Z(s)/Z(s) and then solving for Y(s) and U(s) in terms of Z(s). We also convert back to a differential equation.
We can now choose z and its first two derivatives as our state variables
Now we just need to form the output
Unfortunately, the third derivative of z is not a state variable or an input, so this is not a valid output equation. However, we can represent the term as a sum of state variables and outputs:
and
From these results we can easily form the state space model:
In this case, the order of the numerator of the transfer function was less than that of the denominator. If they are equal, the process is somewhat more complex.
Last edited by a moderator: Apr 19, 2017 | 432 | 1,722 | {"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} | 3.28125 | 3 | CC-MAIN-2018-09 | longest | en | 0.928043 |
https://hirecalculusexam.com/application-of-derivatives-in-economics-2 | 1,720,873,439,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00666.warc.gz | 256,725,434 | 22,263 | # Application Of Derivatives In Economics
## Take My Statistics Tests For Me
This means that we can take the derivative of a company and get the price of its derivatives. We can also take the derivative rate from the market, and get the derivative of that company. Thus, we can get a price for the company and get its price. In many cases, we can do this automatically. In the next section, we will look at the second assumption of the following sections. First, we will show how we can get an estimate of the derivative of an external company. We will begin by looking at the case where the market is on a volatile basis. We will now review the three basic assumptions we have about the market. The first one is that we have a fixed number of financial derivatives. The second one is that the market is a stable one. This is very important for us because we are talking about a market where we can be sure that the market will not be volatile. In other words, we can make a strong statement about the market in terms of the number of financial derivative derivatives. In other situations, we may have a fixed derivative of a fixed number, but this is not the case here. In some situations, we can use anApplication Of Derivatives In Economics – Part 1 by Stéphane Volpe In Part 1 of this series, I look at the problems in the field of economics in the form of questions, and present some suggestions on how to answer them. In this section I will look at some of the problems that arise when looking at the definitions of economic and financial terms in the writings of Stéphanes Volpe. This section will focus on the first one: “The term “capital” in financial terms refers to the value that is attached to a specific asset. This is a term that has been defined by the European Commission and, among other things, the Commission has been under severe criticism for its apparent inefficiency and for the lack of technical clarity on what that means.” A more detailed discussion of this is found in the work of the economist Stephen H. Loeb for the Federal Reserve System, which is based on the concept of the “capital of the future.” In the first part of this series I will look into a number of topics that are of interest to the reader of the book, as they relate to the definition of “capital.
## Pay Someone To Do Accounting Homework
” I will give a brief overview of some of these topics, and I will also cover various other areas of the book. The first topic that I would like to mention is “a measure of financial emergency.” It is a term used by the financial community to describe the financial situation in a given year. The difference between a “capital emergency” and a “steady financial system” is that in a capital emergency it is impossible to know exactly how the stock market will perform in the next year, and what will happen if the stock market crashes. Moreover, the definition of the term “financial emergency” is very specific, and is very difficult to understand. In addition to this, there are several other topics that I would want to add to the book. This is the following: The concept of a “fundamental” asset, which is a financial institution, is defined in the book by the use of the term capital, and this definition works well for the definition of a ‘fundamental asset’ as well. Another term used by financial institutions to refer to the value of a fund, is the “fundamentals of the future” (often called “fundaments”). These are the “future assets” of the financial community. These will be called “future instruments”, and the term ‘fundamentals’ refers to the concept of a fundament. A number of other terms have been used as well in the book, including “saturated” and “stable”, which are the two terms that have been used for the definition in the book. These are referred to as “funds”, “sources of funds”, etc. One of the most important contributions of this book into the field of financial analysis has been the development of a way of describing the that site and the relationship between a financial community and a specific financial institution. The structure of the financial structure of a financial community is described by the book, by the author, and by the way in which the financial community is being described. As I mentioned in Part 1, the definitionApplication Of Derivatives In Economics Why Derivatives? Why Proprietary Imports In The World Briefly, as an alternative to the current “banking” model of banks, the current ‘banking model’ lacks a market for derivatives, and that is why the market is not in the early stages of thinking about derivatives. The market for derivatives is not in it’s infancy, but it is beginning to take off. The market is only beginning to become a market. In the early days of financial markets, derivatives were the first to be sold on credit card debt. The new credit card market of the 1990s saw a new market for derivatives. The credit card market is in the early stage of the market.
## Do My Online Classes
It is in the market for derivatives that the market is in. The markets for derivatives are in the early market stage of the markets, but they are not in the market as such. In the late days of credit cards, derivatives have taken a new form. The market for derivatives will continue to take off until the market for the derivatives market is fully established. How do the markets for derivatives work in a market for credit card debt? The markets for credit card debts are very similar to the markets for debt. In the late days, debt was a very low priority. In fact, the markets for credit cards were very similar to debt. The markets were very similar for debt. When debt was a problem, the market for credit cards was very similar to that for credit cards. In fact there were only two markets for credit: the markets for the credit card debt and the markets for bankruptcy. In the market for debt, debt was the first to take off and not the second to take off, and in the market, the first to go to bankruptcy. What is the market for debts? In terms of credit card debt, the market is very similar to other markets for debt: the market for bankruptcy is very similar. In the markets for debts, debt was used for the first time and only used for the second time. Debt is a debt and credit card debt are the same. In the different markets, the first time the debt was used, the second time it was used. In the first time, the debt was a debt and the second time the debt it used. In fact debt is a debt. The market is very different when it comes to the debt and credit cards. The market of debt is very different. In the debt market, debt is used for the debt.
## Hire An Online Math Tutor Chat
But the debt market in the first place is not a debt market. Why is there no market for these types of debt? In the markets for these types, an intermediate market is used. The debt market is a market for debt. The market that is in the debt market is the market in debt. This market has no market for debt: it is a market in debt, and it is not a market for debts. Are there any consequences for the markets for other types of debt that are used for other types? No. If there is no market for other kinds of debt, then the markets for mortgages are not different. There is a market where the market for these kinds of debt is different. The markets in the market are not different for mortgages. It is not a different market for the credit cards. This is not a difference in the market between the credit card and the financial instruments. I think that in the market of credit cards the market for other types is a different market. The marketplace for other kinds is not different. It is not a new market. It is a market that has no market. For example, there is a market of debt that is not a credit card debt market. The market has no credit card debt that is a credit card. You might say you have a market for other kind of debt that has no credit cards. That is a market with no credit cards that is not debt. If you have a trading market for other type of debt, you have a different market to deal with different types of debt.
## How Much To Pay Someone To Take An Online Class
With credit card debts, you have no credit card. For example, the market in the credit card market has no debt that is very similar in terms of debt. | 1,778 | 8,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.965222 |
https://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=key | 1,624,560,638,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00225.warc.gz | 634,550,733 | 3,855 | Partner with ConvertIt.com
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Conversion Result: ```kilogram = 1 kilogram (mass) ``` Related Measurements: Try converting from "key" to arroba (Mexican arroba), assay ton, Babylonian shekel, catty (Chinese catty), cotton bale (US), crith, denarius (Roman denarius), dram (avoirdupois dram), earth mass, funt (Russian funt), grain (avoirdupois grain), hyl, long quarter, Mexican libra, oz ap (apothecary ounce), quintal, Roman talent, short hundredweight (avoirdupois short hundredweight), talent (Greek talent), uncia (Roman uncia), or any combination of units which equate to "mass" and represent mass. Sample Conversions: key = 3.06 as (Roman as), .02345343 bag, .13778891 bowling ball, .0029395 cotton bale Egypt, 257.21 denarius (Roman denarius), 238.1 dinar (Arabian dinar), .01 doppelzentner, 564.38 dram (avoirdupois dram), 257.21 dram troy (troy dram), 15,432.36 grain (avoirdupois grain), 1,397.86 Greek obolos, 70.92 Israeli shekel mass, 1 kg (kilogram), 308,647.17 mite (English mite), 35.27 oz (ounce), 32.15 oz troy (troy ounce), 643.01 pennyweight troy (troy pennyweight), 2.68 pound ap (apothecary pound), 771.62 scruple (apothecary scruple), .01968413 UK quintal2 (British quintal).
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https://web2.0calc.com/questions/i-have-a-score-of-37-100-on-an-exam | 1,585,961,590,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370518767.60/warc/CC-MAIN-20200403220847-20200404010847-00111.warc.gz | 767,924,495 | 6,005 | +0
# I have a score of 37/100 on an exam.
0
37
1
The average (Mean) score on this exam was 59.82 and the standard deviation was 20.237. What would my new score be if the exam was curved to a new average (Mean) of 80 and a new standard deviation of 10?
Mar 9, 2020
#1
+29249
+2
As follows:
Mar 9, 2020 | 106 | 307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | longest | en | 0.958547 |
http://www.mcqlearn.com/math/g6/finding-coordinates-mcqs.php | 1,529,727,958,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00156.warc.gz | 453,221,270 | 6,970 | Practice finding coordinates MCQs in math quiz for online learning test. Functions and graphs quiz questions has multiple choice questions (MCQ), finding coordinates test to practice as in ordered pair (-2, -3), abscissa is. Answer key help with choices as 3, −2, 2 and −3 problem solving for competitive exam, viva prep, interview questions worksheets. Free math revision notes to practice finding coordinates quiz with MCQs to find questions answers based online learning tests.
MCQ. In the ordered pair (-2, -3), the abscissa is
1. 3
2. −2
3. 2
4. −3
B
MCQ. In the ordered pair (-3, -4), the ordinate or y-ordinate is
1. −4
2. −7
3. −3
4. 3
A
MCQ. The point P in Cartesian plane is located by an ordered pair called
1. (c, b, a)
2. (a, b, c)
3. (a, b)
4. (b, a)
C
MCQ. In an ordered pair of Cartesian plane (a, b), the 'b' is known as
1. x-coordinate
2. y-coordinate
3. abscissa
4. none of the above
B
MCQ. In an ordered pair of Cartesian plane (a, b), the 'a' is known as
1. y-coordinate
2. abscissa
3. x-coordinate
4. both b and c
D | 332 | 1,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2018-26 | latest | en | 0.870027 |
https://forum.allaboutcircuits.com/threads/combining-two-8-bit-reg.148061/ | 1,548,042,068,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583755653.69/warc/CC-MAIN-20190121025613-20190121051613-00041.warc.gz | 507,840,650 | 17,985 | # combining two 8 bit reg
Discussion in 'Programmer's Corner' started by karthi keyan 9, Apr 23, 2018.
1. ### karthi keyan 9 Thread Starter Member
Apr 17, 2018
30
0
I want to calculate the value of forumal
for that i need to take value from two 8bit reg in that of 16bit i want the data of 12MSB. 12MSB i placed in the formula to get the data
My code:
/
#include<pic.h>
#define _XTAL_FREQ 11059200
__CONFIG (0xFF02);
int i,x,y,r;
char data1,data2;
data1=0x07; // 8-bit data from reg 1
data2=0x56; // 8-bit data from reg 2
uint16_t x,y,d3,d4,z;
d3=(data1<<8)|data2; // Combining reg1 and reg2 = 16bit
d4=d3>>4; //Shifting 4 bit left side
void main()
{
r=0.000000000004
x=0x3000;
y=0xA000;
for( i=0;i<255;i++); //delay
while(1)
{
// formula = (((12bit data)- 0x3000)/0xA000)*4pico
z=((d4-x)/y)r; // Forumal
for(i=0;i<255;i++); //delay
}
}
i this code i am getting multiple error i solved few of them rest all i am finding difficulty pls help to solve it
2. ### bogosort Active Member
Sep 24, 2011
223
119
Obvious problems:
• You declare x and y twice (first as int, then as uint16_t).
• You declare r as an int, but define it as a float (and you forget semicolon).
• You declare z as an int, but define it with a division and scale it by a tiny float (also missing '*' symbol in the multiply).
• You have code outside of main that's not part of a function (d3 and d4 definitions).
Note that multiplying by r may exceed the precision of your hardware. You could leave the result unscaled and post-process it outside of the program, or treat it as implicitly scaled.
karthi keyan 9 likes this.
3. ### karthi keyan 9 Thread Starter Member
Apr 17, 2018
30
0
Thank you for your reply. I corrected the some mistakes but still only one error existing in the code.
modified code:
#include<pic.h>
#include<stdio.h>
#define _XTAL_FREQ 11059200
__CONFIG (0xFF02);
int i;
float r;
uint16_t x,y,d3,d4,z;
char d1=0x07; // 8-bit data from reg 1
char d2=0x56; // 8-bit data from reg 2
void main()
{
d3=(d1<<8)|d2; // Combining reg1 and reg2 = 16bit
d4=d3>>4; //Shifting 4 bit left side
r=0.000000000004;
x=0x3000;
y=0xA000;
for( i=0;i<255;i++); //delay
while(1)
{
// formula = (((12bit data)- 0x3000)/0xA000)*4pico
z=((d4-x)/y)*r; // Formula
for(i=0;i<255;i++); //delay
}
}
Error:
1. uint16_t x,y,d3,d4,z; ( "," expected )
4. ### bogosort Active Member
Sep 24, 2011
223
119
The line as written is syntactically correct; maybe you have a typo in your actual source code?
The bigger problem is the multiplication by a tiny float r into a short integer z; you're going to find that z is always 0. Consider what happens when you multiply the largest possible unsigned short int by r:
0xFFFF * 4e-9 = 0.000262
As an integer, that equals zero.
karthi keyan 9 likes this.
5. ### karthi keyan 9 Thread Starter Member
Apr 17, 2018
30
0
Yeah your correct I want measure capacitance in pico farads. It's should be less right. Then what is the solution to calculate those things?
6. ### bogosort Active Member
Sep 24, 2011
223
119
Don't scale the result. In other words, leave z as the capacitance in picoFarads. For example, if you get z = 102, then you know that you have 102 pF of capacitance. The only thing you have to consider is if the range of possible capacitances isn't too large for your 16-bit arithmetic.
7. ### Ian Rogers Active Member
Dec 12, 2012
453
120
I can't see your include for stdint... uint16_t will be defined in a library.
Short cut
typedef unsigned int uint16_t ;
karthi keyan 9 likes this.
8. ### karthi keyan 9 Thread Starter Member
Apr 17, 2018
30
0
Yeah its working..
Thank you | 1,164 | 3,610 | {"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} | 2.703125 | 3 | CC-MAIN-2019-04 | latest | en | 0.766158 |
http://www.indiastudychannel.com/projects/1558-Melting-point-ice-boiling-point-water.aspx | 1,438,644,041,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042990123.20/warc/CC-MAIN-20150728002310-00012-ip-10-236-191-2.ec2.internal.warc.gz | 508,398,630 | 13,313 | # Melting point of ice and boiling point of water
Posted Date: Total Responses: 0 Posted By: C.N.Anantharaman Member Level: Gold Points/Cash: 6
Objective:-
(1)To determine the melting point of ice.
Materials required:-
Ice cubes,blotting paper,250 ml beakers,wire gauze,tripod stand,burner,thermometer,stirer,clamp stand,etc.
Theory:-
Ice is the solid state of water.For ice, the melting point is 0*C.
Melting point: The constant temperature at which a solid changes into a liquid is called melting point of the solid.The melting point of a solid is an indication of the strength of the force of attraction between its particles.
During melting,Seperation of molecules incrases by large amount.Whole of the supplied heat energy is consumed in increasing potential energy of the molecules.The kinetic energy does not increase further.The temperature does not increase so long as the melting continues.
The quantity of heat required to completely change 1 kg of ice into water without any change in temperature is known as the latent heat of fusion(melting) of ice.This amount of heat is 335-kJ
Procedure:-
Take some ice cubes or ice pieces.Dry them with filter paper and quickly put in a beaker.Place the beaker over a wire gauze kept over a tripod stand.Suspend a thermometer with the help of clamp stand so that its bulb remains in the middle of the ice cubes.The cubes are moderately heated by a gas burner and stirred continously to keep a uniform temperature throughout.Note the temperature(t2) of water when oce is completely melted.Record your observations in tabular form.Repeat the experiment three times.
Observations:-
Observation table
S.no Starts melting completely melted
(t1) (t2)
(1) 0*C 0*C
(2) 0*C 0*C
(3) 0*C 0*C
Mean value of t1+t2/2 = 0+0/2= 0/2=0
Melting point of ice is 0*C.
______
Inference:-
Melting point of ice is 0*C.
______
Precautions:-
(1)The bulb of the thermometer should be kept sorrounded with ice cubes.
(2)Ice should be stirred regularly to keep a uniform temperature throughout.
(3)Note the temperature by keeping your eyes in line with the level of mercury.
Objective:-
(2)To determine the boiling point of water.
Materials required:-
Distilled water,hard test tube(boiling testtube),rubber cork with two bores,delivery tube,iron stand with clamp,pieces of pumic stones,250 ml beaker,thermometer
Theory:-
When a liquid(such as water) is heated,its molecules gain energy and their kineti energy increases.On continuous heating a particular temperature is reached when these molecules leave the surface )in the form of vapour) to produce a pressure above the liquid equal to atmospheric pressure.At this stage,the temperature remains stationary(constant) even on heating fur5ther.The stationar6y temperature at which the pressure of the vapours leaving the surface of the liquid equals the atmospheric pressure is called the boiling point of a liquid.Due to the ionic bond,ionic compounds have higher boiling point than covalent compound.
(1)Most of the covalent compounds have low boiling points.
(2)Ionic compounds have high boiling points.
(3)Addition of salt in water rises its boiling point.
Procedure:-
(1)Take about 70 to 100 ml of fresh (distilled) water(not hard water,which contains extra salts) in a hard test tube and add 2-3 small pieces of pumice stone.
(2)Fix a cork with two bores in the mouth of the boiling tube and clamp it with the stand.
(3)Introduce a thermometer in one bore of the rubber cork of the boiling tube and a delivery tube in the second bore.Keep the bulb of the thermometer above about 3-5 cm from the surface of the water.
(4)Place the beaker below the second end of the delivery tube to collect the condensed water.
(5)Heat the tube gently, preferably by rotating the flame. Note the temperature when boiling of water starts.Continue to heat the water remains boiling.Note the constant temperature.
(6)Record your observations in tabular form.
(7)repeat the experiment three times.
Observations:-
(1)Record your observations in a table as given below:-
S.no of readings Starts boliling Continues to boil
t1 t2
(1) 100*C 100*C
(2) 100*C 100*C
(3) 100*C 100*C
Mean value of t1+t2/2=100+100/2 = 200/2 =100*C
i.e. The boiling point of water is 100*C
Inference:-
(1)The boiling point of water is 100*C
Precaution:-
(1)The bulb of the thermometer should be kept about 4-5 cm above the surface of the liquid(water).
(2)Pieces of pumice stone should be added to water before heating to avoid bumping.
(3)Heating of water should be done by rotating the flame.
(4)Note temperature by keeping your eyes in line with the level of mercury.
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More... | 1,239 | 5,008 | {"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-2015-32 | longest | en | 0.88964 |
https://mathoverflow.net/questions/256903/given-two-eigenfunctions-and-eigenvalues-determine-existence-of-eigenvalues-betw | 1,719,094,247,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00382.warc.gz | 349,263,538 | 22,139 | # Given two eigenfunctions and eigenvalues determine existence of eigenvalues between them
Suppose we have two eigenfunctions $f_n(x,y)$ and $f_m(x,y)$ and corresponding eigenvalues $\lambda_n<\lambda_m$ of a differential operator $L$. How can we determine whether there exists another eigenvalue $\lambda_i$ between these given, i.e. $\lambda_n<\lambda_i<\lambda_m$, if the problem is 2D and higher?
I know that one could just count zeros of the eigenfunctions in 1D case, but according to Courant, Hilbert "Methods of Mathematical Physics Vol. 1" $\S\mathrm{VI}.6$, this is not generally true for higher dimensions:
In the case of eigenvalue problems of partial differential equations, arbitrary values of $n$ may exist for which the nodes of the eigenfunctions $u_n$ subdivide the entire fundamental domain into only two subdomains. | 197 | 838 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.775856 |
https://uta-lab.com/interracial/anna-jimskaia-interracial.php | 1,675,935,838,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501555.34/warc/CC-MAIN-20230209081052-20230209111052-00364.warc.gz | 612,022,444 | 6,616 | # Website with math book answers
Website with math book answers is a mathematical instrument that assists to solve math equations. We can solving math problem.
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Interval notation is a method of representing a set of numbers using Intervals. Interval notation solver is a online tool that helps you to solve the problems in Interval Notation. It shows the work by using the properties of Intervals, so you can understand the steps involved in solving the problem. You can also use this tool to check your answers.
One way to solve a problem is by using the process of elimination. This involves looking at all of the possible options and eliminating the ones that are not possible. For example, if you are trying to find out how many books are in a library, you would start by eliminating the options that are not possible. If there are only two books in the library, then you know that the answer is not three or four. You would continue this process until you are left with only one option. This can be a very effective way to solve problems, but it can also be time-consuming.
Web math is a website that provides a variety of resources for students who are struggling with math. The site includes a wide range of topics, from basic arithmetic to more advanced concepts like calculus. In addition, the site provides interactive tools that help students visualize and understand complex concepts. Web math also offers a forum where students can ask questions and get help from other users. The site is free to use, and it does not require registration.
How to solve mode: The mode is the value that appears most often in a set of data. To find the mode, simply order the values from smallest to largest and count how many times each value appears. The value that appears the most is the mode. For example, in the set {1, 2, 2, 3, 3, 4}, the mode is 2 because it appears twice while the other values only appear once. To find the mode of a set of data, follow these steps: 1) Order the values from smallest to largest. 2) Count how many times each value appears. 3) The value that appears the most is the mode.
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https://bitcoin.stackexchange.com/questions/58956/if-a-bitcoin-mining-nounce-is-just-32-bits-long-how-come-is-it-increasingly-diff | 1,718,367,765,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00813.warc.gz | 111,864,401 | 37,052 | # if a Bitcoin mining nounce is just 32 bits long how come is it increasingly difficult to find the winning hash? [duplicate]
I'm learning about mining and the first thing that surprised me is that the nounce part of the algorithm which is supposed to be randomly looped until you get a number smaller than the target hash .. is just 32 bits long. Can you explain why then is it so difficult to loop an unsigned int and how come is it increasingly difficult over time? Thank you.
Edit: It's been suggested this is a duplicate, this is not. The other question asked if you could run out of nounces, that's nothing to do with my question. My question is about the difficulty of Hashing the exact same 4 billion unsigned int entry loops.
• Short answer: With high probability, none of those 2^32 nonces will result in a hash smaller than the target. So then you change something else in the block header and try again. For instance, you can increment a number in the coinbase transaction, which results in a new value for the "Merkle root" hash in the block header. The smaller the target, the more times you are likely to have to do this before success. Commented Aug 31, 2017 at 13:47
• @NateEldredge It looks like you are saying that you do not really need to change the nounce since there is the possibility that none of the 4 billion different values may lead to a winning hash. What makes then a valid 'source material' to hash while attemting to discover a right answer? if you can point me to updated literature about it I will accept that as a right answer. Thanks
– Pol
Commented Aug 31, 2017 at 17:48
• No, I'm not saying that. What you need is a valid 80-byte block header whose hash is below the target value. So you have to keep changing things in the header until you find something that leads to a winning hash. The easiest thing to change is the nonce, so you fix everything else and then try all possible nonces. If you are lucky then one of them works and you win. If not, which happens most of the time, then you change something else in the header and try all nonces for the updated header. Repeat indefinitely until you win. Commented Aug 31, 2017 at 18:08
• There isn't any need for "updated" literature; the block format and hashing algorithm hasn't changed in any relevant way since the initial introduction of Bitcoin. It was always possible that, with everything else in the header fixed, none of the 2^32 possible nonces wins, and this case always had to be handled; it already happens about 36% of the time even at minimum difficulty. Commented Aug 31, 2017 at 18:09 | 605 | 2,595 | {"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 | 3 | CC-MAIN-2024-26 | latest | en | 0.954383 |
https://www.physicsforums.com/threads/proving-constant-curvature.875256/ | 1,653,409,756,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573053.67/warc/CC-MAIN-20220524142617-20220524172617-00190.warc.gz | 1,106,000,392 | 18,836 | # Proving constant curvature
• A
I'm currently on section 5.1 in Wald's book. He is trying to prove that the cosmological principle implies that space has constant curvature.
Given a spacelike hypersurface ##\Sigma_t## for some fixed time ##t##, we say that it is homogeneous if given ##p,q \in \Sigma_t##, there is an isometry, ##\phi##, of the metric ##g## such that ##\phi(p) = q##.
Now at a given point ##p \in \Sigma_t##, ##g## induces a Riemannian metric ##h## on ##\Sigma_t## simply by restricting ##g## to spacelike tangent vectors. The Riemann curvature tensor ##R_{ab}{}^{cd}## (using ##h## to raise the third index) can be viewed as a linear map from ##\mathcal{A}^2(T_p \Sigma_t)## into ##\mathcal{A}^2(T_p \Sigma_t)## (the vector space of antisymmetric ##2##-tensors defined on the tangent space to ##\Sigma_t## at ##p##). Let ##L## denote this linear map. Viewed as a linear map, ##L## is symmetric, or equivalently, self-adjoint. Thus ##T_p \Sigma_t## has an orthonormal basis of eigenvectors of ##L##. If the eigenvalues were distinct then we would be able to construct a preferred tangent vector, violating isotropy. Hence all eigenvalues are the same and ##L = KI## for some constant ##K## and where ##I## is the identity operator. Another way to write this is ##R_{ab}{}^{cd} = K\delta^c{}_{[a} \delta^d{}_{b]}## where the square brackets denote antisymmetrization (recall that the Riemann tensor is antisymmetric in its first two indices, this is why we have the antisymmetric brackets there). Lowering the last two indices gives ##R_{abcd} = K h_{c[a}{}h_{b]d}##.
Now here is the part that is bothering me. Wald says that homogeneity implies that ##K## must be constant, i.e. cannot vary from point to point of ##\Sigma_t##. I get that homogeneity is supposed to mean that everything is the same at each point, but we are trying to prove this mathematically, and we have a mathematical definition of homogeneity. I don't see how our mathematical definition of homogeneity shows that ##K## is constant from point to point.
PeterDonis
Mentor
I don't see how our mathematical definition of homogeneity shows that ##K## is constant from point to point.
An isometry is a transformation that leaves the metric invariant; that implies that it also leaves ##K## invariant, since ##K## is part of the metric. So homogeneity, which requires that there is an isometry taking any spatial point into any other, also requires that ##K## is the same at every spatial point, since if ##K## were different for any pair of points there would not be an isometry taking one to the other.
Sorry but I don't see how ##K## is part of the metric. The only relationship with ##K## and the metric that we have is ##R_{abcd} = K h_{c[a}h_{b]d} ##.
PeterDonis
Mentor
I don't see how ##K## is part of the metric.
The metric encodes all information about the spacetime geometry. ##K## is part of the spacetime geometry. So it's part of the metric. It might not appear explicitly in the expression for the metric tensor; that depends on how you choose coordinates. But it will be uniquely determined by the metric.
Quite honestly, the statement I have just made should be blindingly obvious to you. If it isn't, then you are not prepared for an "A" level thread on this subject matter.
The only relationship with ##K## and the metric that we have is ##R_{abcd} = K h_{c[a}h_{b]d}## .
Note that ##h## is the spatial metric, not the spacetime metric. The spacetime metric is ##g##.
Also, the equation you wrote down here is not the basis for Wald's assertion that homoegeneity implies that ##K## is constant. The basis for that is the mathematical definition of what an isometry is.
Last edited:
JonnyG
Couldn't we use this argument for constant curvature: Let ## {}^3 R_{abc}{}^d## denote the Riemann tensor on a hypersurface ##\Sigma_t##. Since ##R_{abc}{}^d## can be expressed entirely in terms of the spacetime metric ##g##, then ##{}^3 R_{abc}{}^d## can be expressed entirely in terms of the spatial metric ##h##. Since ##h## is position invariant, then ##{}^3R_{abc}{}^d## must be the same everywhere and hence the curvature is constant.
I was thinking about this argument for constant curvature before I started this thread, but it seems that if it were this simple then Wald would have used it. Is it incorrect?
PeterDonis
Mentor
Since ##h## is position invariant
How do you know that ##h## is position invariant?
What I mean by position invariant is that given ##p,q \in \Sigma_t##, there is a diffeomorphism ##\phi## such that ##\phi(p) = q## and for two tangent vectors ##v,w \in T_p \big(\Sigma_t\big)##, ##\phi(v,w) = g\big(d\phi_p v, d\phi_p w\big)##, where ##d\phi_p## is the differential of ##\phi## at ##p##. That is the definition of homogeneous. Since ##h## is obtained by restricting ##g## to ##\Sigma_t##, then this property holds for ##h## as well.
PeterDonis
Mentor
What I mean by position invariant is...That is the definition of homogeneous.
In other words, your definition of "homogeneous" is (a) the same as Wald's (you have basically just restated Wald's definition, that there is an isometry between any two spatial points, in different words, although your restatement seems a little sloppy), and (b) sufficient to prove that homogeneity requires constant ##h## and therefore constant curvature ##K## over a spacelike hypersurface. In other words, you have now answered your own question by showing how homogeneity mathematically implies constant ##K##.
JonnyG
Thanks for the help. | 1,440 | 5,520 | {"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-2022-21 | latest | en | 0.885998 |
https://ai.stackexchange.com/questions/27754/how-would-you-intuitively-but-rigorously-explain-what-the-vc-dimension-is | 1,721,777,058,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00713.warc.gz | 72,257,445 | 43,715 | # How would you intuitively but rigorously explain what the VC dimension is?
The VC dimension is a very important concept in computational/statistical learning theory. However, the first time you read its definition, you may not immediately understand what it really represents or means, as it involves other concepts, such as shattering, hypothesis class, learning and sets. For example, let's take a look at the definition given by Shai Shalev-Shwartz and Shai Ben-David (p. 70)
DEFINITION $$6.5$$ (VC-dimension) The VC-dimension of a hypothesis class $$\mathcal{H}$$, denoted $$\operatorname{VCdim}(\mathcal{H})$$, is the maximal size of a set $$C \subset \mathcal{X}$$ that can be shattered by $$\mathcal{H}$$. If $$\mathcal{H}$$ can shatter sets of arbitrarily large size we say that $$\mathcal{H}$$ has infinite VC-dimension.
Without knowing what a hypothesis class is, or what the specific $$C$$, $$X$$ and $$H$$ in this definition are, it's difficult to understand this definition. Even if you are familiar with what a hypothesis class is (i.e. a set of sets, i.e. our set of functions/hypotheses/models, e.g. the set of all possible neural networks with a specific topology) and you know that $$C$$ and $$X$$ are sets of input points, it should still not be clear what the VC dimesion really is or represents.
So, how would you intuitively and rigorously explain the exact definition of the VC dimension?
The VC dimension represents the complexity (or expressive power, richness, or flexibility) of your model/hypothesis class.
Of course, this is easy to memorize, but it's quite vague. So, I am not looking for vague/general answers. I am looking for answers that rigorously but intuitively describe the mathematical definition of the VC dimension. For example, you could provide an illustration that shows what the VC dimension is, and, in your example (e.g. the XOR problem cannot be solved by a set of lines), you can describe what $$H$$, $$C$$, and $$X$$ are, and how they relate to the typical concepts you will find an introductory course to machine learning, but you should not forget to describe the concept of shattering. If you have other ideas of how to illustrate this concept memnomically, feel free to provide an answer.
Trying to explain the idea of VC to some of my colleagues I've discovered quite an intuitive way of laying out the basic idea. Without going through lots of math and notation as I've done in my other answer.
Imagine a following game between two players $$\alpha$$ and $$\beta$$ :
1. First, player $$\alpha$$ plots $$d=4$$ points on a piece of paper. She may place the points however she likes.
2. Next, player $$\beta$$ marks several of the drawn points.
3. Finally, player $$\alpha$$ should draw a circle such that all the marked points are inside the circle, and all the unmarked points - outside. (Points on the boundary considered "inside".)
The player $$\alpha$$ wins if she can draw such a circle at step #3. The player $$\beta$$ wins if making such circle is impossible.
If you try to analyze this game then you'll notice that the player $$\beta$$ has a winning strategy. For any $$d=4$$ points on a plane, there is always a subset such that player $$\alpha$$ is unable to draw a required circle. ( I don't want to go into the detailed proof of the strategy - it is straightforward, but cumbersome - I've sketched it my other answer). If we now change the number of points to $$d=3$$ then the game suddenly becomes winnable by player $$\alpha$$ - for three points that are not on the same line any subset can be separated by a circle.
The largest number $$d$$ at which the game is winnable by player $$\alpha$$ is called the VC dimension of our classification set. So, in the case of 2D discs (insides of a circle) the VC dimension is 3. If one changes the rules to use rectangles instead of circles, then the maximum number of points winnable by $$\alpha$$ would be 4 - thus, the VC dimension of rectangular classification sets is 4.
Restoring our the mathematical notation, we denote $$\mathcal{X} = \mathbb{R}^2$$ our two-dimensional plane. $$C$$ is the subset of cardinality $$|C| = d$$ that the player $$\alpha$$ selects. And $$\mathcal{H}$$ is a class (a "set-of-sets") of the subsets of $$\mathcal{X}$$ that one should use as a classification boundary. Formally, the statement that the game above is winnable by $$\alpha$$ can be written as:
$$\exists\;C\subset\mathcal{X}.|C|=d \Rightarrow\forall\; A\subset C\; \exists\; B\in\mathcal{H}\;. A = B\,\cap\,A$$
The maximal $$d$$ at which this statement is true would be the VC dimension. (I actually worked backwards from noticing the alternating quantifiers $$\exists\,\forall\,\exists$$ in the VC definition - which is typical in game playing, so I worked back from the definition to make the game above.)
• I think both your answers would benefit from having images of the disks (the "models" in our context) and the points that we want to classify (some configuration of the points you already show in the other answer), although they will be longer. In any case, this answer is interesting.
– nbro
Commented May 12, 2021 at 8:09
Shattered set. First we need a concept of a shattered set. I'll work from a shattered set example in Wikipedia adjusting it to your notation.
The statement that $$\mathcal{H}$$ shatters $$C$$ means that for every subset $$A \subset C$$ there is a set $$B\in\mathcal{H}$$ such that $$B$$ "separates" $$A$$ from $$C \backslash A$$. Writing this formally:
$$\text{shatters}(\mathcal{H},C) = \forall A \subset C\; \exists B\in\mathcal{H}\;.\;A = B\,\cap A$$
As an example consider the set $$C$$ of four points on $$\mathcal{X} = \mathbb{R}^2$$:
$$C = \left\{(0,0); (0,1); (1,0); (1,1)\right\}$$
And the classification class $$\mathcal{H}$$ being all possible 2D discs on $$\mathcal{X} =\mathbb{R}^2$$. (Notice that people use the world "class" here, because we are dealing with "set-of-sets" stuff and that might get tricky). Note that a disk can be represented as the set of all points inside the circle.
Now, it turns out that this $$\mathcal{H}$$ doesn't shatter $$C$$. The counterexample would be the subset $$A$$ of "diagonal" points:
$$A = \left\{(0,0); (1,1)\right\} \subset C$$
There is no 2D disk $$B\in\mathcal{H}$$ (in the context of learning theory, an element/set of the hypothesis class is a hypothesis, which can also be viewed as a function) that satisfies $$A = B\,\cap A$$. Intuitively, this means that you cannot use a 2D disk to classify your pair of points $$A$$ from the rest of the set $$C$$.
The VC dimension of $$\mathcal{H}$$ is the maximal cardinality $$d$$ of the set $$C$$ that it can shatter. For $$d=3$$, we can provide tree points $$C'$$ for which we can easily find all discs that cover all eight possible subsets of $$C'$$:
$$C' = \left\{(0,0); (0,1); (1,0)\right\}$$
Above we've shown that for a particular set $$C$$ of $$d=4$$ points $$\text{shatters}(\mathcal{H}, C)$$ is false. But we need to prove that it is false for all 4-point sets. To prove this, we consider a convex hull of an arbitrary set of four points $$C = \left\{a;b;c;d\right\}$$. In general position, the convex hull is either a triangle or a quad:
In the case of a triangle, we choose the outermost three points as a counterexample set $$A$$. So, with this (labeling) configuration, you cannot find a disk that covers (i.e. classifies) the outermost three points correctly while excluding the point inside the triangle. In the case of a quad, we choose the pair on the longest diagonal. If any three points lay on a single line, then we choose the pair of outermost points.
This sketches a proof that no $$d=4$$ point set can be shattered by $$\mathcal{H}$$, but we've shown that there is a $$d=3$$ set $$C'$$ that can be. Concluding that $$VCdim(\mathcal{H}) = 3$$
Another example is considered on the next page (pg. 71) of the book you've referenced. It again considers 2D plane $$\mathcal{X} =\mathbb{R}^2$$ and the classification class $$\mathcal{H}$$ is all possible axis-aligned rectangles. Authors show a configuration $$C$$ of four points that can be shattered by $$\mathcal{H}$$ on the left of figure 6.1. And then provide a proof that no $$d=5$$ points can be shattered by axis-aligned rectangles. Concluding that VC dimension of their $$\mathcal{H}$$ is four.
Hope these examples help. (BTW, note that, it seems that Deep Learning has quite bad VC dimension but it still works somehow - which is rather puzzling).
• Regarding the VC dimension of neural networks, I am not really an expert in this topic, but you may be interested in my answer here. Essentially, as far as I know, the VC dimension of neural networks is often expressed as a bound rather than a specific number. Given it's often a loose bound, the actual VC dimension may be actually quite big.
– nbro
Commented May 11, 2021 at 18:25
• Moreover, the adversarial examples show that neural networks can actually be fooled. I need to think about how the VC dimension is related to this (I suspect that we are estimating the expected risk is the wrong way), but I think it's misleading to say that "VC dimension of neural networks is low, but neural networks 'work', so the VC dimension is useless", which is a conclusion I've already heard (just to clarify, I had already come across that post on CS SE).
– nbro
Commented May 11, 2021 at 18:25
This may be lacking some rigour, but this is how I have explained it in the past:
The VC Dimension is the maximum number of inputs such that for any subset of these inputs, it is possible for the model to classify the subset as true and the rest as false.
I am aware that this definition still uses the term subset, but in my experience even people who are not familiar with set theory understand the concept of a subset.
The part about this that tends to confuse most people is the notion of a subset unable to be induced by a model. This can be nicely illustrated by picking a really simply hypothesis, such as a closed interval, and showing how it cannot express non-contiguous positive inputs, or using a linear classifier and showing how it cannot express XOR.
• This answer is not really what I was looking for. I was looking for an answer that is rigorous but intuitive, i.e. an answer with an illustration of an example that is described in detail. So, it's ok to just stick to an example to explain the concept, but here you're talking about other concepts (which I personally know, but not everyone will), such as the closed-interval example, but just mentioning them. I also don't think your definition is correct. You're allowed to rearrange the points, but the model should classify them correctly for any possible labelling.
– nbro
Commented May 11, 2021 at 16:43
• So, what I was looking for was, for example, a detailed description of the XOR and how a line cannot solve it (in that example, you would describe what $H$, $C$ and $X$ are, and what it means to shatter, etc.)
– nbro
Commented May 11, 2021 at 16:45 | 2,817 | 11,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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": 85, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-30 | latest | en | 0.924603 |
https://www.clutchprep.com/chemistry/practice-problems/133381/which-reaction-below-represents-the-electron-affinity-of-be-a-be-g-be-g-e-b-be-g | 1,620,814,640,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989693.19/warc/CC-MAIN-20210512100748-20210512130748-00061.warc.gz | 734,725,128 | 32,447 | Periodic Trend: Electron Affinity Video Lessons
Concept: Periodic Trend: Electron Affinity
# Problem: Which reaction below represents the electron affinity of Be?a. Be(g) → Be+(g) + e-b. Be(g) + e- → Be-(g) c. Be+(g) → Be(g) + e-d. Be(g) + e- → Be+(g)e. Be+(g) + e- → Be(g)
###### FREE Expert Solution
We’re being asked to determine the equation representing the electron affinity of Be
Recall that electron affinity is the energy change from the addition of 1 e to a gaseous element/ion
It is represented by this chemical equation:
Atom(g) + e Ion(g), ΔE = –E.A.
87% (135 ratings)
###### Problem Details
Which reaction below represents the electron affinity of Be?
a. Be(g) → Be+(g) + e-
b. Be(g) + e→ Be-(g)
c. Be+(g) → Be(g) + e-
d. Be(g) + e→ Be+(g)
e. Be+(g) + e→ Be(g) | 263 | 789 | {"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-2021-21 | latest | en | 0.564531 |
https://www.cliffsnotes.com/study-guides/algebra/linear-algebra/real-euclidean-vector-spaces/subspaces-of-rn | 1,709,146,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474744.31/warc/CC-MAIN-20240228175828-20240228205828-00629.warc.gz | 721,441,444 | 34,437 | ## Subspaces ofRn
Consider the collection of vectors
The endpoints of all such vectors lie on the line y = 3 x in the x‐y plane. Now, choose any two vectors from V, say, u = (1, 3) and v = (‐2, ‐6). Note that the sum of u and v,
is also a vector in V, because its second component is three times the first. In fact, it can be easily shown that the sum of any two vectors in V will produce a vector that again lies in V. The set V is therefore said to be closed under addition. Next, consider a scalar multiple of u, say,
It, too, is in V. In fact, every scalar multiple of any vector in V is itself an element of V. The set V is therefore said to be closed under scalar multiplication.
Thus, the elements in V enjoy the following two properties:
The sum of any two elements in V is an element of V.
Closure under scalar multiplication:
Every scalar multiple of an element in V is an element of V.
Any subset of R n that satisfies these two properties—with the usual operations of addition and scalar multiplication—is called a subspace of Rn or a Euclidean vector space. The set V = {(x, 3 x): xR} is a Euclidean vector space, a subspace of R2.
Example 1: Is the following set a subspace of R2?
To establish that A is a subspace of R2, it must be shown that A is closed under addition and scalar multiplication. If a counterexample to even one of these properties can be found, then the set is not a subspace. In the present case, it is very easy to find such a counterexample. For instance, both u = (1, 4) and v = (2, 7) are in A, but their sum, u + v = (3, 11), is not. In order for a vector v = (v 1, v 2 to be in A, the second component (v 2) must be 1 more than three times the first component (v 1). Since 11 ≠ 3(3) + 1, (3, 11) ∉ A. Therefore, the set A is not closed under addition, so A cannot be a subspace. [You could also show that this particular set is not a subspace of R 2 by exhibiting a counterexample to closure under scalar multiplication. For example, although u = (1, 4) is in A, the scalar multiple 2 u = (2, 8) is not.]
Example 2: Is the following set a subspace of R3?
In order for a sub set of R 3 to be a sub space of R 3, both closure properties (1) and (2) must be satisfied. However, note that while u = (1, 1, 1) and v = (2, 4, 8) are both in B, their sum, (3, 5, 9), clearly is not. Since B is not closed under addition, B is not a subspace of R 3.
Example 3: Is the following set a subspace of R4?
For a 4‐vector to be in C, exactly two conditions must be satisfied: Namely, its second component must be zero, and its fourth component must be −5 times the first. Choosing particular vectors in C and checking closure under addition and scalar multiplication would lead you to conjecture that C is indeed a subspace. However, no matter how many specific examples you provide showing that the closure properties are satisfied, the fact that C is a subspace is established only when a general proof is given. So let u = (u1, 0, u3, −5 u1) and v = (v1, 0, v3, −5v1) be arbitrary vectors in C. Then their sum,
satisfies the conditions for membership in C, verifying closure under addition. Finally, if k is a scalar, then
is in C, establishing closure under scalar multiplication. This proves that C is a subspace of R4.
Example 4: Show that if V is a subspace of R n, then V must contain the zero vector.
First, choose any vector v in V. Since V is a subspace, it must be closed under scalar multiplication. By selecting 0 as the scalar, the vector 0 v, which equals 0, must be in V. [Another method proceeds like this: If v is in V, then the scalar multiple (−1) v = − v must also be in V. But then the sum of these two vectors, v + (− v) = 0, mnust be in V, since V is closed under addition.]
This result can provide a quick way to conclude that a particular set is not a Euclidean space. If the set does not contain the zero vector, then it cannot be a subspace. For example, the set A in Example 1 above could not be a subspace of R 2 because it does not contain the vector 0 = (0, 0). It is important to realize that containing the zero vector is a necessary condition for a set to be a Euclidean space, not a sufficient one. That is, just because a set contains the zero vector does not guarantee that it is a Euclidean space (for example, consider the set B in Example 2); the guarantee is that if the set does not contain 0, then it is not a Euclidean vector space.
As always, the distinction between vectors and points can be blurred, and sets consisting of points in Rn can be considered for classification as subspaces.
Example 5: Is the following set a subspace of R2
As illustrated in Figure , this set consists of all points in the first quadrant, including the points (x, 0) on the x axis with x > 0 and the points (0, y) on the y axis with y > 0:
Figure 1
The set D is closed under addition since the sum of nonnegative numbers is nonnegative. That is, if (x1, y1) and (x2, y2) are in D, then x1, x2, y1, and y2 are all greater than or equal to 0, so both sums x1 + x2 and y1 + y2 are greater than or equal to 0. This implies that
However, D is not closed under scalar multiplication. If x and y are both positive, then ( x, y) is in D, but for any negative scalar k,
since kx < 0 (and ky < 0). Therefore, D is not a subspace of R 2.
Example 6: Is the following set a subspace of R 2?
As illustrated in Figure , this set consists of all points in the first and third quadrants, including the axes:
Figure 2
The set E is closed under scalar multiplication, since if k is any scalar, then k(x, y) = ( kx, ky) is in E. The proof of this last statement follows immediately from the condition for membership in E. A point is in E if the product of its two coordinates is nonnegative. Since k 2 > 0 for any real k
However, although E is closed under scalar multiplication, it is not closed under addition. For example, although u = (4, 1) and v = (−2, −6) are both in E, their sum, (2, −5), is not. Thus, E is not a subspace of R 2.
Example 7: Does the plane P given by the equation 2 x + y − 3 z = 0 form a subspace of R 3?
One way to characterize P is to solve the given equation for y,
and write
If p 1 = ( x 1, 3 z 1 − 2 x 1, z 1) and p 2 = ( x 2, 3 z 2 − 2 x 2, z 2) are points in P, then their sum,
is also in P, so P is closed under addition. Furthermore, if p = ( x, 3 z − 2 x, z) is a point in P, then any scalar multiple,
is also in P, so P is also closed under scalar multiplication. Therefore, P does indeed form a subspace of R 3. Note that P contains the origin. By contrast, the plane 2 x + y − 3 z = 1, although parallel to P, is not a subspace of R 3 because it does not contain (0, 0, 0); recall Example 4 above. In fact, a plane in R 3 is a subspace of R 3 if and only if it contains the origin.
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REMOVED | 1,892 | 6,805 | {"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} | 4.75 | 5 | CC-MAIN-2024-10 | latest | en | 0.945088 |
https://ncatlab.org/nlab/show/nonunital%20ring | 1,638,260,329,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2021-49/segments/1637964358966.62/warc/CC-MAIN-20211130080511-20211130110511-00445.warc.gz | 470,417,595 | 12,406 | # nLab nonunital ring
Contents
### Context
#### Algebra
higher algebra
universal algebra
# Contents
## Idea
The concept of nonunital ring is like that of ring but without the requirement of the existence of an identity element (“unit” element).
Nonunital rings with homomorphisms between them form the category Rng.
Historically, this was in fact the original meaning of “ring”, and while mostly “ring” has come to mean by default the version with identity element, nonunital rings still play a role (see e.g. the review in Anderson 06) and in some areas of mathematics “nonunital ring” is still the default meaning of “ring”. In particular, non-unital rings may naturally be identified with the augmentation ideals of $\mathbb{Z}$-augmented unital rings, see the discussion below.
###### Remark on terminology
The term “non-unital ring” may be regarded as an example of the “red herring principle”, as a non-unital ring is not in general a ring in the modern sense of the word.
In Bourbaki 6, chapter 1 the term pseudo-ring is used, but that convention has not become established.
Another terminology that has been suggested for “nonunital ring”, and which is in use in part of the literature (e.g. Anderson 06) is “rng”, where dropping the “i” in “ring” is meant to be alluding to the absence of identity elements. This terminology appears in print first in (Jacobson), where it is attributed to Louis Rowen.
Similarly there is, for whatever it’s worth, the suggestion that a ring without negatives, hence a semiring, should be called a rig (although here one may make a technical distinction about the additive identity).
## Definitions
### For rings
Specifically:
###### Definition
A nonunital ring or rng is a set $R$ with operations of addition and multiplication, such that:
• $R$ is a semigroup under multiplication;
• $R$ is an abelian group under addition;
More sophisticatedly, we can say that, just as a ring is a monoid object in Ab, so
###### Definition
A nonunital ring or rng is a semigroup object in Ab.
###### Remark
A non-unital ring may well contain an element that behaves as the identity element, i.e. it may be in the image of the forgetful functor from unital rings to nonunital rings. But if so, then this element is still not part of the defining data and in particular a homomorphism of non-unital rings need not to preserve whatever identity elements may happen to be present.
### For $\mathbb{E}_k$-algebras
nonunital Ek-algebras are discussed in (Lurie, section 5.2.3).
## Properties
### Unitization
###### Definition
Given a non-unital commutative ring $A$, then its unitisation is the commutative ring $F(A)$ obtained by freely adjoining an identity element, hence the ring whose underlying abelian group is the direct sum $\mathbb{Z} \oplus A$ of $A$ with the integers, and whose product operation is defined by
$(n_1, a_1) (n_2, a_2) \coloneqq (n_1 n_2, n_1 a_2 + n_2 a_1 + a_1 a_2) \,,$
where for $n \in \mathbb{Z}$ and $a \in A$ we set $n a \coloneqq \underset{n\;summands}{\underbrace{a + a + \cdots + a}}$.
###### Remark
In the unitization $\mathbb{Z} \oplus A$ we have $(n,0) + (0,a) = (n,a)$ and hence it makes sense to abbreviate $(n,a)$ simply to $n+a$. The product in the unitisation is then fixed by the defining requirement that $1 \cdot a = a$ and by the distributivity law.
###### Remark
One can also think of the unitisation as the quotient of the polynomial ring “$A[x]$” ($A$ “with a generic element adjoined”, i.e. a term algebra) quotiented by the relations $ax - a = 0, xa - a = 0$, so that this $x$ must be a right and left identity for multiplication $a$. Cosets must be represented by expressions of the form $a + nx$; this provides an obvious isomorphism to the above definition, and motivates the multiplication for $\mathbb{Z} \oplus A$ above.
###### Remark
Since $A$ embeds into its unitisation $F(A)$, every rng lives as an ideal in some ring. We can consider $A$-linear actions $A \curvearrowright M$ on abelian groups $M$. Since the endomorphism rings of abelian groups are always unital, the universal property of the unitisation-forgetful adjunction (see below) ensures that there is a unique extension of the action map $A \to \operatorname{End}_{\mathbf{Ab}}(M)$ along $A \hookrightarrow \operatorname{for} \circ F(A)$ to an action map $F(A) \to \operatorname{End}_{\mathbf{Ab}}(M)$. This induces an equivalence $A\text{-}\mathbf{Mod} \simeq F(A)\text{-}\mathbf{Mod},$ so that one may as well study $F(A)$ if one wanted to study $A$ through its module category.
###### Remark
Similar unitisation prescriptions work for non-commutative rings and for nonunital algebras over a fixed base ring, see also at
###### Proposition
Unitisation in def. extends to a functor from $CRng$ to CRing which is left adjoint to the forgetful functor from commutative rings to non-unital commutative rings.
$F \colon CRng \leftrightarrow CRing \colon U \,.$
###### Proof
This is because the definition of any ring homomorphism out of $F(A)= (\mathbb{Z} \oplus A, \cdot)$ is uniquely fixed on the $\mathbb{Z}$-summand.
### Nonunital rings as slices of rings
###### Definition
Write $CRing_{/\mathbb{Z}}$ for the slice category of CRing over the ring of integers (augmented commutative rings). Write
$CRing_{/\mathbb{Z}} \longrightarrow CRng$
for the functor to commutative non-unital rings which sends any $(R \stackrel{\phi}{\to} \mathbb{Z})$ to its augmentation ideal, hence to the kernel of $\phi$
$(R \stackrel{\phi}{\to} \mathbb{Z}) \mapsto ker(\phi) \,.$
###### Proposition
The augmentation ideal-functor in def. is an equivalence of categories whose inverse is given by unitisation, def. , remembering the projection $(\mathbb{Z} \oplus A) \to \mathbb{Z}$:
$CRng \simeq CRing_{/\mathbb{Z}} \,.$
###### Proof
That the functor is fully faithful is to observe that for a ring $R \stackrel{\phi}{\to} \mathbb{Z}$ the fiber $R_{n}$ over $n \in \mathbb{Z}$ is a torsor over the additive group underlying the augmentation ideal $A = R_0 = ker(\phi)$, and moreover it is a pointed torsor, the point being $n$ itself, hence is canonically equivalent to the augmentation ideal $A$, the equivalence being addition by $n$ in $R$. Hence any homomrphism of rings with identity over $\mathbb{Z}$
$\array{ R_1 && \stackrel{f}{\longrightarrow} && R_2 \\ & {}_{\mathllap{\phi_1}}\searrow && \swarrow_{\mathrlap{\phi_2}} \\ && \mathbb{Z} }$
is uniquely fixed by its restriction to the augmentation ideal $ker(\phi_1)$, whose image, moreover, has to be in the augmentation ideal $ker(\phi_2)$.
The identification of non-unital algebras as augmentation ideals of augmented unital algebras is used for instance in (Fresse 06).
###### Remark
In terms of arithmetic geometry, the formally dual statement of prop. is that arithmetic geometry induced by non-unital rings is equivalently ordinary arithmetic geometry under Spec(Z).
###### Remark
The generalization of prop. to nonunital Ek-algebras is (Lurie, prop. 5.2.3.15).
## References
### Nonunital ring theory
The terminology “rng” originates in
• Nathan Jacobson Basic Algebra.
A survey of commutative rng theory is in
• D. Anderson, Commutative rngs, in J. Brewer et al. (eds.) Multiplicative ideal theory in Commutative Algebra, 2006
Discussion of module theory over rngs (with close relation to torsion modules) is in
Discussion of non-unital rings as augmentation ideals of augmented unital rings includes
A definition of algebraic K-theory for nonunital rings is due to
• Daniel Quillen, $K_0$ for nonunital rings and Morita invariance, J. Reine Angew. Math., 472:197-217, 1996.
with further developments (in KK-theory) including
• Snigdhayan Mahanta, Higher nonunital Quillen K’-theory, KK-dualities and applications to topological T-dualities, J. Geom. Phys., 61 (5), 875-889, 2011. (pdf)
Discussion in the context of higher algebra (nonunital Ek-algebras) is in
Last revised on May 25, 2021 at 11:24:33. See the history of this page for a list of all contributions to it. | 2,177 | 8,040 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 61, "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-2021-49 | latest | en | 0.928815 |
http://www.topperlearning.com/forums/home-work-help-19/factorise-p3-125q3-1-15pq-mathematics-polynomials-58878/reply | 1,495,737,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608120.92/warc/CC-MAIN-20170525180025-20170525200025-00299.warc.gz | 673,387,775 | 36,963 | Question
Fri September 07, 2012 By: Pranav Gupta
# Factorise p3-125q3 +1+15pq
Sat September 08, 2012
p3 - 125q3 + 1 + 15pq
= (p)3 + (-5q)3 + 13 -3 × p × (-5q) × 1
= [p + (-5q) + 1] [p2 + (-5q)2 + 12 - p × (-5q) - (-5q) × 1 - 1 × p]
[It is because of a3 + b3 + c3 -3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)]
= (p -5q + 1) (p2 + 25q2 + 1 + 5pq + 5q - p)
Related Questions
Wed May 24, 2017
Wed May 24, 2017 | 229 | 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} | 3.1875 | 3 | CC-MAIN-2017-22 | latest | en | 0.731584 |
https://gmatcompass.com/category/gmat-critical-reasoning/ | 1,721,409,547,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00025.warc.gz | 243,875,764 | 14,305 | ## GMAT Verbal: Think Before You Write
The GMAT Verbal can be tough to crack. While GMAT Quantitative percentiles have been steadily increasing over the years (compare old GMAT Quant Percentiles from 2007 to this current Quant percentile chart), Verbal scores have mostly stagnated, with, for example, a 40 scaled score from 2007 corresponding to same percentile now (again, compare old GMAT…
## GMAT Critical Reasoning vs Data Sufficiency: Two Sides of the Same Coin
Traditional GMAT wisdom suggests that the skills needed for success on the Verbal section overlap little, if at all, with one’s quantitative abilities. Given our educational system and the very fact that the GMAT has separate Quantitative and Verbal sections, such a distinction seems uncontroversial and downright obvious. But, as we all know, the structure…
## Using the LSAT for GMAT Practice: User Beware
As an independent tutor, I pride myself in taking unorthodox approaches to meet my clients’ needs. Sometimes, this requires creativity in how I teach a concept, how I structure a lesson, or how I assign homework to my students. A good GMAT tutor will target your specific needs, and if that requires breaking from convention, then…
## No “Almosts” on GMAT Verbal
So you’ve been studying a couple months for the GMAT. You’ve seen an increase in your Verbal score, but now you’ve suddenly plateaued. When you do a set of 20 Critical Reasoning or Reading Comprehension questions, you consistently get 15 right, but you’ve been stuck at this rate for a couple weeks. But, there’s good…
## GMAT Verbal: Can You Really Improve Your Critical Reasoning?
Conventional GMAT wisdom suggests that you should spend the majority of your time preparing for the Quantitative section. The reasoning behind this claim is largely valid: The Quantitative section tests mathematical reasoning in a highly nuanced way, and before you can even begin to learn and recognize these nuances, you need to brush up on…
## Causality on the GMAT
One of the most common types of arguments you’ll see on the GMAT will be cause-and-effect. A cause-and-effect argument can best be thought of as one arguing that a certain fact or phenomenon directly brings about another one. One of the pitfalls of any causal argument is that the seemingly apparent causal connection might not… | 505 | 2,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-30 | latest | en | 0.898623 |
https://geo.libretexts.org/Bookshelves/Meteorology_and_Climate_Science/Book%3A_Practical_Meteorology_(Stull)/12%3A_Fronts_and_Airmasses | 1,669,488,103,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00534.warc.gz | 310,042,103 | 28,620 | # 12: Fronts and Airmasses
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
A high-pressure center, or high (H), often contains an airmass of well-defined characteristics, such as cold temperatures and low humidity. When different airmasses finally move and interact, their mutual border is called a front, named by analogy to the battle fronts of World War I.
Fronts are usually associated with low-pressure centers, or lows (L, covered in the next chapter). Two fronts per low are most common, although zero to four are also observed. In the Northern Hemisphere, these fronts often rotate counterclockwise around the low center like the spokes of a wheel (Fig. 12.1), while the low moves and evolves. Fronts are often the foci of clouds, low pressure, and precipitation.
In this chapter you will learn the characteristics of anticyclones (highs). You will see how anticyclones are favored locations for airmass formation. Covered next are fronts in the bottom, middle, and top of the troposphere. Factors that cause fronts to form and strengthen are presented. This chapter ends with a special type of front called a dry line.
This page titled 12: Fronts and Airmasses is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Roland Stull via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 688 | 2,390 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.688484 |
https://www.tutorialcup.com/zh-CN/%E8%AE%BF%E9%97%AE/%E5%8A%A8%E6%80%81%E7%BC%96%E7%A8%8B/%E6%A1%A5%E6%A2%81%E5%92%8C%E7%81%AB%E7%82%AC%E9%97%AE%E9%A2%98%E7%A8%8B%E5%BA%8F | 1,624,326,267,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00417.warc.gz | 979,193,768 | 29,355 | # 桥梁和火炬问题程序
## 问题陈述
“桥梁和火炬”问题表明，您需要一定的时间才能跨过桥梁。 由于时间到了，它包含正整数。 随着时间的流逝，我们得到了一座桥梁，一个人需要跨越。 这座桥一次只能允许两个人越过。 他们穿越时手持火炬。 而且因为只有一个火炬。 来自另一边的一个人应返回并把火炬带回到起点。 当两个人越过桥时，他们可以以较慢的人的速度越过。 找到所有人都可以过桥的最短总时间。
## 使用案列
`timeToCross = {1, 2, 3}`
`6`
## 代码
### 桥梁和火炬问题的C ++代码
```#include <bits/stdc++.h>
using namespace std;
int solveBridgeAndTorchProblem(int mask, bool direction, vector<int> &timeToCross, vector<vector<int>> &dp)
{
int n = timeToCross.size();
// if nobody is left to transfer
return 0;
int res = 0;
// transfer people from left to right (from destination to start)
if (direction == 1) {
int minRow = INT_MAX, person;
for (int i = 0; i < n; ++i) {
// choose the person with smallest time to cross bridge
if (transferredMask & (1 << i)) {
if (minRow > timeToCross[i]) {
person = i;
minRow = timeToCross[i];
}
}
}
// now send this person to let and solve for smaller problem
res = timeToCross[person] + solveBridgeAndTorchProblem(mask|(1 << person),direction^1, timeToCross, dp);
}
else {
// __builtin_popcount() counts the bits in mask
for (int i=0;i<n;++i) {
// since there is only a single person on starting side, return him
res = timeToCross[i];
break;
}
}
}
else {
// find the minimum time by solving for each pair
res = INT_MAX;
for(int i=0;i<n;++i) {
// if ith person is not on right side, then do nothing
continue;
// else find another person and try to cross the bridge
for(int j=i+1;j<n;++j) {
// time to cross the bridge for current pair
int val = max(timeToCross[i], timeToCross[j]);
// solve for smaller subproblems
res = min(res, val);
}
}
}
}
}
}
int main()
{
int n;cin>>n;
vector<int> timeToCross(n);
for(int i=0;i<n;i++)cin>>timeToCross[i];
vector<vector<int>> dp(1<<20, vector<int>(2,-1));
cout << solveBridgeAndTorchProblem(mask, 0, timeToCross, dp);
return 0;
}
```
```5
25 6 5 8 4```
`54`
### 桥梁和火炬问题的Java代码
```import java.util.*;
class Main{
static int countBits(int n){
int nn = n;
int cnt = 0;
while(n>0){
if((n&1) == 1)
cnt++;
n = n/2;
}
n = nn;
return cnt;
}
static int solveBridgeAndTorchProblem(int mask, int direction, int timeToCross[], int dp[][])
{
int n = timeToCross.length;
// if nobody is left to transfer
return 0;
int res = 0;
// transfer people from left to right (from destination to start)
if(direction == 1) {
int minRow = Integer.MAX_VALUE, person=0;
for(int i = 0; i < n; ++i) {
// choose the person with smallest time to cross bridge
if((transferredMask & (1 << i)) > 0) {
if (minRow > timeToCross[i]) {
person = i;
minRow = timeToCross[i];
}
}
}
// now send this person to let and solve for smaller problem
res = timeToCross[person] + solveBridgeAndTorchProblem(mask|(1 << person),direction^1, timeToCross, dp);
}
else {
// countBits() counts the bits in mask
for (int i=0;i<n;++i) {
// since there is only a single person on starting side, return him
res = timeToCross[i];
break;
}
}
}
else {
// find the minimum time by solving for each pair
res = Integer.MAX_VALUE;
for(int i=0;i<n;++i) {
// if ith person is not on right side, then do nothing
continue;
// else find another person and try to cross the bridge
for(int j=i+1;j<n;++j) {
// time to cross the bridge for current pair
int val = Math.max(timeToCross[i], timeToCross[j]);
// solve for smaller subproblems
res = Math.min(res, val);
}
}
}
}
}
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] timeToCross = new int[n];
for(int i=0;i<n;i++)
timeToCross[i] = sc.nextInt();
int dp[][] = new int[1<<20][2];
for(int i=0;i<(1<<20);i++){
dp[i][0] = -1;
dp[i][1] = -1;
}
}
}
```
```5
25 6 5 8 4```
`54` | 1,212 | 3,583 | {"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-2021-25 | latest | en | 0.296547 |
http://cboard.cprogramming.com/c-programming/161045-binary-decimal-c-code.html | 1,475,021,605,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661289.41/warc/CC-MAIN-20160924173741-00070-ip-10-143-35-109.ec2.internal.warc.gz | 46,943,205 | 11,706 | # Thread: Binary to decimal c code
1. ## Binary to decimal c code
Hello experts i need a binary to decimal code .
i am interfacing an ADC toa controller to display the the corresponding voltage .
i googled got a lot of but i am facing some confusion please any body provide a sample with logic.
soryy i am not good in c programming
2. This seems like a highly unnecessary code but since you asked, to convert to decimal take the last digit and multiply by 2^0 ,the next digit by 2^1 etc until you reach the MSB then sum up the results. that's the logic. figure out a code.
3. Originally Posted by africanwizz
This seems like a highly unnecessary code but since you asked, to convert to decimal take the last digit and multiply by 2^0 ,the next digit by 2^1 etc until you reach the MSB then sum up the results. that's the logic. figure out a code.
I think that's the reverse of what thannara123 (probably) wants.
To convert a binary number (n) and display it in base10 (decimal) you want to repeatedly get the remainder of n divided by 10 (this is the right-most digit) and print it, divide n by 10, and continue until n is 0. It's the same principle as africanwizz suggested, of course. The thing to look out for is that if you implement exactly as I've described the digits will be printed in reverse.
The maximum number of digits in a (decimal) natural number is floor(log10(n)+1) so you can probably pre-allocate a destination array using malloc and that formula to to work out how much room you'll need, OR if you know the maximum that n can be you can just have an array with floor(log10(n)+1) + 1 elements in it. Store the digits in the array using the above method, terminate the array with \0 (to make it a string) and print it.
Another alternative is sprintf() but depending on what you're doing and how much memory you have, perhaps implementing your own "binary to decimal" function is preferable.
Edit: Yet another alternative is a recursive implementation of your "print" function.
Popular pages Recent additions | 475 | 2,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2016-40 | latest | en | 0.921063 |
http://slideplayer.com/slide/8322598/ | 1,695,470,777,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.7/warc/CC-MAIN-20230923094750-20230923124750-00483.warc.gz | 44,584,145 | 17,594 | MTH 091 Section 10.3 Introduction to Polynomials Section 10.4 Adding and Subtracting Polynomials.
Presentation on theme: "MTH 091 Section 10.3 Introduction to Polynomials Section 10.4 Adding and Subtracting Polynomials."— Presentation transcript:
MTH 091 Section 10.3 Introduction to Polynomials Section 10.4 Adding and Subtracting Polynomials
What Is A Polynomial? A polynomial is an algebraic expression with “counting number” exponents on. Parts of a polynomial: 1.Variables 2.Constants 3.Coefficients Terms are separated by addition and subtraction signs. Polynomials can be classified by type. One term = monomial Two terms = binomial Three terms = trinomial More than three terms = polynomial
More About Polynomials Polynomials can also be classified by degree. The degree is the largest exponent on any term. If the terms have more than one variable, add the exponents on each term to get the degree of that term. A polynomial written in descending order has the terms arranged from greatest exponent to least exponent.
For the polynomial -6x 6 +4x 5 +7x 3 -9x 2 -1, complete the chart below: TermCoefficientDegree 7x 3 -9 6 4
Evaluating A Polynomial Evaluating a polynomial is the same as evaluating an algebraic expression (substitute and simplify). Recall that the order of operations requires that we perform exponents before we perform multiplication.
Examples Find the degree and type of each polynomial: -6y + 4 a + 5a 2 + 3a 3 – 4a 4 7r 2 + 2r – 3r 5 Evaluate each polynomial for x = -1: x 2 + 3x – 4 -2x 3 + 3x 2 – 6
Adding Polynomials: Like Terms Revisited Recall that like terms will have the same variables, with the same exponents on those variables. When we add or subtract like terms, we add or subtract the coefficients only. DO NOT add the exponents!
Subtracting Polynomials Apply “Keep-Change-Change” 1.Keep the first polynomial. 2.Change the subtraction sign between polynomials to an addition sign. 3.Change the signs of all the terms in the second polynomial. 4.Add. | 502 | 2,003 | {"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} | 4.71875 | 5 | CC-MAIN-2023-40 | latest | en | 0.832338 |
https://www.matlabassignmentexperts.com/blog/achieving-precision-and-stability-with-matlab-assignments-on-pid-controller-design.html | 1,726,487,003,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00005.warc.gz | 820,235,174 | 15,720 | Achieving Precision and Stability with MATLAB Assignments on PID Controller Design
July 24, 2023
Joseph Dunn
USA
MATLAB
a skilled PID Controller Design expert, holding a Master's degree in Control Engineering from MIT. With over 8 years of experience, he excels in providing top-notch assistance to students at www.matlabassignmentexperts.com.
In engineering and industrial applications, the Proportional-Integral-Derivative (PID) controller is a popular control algorithm. This feedback control system calculates the difference between the intended setpoint and the actual output of the process or plant on a continuous basis. The system is then brought closer to the desired setpoint by the PID controller using proportional, integral, and derivative actions to modify the control signal.
PID controllers are widely used in many sectors, including process control, robotics, manufacturing, and aerospace. They are renowned for being straightforward, simple to use, and capable of delivering stability and effective performance in a variety of systems.
Understanding the PID Controller
The PID controller's proportional component is its most basic and easily understood component. The proportional gain (Kp), a constant factor, directly scales the error signal. The stronger the response to the error and the quicker the controller acts to reduce it, the higher the proportional gain.
High proportional gains can, however, also cause overshooting, oscillations, and system instability. On the other hand, a low proportional gain may cause a sluggish response and make it challenging to accurately track the setpoint. Finding the ideal balance is crucial when designing a PID controller.
Integral (I) Component
The integral component applies a correction to get rid of any steady-state errors in the system and accounts for the total sum of previous errors. When systematic biases or disturbances result in a constant offset between the setpoint and the process variable, the integral action is especially helpful.
How quickly the accumulated error affects the control signal depends on the integral gain (Ki). Integral windup may occur in the controller if the integral gain is set too high. When the integral term swells out of proportion, it causes the controller to overshoot or oscillate around the setpoint.
Derivative (D) Component
The derivative component accounts for the error's rate of change. It works to anticipate the error's future trend and temper the system's reaction to abrupt changes. The derivative action aids in reducing overshoot and stabilizing the system.
How fast the controller reacts to changes in the error depends on the derivative gain (Kd). A jittery and unstable response can result from the system's noise being amplified by a high derivative gain. In order to create a PID controller that is stable and effective, the derivative gain must be tuned.
Designing a PID Controller in MATLAB
A complete set of tools for designing and analyzing PID controllers is available in MATLAB. To complete MATLAB assignment on Control System Toolbox is specifically made to work with control systems, making it an effective environment for implementing PID controllers.
PID Controller Tuning
To achieve the desired system behavior, PID controller tuning entails choosing appropriate values for the proportional, integral, and derivative gains. There is no one-size-fits-all method for tuning a PID controller; instead, it depends on the characteristics of the particular system and the control goals.
PID controller tuning can be done in a number of ways, as was previously mentioned. The Ziegler-Nichols method, which involves exciting the system with a step input and then adjusting the gains in response to the system's response, is one well-liked technique. PID tuning can be automated using MATLAB's built-in pidTuner and pidtune functions, which can also be used to determine the best PID gains.
Simulink, a graphical modeling and simulation environment from MATLAB, is frequently used for the design of control systems. The P, I, and D components, as well as their connections, are represented using blocks when creating a Simulink model for a PID controller.
The proportional, integral, and derivative components of the error signal can be calculated in Simulink, and they can then be combined to create the control signal. The control loop is then closed by sending the process variable back to the controller after the control signal has been applied to the plant.
You can analyze the PID controller's behavior in various scenarios by simulating the system's response to various inputs and disturbances using Simulink.
PID Autotuning
With the aid of the sophisticated technique known as PID autotuning, MATLAB is able to automatically modify the PID gains in response to the system's response. Autotuning is especially helpful when the system's parameters are ambiguous or dynamic.
The pidTuner and pidtune functions in MATLAB use optimization algorithms to identify the best PID gains that satisfy particular performance requirements. When manual tuning becomes difficult or time-consuming, these functions can be of great help.
Achieving Stability and Performance in Control Systems
The stability and robustness of a control system are strongly correlated with the phase and gain margins. The gain margin denotes the maximum additional gain that the system can tolerate, whereas the phase margin denotes the maximum additional phase shift that the system can tolerate before becoming unstable.
The phase and gain margins of a control system can be calculated using MATLAB's margin function. If a system has both a positive phase margin and a positive gain margin, it is regarded as stable. Better stability and robustness are indicated by higher phase and gain margins.
Stability Analysis
A crucial stage in the design of a control system is stability analysis. It entails examining the system's transfer function or state-space model's poles and zeros. The system's stability is directly influenced by where these poles and zeros are situated in the complex plane.
You can compute the poles and zeros of a control system using MATLAB's pole and zero functions. The root locus plot, produced by the rlocus function, displays how the system's poles alter in response to different PID gains. Engineers can comprehend the stability characteristics and make the necessary adjustments to ensure stability by analyzing the root locus.
Performance Metrics
Performance metrics quantify a control system's behavior and give information about how well it is performing. Typical performance indicators include:
• Rise Time: The amount of time it takes for the system's response to first move from a given lower threshold to a given higher threshold.
• Settling Time: The amount of time it takes for the system's response to stabilize around the setpoint within a given tolerance band.
• Overshoot: The maximum amount by which the system's response exceeds the setpoint before settling is known as overshoot.
• Steady-State Error: After the response has stabilized, the final discrepancy between the setpoint and the system's output.
• These performance metrics can be computed and displayed using MATLAB's step response plots and other analysis tools. Engineers can use this knowledge to adjust the PID gains and enhance the performance of the controller for particular applications.
Common Challenges in MATLAB Assignments on PID Controllers
Although MATLAB is an effective tool for PID controller design, students frequently run into a number of problems when working on assignments on this subject.
Tuning Parameters
Iterative and complex processes can be involved in PID controller tuning. Students must comprehend how each gain affects the system's response and use the appropriate tuning techniques as a result. Additionally, the tuning strategy may change depending on the control goals and dynamics of the system.
Students should experiment with various tuning methods to overcome this difficulty and use MATLAB's autotuning functions to automate the procedure and find the best PID gains for the given system.
Modeling Complex Systems
Because real-world systems are frequently complex, non-linear or multivariable control strategies may be necessary. Students must be familiar with these tools in order to use the advanced functions provided by MATLAB's Control System Toolbox to manage such complexity.
Students who work with complex systems can model the system using transfer functions or MATLAB's state-space representation. The system can be divided into smaller subsystems so that students can create unique controllers for each one before combining them to control the entire system.
Handling Constraints
When designing the PID controller for some applications, physical restrictions on actuators or sensors must be taken into account. It can be difficult to incorporate these constraints into the control system.
Students can overcome this difficulty by using the optimization toolbox in MATLAB to make sure the PID controller operates within the defined constraints. They should also run MATLAB simulations to confirm the controller's effectiveness under constraint conditions.
Conclusion
PID controller design is a fundamental component of control engineering, and MATLAB offers a great platform for investigating and putting different control strategies into practice. PID controllers are crucial for successful control system design because they can help control systems achieve stability and performance.
PID controller design problems, like tuning, stability analysis, and constraint handling, can be effectively handled with the help of MATLAB's Control System Toolbox, Simulink, and optimization functions. Mastering PID controller design in MATLAB opens doors to a world of possibilities in the field of control systems, whether for academic assignments or real-world applications. The future of PID controller design looks even more promising with ongoing enhancements and updates to MATLAB's capabilities, allowing engineers and researchers to successfully tackle control problems that are getting more complex. | 1,844 | 10,205 | {"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-2024-38 | latest | en | 0.893638 |
https://en.wikipedia.org/wiki/1_62_honeycomb | 1,506,065,907,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688671.43/warc/CC-MAIN-20170922055805-20170922075805-00143.warc.gz | 675,686,244 | 21,781 | # E9 honeycomb
(Redirected from 1 62 honeycomb)
In geometry, an E9 honeycomb is a tessellation of uniform polytopes in hyperbolic 9-dimensional space. ${\displaystyle {\bar {T}}_{9}}$, also (E10) is a paracompact hyperbolic group, so either facets or vertex figures will not be bounded.
E10 is last of the series of Coxeter groups with a bifurcated Coxeter-Dynkin diagram of lengths 6,2,1. There are 1023 unique E10 honeycombs by all combinations of its Coxeter-Dynkin diagram. There are no regular honeycombs in the family since its Coxeter diagram is a nonlinear graph, but there are three simplest ones, with a single ring at the end of its 3 branches: 621, 261, 162.
## 621 honeycomb
621 honeycomb
Family k21 polytope
Schläfli symbol {3,3,3,3,3,3,32,1}
Coxeter symbol 621
Coxeter-Dynkin diagram
9-faces 611
{38}
8-faces {37}
7-faces {36}
6-faces {35}
5-faces {34}
4-faces {33}
Cells {32}
Faces {3}
Vertex figure 521
Symmetry group ${\displaystyle {\bar {T}}_{9}}$, [36,2,1]
The 621 honeycomb is constructed from alternating 9-simplex and 9-orthoplex facets within the symmetry of the E10 Coxeter group.
This honeycomb is highly regular in the sense that its symmetry group (the affine E9 Weyl group) acts transitively on the k-faces for k ≤ 7. All of the k-faces for k ≤ 8 are simplices.
This honeycomb is last in the series of k21 polytopes, enumerated by Thorold Gosset in 1900, listing polytopes and honeycombs constructed entirely of regular facets, although his list ended with the 8-dimensional the Euclidean honeycomb, 521.[1]
### Construction
It is created by a Wythoff construction upon a set of 10 hyperplane mirrors in 9-dimensional hyperbolic space.
The facet information can be extracted from its Coxeter-Dynkin diagram.
Removing the node on the end of the 2-length branch leaves the 9-orthoplex, 711.
Removing the node on the end of the 1-length branch leaves the 9-simplex.
The vertex figure is determined by removing the ringed node and ringing the neighboring node. This makes the 521 honeycomb.
The edge figure is determined from the vertex figure by removing the ringed node and ringing the neighboring node. This makes the 421 polytope.
The face figure is determined from the edge figure by removing the ringed node and ringing the neighboring node. This makes the 321 polytope.
The cell figure is determined from the face figure by removing the ringed node and ringing the neighboring node. This makes the 221 polytope.
### Related polytopes and honeycombs
The 621 is last in a dimensional series of semiregular polytopes and honeycombs, identified in 1900 by Thorold Gosset. Each member of the sequence has the previous member as its vertex figure. All facets of these polytopes are regular polytopes, namely simplexes and orthoplexes.
## 261 honeycomb
261 honeycomb
Family 2k1 polytope
Schläfli symbol {3,3,36,1}
Coxeter symbol 261
Coxeter-Dynkin diagram
9-face types 251
{37}
8-face types 241, {37}
7-face types 231, {36}
6-face types 221, {35}
5-face types 211, {34}
4-face type {33}
Cells {32}
Faces {3}
Vertex figure 161
Coxeter group ${\displaystyle {\bar {T}}_{9}}$, [36,2,1]
The 261 honeycomb is composed of 251 9-honeycomb and 9-simplex facets. It is the final figure in the 2k1 family.
### Construction
It is created by a Wythoff construction upon a set of 10 hyperplane mirrors in 9-dimensional hyperbolic space.
The facet information can be extracted from its Coxeter-Dynkin diagram.
Removing the node on the short branch leaves the 9-simplex.
Removing the node on the end of the 6-length branch leaves the 251 honeycomb. This is an infinite facet because E10 is a paracompact hyperbolic group.
The vertex figure is determined by removing the ringed node and ringing the neighboring node. This makes the 9-demicube, 161.
The edge figure is the vertex figure of the edge figure. This makes the rectified 8-simplex, 051.
The face figure is determined from the edge figure by removing the ringed node and ringing the neighboring node. This makes the 5-simplex prism.
### Related polytopes and honeycombs
The 261 is last in a dimensional series of uniform polytopes and honeycombs.
## 162 honeycomb
162 honeycomb
Family 1k2 polytope
Schläfli symbol {3,36,2}
Coxeter symbol 162
Coxeter-Dynkin diagram
9-face types 152, 161
8-face types 142, 151
7-face types 132, 141
6-face types 122, {31,3,1}
{35}
5-face types 121, {34}
4-face type 111, {33}
Cells {32}
Faces {3}
Vertex figure t2{38}
Coxeter group ${\displaystyle {\bar {T}}_{9}}$, [36,2,1]
The 162 honeycomb contains 152 (9-honeycomb) and 161 9-demicube facets. It is the final figure in the 1k2 polytope family.
### Construction
It is created by a Wythoff construction upon a set of 10 hyperplane mirrors in 9-dimensional space.
The facet information can be extracted from its Coxeter-Dynkin diagram.
Removing the node on the end of the 2-length branch leaves the 9-demicube, 161.
Removing the node on the end of the 6-length branch leaves the 152 honeycomb.
The vertex figure is determined by removing the ringed node and ringing the neighboring node. This makes the birectified 9-simplex, 062.
### Related polytopes and honeycombs
The 162 is last in a dimensional series of uniform polytopes and honeycombs.
## Notes
1. ^ Conway, 2008, The Gosset series, p 413
## References
• The Symmetries of Things 2008, John H. Conway, Heidi Burgiel, Chaim Goodman-Strass, ISBN 978-1-56881-220-5 [1]
• Coxeter The Beauty of Geometry: Twelve Essays, Dover Publications, 1999, ISBN 978-0-486-40919-1 (Chapter 3: Wythoff's Construction for Uniform Polytopes)
• Coxeter Regular Polytopes (1963), Macmillan Company
• Regular Polytopes, Third edition, (1973), Dover edition, ISBN 0-486-61480-8 (Chapter 5: The Kaleidoscope)
• Kaleidoscopes: Selected Writings of H.S.M. Coxeter, edited by F. Arthur Sherk, Peter McMullen, Anthony C. Thompson, Asia Ivic Weiss, Wiley-Interscience Publication, 1995, ISBN 978-0-471-01003-6 [2]
• (Paper 24) H.S.M. Coxeter, Regular and Semi-Regular Polytopes III, [Math. Zeit. 200 (1988) 3-45]
Fundamental convex regular and uniform polytopes in dimensions 2–10
Family An Bn I2(p) / Dn E6 / E7 / E8 / E9 / E10 / F4 / G2 Hn
Regular polygon Triangle Square p-gon Hexagon Pentagon
Uniform polyhedron Tetrahedron OctahedronCube Demicube DodecahedronIcosahedron
Uniform 4-polytope 5-cell 16-cellTesseract Demitesseract 24-cell 120-cell600-cell
Uniform 5-polytope 5-simplex 5-orthoplex5-cube 5-demicube
Uniform 6-polytope 6-simplex 6-orthoplex6-cube 6-demicube 122221
Uniform 7-polytope 7-simplex 7-orthoplex7-cube 7-demicube 132231321
Uniform 8-polytope 8-simplex 8-orthoplex8-cube 8-demicube 142241421
Uniform 9-polytope 9-simplex 9-orthoplex9-cube 9-demicube
Uniform 10-polytope 10-simplex 10-orthoplex10-cube 10-demicube
Uniform n-polytope n-simplex n-orthoplexn-cube n-demicube 1k22k1k21 n-pentagonal polytope
Topics: Polytope familiesRegular polytopeList of regular polytopes and compounds | 2,058 | 6,936 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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-2017-39 | latest | en | 0.902578 |
https://www.geeksforgeeks.org/java-program-for-k-th-missing-element-in-sorted-array/?ref=rp | 1,695,732,069,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00715.warc.gz | 860,192,044 | 39,859 | Open In App
# Java Program for k-th missing element in sorted array
Given an increasing sequence a[], we need to find the K-th missing contiguous element in the increasing sequence which is not present in the sequence. If no k-th missing element is there output -1.
Examples :
```Input : a[] = {2, 3, 5, 9, 10};
k = 1;
Output : 1
Explanation: Missing Element in the increasing
sequence are {1,4, 6, 7, 8}. So k-th missing element
is 1
Input : a[] = {2, 3, 5, 9, 10, 11, 12};
k = 4;
Output : 7
Explanation: missing element in the increasing
sequence are {1, 4, 6, 7, 8} so k-th missing
element is 7```
Approach 1: Start iterating over the array elements, and for every element check if the next element is consecutive or not, if not, then take the difference between these two, and check if the difference is greater than or equal to given k, then calculate ans = a[i] + count, else iterate for next element.
## Java
`// Java program to check for``// even or odd``import` `java.io.*;``import` `java.util.*;`` ` `public` `class` `GFG {`` ` ` ``// Function to find k-th`` ``// missing element`` ``static` `int` `missingK(``int` `[]a, ``int` `k,`` ``int` `n)`` ``{`` ``int` `difference = ``0``,`` ``ans = ``0``, count = k;`` ``boolean` `flag = ``false``;`` ` ` ``// iterating over the array`` ``for``(``int` `i = ``0` `; i < n - ``1``; i++)`` ``{`` ``difference = ``0``;`` ` ` ``// check if i-th and`` ``// (i + 1)-th element`` ``// are not consecutive`` ``if` `((a[i] + ``1``) != a[i + ``1``])`` ``{`` ` ` ``// save their difference`` ``difference +=`` ``(a[i + ``1``] - a[i]) - ``1``;`` ` ` ``// check for difference`` ``// and given k`` ``if` `(difference >= count)`` ``{`` ``ans = a[i] + count;`` ``flag = ``true``;`` ``break``;`` ``}`` ``else`` ``count -= difference;`` ``}`` ``}`` ` ` ``// if found`` ``if``(flag)`` ``return` `ans;`` ``else`` ``return` `-``1``;`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `main(String args[])`` ``{`` ` ` ``// Input array`` ``int` `[]a = {``1``, ``5``, ``11``, ``19``};`` ` ` ``// k-th missing element`` ``// to be found in the array`` ``int` `k = ``11``;`` ``int` `n = a.length;`` ` ` ``// calling function to`` ``// find missing element`` ``int` `missing = missingK(a, k, n);`` ` ` ``System.out.print(missing);`` ``}`` ` `}`` ` `// This code is contributed by``// Manish Shaw (manishshaw1)`
Output
`14`
Time Complexity :O(n), where n is the number of elements in the array.
Space Complexity: O(1) as no extra space has been used.
Approach 2:
Apply a binary search. Since the array is sorted we can find at any given index how many numbers are missing as arr[index] – (index+1). We would leverage this knowledge and apply binary search to narrow down our hunt to find that index from which getting the missing number is easier.
Below is the implementation of the above approach:
## Java
`// Java program for above approach``public` `class` `GFG``{` ` ``// Function to find`` ``// kth missing number`` ``static` `int` `missingK(``int``[] arr, ``int` `k)`` ``{`` ``int` `n = arr.length;`` ``int` `l = ``0``, u = n - ``1``, mid; `` ``while``(l <= u)`` ``{`` ``mid = (l + u)/``2``; `` ``int` `numbers_less_than_mid = arr[mid] -`` ``(mid + ``1``);` ` ``// If the total missing number`` ``// count is equal to k we can iterate`` ``// backwards for the first missing number`` ``// and that will be the answer.`` ``if``(numbers_less_than_mid == k)`` ``{` ` ``// To further optimize we check`` ``// if the previous element's`` ``// missing number count is equal`` ``// to k. Eg: arr = [4,5,6,7,8]`` ``// If you observe in the example array,`` ``// the total count of missing numbers for all`` ``// the indices are same, and we are`` ``// aiming to narrow down the`` ``// search window and achieve O(logn)`` ``// time complexity which`` ``// otherwise would've been O(n).`` ``if``(mid > ``0` `&& (arr[mid - ``1``] - (mid)) == k)`` ``{`` ``u = mid - ``1``;`` ``continue``;`` ``}` ` ``// Else we return arr[mid] - 1.`` ``return` `arr[mid] - ``1``;`` ``}` ` ``// Here we appropriately`` ``// narrow down the search window.`` ``if``(numbers_less_than_mid < k)`` ``{`` ``l = mid + ``1``;`` ``}`` ``else` `if``(k < numbers_less_than_mid)`` ``{`` ``u = mid - ``1``;`` ``}`` ``}` ` ``// In case the upper limit is -ve`` ``// it means the missing number set`` ``// is 1,2,..,k and hence we directly return k.`` ``if``(u < ``0``)`` ``return` `k;` ` ``// Else we find the residual count`` ``// of numbers which we'd then add to`` ``// arr[u] and get the missing kth number.`` ``int` `less = arr[u] - (u + ``1``);`` ``k -= less;` ` ``// Return arr[u] + k`` ``return` `arr[u] + k;`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int``[] arr = {``2``,``3``,``4``,``7``,``11``};`` ``int` `k = ``5``;` ` ``// Function Call`` ``System.out.println(``"Missing kth number = "``+ missingK(arr, k));`` ``}``}` `// This code is contributed by divyesh072019.`
Output
`Missing kth number = 9`
Time Complexity: O(logn), where n is the number of elements in the array.
Auxiliary Space: O(1) as no extra space has been used.
Please refer complete article on k-th missing element in sorted array for more details! | 1,965 | 6,141 | {"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} | 3.390625 | 3 | CC-MAIN-2023-40 | latest | en | 0.668274 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/11/lesson/11.4.1/problem/11-107 | 1,575,984,054,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527620.19/warc/CC-MAIN-20191210123634-20191210151634-00263.warc.gz | 393,198,030 | 15,923 | ### Home > APCALC > Chapter 11 > Lesson 11.4.1 > Problem11-107
11-107.
Write the derivative, $\frac { d y } { d x }$, of each equation below. Homework Help ✎
1. $x \sin(y) - 10y^2 = y \ln(x)$
$\sin(y)+x\cos(y)y^\prime-20yy^\prime=y^\prime\ln(x)+\frac{y}{x}$
1. $r = 2 - \cos(θ)$
$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$
$x=r\cos(\theta)=(2-\cos(\theta))\cos(\theta)$
1. $\left\{ \begin{array} { l } { x ( t ) = 4 t - t ^ { 2 } } \\ { y ( t ) = \operatorname { cos } ( 2 t ) } \end{array} \right.$
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
1. $y = x \sec\sqrt { 2 x ^ { 3 } - 4 x }$
$\text{Let }w=\sqrt{2x^3-4x}.$
$y^\prime=\sec(w)+x\sec(w)\tan(w)w^\prime$
What is $w′$ ? Substitute $w$ and $w′$ into the equation in Step 2. | 341 | 768 | {"found_math": true, "script_math_tex": 14, "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} | 4.28125 | 4 | CC-MAIN-2019-51 | longest | en | 0.26137 |
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