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## Solution Droit
cs601 2 assignment solution ( 1). assignment solution mth301 - Virtual University Assignments - VU786 By taking the Partial derivative of f( x) w. r to x from eq- 1.
## Assignment Easy essay
By taking the Partial derivative of f( x) w. r to y from eq- 1.
r to z from eq- 1. Equation for Tangent plane to the surface through P.
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## Solution assignment Wall
Sol Question# 03: Evaluate Solution:. MTH 301: Group Theory. | 5,418 | 21,539 | {"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-2019-09 | latest | en | 0.7969 |
https://mathematica.stackexchange.com/questions/102771/roulette-simulator-for-a-particular-series-of-events | 1,718,205,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00443.warc.gz | 351,029,408 | 41,420 | Roulette simulator for a particular series of events
I'm new to Mathematica, and I would like to simulate the game of roulette.
I want to bet like \\$1 on the opposite color every time four reds or four blacks come up. I would like to know if I will gain or lose money in the long run.
RandomChoice[{17/36, 17/36, 1/18} -> {red, black, green}]
• You'll lose. No matter what "strategy" you pick... you'll lose. Dec 25, 2015 at 3:27
• Which one do you pick when green comes up 4 times in a row? Dec 25, 2015 at 4:22
2 Answers
You go to the casino. You bet a dollar the ball will land on red. If it does you get two dollars if it does not you get zero. Likewise if you bet the ball will land on black. You cannot bet it will land on green.
Use this previous answer to help do the pattern matching. Note to the original poster, I found that using Google and searching for mathematica count subsequence.
g = RandomChoice[{17/36, 17/36, 1/18} -> {red, black, green}, 10^6];
win = {{red, red, red, red, black}, {black, black, black, black, red}};
loss = {{red, red, red, red, red}, {red, red, red, red, green},
{black, black, black, black, black}, {black, black, black, black, green}};
ts = ToString@Row[#, ","] &;
{StringCount[ts@g, ts /@ win, Overlaps -> True],
-StringCount[ts@g, ts /@ loss, Overlaps -> True]}
Whether you are considering overlapping patterns or not isn't specified, but this code is counting those too.
If I run that once I get {47043, -51943} which says I made 47043 winning bets and 51943 loosing bets in that particular 10^6 spins of the wheel.
Note to the original poster. Study this. Look up each function or thing you don't understand in the help system. Things like /@ and @ are also functions that you can find and try to understand. This is a complicated bit of code for someone who is new to Mathematica.
• This does not answer the OP (bets only on observation of 4-run).
– ciao
Dec 25, 2015 at 9:02
• @ciao I think this answers the OP. I think it counts every red 4-run followed by a black as a win, every 4-run of black followed by a red as a win and every other red or black 4-run as a loss. Perhaps I have made a mistake. If you could explain my error I would thank you and try to correct it.
– Bill
Dec 25, 2015 at 18:03
Here's a q-d implementation:
With[{hits =
Cases[Partition[RandomChoice[{17/36, 17/36, 1/18} -> {1, 2, 3}, 1000000], 5, 1], {x_, x_, x_, x_, y_}]},
Count[hits, {x_, x_, x_, x_, y_} /; y != x] - Length@hits]
It will run through 1 million spins, checking for any 4-runs and counting those where the subsequent spin differs, less the count in total (so +1 for "win", -1 for the bet itself).
You will see, as expected, you lose in the long run. Always.
• Of course, this presumes that Mathematica's random number generator is good for a million calls. Dec 25, 2015 at 14:44
• @ciao I think your answer counts red,red,red,red,green and black,black,black,black,green as wins. But the casino will not let you bet on green. That is how they make money. If you change your code to include y!=3 then this might work.
– Bill
Dec 25, 2015 at 16:47
• @ciao I think your answer counts a 4-run of green followed by anything but green as a win and I don't believe you can bet not-green so x!=3 seems required. With both these changes you and I seem to get the same numbers for the same 10^6 sequence.
– Bill
Dec 25, 2015 at 18:00
• @bill s Many years ago there was a spirited debate about the quality of the Wolfram random number generator algorithm. I don't remember all the details but I don't recall anyone being able to show clear flaws.
– Bill
Dec 25, 2015 at 18:02
• @bill s, "MersenneTwister" should do fine if it's only a few million or so… Feb 17, 2016 at 6:39 | 1,087 | 3,721 | {"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-2024-26 | latest | en | 0.928204 |
https://www.codingbroz.com/count-triplets-that-can-form-two-arrays-of-equal-xor-leetcode-solution/ | 1,721,490,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00605.warc.gz | 610,836,266 | 57,498 | # Count Triplets That Can Form Two Arrays of Equal XOR – Leetcode Solution
In this post, we are going to solve the 1442. Count Triplets That Can Form Two Arrays of Equal XOR problem of Leetcode. This problem 1442. Count Triplets That Can Form Two Arrays of Equal XOR is a Leetcode medium level problem. Let’s see the code, 1442. Count Triplets That Can Form Two Arrays of Equal XOR – Leetcode Solution.
## Problem
Given an array of integers `arr`.
We want to select three indices `i`,` j` and `k` where `(0 <= i < j <= k < arr.length)`.
Let’s define `a` and `b` as follows:
• `a = arr[i] ^ arr[i + 1] ^ … ^ arr[j - 1]`
• `b = arr[j] ^ arr[j + 1] ^ … ^ arr[k]`
Note that ^ denotes the bitwise-xor operation.
Return the number of triplets (`i, j` and `k`) Where `a == b`.
### Example 1 :
``````Input: arr = [2,3,1,6,7]
Output: 4
Explanation: The triplets are (0,1,2), (0,2,2), (2,3,4) and (2,4,4)``````
### Example 2 :
``````Input: arr = [1,1,1,1,1]
Output: 10``````
### Constraints
• `1 <= arr.length <= 300`
• `1 <= arr[i] <= 108`
Now, let’s see the code of 1442. Count Triplets That Can Form Two Arrays of Equal XOR – Leetcode Solution.
# Count Triplets That Can Form Two Arrays of Equal XOR – Leetcode Solution
### 1442. Count Triplets That Can Form Two Arrays of Equal XOR – Solution in Java
```class Solution {
public int countTriplets(int[] arr) {
int n = arr.length;
int count = 0;
for(int i=0; i<n; i++){
int xor = arr[i];
for(int k=i+1; k<n; k++){
xor^=arr[k];
if(xor == 0){
count += k-i;
}
}
}
return count;
}
}```
### 1442. Count Triplets That Can Form Two Arrays of Equal XOR – Solution in C++
```class Solution {
public:
int countTriplets(vector<int>& arr) {
int n(arr.size()), ans(0);
int xors[n + 1]; xors[0] = 0;
for (int i = 0; i < n; ++i)
xors[i + 1] = xors[i] ^ arr[i];
for (int i = 0; i < n; ++i)
for (int k = i + 1; k < n; ++k)
if (xors[i] == xors[k+1])
ans += k - i;
return ans;
}
};```
### 1442. Count Triplets That Can Form Two Arrays of Equal XOR – Solution in Python
```class Solution:
def countTriplets(self, arr: List[int]) -> int:
res = 0
for i in range(len(arr)-1):
xor = 0
for j in range(i, len(arr)):
xor ^= arr[j]
if xor == 0:
res += j-i # len-1
return res```
Note: This problem 1442. Count Triplets That Can Form Two Arrays of Equal XOR is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose. | 790 | 2,430 | {"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} | 4.28125 | 4 | CC-MAIN-2024-30 | latest | en | 0.601375 |
https://www.physics.uoguelph.ca/chapter-3-legendre-polynomials | 1,719,115,716,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00296.warc.gz | 833,677,916 | 49,280 | # Chapter 3: Legendre Polynomials
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The material covered in this chapter is also presented in Boas Chapter 12, Sections 2, 5, 6, 7, and 8.
## 3.1 Introduction
The simplest distribution of charge in electrostatics involves a single point charge $q$; this is known as a monopole. Next on the ladder of complexity is a dipole, which consists of a positive charge $q$ and a negative charge $-q$ held at a fixed distance from one another. Next up is the quadrupole, which consists of four charges (two positive, two negative) placed on the corners of a rectangle; this can be thought of as two opposing dipoles next to each other. If we go one more step up the ladder we find the octupole, which consists of eight charges (four positive, four negative) and can be viewed as two opposing quadrupoles placed next to each other. These simple configurations are illustrated in Fig.3.1.
A central and beautiful result of electrostatic theory is the statement that any charge distribution, whether discrete or continuous, can be represented as the superposition of a monopole, dipole, quadrupole, octupole, and so on. The idea, which goes under the name of multipole expansion, is captured by Fig.3.2. To make it precise, and understand what it means specifically, requires the machinery of Legendre polynomials, the topic of this chapter. We shall pick this story up again in Sec.3.8, once we have introduced the required tools.
## 3.2 Legendre Polynomials
The Legendre polynomials $P_\ell(x)$ make up an infinite set of functions of the variable $x$. Each function in the set is given a label $\ell$; this is an integer that begins at $\ell = 0$ and ends at $\ell = \infty$. We therefore have a function $P_0(x)$, another function $P_1(x)$, and an infinite number of additional functions belonging to the set of Legendre polynomials. As the name indicates, all these functions are polynomials in $x$.
The definition of the Legendre polynomials is a little curious and quite different from anything you will have encountered in your mathematical education thus far. We first introduce a function $\Phi(x,h)$ of two variables, known as a generating function. The first variable, $x$, is the same variable that appears as the argument of the Legendre polynomials. The second variable, $h$, is an auxiliary variable with no particular meaning. The generating function is defined by
$$\Phi(x,h) := (1 - 2xh + h^2)^{-1/2}. \tag{3.1}$$
The specific form of the generating function will be motivated once we return to the multipole expansion in Sec.3.8. For the time being we simply accept Eq.(3.1) as a definition, however arbitrary it may seem.
Let us, for the moment, think of $x$ as a fixed parameter in Eq.(3.1), and therefore think of $\Phi$ as a function of a single variable $h$. To focus on this dependence we shall write $\Phi = \Phi(h)$, and momentarily forget about the dependence on $x$. Most functions of $h$ can be expanded as a Taylor expansion in powers of $h$, and this is true of $\Phi(h)$. We write
\begin{align} \Phi(h) &= \Phi(0) + \frac{d\Phi}{dh}\biggr|_{h=0}\, h + \frac{1}{2!} \frac{d^2\Phi}{dh^2}\biggr|_{h=0}\, h^2 + \frac{1}{3!} \frac{d^3\Phi}{dh^3}\biggr|_{h=0}\, h^3 + \cdots \nonumber \\ &= \sum_{\ell=0}^\infty \frac{1}{\ell!} \frac{d^\ell\Phi}{dh^\ell}\biggr|_{h=0}\, h^\ell. \tag{3.2} \end{align}
It is no accident that we use the symbol $\ell$ to denote the summation index in the Taylor expansion, which holds the key to the definition of the Legendre polynomials. Before we come to that, we must restore the $x$-dependence of the generating function. This doesn't change the general appearance of the Taylor expansion, except for the fact that derivatives with respect to $h$ must now be properly written as partial derivatives instead of total derivatives. We therefore have
$$\Phi(x,h) = \sum_{\ell=0}^\infty \frac{1}{\ell!} \frac{\partial^\ell\Phi}{\partial h^\ell}\biggr|_{h=0}\, h^\ell. \tag{3.3}$$
It is important to understand that in this expression, the coefficients of the Taylor expansion, given by the partial derivatives of $\Phi$ with respect to $h$ evaluated at $h=0$, depend on $x$ but are independent of $h$.
The Legendre polynomials are precisely defined to be equal to these coefficients. We write
$$\Phi(x,h) = \sum_{\ell=0}^\infty P_\ell(x)\, h^\ell, \tag{3.4}$$
and comparing with the Taylor expansion, we conclude that
$$P_\ell(x) = \frac{1}{\ell!} \frac{\partial^\ell\Phi}{\partial h^\ell}\biggr|_{h=0}. \tag{3.5}$$
To arrive at this conclusion we relied on the fact that if two expansions $\sum_{\ell} a_\ell\, h^\ell$ and $\sum_{\ell} b_\ell\, h^\ell$ are equal to each other for any value of $h$, then we must have $a_\ell = b_\ell$ for all values of $\ell$. Equation (3.4) is the formal definition of the Legendre polynomials. It captures the main idea that they are identified as the coefficients in the Taylor expansion of the generating function about $h=0$; because the Taylor expansion formally involves an infinite number of terms, we have an infinite number of coefficients, and therefore an infinite set of functions of $x$. Equation (3.5) provides an explicit link with the partial derivatives of $\Phi(x,h)$.
Let us use Eq.(3.5) to calculate the first few polynomials. For $\ell = 0$ we are instructed to take no derivatives, and to evaluate the generating function at $h = 0$. This gives $P_0(x) = 1$; the zeroth polynomial is actually a constant. Moving on to $\ell = 1$, we must differentiate $\Phi$ once with respect to $h$, which yields
$$\frac{\partial \Phi}{\partial h} = (x-h)(1-2xh+h^2)^{-3/2} \tag{3.6}$$
after simplification. Evaluating this at $h=0$ and dividing by $1! = 1$ gives $P_1(x) = x$. For $\ell = 2$ we differentiate $\Phi$ twice, and get
$$\frac{\partial^2 \Phi}{\partial h^2} = (3x^2-1-4xh+2h^2)(1-2xh+h^2)^{-5/2} \tag{3.7}$$
after simplification. Evaluating this at $h=0$ and dividing by $2! = 2$ produces $P_2(x) = \frac{1}{2}(3x^2 - 1)$. We can just keep going like this, and generate any number of polynomials. The calculations, however, become progressively tedious, and as we shall see below, more convenient methods are available to generate Legendre polynomials beyond $\ell = 2$.
Exercise 3.1: Verify the previous expressions for $\partial \Phi/\partial h$ and $\partial^2 \Phi/\partial h^2$.
The first few Legendre polynomials are given by
\begin{align} P_0 &= 1 \tag{3.8a} \\ P_1 &= x \tag{3.8b} \\ P_2 &= \frac{1}{2}(3x^2 - 1) \tag{3.8c} \\ P_3 &= \frac{1}{2}(5x^3 - 3 x) tag{3.8d} \\ P_4 &= \frac{1}{8}(35x^4 - 30x^2 + 3) \tag{3.8e} \\ P_5 &= \frac{1}{8}(63x^5 - 70x^3 + 15x) \tag{3.8f} \\ P_6 &= \frac{1}{16}(231x^6 - 315x^4 + 105x^2 - 5). \tag{3.8g} \end{align}
These are plotted in Fig.3.3
A few properties can be seen to emerge from this partial listing. First, the largest power of $x$ contained in $P_\ell(x)$ is always $x^\ell$ (multiplied by a numerical coefficient). Second, when $\ell$ is even, $P_\ell(x)$ contains only even powers of $x$, starting with $x^\ell$ and ending with $x^0$. Third, when $\ell$ is odd, $P_\ell(x)$ contains only odd powers of $x$, starting with $x^\ell$ and ending with $x$. It follows from the last two observations that $P_\ell(x)$ is an even function of $x$ when $\ell$ is even, and an odd function of $x$ when $\ell$ is odd.
Another property that can gleaned from the listing of Eqs.(3.8) is the fact that the polynomials always evaluate to $1$ when $x=1$:
$$P_\ell(1) = 1. \tag{3.9}$$
The even and odd nature of the polynomials then implies that $P_\ell(-1) = 1$ when $\ell$ is even, and $P_\ell(-1) = -1$ when $\ell$ is odd; these statements can be summarized in the single equation
$$P_\ell(-1) = (-1)^\ell. \tag{3.10}$$
It is actually easy to prove that Eq.(3.9) is true for all Legendre polynomials, not just the first few listed in Eqs.(3.8). This can be done by inserting $x=1$ in the defining relation of Eq.(3.4), taking into account that $\Phi(1,h) = (1-2h+h^2)^{-1/2} = (1-h)^{-1}$. We have
$$\frac{1}{1-h} = \sum_{\ell=0}^\infty P_\ell(1)\, h^\ell, \tag{3.11}$$
and $P_\ell(1)$ is identified as the set of coefficients in the Taylor expansion of the function $(1-h)^{-1}$. But it is known that
$$\frac{1}{1-h} = 1 + h + h^2 + h^3 + \cdots = \sum_{\ell=0}^\infty h^\ell, \tag{3.12}$$
and equality of the two expansions for any $h$ guarantees that $P_\ell(1) = 1$, as claimed.
Exercise 3.2: Verify that the Taylor expansion of $(1-h)^{-1}$ is given by $1 + h + h^2 + h^3 + \cdots$. The easiest way is to multiply both sides of the equation by $(1-h)$ and prove that the right-hand side evaluates to $1$, as required.
## 3.3 Recursion Relations
We shall prove that the Legendre polynomials satisfy the recursion relation
$$\ell P_\ell(x) = (2\ell-1) x P_{\ell-1}(x) - (\ell-1) P_{\ell-2}(x). \tag{3.13}$$
This equation is enormously useful, because it provides us with a very efficient means to generate any number of Legendre polynomials. Suppose, for example, that we have computed $P_0 = 1$ and $P_1 = x$ by exploiting the definition (as we did previously), but that we have no knowledge of the higher-order polynomials. By setting $\ell = 2$ in Eq.(3.13) we can generate $P_2$ with very little effort; the equation gives $2P_2 = 3xP_1 - P_0 = 3x^2 - 1$, or $P_2 =\frac{1}{2}(3x^2 - 1)$. And now that $P_1$ and $P_2$ are known, we can set $\ell =3$ in Eq.(3.13) to obtain $P_3$; we get $3 P_3 = 5x P_2 - 2P_1 = \frac{5}{2}x(3x^2-1) - 2x = \frac{3}{2}(5x^3 - 3x)$, so that $P_3 = \frac{1}{2}(5x^3 - 3x)$, in agreement with the listing of Eqs.(3.8). The recursion relation can be iterated any number of times to generate any number of Legendre polynomials. It is truly a wonderful tool, since the alternative method would involve taking a large number of derivatives of the generating function, which soon becomes unpleasant.
Exercise 3.3: Use the recursion relation to calculate $P_4$, $P_5$, and $P_6$, and make sure that your results agree with Eqs.(3.8).
The recursion relation of Eq.(3.13) can be used to prove that the properties of the Legendre polynomials, identified previously for $\ell \leq 6$ on the basis of Eqs.(3.8), actually hold for all values of $\ell$. The proof is an application of the method of induction.
The first property we identified was that the largest power of $x$ that appears in $P_\ell$ is $x^\ell$. Let us assume that the property is known to hold for all values of $\ell$ up to some limiting value that we denote $\ell^*$; for the listing of Eqs.(3.8) we have that $\ell^* = 6$. We can use the recursion relation to prove that the property continues to hold for the next value of $\ell$. We set $\ell = \ell^*+1$ in Eq.(3.13) and notice that the right-hand side involves $x P_{\ell^*}$ and $P_{\ell^*-1}$, both multiplied by a numerical coefficient. We know that the largest power of $x$ in $P_{\ell^*}$ is $x^{\ell^*}$, and multiplication by $x$ turns this into $x^{\ell^*+1} = x^\ell$. The largest power of the second term is $x^{\ell^*-1} = x^{\ell-2}$, and this is a smaller power than for the first term. The recursion relation therefore reveals that the largest power of $x$ in $P_\ell$ is $x^\ell$ when $\ell = \ell^*+1$. At this stage the property is established for all values of $\ell$ up to $\ell^* + 1$, and additional iterations of the recursion relation allow us to keep on increasing the limiting value of $\ell$. The recursive process has no endpoint, and we can conclude that the property will continue to hold for all values of $\ell$, all the way to infinity.
Exercise 3.4: Use induction and the recursion relation of Eq.(3.13) to convince yourself that for all values of $\ell$, $P_\ell(x)$ is an even (odd) function of $x$ when $\ell$ is even (odd).
The proof of Eq.(3.13) relies on the generating function $\Phi(x,h)$ and the defining relation of Eq.(3.4). The generating function was given explicitly in Eq.(3.1), and its partial derivative with respect to $h$ was calculated in Eq.(3.6). These equations imply
$$(1-2xh+h^2) \frac{\partial \Phi}{\partial h} = (x-h) \Phi, \tag{3.14}$$
and this can be turned into a useful identity by inserting Eq.(3.4) and its derivative with respect to $h$, given by
$$\frac{\partial \Phi}{\partial h} = \sum_{\ell=1}^\infty \ell P_\ell(x)\, h^{\ell-1}; \tag{3.15}$$
we indicate that the sum begins at $\ell=1$ instead of $\ell=0$ as in Eq.(3.4), because the term with $\ell = 0$ obviously vanishes. For the left-hand side of Eq.(3.14) we get
\begin{align} (1-2xh+h^2) \frac{\partial \Phi}{\partial h} &= (1-2xh+h^2) \sum_{\ell=1}^\infty \ell P_\ell\, h^{\ell-1} \nonumber \\ &= \sum_{\ell=1}^\infty \ell P_\ell\, h^{\ell-1} - 2x \sum_{\ell=1}^\infty \ell P_\ell\, h^{\ell} +\sum_{\ell=1}^\infty \ell P_\ell\, h^{\ell+1}. \tag{3.16} \end{align}
We notice that the three sums feature different powers of $h$, and we would like to re-express them so as to consolidate the powers. We leave the first sum alone, but rewrite the second and third sums as
$\sum_{\ell'=1}^\infty \ell' P_{\ell'}\, h^{\ell'}, \qquad \sum_{\ell'=1}^\infty \ell' P_{\ell'}\, h^{\ell'+1},$
denoting the summation index by $\ell'$ instead of $\ell$; this clearly does not alter the value of the sum. In the second sum we write $\ell' = \ell - 1$, replacing the summation index by a new $\ell$, which is not to be confused with the original $\ell$; this gives us
$\sum_{\ell=2}^\infty (\ell-1) P_{\ell-1}\, h^{\ell-1},$
where we must be careful to start the sum at $\ell=2$, the value of $\ell$ that properly corresponds to $\ell'=1$. For the third sum we write $\ell' = \ell -2$ and express it as
$\sum_{\ell=3}^\infty (\ell-2) P_{\ell-2}\, h^{\ell-1}.$
Putting all this together, we have that
$$(1-2xh+h^2) \frac{\partial \Phi}{\partial h} = \sum_{\ell=1}^\infty \ell P_\ell\, h^{\ell-1} - 2x \sum_{\ell=2}^\infty (\ell-1) P_{\ell-1}\, h^{\ell-1} + \sum_{\ell=3}^\infty (\ell-2) P_{\ell-2}\, h^{\ell-1}, \tag{3.17}$$
with all sums featuring the same powers of $h$. Further consolidation is possible if we join the three sums into a single one. For this step we must pay attention to the fact that the sums begin at different values of $\ell$. To work around this we re-express the first sum as
$P_1 + 2P_2\, h + \sum_{\ell=3}^\infty \ell P_\ell\, h^{\ell-1},$
the second sum as
$P_1\, h + \sum_{\ell=3}^\infty (\ell-1) P_{\ell-1}\, h^{\ell-1},$
and leave the third sum alone. Putting all this together and simplifying, we finally arrive at
\begin{align} (1-2xh+h^2) \frac{\partial \Phi}{\partial h} &= P_1 + (2 P_2 - 2x P_1) h \nonumber \\ & \quad \mbox{} + \sum_{\ell=3}^\infty \bigl[ \ell P_\ell - 2(\ell-1) x P_{\ell-1} + (\ell-2) P_{\ell-2} \bigr] h^{\ell-1} \tag{3.18} \end{align}
for the left-hand side of Eq.(3.14).
Exercise 3.5: Make sure that you can reproduce all the steps that led to Eq.(3.18).
Moving on to the right-hand side of Eq.(3.14), we insert Eq.(3.4) and go through similar manipulations. We find
\begin{align} (x-h) \Phi &= (x-h) \sum_{\ell=0}^\infty P_\ell\, h^\ell \nonumber \\ &= x \sum_{\ell'=0}^\infty P_{\ell'}\, h^{\ell'} - \sum_{\ell'=0}^\infty P_{\ell'}\, h^{\ell' + 1} \nonumber \\ &= x \sum_{\ell=1}^\infty P_{\ell-1}\, h^{\ell-1} - \sum_{\ell=2} P_{\ell-2}\, h^{\ell - 1} \nonumber \\ &= x( P_0 + P_1\, h) + x \sum_{\ell=3}^\infty P_{\ell-1}\, h^{\ell-1} - P_0\, h - \sum_{\ell=3} P_{\ell-2}\, h^{\ell - 1} \nonumber \\ &= x P_0 + (x P_1 - P_0)\, h + \sum_{\ell = 3}^\infty \bigl( x P_{\ell-1} - P_{\ell-2} \bigr) h^{\ell - 1}. \tag{3.19} \end{align}
Exercise 3.6: Make sure that you can reproduce all the steps that led to Eq.(3.19).
The hard work is over. In Eqs.(3.19) and (3.18) we have two expansions in powers of $h$ that are known to be equal to each other by virtue of Eq.(3.14). Because equality is guaranteed for any value of $h$, the expansion coefficients of the left-hand side must all be equal to those of the right-hand side. At order $h^0$ this gives us $P_1 = x P_0$, or $x=x$, which is obviously true. At order $h^1$ we get $2P_2 - 2x P_1 = xP_1 - P_0$, or $x^2 -1 = x^2 - 1$. The higher orders are more informative. We get
$$\ell P_\ell - 2(\ell-1) x P_{\ell-1} + (\ell-2) P_{\ell-2} = x P_{\ell-1} - P_{\ell-2}, /tag{3.20}$$
which reduces to the recursion relation of Eq.(3.13) after simplification.
## 3.4 Recursion Relation II
The Legendre polynomials satisfy a number of additional recursion relations,
\begin{align} &P'_{\ell+1} - 2x P'_{\ell} + P'_{\ell-1} = P_\ell, /tag{3.21a} \\ & P'_{\ell+1} - P'_{\ell-1} = (2\ell+1) P_\ell, \tag{3.21b} \\ & P'_{\ell+1} - x P'_{\ell} = (\ell+1) P_\ell, \tag{3.21c}\\ & P'_{\ell-1} - x P'_{\ell} = -\ell P_\ell, \tag{3.21d} \\ & (1-x^2) P'_\ell = \ell P_{\ell-1} - \ell x P_\ell, \tag{3.21e} \end{align}
in which a prime indicates differentiation with respect to $x$. These relations, unlike the original one of Eq.(3.13), feature derivatives of the Legendre polynomials.
We establish Eq.(3.21a) by following steps very similar to those outlined in Sec.3.3. We again begin with the generating function $\Phi(x,h) = (1-2xh+h^2)^{-1/2}$, but this time we differentiate it with respect to $x$, which yields $\partial \Phi/\partial x = h (1-2xh+h^2)^{-3/2}$. This gives rise to the identity
$$(1-2xh+h^2) \frac{\partial \Phi}{\partial x} = h \Phi, \tag{3.22}$$
which shall play the same role here as Eq.(3.14) in the derivation of Eq.(3.13). Inserting Eq.(3.4) on the left-hand side and going through the familiar manipulations, we get
\begin{align} (1-2xh+h^2) \frac{\partial \Phi}{\partial x} &= (1-2xh+h^2) \sum_{\ell=0}^\infty P'_\ell\, h^\ell \nonumber \\ & = \sum_{\ell'=0}^\infty P'_{\ell'}\, h^{\ell'} -2 x \sum_{\ell=0}^\infty P'_\ell\, h^{\ell+1} + \sum_{\ell'=0}^\infty P'_{\ell'}\, h^{\ell'+2} \nonumber \\ & = \sum_{\ell=-1}^\infty P'_{\ell+1}\, h^{\ell+1} -2 x \sum_{\ell=0}^\infty P'_\ell\, h^{\ell+1} + \sum_{\ell=1}^\infty P'_{\ell-1}\, h^{\ell+1} \nonumber \\ & = P'_0 + P'_1\, h + \sum_{\ell=1}^\infty P'_{\ell+1}\, h^{\ell+1} - 2 x P'_0\, h - 2 x \sum_{\ell=1}^\infty P'_\ell\, h^{\ell+1} + \sum_{\ell=1}^\infty P'_{\ell-1}\, h^{\ell+1} \nonumber \\ & = P'_0 + (P'_1 - 2x P'_0) h + \sum_{\ell=1}^\infty \bigl( P'_{\ell+1} - 2x P'_{\ell} + P'_{\ell-1} \bigr) h^{\ell + 1}. \tag{3.23} \end{align}
For the right-hand side we get
$$h \Phi = h \sum_{\ell =0}^\infty P_{\ell}\, h^{\ell} = \sum_{\ell =0}^\infty P_{\ell}\, h^{\ell +1} = P_0\, h + \sum_{\ell =1}^\infty P_{\ell}\, h^{\ell +1}. \tag{3.24}$$
Equating both sides informs us that $P'_0 = 0$, which we already knew, as well as $P'_1 - 2x P'_0 = P_0$, which merely states that $1=1$. The higher-order terms return Eq.(3.21a).
Exercise 3.7: Make sure that you can reproduce all the steps involved in the derivation of Eq.(3.21a).
To establish Eq.(3.21b) we return to Eq.(3.13), which we write as
$$\ell ' P_{\ell '} = (2\ell '-1) x P_{\ell '-1} - (\ell '-1) P_{\ell '-2} \tag{3.25}$$
In this we set $\ell' = \ell + 1$, and get
$$(2\ell+1) x P_\ell = (\ell+1) P_{\ell+1} + \ell P_{\ell - 1} \tag{3.26}$$
after simplification and a slight re-arrangement. We next differentiate with respect to $x$ and multiply by 2:
$$2(2\ell+1) x P'_\ell + 2(2\ell+1) P_\ell = 2(\ell+1) P'_{\ell+1} + 2\ell P'_{\ell-1}. \tag{3.27}$$
For the final step we multiply Eq.(3.21a) by $2\ell+1$,
$$2(2\ell+1) x P'_{\ell} + (2\ell+1) P_\ell = (2\ell+1) P'_{\ell+1} + (2\ell+1) P'_{\ell-1}, \tag{3.28}$$
and subtract the two equations. This leaves us with Eq.(3.21b) after a bit of simplification.
Exercise 3.8: Go through all the steps involved in the derivation of Eq.(3.21b).
Equation (3.21c) follows immediately after adding Eq.(3.21a) to Eq.(3.21b) and dividing by $2$. To get Eq.(3.21d) we subtract the equations instead. Finally, to establish Eq.(3.21e) we replace $\ell$ by $\ell - 1$ in Eq.(3.21c),
$$P'_\ell - x P'_{\ell-1} = \ell P_{\ell-1}, /tag{3.29}$$
multiply Eq.(3.21d) by $x$,
$$-x^2 P'_\ell + x P'_{\ell-1} = -\ell x P_\ell, \tag{3.30}$$
and add the results.
Exercise 3.9: You know what to do.
## 3.5 Legendre's Equation
The Legendre polynomials satisfy the second-order differential equation
$$(1-x^2) P''_\ell - 2x P'_\ell + \ell(\ell+1) P_\ell = 0. \tag{3.31}$$
This can also be expressed in the alternative form
$$\frac{d}{dx} \biggl[ (1-x^2) \frac{dP_\ell}{dx} \biggr] + \ell(\ell+1) P_\ell = 0, \tag{3.32}$$
as you can quickly verify by expanding the first term on the left-hand side of the equation. Either form of the differential equation is known as Legendre's equation. It is a very important equation of mathematical physics, with applications in many different fields.
Exercise 3.10: Verify that $P_0$, $P_1$, $P_2$, and $P_3$ are all solutions to Eq.(3.31).
To arrive at Eq.(3.31) we begin with Eq.(3,21e),
$$(1-x^2) P'_\ell - \ell P_{\ell-1} + \ell x P_\ell = 0,/tag{3.33}$$
which we differentiate with respect to $x$. This gives
$$(1-x^2) P''_\ell - 2x P'_\ell - \ell P'_{\ell-1} + \ell x P'_\ell + \ell P_\ell = 0, /.tag{3.34}$$
and we eliminate the term involving $P'_{\ell-1}$ by making use of Eq.(3.21e). A bit of simplification returns Legendre's equation.
Exercise 3.11: Make sure you can recover Eq.(3.31) from Eq.(3.21e).
In many textbook presentations of the Legendre polynomials, the differential equation is stated first (with whatever motivation), and the polynomials $P_\ell(x)$ are shown to be solutions to that differential equation. We shall have more to say about this point of view in Sec.3.9. For our purposes here we have preferred to define the polynomials first, with the help of the generating function, and to derive Legendre's equation as a consequence of the definition.
## 3.6 Orthogonal Functions
We now interrupt the development of the main topic (Legendre polynomials) to introduce the concept of orthogonal functions. The notion of orthogonality is familiar in the context of vectors; here we wish to generalize it to the context of functions.
In ordinary, three-dimensional space, two vectors $\boldsymbol{A}$ and $\boldsymbol{B}$ are orthogonal when
$$0 = \boldsymbol{A} \cdot \boldsymbol{B} = \sum_{j=1}^3 A_j B_j; \tag{3.35}$$
we make use of the component notation for vectors, writing, for example, $A_1 := A_x$, $A_2 := A_y$, and $A_3 := A_z$. It is easy to generalize this notion of orthogonality to a higher-dimensional space. Indeed, in an $n$-dimensional space, the vectors $\boldsymbol{A}$ and $\boldsymbol{B}$, each possessing $n$ components, would be declared orthogonal when
$$0 = \sum_{j=1}^n A_j B_j. \label{eq3:vector_ortho} \tag{3.36}$$
This expression can be used to motivate a generalization from vectors to functions.
Let us imagine replacing the vector index $j$, which jumps discontinuously as it runs from $j=1$ to $j = n$, by a continuous variable $x$. At the same time, let us imagine replacing the vector components $A_j$ and $B_j$ by the values of two functions, $A(x)$ and $B(x)$. Just as the finite number of distinct values of $j$ have been replaced by an infinity of values for $x$, the finite number of vector components $A_j$ have been replaced by an infinity of values for the function $A(x)$. In this generalization, it seems appropriate to replace the discrete sum over $j$ by a continuous sum over the variable $x$. This, of course, defines an integral, and the functions $A(x)$ and $B(x)$ shall be declared to be orthogonal when $0 = \int A(x) B(x)\, dx$.
A final ingredient is required to make this generalized notion of orthogonality well defined: we must specify the domain of integration. We shall thus adopt the following definition: Two functions $A(x)$ and $B(x)$, defined over an interval going from $x=a$ to $x=b$, are orthogonal on that interval when
$$\int_a^b A(x) B(x)\, dx = 0. \tag{3.37}$$
The specification of the interval is important. In general, two functions that are orthogonal on a given interval will not be orthogonal on other intervals. The integral
$$N[A] := \int_a^b \bigl[ A(x) \bigr]^2\, dx \tag{3.38}$$
is known as the norm of the function $A(x)$ on the interval $(a,b)$, by analogy with the norm of a vector $\boldsymbol{A}$, defined by $|\boldsymbol{A}|^2 = \sum_j A_j A_j$.
This notion of functional orthogonality can be generalized further. It can, for example, be extended to functions of more than one variable. It can also be extended to complex functions. In this case we would say that two complex functions $f(x)$ and $g(x)$ are orthogonal on an interval $(a, b)$ when
$$\int_a^b f^*(x) g(x)\, dx = 0, \tag{3.39}$$
where the asterisk indicates complex conjugation. Which of two functions is complex conjugated is a matter of choice; the outcome is independent of this choice. The reason for complex conjugation is that in the case of a complex function, the norm is naturally defined by
$$N[f] := \int_a^b |f(x)|^2\, dx, \tag{3.40}$$
where $|f|^2 = f^* f$, so as to necessarily return a real and positive number.
It is also possible to make the notion of functional orthogonality more abstract by altering the integral criterion. A frequently encountered generalization of Eq.(3.37) is
$$N[f] := \int_a^b |f(x)|^2\, dx, \tag{3.41}$$
where $|f|^2 = f^* f$, so as to necessarily return a real and positive number.
## 3.7 Orthogonality of Legendre Polynomials
Legendre polynomials form a set of orthogonal functions on the interval $(-1,1)$. We shall indeed prove that
$$\int_{-1}^1 P_\ell(x) P_{\ell'}(x)\, dx = 0 \tag{3.42}$$
when $\ell \neq \ell'$. When $\ell = \ell'$ the integral does not vanish, and we shall see instead that
$$N_\ell := N[P_\ell] = \int_{-1}^1 \bigl[ P_\ell(x) \bigr]^2\, dx =\frac{2}{2\ell + 1}; \tag{3.43}$$
this is the norm of the function $P_\ell(x)$ on the interval $(-1,1)$. Equations (3.42) and (3.43) can be summarized by the single statement
$$\int_{-1}^1 P_\ell(x) P_{\ell'}(x)\, dx = \frac{2}{2\ell + 1}\, \delta_{\ell\ell'}, \tag{3.44}$$
where $\delta_{\ell\ell'}$ is our old friend the Kronecker delta, equal to $1$ when $\ell=\ell'$, and to $0$ when $\ell\neq \ell'$.
To establish Eq.(3.42) we proceed from Legendre's equation, in the form given by Eq.(3.32). We first multiply the equation by $P_{\ell'}$ and get
$$P_{\ell'} \frac{d}{dx} \biggl[ (1-x^2) \frac{dP_{\ell}}{dx} \biggr] + \ell(\ell+1) P_{\ell} P_{\ell'} = 0. \tag{3.45}$$
We next interchange $\ell$ and $\ell'$ and produce a different version of the same equation,
$$P_{\ell} \frac{d}{dx} \biggl[ (1-x^2) \frac{dP_{\ell'}}{dx} \biggr] + \ell'(\ell'+1) P_{\ell} P_{\ell'} =0. \tag{3.46}$$
The next step is to subtract the second equation from the first,
$$P_{\ell'} \frac{d}{dx} \biggl[ (1-x^2) \frac{dP_{\ell}}{dx} \biggr] - P_{\ell} \frac{d}{dx} \biggl[ (1-x^2) \frac{dP_{\ell'}}{dx} \biggr] + \bigl[ \ell(\ell+1) - \ell'(\ell'+1) \bigr] P_{\ell} P_{\ell'} = 0,\tag{3.47}$$
and to write the final result as
$$\frac{d}{dx} \biggl[ P_{\ell'} (1-x^2) \frac{dP_{\ell}}{dx} - P_{\ell} (1-x^2) \frac{dP_{\ell'}}{dx} \biggr] + \bigl[ \ell(\ell+1) - \ell'(\ell'+1) \bigr] P_{\ell} P_{\ell'} = 0, \tag{3.48}$$
with the first two terms consolidated into a total derivative.
Exercise 3.12: Verify that the total derivative in Eq.(3.48) is equal to the first two terms of the preceding equation.
The result displayed in Eq.(3.48) is a direct consequence of Legendre's equation. It is this equation that will lead us to the orthogonality property of the polynomials, and will reveal why the interval had to be $(-1,1)$ and not some other interval. The idea is to integrate Eq.(3.48) over an arbitrary interval $(a,b)$, and see what choice of interval gets us something useful. Integration yields
$$\biggl[ P_{\ell'} (1-x^2) \frac{dP_{\ell}}{dx} - P_{\ell} (1-x^2) \frac{dP_{\ell'}}{dx} \biggr] \biggr|^{b}_{a} + \bigl[ \ell(\ell+1) - \ell'(\ell'+1) \bigr] \int_a^b P_{\ell} P_{\ell'}\, dx = 0, \tag{3.49}$$
and we observe that in general, the boundary terms at $x=a$ and $x=b$ do not vanish; this prevents us from saying anything useful about the remaining integral. But we see that a judicious choice of interval allows us to eliminate the boundary terms: the factor $(1-x^2)$ vanishes at both $x=-1$ and $x=1$, and choosing $(-1,1)$ will force the boundary terms to vanish. With this choice of interval we now have that
$$\bigl[ \ell(\ell+1) - \ell'(\ell'+1) \bigr] \int_{-1}^{1} P_{\ell} P_{\ell'}\, dx = 0. \tag{3.50}$$
When $\ell = \ell'$ the factor in square brackets vanishes, and the equation reduces to $0 = 0$. But when $\ell \neq \ell'$ we obtain the statement of Eq.(3.42), and confirm that distinct Legendre polynomials are orthogonal on $(-1,1)$.
To derive Eq.(3.43) we begin\footnote{This method to derive Eq.(3.13) is not found in standard textbooks. It was extracted from Sec.4.5 of Special functions and their applications, by N.N.\ Lebedev (Dover Publications, 1972).} with the recursion relation of Eq.(3.13), which we write as
$$\ell P_\ell - (2\ell-1) x P_{\ell-1} + (\ell-1) P_{\ell-2} = 0. \tag{3.51}$$
We produce an alternative form by replacing $\ell$ by $\ell+1$,
$$(\ell+1) P_{\ell+1} - (2\ell+1) x P_{\ell} + \ell P_{\ell-1} = 0. \tag{3.52}$$
Next we multiply the first equation by $(2\ell+1) P_\ell$,
$$\ell (2\ell+1) \bigl( P_\ell \bigr)^2 - (2\ell-1) (2\ell+1) x P_{\ell-1} P_{\ell} + (\ell-1) (2\ell+1) P_{\ell-2} P_{\ell} = 0,\tag{3.53}$$
and multiply the second equation by $(2\ell-1) P_{\ell-1}$,
$$(\ell+1) (2\ell-1) P_{\ell-1} P_{\ell+1} - (2\ell-1) (2\ell+1) x P_{\ell-1} P_{\ell} + \ell (2\ell-1) \bigl(P_{\ell-1})^2 = 0. \tag{3.54}$$
We now subtract the last equation from the preceding one, and cancel out the common terms involving $x P_{\ell-1} P_{\ell}$. This gives
\begin{align} & \ell (2\ell+1) \bigl( P_\ell \bigr)^2 + (\ell-1) (2\ell+1) P_{\ell-2} P_{\ell} \nonumber \\ & \mbox{} - (\ell+1) (2\ell-1) P_{\ell-1} P_{\ell+1} \tag{3.55} \end{align}
In the next step we integrate the equation from $x=-1$ to $x=1$, and invoke the orthogonality of $P_{\ell-2}$ and $P_\ell$ to eliminate the second term, as well as the orthogonality of $P_{\ell-1}$ and $P_{\ell+1}$ to discard the third term. We are left with
$$\ell (2\ell+1) \int_{-1}^1 \bigl( P_\ell \bigr)^2\, dx - \ell (2\ell-1) \int_{-1}^1 \bigl(P_{\ell-1})^2\, dx = 0, \tag{3.56}$$
a relation between the norms of $P_{\ell}$ and $P_{\ell-1}$. In the notation of Eq.(3.43), the relation is
$$N_\ell = \frac{2\ell-1}{2\ell+1} N_{\ell-1}, \tag{3.57}$$
and it takes the form of a recursion relation for $N_\ell$.
To obtain $N_\ell$ we must solve the recursion relation. First, let us set $\ell = 1$ in Eq.(3.57), to get $N_1 = \frac{1}{3} N_0$. If we next set $\ell=2$, we get $N_2 = \frac{3}{5} N_1 = \frac{1}{5} N_0$; notice that the factors of $3$ have cancelled out. Something like this happens again in the next step: With $\ell = 3$ we have that $N_3 = \frac{5}{7} N_2 = \frac{1}{7} N_0$, and this time it is the factors of $5$ that cancel out. A little thought will convince you that this phenomenon keeps on happening, so that after $\ell$ iterations of the recursion relation, we eventually obtain
$$N_\ell = \frac{1}{2\ell+1} N_0. \tag{3.58}$$
We almost have our answer, but we are still missing a value for $N_0$. But this is easy to get: we just calculate it directly as $N_0 = \int_{-1}^1 (P_0)^2\, dx = \int_{-1}^1 dx = 2$. We finally arrive at
$$N_\ell = \frac{2}{2\ell+1}, \tag{3.59}$$
which is the same statement as in Eq.(3.43).
## 3.8 Multipole Expansion
We may now return to the discussion initiated in Sec.3.1. We wish to make precise the idea captured by Fig.3.4, that any charge distribution, whether discrete or continuous, can be represented as the superposition of a monopole, dipole, quadrupole, octupole, and so on. This is the multipole expansion of electrostatics.
For this purpose we examine the electrostatic potential $V$ created by an arbitrary distribution of charge. This, you will recall, is related to the electric field by $\boldsymbol{E} = -\boldsymbol{\nabla} V$, so that the electric field can easily be constructed once we have the potential. You will also recall that the potential is given by
$$V(\boldsymbol{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\boldsymbol{r'})}{|\boldsymbol{r}-\boldsymbol{r'}|}\, dV', \tag{3.60}$$
where $\boldsymbol{r}$ designates the position at which the potential is evaluated, $\rho$ is the charge density, a function of position $\boldsymbol{r'}$ inside the distribution, and $dV' := dx' dy' dz'$ is the volume element associated with the variables contained in $\boldsymbol{r'}$. The situation is depicted in Fig.3.4.
To aid with the calculation we introduce some convenient notation. We let $r := |\boldsymbol{r}|$ be the length of the vector $\boldsymbol{r}$, $r' := |\boldsymbol{r'}|$ be the length of the vector $\boldsymbol{r'}$, and we introduce the unit vectors $\boldsymbol{n} := \boldsymbol{r}/r$ and $\boldsymbol{n'} := \boldsymbol{r'}/r'$. The angle between the vectors $\boldsymbol{r}$ and $\boldsymbol{r'}$ is denoted $\gamma$, and we have that $\boldsymbol{n} \cdot \boldsymbol{n'} = \cos\gamma$. For this discussion we shall assume that $r'$ is everywhere smaller than $r$, so that $V$ is evaluated outside the charge distribution.
The integral for the potential features $|\boldsymbol{r}-\boldsymbol{r'}|$, the distance between the points associated with $\boldsymbol{r}$ and $\boldsymbol{r'}$. The square of this is
\begin{align} |\boldsymbol{r}-\boldsymbol{r'}|^2 &= (\boldsymbol{r}-\boldsymbol{r'}) \cdot (\boldsymbol{r}-\boldsymbol{r'}) \nonumber \\ &= \boldsymbol{r} \cdot \boldsymbol{r} - 2 \boldsymbol{r} \cdot \boldsymbol{r'} + \boldsymbol{r'} \cdot \boldsymbol{r'} \nonumber \\ &= r^2 - 2rr' \cos\gamma + r^{\prime 2} \nonumber \\ &= r^2 \bigl[ 1 - 2 \cos\gamma\, (r'/r) + (r'/r)^2 \bigr],\tag{3.61} \end{align}
and we see that the function that appears inside the integral can be expressed as
$$\frac{1}{|\boldsymbol{r}-\boldsymbol{r'}|} = \frac{1}{r} \bigl[ 1 - 2 \cos\gamma\, (r'/r) + (r'/r)^2 \bigr]^{-1/2}. \tag{3.62}$$
It is this expression that reveals the link with the Legendre polynomials. Compare it with Eq.(3.1), which specifies the generating function $\Phi(x,h)$. Do you see the relation? Inspection reveals that if we set $x = \cos\gamma$ and $h = r'/r$, then
$$\frac{1}{|\boldsymbol{r}-\boldsymbol{r'}|} = \frac{1}{r} \Phi(x,h). \tag{3.63}$$
Incorporating Eq.(3.4), which defines the Legendre polynomials, we arrive at the interesting identity
$$\frac{1}{|\boldsymbol{r}-\boldsymbol{r'}|} = \frac{1}{r} \sum_{\ell=0}^\infty P_\ell(\cos\gamma)\, (r'/r)^\ell = \sum_{\ell=0}^\infty \frac{r^{\prime \ell}}{r^{\ell+1}}\, P_\ell(\cos\gamma), \tag{3.64}$$
which relates the distance function to the Legendre polynomials, expressed as functions of $x = \cos\gamma$. It is important to understand that the right-hand side of the equation depends on the variables contained in $\boldsymbol{r'}$ through both $r'$ and $\cos\gamma$.
We can evaluate the potential by inserting Eq.(3.64) into Eq.(3.60). We get
\begin{align} V(\boldsymbol{r}) &= \frac{1}{4\pi\epsilon_0} \int \rho(\boldsymbol{r'}) \Biggl[ \sum_{\ell=0}^\infty\frac{r^{\prime \ell}}{r^{\ell+1}}\, P_\ell(\cos\gamma) \Biggr]\, dV' \nonumber \\ &= \sum_{\ell=0}^\infty \frac{1}{4\pi\epsilon_0} \frac{1}{r^{\ell+1}} \int \rho(\boldsymbol{r'}) r^{\prime \ell} P_\ell(\cos\gamma)\, dV', \tag{3.65} \end{align}
where we took out everything that doesn't belong inside the integral. We put this in the final form of
$$V = \sum_{\ell=0}^\infty V_\ell, \tag{3.66}$$
with
$$V_\ell(\boldsymbol{r}) = \frac{1}{4\pi\epsilon_0} \frac{1}{r^{\ell+1}} \int \rho(\boldsymbol{r'}) r^{\prime \ell} P_\ell(\cos\gamma)\, dV'. \tag{3.67}$$
In the field's lingo, Eq.(3.66) is the multipole expansion of the electrostatic potential $V$, and each term $V_\ell$ in the infinite sum is the potential of a specific multipole. The term with $\ell = 0$ is known as the monopole term, the one with $\ell = 1$ is the dipole term, the quadrupole term corresponds to $\ell = 2$, the octupole term to $\ell = 3$, and so on. The correspondence with the physical multipoles of Fig.3.1 will be made clear as we explore this in more detail.
Let us examine the monopole term in Eq.(3.66). We set $\ell = 0$ in Eq.(3.67), recall that $P_0(\cos\gamma) = 1$, and get
$$V_0 = \frac{1}{4\pi\epsilon_0} \frac{1}{r} \int \rho\, dV'. \tag{3.68}$$
The integral is recognized as $q$, the total charge of the distribution, and we conclude that
$$V_0 = \frac{1}{4\pi\epsilon_0} \frac{q}{r}, \qquad q := \int \rho(\boldsymbol{r'})\, dV'. \tag{3.69}$$
You have seen this before: it is the potential of a single point charge $q$ situated at the origin of the coordinate system. And this is what we called a monopole in Fig.3.1.
Turn next to the dipole term in Eq.(3.66). This time we set $\ell = 1$ in Eq.(3.67), recall that $P_1(\cos\gamma) = \cos\gamma$, and get
$$V_1 = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int \rho\, r' \cos\gamma\, dV'. \tag{3.70}$$
We make this more explicit by writing $r' \cos\gamma = r' (\boldsymbol{n'} \cdot \boldsymbol{n}) = \boldsymbol{r'} \cdot \boldsymbol{n}$, and we take the vector $\boldsymbol{n}$ out of the integral, because it does not depend on the variables contained in $\boldsymbol{r'}$. This gives
$$V_1 = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \boldsymbol{n} \cdot \int \rho\, \boldsymbol{r'}\, dV', \tag{3.71}$$
and we put this into its final form by introducing the notation $\boldsymbol{p}$ for the integral. The dipole term becomes
$$V_1 = \frac{1}{4\pi\epsilon_0} \frac{\boldsymbol{n} \cdot \boldsymbol{p}}{r^2}, \qquad \boldsymbol{p} := \int \rho(\boldsymbol{r'})\, \boldsymbol{r'}\, dV', \tag{3.72}$$
and the vector $\boldsymbol{p}$ is known as the {\it dipole moment} of the charge distribution. You will recall that the potential of a physical dipole, as depicted in Fig.3.1, has precisely the form given by $V_1$, with $\boldsymbol{p}$ given by the product of the positive charge with the displacement vector from the negative to the positive charges. The expression of Eq.(3.72), however, applies to an arbitrary charge distribution, and in this context $\boldsymbol{p}$ is defined by the integral. The link with Fig.3.1 is nevertheless clear, because to the $\boldsymbol{p}$ given by the integral we can always associate a physical dipole with the same dipole moment. The multipole expansion is beginning to make sense.
Let us consider one more example, that of the quadrupole term --- we could go on and on, but this shall do. We set $\ell = 2$ in Eq.(3.67), write $P_2(\cos\gamma) = \frac{1}{2} (3\cos^2\gamma-1)$, and get
$$V_2 = \frac{1}{4\pi\epsilon_0} \frac{1}{r^3} \frac{1}{2} \int \rho\, r^{\prime 2} (3\cos^2\gamma-1)\, dV'. \tag{3.73}$$
The factor inside the integral is
$$r^{\prime 2}\bigl[ 3(\boldsymbol{n'} \cdot \boldsymbol{n})^2 - 1 \bigr] =3 (\boldsymbol{r'} \cdot \boldsymbol{n})^2 - r^{\prime 2}, \tag{3.74}$$
and to produce the simplest form for the integral we introduce the component notation $n_j$ and $r'_j$ for the vectors $\boldsymbol{n}$ and $\boldsymbol{r'}$, respectively. To be clear, we note that in the component notation, $n_1 = x/r$, $n_2 = y/r$, and $n_3 = z/r$, while $r'_1 = x'$, $r'_2 = y'$, and $r'_3 = z'$. We write $\boldsymbol{n} \cdot \boldsymbol{r'} = \sum_j n_j r'_j$, so that
$$(\boldsymbol{n} \cdot \boldsymbol{r'})^2 = \Biggl( \sum_{j=1}^3 n_j r'_j \Biggr) \Biggl( \sum_{k=1}^3 n_k r'_k \Biggr) = \sum_{j=1}^3 \sum_{k=1}^3 n_j n_k r'_j r'_k. \tag{3.75}$$
As a convenient trick, we also write
$$r^{\prime 2} = \boldsymbol{r'} \cdot \boldsymbol{r'} = \sum_{j=1}^3 r'_j r'_j = \sum_{j=1}^3 \sum_{k=1}^3 \delta_{jk}r'_j r'_k\tag{3.76}$$
so that both terms inside the integral can be expressed as a double sum. With all this we have
$$3 (\boldsymbol{r'} \cdot \boldsymbol{n})^2 - r^{\prime 2} = \sum_{j=1}^3 \sum_{k=1}^3 (3 n_j n_k - \delta_{jk}) r'_j r'_k, \tag{3.77}$$
and making the substitution inside the integral yields
$$\int \rho\, r^{\prime 2} (3\cos^2\gamma-1)\, dV' = \sum_{j=1}^3 \sum_{k=1}^3 (3 n_j n_k - \delta_{jk}) \int \rho\, r'_j r'_k\, dV'. \tag{3.78}$$
The quadrupole term in the potential can therefore be expressed as
$$V_2 = \frac{1}{4\pi\epsilon_0} \frac{1}{r^3} \frac{1}{2} \sum_{j=1}^3 \sum_{k=1}^3 (3 n_j n_k - \delta_{jk}) Q_{jk}, \qquad Q_{jk} := \int \rho(\boldsymbol{r'})\, r'_j r'_k\, dV'. \tag{3.79}$$
The quantities $Q_{jk}$ form a tensor, and they are collectively known as the quadrupole moment of the charge distribution. To be clear, each index $j$ and $k$ represents a possible choice among $x$, $y$, and $z$; for example, $Q_{xz} = \int \rho\, x' z'\, dV'$. The link with Fig.3.1 is perhaps less obvious in the quadrupole case, but it can be shown that to any quadrupole-moment tensor $Q_{jk}$ we can always associate a system of four point charges (two positive, and two negative) arranged on a rectangle, as shown in the figure. The precise geometry and orientation of the rectangle in three-dimensional space corresponds to the precise collection of components $\{ Q_{xx}, Q_{xy}, Q_{xz}, Q_{yy}, Q_{yz}, Q_{zz} \}$, and the electrostatic potential of this physical quadrupole will have precisely the form displayed in Eq.(3.79).
The multipole expansion of Eq.(3.66) is exact. It shows that the potential outside any distribution of charge is the sum of a monopole term, a dipole term, a quadrupole term, an octupole term, and so on. To each term we can associate a physical multipole, and we have made precise the idea introduced at the beginning of the chapter and illustrated in Fig.3.2, that any charge distribution can be represented as a superposition of multipoles. This is a beautiful result, and a powerful idea. But it is also a practical tool, because of an observation that we have yet to make. Examine Eq.(3.67) closely, and notice that the $\ell$-pole term in the multipole expansion of the potential comes with a factor of $1/r^{\ell+1}$. Schematically, therefore, the expansion takes the form of
$$V = \frac{\text{monopole}}{r} + \frac{\text{dipole}}{r^2} + \frac{\text{qudrupole}}{r^3} + \frac{\text{octupole}}{r^4} + \cdots. \tag{3.80}$$
For an exact representation of $V$ we need an infinite number of terms in the multipole expansion. But when $r$ is sufficiently large, we may well be satisfied with an approximation that retains only a finite number of terms. This approximate potential is much more economical, and the essential details of the charge distribution can be adequately captured by a finite number of multipole moments. At the crudest level the potential would be approximated by $\text{monopole}/r$, and the only aspect of the charge distribution that survives this description is the total charge $q$, a single number. For a better approximation we would also include $\text{dipole}/r^2$, and three more quantities --- the dipole moment $\boldsymbol{p}$ --- would be required to capture the essentials of the charge distribution. An adequate approximation may well require the inclusion of a few more terms, but the main message remains: In any truncation of the multipole expansion, the relevant aspects of the charge distribution are completely captured by a finite number of quantities packaged into multipole moments.
## 3.9 Legendre Functions
As was mentioned back in Sec.3.5, textbook presentations of Legendre polynomials often start with the differential equation
$$(1-x^2) y'' - 2x y' + \ell (\ell +1) y = 0, \tag{3.81}$$
and the polynomial $P_\ell(x)$ is shown to be a solution to the equation. This approach is analogous to one in which we would define $\sin x$ to be a solution to the differential equation $y'' + y = 0$. But we know that this equation admits a second solution, $y = \cos x$. Should we not expect Eq.(3.81) to also possess a second solution? After all, it is a known fact of mathematics that second-order differential equations admit two independent solutions.
The answer is in the affirmative. For each integer $\ell$, Eq.(3.81) admits a second solution, denoted $Q_\ell(x)$ and known as a Legendre function of the second kind. These new solutions are {\it not} polynomials. The first few are given by
\begin{align} Q_0 &= \frac{1}{2} \ln \frac{1+x}{1-x}, \tag{3.82a}\\ Q_1 &= \frac{1}{2} x \ln \frac{1+x}{1-x} - 1, \tag{3.82b}\\ Q_2 &= \frac{1}{4} (3x^2 - 1) \ln \frac{1+x}{1-x} - \frac{3}{2} x, \tag{3.82c} \end{align}
and you can see that they contain logarithms and are definitely not simple polynomials. In fact you can see more than this. Unlike the polynomials $P_\ell(x)$, which are finite everywhere on the interval $(-1,1)$, the functions $Q_\ell(x)$ are all singular at $x = 1$ and $x = -1$, thanks specifically to the logarithms. This is the main distinguishing property between the two classes of solutions to Eq.(3.81).
Exercise 3.13: Verify that $Q_0$, $Q_1$, and $Q_2$ are solutions to Eq.(3.81).
The differential equation (3.81) can be generalized by letting $\ell$ differ from an integer. In this case we would write it as
$$(1-x^2) y'' - 2x y' + \lambda(\lambda+1) y = 0,\tag{3.83}$$
and reserve the notation $\ell$ for integers. Like any other second-order differential equation, this one admits two independent solutions. The first is denoted $P_\lambda(x)$, and is known as a Legendre function of the first kind. The second is denoted $Q_\lambda(x)$, and is known, naturally enough, as a Legendre function of the second kind. In general, the functions $P_\lambda(x)$ are finite at $x=1$, but they blow up at $x=-1$ unless $\lambda$ is an integer (and the functions reduce to the familiar polynomials). The functions $Q_\lambda(x)$ are singular at both $x=-1$ and $x=1$.
This generalized view of Legendre's equation reveals that Eq.(3.81) is distinguished, in that it admits a class of solutions that is regular everywhere on the interval $(-1,1)$, namely our good old Legendre polynomials. The mere change from an integer $\ell$ to an noninteger $\lambda$ completely spoils this property.
## 3.10 Practice Problems
1. (Boas Chapter 12, Section 5, Problem 4) Derive the identity $(x-h) \partial\Phi/\partial x = h \partial\Phi/\partial h$, and find the recursion relation that follows from it.
2. (Boas Chapter 12, Section 5, Problem 9) Express $3x^2 + x - 1$ as a linear combination of Legendre polynomials.
3. (Boas Chapter 12, Section 5, Problem 10) Express $x^4$ as a linear combination of Legendre polynomials.
4. (Boas Chapter 12, Section 5, Problem 12) Express $7x^4 - 3x + 1$ as a linear combination of Legendre polynomials.
5. (Boas Chapter 12, Section 6, Problem 2) Show that the functions $e^{in\pi x/L}$, with $n = 0, \pm 1, \pm 2, \cdots$, are a set of orthogonal functions on $(-L,L)$.
6. (Boas Chapter 12, Section 6, Problem 3) Show that the functions $x^2$ and $\sin x$ are orthogonal on $(-1,1)$.
7. (Boas Chapter 12, Section 6, Problem 5) Evaluate $\int_{-1}^1 P_0(x) P_2(x)\, dx$ directly to show that $P_0(x)$ and $P_2(x)$ are orthogonal on $(-1,1)$.
8. (Boas Chapter 12, Section 7, Problem 5) Calculate $\int_{-1}^1 P_\ell(x)\, dx$ for all integers $\ell$.
9. (Boas Chapter 12, Section 8, Problem 2) Calculate $\int_{-1}^1 [P_2(x)]^2\, dx$ directly, and show that it agrees with Eq.(3.43). Repeat for $P_3(x)$.
## 3.11 Challenge Problems
1. Find an explicit expression for $P_\ell(0)$. You may wish to involve the generating function $\Phi(x,h)$ and make use of the binomial series
\begin{align} (1+z)^\alpha &= 1 + \alpha z + \frac{\alpha(\alpha-1)}{2!} z^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!} z^3 + \cdots \nonumber \\ &= \sum_{n=0}^\infty \frac{\alpha(\alpha-1) \cdots (\alpha-n+1)}{n!} z^n. \nonumber \end{align}
Simplify your result by expressing $P_\ell(0)$ in terms of two double factorials. Verify your result with the special cases $\ell = 2, 4, 6$.
2. Evaluate the integral
$\int_{-1}^1 x P_\ell(x) P_{\ell'}(x)\, dx$
for arbitrary values of $\ell$ and $\ell'$. Verify your result with the special cases $\ell = 2$ and $\ell' = 0, 1, 2, 3, 4$.
3. The octupole ($\ell=3$) term in a multipole expansion of the electrostatic potential is given by
$V_3 = \frac{1}{4\pi\epsilon_0} \frac{1}{r^4} \int\rho(\boldsymbol{r'})\, r^{\prime 3} P_3(\cos\gamma)\, dV',$
where $\cos\gamma := \boldsymbol{n} \cdot \boldsymbol{n'}$, with $\boldsymbol{n} := \boldsymbol{r}/r$ and $\boldsymbol{n'} := \boldsymbol{r'}/r'$. Express this octupole potential in terms of the octupole moment tensor of the charge distribution,
$Q_{ijk} := \int \rho(\boldsymbol{r'})\, r'_i r'_j r'_k\, dV',$
where $r'_i$ are the components of the position vector $\boldsymbol{r'}$. Each index $i$, $j$, and $k$ ranges over the values $x$, $y$, and $z$. | 16,362 | 48,622 | {"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": 2, "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": 12, "equation": 72, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-26 | latest | en | 0.916461 |
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Master Course. MSc Bioinformatics for Health Sciences H15: Algorithms on strings and sequences Xavier Messeguer Peypoch (http://www.lsi.upc.es/~alggen) Dep. de Llenguatges i Sistemes Informàtics CEPBA-IBM Research Institute Universitat Politècnica de Catalunya. Contents.
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Master Course
MSc Bioinformatics for
Health Sciences
H15: Algorithms on strings and sequences
Xavier Messeguer Peypoch (http://www.lsi.upc.es/~alggen)
Dep. de Llenguatges i Sistemes Informàtics
CEPBA-IBM Research Institute
Universitat Politècnica de Catalunya
Contents
1. (Exact) String matching of one pattern
2. (Exact) String matching of many patterns
3. Extended string matching and regular expressions
4. Approximate string matching (Dynamic programming)
5. Pairwise and multiple alignment
6. Suffix trees
Master Course
Second lecture:
First part:
Extended string matching
Extended string matching
There are classes of characters represented by one
Symbol. For instace the IUPAC code for the
DNA alphabet is:
R = {G,A} Y = {T,C} K = {G,T} M = {A,C} S = {G,C} W = {A,T}
B = {G,T,C } D = {G,A,T} H = {A,C,T} V = {G,C,A} N = {A,G,C,T} (any)
1. Classes of characters in the tetx.
There are characters in the text that
represent sets of simbols
2. Classes of characters in the pattern.
There are characters in the text that
represent sets of simbols
Classes in the text
Algorismes més eficients (Navarro & Raffinot)
| |
64
32
16
Horspool
8
BOM
BNDM
4
Long. patró
2
w
2 4 8 16 32 64 128 256
Classes in the text :Horspool example
A 4
C 5
G 2
T 1
R ?
N ?
Given the pattern ATGTA
the shift table is:
Classes in the text :Horspool example
A 4
C 5
G 2
T 1
R 2
N ?
Suposem que el patró és ATGTA
La taula de salts seria:
Classes in the text :Horspool example
Given the taxt :
G T A R T R N A A G G A …
A T G T A
A T G T A
A T G T A
A 4
C 5
G 2
T 1
R 2
N 1
Given the pattern ATGTA
and the shift table:
Classes in the text :Horspool example
IGiven the text :
G T A R T R N A A G G A ...
A T G T A
A T G T A
A T G T A
A T G T A
A 4
C 5
G 2
T 1
R 2
N 1
Given the pattern ATGTA
and the shift table:
Classes in the text
Algorismes més eficients (Navarro & Raffinot)
BNDM : Backward Nondeterministic Dawg Matching
| |
BOM : Backward Oracle Matching
64
32
16
Horspool
8
BOM
BNDM
4
Long. patró
2
w
2 4 8 16 32 64 128 256
Alg. Cerca exacta d’un patró (text on-line)
Algorismes més eficients (Navarro & Raffinot)
BNDM : Backward Nondeterministic Dawg Matching
| |
BOM : Backward Oracle Matching
64
32
16
Horspool
8
BOM
BNDM
4
Long. patró
2
w
2 4 8 16 32 64 128 256
Classes in the text: BOM
Com fa la comparació?
Text :
Patró :
Autòmata: Factor Oracle
Com es determina la següent posició de la finestra?
Comproba si el sufix és factor del patró
Però primer analitzem com fa la comparació…
Classes in the text: BOM example
G
T
A
G
T
T
A
G
T
A
I la cerca sobre el text :
G T A R T R N A A T G…
Com fa la comparació?
Es construeix l’autòmata del patró invers: Suposem que el patró és ATGTATG
A T G T A T G
No és possible cap millora!
Alg. Cerca exacta de molts patrons
8
| |
(5 mots)
Wu-Manber
4
SBOM
Long. mínima
2
5 10 15 20 25 30 35 40 45
8
Wu-Manber
(10 mots)
(100 mots)
4
SBOM
8
Wu-Manber
Ad AC
2
SBOM
4
5 10 15 20 25 30 35 40 45
Ad AC
2
5 10 15 20 25 30 35 40 45
Wu-Manber
8
(1000 mots)
SBOM
4
Ad AC
2
5 10 15 20 25 30 35 40 45
Classes in the text: Set Horspool
G
T
A
T
A
T
G
G
T
A
T
A
A
T
A
A
Search for the patterns ATGTATG,TATG,ATAAT,ATGTG
In the text: ARTGNCTATGTGACA…
<it’s not possible any improvment!
Classes in the text
8
| |
(5 mots)
Wu-Manber
4
SBOM
Long. mínima
2
5 10 15 20 25 30 35 40 45
8
Wu-Manber
(10 mots)
(100 mots)
4
SBOM
8
Wu-Manber
Ad AC
2
SBOM
4
5 10 15 20 25 30 35 40 45
Ad AC
2
5 10 15 20 25 30 35 40 45
Wu-Manber
8
(1000 mots)
SBOM
4
Ad AC
2
5 10 15 20 25 30 35 40 45
Classes in the pattern
Algorismes més eficients (Navarro & Raffinot)
| |
64
32
16
Horspool
8
BOM
BNDM
4
Long. patró
2
w
2 4 8 16 32 64 128 256
Classes in the text
8
| |
(5 mots)
Wu-Manber
4
SBOM
Long. mínima
2
5 10 15 20 25 30 35 40 45
8
Wu-Manber
(10 mots)
(100 mots)
4
SBOM
8
Wu-Manber
Ad AC
2
SBOM
4
5 10 15 20 25 30 35 40 45
Ad AC
2
5 10 15 20 25 30 35 40 45
Wu-Manber
8
(1000 mots)
SBOM
4
Ad AC
2
5 10 15 20 25 30 35 40 45
Alg. Cerca exacta de molts patrons
8
| |
(5 mots)
Wu-Manber
4
SBOM
Long. mínima
2
5 10 15 20 25 30 35 40 45
8
Wu-Manber
(10 mots)
(100 mots)
4
SBOM
8
Wu-Manber
Ad AC
2
SBOM
4
5 10 15 20 25 30 35 40 45
Ad AC
2
5 10 15 20 25 30 35 40 45
Wu-Manber
8
(1000 mots)
SBOM
4
Ad AC
2
5 10 15 20 25 30 35 40 45
Alg. Cerca exacta de molts patrons
8
| |
(5 mots)
Wu-Manber
4
SBOM
Long. mínima
2
5 10 15 20 25 30 35 40 45
8
Wu-Manber
(10 mots)
(100 mots)
4
SBOM
8
Wu-Manber
Ad AC
2
SBOM
4
5 10 15 20 25 30 35 40 45
Ad AC
2
5 10 15 20 25 30 35 40 45
Wu-Manber
8
(1000 mots)
SBOM
4
Ad AC
2
5 10 15 20 25 30 35 40 45
Master Course
Second lecture:
Second part:
Regular expressions matching
Expressions regulars
Una expressió regular ℛés una cadena sobre
ΣU { ε, |, · , * , (, ) } definida recursivament com:
ε és una expressió regular
Un caràcter de Σés una expressió regular
( ℛ ) és una expressió regular
ℛ1 ·ℛ2és una expressió regular
ℛ1 |ℛ2és una expressió regular
ℛ *és una expressió regular
Llenguatge regular
El llenguatge representat per una expressió regular és el conjunt dels mots que es poden construir a partir de l’expressió regular.
El problema de buscar una expressió regular dins el text és el de buscar tots els factors que pertanyen al respectiu llenguatge regular.
Master Course
Second lecture:
Third part:
Approximate string matching
Approximate string matching
For instance, given the sequence
CTACTACTACGTCTATACTGATCGTAGCTACTACATGC
search for the pattern ACTGA allowing one error…
… but what is the meaning of “one error”?
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)= d(ACT,AC)= d(ACT,C)=
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= 3 d(AC,ATC)=1 d(ACTTG,ATCTG)=2
Edit distance and alignment of strings
The Edit distance is related with the best alignment of strings
Given
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACTTG,ATCTG)=2
which is the best alignment in every case?
• ACT and ACT : ACT
ACT
• ACT and AC: ACT
AC-
ACTTG and ATCTG:
ACTTG
ATCTG
ACT - TG
A - TCTG
Edit distance and alignment of strings
But which is the distance between the strings
ACGCTATGCTATACG and ACGGTAGTGACGC?
… and the best alignment between them?
1966 was the first time this problem was discussed…
and the algorithm was proposed in 1968,1970,…
using the technique called “Dynamic programming”
Approximate string matching
For instance, given the sequence
CTACTACTACGTCTATACTGATCGTAGCTACTACATGC
search for the pattern ACTGA allowing one error…
… but what is the meaning of “one error”?
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)= d(ACT,AC)= d(ACT,C)=
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Approximate string matching
For instance, given the sequence
CTACTACTACGTCTATACTGATCGTAGCTACTACATGC
search for the pattern ACTGA allowing one error…
… but what is the meaning of “one error”?
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)= d(ACT,AC)= d(ACT,C)=
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= 3 d(AC,ATC)=1 d(ACTTG,ATCTG)=2
Edit distance and alignment of strings
The Edit distance is related with the best alignment of strings
Given
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACTTG,ATCTG)=2
which is the best alignment in every case?
• ACT and ACT : ACT
ACT
• ACT and AC: ACT
AC-
ACTTG and ATCTG:
ACTTG
ATCTG
ACT - TG
A - TCTG
Edit distance and alignment of strings
But which is the distance between the strings
ACGCTATGCTATACG and ACGGTAGTGACGC?
… and the best alignment between them?
1966 was the first time this problem was discussed…
and the algorithm was proposed in 1968,1970,…
using the technique called “Dynamic programming”
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= 3 d(AC,ATC)=1 d(ACTTG,ATCTG)=2
Edit distance and alignment of strings
The Edit distance is related with the best alignment of strings
Given
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACTTG,ATCTG)=2
which is the best alignment in every case?
• ACT and ACT : ACT
ACT
• ACT and AC: ACT
AC-
ACTTG and ATCTG:
ACTTG
ATCTG
ACT - TG
A - TCTG
Edit distance and alignment of strings
But which is the distance between the strings
ACGCTATGCTATACG and ACGGTAGTGACGC?
… and the best alignment between them?
1966 was the first time this problem was discussed…
and the algorithm was proposed in 1968,1970,…
using the technique called “Dynamic programming”
Edit distance and alignment of strings
C T A C T A C T A C G T
A
C
T
G
A
Edit distance and alignment of strings
C T A C T A C T A C G T
A
C
T
G
A
Edit distance and alignment of strings
C T A C T A C T A C G T
A
C
T
G
A
The cell contains the distance between AC and CTACT.
Approximate string matching
For instance, given the sequence
CTACTACTACGTCTATACTGATCGTAGCTACTACATGC
search for the pattern ACTGA allowing one error…
… but what is the meaning of “one error”?
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)= d(ACT,AC)= d(ACT,C)=
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= 3 d(AC,ATC)=1 d(ACTTG,ATCTG)=2
Edit distance and alignment of strings
The Edit distance is related with the best alignment of strings
Given
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACTTG,ATCTG)=2
which is the best alignment in every case?
• ACT and ACT : ACT
ACT
• ACT and AC: ACT
AC-
ACTTG and ATCTG:
ACTTG
ATCTG
ACT - TG
A - TCTG
Edit distance and alignment of strings
But which is the distance between the strings
ACGCTATGCTATACG and ACGGTAGTGACGC?
… and the best alignment between them?
1966 was the first time this problem was discussed…
and the algorithm was proposed in 1968,1970,…
using the technique called “Dynamic programming”
Approximate string matching
For instance, given the sequence
CTACTACTACGTCTATACTGATCGTAGCTACTACATGC
search for the pattern ACTGA allowing one error…
… but what is the meaning of “one error”?
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)= d(ACT,AC)= d(ACT,C)=
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Approximate string matching
For instance, given the sequence
CTACTACTACGTCTATACTGATCGTAGCTACTACATGC
search for the pattern ACTGA allowing one error…
… but what is the meaning of “one error”?
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)= d(ACT,AC)= d(ACT,C)=
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= 3 d(AC,ATC)=1 d(ACTTG,ATCTG)=2
Edit distance and alignment of strings
The Edit distance is related with the best alignment of strings
Given
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACTTG,ATCTG)=2
which is the best alignment in every case?
• ACT and ACT : ACT
ACT
• ACT and AC: ACT
AC-
ACTTG and ATCTG:
ACTTG
ATCTG
ACT - TG
A - TCTG
Edit distance and alignment of strings
But which is the distance between the strings
ACGCTATGCTATACG and ACGGTAGTGACGC?
… and the best alignment between them?
1966 was the first time this problem was discussed…
and the algorithm was proposed in 1968,1970,…
using the technique called “Dynamic programming”
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= d(AC,ATC)= d(ACTTG,ATCTG)=
Edit distance
Indel
We accept three types of errors:
1. Mismatch: ACCGTGAT ACCGAGAT
2. Insertion: ACCGTGAT ACCGATGAT
3. Deletion: ACCGTGAT ACCGGAT
The edit distance d between two strings is the
minimum number of
substitutions,insertions and deletions
needed to transform the first string into the second one
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACT,C)=2
d(ACT,)= 3 d(AC,ATC)=1 d(ACTTG,ATCTG)=2
Edit distance and alignment of strings
The Edit distance is related with the best alignment of strings
Given
d(ACT,ACT)=0 d(ACT,AC)=1 d(ACTTG,ATCTG)=2
which is the best alignment in every case?
• ACT and ACT : ACT
ACT
• ACT and AC: ACT
AC-
ACTTG and ATCTG:
ACTTG
ATCTG
ACT - TG
A - TCTG
Edit distance and alignment of strings
But which is the distance between the strings
ACGCTATGCTATACG and ACGGTAGTGACGC?
… and the best alignment between them?
1966 was the first time this problem was discussed…
and the algorithm was proposed in 1968,1970,…
using the technique called “Dynamic programming”
Edit distance and alignment of strings
C T A C T A C T A C G T
A
C
T
G
A
Edit distance and alignment of strings
C T A C T A C T A C G T
A
C
T
G
A
Edit distance and alignment of strings
C T A C T A C T A C G T
A
C
T
G
A
The cell contains the distance between AC and CTACT.
Edit distance and alignment of strings
C T A C T A C T A C G T
A
C
T
G
A
?
Edit distance and alignment of strings
C T A C T A C T A C G T
0
A
C
T
G
A
?
Edit distance and alignment of strings
C T A C T A C T A C G T
0 1
A
C
T
G
A
?
-
C
Edit distance and alignment of strings
C T A C T A C T A C G T
0 1 2
A
C
T
G
A
?
- -
CT
Edit distance and alignment of strings
C T A C T A C T A C G T
0 1 2 3 4 5 6 7 8 …
A
C
T
G
A
- - - - - -
CTACTA
Edit distance and alignment of strings
C T A C T A C T A C G T
0 1 2 3 4 5 6 7 8 …
A ?
C ?
T ?
G
A
Edit distance and alignment of strings
C T A C T A C T A C G T
0 1 2 3 4 5 6 7 8 …
A 1
C 2
T 3
G…
A
ACT
- - -
Edit distance and alignment of strings
-
C
C
C
C
-
BA(AC,CTA)
BA(A,CTA)
BA(A,CTAC)
C T A C T A C T A C G T
0 1 2 3 4 5 6 7 8 …
A 1
C 2
T 3
G
A
C T A C T A C T A C G T
A
C
T
G
A
d(AC,CTA)+1
d(A,CTA)
BA(AC,CTAC)= best
d(AC,CTAC)=min
d(A,CTAC)+1
Edit distance and alignment of strings
Connect to
http://alggen.lsi.upc.es/docencia/ember/leed/Tfc1.htm
and use the global method.
Edit distance and alignment of strings
How this algorithm can be applied
to the approximate search?
to the K-approximate string searching?
K-approximate string searching
C T A C T A C T A C G T A C T G G T G A A …
A
C
T
G
A
This cell …
K-approximate string searching
C T A C T A C T A C G T A C T G G T G A A …
A
C
T
G
A
This cell gives the distance between (ACTGA, CT…GTA)…
…but we only are interested in the last characters
K-approximate string searching
C T A C T A C T A C G T A C T G G T G A A …
A
C
T
G
A
This cell gives the distance between (ACTGA, CT…GTA)…
…but we only are interested in the last characters
Master Course
Second lecture:
Fourth part:
Pairwise and multiple alignment
Bioinformatics
Pairwise and multiple alignment
Pairwise alignment
+
-
s(A,CTAC)-2
s(AC,CTACT)=maximum s(A,CTA) 1
s(AC,CTA)-2
Edit distance:
match=0 mismatch=1 indel=1
d(A,CTAC)+1
d(AC,CTACT)=minimum d(A,CTA)….+1
d(AC,CTA)+1
Similarity:
match=1 mismatch=-1 indel=-2
Pairwise alignment
Connect to
http://alggen.lsi.upc.es
Links to TEACHING EMBER LePA
Pairwise to multiple alignment
S2
A
C
A
-1
S3
__
S1
What happens with three strings?
Let n be their lenght, then the cost becomes
O(n3)
O(23)
O(32)
And with k strings?
O(nk 2k k2)
Multiple alignment
Programs of multialignment use different heuristics:
• Clustal (Progressive alignment)
http://www.ebi.ac.uk/clustalw
• TCoffee (Progressive alignment + data bases)
http://igs-server.cnrs-mrs.fr/Tcoffee_cgi/index.cgi
• HMM (Hidden Markov Models)
Multiple alignment
Connect to
http://alggen.lsi.upc.es/
and follow the links TEACHING EMBER. | 7,434 | 22,026 | {"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-2017-43 | latest | en | 0.64562 |
http://softmath.com/algebra-software-3/converting-a-mixed-fraction-to.html | 1,547,995,737,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583722261.60/warc/CC-MAIN-20190120143527-20190120165527-00107.warc.gz | 203,081,746 | 12,999 | English | Español
# Try our Free Online Math Solver!
Online Math Solver
Depdendent Variable
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Number of inequalities to solve: 23456789
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Author Message
Ielke
Registered: 20.12.2001
From: NL, Bussum
Posted: Tuesday 07th of Mar 07:35 Hey, This morning I began working on my mathematics homework on the topic Basic Math. I am currently unable to finish the same because I am unfamiliar with the fundamentals of 3x3 system of equations, y-intercept and simplifying expressions. Would it be possible for anyone to aid me with this?
oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Wednesday 08th of Mar 07:36 Hi, I think that I can to help you out. Have you ever tried out a program to assist you with your algebra assignments? Some time ago I was also stuck on a similar issues like you, but then I came across Algebrator. It helped me a lot with converting a mixed fraction to a decimal calculator and other math problems, so since then I always count on its help! My algebra grades got better thanks to the help of Algebrator.
Vild
Registered: 03.07.2001
From: Sacramento, CA
Posted: Friday 10th of Mar 09:05 Yes I agree, Algebrator is a really useful product . I bought it a few months back and I can say that it is the main reason I am passing my math class. I have recommended it to my friends and they too find it very useful. I strongly recommend it to help you with your math homework.
Registered: 17.08.2003
From:
Posted: Friday 10th of Mar 12:37 Wow! This sounds tempting. I would like to try the program. Is it expensive? Where can I find it?
Jrahan
Registered: 19.03.2002
From: UK
Posted: Sunday 12th of Mar 08:10 Here https://softmath.com/links-to-algebra.html. Happy problem solving! | 658 | 2,428 | {"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} | 2.59375 | 3 | CC-MAIN-2019-04 | latest | en | 0.9064 |
https://www.projectrhea.org/rhea/index.php/ECE-QE_AC2-2016 | 1,582,514,930,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00218.warc.gz | 823,089,327 | 8,389 | 2016 AC-2 P1. (a) $X=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix} y \\ \dot{y} \end{bmatrix}$
$\begin{cases} \dot{x}=\begin{bmatrix} \dot{x_1} \\ \dot{x_2} \end{bmatrix}=\begin{bmatrix} x_2 \\ -2x_1 x_2-u x_1+2u \end{bmatrix}\\ y=x_1 \end{cases}$
(b) $u \equiv 2$
$\dot{x} =\begin{bmatrix} x_2 \\ -2x_1 x_2-2x_1+4 \end{bmatrix}=\begin{bmatrix} x_2 \\ -2x_1 (x_2+1)+4 \end{bmatrix}$
let $\begin{cases} -2x_1 (x_2+1)+4=0 \\ x_2=0 \end{cases} \Rightarrow \begin{cases} -2x_1 +4=0 \\ x_2=0 \end{cases} \Rightarrow \begin{cases} x_1=2 \\ x_2=0 \end{cases}$
$\therefore The \; equilibrum\; point\; is \;x_e=\begin{bmatrix} 2 \\ 0 \end{bmatrix}$
(c) $u \equiv 2 \quad x_e=\begin{bmatrix} 2 \\ 0 \end{bmatrix}, \quad let \;x=f(x)$
The Jacobin of $\dot{x}$ is: \begin{align} Df(x)= \begin{bmatrix} 0 & 1 \\ -2x_1-2 & -2x_1 \end{bmatrix} \end{align}
The linear dynamics around $x_e$ is $\frac{d}{dt}f(x)=\begin{bmatrix} 0 & 1 \\ -2 & -4 \end{bmatrix} f(x)$
which is stable, locally stable at $x_e$.
P2. i) $x[k+1]=A x[k]$
$y[k]=\begin{bmatrix} -1 & 1 \end{bmatrix}x[k]$
let $x_[k]=\begin{bmatrix} a\\ b \end{bmatrix} \quad y[0]=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}=1$
$-a+b=1$
$y[1]=\begin{bmatrix} -1 & 1 \end{bmatrix}x[1]=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix}x[0]=0$
$x[1]=Ax[0]$
$3a+2b=0$
$\therefore a=-\frac{2}{5} \quad b=\frac{3}{5}$
$x[0]=\begin{bmatrix} -\frac{2}{5}\\ \frac{3}{5} \end{bmatrix}$
ii)$x[0]=\begin{bmatrix} a\\ b \end{bmatrix} \quad x[2]=Ax[1]=A^2x[0]$
$A^2=0 \quad x[2]=0$
$y[2]=[-1 \quad 1]\quad x[2]=0 \quad y[1]=[-1 \quad 1] \quad x[1]=1$
$[-1\quad 1]\quad \begin{bmatrix} 2 &4\\ -1&2 \end{bmatrix}\quad\begin{bmatrix} a\\ b \end{bmatrix}=1$
we only have -3a-2b=1,so we can't uniquely determine a,b.
P3 (a)$\lambda I-A=\begin{bmatrix} \lambda+2&-4\\ 1&\lambda-2 \end{bmatrix}$
$\lambda_1=\lambda_2=0$
$\begin{bmatrix} -2-\lambda_1 & 4\\ -1 & 2-\lambda_1 \end{bmatrix}\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$
$\begin{cases} -2u_1-\lambda_1u_1+4u_2=0\\ -u_1+2u_2-\lambda_1u_2=0 \end{cases}$
$u_1=2u_2$
$\therefore eigenvector \begin{bmatrix} u_1\\ u_2 \end{bmatrix}=\begin{bmatrix} 2\\ 1 \end{bmatrix}$
$J=MAM^{-1}$
(b)$e^{At}=L^{-1}\begin{bmatrix} (SI-A)^{-1} \end{bmatrix}=L^{-1}\begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix}$
$L^{-1}\begin{bmatrix} \frac{s-2}{s^2} \end{bmatrix}= L^{-1}\begin{bmatrix} \frac{1}{s}-\frac{2}{s^2} \end{bmatrix}=1-2t$
$L^{-1}\begin{bmatrix} \frac{4}{s^2} \end{bmatrix}=4t$
$L^{-1}\begin{bmatrix} \frac{-1}{s^2} \end{bmatrix}=-t$
$L^{-1}\begin{bmatrix} \frac{s+2}{s^2} \end{bmatrix}=L^{-1}\begin{bmatrix} \frac{1}{s}+\frac{2}{s^2} \end{bmatrix}=1+2t$
$e^{At}=\begin{bmatrix} 1-2t & 4t\\ -t & 1+2t \end{bmatrix}$
(c)T(s)=$C(SI-A)^{-1}B$
=$[-1 \quad 1] \quad \begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix} \quad \begin{bmatrix} 2\\ 1 \end{bmatrix}$
$=\frac{\begin{bmatrix} -1 & 1 \end{bmatrix}}{s^2}\begin{bmatrix} 2s-4+4 \\ -2+s+2 \end{bmatrix}$
$T(s)=\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} \frac{2}{s} \\ \frac{1}{s} \end{bmatrix}=\frac{-2}{s}+\frac{1}{s}$
$T(s)=\frac{-1}{s}$
pole is at s=0.
$\therefore marginally \; stable.$
(d) Given $\dot{x}=Ax+Bu$
$Ax=\begin{bmatrix} -2 & u \\ -1 & 2 \end{bmatrix} x,\quad Bu=\begin{bmatrix} 2 \\ 1 \end{bmatrix} u$
$C=\begin{bmatrix} B & AB & A^2B \end{bmatrix}=\begin{bmatrix} 2 & 0 \\ 1 & 0 \end{bmatrix}$
$\therefore not \; controllable$
To make $x(1)=\begin{bmatrix} -u & -2 \end{bmatrix}$
corres ponding characteristic equation.
$(s+u)(s+2)=0$
$s^2+6s+8=0$
$u=-kx$
$\dot{x}=(A-Bk) x$
$\begin{vmatrix} SI-(A-Bk) \end{vmatrix}=0$
$A-BK=\begin{bmatrix} -2 & 4\\ -1 & 2 \end{bmatrix}-\begin{bmatrix} 2K_1 \\ 1K_2 \end{bmatrix}$
$\begin{vmatrix} S+2+2k_1 & -4 \\ 1+k_2 & S-2 \end{vmatrix}=0 \qquad \begin{bmatrix} A-Bk \end{bmatrix}=\begin{bmatrix} -2-2k_1 & 4 \\ -1-k_2 & 2 \end{bmatrix}$
$(S+2+2k_1)(S-2)+4(1+k_2)=0$
$2k_1=6$
$k_1=3$
$uk_2=20$
$k_2=5$
$\therefore$ control $u_{(t)}=-\begin{bmatrix} k_1 & k_2 \end{bmatrix} x$
or $u_{(t)}=-\begin{bmatrix} 3 & 5 \end{bmatrix} x$
our goal is to have $x_{(1)}= -\begin{bmatrix} 1 & 2 \end{bmatrix}$
$(s-1)(s+2)=0$
$2k_1=-3$
$k_1=-\frac{3}{2}$
$uk_2=-4$
$2k_1=-3$
$k_2=-1$
$\therefore u_{(t)}=-\begin{bmatrix} k_1 & k_2 \end{bmatrix} x =\begin{bmatrix} -\frac{3}{2} & -1 \end{bmatrix} x$
e) consider $\dot{x}=A_{C1}X$
$A_{C1}=A-Bk$
$s^2+2k_1s-uk_1+uk_2=0$
$1+ \frac{C_t(s)}{t(s)}=0$
$\frac{C_t(s)}{t(s)}=s^2+2k_1s-uk_1+uk_2-1$
$\lim_{s \to 0} s\frac{C_t(s)}{t(s)}=\lim_{s \to 0} s(s^2+2k_1s-uk_1+uk_2-1)$
Final value is zero and no such controller is possible to design.
But for Bounded time period,it is possible to have solution of x(t). As we have already seen in previous problem.
$\alpha_1+\alpha_2=2k_1, \quad \alpha_1\alpha_2=-uk_1+uk_2.$
$(s+\alpha_1)(s+\alpha_2)=0$
$\begin{cases} \alpha^2+(\alpha_1+\alpha_2)s+\alpha_1\alpha_2=0 \\ s^2+2k_1s-uk_1+uk_2=0 \end{cases}$
(f) $\dot{x}=\begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix} x+\begin{bmatrix} 2 \\ 1 \end{bmatrix} u$
$y=\begin{bmatrix} -1 & 1 \end{bmatrix} x \quad D=[0]$
$N=\begin{bmatrix} C \\ CA \end{bmatrix}=\begin{bmatrix} -1 & 1 \\ 1 & -2 \end{bmatrix}$
The system is observable.
(g) $\frac{Y(S)}{U(S)}=C[SI-A]^{-1}B+D$
$SI-A=\begin{bmatrix} S & 0 \\ 0 & S \end{bmatrix}-\begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix}$
$[SI-A]^{-1}=\frac{1}{S^2-4+4}\begin{bmatrix} S-2 & 4 \\ -1 & S+2 \end{bmatrix}$
$\frac{Y(s)}{u(s)}==\begin{bmatrix} -1 & 1 \end{bmatrix}\begin{bmatrix} \frac{s-2}{s^2} & \frac{4}{s^2} \\ \frac{-1}{s^2} & \frac{s+2}{s^2} \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix}$
$H (s)= \frac{-1}{s}$ marginally stable ,not BIBO.
(h) Cant not resolve.
## Alumni Liaison
Basic linear algebra uncovers and clarifies very important geometry and algebra.
Dr. Paul Garrett | 2,912 | 6,022 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-10 | latest | en | 0.115072 |
https://ifcglobalnetwork.com/qa/what-number-means-evil.html | 1,606,270,470,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00051.warc.gz | 323,174,162 | 9,279 | # What Number Means Evil?
## What number symbolizes death?
Because 4 is generally a practical, material number, few superstitions are associated with it.
An exception is in China, where 4 is unlucky because she (“four”) and shi (“death”) sound similar..
## Is 9 a divine number?
The number 9 is revered in Hinduism and considered a complete, perfected and divine number because it represents the end of a cycle in the decimal system, which originated from the Indian subcontinent as early as 3000 BC.
## What animal symbolizes peace?
doveThe use of a dove as a symbol of peace originated with early Christians, who portrayed baptism accompanied by a dove, often on their sepulchres.
## What is nine as a number?
Nine is the Arabic number which comes after 8 and before 10. It is an odd number, and is the highest single-digit number. It is also a square number. In Roman numerals, nine can be written as IX.
## Is 7 a lucky number in China?
The number 7 (七, pinyin: qī) in Mandarin sounds like “even” in Mandarin (齊, pinyin: qí), so it is a good number for relationships. It also sounds like “arise” (起, pinyin: qǐ) and “life essence” (氣, pinyin: qì) in Mandarin. Seven can also be considered an unlucky number since the 7th month (July) is a “ghost month”.
## Why is 12 a powerful number?
Twelve is the smallest abundant number, since it is the smallest integer for which the sum of its proper divisors (1 + 2 + 3 + 4 + 6 = 16) is greater than itself. Twelve is a sublime number, a number that has a perfect number of divisors, and the sum of its divisors is also a perfect number.
## Is 3 a bad number?
The number 3 can be unlucky as well depending on the situation and use. For example, gifting to friends or to couples seldom contains the number 3 in any form of association. Three is pronounced ‘san’ which is similar to the word that means ‘to part ways’.
## Is death a symbol?
The human skull is an obvious and frequent symbol of death, found in many cultures and religious traditions. … Less blunt symbols of death frequently allude to the passage of time and the fragility of life, and can be described as memento mori; that is, an artistic or symbolic reminder of the inevitability of death.
## What is the Greek symbol for death?
In classical Athens, it was used as an abbreviation for the Greek θάνατος (thanatos, “death”) and as it vaguely resembles a human skull, theta was used as a warning symbol of death, in the same way that skull and crossbones are used in modern times.
## What animal represents grief?
Marc Bekoff, a scientist, has spent his time researching emotions in animals, including grief. Combined with other research, the following animals have been seen to grieve: wolves, chimpanzees, magpies, elephants, dolphins, otters, geese, sea lions, and many more.
## Is 8 a lucky number?
The number eight is considered to be a lucky number in Chinese and other Asian cultures. Eight (八; accounting 捌; pinyin bā) is considered a lucky number in Chinese culture because it sounds like the word meaning to generate wealth (發(T) 发(S); Pinyin: fā). Property with the number 8 may be valued greatly by Chinese.
## Why is there no 4th floor in hospitals?
Floor 4 is missing because of the very similar pronunciation of “four” and “death” in Mandarin Chinese. Floor 13 is missing due to triskaidekaphobia. Floor 14 is missing because 4 is included in 14. Note that there is a “negative first” floor.
## What is the number before eleven?
11 (eleven) is the natural number following 10 and preceding 12. It is the first repdigit.
## Why is 888 lucky?
366, 666, 888 & 1666 These are the lucky number for Chinese New Year, or specifically the quantity of cash Chinese usually put in the Red Envelope. It is a blessing from the elders to the younger generation, which hopes that they will have a smooth journey in life.
## Is 18 a lucky number?
As lucky or unlucky number In Chinese tradition, 18 pronounced 十八 (shí bā) and is considered a lucky number due to similarity with 實發 (shì fā) ‘definitely get rich’, ‘to get rich for sure’.
## Is 13 a lucky number in China?
China: 13 is traditionally considered a lucky number in China. This is because its pronunciation sounds similar to the words for “assured growth” or “definitely vibrant”.
## What is God’s number?
Political significance The number refers to a triumph of “God’s number” 7 over the Devil’s number 666. On the AWB flag, the numbers are arranged in a triskelion shape, resembling the Nazi hakenkreuz.
## Why is 4 unlucky?
But the number four is considered unlucky because it sounds a lot like the word for “death,” and as a result Chinese buildings often lack a fourth floor (just as American buildings sometimes skip the 13th). Likewise, Chinese drivers avoid license plates ending in four. | 1,148 | 4,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-50 | latest | en | 0.946336 |
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1. Change 1.569 to a percentA) 15.69%B) 1.569%C) 1569%D) 156.9%2. Change 1569 % to a decimalA) 1569B) 1.569C) 15.69D) 156.93. Change .501 to a percentA) 501%B) 50.1%C) 105%D) 5.01%4. Change 50.1% to a decimalA) .105B) 1.05C) 112D) .5015. Change 96.115 to a percentA) 961.15%B) 9.6115%C) 96.115%D) 9611.5%6. Change 72.34% to a decimalA) 72.34B) 7234C) .7234D) 7.2347. Change .024 to a percentA) 24%B) 2.4%C) .024%D) 240%8. Change 265% to a decimalA) 26.5B) .265C) 265D) 2.659. Change .225 to a percentA) 22.5%B) 2.25%C) .225%D) 225%10. Change 50% to a decimalA) 5.0B) 50C) .50D) 550
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A210412 Number of (n+1)X8 0..3 arrays with every 2X2 subblock having three or four distinct values, and new values 0..3 introduced in row major order 1
13035456, 84204544256, 525218223843536, 3289463792785895520, 20592266925119437726832, 128916198765144572103220336, 807064033391096001306908915584, 5052529194034453788657081337165632 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 7 of A210413 LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 EXAMPLE Some solutions for n=4 ..0..0..0..0..0..0..1..2....0..0..0..0..0..0..0..0....0..0..1..2..0..3..3..1 ..1..2..1..3..1..3..3..1....1..2..3..2..1..2..3..2....2..1..3..0..1..0..1..0 ..1..3..0..1..0..0..1..0....3..3..0..3..0..0..1..2....3..3..2..3..0..2..3..0 ..0..1..1..3..3..1..3..1....0..2..0..2..1..2..1..3....0..1..0..2..2..3..0..1 ..1..3..2..1..2..1..0..3....1..3..2..1..3..0..3..2....1..2..3..3..0..3..1..3 CROSSREFS Sequence in context: A256734 A237034 A210182 * A048937 A257269 A283274 Adjacent sequences: A210409 A210410 A210411 * A210413 A210414 A210415 KEYWORD nonn AUTHOR R. H. Hardin Mar 21 2012 STATUS approved
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Last modified October 16 21:10 EDT 2019. Contains 328103 sequences. (Running on oeis4.) | 605 | 1,582 | {"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-2019-43 | latest | en | 0.568885 |
https://www.khanacademy.org/math/cc-fourth-grade-math/cc-4th-fractions-topic/cc-4th-mult-whole-number-frac/e/understanding-multiplying-fractions-and-whole-numbers | 1,474,912,217,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738660871.1/warc/CC-MAIN-20160924173740-00173-ip-10-143-35-109.ec2.internal.warc.gz | 942,649,224 | 48,628 | # Multiply fractions and whole numbers 3
Practice seeing how one whole-number-times-fraction problem is the same as another. Find equivalent multiplication expressions.
### Problem
Complete the equation.
times, start fraction, 4, divided by, 2, end fraction, space, equals, space, 28, times, start fraction, 1, divided by, 2, end fraction | 82 | 342 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "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-2016-40 | longest | en | 0.795156 |
http://electronics.stackexchange.com/questions/tagged/measurement+multimeter | 1,386,832,371,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164567400/warc/CC-MAIN-20131204134247-00023-ip-10-33-133-15.ec2.internal.warc.gz | 62,805,891 | 12,037 | # Tagged Questions
41 views
### Current measurement maximum time
On my multimeter I have two current measurement connection. One for the mA and one for the A. On the ampere one there's a time limitation but not on the mA one. Also they mention in the manual a kind ...
52 views
### Multimeter resistance measure
I've found this question but they don't talk about damage (Multimeter ranges) If I measure a 20KOhm resistance an I'm for example in the 200Ohm range, can this damage my multimeter ?
55 views
### Using a DMM to measure output isolation/crosstalk
I'm making some PCB's that have some DC to DC converters on them with three 5 Volt converters, one 15 volt converter, and a converter with +/- 12 volts DC. On the PCB the output pins are in a ...
366 views
### What color is positive on a multimeter?
I bought a multimeter with attached leads. I would think red is positive but when I take a reading it only works if I use black as positive. Was it assembled wrong? There's nothing in the manual. It's ...
183 views
### A device that runs on AC current it's voltage inside is measured by always turning the multimeter's knob to AC voltage:same for DC right?
I've bought a cheap multimeter and it's not automatic. I've to manually rotate the knob. I live in North America. AC current is used in my home. Does this mean I can always measure voltage by turning ...
1k views
### How does autoranging work in a multimeter? What is the circuit?
I was curious how this is done. It seems you would have to vary the voltage and current to measure resistance from 1ohm to 10Mohm.
310 views
### What's the low voltage displayed by a voltmeter when only one of its probes is connected?
Recently I was toying with a digital multimeter and observed the following. The multimeter was set to measuring AC voltage with the upper limit of 750 volts. It has two probes which I will call "red" ... | 443 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-48 | latest | en | 0.958608 |
http://at.metamath.org/mpegif/mmtheorems161.html | 1,490,814,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191353.5/warc/CC-MAIN-20170322212951-00563-ip-10-233-31-227.ec2.internal.warc.gz | 29,551,889 | 32,125 | Home Metamath Proof ExplorerTheorem List (p. 161 of 309) < Previous Next > Browser slow? Try the Unicode version.
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Theorem List for Metamath Proof Explorer - 16001-16100 *Has distinct variable group(s)
TypeLabelDescription
Statement
Theoremmvrf 16001 The power series variable function is a function from the index set to elements of the power series structure representing for each . (Contributed by Mario Carneiro, 29-Dec-2014.)
mPwSer mVar
Theoremmvrf1 16002 The power series variable function is injective if the base ring is nonzero. (Contributed by Mario Carneiro, 29-Dec-2014.)
mPwSer mVar
Theoremmvrcl2 16003 A power series variable is an element of the base set. (Contributed by Mario Carneiro, 29-Dec-2014.)
mPwSer mVar
Theoremreldmmpl 16004 The multivariate polynomial constructor is a proper binary operator. (Contributed by Mario Carneiro, 21-Mar-2015.)
mPoly
Theoremmplval 16005* Value of the set of multivariate polynomials. (Contributed by Mario Carneiro, 7-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly mPwSer s
Theoremmplbas 16006* Base set of the set of multivariate polynomials. (Contributed by Mario Carneiro, 7-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly mPwSer
Theoremmplelbas 16007 Property of being a polynomial. (Contributed by Mario Carneiro, 7-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly mPwSer
Theoremmplval2 16008 Self-referential expression for the set of multivariate polynomials. (Contributed by Mario Carneiro, 7-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly mPwSer s
Theoremmplbasss 16009 The set of polynomials is a subset of the set of power series. (Contributed by Mario Carneiro, 7-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly mPwSer
Theoremmplelf 16010* An polynomial is defined as a function on the coefficients. (Contributed by Mario Carneiro, 7-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly
Theoremmplsubglem 16011* If is an ideal of sets (a nonempty collection closed under subset and binary union) of the set of finite bags (the primary applications being and for some ), then the set of all power series whose coefficient functions are supported on an element of is a subgroup of the set of all power series. (Contributed by Mario Carneiro, 12-Jan-2015.)
mPwSer SubGrp
Theoremmpllsslem 16012* If is an ideal of subsets (a nonempty collection closed under subset and binary union) of the set of finite bags (the primary applications being and for some ), then the set of all power series whose coefficient functions are supported on an element of is a linear subspace of the set of all power series. (Contributed by Mario Carneiro, 12-Jan-2015.)
mPwSer
Theoremmplsubg 16013 The set of polynomials is closed under addition, i.e. it is a subgroup of the set of power series. (Contributed by Mario Carneiro, 8-Jan-2015.)
mPwSer mPoly SubGrp
Theoremmpllss 16014 The set of polynomials is closed under scalar multiplication, i.e. it is a linear subspace of the set of power series. (Contributed by Mario Carneiro, 7-Jan-2015.)
mPwSer mPoly
Theoremmplsubrglem 16015* Lemma for mplsubrg 16016. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPwSer mPoly
Theoremmplsubrg 16016 The set of polynomials is closed under multiplication, i.e. it is a subring of the set of power series. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPwSer mPoly SubRing
Theoremmpl0 16017* The zero polynomial. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmpladd 16018 The addition operation on multivariate polynomials. (Contributed by Mario Carneiro, 9-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly
Theoremmplmul 16019* The multiplication operation on multivariate polynomials. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly g
Theoremmpl1 16020* The identity element of the ring of polynomials. (Contributed by Mario Carneiro, 10-Jan-2015.)
mPoly
Theoremmplsca 16021 The scalar field of a multivariate polynomial structure. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly Scalar
Theoremmplvsca2 16022 The scalar multiplication operation on multivariate polynomials. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly mPwSer
Theoremmplvsca 16023* The scalar multiplication operation on multivariate polynomials. (Contributed by Mario Carneiro, 9-Jan-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPoly
Theoremmplvscaval 16024* The scalar multiplication operation on multivariate polynomials. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmvrcl 16025 A power series variable is a polynomial. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly mVar
Theoremmplgrp 16026 The polynomial ring is a group. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmpllmod 16027 The polynomial ring is a left module. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmplrng 16028 The polynomial ring is a ring. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmplcrng 16029 The polynomial ring is a commutative ring. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmplassa 16030 The polynomial ring is an associative algebra. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly AssAlg
Theoremressmplbas2 16031 The base set of a restricted polynomial algebra consists of power series in the subring which are also polynomials (in the parent ring). (Contributed by Mario Carneiro, 3-Jul-2015.)
mPoly s mPoly SubRing mPwSer
Theoremressmplbas 16032 A restricted polynomial algebra has the same base set. (Contributed by Mario Carneiro, 3-Jul-2015.)
mPoly s mPoly SubRing s
Theoremressmpladd 16033 A restricted polynomial algebra has the same addition operation. (Contributed by Mario Carneiro, 3-Jul-2015.)
mPoly s mPoly SubRing s
Theoremressmplmul 16034 A restricted polynomial algebra has the same multiplication operation. (Contributed by Mario Carneiro, 3-Jul-2015.)
mPoly s mPoly SubRing s
Theoremressmplvsca 16035 A restricted power series algebra has the same scalar multiplication operation. (Contributed by Mario Carneiro, 3-Jul-2015.)
mPoly s mPoly SubRing s
Theoremsubrgmpl 16036 A subring of the base ring induces a subring of polynomials. (Contributed by Mario Carneiro, 3-Jul-2015.)
mPoly s mPoly SubRing SubRing
Theoremsubrgmvr 16037 The variables in a subring polynomial algebra are the same as the original ring. (Contributed by Mario Carneiro, 4-Jul-2015.)
mVar SubRing s mVar
Theoremsubrgmvrf 16038 The variables in a polynomial algebra are contained in every subring algebra. (Contributed by Mario Carneiro, 4-Jul-2015.)
mVar SubRing s mPoly
Theoremmplmon 16039* A monomial is a polynomial. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmplmonmul 16040* The product of two monomials adds the exponent vectors together. For example, the product of with is , where the exponent vectors and are added to give . (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly
Theoremmplcoe1 16041* Decompose a polynomial into a finite sum of monomials. (Contributed by Mario Carneiro, 9-Jan-2015.)
mPoly g
Theoremmplcoe3 16042* Decompose a monomial in one variable into a power of a variable. (Contributed by Mario Carneiro, 7-Jan-2015.)
mPoly mulGrp .g mVar
Theoremmplcoe2 16043* Decompose a monomial into a finite product of powers of variables. (The assumption that is a commutative ring is not strictly necessary, because the submonoid of monomials is in the center of the multiplicative monoid of polynomials, but it simplifies the proof.) (Contributed by Mario Carneiro, 10-Jan-2015.)
mPoly mulGrp .g mVar g
Theoremmplbas2 16044 An alternative expression for the set of polynomials, as the smallest subalgebra of the set of power series that contains all the variable generators. (Contributed by Mario Carneiro, 10-Jan-2015.)
mPoly mPwSer mVar AlgSpan
Theoremltbval 16045* Value of the well-order on finite bags. (Contributed by Mario Carneiro, 8-Feb-2015.)
bag
Theoremltbwe 16046* The finite bag order is a well-order, given a well order of the index set. (Contributed by Mario Carneiro, 2-Jun-2015.)
bag
Theoremreldmopsr 16047 Lemma for ordered power series. (Contributed by Stefan O'Rear, 2-Oct-2015.)
ordPwSer
Theoremopsrval 16048* The value of the "ordered power series" function. This is the same as mPwSer psrval 15942, but with the addition of a well-order on we can turn a strict order on into a strict order on the power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.)
mPwSer ordPwSer bag sSet
Theoremopsrle 16049* An alternative expression for the set of polynomials, as the smallest subalgebra of the set of power series that contains all the variable generators. (Contributed by Mario Carneiro, 8-Feb-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPwSer ordPwSer bag
Theoremopsrval2 16050 Self-referential expression for the ordered power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.)
mPwSer ordPwSer sSet
Theoremopsrbaslem 16051 Get a component of the ordered power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.) (Revised by Mario Carneiro, 2-Oct-2015.)
mPwSer ordPwSer Slot
Theoremopsrbas 16052 The base set of the ordered power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.) (Revised by Mario Carneiro, 30-Aug-2015.)
mPwSer ordPwSer
Theoremopsrplusg 16053 The addition operation of the ordered power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.) (Revised by Mario Carneiro, 30-Aug-2015.)
mPwSer ordPwSer
Theoremopsrmulr 16054 The multiplication operation of the ordered power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.) (Revised by Mario Carneiro, 30-Aug-2015.)
mPwSer ordPwSer
Theoremopsrvsca 16055 The scalar product operation of the ordered power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.) (Revised by Mario Carneiro, 30-Aug-2015.)
mPwSer ordPwSer
Theoremopsrsca 16056 The scalar ring of the ordered power series structure. (Contributed by Mario Carneiro, 8-Feb-2015.) (Revised by Mario Carneiro, 30-Aug-2015.)
mPwSer ordPwSer Scalar
Theoremopsrtoslem1 16057* Lemma for opsrtos 16059. (Contributed by Mario Carneiro, 8-Feb-2015.)
ordPwSer Toset mPwSer bag
Theoremopsrtoslem2 16058* Lemma for opsrtos 16059. (Contributed by Mario Carneiro, 8-Feb-2015.)
ordPwSer Toset mPwSer bag Toset
Theoremopsrtos 16059 The ordered power series structure is a totally ordered set. (Contributed by Mario Carneiro, 10-Jan-2015.)
ordPwSer Toset Toset
Theoremopsrso 16060 The ordered power series structure is a totally ordered set. (Contributed by Mario Carneiro, 10-Jan-2015.)
ordPwSer Toset
Theoremopsrcrng 16061 The ring of ordered power series is commutative ring. (Contributed by Mario Carneiro, 10-Jan-2015.)
ordPwSer
Theoremopsrassa 16062 The ring of ordered power series is an associative algebra. (Contributed by Mario Carneiro, 29-Dec-2014.)
ordPwSer AssAlg
Theoremmplrcl 16063 Reverse closure for the polynomial index set. (Contributed by Stefan O'Rear, 19-Mar-2015.) (Revised by Mario Carneiro, 30-Aug-2015.)
mPoly
Theoremmplelsfi 16064 A polynomial treated as a coefficient function has finitely many nonzero terms. (Contributed by Stefan O'Rear, 22-Mar-2015.)
mPoly
Theoremmvrf2 16065 The power series/polynomial variable function maps indices to polynomials. (Contributed by Stefan O'Rear, 8-Mar-2015.)
mPoly mVar
Theoremmplmon2 16066* Express a scaled monomial. (Contributed by Stefan O'Rear, 8-Mar-2015.)
mPoly
Theorempsrbag0 16067* The empty bag is a bag. (Contributed by Stefan O'Rear, 9-Mar-2015.)
Theorempsrbagsn 16068* A singleton bag is a bag. (Contributed by Stefan O'Rear, 9-Mar-2015.)
Theoremmplascl 16069* Value of the scalar injection into the polynomial algebra. (Contributed by Stefan O'Rear, 9-Mar-2015.)
mPoly algSc
Theoremmplasclf 16070 The scalar injection is a function into the polynomial algebra. (Contributed by Stefan O'Rear, 9-Mar-2015.)
mPoly algSc
Theoremsubrgascl 16071 The scalar injection function in a subring algebra is the same up to a restriction to the subring. (Contributed by Mario Carneiro, 4-Jul-2015.)
mPoly algSc s mPoly SubRing algSc
Theoremsubrgasclcl 16072 The scalars in a polynomial algebra are in the subring algebra iff the scalar value is in the subring. (Contributed by Mario Carneiro, 4-Jul-2015.)
mPoly algSc s mPoly SubRing
Theoremmplmon2cl 16073* A scaled monomial is a polynomial. (Contributed by Stefan O'Rear, 8-Mar-2015.)
mPoly
Theoremmplmon2mul 16074* Product of scaled monomials. (Contributed by Stefan O'Rear, 8-Mar-2015.)
mPoly
Theoremmplind 16075* Prove a property of polynomials by "structural" induction, under a simplified model of structure which loses the sum of products structure. The commutativity condition is stronger than strictly needed. (Contributed by Stefan O'Rear, 11-Mar-2015.)
mVar mPoly algSc
Theoremmplcoe4 16076* Decompose a polynomial into a finite sum of scaled monomials. (Contributed by Stefan O'Rear, 8-Mar-2015.)
mPoly g
10.10.2 Polynomial evaluation
Theoremevlslem4 16077* The support of a tensor product of ring element families is contained in the product of the supports. (Contributed by Stefan O'Rear, 8-Mar-2015.)
Theorempsrbagsuppfi 16078* Finite bags have finite nonzero-support. (Contributed by Stefan O'Rear, 9-Mar-2015.)
Theorempsrbagev1 16079* A bag of multipliers provides the conditions for a valid sum. (Contributed by Stefan O'Rear, 9-Mar-2015.)
.g CMnd
Theorempsrbagev2 16080* Closure of a sum using a bag of multipliers. (Contributed by Stefan O'Rear, 9-Mar-2015.)
.g CMnd g
Theoremevlslem2 16081* A linear function on the polynomial ring which is multiplicative on scaled monomials is generally multiplicative. (Contributed by Stefan O'Rear, 9-Mar-2015.)
mPoly
10.10.3 Univariate Polynomials
Syntaxcps1 16082 Univariate power series.
PwSer1
Syntaxcv1 16083 The base variable of a univariate power series.
var1
Syntaxcpl1 16084 Univariate polynomials.
Poly1
Syntaxces1 16085 Evaluation in a subring.
evalSub1
Syntaxce1 16086 Evaluation of a univariate polynomial.
eval1
Syntaxcco1 16087 Convert a multivariate polynomial representation to univariate.
coe1
Syntaxctp1 16088 Convert a univariate polynomial representation to multivariate.
toPoly1
Definitiondf-psr1 16089 Define the algebra of univariate power series. (Contributed by Mario Carneiro, 29-Dec-2014.)
PwSer1 ordPwSer
Definitiondf-vr1 16090 Define the base element of a univariate power series (the element of the set of polynomials and also the in the set of power series). (Contributed by Mario Carneiro, 8-Feb-2015.)
var1 mVar
Definitiondf-ply1 16091 Define the algebra of univariate polynomials. (Contributed by Mario Carneiro, 9-Feb-2015.)
Poly1 PwSer1s mPoly
Definitiondf-evls1 16092* Define the evaluation map for the univariate polynomial algebra. The function evalSub1 makes sense when is a ring and is a subring of , and where is the set of polynomials in Poly1. This function maps an element of the formal polynomial algebra (with coefficients in ) to a function from assignments to the variable from into an element of formed by evaluating the polynomial with the given assignment. (Contributed by Mario Carneiro, 12-Jun-2015.)
evalSub1 evalSub
Definitiondf-evl1 16093* Define the evaluation map for the univariate polynomial algebra. The function eval1 makes sense when is a ring, and is the set of polynomials in Poly1. This function maps an element of the formal polynomial algebra (with coefficients in ) to a function from assignments to the variable from into an element of formed by evaluating the polynomial with the given assignment. (Contributed by Mario Carneiro, 12-Jun-2015.)
eval1 eval
Definitiondf-coe1 16094* Define the coefficient function for a univariate polynomial. (Contributed by Stefan O'Rear, 21-Mar-2015.)
coe1
Definitiondf-toply1 16095* Define a function which maps a coefficient function for a univariate polynomial to the corresponding polynomial object. (Contributed by Mario Carneiro, 12-Jun-2015.)
toPoly1
Theorempsr1baslem 16096 The set of finite bags on is just the set of all functions from to . (Contributed by Mario Carneiro, 9-Feb-2015.)
Theorempsr1val 16097 Value of the ring of univariate power series. (Contributed by Mario Carneiro, 8-Feb-2015.)
PwSer1 ordPwSer
Theorempsr1crng 16098 The ring of univariate power series is a commutative ring. (Contributed by Mario Carneiro, 8-Feb-2015.)
PwSer1
Theorempsr1assa 16099 The ring of univariate power series is an associative algebra. (Contributed by Mario Carneiro, 8-Feb-2015.)
PwSer1 AssAlg
Theorempsr1tos 16100 The ordered power series structure is a totally ordered set. (Contributed by Mario Carneiro, 2-Jun-2015.)
PwSer1 Toset Toset
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Copyright terms: Public domain < Previous Next > | 7,845 | 22,982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-13 | latest | en | 0.649703 |
https://pubs.nctm.org/browse?access=all&pageSize=10&sort=relevance&t_0=geometry_1&t_3=grades_pk-2_0&t_4=measurement | 1,669,528,344,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00077.warc.gz | 518,385,140 | 31,999 | # Browse
## You are looking at 1 - 10 of 39 items for :
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## Leveraging Read Alouds for Mathematical Connections
This article presents examples of how early childhood educators (prek-2nd grade) might use their daily read alouds as a vehicle for increasing mathematical talk and mathematical connections for their students.
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The Asked & Answered department shares excerpts from discussion threads on the online MyNCTM community. In this issue, featured threads highlight responses to members' questions regarding 1st grade number sense, multiplication and division of fractions, issues of definition and precision related to circles, and the value of rationalizing denominators.
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## 5 Ways to Improve Children's Understanding of Length Measurement
These activities can support elementary school teachers in building students' conceptions of measurement.
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## A critical focus on the M in STEAM
Contributors to the iSTEM (Integrating Science, Technology, Engineering, and Mathematics) department share ideas and activities that stimulate student interest in the integrated fields of science, technology, engineering, and mathematics (STEM) in K–grade 6 classrooms. The authentic STEAM project described here was born of a critical need of one child in the community. Using the Design Thinking framework, a class of fourth graders embarked on what was arguably the most meaningful school project of their lives. We place an explicit focus on the M in STEAM.
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## Clocks: For more than telling time
Postscript items are designed as rich “grab-and-go” resources that any teacher can quickly incorporate into their classroom repertoire with little effort and maximum impact. In this article, classroom clocks are used as an effective tool to support student understanding of basic number, fraction, and geometry concepts.
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## Learning Mathematics through Minecraft
The Common Core State Standards can be taught with Minecraft, an interactive creative Lego®-like game. Integrating Science, Technology, Engineering, and Mathematics (iSTEM) authors share ideas and activities that stimulate student interest in the integrated fields of science, technology, engineering, and mathematics (STEM) in K—grade 6 classrooms.
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## Classroom Photographs: Reframing What and How We Notice
Use these three activities as professional learning community tools to support powerful conversations.
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## Untangling Geometric Ideas
This geometry lesson uses the work of abstract artist Wassily Kandinsky as a springboard and is intended to promote the conceptual understanding of mathematics through problem solving, group cooperation, mathematical negotiations, and dialogue.
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## We are growing
Math by the Month is a regular department of the journal. It features collections of short activities focused on a monthly theme. These articles aim for an inquiry or problem-solving orientation that includes at least four activities each for grade bands K—2, 3—4, and 5—6. In this issue, the problems capitalize on the natural curiosity of children to explore measurement.
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## Mathematics in Kindergarten Classrooms
This department publishes brief news articles, announcements, and guest editorials on current mathematics education issues that stimulate the interest of TCM readers and cause them to think about an issue or consider a specific viewpoint about some aspect of mathematics education. | 686 | 3,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.931237 |
https://www.skipperwbreeders.com/interesting/how-to-read-a-horse-racing-program.html | 1,606,822,360,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00069.warc.gz | 842,842,901 | 20,916 | # How to read a horse racing program
## How do you read a horse racing form?
Form runs from left to right, with the oldest races on the left and the most recent on the right.
1. The numbers 1-9 indicate the position the horse finished in the race.
2. The number 0 indicates that the horse finished outside the first 9.
3. The symbol – separates racing seasons.
## What do they say at the start of a horse race?
Horse Racing MetaphorsExpressionSports UsageA head startAlso ranWhen the result of a race is given, only the first 3 placed horses are announced. The only information given about the other poor horses in the race is that they ‘also ran’.And they’re offFrequently the words of a commentator when a race starts.Ещё 37 строк
## How do you read horse racing odds?
Odds are simply the way prices and payouts are shown at a horse track. The numbers displayed as 4-7 or 2-5 tell you what you pay and how much you get back if the horse you bet on wins. The first number tells you how much you could win, the second number is the amount you bet.
## How can I learn horse racing?
How to study the form: our 3-step guide
1. Establish the race distance, race conditions and track conditions. …
2. Look at a horse’s recent form and try and place where the horse is at in its preparation relative to the race conditions. …
3. Find something that recommends a horse to you.
they have won
## What’s the best bet in horse racing?
Types of Horse Racing Wagers (and Your Chances of Winning)Bet TypeYour Chances of WinningSuggested Plays (Based upon a \$100 Bankroll)ShowVery good\$6 per horsePlaceGood\$5 per horseWinAverage\$4 per horseQuinellaAverage\$2 quinella box using three horses costs \$6Ещё 7 строк
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## What is it called when you pick 4 horses?
Trifecta — Pick three horses. If they finish 1st, 2nd and 3rd, in exact order, you win. Superfecta — Pick four horses. If they finish 1st, 2nd, 3rd and 4th, in exact order, you win.
## Are racing horses cruel?
Behind the romanticized façade of Thoroughbred horse racing is a world of injuries, drug abuse, gruesome breakdowns, and slaughter. While spectators show off their fancy outfits and sip mint juleps, horses are running for their lives.
## What is it called when you bet on 4 horses?
Trifecta is the same as Exacta, but this time you are choosing three horses. They need to finish in the exact specified order for you to win. Superfecta is the same as Trifecta, but you’re betting on four horses to finish in the exact order.
## What are 7 to 2 odds?
When horse racing odds are shown in the form of 7-2, 5-1, etc, it expresses the amount of profit to the amount invested. So odds of 7-2 mean that for every \$2 invested, the punter gets \$7 profit in return. This means when you bet \$2, the total return if the bet is successful is \$9.
## What does 20 to 1 odds pay?
For example, 6-5 means you will get \$6 in profit for every \$5 you wager, while 20-1 means you get \$20 in profit for every \$1 you wager. In the latter example, a bet of \$2 means you would get \$42 back for a winning wager.
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## What does 4 to 5 odds mean in horse racing?
Traditional Odds in Online Horse Betting
When odds are shown in the traditional fractions, i.e. 4-1 or 9-2, it shows the amount of profit there is to be gained versus the stake. So if you bet a horse at 4-1 for \$5 and it wins, you will return \$25 (\$5 x 4 + original stake).
## Are horses faster on dirt or turf?
Grass is a more slippery surface than dirt and harder too when dried. But it is also easier on a horse’s feet and leg because of the cushion underneath their feet. But it is also a much faster surface and speed horses tends to run faster than what their conditions allows them to.
## What does horse form mean?
In horse racing, the form of a horse is a record of significant events, mainly its performance in previous races. The form may identify the horse’s sire, dam and wider pedigree.
21 hours ago | 1,011 | 4,058 | {"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-2020-50 | longest | en | 0.906436 |
http://www2.nzmaths.co.nz/Families/NumFramework.aspx | 1,531,756,261,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00341.warc.gz | 583,477,648 | 4,078 | What is the Number Framework?
The Number Framework is intended to help teachers, parents and students understand the stages of learning of number knowledge and understanding.
There are two sections to the Number Framework. The Strategy section describes the processes students use to solve problems involving numbers - how they work things out. The Knowledge section describes the key items about number that children know and can recall quickly.
The two sections are linked, with children requiring knowledge to improve their strategies, and using strategies to develop new knowledge.
The Strategy Section
The Strategy section of the Number Framework describes a series of stages that children progress through as they develop their understanding of a range of strategies for solving number problems. There are eight stages altogether, with the first three often grouped together:
• Stage 0-3: Counting from One - children can solve problems by counting from one, either using materials or in their head.
• Stage 4: Advanced Counting - children can solve problems by counting in ones, or by skip counting, starting from numbers other than one.
• Stage 5: Early Additive - children can solve simple problems by splitting up and adding together the numbers in their head.
• Stage 6: Advanced Additive - children use a range of different methods to solve more challenging problems in their head.
• Stage 7: Advanced Multiplicative - children use a range of different methods to solve multiplication and division problems in their head.
• Stage 8: Advanced Proportional - children can solve complicated problems involving fractions, decimals and percentages using a combination of methods.
There are three areas, or 'domains' within the Strategy section, which describe a child's ability to solve different types of problems (additive, multiplicative and proportional). Your child is likely to be learning a broad range of strategies in their classroom mathematics programme. One of the ways that you can most easily support them is to help them develop the knowledge that they will need to be able to use these strategies.
The Knowledge Section
The Knowledge section is usually broken down into five areas, referred to as 'domains': Numeral Identification, Number Sequence and Order, Grouping/Place Value, Basic Facts, and Written Recording. We have provided a collection of suggested activities that parents and families can use, grouped under three domains:
• Number Identification and Order - activities to help children learn to read numbers and know the order of numbers.
• Place Value - activities to help children learn how 10s, 100s, 1000s, tenths, hundredths, thousandths etc are used.
• Number Facts - activities that will help children learn their addition, subtraction, multiplication and division facts.
The activities in each domain are grouped by the stage of development they are most suitable for. To choose the appropriate stage, either use the brief descriptions under the Strategy Section above, ask your child's teacher, or choose a range of activities to see which are suitable.
Click to select activities by stage and domain. | 595 | 3,158 | {"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-2018-30 | latest | en | 0.945521 |
https://www.doityourself.com/stry/multiple-dial-vs-single-dial-combination-padlock | 1,716,671,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00239.warc.gz | 639,531,507 | 67,188 | # Multiple Dial vs Single Dial Combination Padlock
People need a padlock, such as combination padlock, to secure their private rooms, lockers, cabinets, boxes and other storage areas. One of the best choices for a padlock is the one with a combination lock.
This type of lock has a series of letters, symbols, or numbers that are used to unbolt it. These series of numbers are manipulated by rotating the dial with inscribed symbols. The dial interrelates with several discs or cams to match a certain series of symbols that will unlock the locking mechanism.
A combination padlock is sold as a mechanical manually manipulated locking system or as an electronic gadget. Mathematicians often disagree on the terminology given for a combination lock, as it does not need a combination of numbers or values for it to unlock.
The term preferred for describing the set of numbers, letters or symbols to unbolt the locking system is sequence. There are two major types of combination locks and they are the multiple dial and the single dial combination padlock.
This type of padlock is one of the simplest designs. It is often used in briefcases, travel bags, lockers, and bicycles, and uses several rotating cams with indentations cut into it. A pin with many grooves that hooks into the cams attaches this type of lock.
When the indentations of the cams align with the pin’s teeth or grooves, it will unlock. Although it may have a multiple dial, it is but considered as one of the least secure types of combination padlock. Even without any idea on what the correct combination or sequence of symbols were used, it can still be easily opened.
This is perhaps due to the degree of machine work, whether it was created precisely or that it was made haphazardly. But to open it even without knowing the certain code is easy. The discs are constantly being rotated while the pin is being pulled until certain clicking sounds are heard. This means that the teeth have settled on the indentations. Buyers are cautioned not to use this type of lock if the material being secured is highly confidential. | 425 | 2,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-22 | latest | en | 0.963389 |
https://www.studypool.com/discuss/1034398/which-linear-inequality-is-represented-by-the-graph-3?free | 1,481,198,664,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00448-ip-10-31-129-80.ec2.internal.warc.gz | 1,007,006,335 | 14,145 | ##### Which linear inequality is represented by the graph?
Algebra Tutor: None Selected Time limit: 1 Day
Jun 12th, 2015
The line on the edge passes through (-1/2,0) and (0,1)
Slope of the line is (1-0)/(0-(-1/2)) =2
The equation of the line is y-1 =2(x-0)
Or
y=2x+1
Hence first, second and fourth options are ruled out.
We are interested in y > 2x+1
It must exclude these points.
Thus
Y > 2x+1
Third option is correct.
Jun 12th, 2015
...
Jun 12th, 2015
...
Jun 12th, 2015
Dec 8th, 2016
check_circle | 183 | 516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-50 | longest | en | 0.926179 |
https://www.tropicalbuild.com.au/questions-about-construction/quick-answer-floor-joists-are-typically-what-size-in-residential-construction.html | 1,632,365,778,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057416.67/warc/CC-MAIN-20210923013955-20210923043955-00125.warc.gz | 1,037,694,569 | 20,830 | ## What size are floor joists in residential construction?
Looking at this table you will see there is a choice in the size of floor joist (2 X 6, 2 X 8, 2 X 10 or 2 X 12) and there is a choice in the joist spacing (12″, 16″ or 14″). The floor joist spacing is the distance between the centers of any two installed joists.
## What size should floor joists be?
The spacing of floor joists is just one of the components used to determine the minimum size of the floor joist. Per the prescriptive tables found in Chapter 5 of the International Residential Code (IRC), the standard floor joist spacing used is 12, 16, 19.2, and 24 inches on center.
## What is the minimum size of a floor joist?
3.2. Floor joists shall have a bearing support length of not less than 11/2 inches (38 mm) for exterior wall supports and 31/2 inches (89 mm) for interior wall supports. Tracks shall be not less than 33 mils (0.84 mm) thick except where used as part of a floor header or trimmer in accordance with Section R505. 3.8.
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## What size wood is used for floor joists?
Lumber graded as #2 is the most common choice for floor joists and other framing lumber.
## Can I use 2×6 for floor joists?
How do I keep them even? In general terms, joists spaced 16 inches on center can span 1.5 times in feet their depth in inches. A 2×8 up to 12 feet; 2×10 to 15 feet and 2×12 to 18 feet. 2×6 joists should only be used on ground-level decks that do not require, and will not provide for, any guards.
## How far can a 2×6 joist span without support?
2-grade 2×6 joists can span up to 10 feet 9 inches from beam to beam when spaced the standard 16 inches apart with a maximum live load of 30 inches per square foot. In comparison, No. -1 grade lumber can span slightly further to 10 feet 11 inches under the same parameters.
## How much weight can a 2×6 floor joist hold?
The larger the deck, the larger the joists. how much weight will a 2×6 floor hold? For example, a properly designed office floor can support 50 pounds per square foot. This may seem light, but this is 50 pounds over each and every square foot of floor space.
## What size floor joist do I need for a 14 foot span?
Max. Live Load 60 lbs/ft2 (2873 N/m2)
Maximum Span ( ft – in)
Nominal Size (inches) Joist Spacing Center to Center (inches) Lumber Grade
2 x 12 24 13′ – 2″
2 x 14 12 20′ – 10″
16 18′ – 0″
10
## How do I calculate how many joists I need?
To estimate them 12 inches from centers add 1 to the number of feet in length of one wall on which the joists are placed. For example, suppose a building is 32 feet long, and the joists are placed 12 inches from centers. We simply add 1 to 32, which makes 33 the number of joists required for one span.
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## What is code for floor joist?
Floor Joist Building Code Requirements The joist must have a minimum of 1 1/2″ riding on the wall or beam. If the joists are lapped, then they must be lapped at least 3″. Maximum end notches for 2 x 8, 2 x 10, and 2 x 12 respectively are 1 13/16″, 2 5/16″, and 2 13/16″.
## What goes under floor joists?
What Is a Subfloor? Subflooring provides a base for finish flooring and also serves as a platform during construction. It may be made of boards laid either at right angles or diagonally across joists. Or the subfloor may be made of plywood or other panel products that are laid perpendicular to the joists.
## Can I use 2×4 for floor joists?
for a real house no you cannot, 2×4 will not have the structural strength to support a load as a floor house in a house. There are also 2×4 truses that are used as floor joists. Floor joists are normally 2×12 so all the floors are the proper height. | 1,040 | 3,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | latest | en | 0.909054 |
https://betterlesson.com/lesson/573860/solving-linear-inequalities | 1,545,035,124,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828448.76/warc/CC-MAIN-20181217065106-20181217091106-00057.warc.gz | 533,515,363 | 26,064 | Solving Linear Inequalities
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Objective
SWBAT solve linear inequalities and represent their solution on a number line.
Big Idea
Students will apply their knowledge of multi-step equations to solve linear inequalities.
Do Now
10 minutes
Today's Do Now will take about 5 minutes. My students are asked to order a set of numbers from least to greatest. The majority of my students can easily complete this activity , but some may struggle to remember place value. As a visual aid, my students can use the number line diagram on the top of their paper.
After 5 minutes, we will review the answers as a whole group. I will model the answers with a number line, as students call out their responses aloud. Graphing on and reading from number lines will be used heavily throughout this unit, so it is important to review this exercise in its entirety.
Next, a student volunteer will read today's objective, "SWBAT solve linear inequalities and represent their solution on a number line." I will ask students to define inequality, both in a real life sense, and, in mathematics.
Introduction
15 minutes
During this section of the lesson, I ask my students to complete a sorting activity. The task gets students thinking about the truth value of a statement as we begin this unit on solving inequalities. The class will have 5 minutes to sort the cards into two separate piles, one for true statements and one for false statements. We will then review statements on the cards aloud.
Teacher Note: The cards for the true false sort must be cut out before class begins.
Guided Notes + Practice
25 minutes
As we move from numerical examples into algebraic expressions, students will follow along at their seats using these Guided Notes and this Presentation.
Slide 3: I will ask students to call out all of the speeds that can be driven on this road, without the risk of a speeding ticket. As students call out speeds, I will write them on the screen. I will then summarize all of our answers by writing x ≤ 40. I will ask students it they agree or disagree with the inequality that I created using a thumb up or a thumb down.
Slide 4: Next I will ask students to call out heights of people that would be allowed to board this ride. I will again record the values that students call out on the board, and then summarize these responses as x ≥ 54.
Slide 6: Students will write the corresponding inequality symbol below its name. Students will also record which inequalities should be represented on a number line with an open circle, and which should be represented with a closed circle. We will then discuss the relationship between the the shading of a circle with its actual meaning.
I will then give the class a few minutes to graph the 4 inequalities on bottom of their Guided Notes. Many students will incorrectly graph the inequality symbols by simply drawing the arrow the same direction as the inequality sign. To combat this, I will ask students to use test points to verify that their shading is correct.
Slide 6: We will solve these four problems, highlighting their similarities and differences. For the two inequality problems, I will ask students to populate a list of different solutions that would keep the the inequality true.
I will tell students to solve the inequalities on Slide 9 independently, but to check their answer with a test point. Students will realize that two of their solution sets are incorrect, even though they have not made any errors while solving. I will ask students to examine the four problems, and to find a similarity between the two problems with incorrect solution sets. This will lead students to the Main Idea prompt on their notes:
To keep a true statement when solving an inequality with the multiplication of a negative number and the division of a negative number, you must reverse the inequality.
Partner Practice: Versatiles
20 minutes
Students will practice solving linear inequalities using the ETA Hand to Mind Versatiles. Students will match correct responses to the numbered tiles in the black VersaTile case. If you do not have a VersaTiles classroom set, the assignment can still be completed by having students match questions and responses with pencil and paper.
Closing
10 minutes
As we close, I will ask students to Turn and Talk with a neighbor to decide if the solution of a linear inequality is the same as a linear equations with infinite solutions.
Students will then complete an Exit Card. | 915 | 4,520 | {"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} | 4.25 | 4 | CC-MAIN-2018-51 | latest | en | 0.919239 |
https://math.stackexchange.com/questions/2874849/p-is-an-odd-prime-then-prove-that-there-is-no-group-with-exactly-p-elements | 1,702,151,823,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00223.warc.gz | 416,354,907 | 37,514 | # $p$ is an odd prime, then prove that there is no group with exactly $p$ elements of order $p$.
Question
If $p$ is an odd prime, then prove that there is no group with exactly $p$ elements of order $p$.
Attempt
Assume that such a group $G$ exists.
If $x_1 \in G$ and if $|x_1|=p$ then $|{x_1}^{-1}|=p$
hence such elements occur in pairs,
$p$ being an odd prime implies $x=x^{-1}$ for some element x whose order is $p$. Otherwise there would be even number of elements of order $p$
$x=x^{-1}$
$\implies x^2=e$
$\implies |x|=2$
$\implies 2$ is an odd prime.
• The proof looks good to me. Aug 7, 2018 at 11:39
– Did
Aug 7, 2018 at 11:41
• Your proof is correct. Aug 7, 2018 at 11:59
• If $x^2=e$ then can't $x=e$ and hence $|x|=1$? The overall proof structure is still correct, but that last conclusion is a little bit overgeneral :) Aug 7, 2018 at 12:29
• @postmortes In the hypothesis, $|x|=p>1$. Aug 7, 2018 at 14:07
Other approach, pick an element of given ones of order $p$. But then all non trivial powers of this element give rise to elements of order $p$, hence there are at least $p-1$ of them. If $G$ has precisely $p$ elements of order $p$, then there must be precisely one other, giving also rise to $p-1$ elements of order $p$. Hence $2(p-1)=p$, so $p=2$ a contradiction. By the way: there are no groups (finite or infinite) with exactly $2$ elements of order $2$.
• How cauchy gurantees the existence of element of order $p$, When the Order of the group is not given how can you use cauchy's theorem. It is also possible that the Group is Infinite order. Aug 7, 2018 at 16:04
• Yes my approach is certainly for finite groups because of Cauchy. Should not have mentioned that. It is not clear from your post that it concerns also infinite groups. But my argument also applies to infinite groups. We do not need mr Cauchy since the elements of order $p$ are a given. Aug 7, 2018 at 16:11
• Your also did not mention that the group might be infinite. So why all this fuzz? If the group is finite then the existence of elements of order $p$ implies that the order of $G$ is divisible by $p$. Aug 7, 2018 at 16:17 | 649 | 2,133 | {"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} | 3.8125 | 4 | CC-MAIN-2023-50 | latest | en | 0.905904 |
http://ck5.com/forums/threads/bds-6-spring-rate-is-this-right-updated.82779/ | 1,477,224,355,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719273.37/warc/CC-MAIN-20161020183839-00443-ip-10-171-6-4.ec2.internal.warc.gz | 50,981,889 | 12,041 | BDS 6" spring rate.....is this right?? **UPDATED**
Discussion in 'The Garage' started by 88K5Jimmy, Sep 4, 2003.
1. 88K5Jimmy1/2 ton status
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BDS 6\" spring rate.....is this right?? **UPDATED**
I am using this formula provided by Greg72 (Thanks):
Leaf Spring Rate=(WN/12) x (1000T/L)^3
W=width of leaves
N=number of leavbes in the pack (minus overload)
T=total thickness of pack (minus overload)/total # of leaves
L=length of spring eye-to-eye
12=constant for leaf spring
Here are my calculations:
W=2.5
N=5
T=.4
L=45.5
After putting my numbers into this equation I get a whopping number of 707.741 /forums/images/graemlins/shocked.gif
I then called ORD to see what the ideal spring rate is for a K5 and they said 250-300.
I haven't been able to confirm if my number is right because BDS's contact info is down on their website.
The reason I calculated my spring rate was because of the ride was so rough. I m thinking about taking the overload spring off the bottom of the spring pack and seeing if that makes it ride any better.
2. heavy4x41/2 ton status
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Re: BDS 6\" spring rate.....is this right??
No way can you use a standard formula and get the spring rate right on...there are just too many variables (metal variances, teflon pads, what kind of friction/contact the leafs have on the others). You would get a more accurate number by standing on one of the leaf springs while it stands upside-down on the ground, measuring how much it deflects with your weight, and extrapulating that out to 1". Probably be closer than any formula will get you.
I e-mailed BDS a while back and their front 4" lift springs are 425 lb/in.
3. Greg72"Might As Well..."Staff MemberSuper Moderator
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Re: BDS 6\" spring rate.....is this right??
The only thing I notice is that the description for "T" is not right.
T = thickness of 1 leaf (in inches)
You could measure a 4 leaf pack and get 1.6" then divide that by 4 to get a "T" value of .40" however.....
Don't feel bad about the numbers, no matter WHO you ask (including spring manufacturers) are all going to tell you different ways to verify the spring rate.
The idea about using a known weight to deflect the spring (and measuring the deflection) is another good way. The hardest part is finding a "repeatable" way to measure.
4. 88K5Jimmy1/2 ton status
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Re: BDS 6\" spring rate.....is this right??
[ QUOTE ]
The only thing I notice is that the description for "T" is not right.
T = thickness of 1 leaf (in inches)
You could measure a 4 leaf pack and get 1.6" then divide that by 4 to get a "T" value of .40" however.....
[/ QUOTE ]
That's what I did. My spring pack thickness is 2"/5 leaves = .4
I just got off the phone with BDS and they said the spring rates for their leaf springs weren't available. They said they didn't have them anymore. That doesn't make any sense.
Next step is to email them I guess /forums/images/graemlins/tongue.gif
5. marv_springer1/2 ton status
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Re: BDS 6\" spring rate.....is this right??
[ QUOTE ]
The idea about using a known weight to deflect the spring (and measuring the deflection) is another good way. The hardest part is finding a "repeatable" way to measure.
[/ QUOTE ]
I've experimented w/ this....
Get some good heavy beer drinkin' buds over, setup the spring on the floor upside down (like this /\ ) with a shackle on one end. Measure the free arch. Then ask the Big Man to gracefully stand on the spring in the center. Then (this is where everthing can come crashing down /forums/images/graemlins/eek.gif if your not careful) measure how much it's height has been reduced. The divide the Big Man's weight by the change in height. /forums/images/graemlins/deal.gif
[ QUOTE ]
I just got off the phone with BDS and they said the spring rates for their leaf springs weren't available. They said they didn't have them anymore. That doesn't make any sense.
[/ QUOTE ]
Ha...! /forums/images/graemlins/rotfl.gif
Marv
6. 88K5Jimmy1/2 ton status
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Re: BDS 6\" spring rate.....is this right??
So I got an email back from BDS:
Approx. 500 lbs/in.
So somewhere my calculations were a bit off, lol. Does anybody know the spring rate of stock front springs? | 1,259 | 4,621 | {"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-2016-44 | longest | en | 0.92107 |
https://brainmass.com/statistics/regression-analysis/anova-for-regression-analysis-224078 | 1,709,476,743,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00383.warc.gz | 141,997,186 | 7,407 | Purchase Solution
# ANOVA for regression analysis
Not what you're looking for?
13.3 Refer to the ANOVA table for this regression. (a) State the degrees of freedom for the F test for
overall significance. (b) Use Appendix F to look up the critical value of F for α = .05. (c) Calculate
the F statistic. Is the regression significant overall? (d) Calculate R2 and R2
adj, showing your
formulas clearly. PlasmaTV
Source d.f. SS MS
Regression 4 259, 412 64,853
Error 45 224, 539 4,990
Total 49 483,951
13.4 Refer to the ANOVA table for this regression. (a) State the degrees of freedom for the F test for
overall significance. (b) Use Appendix F to look up the critical value of F for α = .05. (c) Calculate
the F statistic. Is the regression significant overall? (d) Calculate R2 and R2
adj, showing your
formulas clearly.
Source d.f. SS MS
Regression 3 1,196,410 398,803
Error 26 379,332 14,590
Total 29 1,575,742
( I think this is the chart I am suppose to refer)
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
AYNOR 4 51 12.75 0.916667
LORIS 4 46 11.5 1.666667
LANDER 5 85 17 0.5
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 76.25 2 38.125 39.10256 1.873E-05 4.102816
Within Groups 9.75 10 0.975
Total 86 12
##### Solution Summary
The solution provides step by step method for the calculation of ANOVA for regression analysis . Formula for the calculation and Interpretations of the results are also included.
##### Measures of Central Tendency
This quiz evaluates the students understanding of the measures of central tendency seen in statistics. This quiz is specifically designed to incorporate the measures of central tendency as they relate to psychological research.
##### Terms and Definitions for Statistics
This quiz covers basic terms and definitions of statistics.
##### Know Your Statistical Concepts
Each question is a choice-summary multiple choice question that presents you with a statistical concept and then 4 numbered statements. You must decide which (if any) of the numbered statements is/are true as they relate to the statistical concept.
##### Measures of Central Tendency
Tests knowledge of the three main measures of central tendency, including some simple calculation questions. | 592 | 2,275 | {"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} | 3.953125 | 4 | CC-MAIN-2024-10 | latest | en | 0.813714 |
https://mathhelpboards.com/threads/patricks-question-at-yahoo-answers-first-fundamental-theorem-of-calculus.3883/ | 1,642,917,759,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00243.warc.gz | 432,269,059 | 14,481 | # Patrick's question at Yahoo! Answers (First fundamental theorem of Calculus)
MHB Math Helper
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Hello Patrick,
The problem is:
Find $\dfrac{f''(2)}{\pi}$ if $f(x)=\displaystyle\int_1^x\sin (\pi t^2)\;dt$. Enter your answer as an integer.
Solution. Using the First fundamental theorem of Calculus, $f'(x)=\sin (\pi x^2)$. Deriving again, $f''(x)=2\pi x\cos (\pi x^2)$ so, $$\dfrac{f''(2)}{\pi}=\frac{4\pi\cos(4\pi)}{\pi}=4\cos(4\pi)=4$$ | 177 | 502 | {"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} | 3.546875 | 4 | CC-MAIN-2022-05 | latest | en | 0.691876 |
https://www.ifr-magazine.com/technique/fuel-blunders/ | 1,726,218,562,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00487.warc.gz | 750,354,719 | 28,294 | Fuel Blunders
Perhaps you’ve noticed that we’ve been combing the ASRS reports to see what we could learn. One topic that was surprisingly prevalent was fuel and fueling errors.
0
These ASRS reports show many ways to get something simple wrong. You’ll see how creative errors can sneak in and deprive us of the fuel that keeps us airborne.
Flight Planning
A pilot encountered 32-35 knot headwinds at cruise. At 2.5 hours airborne, the pilot decided to land for fuel because only eight gallons remained. Fifteen minutes later, the engine failed. He landed safely on a county road and found the tanks were dry.
After a review, the airplane’s fuel capacity was less than the expected 40 at 35-37 gallons, and the fuel gauge was too optimistic. The PIC wrote, “Flight planning will be better done and executed to prevent such an incident from happening again.”
An instructor and student planned a trip in three legs. The first leg burned 21 gallons, leaving 29 gallons plus a 10-gallon reserve to reach their second stop, where they would refuel. About 50 NM out, the engine began to run rough from a dry tank. They changed tanks to use some of their reserve. The engine failed as they turned base leg. They, too, landed on a highway.
The aircraft checked out, but they learned the fuel capacity was 48.2, not 50 usable gallons. Sound familiar? They encountered an unexpected headwind on the second leg and recalculated their fuel needs. Later they found a simple math error. Arithmetic skills sometimes atrophy in the cockpit. If you run numbers while airborne, be doubly sure of your arithmetic.
Alternators
Bad alternators can cause fuel starvation because electricity runs many fuel pumps.
A multi-engine pilot transferred fuel after two hours in flight but was unaware there was insufficient power to run the pump. An hour later, the entire electrical system failed, leaving him unable to transfer fuel from the auxiliary tanks into the empty mains. The low voltage also weakened all the avionics.
On final approach, the tower told him to go around. The pilot replied, “unable,” and landed. Good thing, too, because the engines failed as he pulled up to a maintenance ramp. The mechanic found a failed alternator and a depleted battery.
It might be challenging to detect a weakened electrical system from the cockpit, stressing that it is important to monitor all the gauges.
Fueling
On an IFR cross-country flight, the pilot of a Piper Apache saw oil coming out of the right engine. He shut it down and informed ATC that he needed to divert and receive priority handling. He landed safely.
Upon examination by a mechanic, one of the nuts on a recently installed vacuum pump was loose. The mechanic who installed the pump didn’t tighten it sufficiently. After repair the pilot departed IFR to his destination.
In “the duress of the moment,” the pilot thought he had enough fuel to get there plus a 90-minute reserve. However, he neglected to include the climb fuel to 12,000 feet. Over an airport, both engines began to run rough despite turning on the fuel pumps and switching tanks. He again requested priority handling and landed at the airport below.
Upon inspection, 16 gallons remained in all four tanks. Said he, “In the stress and haste of the moment, my mistake was not refueling before I left.”
A recurring theme in fueling mishaps is a distraction that disrupts the orderly flow of events. Try to manage interruptions and keep perspective.
A multi-engine instructor was giving dual to a commercial, multi-engine pilot on a cross-country flight. After landing, they went into town while the FBO topped all four tanks. During preflight, the instructor checked the left side for fuel; the trainee checked the right. They agreed that all four tanks were full. The fuel receipt read 26 gallons which was about right.
About halfway back to their home airport, the right engine failed. They diverted to a third airport.
The right main tank was empty. The 26 gallons went into the auxiliary and left main tanks. The right tank had ten gallons which got them halfway home. Had the trainee properly checked the right tank, he would have caught that the FBO failed to fill it. Perhaps he saw what he expected to see.
The instructor spelled out corrective actions. First, the PIC should always supervise fueling to prevent confusion and errors. Second, the PIC should always visually check the fuel tanks and never trust even another experienced pilot. Third, a gauge that reads significantly different from the visual check calls for a recheck.
He emphasized the need to stay proficient on engine failure procedures and have a VMC alternate. The result was an uneventful approach and landing.
Jet-A Blues
The need to supervise fuelers is evident from this pilot’s experience. After arriving at an airport, he requested 50 gallons of avgas—25 per side. On his initial departure climb, the cylinder head temperature on both engines spiked to the red line. He verified the mixtures were full rich, with airspeed and deck angle adjusted for better cooling.
He made an immediate 180-degree turn and landed. Afterward, he spoke with the lineman about fueling. He confirmed the aircraft had been misfueled with Jet-A.
One thing missing here is how one can mistake Jet-A for avgas during preflight. They smell, look, and feel different.
Distractions Aplenty
A twin Bonanza pilot ordered fuel, instructing the fuelers to top off both inboard tanks and add 20 gallons per side in the auxiliary tanks. He showed the line person each filler location and confirmed the fuel type as 100LL. He paid for the fuel but received no receipt due to a change in the FBO’s payment system.
The next day, using a calibrated dipstick, he discovered the left auxiliary tank was full. Before he could check the right side, his young child needed to use the restroom. Afterward, he noted the fuel quantity in each tank and sumped each.
He then did a weight and balance and found that he needed to reduce the weight to within maximum gross weight. The pilot removed some items and got it down. He flew uneventfully to a stopover airport but bought no fuel and didn’t dip the tanks.
The pilot departed for a third airport. About 25 minutes into the flight, the right engine lost power. Observing no fuel pressure, he turned on the boost pump. Fuel pressure and the engine returned to normal. He figured the engine-driven pump could have failed, but soon after the engine began surging again. Switching to the right AUX tank didn’t help. Having completed the checklist, he shut the engine down. Center suggested a diversion airport, where he landed safely.
Postflight, he found the right main tank dry. He thought he restarted his preflight in the wrong place and never checked the tank.
Like the Swiss cheese risk model, six holes fell into line. First and second, the lineman filled the tanks, but the pilot did not check the tanks as usual. Third, the missing receipt would have shown the wrong quantity. Fourth and fifth are the double distractions during preflight and having to remove items to get within weight. Sixth, buying fuel at his first stop would have revealed the dry tank, or seventh, failing to check the tanks there per his usual procedure before departure even though he added no fuel.
By his count, this conscientious pilot deviated three times from his usual habits.
Takeaways: Inspect fuel immediately after fueling. Don’t restart a preflight in the middle; begin it again. Dip all fuel fillers before every takeoff.
Leaks
Approaching his destination, the multi-engine pilot switched tanks from the AUX to the main tanks. About 20 minutes later, the left engine started to surge. All engine monitoring instruments showed normal values. The engine returned to standard cruise profile for about 15 to 20 seconds, then surged violently. Since the aircraft was flying erratically, he decided to cage the engine. Thinking of fuel contamination, he switched the left engine back to the AUX tank. The left engine continued to surge, then stopped.
Contacting Approach, he requested vectors to the nearest airport, which was his destination. A subsequent inspection by an AMT found a leak around the filler cap on the left main tank, which allowed the low pressure air over the wing to suction all the fuel out of the tank. The engine’s main filler cap is out of the pilot’s view. The power levers obstruct the fuel gauges, so his initial scan missed the empty left tank.
Critically, the pilot decided to troubleshoot the problem first because the engine ran smoothly for 3½ hours. This weighed heavily in the pilot’s decision to troubleshoot first. Had he scanned the left tank gauge, he would have known to shut the engine down immediately.
Many multi-engine pilots prefer basic troubleshooting before shutting an engine down unless the problem is obvious, like leaking oil or an empty tank.
Gauges
In the regulations, §23.2430(a)(4) says that each fuel system must provide the flight crew with a means to determine the total useable fuel available and provide uninterrupted supply of that fuel when the system is correctly operated, accounting for fuel fluctuations. Further, § 91.205 requires a “fuel gauge indicating the quantity of fuel in each tank.”
Fuel gauges are notoriously unreliable in older aircraft. Modern aircraft with electronic gauges or totalizers are much more accurate. The scenarios below illustrate the best and worst of older gauges.
Fifteen miles from his destination, the engine failed. Switching pumps and tanks failed. The pilot landed on a highway.
The cause of this fuel exhaustion accident was engine blowby. Blowby results when the compressed fuel-air mixture gets past the piston rings into the crankcase ventilation, usually due to worn piston rings or cylinders. The result was abnormally high fuel consumption at 11.1 gph. The pilot wished he had stopped for fuel before the tanks got “low to empty.” His words speak of poor judgment, but the gauges were correct.
The fuel gauges in this Cherokee Six were reputedly accurate within a gallon. It held 47 gallons for a 12-gallon flight.
The departure was on the right main tank, holding 11 gallons. At 500 feet in a climbing left turn, the engine failed. An instructor in the right seat set the fuel selector to the right tip tank containing 17 gallons, restoring power.
The right tank was empty, but the supposedly accurate gauge indicated 11 gallons. The pilot said he would believe the observed fuel level over the fuel gauges in the future. Ya think?
Minimum Fuel
The Cessna 320 entered the traffic pattern with 40 minutes of fuel. His first approach failed because he was sandwiched between two military jets. The tower told him to go around.
Entering the downwind again, the Cessna was at minimum fuel, and the pilot advised the tower, who promised to get him down.
The jet he followed took too long to clear the runway, and the tower issued another go-around. The Cessna pilot called “unable” and landed, but the runway was clear when he landed.
The tower and the Cessna apologized to each other. The pilot was uncomfortable doing another lap in the pattern. He decided to land “rather than risk a worse situation.”
An Extra 300 pilot entered the nontower traffic pattern to land behind a Cherokee and a Cessna 182, flying very wide downwinds. The Extra was forced to fly two 360-degree circles to increase spacing. Behind the Cessna, he flew S-turns on final but was forced to go around anyway. Then the Cessna announced it was disabled on the runway. Lacking the fuel to divert, he declared a fuel emergency on the CTAF and landed on the parallel taxiway.
The Extra pilot stated that flight schools don’t comply with airport rules that require tighter patterns and said the FAA should push schools to fly proper patterns.
In several ASRS reports, pilots refused to go around. They landed on runways, taxiways, and even in the grass adjacent to an unusable runway. Do what you must to prevent an accident, and don’t let anyone else, even ATC, fly your airplane.
Assert your PIC authority as needed to do the safest thing. Practice lost engine procedures and have an alternate airport ready. Fuel is one place where paranoia is a good thing.
Fred Simonds, CFII, has had two fuel leaks—one on the ground and one airborne. | 2,596 | 12,355 | {"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-2024-38 | latest | en | 0.966152 |
https://nbviewer.org/github/donnemartin/interactive-coding-challenges/blob/master/recursion_dynamic/n_pairs_parentheses/n_pairs_parentheses_solution.ipynb | 1,638,282,585,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359037.96/warc/CC-MAIN-20211130141247-20211130171247-00459.warc.gz | 492,027,336 | 6,281 | This notebook was prepared by Rishi Rajasekaran. Source and license info is on GitHub.
# Solution Notebook¶
## Constraints¶
• Is the input an integer representing the number of pairs?
• Yes
• Can we assume the inputs are valid?
• No
• Is the output a list of valid combinations?
• Yes
• Should the output have duplicates?
• No
• Can we assume this fits memory?
• Yes
## Test Cases¶
* None -> Exception
* Negative -> Exception
* 0 -> []
* 1 -> ['()']
* 2 -> ['(())', '()()']
* 3 -> ['((()))', '(()())', '(())()', '()(())', '()()()']
# Algorithm¶
Let l and r denote the number of left and right parentheses remaining at any given point. The algorithm makes use of the following conditions applied recursively:
• Left braces can be inserted any time, as long as we do not exhaust them i.e. l > 0.
• Right braces can be inserted, as long as the number of right braces remaining is greater than the left braces remaining i.e. r > l. Violation of the aforementioned condition produces an unbalanced string of parentheses.
• If both left and right braces have been exhausted i.e. l = 0 and r = 0, then the resultant string produced is balanced.
The algorithm can be rephrased as:
• Base case: l = 0 and r = 0
• Add the string generated to the result set
• Case 1: l > 0
• Add a left parenthesis to the parentheses string.
• Recurse (l - 1, r, new_string, result_set)
• Case 2: r > l
• Add a right parenthesis to the parentheses string.
• Recurse (l, r - 1, new_string, result_set)
Complexity:
• Time: O(4^n/n^(3/2)), see Catalan numbers - 1, 1, 2, 5, 14, 42, 132...
• Space complexity: O(n), due to the implicit call stack storing a maximum of 2n function calls)
## Code¶
In [1]:
class Parentheses(object):
def find_pair(self, num_pairs):
if num_pairs is None:
raise TypeError('num_pairs cannot be None')
if num_pairs < 0:
raise ValueError('num_pairs cannot be < 0')
if not num_pairs:
return []
results = []
curr_results = []
self._find_pair(num_pairs, num_pairs, curr_results, results)
return results
def _find_pair(self, nleft, nright, curr_results, results):
if nleft == 0 and nright == 0:
results.append(''.join(curr_results))
else:
if nleft >= 0:
self._find_pair(nleft-1, nright, curr_results+['('], results)
if nright > nleft:
self._find_pair(nleft, nright-1, curr_results+[')'], results)
## Unit Test¶
In [2]:
%%writefile test_n_pairs_parentheses.py
import unittest
class TestPairParentheses(unittest.TestCase):
def test_pair_parentheses(self):
parentheses = Parentheses()
self.assertRaises(TypeError, parentheses.find_pair, None)
self.assertRaises(ValueError, parentheses.find_pair, -1)
self.assertEqual(parentheses.find_pair(0), [])
self.assertEqual(parentheses.find_pair(1), ['()'])
self.assertEqual(parentheses.find_pair(2), ['(())',
'()()'])
self.assertEqual(parentheses.find_pair(3), ['((()))',
'(()())',
'(())()',
'()(())',
'()()()'])
print('Success: test_pair_parentheses')
def main():
test = TestPairParentheses()
test.test_pair_parentheses()
if __name__ == '__main__':
main()
Overwriting test_n_pairs_parentheses.py
In [3]:
%run -i test_n_pairs_parentheses.py
Success: test_pair_parentheses | 846 | 3,127 | {"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} | 3.8125 | 4 | CC-MAIN-2021-49 | latest | en | 0.643264 |
https://www.experts-exchange.com/questions/27346272/Finding-the-highest-date-in-Column-B-for-all-occurrences-of-a-specified-string-in-Column-D.html | 1,521,335,093,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645405.20/warc/CC-MAIN-20180317233618-20180318013618-00426.warc.gz | 799,334,663 | 17,574 | # Finding the highest date in Column B for all occurrences of a specified string in Column D
I need a formula or function that finds all matches for a given string in one column, then looks at the corresponding dates in another column and returns the highest date that is NOT higher than the date in G1.
Please take a look at the attached workbook. I had a formula that seemed to work but apparently not as you can see in the workbook.
Thanks,
John
FindTailByDate-v2.xls
LVL 1
###### Who is Participating?
Commented:
Hello John,
Your existing LOOKUP formula will work accurately only if the data is sorted (ascending) by the date column, as it finds the last match for your criteria that will also be the MAX if sorted....
If unsorted then I recommend this "array formula"
=INDEX(LKP_Tails,MATCH(1,(LKP_EIS=MAX(IF((LKP_SN=F5)*(LKP_EIS<=CutOffDate),LKP_EIS)))*(LKP_SN=F5),0))
confirmed with CTRL+SHIFT+ENTER
see attached
regards, barry
27346272.xls
0
Commented:
=MIN(CutOffDate,MAX(IF(LKP_SN=F5,LKP_EIS)))
0
Commented:
Hello John,
Generically you can use a formula like this
=MAX(IF(A1:A100=D1,IF(B1:B100<=G1,B1:B100)))
Array entered.....
where strings are in A and dates in B with specific string to match in D1
regards, barry
0
Commented:
Hello John, are you looking for a date? That's what your description says but your formula in J5 is returning a value from column C...not a date?
Assuming you do want a date then using the setup I suggested above but adjusted for your specific ranges you get
=MAX(IF(LKP_SN=F5,IF(LKP_EIS<=G1,LKP_EIS)))
format result cell as
mm/dd/yyyy;;
....then if there is no match you'll get a blank
Stephen's formula will give the same result in this case and possibly others but if all the dates for the qualifying rows were greater than the cutoff date my formula returns a blank/zero while Stephen's returns the cutoff date
0
Reliability Business Tools Analyst IIAuthor Commented:
My typo, sorry about that. What I'm looking for is the value in Column C corresponding to the highest date (in Column B) associated with the three 27613's in Column D.
I tried customizing your formulas, both of which work, but I couldn't figure it out. Could you re-post the workbook with a working example?
Thanks!
John
0
Reliability Business Tools Analyst IIAuthor Commented:
Thanks Barry, that really does the trick. I wanted to give Stephen some points for giving me the solution I inadvertently asked for!
- John
0
Commented:
Thanks John!
0
Question has a verified solution.
Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.
Have a better answer? Share it in a comment. | 659 | 2,670 | {"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-2018-13 | latest | en | 0.88701 |
https://exceladept.com/understanding-operators-in-excel/ | 1,721,900,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00228.warc.gz | 216,105,405 | 25,474 | # Understanding Operators In Excel
## Key Takeaway:
• Excel operators are symbols or characters that perform calculations, compare values, manipulate text, and make logical decisions based on true or false conditions.
• The five main types of operators in Excel are arithmetic operators, comparison operators, text operators, reference operators, and logical operators. Each type serves a specific purpose and can be used together to create complex formulas and expressions.
• To use operators in Excel, you must first understand the order of operations, which determines the sequence in which calculations are performed. You can also use parentheses to override the default order and control the evaluation of formulas.
Confused about how to use operators in Excel? You’re not alone! In this article, we’ll give you the breakdown on how to understand and use them correctly for more efficient data analysis.
## The Different Types of Operators
To get an understanding of Excel operators, you need to take a deeper look. From arithmetic to comparison, text, reference and logical, each type has its own benefits. Break it down and learn their advantages. Then decide which ones are best for your Excel needs.
### Arithmetic Operators
Understanding the calculations in Excel requires an awareness of different arithmetic functionality. Symbolic designators, such as “+”, “-“, “/”, and “*” are known as Arithmetic Operators. These operators work on numerical values to perform basic functions including addition, subtraction, multiplication, and division. By using these operators whilst incorporating cell values, a user can create a variety of macro-level functions.
For instance, typing “=5+2” results in the return of “7,” whereas typing “=5/2” follows the rules of division calculations returning the value 2.5. Similarly, “=6%4” returns “2“, factorizing out the remaining two when dividing six by four.
It is also crucial to be mindful that BODMAS operations apply to all arithmetic functionalities performed within a spreadsheet cell unless otherwise stated. This rule mandates calculations to be resolved in brackets before proceeding with any other function or operation.
Regarding historical development, arithmetic operators have been acknowledged since the early days of computers and programming languages in conjunction with coding exercises for handheld calculators and table processors.
Comparing the similarities between Excel operators and exes? At least with operators, you know what you’re getting into.
### Comparison Operators
To compare data in Excel, we use specific symbols called Comparison Operators. These symbols are used to compare values and determine if they fulfill a specific criterion. For instance, we can use these operators to compare numbers, dates, text strings, etc.
Below is a table showcasing the various Comparison Operators available in Excel:
Operator Meaning Example
= Equal to A1 = B1
> Greater than B2 > C2
< Less Than D1 < E1
>= Greater than or equal to F2 >= G2
<= Less than or equal to H1 <= I1
<> Not equal to J2 <> K2
Interestingly, these operators can also be combined with logical formulas like AND and OR operators to make complex comparisons.
It’s noteworthy that these comparison operators can only be used on numeric values in some cases. In other scenarios, we need to use Text Operators like “Contains” or “Begins With” for comparing text data specifically.
According to an article by the Microsoft team titled “Using Comparison Operators“, it’s easier for users of all levels of experience to carry out operations using Comparison Operators.
Manipulating text in Excel is like trying to untangle headphone wires – it takes patience, precision, and a lot of swear words.
### Text Operators
Text: Text Manipulation Methods in Excel
Text manipulation methods in Excel refer to the different types of operators used to manipulate text. These operators are extremely useful for cleaning up and processing a large amount of data.
Here are some common text operators you can use:
• Concatenation Operator: This operator allows you to join two or more pieces of text into one cell using the “&” symbol.
• Left, Right and Mid Operators: These operators allow you to extract specific sections of text from a cell based on the position of the characters within it.
• LEN Function: This function calculates the number of characters in a cell.
• LOWER and UPPER Functions: These functions convert text into lower or uppercase format respectively.
• Search and Replace Operators: The search operator helps locate specific characters or words within cells, while replace allows you to substitute these characters with new ones.
• Txt Function: This function enables users to display numeric values as a text string with specified formatting codes.
When using these operators, keep in mind that they are case-sensitive, so ensure that any exact matches are capitalized or lowercase accordingly.
It is essential to understand how these various text manipulation methods can simplify your work when dealing with large data sets. They not only save time but also reduce errors and increase productivity.
According to Microsoft, Excel is still one of the most widely used software programs worldwide, and its extensive features make it an incredibly versatile tool for businesses across many industries.
Reference operators in Excel are like a GPS for your data, guiding you to the exact cell you need with precision… or leading you to a dead end if you’re not careful.
### Reference Operators
Operators for referencing in Excel are vital for completing complex calculations and organizing data. These operators help link multiple cells together, facilitating users’ access to the information they require, leading to better decision making.
Operator Description
: Creates a range between two cells.
, Selects non-adjacent cells for merging or editing at once.
= Sets a formula or function to perform a task.
Reference operators may also include the use of symbols like “\$” and “&” that alter referenced cell values to adhere to user needs. It is crucial to understand these operators fully when performing complex formulas.
History-wise, referencing operators came into existence with the introduction of spreadsheets, which could store numerical data and perform calculations on it. As personal computing expanded its reach worldwide in the latter part of the 20th century, referencing operators became standard aspects of spreadsheet software like Microsoft Excel and Lotus123.
Logical operators in Excel: where the computer finally gets to make its own decisions.
### Logical Operators
In Excel, operators are used to perform various calculations and comparisons between values. One type of operator is known as Logical Operations. These operations are used to evaluate whether a condition is True or False.
Logical Operators can be categorised into three types – AND, OR and NOT. Using the AND operator results in True if both conditions are met. On the other hand, using the OR operator results in True if either condition is met. Lastly, the NOT operator changes a true condition to false, and vice versa.
Using Logical Operators in Excel helps automate workflows by allowing more complex calculations and comparisons between values. Remember to use brackets when working with multiple logical operators to ensure correct evaluation of conditions.
Overall, understanding Logical Operators can help improve problem-solving abilities and increase efficiency when working with data in Excel. Get ready to unleash the power of operators in Excel and make those pesky calculations bow down to your will.
## How to Use Operators in Excel
In Excel, mastering the use of operators is essential for efficient calculations and data analysis. Here is a concise guide on how to leverage operators in Excel:
1. Arithmetic Operators: Excel supports the standard arithmetic operations of addition (+), subtraction (-), multiplication (*), and division (/). Operators can also be combined to create complex expressions.
2. Comparison Operators: To compare values in Excel, you can use the comparison operators such as less than (<), greater than (>), equal to (=), not equal to (<>), less than or equal to (<=), and greater than or equal to (>=).
3. Logical Operators: Excel has three logical operators – AND, OR, and NOT. These operators can be used to create complex logical expressions.
4. Concatenation Operator: In Excel, the ampersand (&) operator can be used to combine strings of text from different cells.
5. Range Operators: The colon (:) operator is used to select a range of cells. It can also be combined with other operators to perform complex calculations.
Notably, when using operators in Excel, it is crucial to consider the order of operations to ensure accuracy.
In addition, operators can be used in Excel’s conditional formatting feature to highlight specific data.
A successful data analyst once shared that understanding operators in Excel was the stepping stone to advancing in their career. It allowed them to create complex calculations efficiently and improved their data analysis skills.
## Common Mistakes to Avoid When Using Operators in Excel
In Excel, errors can occur when using operators. To avoid these mistakes, it is important to understand how operators work and their proper usage. Here are three common mistakes to avoid when using operators in Excel:
• Incorrect Operator Usage: Using the wrong operator or using an operator incorrectly can compromise the accuracy and integrity of your data.
• Not Following Order of Operations: When performing multiple calculations, not following the correct order of operations can lead to incorrect results.
• Using Blank Cells: Operators can produce errors if blank cells are used in calculations without taking into account the potential impact of those cells on the formula.
To avoid these mistakes, make sure to double-check the operator being used, follow the order of operations, and consider the impact of blank cells on calculations. These strategies will help ensure accuracy and reliability in your data analysis.
In addition to these common mistakes, it is also important to be aware of the potential for errors when working with large data sets or complex calculations. Taking the time to carefully review your work and troubleshoot any issues that arise can save time and frustration in the long run.
Remember that accuracy is key when working with Excel data, so take the necessary steps to avoid errors and ensure the best possible outcomes for your projects.
Make sure to stay up-to-date on the latest Excel best practices and seek out resources for continued learning and improvement. Don’t let the fear of missing out on important techniques and strategies hold you back from achieving success in Excel.
## Some Facts About Understanding Operators in Excel:
• ✅ Operators in Excel are symbols used to perform calculations on cells and values. (Source: Microsoft)
• ✅ There are several types of operators in Excel, including arithmetic, comparison, logical, and reference operators. (Source: Excel Easy)
• ✅ Using operators can save time and simplify complex calculations in Excel. (Source: Techwalla)
• ✅ It’s important to understand the order of operations (PEMDAS) when using multiple operators in a single formula in Excel. (Source: Lifewire)
• ✅ Excel allows for the use of custom operators through the creation of custom functions using Visual Basic for Applications (VBA). (Source: Ablebits)
## FAQs about Understanding Operators In Excel
### What are operators in Excel?
Operators in Excel are symbols or characters that perform various arithmetic or logical operations, such as addition, subtraction, multiplication, division, and comparison. They are used to perform calculations and manipulate data in Excel formulas.
### What types of operators are used in Excel?
Excel supports various types of operators, including arithmetic operators (+,-,*,/), comparison operators (=,>,<,>=,<=,<>) and logical operators (AND, OR, NOT). For example, the equals sign (=) is used to assign values to a cell and comparison operators are used to compare two values.
### How do I use arithmetic operators in Excel?
To use arithmetic operators in Excel, simply enter them between two values in a formula. For example, to add two values, enter ‘value1 + value2’ in a cell. Excel will automatically calculate the result and display it in the cell.
### How do I use comparison operators in Excel?
Comparison operators are used to compare two values in Excel. To use these operators, you can enter them in a cell with two values separated by the operator. For example, to check if a value is equal to another value, enter ‘value1 = value2’ in a cell.
### What are logical operators in Excel?
Logical operators in Excel are used to combine multiple conditions in a formula. For example, the AND and OR operators can be used to check if multiple conditions are met before a calculation is performed. The NOT operator can be used to negate a condition.
### Can I use multiple operators in a single formula in Excel?
Yes, you can use multiple operators in a single formula in Excel. Excel uses the order of operations (PEMDAS) to calculate the formula, so it’s important to specify the correct operator precedence. You can also use parentheses to group certain parts of the formula and ensure that they are calculated first. | 2,569 | 13,506 | {"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} | 3.234375 | 3 | CC-MAIN-2024-30 | latest | en | 0.895463 |
https://manual.cs50.io/3/fmaf | 1,725,894,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00058.warc.gz | 358,441,502 | 4,698 | # NAME
fma, fmaf, fmal - floating-point multiply and add
# LIBRARY
Math library (`libm`, `-lm`)
# SYNOPSIS
``````#include <math.h>
double fma(double x, double y, double z);
float fmaf(float x, float y, float z);
long double fmal(long double x, long double y, long double z);``````
Feature Test Macro Requirements for glibc (see feature_test_macros(7)):
fma(), fmaf(), fmal():
`` _ISOC99_SOURCE || _POSIX_C_SOURCE >= 200112L``
# DESCRIPTION
These functions compute `x` * `y` + `z`. The result is rounded as one ternary operation according to the current rounding mode (see fenv(3)).
# RETURN VALUE
These functions return the value of `x` * `y` + `z`, rounded as one ternary operation.
If `x` or `y` is a NaN, a NaN is returned.
If `x` times `y` is an exact infinity, and `z` is an infinity with the opposite sign, a domain error occurs, and a NaN is returned.
If one of `x` or `y` is an infinity, the other is 0, and `z` is not a NaN, a domain error occurs, and a NaN is returned.
If one of `x` or `y` is an infinity, and the other is 0, and `z` is a NaN, a domain error occurs, and a NaN is returned.
If `x` times `y` is not an infinity times zero (or vice versa), and `z` is a NaN, a NaN is returned.
If the result overflows, a range error occurs, and an infinity with the correct sign is returned.
If the result underflows, a range error occurs, and a signed 0 is returned.
# ERRORS
See math_error(7) for information on how to determine whether an error has occurred when calling these functions.
The following errors can occur:
Domain error: `x` * `y` + `z`, or `x` * `y` is invalid and `z` is not a NaN
An invalid floating-point exception (FE_INVALID) is raised.
Range error: result overflow
An overflow floating-point exception (FE_OVERFLOW) is raised.
Range error: result underflow
An underflow floating-point exception (FE_UNDERFLOW) is raised.
These functions do not set `errno`.
# ATTRIBUTES
For an explanation of the terms used in this section, see attributes(7).
Interface Attribute Value fma(), fmaf(), fmal() Thread safety MT-Safe
# STANDARDS
C11, POSIX.1-2008.
# HISTORY
glibc 2.1. C99, POSIX.1-2001. | 605 | 2,154 | {"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} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.787005 |
https://math.stackexchange.com/questions/3893122/does-the-computation-of-the-inverse-function-of-a-set-that-is-not-in-the-image-o | 1,718,383,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861567.95/warc/CC-MAIN-20240614141929-20240614171929-00871.warc.gz | 338,137,968 | 35,196 | # Does the computation of the inverse function of a set that is not in the image of $f$ makes sense?
I am studying measure theory, and I was wondering if the computation of the inverse function $$f^{-1}$$ of a set that is not in the image of $$f$$ makes sense. The fact is that all the functions that I have seen in my course are defined $$(\Omega,F)\rightarrow (\mathbb{R},B(\mathbb{R}))$$. But for example, let us consider the logistic function $$f(x)=\frac{1}{1-e^{-x}}$$, would that function be defined on $$(\mathbb{R},B(\mathbb{R}))\rightarrow ([0,1],B([0,1])$$ or in $$(\Omega,F)\rightarrow (\mathbb{R},B(\mathbb{R}))$$? In the latter case, what would happened for the sets that are not in the image of $$f$$?.
• $f^{-1}(Y)=\{\,x\in X\mid f(x)\in Y\,\}$ will simply be empty if $Y$ is disjoint from the image of $f$. Commented Nov 3, 2020 at 21:44
Note that here $$f^{-1}$$ does not necessarily refer to the inverse function. Indeed, $$f$$ might not admit an inverse. Instead, $$f^{-1}$$ denotes the pre-image under $$f$$. I.e. $$f^{-1}(A)=\{x\in\Omega|\;f(x)\in A\}$$ In the case where $$A\cap f(\Omega)=\emptyset,$$ you'll simply conclude that $$f^{-1}(A)=\emptyset$$.
Note that if $$f$$ is injective and $$A\subseteq f^{-1}(\Omega)$$, then the two notions (pre-images and images under the inverse function) agree and thus, the difference is pedantic at best. However, the pre-image always makes sense for any function and thus, is a better concept in general. | 444 | 1,472 | {"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": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.876814 |
https://math.stackexchange.com/questions/2667222/identity-for-lower-central-series | 1,571,659,839,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987773711.75/warc/CC-MAIN-20191021120639-20191021144139-00547.warc.gz | 603,695,470 | 30,712 | # Identity for Lower Central Series
Let $G$ be a group and $N \trianglelefteq G$ a normal subgroup. Consider the lower central series
$$G = \gamma_0 G \trianglerighteq \gamma_1 G \trianglerighteq ... \trianglerighteq \gamma_n G= { 1 }$$
terminating in the trivial subgroup after finitely many steps, where $\gamma_{i+1}G := [\gamma_iG, G]$.
How to see that following equation holds:
$$\gamma_i (G/N) = (\gamma_i G)N / N$$
Using induction on $i$ I get only $$\gamma_{i+1} (G/N) = [\gamma_i (G/N), G/N] = [(\gamma_i G)N / N, G/N] \subset [\gamma_i (G), G]N/N = (\gamma_{i+1} G)N / N$$
but how to get an inclusion in other direction?
• You should change the $\zeta$ into $\gamma$, otherwise you are mixing the standard notation for upper and lower central series. – Nicky Hekster Feb 26 '18 at 9:47
• The inclusion in the middle looks like an equality to me. A commutator $[a,b]N$ in $[\gamma_i(G),G)]N/N$ and be written as $[aN,bN]$. – Derek Holt Feb 26 '18 at 10:30 | 321 | 972 | {"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} | 3.28125 | 3 | CC-MAIN-2019-43 | latest | en | 0.799188 |
http://mathhelpforum.com/calculus/81972-slope-field-differential-equations-print.html | 1,529,421,524,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863043.35/warc/CC-MAIN-20180619134548-20180619154548-00109.warc.gz | 204,503,065 | 3,690 | # Slope Field and Differential Equations
• Apr 2nd 2009, 10:02 AM
premed11
Slope Field and Differential Equations
Ok so I have a calc exam tmrw and I am having a hard time with slope fields. My professor told us that on the exam he is going to give us the slope field (graph) and we are going to have to pick one differential equation from a set as the one that matches the graph.
SO my question is: how can you tell which slope field pretains to what graph? What are the rules in figuring that out?
• Apr 2nd 2009, 11:33 AM
coolguy99
Well, what do you know about derivatives? If a derivative is positive, what is the original function doing? If a derivative is negative, what is the original function doing? Or if the derivative is 0?
• Apr 2nd 2009, 12:17 PM
premed11
Quote:
Originally Posted by coolguy99
Well, what do you know about derivatives? If a derivative is positive, what is the original function doing? If a derivative is negative, what is the original function doing? Or if the derivative is 0?
A positive derivative means that the function is increasing
A negative derivative means that the function is decreasing
A zero derivative means that the function has some special behaviour at the given point. (max/min)
for the second derivative:
if it is positive then it is concave up
if it negative then it is concave down
Even when I consider the above rules...I cant seem to figure it out.
For example: y'=1+y^2 the graph-answer given in the book is the same that my calculator gives but if I had to say why I chose that graph in words ( without saying "thats what the calculator had!") I would be lost.
• Apr 2nd 2009, 01:40 PM
coolguy99
If they give you a slope field, then you just need to correlate the positive/negative values of the slope to their coordinates on the real equation. If the derivative is negative from x=1 to x=7, find an equation that is decreasing on those values. etc
if you have a more specific question or example I could help you out better haha.
• Apr 2nd 2009, 02:05 PM
premed11
Quote:
Originally Posted by coolguy99
If they give you a slope field, then you just need to correlate the positive/negative values of the slope to their coordinates on the real equation. If the derivative is negative from x=1 to x=7, find an equation that is decreasing on those values. etc
if you have a more specific question or example I could help you out better haha.
lol ok.
So lets say I have y'=sin(x) and I am able to tell what graph it corresponds to by using my calculator. Then if I had to explain why I chose that graph what would I say?
Same thing for y'=1+y^2.
I just want to get an idea of how you would reason your answer.
• Apr 2nd 2009, 02:17 PM
coolguy99
You would explain the interval on which the derivative is positive/negative, then correlate it to the graph you chose by saying that it is increasing/decreasing on those same intervals, as well as noting relative maximums and relative minimums appearing at their respective places in relation to the derivative's cross-over points.
• Apr 2nd 2009, 02:34 PM
premed11
Quote:
Originally Posted by coolguy99
You would explain the interval on which the derivative is positive/negative, then correlate it to the graph you chose by saying that it is increasing/decreasing on those same intervals, as well as noting relative maximums and relative minimums appearing at their respective places in relation to the derivative's cross-over points.
Ok I totally understand what you are saying about the y'=sin(x) graph. I checked the slope field and it was just like you described it. But when I tried to do the other equations once again I failed. lol I dont even know why such a small thing is confusing me, I have an A in the class!
So it if was something like y'=4-y how would you explain it? | 914 | 3,793 | {"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-2018-26 | latest | en | 0.948658 |
http://www.askmehelpdesk.com/2784252-post3.html | 1,369,201,806,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701409268/warc/CC-MAIN-20130516105009-00070-ip-10-60-113-184.ec2.internal.warc.gz | 326,712,526 | 5,995 | View Single Post
jcaron2 Posts: 983, Reputation: 1034 Senior Member #3 Apr 26, 2011, 10:40 AM
The length of the L.R. Is 4 times the distance from the focus to the vertex. It's centered on the focus, and it's oriented perpendicular to the line connecting the focus and the vertex. Since the focus and vertex fall in line vertically in this case, that means the L.R. Is horizontal.
So what's the distance between (0,2) and (0,0)? The length of the L.R. Will be four times that length. So find two points who lie on a horizontal line, are centered at (0,2), and are spaced apart the appropriate length of the L.R.
Reply with your answer, and we'll tell you if you got it right or give you more guidance if you need it. | 188 | 718 | {"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-2013-20 | latest | en | 0.938862 |
https://math.stackexchange.com/questions/931488/for-the-given-quadratic-equation-find-the-value-of-p | 1,576,358,785,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541294513.54/warc/CC-MAIN-20191214202754-20191214230754-00515.warc.gz | 447,167,985 | 27,813 | # For the given quadratic equation find the value of p
For the equation 3x^2 + px + 3 = 0 , p>0, if one of the roots is square of the other, then p is equal to?
Solving the equation, i get the value of p as -6 but the question states that p>0. Is there another way of solving this or is the question wrong?
• I agree with you; I think the question has a mistake. – user84413 Sep 14 '14 at 21:39
• @user84413: You can consider complex roots and then $p>0$ is possible. – gammatester Sep 15 '14 at 7:52
• @gammatester Thanks - I should have realized that. – user84413 Sep 15 '14 at 22:01
The value is $p=3$.
Proof: Let $a, a^2$ be the roots of the equation. Then you get $$3x^2+px+3 = 3(x-a)(x-a^2) = 3x^2-3(a^2+a)x+a^3,$$ So you first have to solve $a^3=1$. This gives the three solutions (the three cube-roots of unity) $$a_1=1,\quad a_2=-\frac{1}{2}+\frac{i}{2}\sqrt{3},\quad a_3=-\frac{1}{2}-\frac{i}{2}\sqrt{3}$$ which in turn give the possible values for $p$ $$p_1 = -3(a_1^2+a_1) = -6$$ $$p_2 = -3(a_2^2+a_2) = 3$$ $$p_3 = -3(a_3^2+a_3) = 3$$ $p_1\;$ is your negative value, but $p_2=p_3=3\;$ is positive and therefore $p=3\;$ is the answer to the question.
• Wow!Thanks a lot gammatester :) – Mohit Singh Sep 15 '14 at 14:02 | 468 | 1,236 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-51 | latest | en | 0.836444 |
https://codegolf.stackexchange.com/questions/25314/predict-where-the-man-will-go/25372 | 1,566,558,217,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027318375.80/warc/CC-MAIN-20190823104239-20190823130239-00136.warc.gz | 423,293,222 | 39,508 | Predict where the man will go
A man lives in the north-west corner (0, 0) of a town with height h and width w . Everyday he walks from his home to the border (?, w) or (h, ?). In the following example, the man goes to (3, 3) today.
(0, 0) +--+ + + . (0, 4)
|
+ +--+--+ .
|
+ + + + .
|
(3, 0) . . . . . (3, 4)
The man records a bit at each points (+ in example above). Every time he reaches a point, he goes east if the bit is 1 and south otherwise. The bit is flipped after he leaves. For example:
Day 1: 1--0 1 1 Day 2: 0 1 1 1 Day 3: 1--1--1--1-- Day 4: 0 0 0 0
| | |
0 1--0 0 0 0 1 0 1 0 1 0 1--0 1 0
| | |
1 0 1--0 1--0 0 1 0 1 0 1 0 1--0 1
| | |
Destination: (3, 3) Destination: (3, 1) Destination: (0, 4) Destination: (3, 2)
Given the size of the town and the man's record, calculate the man's destination after n days.
Input:
In the first line are three integers, h, w and n.
In the following h lines are w integers, denoting the man's record.
h <= 1000, w <= 1000, n <= 1000000000
Output:
Two integers, denoting the man's destination after n days.
Sample Input:
3 4 3
1 0 1 1
0 1 0 0
1 0 1 0
Sample Output:
0 4
Sample Code:
#include <iostream>
using namespace std;
bool d[1000][1000];
int main(){
int h, w, n;
cin >> h >> w >> n;
for(int i = 0; i < h; i++)
for(int j = 0; j < w; j++)
cin >> d[i][j];
int i, j;
while(n--)
for(i = 0, j = 0; i < h && j < w;){
bool &b = d[i][j];
d[i][j] ? j++ : i++;
b = !b;
}
cout << i << " " << j << endl;
}
Scoring:
• Lowest byte count in UTF-8 wins.
• If the running time of your code is independent of n, reduce your score by 50%.
• Don't just calculate the results of all 1000000000 days or do anything similarly stupid to get this bonus. Find an efficient algorithm!
• 2 things I dont understand. The output, sometimes you use 0 index other times you dont. How does that work? Should it be like border+1? Second thing is the second line with scoring. How do you mean that? – Teun Pronk Apr 4 '14 at 9:58
• Day 4 should output 3,2 right? – Teun Pronk Apr 4 '14 at 10:26
• If, no matter what n is, my code calculates the results of all 1000000000 days, then output the result of n, do I still get the -50% bonus? – ace Apr 4 '14 at 10:48
• @ace now you put it like that, it does make sense doesnt it? Thanks for that :P – Teun Pronk Apr 4 '14 at 10:59
• @TeunPronk Yes. It's my fault. – johnchen902 Apr 4 '14 at 11:14
Delphi XE3 (437 bytes|| 897874 without bonus counted)
When thinking about how to solve this with the bonus I thought of the following.
If you walk 4 days cell 0,0 is changed 4 times. The cell on its right is changed twice aswell as the cell beneath it.
If there is an uneven number of days and the number in the cell starts with 1 the cell on the right gets one more than the cell beneath, and the other way around if the cell is 0.
By doing this for every cell you can see if the end value should be changed by: Cell was changed X times. if X mod 2>0 then change the cell.
Results in the following code:
{Whispers at JohnChen902} do I get your upvote now? :P
uses SysUtils,Classes,idglobal;var a:TArray<TArray<byte>>;b:TArray<TArray<int64>>;h,w,x,y,t:int16;n:int64;s:string;r:TStringList;tra:byte;begin r:=TStringList.Create;readln(h,w,n);h:=h-1;w:=w-1;for y:=0to h do begin readln(s);r.Add(StringReplace(s,' ','',[rfReplaceAll]));end;SetLength(a,h);SetLength(b,h);for y:=0to h do begin SetLength(a[y],w);SetLength(b[y],w);for x:=1to Length(r[y])do a[y][x-1]:=Ord(r[y][x])-48;end;b[0][0]:=n-1;for Y:=0to h do for X:=0to w do begin t:=b[y][x];if x<w then b[y][x+1]:=b[y][x+1]+iif((t mod 2=1)and(a[y][x]=1),(t div 2)+1,t div 2);if y<h then b[y+1][x]:=b[y+1][x]+iif((b[y][x]mod 2=1)and(a[y][x]=0),(t div 2)+1,t div 2);end;for Y:=0to h do for X:=0to w do if b[y][x]mod 2=1then a[y][x]:=iif(a[y][x]=1,0,1);y:=0;x:=0;repeat a[y][x]:=iif(a[y][x]=1,0,1);if a[y][x]=1then inc(y) else inc(x);until(y>h)or(x>w);write(Format('%d %d',[y,x]));end.
Ungolfed
uses
SysUtils,Classes,idglobal;
var
a:TArray<TArray<byte>>;
b:TArray<TArray<int64>>;
h,w,x,y,t:int16;
n:int64;
s:string;
r:TStringList;
tra:byte;
begin
r:=TStringList.Create;
h:=h-1;w:=w-1;
for y:=0to h do
begin
end;
SetLength(a,h);
SetLength(b,h);
for y:=0to h do
begin
SetLength(a[y],w);
SetLength(b[y],w);
for x:=1to Length(r[y])do
a[y][x-1]:=Ord(r[y][x])-48;
end;
b[0][0]:=n-1;
for Y:=0to h do
for X:=0to w do
begin
t:=b[y][x];
if x<w then
b[y][x+1]:=b[y][x+1]+iif((t mod 2=1)and(a[y][x]=1),(t div 2)+1,t div 2);
if y<h then
b[y+1][x]:=b[y+1][x]+iif((b[y][x]mod 2=1)and(a[y][x]=0),(t div 2)+1,t div 2);
end;
for Y:=0to h do
for X:=0to w do
if b[y][x]mod 2=1then
a[y][x]:=iif(a[y][x]=1,0,1);
y:=0;x:=0;
repeat
a[y][x]:=iif(a[y][x]=1,0,1);
if a[y][x]=1then
inc(y)
else
inc(x);
until(y>h)or(x>w);
write(Format('%d %d',[y,x]));
end.
• You haven't get my vote yet. I was eating dinner. (Upvoted) – johnchen902 Apr 4 '14 at 13:46
C++ 213 bytes * 0.5 = 106.5
Here is my solution. It's similar to user2357112's solution, but there are several difference:
• First, I dispatch visiting times to the right and bottom, instead of compute them from the top and left.
• Second, I do everything (reading input, dispatching, tracking the man's location) simultaneously.
• Third, I keep only one row of memory.
#include <iostream>
int o[1001],h,w,r,c,i,j,t,u;int main(){std::cin>>h>>w>>*o;for(;i<h;i++)for(j=0;j<w;)std::cin>>t,u=o[j],o[j]/=2,u%2&&o[j+t]++,r-i|c-j||((u+t)%2?r:c)++,o[++j]+=u/2;std::cout<<r<<" "<<c<<"\n";}
Here is the ungolfed version:
#include <iostream>
using namespace std;
int o[1001];
int main(){
int h, w, n;
cin >> h >> w >> n;
o[0] = n;
int r = 0, c = 0;
for(int i = 0; i < h; i++)
for(int j = 0; j < w; j++){
bool t;
cin >> t;
int u = o[j];
o[j + 1] += u / 2;
o[j] = u / 2;
if(u % 2)
(t ? o[j + 1] : o[j])++;
if(r == i && c == j)
((u + t) % 2 ? r : c)++;
}
cout << r << " " << c << endl;
}
• These three differences make things much terser. We can shorten the indexing and combine several redundant data structures. The logic for pushing visits forward turns out to be much shorter than the logic for pulling visits from previous cells. Horizontal boundary conditions are handled simply by extending the data structure an extra space to the right, and vertical boundary conditions aren't an issue. – user2357112 Apr 5 '14 at 5:47
• I've upvoted your answer and incorporated the concepts into my own code. So far, they've taken 84 bytes out of my solution, an improvement of 30%. – user2357112 Apr 5 '14 at 5:50
• I suspect you might be able to save some bytes by not doing --*o;, and instead switching which case you move the guy down and which case you move the guy to the right. – user2357112 Apr 5 '14 at 5:59
• @user2357112 Implemented, but code length increase due to a previous mistake (It should have been 218 bytes). – johnchen902 Apr 5 '14 at 8:20
Python, 177 bytes
My first try ever in Code Golfing, so sorry if I got something wrong here! Code used to grab the input based on user2357112's code.
l=lambda:map(int,raw_input().split())
h,w,n=l()
m=[l() for i in[1]*h]
while n>0:
n-=1;x=y=0
while x!=w and y!=h:
if m[y][x]>0:m[y][x]=0;x+=1
else:m[y][x]=1;y+=1
print y,x
Input:
3 4 3
1 0 1 1
0 1 0 0
1 0 1 0
Output:
0 4
R, 196 bytes * 0.5 = 98
f=function(h,w,n,x){I=J=rep(1,n);for(i in 1:h)for(j in 1:w){M=which(I==i&J==j);N=length(M);if(N){z=seq(1,N,by=2);if(x[i,j])z=-z;f=M[-z];s=M[z];I[f]=i;J[f]=j+1;I[s]=i+1;J[s]=j}};cat(I[n]-1,J[n]-1)}
Ungolfed:
f=function(h,w,n,x) {
I = J = rep(1,n)
for(i in 1:h) for(j in 1:w) {
M = which(I==i&J==j)
N = length(M)
if (N) {
z = seq(1,N,by=2)
if (x[i,j]) z = -z
f = M[-z]
s = M[z]
I[f] = i
J[f] = j+1
I[s] = i+1
J[s] = j
}
}
cat(I[n]-1, J[n]-1)
}
Usage:
f(3,4,4,matrix(c(1,0,1,0,1,0,1,0,1,1,0,0),3))
3 2 | 2,990 | 8,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-35 | longest | en | 0.760002 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition-anton/chapter-2-the-derivative-2-7-implicit-differentiation-exercises-set-2-7-page-166/33 | 1,575,636,157,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540488620.24/warc/CC-MAIN-20191206122529-20191206150529-00168.warc.gz | 723,982,826 | 12,811 | ## Calculus, 10th Edition (Anton)
$\dfrac{d\omega}{d\lambda}=-\dfrac{b^2\lambda}{a^2\omega}$
In order to derivate this function you have to apply implicit differentation method. First, take the function to it's $f(\lambda)=0$ form $a^2\omega^2+b^2\lambda^2-1=0$ Then derivate the whole equation. Rember the put $\omega'$ every time you derivate $\omega$ $2a^2\omega \omega^2 + 2b^2\lambda=0$ Solve for a' and you have the answer $\omega' (2a^2\omega)=-2b^2\lambda$ $\omega '=-\dfrac{b^2\lambda}{a^2\omega}$ | 187 | 507 | {"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} | 3.9375 | 4 | CC-MAIN-2019-51 | latest | en | 0.538787 |
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SailBoat Speed
How do you determine speed of a sailboat before you buy it? When I say speed of a sailboat I mean in comparison to wind speed. What can you do to enhance the performance of a sailboat?
Thank you
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Check the polar diagrams. That will give you a rough idea of how a well-sailed boat will perform under various wind speeds at various points of sail.
Another way to compare boats, at least on a rough basis, is to look at their PHRF ratings. The higher the rating, the slower the boat.
There really isn't a lot you can do to enhance the performance of a boat. Other than keeping excess weight off the boat, fairing and smoothing the bottom and keeping it free of growth, keeping your sails in good shape, etc... but the basic speed of a given boat won't change much.
Some designs are just going to be much faster than others... Multihulls are a good example of that... trimarans, and to some degree catamarans, are going to be faster than their monohull counterparts.
Sailingdog
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Telstar 28
New England
You know what the first rule of sailing is? ...Love. You can learn all the math in the 'verse, but you take
a boat to the sea you don't love, she'll shake you off just as sure as the turning of the worlds. Love keeps
her going when she oughta fall down, tells you she's hurting 'fore she keens. Makes her a home.
—Cpt. Mal Reynolds, Serenity (edited)
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Last edited by sailingdog; 04-22-2010 at 10:20 AM.
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Quote:
Originally Posted by Sailboat101 View Post
How do you determine speed of a sailboat before you buy it? When I say speed of a sailboat I mean in comparison to wind speed.
Wow! What a far-reaching question! But, let me start off the discussion, and there'll be more contributions from others.
Generally, sailboats have displacement hulls, and their speed capabilities are traditionally thought to be limited by their waterline length, but that's not exactly true. Sailboats can, and often do, exceed their theoretical hull speed, especially when sailing off the wind. Therefore, you can get a very limited idea of a boat's speed capability by determining its hull speed, which can be calculated by using the following formula: Hull Speed = 1.34 x the square root of the LWL (length of the waterline)
However, the length of the waterline doesn't tell the whole story. Some sailboats with a waterline of a certain length will be faster than other sailboats with the same waterline length, due to other design variables, such as the beam (width) of the boat, and the shape of the keel or rudder, the amount of sail area, and the displacement (weight) of the boat, among other things.
The differences in the design features of a sailboat can cause one boat to exert much more drag than another boat as it moves through the water, reducing it's speed. Design differences can also enable one boat to plane more easily than another boat, which greatly reduces the amount of drag and increases speed.
Quote:
What can you do to enhance the performance of a sailboat?
The "performance" of a sailboat includes not only it's speed capability, but also it's ability to sail close to the wind, among other things. I assume your question refers to speed.
Generally, you can enhance a boat's speed by maximizing the ability of it's sails to generate driving force, and by minimizing drag. You should use the biggest sails that the boat can carry efficiently, and trim them well. You can reduce drag by placing the crew's weight in a position that will usually keep the boat as upright as possible, and will result in the least amount of wetted surface. There are a gazillion other factors that often take a lifetime to learn, but I'll let others try to think of them all.
You might ask, why would anyone use design elements in a boat that don't maximize it's speed? The reason is because speed isn't the only quality that people find desirable in a sailboat. Sailboat's are designed for a particular purpose. Some are designed for all-out speed and pointing ability, and some are designed for safe, reasonably fast and comfortable cruising. People buy the boat that best meets their needs.
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Quote:
Originally Posted by Sailormon6 View Post
Generally, sailboats have displacement hulls, and their speed capabilities are traditionally thought to be limited by their waterline length, but that's not exactly true. Sailboats can, and often do, exceed their theoretical hull speed, especially when sailing off the wind. Therefore, you can get a very limited idea of a boat's speed capability by determining its hull speed, which can be calculated by using the following formula: Hull Speed = 1.34 x the square root of the LWL (length of the waterline)....
That's a good point to make out diferent types of cruising boats, in what concerns speed:
There are some sailingboats that downwind are pratically limited to their hull speed, other that can go easely 2 or 3k over hull speed (allmost all modern designs) and there are some new generation fast light cruising boats that can sail safely downwind at almost two times their hull speed.
Regards
Paulo
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In addition to the US Sailing's 'polar' diagrams, you can also get such info from the US Sailing "VPP programing" - will differentiate between different equipment, differing sails used, etc. etc.
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So I looked up data PHRF data on the two boats in question. I nearly fell
over when I found that the Bahama Islander Older and Heavier boat is faster...
Is this right?
Class\Type Lowest Handicap Highest Handicap Average Handicap ISLANDER BAHAMA 24 246 270 258 ISLANDER 24 BAHAMA 258 273 264 CATALINA 22 FK 249 279 267 CATALINA 22 ODR 270 276 270 CATALINA 22 SK 263 279 270 CATALINA 22 WK 267 285 273
Quote:
Originally Posted by sailingdog View Post
Check the polar diagrams. That will give you a rough idea of how a well-sailed boat will perform under various wind speeds at various points of sail.
Another way to compare boats, at least on a rough basis, is to look at their PHRF ratings. The higher the rating, the slower the boat.
There really isn't a lot you can do to enhance the performance of a boat. Other than keeping excess weight off the boat, fairing and smoothing the bottom and keeping it free of growth, keeping your sails in good shape, etc... but the basic speed of a given boat won't change much.
Some designs are just going to be much faster than others... Multihulls are a good example of that... trimarans, and to some degree catamarans, are going to be faster than their monohull counterparts.
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A larger handicap number, means a slower boat.
PHRF numbers as I recall are based on a triangle (upwind, downwind, beam wind) and also locally adjusted for local wind speed range, so they are not an absolute comparison, they must be taken in context. Some boats, usually heavier ones, will be disadvantaged in light air but become relatively faster in bad weather when they can stay upright and there's plenty of wind. (Again, within limits.)
To make any boat faster, you clean the bottom. You get rid of excess weight. And you add sail area, and new sails as opposed to blown out old sails. And, of course, you learn to trim the sails for maximum speed made good on course.
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Hello,
Those boats are basically all the same speed. PHRF is good for determining differences in speed, but for the information to be useful, you need to have a LARGE difference in the number. If one boat was 150 and other 100, that would mean something, or one was over 250 and other under 200. The difference between 258 and 270 is basically nothing.
This assumes you are NOT racing the boat. If you are racing, then every second counts. For cruising, the difference between the two boats might be 5 minutes after a few hours of sailing.
Barry
Barry Lenoble
Deep Blue C, 2002 C&C 110
Mt. Sinai, NY
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I had an Islander Bahama 24 for a year and sailed it a lot. It was great in over 5 knots. Under that it was sooooo slow. Not a good light air boat, lots of skin friction, low wavemaking resistance.
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The B24 has a longer waterline and thus a higher hull speed. But they are only about 10 secs per mile apart which isn't much. Esssentially equivalent speed.
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posted by on .
The perimeter of the polygon below is 9x-1. Find the expression for the missing side length.
There are 5 expressions: 3x-4, x-5, x+1, 2x+3, and ?. I got ?=2x-6. I added all the equation and minus it from 9x-1.
• math - ,
2x+4 | 90 | 252 | {"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-2017-22 | latest | en | 0.814546 |
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Division isn’t symmetric to multiplication or even subtraction regarding difficulty – it is probably the hardest of all math operations. That’s why children often feel uncomfortable about dividing two-digit and three-digit numbers. While division is reverse multiplication, schools don’t provide division tables. They don’t require students to learn division facts by heart.
Another difficulty is that division is based on the knowledge of previous math concepts. It requires you to apply subtraction, addition, and multiplication.
Division With Remainders Worksheet Grade 4
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Book a free lesson with a math tutor or download a worksheet for independent study | 527 | 2,738 | {"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-2024-33 | latest | en | 0.907565 |
https://techweb.rohm.com/product/motor/brushless-mortor/brushless-mortor-basic/760/ | 1,701,635,317,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00665.warc.gz | 626,620,155 | 11,730 | # Methods of Sensorless 120° Commutation Driving Startup 2: Startup on Detection of Permanent Magnet Stopped Position
2023.08.30
・The method of detecting the stopped position of the permanent magnet to start the motor avoids the problems of reverse rotation and low-torque startup of the method in which the induced voltage from synchronous operation is detected for startup, and so reduces the time needed for startup.
・To detect the stopped position of the permanent magnet, current is passed in six patterns over short intervals during which the motor does not rotate, and the pattern for which the power supply current is largest (or smallest) is determined.
In the previous article, among the two basic methods used for sensorless starting of a 3-phase full-wave brushless motor, the method in which an induced voltage due to synchronous operation is detected to start the motor was explained. In this article, the method of detecting the stopped position of the permanent magnet to start the motor is explained.
## Problems with the Method of Startup by Detecting the Induced Voltage Due to Synchronous Operation
The previous article explained a number of problems with the method of starting a motor by detecting the induced voltage due to synchronous operation. The method explained here, in which the stopped position of the permanent magnet is detected to start the motor, is a method that resolves these problems. For reference, we shall once again state these problems.
＜Problems with the Method of Starting a Motor by Detecting the Induced Voltage Due to Synchronous Operation＞
• ・A synthetic magnetic field is created regardless of the position of the permanent magnet, so that depending on the motor state, a reverse-direction torque may act, and depending on the stopped position of the permanent magnet, more time may be required for startup.
• ・Ordinarily, the positional relationship between the permanent magnet and a synthetic magnetic field to obtain a large torque is a difference of 90°, but because the synthetic magnetic field is created regardless of the permanent magnet position, the starting angle is for example 60° or 70°, so that the desired large torque is not obtained.
## Methods of Sensorless 120° Commutation Driving Startup 2: Startup on Detection of Permanent Magnet Stopped Position
The following is an explanation of the method of motor startup by detecting the stopped position of the permanent magnet, which resolves the above-described problems.
The first diagram shows the state in which the permanent magnet (rotor) used in the explanation is stopped. The magnet is stopped with the S pole at the 3 o’clock position and the N pole at the 9 o’clock position.
Next, the circuit diagram is a schematic diagram used to explain the principles and operation involved in detection of the permanent magnet position. Using the outputs A1, A2, A3, the coils are energized such that current flows through the coils in the six patterns ① to ⑥. This energizing is for a short time duration, such that the permanent magnet does not rotate.
The energizing waveforms and current waveforms (IVM) for the six patterns are shown in the waveform diagram. A1 to A3 output voltages corresponding to the energization patterns ① to ⑥. The power supply current is of varying sizes for each of these patterns. This is because the current for each pattern differs depending on the position of the permanent magnet; in this method, the difference in currents is used to detect the position of the permanent magnet.
A specific example is explained. In ③, energization results in the S pole at coil 2 and the N pole at coil 3. The permanent magnet S pole opposes coil 2, and the permanent magnet N pole opposes coil 3, so that magnetic polarization of the coils is impeded. Hence the current rise is most gradual, and the current is small.
However, ⑥ is the opposite of ③; energization causes an N pole to appear at coil 2 and an S pole at coil 3. Because the permanent magnet S pole opposes coil 2 and the permanent magnet N pole opposes coil 3, the coil magnetic polarization is promoted. As a result, the current rise is most rapid, and the current is large.
In other words, by determining the energization pattern resulting in the largest or the smallest current, the permanent magnet position can be detected.
This method is explained in somewhat more detail using a circuit block example and operation waveform diagram for a specific driver.
The circuit block for this method 2 is basically the circuit block of method 1 for detecting the induced voltage from synchronous operation to start the motor, to which is added a circuit for generating the six energization patterns and detecting the power supply currents, comparing the currents, and generating an initial driving pattern. Some parts are omitted, but the portions surrounded by blue are the outputs A1 to A3.
(1) The six patterns generated by the position detection pattern generation block are sent to a driving basic waveform synthesis block, and energization is performed by A1 to A3. The detected power supply currents are converted into voltages by a current detection resistor and an amplifier, and based on (2) a sample and hold circuit and (3) a current comparison/maximum value pattern detection circuit, (4) an initial driving pattern generation block generates an initial driving pattern based on the permanent magnet position, and returns the pattern to the driving basic waveform synthesis block, and driving begins.
Short pulses are shown immediately after power-on in the output voltage waveforms A1 to A3 in the operation waveform diagrams. This represents energization of the six patterns for position detection. As explained above, the energization is over a short time, and so in contrast with the other waveforms, the time is very short.
Immediately after power-on, the permanent magnet position is detected, and based on the result, an initial driving pattern is used for driving, so that the problems of reverse rotation and low-torque startup that occurred in the method of startup by detecting an induced voltage are avoided, and the motor can be rotated by a large torque from the very beginning.
Operation after startup basically relies on the same process as in the method of startup by detecting an induced voltage. The timing for switching the driving pattern during initial driving is, similarly to the method of startup by detecting an induced voltage, based on ST_CLK.
However, rotation is begun using a large torque from the start, so that a sufficient induced voltage is obtained using several patterns (in these waveform diagrams, four ST_CLK signals), and steady-state driving is begun using an induced voltage. That is, the problem with the method of startup by detecting an induced voltage, that time is required for startup, is alleviated.
## 【Download Documents】Overview of Brushed DC Motor Structure, Operating Principle, Drive Method, Characteristics, Features, and Applications
Basic overview of brushed motors, including structure, principle of operation, drive method, and features. | 1,412 | 7,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-50 | latest | en | 0.926407 |
ayumilovemaple.wordpress.com | 1,526,936,240,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00027.warc.gz | 530,830,365 | 21,807 | Home > Technical FAQ > [Validation] Singapore NRIC Number Verification
## [Validation] Singapore NRIC Number Verification
Title [Validation] Singapore NRIC Number Verification
Written By samliew
Regarding the NRIC math behind the letter, it’s called a checksum. This is to help check whether the actual IC numbers are valid and not just random numbers.
The math goes like this:
1) Take for example I want to test the NRIC number S1234567. The first digit you multiply by 2, second multiply by 7, third by 6, fourth by 5, fifth by 4, sixth by 3, seventh by 2. Then you add the totals together. So,1×2+2×7+3×6+4×5+5×4+6×3+7×2=106.
2) If the first letter of the NRIC starts with T or G, add 4 to the total.
3) Then you divide the number by 11 and get the remainder. 106/11=9r7
4) You can get the alphabet depending on the IC type (the first letter in the IC) using the code below:
If the IC starts with S or T: 0=J, 1=Z, 2=I, 3=H, 4=G, 5=F, 6=E, 7=D, 8=C, 9=B, 10=A
If the IC starts with F or G: 0=X, 1=W, 2=U, 3=T, 4=R, 5=Q, 6=P, 7=N, 8=M, 9=L, 10=K
.
Credits
Credits to samliew for NRIC number verification information
Singapore NRIC Validator
Singapore NRIC Generator
.
1. 15/04/2012 at 8:57 am
4) You can get the alphabet depending on the IC type (the first letter in the IC) using the code below: what does this mean?
2. 15/04/2012 at 8:53 am
4) You can get the alphabet depending on the IC type (the first letter in the IC) using the code below:
3. 23/05/2009 at 11:12 am
So that you cannot create (too many) multiple accounts. Obviously that won’t work since we have NRIC generators e.g. mine.
4. 08/05/2009 at 10:44 am
v8GMuJ comment3 ,
5. 13/03/2009 at 4:00 am
i know this tread is old, but i’m looking for a reason why Maplestory SEA wants our I/C numbers. i don’t like to give that kind of information online. and it seems i can’t play on Maplestory Global. Do you have any idea why they want I/C numbers?
6. 11/10/2008 at 4:21 am | 636 | 1,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-22 | latest | en | 0.780177 |
https://electricaliq.com/lvdt-working-principle-applications-and-advantages/ | 1,702,161,540,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00483.warc.gz | 255,688,681 | 14,465 | # LVDT- Working Principle, Applications and Advantages
An LVDT, or Linear Variable Differential Transformer, is a type of transducer that converts linear displacement, the movement of an object along a straight line, into an electrical signal. This signal, typically a voltage or current, is proportional to the displacement of the object, making LVDTs highly accurate and reliable displacement measurement devices.
### Working Principle of LVDTs
The operation of LVDTs is based on the principle of mutual inductance, a phenomenon in which a changing current in one coil induces a voltage in another coil located nearby. An LVDT consists of three coils: a primary coil and two secondary coils. The primary coil is excited with an alternating current (AC) signal, creating a magnetic field around it. The two secondary coils are wound on a cylindrical core and are connected in series opposition.
As the core moves along the axis of the LVDT, the magnetic field coupling between the primary coil and each secondary coil changes. This change in coupling induces voltages in the secondary coils. The voltage difference between the secondary coils is proportional to the displacement of the core from the null position, the point where the core is equidistant from both secondary coils.
### LVDT Components
An LVDT comprises several key components that work together to achieve precise displacement measurement:
1. Primary Coil: The primary coil is the source of the AC excitation signal. It is typically wound on a bobbin and is positioned between the two secondary coils.
2. Secondary Coils: The secondary coils are responsible for generating the electrical signal proportional to the displacement of the core. They are wound on a cylindrical core and are connected in series opposition.
3. Core: The core, a movable element, is positioned within the coils and is mechanically coupled to the object whose displacement is being measured. As the object moves, the core moves along the axis of the LVDT, altering the magnetic coupling between the coils.
4. Signal Conditioning Circuit: The signal conditioning circuit amplifies and processes the voltage difference between the secondary coils, converting it into a standardized electrical signal suitable for further processing or transmission.
### Applications of LVDTs:
LVDTs find widespread applications in various industries and settings due to their high accuracy, reliability, and non-contact operation:
1. Position Measurement: LVDTs are extensively used in position measurement systems, particularly in hydraulic and pneumatic systems, to monitor the movement of pistons, actuators, and other linear components.
2. Displacement Measurement: LVDTs are employed in a variety of displacement measurement applications, including machine tool control, structural monitoring, and vibration analysis.
3. Force Measurement: LVDTs can be used to measure force by converting it into displacement using a spring or diaphragm. They are commonly used in load cells and pressure transducers.
4. Liquid Level Measurement: LVDTs can be utilized to measure liquid level by attaching a float to the core. As the liquid level changes, the float moves, causing the core to displace and generating a corresponding electrical signal. | 620 | 3,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.94176 |
https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-connecting-concepts-through-application/chapter-6-logarithmic-functions-6-3-graphing-logarithmic-functions-6-3-exercises-page-511/43 | 1,537,896,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267161902.89/warc/CC-MAIN-20180925163044-20180925183444-00395.warc.gz | 768,293,955 | 12,946 | ## Intermediate Algebra: Connecting Concepts through Application
$g^{-1}(x) = 9^{x}$
$g(x) =\log_9 x$ Let $g(x) = y$ $y = \log_9 x$ Swap the $x$ and $y$ variable to find the inverse: $x =\log_9 y$ $y =9^{x}$ $g^{-1}(x) = 9^{x}$ | 96 | 228 | {"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} | 3.78125 | 4 | CC-MAIN-2018-39 | longest | en | 0.508651 |
https://pastebin.com/udH9cXUm | 1,516,306,051,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00197.warc.gz | 737,248,541 | 7,478 | • API
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# Rationals
a guest Sep 15th, 2012 34 Never
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1. public class Rational {
2.
3. // Creating instance variables for the numerator and denominator
4. private long numerator;
5. private long denominator;
6.
7. // Constructor that creates a new Rational object, making sure the denominator is not 0 or negative.
8. public Rational (long numer, long denom){
9.
10. if (denom == 0){
11. throw new RuntimeException("Error, can't have a denominator of 0.");
12. }
13.
14. if (denom <0){
15. denom = -denom;
16. numer = -numer;
17. }
18.
19. long greatestCommonDivisor = gcd(numer, denom);
20. numerator = numer / greatestCommonDivisor;
21. denominator = denom / greatestCommonDivisor;
22.
23. }
24.
25. // This method takes two rational numbers and adds them together
26. public Rational plus(Rational b){
27. Rational answer = new Rational (this.numerator * b.denominator + b.numerator * this.denominator, b.denominator * this.denominator);
28.
30. }
31.
32. //This method takes two rational numbers and subtracts them
33. public Rational minus (Rational b){
34. Rational answer = new Rational (this.numerator * b.denominator - b.numerator * this.denominator, b.denominator * this.denominator);
35.
36.
38. }
39.
40. //This method takes two rational numbers and multiplies them
41. public Rational times (Rational b){
42. Rational answer = new Rational (this.numerator * b.numerator, this.denominator * b.denominator);
43.
45. }
46.
47.
48.
49. // This method uses Euclid's algorithm which calculates the greatest common divisor
50. public static long gcd(long p, long q){
51. if (q==0) return p;
52. long r = p % q;
53. return gcd(q, r);
54. }
55.
56. // This method produces a string representation of values
57. public String toString(){
58. if (denominator == 1)
59. return numerator + "";
60. else
61. return numerator + "/" + denominator;
62. }
63.
64.
65. // Test client
66. public static void main (String args[]){
67. Rational p = new Rational (1, 6);
68. Rational q = new Rational (4, 8);
69.
70. Rational r = p.minus(q);
71. System.out.println(r);
72. }
73.
74.
75. }
RAW Paste Data
Top | 671 | 2,851 | {"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} | 3.34375 | 3 | CC-MAIN-2018-05 | latest | en | 0.342201 |
https://difftech.herokuapp.com/addition-vs-division/ | 1,619,185,880,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594808.94/warc/CC-MAIN-20210423131042-20210423161042-00449.warc.gz | 327,632,330 | 271,986 | Aspects
vs
Addition is a mathematical operation that represents combining collections of objects together into a larger collection
Division
In mathematics, division (÷) is an arithmetic elementary operation.
Faster slower operation
Example
"It is true that division and modulo a division operation is slower than addition"
from question
Execution time of different operators
"With careful optimization however you can make addition 61 times faster than division"
from question
How expensive are IEEE 754 double operations in respect to each other on Intel I7 chips?
"For example an addition is typically much faster than a division"
from question
How long does each machine language instruction take to execute?
"For example on most 32 bit systems 64-bit addition is faster than 32-bit division modulo"
from question
Is there any way to find arithmetic mean "better" than sum()/N?
"Performing addition on this slightly larger type will pretty much always be faster than doing division or modulo on the type itself"
from question
Is there any way to find arithmetic mean "better" than sum()/N?
"An addition is faster than a division and a multiplication"
from question
How do I use applyLinearImpulse based on rotation in Corona / Lua
"As far as i know the division is more complex and slower than other operations like addition so is my code incorrect then"
from question
Comparing the time requirements of addition and division operation in GPU (CUDA)
"2.the division by 2 can be done by bit - shift operation is it really slower than addition"
from question
Is fibonacci search faster than binary search?
" think it is a few cycles slower than addition but yes division is very slow compared to the others;division takes significantly longer and unlike the other 3 operations division is not pipelined"
from question
Floating Point Div/Mul > 30 times slower than Add/Sub?
"I need to find out that how much division operation is faster than addition operation in a gpu"
from question
Comparing the time requirements of addition and division operation in GPU (CUDA)
"The compiler could be done via division which is much slower than addition or the compiler could be translated into a bitwise and operation as well and end up being just as fast as the version"
from question
Is i=(i+1)&3 faster than i=(i+1)%4
"A simple branch assignment is way cheaper than addition and division"
from question
Imparting a number's sign onto another number?
Higher precedence subtraction
Example
"The term is apparently not an exact measurement as it is clear that a double-precision floating-point operation is going to take longer than a single-precision one and multiplication and division are going to take longer than addition and subtraction"
from question
FLOPS what really is a FLOP
"Since division has a higher precedence than addition 5 2 gets evaluated as a integer division returning 2 as an integer"
from question
Explain "accumulated round-off error" for double values
"The division operator has a higher precedence than the addition operator + so you need to enclose the sum with brackets before dividing"
from question
Java How to calculate the average of 3 bowling scores
"Multiplication and division are higher precedence than addition so they get done first - before the implicit conversion to string for concatenation"
from question
Javascript subtraction returns NaN
"Multiplication and division operators have higher precedence than addition and subtraction in c++ same as in scientific notation"
from question
C++ Converting Fahrenheit to Celsius
"Finally we all know that multiplication and division have higher precedence than addition and subtraction so we can remove the extraneous parentheses so this turns into"
from question
C++ using ceil and length outputs incorrect value
"Note that division has a higher precedence than addition"
from question
Js bug in the average function
"Division has higher precedence than addition"
from question
How to print output in java
"Division has a higher precedence than addition or subtraction so it s really this"
from question
I don't know about pre and post incrementation
"Division has a higher precedence than addition ergo"
from question
C++ order of evaluation: division vs addition
"Multiplication and division have a higher precedence than addition and subtraction"
from question
Precedence of && over ||
from question
Operator Precedence with Scala Parser Combinators
"The division operator has a higher precendence than the addition operator so your function is calculating 1 1 + e -x"
from question
Sigmoid function in vb
"This happens because the division operator has higher precedence than the + addition operator"
from question
Math operation in a While
"The division operator has a higher order precedence as the addition operator"
from question
B-Spline recursive definition in C#
"Knuth writes that fibonacci search is preferable on some computers because it involves only addition and subtraction not division by 2. but almost all computers use binary arithmetic in which division by 2 is simpler than addition and subtraction"
from question
Is golden section search better than binary search?
"I don t know how division modulo both works but division modulo both s much more complex than addition subtraction or even multiplication"
from question
What is the performance impact of using int64_t instead of int32_t on 32-bit systems?
"Usually simple operations like addition subtraction and multiplication are very fast;division is a bit slower"
from question
How do I measure the FLOPS my C# app uses?
"Since most processors can do an addition comparison or multiplication in a single cycle those are all counted as one flop;but division always takes longer"
from question
What is FLOP/s and is it a good measure of performance?
"The addition and subtraction is okay because the types of a and b force them to be performed using floating point arithmetic - but because division binds more tightly than addition and subtraction it s like using the brackets above only the immediate operands are considered"
from question
To return a double, do I have to cast to double even if types are double in c#?
"The division has higher precedence than the addition so what you re calculating is sumaverage1+ sumaverage2 5 which is integer division which is probably not what you want"
from question
User enter 5 characters and averge sums out the 5 numbers in C
"Multiplication and division have higher priority than addition and subtraction"
from question
How do I do arithmetic operations specified in strings?
Difference expensive matrix
Example
"If you are doing physical simulations things like division or square roots are going to be way more expensive than addition"
from question
Precision error in summation addition/subtraction (of mass)
"Generally the division is more costly than addition i think but not much difference in this case"
from question
32bit floating division is not as slow as I expected
"The first difference is that division is much more expensive than addition"
from question
Why is Sieve of Eratosthenes more efficient than the simple "dumb" algorithm?
"Division gets really bad;interestingly the matrix addition is not much difference at all"
from question
What is the performance impact of using int64_t instead of int32_t on 32-bit systems?
"Division is much more expensive than multiplication;in addition calculating an approximate reciprocal is much more simd-friendly since there are usually reciprocal estimate instructions that provide a starting point which can be refined by the newton-raphson method"
from question
Decrease in instructions retired after loop Unrolling
Time example cycles
Example
"Best example the division it an an addition are both o 1 but usually the division takes far more cycles time to execute than the addition"
from question
Big-O of division
"I remember it says something like division takes much much more time than addition"
from question
Complexity of basic operations: Addition subtraction multiplication division greater equal
Others
Example
For multiplication the technique described at is a reasonably easy thing to implement and is better than serial addition;division is more complex in general but a good place to start is
from question
64 bit arithmetic with 16 bit word length(store 64 bit number across 4 words)
And division has larger complexity than addition
from question
Why is Sieve of Eratosthenes more efficient than the simple "dumb" algorithm?
So even disregarding the trial division is more expensive than addition and multiplication we see that the number of operations the sieve requires is much smaller than the number of operations required by trial division if the limit is not too small
from question
Why is Sieve of Eratosthenes more efficient than the simple "dumb" algorithm?
Think of long division - you do a series of subtract - shift operations and you don t know what you need to do next until you have completed the previous part of the operation;for addition it is easier to see how you could achieve a complete operation in one cycle
from question
An attempt to understand what a Clock cycle is through example
In addition to that the crossing off may be less work than a division don t know about python it is for c arrays
from question
Why do two algorithms for finding primes differ in speed so much even though they seem to do the same number of iterations?
I think division by a power of 10 other than 10 9 would be somewhat cheap but would require an actual division on each limb and propagating the remainder to the next limb;extended-precision addition is somewhat more expensive this way than with binary limbs because i have to generate the carry-out manually with a compare unsigned comparison
from question
Integer division algorithm
Use float a b instead or add a from __future__ import division to the top of your file;tentative conclusion using a for-loop and simple addition method1 is a lot faster than any of the list comprehension methods for this example
from question
Finding the average of non-zero values in a python dictionary
##### Back to Home
Data comes from Stack Exchange with CC-BY-SA-4.0 | 2,037 | 10,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-17 | latest | en | 0.953356 |
https://oeis.org/A152934 | 1,709,516,092,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00396.warc.gz | 426,407,364 | 4,431 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A152934 Number of sets (in the Hausdorff metric geometry) at each location between two sets defining a polygonal configuration consisting of two m-gonal polygonal components chained with string components of length 3 as m varies. 47
289, 1962, 13429, 92025, 630730, 4323069, 29630737, 203092074, 1392013765, 9541004265, 65395016074, 448224108237, 3072173741569, 21056992082730, 144326770837525, 989230403779929, 6780286055621962, 46472771985573789, 318529117843394545, 2183231052918188010 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,1 LINKS Table of n, a(n) for n=2..21. S. Schlicker, L. Morales, and D. Schultheis, Polygonal chain sequences in the space of compact sets, J. Integer Seq. 12 (2009), no. 1, Article 09.1.7, 23 pp. Index entries for linear recurrences with constant coefficients, signature (8, -8, 1). FORMULA Conjectures from Colin Barker, Jul 09 2020: (Start) G.f.: x^2*(289 - 350*x + 45*x^2) / ((1 - x)*(1 - 7*x + x^2)). a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>4. (End) MAPLE with(combinat): a := proc(n) local aa, b, c, d, lambda, delta, R, S, F, L, k, l: k:=2: l:=3: F := t -> fibonacci(t): L := t -> fibonacci(t-1)+fibonacci(t+1): aa := (n, l) -> L(2*n)*F(l-2)+F(2*n+2)*F(l-1): b := (n, l) -> L(2*n)*F(l-1)+F(2*n+2)*F(l): c := (n, l) -> F(2*n+2)*F(l-2)+F(n+2)^2*F(l-1): d := (n, l) -> F(2*n+2)*F(l-1)+F(n+2)^2*F(l): lambda := (n, l) -> (d(n, l)+aa(n, l)+sqrt((d(n, l)-aa(n, l))^2+4*b(n, l)*c(n, l)))*(1/2): delta := (n, l) -> (d(n, l)+aa(n, l)-sqrt((d(n, l)-aa(n, l))^2+4*b(n, l)*c(n, l)))*(1/2): R := (n, l) -> ((lambda(n, l)-d(n, l))*L(2*n)+b(n, l)*F(2*n+2))/(2*lambda(n, l)-d(n, l)-aa(n, l)): S := (n, l) -> ((lambda(n, l)-aa(n, l))*L(2*n)-b(n, l)*F(2*n+2))/(2*lambda(n, l)-d(n, l)-aa(n, l)): simplify(R(n, l)*lambda(n, l)^(k-1)+S(n, l)*delta(n, l)^(k-1)); end proc; CROSSREFS Cf. A152927, A152928, A152929, A152930, A152931, A152932, A152933, A152935. Sequence in context: A218766 A188186 A112077 * A332737 A156575 A296404 Adjacent sequences: A152931 A152932 A152933 * A152935 A152936 A152937 KEYWORD nonn AUTHOR Steven Schlicker, Dec 15 2008 STATUS approved
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Last modified March 3 20:15 EST 2024. Contains 370512 sequences. (Running on oeis4.) | 1,017 | 2,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-10 | latest | en | 0.479061 |
https://www.mathworks.com/matlabcentral/answers/58634-setting-ticks-in-a-colorbar?requestedDomain=www.mathworks.com&nocookie=true | 1,516,581,869,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890928.82/warc/CC-MAIN-20180121234728-20180122014728-00601.warc.gz | 943,261,571 | 14,901 | Setting Ticks in a Colorbar
Natalya (view profile)
on 12 Jan 2013
Latest activity Commented on by Nicholas DeWind
Nicholas DeWind (view profile)
on 9 Nov 2015
How do I set the ticks in a color bar to be at specific values?
For example, when I just use the command colorbar('h'), Matlab automatically generates a colorbar with ticks at -10, -5, 0,5,10. I would like the ticks to be at -12 -9,-6,-3,0,3,6,9,12. I tried:
cbh=colorbar('h');
set(cbh,'XTick',[-12:3:12])
This didn't change anything.
set(cbh,'XTickLabel',{'-12','-9','-6','-3','0','3','6','9','12'})
just relabeled the ticks, but did not change their location (i.e. the value of -10 got labeled '-12')
I also tried this:
cbh=colorbar('h');
cy=get(cbh,'XTick');
set(cy,[-12:3:12])
This set ticks at smaller intervals and labeled them from -70 to 10... I do not understand why.
Does anyone have any other suggestions?
Thanks!
Nicholas DeWind
Nicholas DeWind (view profile)
on 9 Nov 2015
Use YTick:
cbh=colorbar('h');
set(cbh,'YTick',[-12:3:12])
Jan Simon (view profile)
on 12 Jan 2013
Matlab's colorbar command creates an image object. See:
```cbh = colorbar('h');
get(get(cbh, 'Children'))
>> ...
CData = [ (1 by 64) double array]
DataMapping = direct
XData = [1.5 64.5]
YData = [0 1]
...
Type = image
```
When you want to change the ticks from -12:3:12, I guess you want 25 different colors in the colorbar.
```AxesH = axes('CLim', [-12, 12]);
cbh = colorbar('peer', AxesH, 'h', ...
'XTickLabel',{'-12','-9','-6','-3','0','3','6','9','12'}, ...
'XTick', -12:3:12)
``` | 506 | 1,557 | {"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} | 2.703125 | 3 | CC-MAIN-2018-05 | latest | en | 0.768338 |
https://wikimili.com/en/Wavelet_transform | 1,571,476,296,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692723.54/warc/CC-MAIN-20191019090937-20191019114437-00332.warc.gz | 765,426,388 | 26,167 | # Wavelet transform
Last updated
In mathematics, a wavelet series is a representation of a square-integrable (real- or complex-valued) function by a certain orthonormal series generated by a wavelet. This article provides a formal, mathematical definition of an orthonormal wavelet and of the integral wavelet transform.
Mathematics includes the study of such topics as quantity, structure (algebra), space (geometry), and change. It has no generally accepted definition.
In mathematics, a real number is a value of a continuous quantity that can represent a distance along a line. The adjective real in this context was introduced in the 17th century by René Descartes, who distinguished between real and imaginary roots of polynomials. The real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers, such as 2. Included within the irrationals are the transcendental numbers, such as π (3.14159265...). In addition to measuring distance, real numbers can be used to measure quantities such as time, mass, energy, velocity, and many more.
A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. Because no real number satisfies this equation, i is called an imaginary number. For the complex number a + bi, a is called the real part, and b is called the imaginary part. Despite the historical nomenclature "imaginary", complex numbers are regarded in the mathematical sciences as just as "real" as the real numbers, and are fundamental in many aspects of the scientific description of the natural world.
## Definition
A function ${\displaystyle \scriptstyle \psi \,\in \,L^{2}(\mathbb {R} )}$ is called an orthonormal wavelet if it can be used to define a Hilbert basis, that is a complete orthonormal system, for the Hilbert space ${\displaystyle \scriptstyle L^{2}\left(\mathbb {R} \right)}$ of square integrable functions.
In linear algebra, two vectors in an inner product space are orthonormal if they are orthogonal and unit vectors. A set of vectors form an orthonormal set if all vectors in the set are mutually orthogonal and all of unit length. An orthonormal set which forms a basis is called an orthonormal basis.
The mathematical concept of a Hilbert space, named after David Hilbert, generalizes the notion of Euclidean space. It extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions. A Hilbert space is an abstract vector space possessing the structure of an inner product that allows length and angle to be measured. Furthermore, Hilbert spaces are complete: there are enough limits in the space to allow the techniques of calculus to be used.
In mathematics, a square-integrable function, also called a quadratically integrable function or function, is a real- or complex-valued measurable function for which the integral of the square of the absolute value is finite. Thus, square-integrability on the real line is defined as follows.
The Hilbert basis is constructed as the family of functions ${\displaystyle \scriptstyle \{\psi _{jk}:\,j,\,k\,\in \,\mathbb {Z} \}}$ by means of dyadic translations and dilations of ${\displaystyle \scriptstyle \psi \,}$,
The dyadic transformation is the mapping
In Euclidean geometry, a translation is a geometric transformation that moves every point of a figure or a space by the same distance in a given direction.
In operator theory, a dilation of an operator T on a Hilbert space H is an operator on a larger Hilbert space K, whose restriction to H composed with the orthogonal projection onto H is T.
${\displaystyle \psi _{jk}(x)=2^{\frac {j}{2}}\psi \left(2^{j}x-k\right)\,}$
for integers ${\displaystyle \scriptstyle j,\,k\,\in \,\mathbb {Z} }$.
If under the standard inner product on ${\displaystyle \scriptstyle L^{2}\left(\mathbb {R} \right)}$,
${\displaystyle \langle f,g\rangle =\int _{-\infty }^{\infty }f(x){\overline {g(x)}}dx}$
this family is orthonormal, it is an orthonormal system:
{\displaystyle {\begin{aligned}\langle \psi _{jk},\psi _{lm}\rangle &=\int _{-\infty }^{\infty }\psi _{jk}(x){\overline {\psi _{lm}(x)}}dx\\&=\delta _{jl}\delta _{km}\end{aligned}}}
where ${\displaystyle \scriptstyle \delta _{jl}\,}$ is the Kronecker delta.
In mathematics, the Kronecker delta is a function of two variables, usually just non-negative integers. The function is 1 if the variables are equal, and 0 otherwise:
Completeness is satisfied if every function ${\displaystyle \scriptstyle f\,\in \,L^{2}\left(\mathbb {R} \right)}$ may be expanded in the basis as
${\displaystyle f(x)=\sum _{j,k=-\infty }^{\infty }c_{jk}\psi _{jk}(x)}$
with convergence of the series understood to be convergence in norm. Such a representation of f is known as a wavelet series. This implies that an orthonormal wavelet is self-dual.
In mathematics, a dual wavelet is the dual to a wavelet. In general, the wavelet series generated by a square integrable function will have a dual series, in the sense of the Riesz representation theorem. However, the dual series is not itself in general representable by a square integrable function.
The integral wavelet transform is the integral transform defined as
${\displaystyle \left[W_{\psi }f\right](a,b)={\frac {1}{\sqrt {|a|}}}\int _{-\infty }^{\infty }{\overline {\psi \left({\frac {x-b}{a}}\right)}}f(x)dx\,}$
The wavelet coefficients${\displaystyle \scriptstyle c_{jk}}$ are then given by
${\displaystyle c_{jk}=\left[W_{\psi }f\right]\left(2^{-j},k2^{-j}\right)}$
Here, ${\displaystyle \scriptstyle a\;=\;2^{-j}}$ is called the binary dilation or dyadic dilation, and ${\displaystyle \scriptstyle b\;=\;k2^{-j}}$ is the binary or dyadic position.
## Principle
The fundamental idea of wavelet transforms is that the transformation should allow only changes in time extension, but not shape. This is affected by choosing suitable basis functions that allow for this.[ how? ] Changes in the time extension are expected to conform to the corresponding analysis frequency of the basis function. Based on the uncertainty principle of signal processing,
${\displaystyle \Delta t\Delta \omega \geqq {\frac {1}{2}}}$
where t represents time and ω angular frequency (ω = 2πf, where f is temporal frequency).
The higher the required resolution in time, the lower the resolution in frequency has to be. The larger the extension of the analysis windows is chosen, the larger is the value of ${\displaystyle \scriptstyle \Delta t}$[ how? ].
When Δt is large,
2. Good frequency resolution
3. Low frequency, large scaling factor
When Δt is small
1. Good time resolution
3. High frequency, small scaling factor
In other words, the basis function Ψ can be regarded as an impulse response of a system with which the function x(t) has been filtered. The transformed signal provides information about the time and the frequency. Therefore, wavelet-transformation contains information similar to the short-time-Fourier-transformation, but with additional special properties of the wavelets, which show up at the resolution in time at higher analysis frequencies of the basis function. The difference in time resolution at ascending frequencies for the Fourier transform and the wavelet transform is shown below.
This shows that wavelet transformation is good in time resolution of high frequencies, while for slowly varying functions, the frequency resolution is remarkable.
Another example: The analysis of three superposed sinusoidal signals ${\displaystyle \scriptstyle y(t)\;=\;\sin(2\pi f_{0}t)\;+\;\sin(4\pi f_{0}t)\;+\;\sin(8\pi f_{0}t)}$ with STFT and wavelet-transformation.
## Wavelet compression
Wavelet compression is a form of data compression well suited for image compression (sometimes also video compression and audio compression). Notable implementations are JPEG 2000, DjVu and ECW for still images, CineForm, and the BBC's Dirac. The goal is to store image data in as little space as possible in a file. Wavelet compression can be either lossless or lossy. [1]
Using a wavelet transform, the wavelet compression methods are adequate for representing transients, such as percussion sounds in audio, or high-frequency components in two-dimensional images, for example an image of stars on a night sky. This means that the transient elements of a data signal can be represented by a smaller amount of information than would be the case if some other transform, such as the more widespread discrete cosine transform, had been used.
Discrete wavelet transform has been successfully applied for the compression of electrocardiograph (ECG) signals [2] In this work, the high correlation between the corresponding wavelet coefficients of signals of successive cardiac cycles is utilized employing linear prediction.
Wavelet compression is not good for all kinds of data: transient signal characteristics mean good wavelet compression, while smooth, periodic signals are better compressed by other methods, particularly traditional harmonic compression (frequency domain, as by Fourier transforms and related).
See Diary Of An x264 Developer: The problems with wavelets (2010) for discussion of practical issues of current methods using wavelets for video compression.
### Method
First a wavelet transform is applied. This produces as many coefficients as there are pixels in the image (i.e., there is no compression yet since it is only a transform). These coefficients can then be compressed more easily because the information is statistically concentrated in just a few coefficients. This principle is called transform coding. After that, the coefficients are quantized and the quantized values are entropy encoded and/or run length encoded.
A few 1D and 2D applications of wavelet compression use a technique called "wavelet footprints". [3] [4]
## Comparison with Fourier transform and time-frequency analysis
TransformRepresentationInput
Fourier transform ${\displaystyle {\hat {f}}(\xi )=\int _{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }\,dx}$ξ, frequency
Time-frequency analysis ${\displaystyle X(t,f)}$t, time; f, frequency
Wavelet transform${\displaystyle X(a,b)={\frac {1}{\sqrt {a}}}\int _{-\infty }^{\infty }{\overline {\Psi \left({\frac {t-b}{a}}\right)}}x(t)\,dt}$a, scaling; b, time
Wavelets have some slight benefits over Fourier transforms in reducing computations when examining specific frequencies. However, they are rarely more sensitive, and indeed, the common Morlet wavelet is mathematically identical to a short-time Fourier transform using a Gaussian window function. [5] The exception is when searching for signals of a known, non-sinusoidal shape (e.g., heartbeats); in that case, using matched wavelets can outperform standard STFT/Morlet analyses. [6]
## Other practical applications
The wavelet transform can provide us with the frequency of the signals and the time associated to those frequencies, making it very convenient for its application in numerous fields. For instance, signal processing of accelerations for gait analysis, [7] for fault detection, [8] for design of low power pacemakers and also in ultra-wideband (UWB) wireless communications. [9]
1. Discretizing of the c-τ-axis
Applied the following discretization of frequency and time:
{\displaystyle {\begin{aligned}c_{n}&=c_{0}^{n}\\\tau _{m}&=m\cdot T\cdot c_{0}^{n}\end{aligned}}}
Leading to wavelets of the form, the discrete formula for the basis wavelet:
${\displaystyle \Psi (k,n,m)={\frac {1}{\sqrt {c_{0}^{n}}}}\cdot \Psi \left[{\frac {k-mc_{0}^{n}}{c_{0}^{n}}}T\right]={\frac {1}{\sqrt {c_{0}^{n}}}}\cdot \Psi \left[\left({\frac {k}{c_{0}^{n}}}-m\right)T\right]}$
Such discrete wavelets can be used for the transformation:
${\displaystyle Y_{DW}(n,m)={\frac {1}{\sqrt {c_{0}^{n}}}}\cdot \sum _{k=0}^{K-1}y(k)\cdot \Psi \left[\left({\frac {k}{c_{0}^{n}}}-m\right)T\right]}$
2. Implementation via the FFT (fast Fourier transform)
As apparent from wavelet-transformation representation (shown below)
${\displaystyle Y_{W}(c,\tau )={\frac {1}{\sqrt {c}}}\cdot \int _{-\infty }^{\infty }y(t)\cdot \Psi \left({\frac {t-\tau }{c}}\right)\,dt}$
where c is scaling factor, τ represents time shift factor
and as already mentioned in this context, the wavelet-transformation corresponds to a convolution of a function y(t) and a wavelet-function. A convolution can be implemented as a multiplication in the frequency domain. With this the following approach of implementation results into:
• Fourier-transformation of signal y(k) with the FFT
• Selection of a discrete scaling factor ${\displaystyle c_{n}}$
• Scaling of the wavelet-basis-function by this factor ${\displaystyle c_{n}}$ and subsequent FFT of this function
• Multiplication with the transformed signal YFFT of the first step
• Inverse transformation of the product into the time domain results in YW${\displaystyle (c,\tau )}$ for different discrete values of τ and a discrete value of ${\displaystyle c_{n}}$
• Back to the second step, until all discrete scaling values for ${\displaystyle c_{n}}$are processed
There are many different types of wavelet transforms for specific purposes. See also a full list of wavelet-related transforms but the common ones are listed below: Mexican hat wavelet, Haar Wavelet, Daubechies wavelet, triangular wavelet.
## Related Research Articles
In mathematics, Fourier analysis is the study of the way general functions may be represented or approximated by sums of simpler trigonometric functions. Fourier analysis grew from the study of Fourier series, and is named after Joseph Fourier, who showed that representing a function as a sum of trigonometric functions greatly simplifies the study of heat transfer.
A wavelet is a wave-like oscillation with an amplitude that begins at zero, increases, and then decreases back to zero. It can typically be visualized as a "brief oscillation" like one recorded by a seismograph or heart monitor. Generally, wavelets are intentionally crafted to have specific properties that make them useful for signal processing. Using a "reverse, shift, multiply and integrate" technique called convolution, wavelets can be combined with known portions of a damaged signal to extract information from the unknown portions.
In mathematics, the Haar wavelet is a sequence of rescaled "square-shaped" functions which together form a wavelet family or basis. Wavelet analysis is similar to Fourier analysis in that it allows a target function over an interval to be represented in terms of an orthonormal basis. The Haar sequence is now recognised as the first known wavelet basis and extensively used as a teaching example.
The Fourier transform (FT) decomposes a function of time into its constituent frequencies. This is similar to the way a musical chord can be expressed in terms of the volumes and frequencies of its constituent notes. The term Fourier transform refers to both the frequency domain representation and the mathematical operation that associates the frequency domain representation to a function of time. The Fourier transform of a function of time is itself a complex-valued function of frequency, whose magnitude (modulus) represents the amount of that frequency present in the original function, and whose argument is the phase offset of the basic sinusoid in that frequency. The Fourier transform is not limited to functions of time, but the domain of the original function is commonly referred to as the time domain. There is also an inverse Fourier transform that mathematically synthesizes the original function from its frequency domain representation.
The short-time Fourier transform (STFT), is a Fourier-related transform used to determine the sinusoidal frequency and phase content of local sections of a signal as it changes over time. In practice, the procedure for computing STFTs is to divide a longer time signal into shorter segments of equal length and then compute the Fourier transform separately on each shorter segment. This reveals the Fourier spectrum on each shorter segment. One then usually plots the changing spectra as a function of time.
In signal processing, a finite impulse response (FIR) filter is a filter whose impulse response is of finite duration, because it settles to zero in finite time. This is in contrast to infinite impulse response (IIR) filters, which may have internal feedback and may continue to respond indefinitely.
In mathematics, the continuous wavelet transform (CWT) is a formal tool that provides an overcomplete representation of a signal by letting the translation and scale parameter of the wavelets vary continuously.
In numerical analysis and functional analysis, a discrete wavelet transform (DWT) is any wavelet transform for which the wavelets are discretely sampled. As with other wavelet transforms, a key advantage it has over Fourier transforms is temporal resolution: it captures both frequency and location information.
In mathematics, the discrete-time Fourier transform (DTFT) is a form of Fourier analysis that is applicable to a sequence of values.
In digital signal processing, upsampling, expansion, and interpolation are terms associated with the process of resampling in a multi-rate digital signal processing system. Upsampling can be synonymous with expansion, or it can describe an entire process of expansion and filtering (interpolation). When upsampling is performed on a sequence of samples of a signal or other continuous function, it produces an approximation of the sequence that would have been obtained by sampling the signal at a higher rate. For example, if compact disc audio at 44,100 samples/second is upsampled by a factor of 5/4, the resulting sample-rate is 55,125.
The Gabor transform, named after Dennis Gabor, is a special case of the short-time Fourier transform. It is used to determine the sinusoidal frequency and phase content of local sections of a signal as it changes over time. The function to be transformed is first multiplied by a Gaussian function, which can be regarded as a window function, and the resulting function is then transformed with a Fourier transform to derive the time-frequency analysis. The window function means that the signal near the time being analyzed will have higher weight. The Gabor transform of a signal x(t) is defined by this formula:
Continuous wavelets of compact support can be built [1], which are related to the beta distribution. The process is derived from probability distributions using blur derivative. These new wavelets have just one cycle, so they are termed unicycle wavelets. They can be viewed as a soft variety of Haar wavelets whose shape is fine-tuned by two parameters and . Closed-form expressions for beta wavelets and scale functions as well as their spectra are derived. Their importance is due to the Central Limit Theorem by Gnedenko and Kolmogorov applied for compactly supported signals [2].
In the mathematics of signal processing, the harmonic wavelet transform, introduced by David Edward Newland in 1993, is a wavelet-based linear transformation of a given function into a time-frequency representation. It combines advantages of the short-time Fourier transform and the continuous wavelet transform. It can be expressed in terms of repeated Fourier transforms, and its discrete analogue can be computed efficiently using a fast Fourier transform algorithm.
In functional analysis, a Shannon wavelet may be either of real or complex type. Signal analysis by ideal bandpass filters defines a decomposition known as Shannon wavelets. The Haar and sinc systems are Fourier duals of each other.
Overcompleteness is a concept from linear algebra that is widely used in mathematics, computer science, engineering, and statistics. It was introduced by R. J. Duffin and A. C. Schaeffer in 1952.
The wavelet transform modulus maxima (WTMM) is a method for detecting the fractal dimension of a signal.
Fractional wavelet transform (FRWT) is a generalization of the classical wavelet transform (WT). This transform is proposed in order to rectify the limitations of the WT and the fractional Fourier transform (FRFT). The FRWT inherits the advantages of multiresolution analysis of the WT and has the capability of signal representations in the fractional domain which is similar to the FRFT.
In mathematics, in functional analysis, several different wavelets are known by the name Poisson wavelet. In one context, the term "Poisson wavelet" is used to denote a family of wavelets labeled by the set of positive integers, the members of which are associated with the Poisson probability distribution. These wavelets were first defined and studied by Karlene A. Kosanovich, Allan R. Moser and Michael J. Piovoso in 1995–96. In another context, the term refers to a certain wavelet which involves a form of the Poisson integral kernel. In a still another context, the terminology is used to describe a family of complex wavelets indexed by positive integers which are connected with the derivatives of the Poisson integral kernel.
Wavelet packet bases are designed by dividing the frequency axis in intervals of varying sizes. These bases are particularly well adapted to decomposing signals that have different behavior in different frequency intervals. If has properties that vary in time, it is then more appropriate to decompose in a block basis that segments the time axis in intervals with sizes that are adapted to the signal structures.
## References
• Meyer, Yves (1992). Wavelets and Operators. Cambridge: Cambridge University Press. ISBN 0-521-42000-8.
• Chui, Charles K. (1992). An Introduction to Wavelets. San Diego: Academic Press. ISBN 0-12-174584-8.
• Akansu, Ali N.; Haddad, Richard A. (1992). Multiresolution Signal Decomposition: Transforms, Subbands, Wavelets. San Diego: Academic Press. ISBN 978-0-12-047141-6.
1. JPEG 2000, for example, may use a 5/3 wavelet for lossless (reversible) transform and a 9/7 wavelet for lossy (irreversible) transform.
2. A. G. Ramakrishnan and S. Saha, "ECG coding by wavelet-based linear prediction," IEEE Trans. Biomed. Eng., Vol. 44, No. 12, pp. 1253-1261, 1977.
3. N. Malmurugan, A. Shanmugam, S. Jayaraman and V. V. Dinesh Chander. "A New and Novel Image Compression Algorithm Using Wavelet Footprints"
4. Ho Tatt Wei and Jeoti, V. "A wavelet footprints-based compression scheme for ECG signals". Ho Tatt Wei; Jeoti, V. (2004). "A wavelet footprints-based compression scheme for ECG signals". 2004 IEEE Region 10 Conference TENCON 2004. A. p. 283. doi:10.1109/TENCON.2004.1414412. ISBN 0-7803-8560-8.
5. Bruns, Andreas (2004). "Fourier-, Hilbert- and wavelet-based signal analysis: are they really different approaches?". Journal of Neuroscience Methods. 137 (2): 321–332. doi:10.1016/j.jneumeth.2004.03.002. PMID 15262077.
6. Krantz, Steven G. (1999). A Panorama of Harmonic Analysis. Mathematical Association of America. ISBN 0-88385-031-1.
7. "Novel method for stride length estimation with body area network accelerometers", IEEE BioWireless 2011, pp. 79-82
8. Liu, Jie (2012). "Shannon wavelet spectrum analysis on truncated vibration signals for machine incipient fault detection". Measurement Science and Technology. 23 (5): 1–11. doi:10.1088/0957-0233/23/5/055604.
9. Akansu, A. N.; Serdijn, W. A.; Selesnick, I. W. (2010). "Emerging applications of wavelets: A review" (PDF). Physical Communication. 3: 1. doi:10.1016/j.phycom.2009.07.001. | 5,503 | 23,593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 32, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2019-43 | latest | en | 0.922304 |
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Article by: Francisco Javier Delatorre | Last update: April 10, 2022
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Rectangles, which have all four equal angles. Rhombuses, which have four equal sides. Squares, which have four equal angles and four equal sides. Parallelograms proper, that is, those that are not rectangles, rhombuses, or squares, are also called rhomboids.
## What is the difference and similarity between quadrilaterals?
Remember: Two quadrilaterals are similar if they have the same shape, but not necessarily the same size. When reducing or enlarging a quadrilateral, the result is another quadrilateral similar to the first one, in addition its angles are equal.
## What are the types of quadrilaterals?
Parallelograms: is a quadrilateral that has the two pairs of opposite sides parallel and the opposite angles equal. Square: Quadrilateral whose sides and angles are equal. Rectangle: it has four equal angles (90º) and equal sides two by two, the adjacent sides being different.
## What are the characteristics of a quadrilateral?
elements of a quadrilateral
4 vertices: points of intersection of the sides that make up the quadrilateral. 4 sides: segments that join the adjoining vertices. 2 diagonals: segments whose ends are two non-contiguous vertices. 4 interior angles: the one determined by two contiguous sides.
## How are quadrilaterals and their characteristics classified?
Classification of quadrilaterals A quadrilateral is a polygon with four sides and four vertices. According to the relationships established between their sides and between their angles, three types of quadrilaterals can be distinguished: parallelograms, trapezoids and trapezoids.
29 related questions found
### How are quadrilaterals classified and what are their characteristics?
Squares: If they have the four angles and the four equal sides. Rectangles: If they have their four equal angles and the equal opposite sides. Rhomboids: If they have their opposite angles and their opposite sides equal. Rhombuses: If they have four equal sides and equal opposite angles.
### Which and how many are the quadrilaterals?
Quadrilaterals – Square, rectangle, rhombus, trapezoid, parallelogram.
### What is congruence of quadrilaterals?
Quadrilaterals are polygons, geometric figures made up of straight lines that enclose a finite portion of the plane, whose only characteristic is that it has four sides. A quadrilateral is a plane figure made up of four sides that intersect two by two.
### What is a quadrilateral and examples?
The quadrilateral is a geometric figure, specifically a polygon, made up of four sides, four angles and four vertices. It should be noted that a polygon is a closed two-dimensional figure made up of a finite number of consecutive segments.
### Which figure has 4 congruent sides?
Opposite sides of a parallelogram are of equal length, (congruent). Opposite angles of a parallelogram are equal in measure. The angles of any two neighboring vertices are supplementary (add to 180°). The sum of the interior angles of any parallelogram is always equal to 360°.
### What are the quadrilaterals for elementary children?
Quadrilaterals are polygons with four sides and the sum of their interior angles equals 360°. Quadrilaterals have three main classifications: parallelograms, trapezoids, and trapezoids. They are the quadrilaterals that have parallel sides two by two.
### How are quadrilaterals classified according to their sides and angles?
Square: Four equal sides and four equal angles. Rhombus: They have four equal sides and only opposite angles are equal. Rectangle: It only has opposite sides equal and they have four equal angles. Rhomboid: Only opposite sides are equal and only opposite angles are equal.
### How are triangles classified according to their angles?
Classification according to their angles:
Acute triangle: has all three acute angles. Right triangle: it has a right angle. Obtuse triangle: has an obtuse angle.
### Which figures have congruent diagonals?
A rectangle has all the properties of a parallelogram, plus the following: Diagonals are congruent.
Opposite angles are congruent; Opposite sides are congruent; Adjacent angles are supplementary; Diagonals bisect each other.
### Which figure has four equal sides and four right angles?
A square is a quadrilateral with 4 equal sides and 4 right angles.
### What are the figures that have 4 right angles?
The rectangle has all four right angles. By definition, it has a right angle. Because it is a parallelogram, its opposite is also a right angle; and the other two angles, which are supplementary to the previous two, add up to 180º.
### What are the quadrilaterals that have 4 equal angles?
Rectangles, which have all four equal angles. Rhombuses, which have four equal sides. Squares, which have four equal angles and four equal sides.
### What is the figure that has all its equal sides?
Regular polygons are those that have all their sides and angles equal. Irregular polygons are those that do not meet these two conditions. The main characteristics of all regular polygons are: ● All their sides measure the same. All its interior angles have the same measure.
### What does congruent diagonals mean?
In mathematics, two geometric figures are congruent if they have the same dimensions and the same shape regardless of their position or orientation, that is, if there is an isometry that relates them: a transformation that can be translation, rotation or reflection.
### What are congruent and perpendicular diagonals?
In this case, the measures of the four sides are equal (they are congruent) and the two diagonals are perpendicular to each other (they intersect at 90°).
### What are the perpendicular diagonals?
Perpendicular Diagonals in Rhombuses
The diagonals of a rhombus not only bisect each other (since they are parallelograms), but their diagonals also meet at right angles. In other words, the diagonals are perpendicular.
### What is the classification of angles?
types of angles
Acute angle: Measures less than 90° and more than 0°. Right angle: It measures 90° and its sides are always perpendicular to each other. In this blog post you can learn all about right angles. Obtuse Angle: Greater than 90° but less than 180°.
### What are the 6 types of triangles?
1 According to its sides: Equilateral triangle. Three equal sides. Isosceles triangle. Two equal sides. Scalene triangle. Three unequal sides. … 2 According to its angles: Acute triangle. Three acute angles. Right triangle. A right angle. The longest side is the hypotenuse. The shorter sides are the legs.
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# 17_hwc1SolnsODDA - (The polynomial transformation is...
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(The polynomial transformation is multiply by x 2 , then differentiate twice). Is f linear? If so, find the corresponding matrix. If not, explain why not. SOLUTION As we saw in exercise 1.15, since d 2 d x 2 ( x 2 ( a + bx + cx 2 )) = 2 a + 6 bx + 12 cx 2 , the formula for f is f a b c = 2 a 6 b 12 c . Since the variables enter only in the first power – not zeroth, second, or anything else, this function is homogeneous and additive, and so it is linear. Using Theorem 2, and computing f ( e 1 ), f ( e 2 ) and f ( e 3 ) we find that its matrix is 2 0 0 0 6 0 0 0 12 . 2.33 Define a transformation f from IR 3 to IR 4 as follows. Define f a b c = s t u v where p ( x ) = a + bx + cx 2 and s = p (1) t = p (2) u = p (3) v = p (4) . Is
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https://zong-music.com/ha2ten/greater-than-or-equal-to-sign-9b504e | 1,708,667,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474361.75/warc/CC-MAIN-20240223053503-20240223083503-00454.warc.gz | 1,143,670,405 | 8,206 | # greater than or equal to sign
For example, x ≥ -3 is the solution of a certain expression in variable x. Select Symbol and then More Symbols. For example, the symbol is used below to express the less-than-or-equal relationship between two variables: ≥. "Greater than or equal to", as the suggests, means something is either greater than or equal to another thing. is less than > > is greater than ≮ \nless: is not less than ≯ \ngtr: is not greater than ≤ \leq: is less than or equal to ≥ \geq: is greater than or equal to ⩽ \leqslant: is less than or equal to ⩾ 923 Views. Use the appropriate math symbol to indicate "greater than", "less than" or "equal to" for each of the following: a. Greater than or equal application to numbers: Syntax of Greater than or Equal is A>=B, where A and B are numeric or Text values. With Microsoft Word, inserting a greater than or equal to sign into your Word document can be as simple as pressing the Equal keyboard key or the Greater Than keyboard key, but there is also a way to insert these characters as actual equations. For example, 4 or 3 ≥ 1 shows us a greater sign over half an equal sign, meaning that 4 or 3 are greater than or equal to 1. In such cases, we can use the greater than or equal to symbol, i.e. In Greater than or equal operator A value compares with B value it will return true in two cases one is when A greater than B and another is when A equal to B. Rate this symbol: (3.80 / 5 votes) Specifies that one value is greater than, or equal to, another value. This symbol is nothing but the "greater than" symbol with a sleeping line under it. Less Than or Equal To (<=) Operator. “Greater than or equal to” and “less than or equal to” are just the applicable symbol with half an equal sign under it. Greater Than or Equal To: Math Definition. 2 ≥ 2. But, when we say 'at least', we mean 'greater than or equal to'. The less than or equal to symbol is used to express the relationship between two quantities or as a boolean logical operator. "Greater than or equal to" is represented by the symbol " ≥ ≥ ". Solution for 1. The greater-than sign is a mathematical symbol that denotes an inequality between two values. In an acidic solution [H]… Greater than or Equal in Excel – Example #5. Here a could be greater … Examples: 5 ≥ 4. The sql Greater Than or Equal To operator is used to check whether the left-hand operator is higher than or equal to the right-hand operator or not. Category: Mathematical Symbols. When we say 'as many as' or 'no more than', we mean 'less than or equal to' which means that a could be less than b or equal to b. Select the Greater-than Or Equal To tab in the Symbol window. use ">=" for greater than or equal use "<=" for less than or equal In general, Sheets uses the same "language" as Excel, so you can look up Excel tips for Sheets. Copy the Greater-than Or Equal To in the above table (it can be automatically copied with a mouse click) and paste it in word, Or. Finding specific symbols in countless symbols is obviously a waste of time, and some characters like emoji usually can't be found. Graphical characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has no crossing lines. Select the Insert tab. If left-hand operator higher than or equal to right-hand operator then condition will be true and it will return matched records. Sometimes we may observe scenarios where the result obtained by solving an expression for a variable, which are greater than or equal to each other. As we saw earlier, the greater than and less than symbols can also be combined with the equal sign. 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That denotes an inequality between two values, Has no crossing greater than or equal to sign we say 'at least ', can... | 2,698 | 12,236 | {"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} | 3.578125 | 4 | CC-MAIN-2024-10 | latest | en | 0.920572 |
https://stats.stackexchange.com/questions/472974/green-poop-leafy-greens-and-probability-of-disease-how-can-i-formalize-this-re | 1,701,629,827,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00809.warc.gz | 596,177,045 | 46,963 | # Green poop, leafy greens and probability of disease, how can I formalize this reasoning?
Read everything before making your judgment, this is a serious probability question, it is not a joke.
Today when I woke up and went to do my usual business, my poop was green. I got worried, and very nervous, and thought that I might have a serious disease.
Mathematically, we can say that after I saw my poop was green, I updated my probability
$$P(\text{disease}|\text{green poop}, \text{everything else I did yesterday}) = \text{high}$$
So continuing our story, I was super worried and then searched the internet for "what causes green poop." Then I learned that eating leafy greens causes green poop! And yesterday I did eat leafy greens. So now, after this information, my new assessment was:
$$P^*(\text{disease}|\text{green poop}, \text{everything else I did yesterday}) = \text{low}$$
Where I'm using $$P^*$$ here to represent my updated probability measure. But this is the trick where I am at a loss, so here goes my question: how can I formalize mathematically my reasoning above? After all, all the evidence was already avaialble to me, I already knew I had eaten leafy greens. What I did not know was that they could have been an explanation for the green poop. Can you formalize mathematically exactly what type of updating I did to move from high to low probability?
I use the following binary variables:
• Poop is green: G
• Am sick: D
• Ate leafy greens: L
First, let's see how you can reach $$P(D=1|G=1) = 0.8$$. While you "knew" that you had eaten leafy greens and that it could cause green poop, when you thought about it first, you only considered a disease as a potential cause. That is, you only had in mind the probabilistic graph D -> G in mind, meaning $$P(D,G) = P(D)P(G|D)$$. For example, $$P(D=1) = 0.1$$ (you felt fine other than the poop), and $$P(G=1|D=1)$$ is also low (you know very little diseases that cause green poop), therefore $$P(D=1,G=1)$$ is pretty low. So how come you have $$P(D=1|G=1)=0.8$$? The alternative $$P(D=0|G=1)$$ is even lower: yes, $$P(D=0)=0.9$$ is high, yet having green poop while not being sick is extremely extremely unlikely (because most days, I am fine, yet my poop is not green)! You can check that by fixing actual probabilities.
Now when you learn or are reminded about leafy greens on the internet, you update your graph and add a potential cause "leafy greens". Formally, $$P(D,G,L) = P(L) P(D) P(G|D,L)$$. Now, because $$P(L)=1$$ (I know for sure I ate greens yesterday) and $$P(G=1|D=d,L=1)$$ for any $$d$$ is high: that's what I was "reminded" about on the internet: sick or not, leafy greens cause green poop.
By Bayes rules, $$P(D|G,L) \propto P(D) P(L) P(G|D,L)$$ and by fixing concrete probabilities you will find a low probability of disease thanks to the high $$P(G=1|D=d,L=1)$$.
That's an instance of explaining away: in the V-shaped graph, when you fix the value of the effect (G), the two causes are now dependent (D and L are dependent given G). The observation that one of the cause is present will decrease the probability of the other (in our case, drastically) and vice versa: if one cause is not present, the probability of the other cause will go up (in our case, you didn't eat leafy greens so you'd still think you are sick with high probability).
I tried to find a good reference for explaining away but did not. Pearl's automobile example seems to be frequently given, for example here.
## Relating this to Ben's answer
Yes, I did change the model by adding an edge in the graph, and it is not a fully "Bayesian" formalisation of the problem. I am reasoning like a scientist who incrementally builds a Bayesian model.
Your want to model your own thought process: you know that leafy greens are a relevant cause that you used to ignore, and therefore you want to put the variable I in the graph. Thanks to Ben's answer, you realize that the probabilistic graph of causes can be encoded in a very flexible way, where every possible cause can have no to a huge influence on the inference you are trying to draw, via these "gating" variables like I. I think that you were looking for Ben's answer, actually.
However, I want to point out that even though Ben's fully Bayesian model might (might only, see next paragraph) be a good (although HUGE) model for "thought processes", it does not reflect scientific elaboration of models. Imagine that I is binary, 1 if L causes G and 0 otherwise. A Bayesian scientist needs to put a prior over I, and in doing so, should think about whether L causes G. But as you said, you did not learn that $$I=1$$ on the internet; you were merely reminded about it. So if you had thought about it, you would have put a very probable I as a prior. In that case, you see that there is no updating going on and you just recover the analysis I provided with the second model. On the contrary, if you did not think about the cause, you would have built the first model I presented. In other words, if the Bayesian scientist is not fully satisfied with his model, he needs to build another one and his approach is not "fully Bayesian" (in the extreme, formal and dogmatic sense of the term).
Most importantly, I am still puzzled by Ben's answer, though, because he did not specify the prior over I. If we are modelling thought processes, we could see beliefs of an individual as continually updated throughout his life. For Ben's answer to be fully complete and convincing, we need the "prior" probability (before seeing the information on the internet) $$P(I=1)$$ to be low. Why would it be the case? I don't think the individual has been exposed to evidence for that in his life. There is something wrong.
Therefore, I am more inclined to imagine that we do approximate Bayesian inference in our heads with very partial graphs that are "instantiated" by extracting pieces of a "full knowledge graph" in an imperfect way.
I am very curious to hear Ben's opinion on that. There are probably tons of resources discussing the problem (maybe in the "objective vs subjective" or "Bayesian vs frequentist" debates?), but I'm not an expert.
• Thanks, I like this answer, but should we use different probability measures for the two models then? It seems you are suggesting that before learning the information I assumed that L was independent of G, and after learning the information I updated my model? Jun 21, 2020 at 4:22
• Please see my edit! Jun 21, 2020 at 17:05
It seems to me you are looking at the Bayes's theorem and in particular at the prior probability.
Your data ($$green\;poop, \; etc$$) is the same before and after checking the internet. However, initially, your prior probability is either neutral or in favour of disease since green poop is odd. After checking the internet your prior shifts in favour of not-disease and that updates the posterior towards $$P(disease|green\,poop,\; etc)=low$$. Mathematically, I guess you could use a beta distribution to model your prior belief more or less strongly in favour or against the disease.
This kind of problem can be handled using Bayesian analysis, but it requires a bit of care. The tricky bit here is that there is a distinction between the conditioning event "ate leafy greens" and the other conditioning event "information showing that eating leafy greens causes green poo". You already know you ate leafy greens in both scenarios, so that conditioning event is not what is changing your probability. Rather, it is the additional information you have obtained from your internet search that is telling you that leafy greens cause green poo, and therefore lead you to reduce your inferred probability of disease.
To simplify this analysis, I will assume that the only relevant conditioning event from the previous day is that you ate leafy greens (i.e., the event "ate leafy greens" will be equivalent to "everything I did yesterday). This removes explicit conditioning on the remainder of what happened that day. I will use the following events:
\begin{align} \mathcal{D} & & & \text{Disease}, \\[6pt] \mathcal{G} & & & \text{Green poop}, \\[6pt] \mathcal{L} & & & \text{Ate leafy greens}, \\[6pt] \mathcal{I} & & & \text{Information showing that } \mathcal{L} \text{ causes } \mathcal{G}. \\[6pt] \end{align}
The circumstance you are describing is that $$\mathbb{P}(\mathcal{D}|\mathcal{G} \cap \mathcal{L})$$ is high but $$\mathbb{P}(\mathcal{D}|\mathcal{G} \cap \mathcal{L} \cap \mathcal{I})$$ is low (i.e., the addition of the new information lowers the probability that you have a disease). There are many reasonable ways that you could be led to this outcome, but a general structure would look like the DAG below. Disease can cause green poo, but it can also be cauesd by eating leafy greens. (The joint path for the latter depends on the fact that the causal pathway from leafy greens to green poo is not known unless you obtain the information to that effect.)
In this case, the effect of gaining the information that relates eating leafy greens with green poo is that it "opens the pathway" at the bottom of the DAG, and thereby provides an alternative reason to believe that green poo could occur in the absence of a disease. This leads you to lower the conditional probability of disease accordingly. It would be possible to formalise this analysis further by giving some appropriate probability values to the various events of interest, but I will not pursue that level of detail. Hopefully this structural discussion assists you in understanding the nature of the inference you are making. Suffice to say, your reduction in the inferred probability of disease is a sensible conclusion from the additional conditioning information you obtained.
• Thanks, this seems to be in the right direction, although I'm not sure why "I" can be considered an event. It seems that I'm modifying my Bayesian model, and not actually conditioning on an event, no? Jun 21, 2020 at 4:21
• Is your answer similar to bomzh, in a sense that I here is actually "updating my graph"? Jun 21, 2020 at 4:24
• If you do not have the event $\mathcal{I}$ under consideration then when you are "updating your model" you are actually creating a new model with different probabilities, which means that your inference is not occurring within a single Bayesian analysis.
– Ben
Jun 21, 2020 at 5:20
$$statistics \neq mathematics$$
We can mathematically express probabilities (like you did two times) but they are not the real probabilities and instead only probabilities according to some model.
So a probability expression has a "probability" to fail. By how much... that depends on the quality of the model.
If your model is considered good (which is not well expressed mathematically), such that the effect of the bias of your model, having an influence on the discrepancy between calculations and reality, is negligible in comparison to the random error/variation occuring within the model, then we may consider the inaccuracies of the model negligible.
In your example we could say that your first model was not very accurate, and that is why it's result is so different from the more accurate second model. There is no contradiction.
Probabilities obtained from models, like p-values or posterior densities, are not real probabilities, and only a reflection of the real situation. These reflections can be distorted to various extents. This distortion is almost never the subject of the (mathematical) considerations/models. | 2,714 | 11,515 | {"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": 25, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-50 | longest | en | 0.963991 |
https://proofwiki.org/wiki/Definition_talk:Real_Number/Real_Number_Line | 1,596,475,617,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735823.29/warc/CC-MAIN-20200803170210-20200803200210-00186.warc.gz | 476,623,162 | 11,551 | Definition talk:Real Number/Real Number Line
The question is:
"Okay, this gives a list of proofs, but what exactly is this "real number line" defined as?"
It's just a list of proofs justifying the identification of the real number line with the geometrical entity. Feel free to put it into a format that suits our site better - I'm not sure how best to structure it. --prime mover (talk) 09:18, 3 November 2012 (UTC)
I was always taught to call it the Real Number System. It pleased me that this coincided with Definition:Algebraic System. I have recently become aware of the many proofs with "induces" in their title too. (I believe according to LF this would just be a shift from one point to another in the gigantic landscape of Magmas of Sets).
I would like to know PM when you think the verb "induces" is appropriate in a title (the power set is involved a lot isn't it?). I could take \mathbb{R} and equip it with the discrete topology and discrete metric but that's not really an interesting result because you can do it with any.
Btw, on a completely pointless note some of the titles capitalise induce others don't. --Jshflynn (talk) 11:01, 3 November 2012 (UTC)
The trouble with "real number system" is that "system" is one of those lazy but clever-sounding words which is frequently used instead of the specific word that is really required. A "system", as I was taught in one of my courses in higher education, is "anything" made out of bits which "interact somehow". That is of course completely general, and "general" here just means "vague". Thus I would generally discourage the word "system" unless there is a specific meaning to be exploited. As such, the term Definition:Algebraic System is almost as vague - and in this particular context that is too vague to be of any use.
Let me try to explain. The real numbers under addition form an Definition:Abelian Group; under addition and multiplication form a Definition:Field (Abstract Algebra); with the Definition:Positivity Property added on top of that they form an Definition:Ordered Field (a.k.a. totally ordered field but that's a tautology); with the conventional metric forms a Definition:Metric Space. My view is that each of these properties is relevant at a particular level of abstraction: sometimes you need the group properties, sometimes the field properties, and (in the context of topology and metric spaces, which we're currently tightening up) sometimes you need its metric space properties. In this particular instance, the intention of "real number line" is to exploit its ability to be treated as a mathematical model of an infinite straight line in geometrical space - for which purpose it needs to be considered as a Definition:Euclidean Space, that is, as a Definition:Metric Space.
Yes, the Real Number line is an Definition:Algebraic System, but it's a close call, because the Definition:Distance Function is only a "finitary operation on $\R$" by default - a Definition:Metric Space whose domain is not the Real Number Line is (technically) not an algebraic system because $d: M \times M \to \R$ is not an operation (finitary or otherwise) on $M$, unless $M$ is itself $\R$. --prime mover (talk) 11:48, 3 November 2012 (UTC)
Oh, and I'm not discussing the word "induce" here for obvious reasons. --prime mover (talk) 11:50, 3 November 2012 (UTC)
Firstly, I am in complete agreement with you about the overused word "system". Secondly, I understand now that as you equip a set with these relations and operations that others are not mentioned because they are irrelevant in context.
Now on the topic in question I would like to raise a point that may or may not change your mind completely. Looking at Richard S. Millman, George D. Parker: Geometry: A metric approach with models (1990) there is a set theoretic definition of a line:
Let $S$ be a set and $L$ be a set of subsets of $S$ such that $\forall l \in L: |l|>1$ and $\forall x, y \in S: \exists l \in L: x,y \in l$ ($S$ is the set of points and the elements of $L$ are called lines).
In Cantor-Dedekind Hypothesis, Euclid's postulates and the non-mathematical words "right" and "left" are used. So it may be worth taking a different approach (or at least mentioning this different approach). --Jshflynn (talk) 00:11, 5 November 2012 (UTC)
Why should that change my mind about use of the word "system"? --prime mover (talk) 06:09, 5 November 2012 (UTC)
Pardon my ambiguity. The topic in question (that I thought I might change your mind about) was the geometrical interpretation of the real number line.
• The book I mentioned bases geometry on the theory of metric spaces. Lines, triangles, betweeness and other concepts are defined in it in a much more powerful way. The set of real numbers under this framework can be proved to be a line.
• In Definition:Axiom it is stated that one of the ultimate goals of PW is to base mathematics on a handful of axioms and by taking this approach you can have geometry be an extension of set theory.
• While I think the definitions of Euclid should be preserved I think this site should not define line in the way given as mathematics is more powerful than that now.
If you disagree I will not mind. Excessive amounts of studying formal languages have made me distrust geometry to the point of impracticality. --Jshflynn (talk) 10:03, 5 November 2012 (UTC)
I'm sorry, you've completely lost me. I have studied the above paragraph to which you responded with something about me changing my mind, and I'm at a loss to work out what you are specifically referring to. --prime mover (talk) 13:47, 5 November 2012 (UTC)
No worries. I will put up a demonstration of it so you can tell me whether or not it goes against ProofWiki.
Turns out I have an ebook version (unspeakably violated by the reader who scanned his copy with underlining, circling etc. etc.) and I read the first few paragraphs. It seems to be an interesting approach to geometry. However, I don't really like geometry (my intuition for it sucks as soon as the plane is left) so I'll leave others to maybe cover its content on PW some day. --Lord_Farin (talk) 10:40, 5 November 2012 (UTC) | 1,468 | 6,148 | {"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} | 3.40625 | 3 | CC-MAIN-2020-34 | latest | en | 0.973202 |
https://www.physicsforums.com/threads/temperature-and-pressure-problem.162152/ | 1,585,674,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00086.warc.gz | 1,121,298,563 | 14,179 | # Temperature and pressure problem
I know that in a closed container, pressure increases as temperature increases, but I am having trouble trying to explain it mathematically. I did some searching around and came up with this formula: PV = nRT. I am having some trouble trying to apply this to my scenario;
closed volume - 496.5 cu-inche (~.5 L)
Temperature - 120* F
Pressure - 6 psi (~.41 bar)
The volume remains constant. What would the pressure be if the temperature increased to 900* F?
Related Introductory Physics Homework Help News on Phys.org
russ_watters
Mentor
All you need to do is rearrange the equation and drop out the constants. I'm not a big fan of derivations, but....
Since n, V, and R are constants...
P/T=Constant
P1/T1=P2/T2
There are many forms for the ideal gas equation... | 198 | 802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2020-16 | longest | en | 0.928556 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=19&t=34602 | 1,606,469,694,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141191511.46/warc/CC-MAIN-20201127073750-20201127103750-00240.warc.gz | 372,208,120 | 11,208 | ## Heisenberg Indeterminacy Post Assessment #20
$\Delta p \Delta x\geq \frac{h}{4\pi }$
Julia Lindner 1I
Posts: 35
Joined: Fri Sep 28, 2018 12:17 am
### Heisenberg Indeterminacy Post Assessment #20
"Use the above uncertainty in velocity to calculate the electron's uncertainty in kinetic energy. Then calculate the uncertainty in kinetic energy per mole of electrons (that is, per mole of hydrogen atoms). Comment on your value."
How do you find uncertainty in kinetic energy? As far as I know this was not mentioned in either the lectures or the textbook.
Blake Salfer 1B
Posts: 30
Joined: Fri Sep 28, 2018 12:18 am
### Re: Heisenberg Indeterminacy Post Assessment #20
all you have to do is plug in the velocity found in the problem above into the equation Ek=1/2mv^2 and you will get the uncertainty in KE
Grace Kim 1J
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am
Been upvoted: 1 time
### Re: Heisenberg Indeterminacy Post Assessment #20
You use the answer of the velocity you found in problem #19. Then you want to use the kinetic energy formula. Make sure you convert the answer you get to energy per mol of electrons (multiply the answer with 6.022x10^23e/mol). When are you asked to find the uncertainty in kinetic energy, just remember the formula (Ek=1/2mv^2). | 350 | 1,278 | {"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": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-50 | latest | en | 0.841365 |
http://forums.wolfram.com/mathgroup/archive/2011/Mar/msg00244.html | 1,716,998,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059246.67/warc/CC-MAIN-20240529134803-20240529164803-00122.warc.gz | 11,037,177 | 8,385 | Re: FindFit power law problem
• To: mathgroup at smc.vnet.net
• Subject: [mg117043] Re: FindFit power law problem
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Tue, 8 Mar 2011 05:37:14 -0500 (EST)
```On 3/7/11 at 5:48 AM, jgchilders at mailaps.org (Greg Childers) wrote:
>I'm having a problem with FindFit and a power law problem. Here's
>the data:
>data = {{1004, 0.003977}, {9970, 0.006494}, {100000, 0.012921},
>{1001000, .059795}}
>I'm wanting to fit it to a function of the form y = a x^b, and
>determine the best value of b. When entered into Excel, it returns
>the exponent b = 0.383. However, Mathematica gives
>FindFit[data, a x^b, {a, b}, x] {a->0.0000145749, b->0.601807}
>A graph of these values overlaid on the original data simply didn't
>look right. Another way to find the exponent b is to take the log
>of both sides and do a linear fit:
>Fit[Log[10, data], {1, x}, x] -3.64936 + 0.383174 x
>And sure enough the exponent is 0.383 in agreement with Excel. Why
>does FindFit give a different value?
The short answer is FindFit is trying to *correctly* solve the
following problem,
y = a x^b + error
where the error term is assumed to be normally distributed.
Excel most definitely is not doing this. Excel does precisely
what you did by taking the Log of your data then estimating the
parameters. Note, when you take the Log you have
Log[y] = Log[a] + Log[x^b + error]
not
Log[y] = Log[a] + b Log[x] + error
It is this last problem Excel solves and what FindFit solves
when you first take the Log of your data array then use FindFit.
An additional problem is seen when you do ListLogLogPlot[data].
That is, the resulting plot shows curvature indicating y = a x^b
isn't the underlying model for your data.
The curvature suggests a model like y = a(x+b)^c would better
fit your data. When I use this model I get
In[6]:= FindFit[data, a*(x + b)^c, {a, b, c}, x]
Out[6]= {a -> 2.14496457174612*^-6, b -> 32107.42765117816,
c -> 0.7391222616795872}
and on checking
In[7]:= model = a (x + b)^c /. %
Out[7]= 2.14496*10^-6 (x+32107.4)^0.739122
The plot that results from
LogLogPlot[model, {x, 1000, 1.5 10^6},
Epilog -> {PointSize[.01], Point[Log[data]]}]
seems to be in good agreement with the data
Note, I am not claiming this is a correct model for your data.
It is merely the first model that came to mind after looking a a
log log plot of your data points. And since you only have 4 data
points and this model has three free parameters, it is virtually
guaranteed to show a better fit to your data regardless of
whether it matches the "true" model or not. I've put true in
quotes here since if this data is experimental data, the "true"
model is often unknown.
```
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• Next by thread: Re: FindFit power law problem | 895 | 2,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.829029 |
http://mechonomic.blogspot.com/2012/01/quarterly-report-of-hewlett-packard.html | 1,550,442,037,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247482788.21/warc/CC-MAIN-20190217213235-20190217235235-00631.warc.gz | 164,666,517 | 17,287 | ## 1/21/12
### Quarterly report: the performance of share price model for Hewlett Packard
This is quarterly report of the performance of our share price model. Hewlett Packard (NYSE:HPQ) provides a good example of a successful share price prediction at a several month horizon. We have already published our predictions at a four month horizon four times (July 2010, January 2011, March 2011, July 2011, and September 2011). All predictions were based on our concept of share pricing as decomposition into a weighted sum of two CPI components. We calculated the evolution of the monthly closing price (adjusted for dividends and splits). Here we test and update the model using data through December 2011. The model is still accurate and robust.
Originally, the long term model for HPQ share price was defined by the index of food without beverages (FB) and that of rent of primary residency (RPR). The former CPI component led the share price by 4 months and the latter one led by 5 months. Figure 1 depicts the overall evolution of both involved indices through December 2011 (this is the reason of the time lead increase by 1 month relative to previous models where CPI were not contemporaneous). Below we present four best-fit 2-C models for HPQ(t) obtained at different times:
HPQ(t) = -3.20FB(t-4) + 2.91RPR(t-5) + 3.64(t-1990) - 50.82, July 2010
HPQ(t) = -3.34FB(t-4) + 3.41RPR(t-5) + 0.51(t-1990) - 85.44, June 2011
HPQ(t) = -3.46FB(t-4) + 3.68RPR(t-5) – 0.72(t-1990) - 99.88, September 2011
HPQ(t) = -3.40FB(t-5) + 3.60RPR(t-6) – 0.57(t-1990) – 97.72, December 2011
where HPQ(t) is the price in US dollars, t is calendar time. All coefficients have been slightly drifting but very close. This process expresses the trade-off between the linear trend in the difference between the defining CPIs and the time trend term in the above equtions.
The predicted curves are shown in Figure 2 (March, September, and December 2011). From Figure 2, we predict the price to fall to \$20 in the first quarter of 2012 and then rise to \$25 in Q2.
Figure 1. Evolution of the price of FB and RPR.
Figure 2. Observed and predicted HPQ share prices in March, September, and December 2011. The contemporaneous prediction is shown by solid red line. High and low prices are shown by dashed lines. | 636 | 2,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-09 | latest | en | 0.945576 |
https://www.justintools.com/unit-conversion/area.php?k1=centiares&k2=square-leagues-US-STATUTE | 1,632,485,679,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00004.warc.gz | 836,201,934 | 27,848 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# AREA Units Conversioncentiares to square-leagues-US-STATUTE
1 Centiares
= 4.290006866585E-8 Square Leagues US STATUTE
Category: area
Conversion: Centiares to Square Leagues US STATUTE
The base unit for area is square meters (Non-SI/Derived Unit)
[Centiares] symbol/abbrevation: (ca)
[Square Leagues US STATUTE] symbol/abbrevation: (sq leag [US])
How to convert Centiares to Square Leagues US STATUTE (ca to sq leag [US])?
1 ca = 4.290006866585E-8 sq leag [US].
1 x 4.290006866585E-8 sq leag [US] = 4.290006866585E-8 Square Leagues US STATUTE.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Centiares (ca) is equal to 1 square-meters, while 1 Square Leagues US STATUTE (sq leag [US]) = 23309986 square-meters.
1 Centiares to common area units
1 ca = 1 square meters (m2, sq m)
1 ca = 10000 square centimeters (cm2, sq cm)
1 ca = 1.0E-6 square kilometers (km2, sq km)
1 ca = 10.763915051182 square feet (ft2, sq ft)
1 ca = 1550.0031000062 square inches (in2, sq in)
1 ca = 1.1959900463011 square yards (yd2, sq yd)
1 ca = 3.8610215859254E-7 square miles (mi2, sq mi)
1 ca = 1550003100.0062 square mils (sq mil)
1 ca = 0.0001 hectares (ha)
1 ca = 0.00024710516301528 acres (ac)
Centiaresto Square Leagues US STATUTE (table conversion)
1 ca = 4.290006866585E-8 sq leag [US]
2 ca = 8.58001373317E-8 sq leag [US]
3 ca = 1.2870020599755E-7 sq leag [US]
4 ca = 1.716002746634E-7 sq leag [US]
5 ca = 2.1450034332925E-7 sq leag [US]
6 ca = 2.574004119951E-7 sq leag [US]
7 ca = 3.0030048066095E-7 sq leag [US]
8 ca = 3.432005493268E-7 sq leag [US]
9 ca = 3.8610061799265E-7 sq leag [US]
10 ca = 4.290006866585E-7 sq leag [US]
20 ca = 8.58001373317E-7 sq leag [US]
30 ca = 1.2870020599755E-6 sq leag [US]
40 ca = 1.716002746634E-6 sq leag [US]
50 ca = 2.1450034332925E-6 sq leag [US]
60 ca = 2.574004119951E-6 sq leag [US]
70 ca = 3.0030048066095E-6 sq leag [US]
80 ca = 3.432005493268E-6 sq leag [US]
90 ca = 3.8610061799265E-6 sq leag [US]
100 ca = 4.290006866585E-6 sq leag [US]
200 ca = 8.58001373317E-6 sq leag [US]
300 ca = 1.2870020599755E-5 sq leag [US]
400 ca = 1.716002746634E-5 sq leag [US]
500 ca = 2.1450034332925E-5 sq leag [US]
600 ca = 2.574004119951E-5 sq leag [US]
700 ca = 3.0030048066095E-5 sq leag [US]
800 ca = 3.432005493268E-5 sq leag [US]
900 ca = 3.8610061799265E-5 sq leag [US]
1000 ca = 4.290006866585E-5 sq leag [US]
2000 ca = 8.58001373317E-5 sq leag [US]
4000 ca = 0.0001716002746634 sq leag [US]
5000 ca = 0.00021450034332925 sq leag [US]
7500 ca = 0.00032175051499387 sq leag [US]
10000 ca = 0.0004290006866585 sq leag [US]
25000 ca = 0.0010725017166462 sq leag [US]
50000 ca = 0.0021450034332925 sq leag [US]
100000 ca = 0.004290006866585 sq leag [US]
1000000 ca = 0.04290006866585 sq leag [US]
1000000000 ca = 42.90006866585 sq leag [US]
(Centiares) to (Square Leagues US STATUTE) conversions | 1,287 | 3,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2021-39 | latest | en | 0.761906 |
http://mathhelpforum.com/calculus/180821-find-centroid-c-o-m-thin-plane.html | 1,529,861,469,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00385.warc.gz | 194,361,064 | 9,660 | # Thread: find the centroid (C.O.M) of a thin plane
1. ## find the centroid (C.O.M) of a thin plane
hi,
find the center of mass (or centroid) of a thin plate of constant density covering the given region.
(a) The region bounded by y = x^2 and y = 4.
I can't seem to get this question... I tried to find the moment about the y axis:
\int px(f1 - f2)dx
4,0\int x((x^2)-4)dx
and I got an answer of 32
then I did the M:
4,0\int ((x^2)-4)
and got an answer of 16/3
2. \int px(f1 - f2)dx
4,0\int x((x^2)-4)dx
and I got an answer of 32
It's a little hard to read this. Do you mean $\displaystyle \int_0^4x(x^2-4)\,dx=32$? You can write this formula as follows: $$\int_0^4x(x^2-4)\,dx=32$$. This is correct, but I don't see how it helps. The x-coordinate is $\displaystyle \int_{-2}^2x(4-x^2)\,dx=0$.
then I did the M:
4,0\int ((x^2)-4)
and got an answer of 16/3
It should be $\displaystyle \int_{-2}^2(4-x^2)\,dx=32/3$.
First, I calculated the areas under the parabola and found the area of the plate by subtracting those areas from 16. Second, the x-coordinate of the centroid is 0 because of the symmetry. (You can still calculate it using the general method, though.) Finally, the width of the plate at height y is $\displaystyle 2\sqrt{y}$, so the y-cooldinate of the centroid is $\displaystyle 1/M\int_0^42y\sqrt{y}\,dy$. My answer is 12/5. | 471 | 1,346 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.9375 | 4 | CC-MAIN-2018-26 | latest | en | 0.879155 |
https://www.experts-exchange.com/questions/20167682/Make-a-grid-with-selectable-squares.html | 1,524,534,816,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946314.70/warc/CC-MAIN-20180424002843-20180424022843-00210.warc.gz | 779,392,439 | 24,091 | Solved
# Make a grid with selectable squares
Posted on 2001-08-13
Medium Priority
940 Views
Hi, i'm quite a newbie java programmer, and i got a quite simple question (with a long answer)... I got an idee to make a turn based strategie game (already made an rpg), but i need a grid with about 25x25 squares, and every square needs some properties. I tried some newbie stuff but i'm afraight i'll have to use some kind of difficult math line that i have never heard of.
So if you could just explain me how i have to tell my program over wich square my mouse is when i press it, i should be able to continue on my own. I Painted a grid and placed some buttons above it already, but that's all i have done so far...
It may take some of your time so that's why i filled in 150 points.
tnx,
Vincent Sels
0
Question by:Vincent_Sels
• 5
• 2
LVL 1
Accepted Solution
lawrie earned 450 total points
ID: 6381247
The most flexible way would be to make a 'Cell' class, that takes care of each square. This could include drawing it (assuming later it will be more sophisticated than just a square) and knowing what size it is .. my suggestion is to have an java.awt.Rectangle class for each Cell that defines its location and size.
Then on your mouse click events, you will get the x & y coord of your mouse (event.getX() & event.getY()) and just test to see if you are in each Cell with Rectangle.contains(new Point(x, y))
This would be best if the board will not be just a big grid (eg some cells are different widths). If it will be (make sure this won't change later) perhaps just some maths to check which cell the x & y of the mouse click event occurred in.
If all cells the same, something like this would work
clickedColumn = x / (boardWidth / numColumns);
clickedRow = y / (boardHeight / numRows);
Hope this helps
0
Author Comment
ID: 6381347
tnx i'll try some more and let you know if i was able to continue...
0
Author Comment
ID: 6381658
OK this is how i'm gonna do it: subclass rectangle so i can add extra options, add all those 'rects' (in an array field) to the applet, and draw lines around them. Than i'll just use the contains method; that should do it...
but i encountered a problem... see for yourself:
extracts from class WorldMap:
Rect[] field = new Rect[400];
{
for (int y=0; y<20; y++)
{
for (int x=0; x<20; x++)
{
field[(y*20)+x] = new Rect(3+ x*25, 30+ y*25, 25, 25); //here's the problem...
field[(y*20)+x].nr = y*20 + x;
}
}
}
public class Rect extends java.awt.Rectangle
{
stuff
}
when I try to create a new rect the compiler sais he cant find constructor Rect(int, int, int, int) in class Rect... but there certainly is one in class Rectangle. Do I do something wrong ? Do I have to make a new constructor ?
And btw, will it work that way or would you do it completely different ?
0
LVL 1
Expert Comment
ID: 6381842
If you have no constructors for Rect, will only get the default constructor, and hence the default Rectangle constructor. add the following to Rect, and it should fix things
public Rect(int x, int y, int width, int height)
{
super(x, y, width, height);
}
What you propose would be the most generic way of doing things, offering flexibility to have different sizes or shapes later, so it would be the method I would go for personally. The only drawback would be having to search through all 400 cells to find a click, which will be a bit slower, but shouldn't have a great impact as long as you don't do many lookups unnecessarily (eg don't find it again if you have already found the required cell and could save which one it was instead of recalculating)
0
LVL 92
Expert Comment
ID: 6382453
If you simply make each of your cells a component then you can add mouse motion listeners to each cell, removing the need for any calculations to work out which cell the mouse is over.
0
Author Comment
ID: 6383602
Lawrie, I used that constructor and now both classes compile but unfortunately the whole thing won't work... When, for example, I try to display the string 'it worked' in field[200] (the rect in the middle), and draw a line from field[1].x, field[1].y to field[399].x, field[399].y nothing at all shows up ! I checked my formulas (field[(y*20)+x] = new Rect(3+ x*25, 30+ y*25, 25, 25);) to place the rectangles and I just can't find any mistakes :/
(the +3 or +30 i add to the place were the rectangle should be construted, is because it has to start at point (3,30) in stead of 0,0. With a canvas I would probably be able to avoid this but I just couldn't get the canvas in the right position so I decided to do it without...)
regards,
BlackDeath
0
Author Comment
ID: 6383991
Ok i'll just give the whole program... find out what's wrong because i really don't see the problem. I included some actions to trace the error but I don't see why it happens there.
import java.awt.*;
import java.lang.Math;
public class WorldMap extends java.applet.Applet
{
/*////////////////////////////////////////*/
/*/ Variables /*/
/*////////////////////////////////////////*/
/******************\
* LAYOUT VARIABLES *
\******************/
Panel buttons = new Panel();
Button b_endTurn = new Button("END TURN");
Button b_money = new Button("money");
Button b_resources = new Button ("resources");
Button b_production = new Button ("production");
Button b_research = new Button ("research");
Button b_bases = new Button ("bases");
Button b_groups = new Button ("groups");
/*********************\
* SELECTION VARIABLES *
\*********************/
Rect[] field = new Rect[400];
Point mousePos;
boolean groupSelected;
/*****************\
* OTHER VARIABLES *
\*****************/
/*////////////////////////////////////////*/
/*/ Functions /*/
/*////////////////////////////////////////*/
{
for (int y=0; y<20; y++)
{
for (int x=0; x<20; x++)
{
field[(y*20)+x] = new Rect(3+ x*25, 30+ y*25, 25, 25);
field[(y*20)+x].nr = y*20 + x;
}
}
}
public void init()
{
/******************\
* ARRANGING LAYOUT *
\******************/
buttons.setLayout(new GridLayout(1, 7, 0, 0));
}
public void paint(Graphics g)
{
/**************\
* DRAWING GRID *
\**************/
///////////////////////TEST///////////////////////
else
//////////////////////////////////////////////////
g.setColor(Color.black);
g.drawLine(3, 30, 503, 30);
g.drawLine(3, 30, 3, 530);
g.drawLine(503, 30, 503, 530);
g.drawLine(3, 530, 503, 530);
///////////////////////TEST///////////////////////
if (field[1].x != 0 && field[1].y != 0 && field[499].x != 0 && field[499].y != 0)
g.drawString("field[1].x = " + field[1].x + ". field[1].y = " + field[1].y + ". field[499].x = " +
field[499].x + ". field[499].y = " + field[499].y + ".", 50, 50);
else
g.drawString("1 or more values = 0...", 50, 50);
//////////////////////////////////////////////////
}
public boolean mouseDown(Event evt, int x, int y)
{
return true;
}
}
import java.awt.Rectangle;
public class Rect extends java.awt.Rectangle
{
Rect(int x, int y, int widht, int height)
{
super();
}
int nr;
boolean explored;
boolean isSite;
boolean isResource;
boolean isMountain;
boolean isForest;
boolean groupPresent;
int sitePlace;
int resPerTurn;
int resTurns;
int numSite;
int numResource;
int numGroup;
}
0
Author Comment
ID: 6385882
Ok I got a lot further now; I have the cells and they can contain new properties, they are selectable and easy editable... just what I wanted.
I'll accept on of your answers lawrie, because you first mentioned the rectangle system... thanks again !
If I encounter other problems I'll post them here to.
0
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Have a better answer? Share it in a comment. | 1,999 | 7,779 | {"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-2018-17 | latest | en | 0.936686 |
https://community.tableau.com/message/1033817 | 1,585,648,605,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00504.warc.gz | 405,136,661 | 32,684 | 10 Replies Latest reply on Feb 3, 2020 11:33 AM by Bryce Larsen
# Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Hello Tableau Community!
So I've looked far and wide for an answer to this question, but I have not come across the exact answer I'm looking for. My data looks like the following:
Customer IDEvent Type
123
Event B
123Event E
123Event B
123Event C
234Event A
234Event A
345Event D
456Event D
456Event E
567Event A
567Event C
567Event A
678Event D
678Event E
789Event A
789Event C
890Event E
890Event D
890Event D
What I need to do is count the number of Customer IDs by the distinct combination of Event Types that puts the Event Types in alphabetical order. So when I create the calculated field or aggregation, the output I'm looking for should be this:
Number of Customer IDsUnique Combination of Event Type
1Event B, Event C, Event E
1Event A
1Event D
2Event A, Event C
3Event D, Event E
The calculated field would concatenate Event B, Event C, Event E into one row and count 1 (Customer ID 123); Event A into one row and count 1 (Customer ID 234); Event D into one row and count 1 (Customer ID 345); Event A, Event C into one row and count 2 (Customer ID 456 and 789); and Event D, Event E into one row and count 3 (Customer ID 567, 678, and 890).
I've thought that I can maybe do this using a Level of Detail expression incorporating PREVIOUS_VALUE(), LOOKUP(), or something else, but I have not been successful in my attempts. Can anyone help?
I've looked at the following articles for reference:
Thank you for your future help!
• ###### 1. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Without having an indication of the number of Event Types or the number of occurrences per Customer ID perhaps this can be solved mathematically? Would it be possible to assign, for example, 1 to Type A, 10 to Type B, 100 to Type C? You may be able to sum those values to work out which Event Type combination each customer sits in. For example 11 (1 + 10) would mean Type A and Type B, 101 (1 + 100) would mean Type A and Type C, etc.
The feasibility would depend on the number of event types - too many and the numbers would get too large, and the number of times an event can happen per customer id. For example if it's possible for Type A to occur 11 times then a value returned of 11 could either mean Type A and Type B or 11 Types As.
Note there would be more complex calculations available to get around the summation challenges.
Hope you understand what I mean with the above possibility. Happy to clarify if not clear.
• ###### 2. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Hi Kyle,
This was difficult! But I think I ended up getting to what you needed:
This was achieved using a combination of LOD expressions and Table Calculations. I have a meeting shortly, so I can't go into too much detail at the moment, but wanted to post in case you want to dig in.
Please note that there is some hidden data (on Max Event Per Customer). It's used during some of the table calculations but ends up needing to be hidden to get the viz you're looking for.
Best,
Bryce
1 of 1 people found this helpful
• ###### 3. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Hey Andrew,
Thank you for the response! Unfortunately there can be an unlimited number of events per customer. For example, Customer 123 could have 10 Event A rows, 20 Event B rows, and 30 Event C rows. While assigning mathematical values would work if they were limited to having a certain number of events per Event Type, they are not.
Let me know if that makes sense, but I think understood and addressed your suggestion correctly.
• ###### 4. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Hey Bryce,
Thanks for sending this along! This is really helpful. Is there anyway you can post instructions with steps on how you did it? I've replicated the same thing in my internal workbook, but whenever I get down to the table with just [Customer Running Count] and [Events], it only shows me the [Customer Running Count] as being equal to 1, when I know there are more values than that. I'm not sure if order matters when you built out and applied these different calculated fields and table calculations.
Thanks!
Kyle
• ###### 5. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Hi Kyle,
Do you have all the necessary fields on your detail pane still?
Here are the steps I went through (after failing several other ways I must say!).
1. Calculate total events per Customer (Events Per Customer LOD)
• {fixed [Customer ID]: COUNTD([Event Type])}
2. Calculate the last/max event per Customer (Max Event Per Customer)
• {fixed [Customer ID], [Event Type]: MAX([Event Type] = {fixed [Customer ID]: MAX([Event Type])})}
3. Drag all of these onto a worksheet.
4. Create the concatenated Event Type field (Events)
• IF FIRST()=0 THEN MIN([Event Type]) ELSE PREVIOUS_VALUE(MIN([Event Type])) + ", " + MIN([Event Type]) END
• Partition it by Customer ID
5. Create the First Customer of Events field.
• IF MAX([Max Event Per Customer]) THEN FIRST()=0 END
• This is partitioned by everything except Customer ID.
6. Then I made the Customer Running Count field
7. And I added a Last Customer Per Event field in order to hide unnecessary rows
8. Similar, I right clicked on False under Max Event Per Customer and selected 'Hide' to clean it up.
Now...having said all of that...there's an error with this approach I've just confirmed. I added a Customer, 999, with Events A, C, and E and this then returns the wrong count...
I'll have to try something else.
• ###### 6. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
OK!
I'm back. Steps 1-4 are the same.
1. Calculate total events per Customer (Events Per Customer LOD)
• {fixed [Customer ID]: COUNTD([Event Type])}
2. Calculate the last/max event per Customer (Max Event Per Customer)
• {fixed [Customer ID], [Event Type]: MAX([Event Type] = {fixed [Customer ID]: MAX([Event Type])})}
3. Drag all of these onto a worksheet.
4. Create the concatenated Event Type field (Events). Partitioned by Customer ID (everything except Customer ID checked).
• IF FIRST()=0 THEN MIN([Event Type]) ELSE PREVIOUS_VALUE(MIN([Event Type])) + ", " + MIN([Event Type]) END
5. Create a Window Calc (Events per Customer TC). This is also Partitioned by Customer ID.
• WINDOW_MAX([Events])
6. Create a flag for a single Customer Event String
7. Next, create the # Customers field
• IF [Events per Customer TC]<>LOOKUP([Events per Customer TC],-1) OR ISNULL(LOOKUP([Events per Customer TC],-1)) THEN 1 ELSE PREVIOUS_VALUE(SUM(1))+1 END
8. Lastly, I made a flag for which rows to keep:
• [Events per Customer TC]<>LOOKUP([Events per Customer TC],1) OR LAST()=0
• This uses every Dimension (all of them checked)
Final result:
You can see this now shows the combination of Event A, C, and E for Customer 999 that I added. Again, please note that I right clicked on Max Event Per Customer and hid the False rows.
Best,
Bryce
• ###### 7. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Hey Bryce Larsen
Thanks so much for all of this, this is incredible work that you have put together and you clearly have put a tremendous amount of thought into it.
While I see it working for this dataset, I'm having trouble with getting it to work for the larger more complex datasets I am working with.
I've attached a scrubbed example of what I'm working with, so I'm hoping you can take a look at it to see if you can get it to work for you, as I'm nailing a few steps but then am struggling with the [# Customers] and [Rows to Keep]. Additionally, do you use the [Single Customer Flag] field? I know you mentioned to build that, but I don't see it included on the visuals in your workbook. Check out the attachment to this reply for a larger a dataset to see if you can solve the problem for that too. I'm not sure why it is struggling with this large of a dataset as this process you built out should scale.
Thanks so much for all of your hard work already too - can't tell you how educational and helpful this has been.
Best,
Kyle
• ###### 8. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Definitely proving tricky! Out of curiosity, what's your data source?
I'm just curious if it's easier doing some of this at the data source layer unless there's a reason you need this dynamic (date filters?).
Best,
Bryce
• ###### 9. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
My data source is an Amazon Athena database where I'm using Custom SQL to pull it. Ideally, I could be able to filter on this data using dates for when the events happened too. There is a date column I'm not including that is to the right of that data - I can include it if you think it would help?
• ###### 10. Re: Concatenating Unique Text Values in Different Rows for a Dimension at a Higher Level of Detail
Gotcha. I just wanted to confirm - always easier to concatenate at data source if possible, but not if you want it dynamic (unless it was a stored procedure or something).
I've shared this to twitter as well in case others find time to help out before I can! | 2,289 | 9,590 | {"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} | 3.125 | 3 | CC-MAIN-2020-16 | latest | en | 0.884794 |
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GuruFocus has detected 2 Warning Signs with BBX Capital Corp \$BBX.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
BBX Capital Corp (NYSE:BBX)
Net-Net Working Capital
\$4.83 (As of Sep. 2016)
In calculating the Net-Net Working Capital (NNWC), Benjamin Graham assumed that a companys accounts receivable is only worth 75% its value, its inventory is only worth 50% of its value, but its liabilities have to be paid in full. In addition, Graham believed that preferred stock belongs on the liability side of the balance sheet, not as part of capital and surplus. This is a conservative way of estimating the companys value.
BBX Capital Corp's net-net working capital per share for the quarter that ended in Sep. 2016 was \$4.83.
Definition
BBX Capital Corp's Net-Net Working Capital (NNWC) per share for the fiscal year that ended in Dec. 2015 is calculated as
Net-Net Working Capital Per Share (A: Dec. 2015 ) = (Cash And Cash Equivalents + 0.75 * Acct. Receivable + 0.5 * Inventory - Total Liabilities - Preferred Stock) / Shares Outstanding = (69.04 + 0.75 * 13.732 + 0.5 * 0 - 58.701 - 0) / 16.39 = 1.26
BBX Capital Corp's Net-Net Working Capital (NNWC) per share for the quarter that ended in Sep. 2016 is calculated as
Net-Net Working Capital Per Share (Q: Sep. 2016 ) = (Cash And Cash Equivalents + 0.75 * Acct. Receivable + 0.5 * Inventory - Total Liabilities - Preferred Stock) / Shares Outstanding = (120.266 + 0.75 * 14.823 + 0.5 * 0 - 51.147 - 0) / 16.62 = 4.83
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
In calculating the Net-Net Working Capital (NNWC), Benjamin Graham assumed that a companys accounts receivable is only worth 75% its value, its inventory is only worth 50% of its value, but its liabilities have to be paid in full.
In addition, Graham believed that preferred stock belongs on the liability side of the balance sheet, not as part of capital and surplus. In "Security Analysis", preferred stock is dubbed "an imperfect creditorship position" that is best placed on the balance sheet alongside funded debt.
This is a conservative way of estimating the companys value.
Explanation
One research study, covering the years 1970 through 1983 showed that portfolios picked at the beginning of each year, and held for one year, returned 29.4 percent, on average, over the 13-year period, compared to 11.5 percent for the S&P 500 Index. Other studies of Grahams strategy produced similar results.
Benjamin Graham looked for companies whose market values were less than two-thirds of their net-net value. They are collected under our Net-Net screener. GuruFocus also publishes a monthly Net-Net newsletter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
BBX Capital Corp Annual Data
Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 NNWC -2,390.07 -2,574.04 -2,310.18 -405.32 -295.35 -186.24 -10.62 -5.01 -0.88 1.26
BBX Capital Corp Quarterly Data
Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 NNWC -2.42 -0.46 -0.88 -1.50 0.03 -0.34 1.26 1.26 3.20 4.83
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GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 934 | 3,498 | {"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-2017-17 | latest | en | 0.904475 |
https://csbnews.org/who-has-the-queen-part-v-by-frank-stewart/?lang=en | 1,726,063,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651387.63/warc/CC-MAIN-20240911120037-20240911150037-00169.warc.gz | 166,612,697 | 19,905 | # Who has the Queen? Part V by Frank Stewart
0
169
Source: 2006 ACBL Bulletins
Test your queen-placing in the following problems. Assume IMP scoring.
West leads a low diamond, and East wins with the jack. Cashes the A and K and leads the K. You ruff. Who has the Q?
East had the A, K and A, K and J. That’s 15 points. and he is also marked with a club honor. Since if West had the K Q, his opening would have been a high club. So West has the Q. East should have been less forthcoming about his diamond holding. If he leads the A at the second trick. South won’t be sure which defender has the king.
West leads the 2, and East takes the queen and ace, cashes the A and exits with a heart.West follows. You continue with a diamond to the ace, both defenders playing low. On the next diamond East plays low again. Who has the queen?
A count of high-card points is inconclusive. East had the A Q and A but could have opened with either the K or with the Q and J. Instead, draw an inference from the defense. If West had a singleton diamond, he could have beaten the contract by leading it. When East got in with the A, he’d give West a diamond ruff; and the defense would take two spades. Since even an inexperienced West would have led a side singleton, especially from a weak hand, go up with the K, expecting West to have held Q 7.
West leads the 2, dummy plays low, and East wins with the jack and shifts to the J. How do you play the trumps?
East has the A J and J, and you must assume he has the A so, place West with the Q since East didn’t open. | 393 | 1,562 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.980612 |
https://www.coursehero.com/file/22322/Final-1/ | 1,529,619,539,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864300.98/warc/CC-MAIN-20180621211603-20180621231603-00420.warc.gz | 795,792,155 | 87,622 | # Final 1 - Physics 132 Final Exam Winter 2006 Name Time...
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AJM:1/1/07 Score /100 Physics 132 Final Exam Winter 2006 Name Time allowed: 125 minutes. You may use two sheets of notes (8-1/2 x 11, both sides) and a calculator. No brimmed hats; no communication devices. Work the problems on separate sheets of paper and staple this sheet to the front. Read each problem carefully and be sure to pay attention to any hints that are provided. There is no need to be “wordy” like I ask you to be on homework, but you must show enough work or give enough of an explanation to show me that you understand what you are doing. I give no credit for unsupported answers. Partially correct solutions will get partial credit, if I can figure out what you are doing, so use plenty of space and try your best to be reasonably neat and clear. All numerical answers must be given with an appropriate number of significant digits and appropriate , simplified units. Check your answers for physical reasonableness ; I deduct a small number of points for ridiculous answers that go uncommented upon. 1. The pipe organ at the Atlantic City Convention Hall features a 64' 9" long pipe called the “Diaphone Profunda” that operates as an open pipe producing a pitch that is supposed to be four octaves below “middle C.” a) [15] Assuming that the speed of sound is 340 m/s, what is the fundamental frequency of an open pipe of that length?
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Fixed Costs
If you're running a business, you most likely have fixed costs. How? Imagine that you run a bakery where you make 500 pieces of cake each week. You realize that business is getting unusually busy, so you buy more flour and increase the pieces of cake to 600 each week. In the scenario we have just described, the cost of flour changed since you increased the amount of flour used. However, notice that you did not increase the number of ovens or the size of the bakery. That's because they are fixed costs, and you didn't have to change them when you increased the quantity of cakes baked. Let's dive right into the topic of fixed costs to learn more. We will calculate fixed costs using the fixed cost formula and use examples of different fixed cost types to understand this concept better. Read on!
Fixed Costs Definition
Fixed costs are the costs that do not change when the quantity of output changes, and they only go away when the business fails or closes down.
Fixed costs are the costs that do not change when the quantity of output changes, and they only go away when the business fails or closes down.
Now that you know the definition, let's explain it with an example.
A shoemaker pays $500 to acquire a shoe-making machine. He then pays$40 or $50 for leather to make shoes, depending on the level of demand each week. From the above example, the$500 paid to acquire the shoe-making machine is a fixed cost because it does not change regardless of the quantity of shoes the shoemaker wants to make.
Fixed Costs Economics
To better understand fixed costs in economics, a series of related concepts must also be explained. Just as a refresher, fixed costs are those costs that do not change with output. So, what are the costs that do change when output changes? They are variable costs!
Variable costs are costs that change when the quantity of output changes.
You must be wondering why we brought variable costs into this when it's about fixed costs. It's because variable costs combine with fixed costs to make total costs. Therefore, each firm needs to consider both types of costs to get an overall picture of how much it is spending on production.
Total costs are the overall economic production cost and are a combination of variable costs and fixed costs.
All the costs discussed fall under the umbrella of production costs, which refer to the costs a business incurs to employ the factors of production for its business processes.
Production costs are the costs a business incurs to employ production factors for its business processes.
It is important to note that fixed costs often only exist in the short run, and in the long run, all costs can change.
Now, let's see how all these fit in an example.
Gail rented a shop for $700 for her cake business. She also spends$100 a week on employees and cake ingredients to make 50 pieces of cake each week. If Gail wants to increase the quantity of cake her business bakes, she will only have to spend more on employees and cake ingredients. She will not have to spend more on rent.
In the above example, Gail's fixed cost is $700, whereas her variable cost is$100.
Note that Gail's fixed cost will not change even if she decides not to bake any cake at all.
Fixed Costs Formula
For the fixed costs formula, we simply subtract variable costs from the total cost. Mathematically, we can write this as:
$$FC=TC-VC$$
Where FC represents fixed costs, TC represents total costs, and VC represents variable costs.
Let's try an example now.
With a total cost of $300 and a variable cost of$250, what is the fixed cost?
Solution:
Using
$$FC=TC-VC$$
We have
$$FC=300-250$$
$$FC=50$$
Fixed Costs Curve
We can show fixed costs on the production costs graph of a firm. This graph is plotted with cost on the vertical axis and quantity of output on the horizontal axis. Because fixed costs do not change even when the quantity of output changes, it is a flat horizontal line on the graph. Figure 1 illustrates this.
Fig. 1 - Fixed Costs Curve
As shown in Figure 1, fixed costs remain the same regardless of quantity produced, whereas variable costs increase as the quantity produced increases. This causes the total costs to increase at the rate at which the variable costs increase.
Read our article on Sunk Costs to learn about another type of cost.
How to calculate fixed costs?
The best way to learn how to calculate fixed costs it to work on an example
If a firm has a total cost of $400 and a variable cost of$300, what is the fixed cost of the firm?
Solution:
Using
$$FC=TC-VC$$
We have
$$FC=400-300$$
$$FC=100$$
Want to know how a firm makes profits after incurring all these costs? Read our article on Revenue vs Profit.
Fixed Costs Examples
Examples of fixed costs include rent, salaries, insurance and loan payments.
Insurance
For example, a business that pays \$1,000 monthly for liability insurance will have to pay that amount regardless of how much they produce or sell. The insurance coverage amount may change based on the business's needs, but the premium amount will remain fixed.
Fixed Costs - Key takeaways
• Fixed costs are the costs that do not change when the quantity of output changes, and they only go away when the business fails or closes down.
• Variable costs are costs that change when the quantity of output changes.
• Total cost is the overall economic production cost and is a combination of variable costs and fixed costs.
• The formula for fixed cost is $$FC=TC-VC$$
• Average costs divide the cost by the quantity of output.
• Examples of fixed costs include rent, salaries, insurance and loan payments.
Flashcards in Fixed Costs 10
Learn with 10 Fixed Costs flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What are fixed costs?
Fixed costs are the costs that do not change when the quantity of output changes, and they only go away when the business fails or closes down.
What are fixed costs examples?
An example of fixed cost is rent or equipment cost.
How do fixed costs differ from variable costs?
Fixed costs don't change in the short run, but variable costs change in the short run.
Is Depreciation a fixed cost?
Depreciation is a fixed cost.
How to find fixed cost?
The formula for fixed cost is Fixed Cost =Total Cost - Variable Cost
Is rent a fixed cost?
Yes, rent is a fixed cost.
Test your knowledge with multiple choice flashcards
Total cost is the overall economic production cost and consists of only variable costs.
Fixed cost changes in the short run.
In the long run, variable costs become fixed.
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• Checked by StudySmarter Editorial Team | 1,581 | 7,547 | {"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} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.958497 |
https://scribesoftimbuktu.com/simplify-5y-82y-9/ | 1,675,163,064,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00324.warc.gz | 538,107,950 | 13,158 | # Simplify (5y)/8+(2y)/9
5y8+2y9
To write 5y8 as a fraction with a common denominator, multiply by 99.
5y8⋅99+2y9
To write 2y9 as a fraction with a common denominator, multiply by 88.
5y8⋅99+2y9⋅88
Write each expression with a common denominator of 72, by multiplying each by an appropriate factor of 1.
Multiply 5y8 and 99.
5y⋅98⋅9+2y9⋅88
Multiply 8 by 9.
5y⋅972+2y9⋅88
Multiply 2y9 and 88.
5y⋅972+2y⋅89⋅8
Multiply 9 by 8.
5y⋅972+2y⋅872
5y⋅972+2y⋅872
Combine the numerators over the common denominator.
5y⋅9+2y⋅872
Simplify the numerator.
Factor y out of 5y⋅9+2y⋅8.
Factor y out of 5y⋅9.
y(5⋅9)+2y⋅872
Factor y out of 2y⋅8.
y(5⋅9)+y(2⋅8)72
Factor y out of y(5⋅9)+y(2⋅8).
y(5⋅9+2⋅8)72
y(5⋅9+2⋅8)72
Multiply 5 by 9.
y(45+2⋅8)72
Multiply 2 by 8.
y(45+16)72 | 412 | 755 | {"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} | 4.09375 | 4 | CC-MAIN-2023-06 | latest | en | 0.715758 |
https://www.jiskha.com/display.cgi?id=1248464757 | 1,503,325,989,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886108709.89/warc/CC-MAIN-20170821133645-20170821153645-00659.warc.gz | 912,924,799 | 4,565 | # chemistry
posted by .
A microtube is found to have an ungraduated portion that will contain 0.26 g of water. Upon setting up the experimental apparatus as described in the experiment, a microtube is filled to 0.01 ml with water at 4ºC. At a temperature of 60ºC, the volume of head space within the microtube has increased to 0.11 ml. What is the vapour pressure of water at this temperature?
Here is my attempted work. What am I doing incorrect?
n = m x M
= 0.26g x 18.01 g/mol
= 4.6826 mols
PV = nRT
Rearrange to get:
P = nRT/V
= (4.6826 mols)(8.3145 J/K⋅mol)(60ºC)/ 0.11 mL
= 21236.44 atm
Therefore, the vapour pressure of water at this temperature is 21236.44 atm.
• chemistry -
You may have other mistakes, also, but 60 degrees C must be substituted into the equation as Kelvin and volume as liters. And if you want the answer to be in atm, then R must be 0.08205.
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More Similar Questions | 868 | 3,316 | {"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} | 2.828125 | 3 | CC-MAIN-2017-34 | latest | en | 0.929353 |
https://mail.python.org/pipermail/python-list/2001-February/082506.html | 1,566,681,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00463.warc.gz | 541,101,089 | 2,265 | # do...while loops
Greg Jorgensen gregj at pobox.com
Tue Feb 6 21:25:50 CET 2001
```In article <3A802319.46EBCA35 at cybermesa.com>, Jay O'Connor
<joconnor at cybermesa.com> wrote:
> Umm..then how about a counter? Would this work?
>
> x = 100
> while x:
> print x
> x -= 1
>
> Yeah, just tried it...it works.
>
> Umm...anyone really like it? :)
That's a common C idiom, but I've always disliked it because x is
either a logical or arithmetic value depending on which statement you
are looking at. I prefer:
while x > 0:
...
x -= 1
To me this has several advantages:
1. There is absolutely no question about what the loop is doing:
counting down to zero, not waiting for some flag named x to be false.
2. If x is initially negative the loop won't even get started, whereas
x = -1
...
while x:
...
x -= 1
Will loop for a long time and then crash.
3. The explicit comparison will still work if x is changed from an
integer to a floating point type. This kind of bug is particularly hard
to find:
x = (((11.0 / 3.0) / 3.0) * 3.0) * 3.0 # 10.999999999999998
...
while x:
print x
x -= 1
x will never be exactly equal to 0 because of floating-point
representation and rounding errors. The example above will even print
the sequence 11.0, 10.0, 9.0, ... 1.0, so it may not be immediately
obvious why the loop doesn't terminate.
--
Greg Jorgensen
Portland, Oregon, USA
gregj at pobox.com
Sent via Deja.com
http://www.deja.com/
``` | 443 | 1,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | longest | en | 0.881647 |
https://www.aptech.com/questions/create-multiple-vectors-in-a-loop/ | 1,680,240,359,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949573.84/warc/CC-MAIN-20230331051439-20230331081439-00306.warc.gz | 735,420,072 | 27,737 | # create multiple vectors in a loop
Hello,
I am running Gauss 12 and would like to create multiple vectors in a loop. For simplicity, assume the following situation, the code of which is obviously not working:
for i (1,4,1);
x _i = zeros(10,1);
endfor;
The goal would thus be to get four (10,1) zero-vectors labeled as x_1,x_2,.... Is there an easy way to fix this in Gauss?
Etiënne
1
You can create new GAUSS global variables with the GAUSS function varput. Let's first go through a simple example of varput usage. If we want to create a new GAUSS global variable named phi and set it equal to 3, we would usually write this line of GAUSS code:
```phi = 3;
```
We can also accomplish the same thing using the varput function. varput will create (or "put") a new variable in your GAUSS workspace.
```//alternative to: phi = 3;
ret = varput(3, "phi");
```
Now that we can create a new GAUSS variable from a string, we need a way to turn the loop counter from a number into a string and then combine strings. In GAUSS 14 and newer, you can use ntos to turn a number into a string, like this:
```i = 1;
i_string = ntos(i);
```
In older versions of GAUSS, you will need to use a combination of ftocv and cvtos, or ftos which is slightly more complicated. Here is an example:
```i = 1;
i_string = cvtos(ftocv(i, 0, 0));
```
Now that we have converted our number to a string, we can combine it with another string using the GAUSS string combination operator, \$+:
```var_name = "x_" \$+ i_string;
```
Putting all of this together, would look like this (for GAUSS 14 or newer):
```for i(1, 4, 1);
ret = varput(zeros(10,1), "x_" \$+ ntos(i));
endfor;
```
or like this (for older versions of GAUSS):
```for i(1, 4, 1);
ret = varput(zeros(10,1), "x_" \$+ cvtos(ftocv(i, 0, 0)));
endfor;
```
However, I would strongly recommend that you simply create one x matrix and reference the different columns using indexing, like this:
```x = zeros(10, 4);
//reference the first column of 'x'
first_var = x[.,1];
```
The code will almost always be much simpler to read and also execute faster. If you are not sure how you might incorporate this second method into your specific code, I would encourage you to post that as a question. Thinking about your model in terms of a matrix and indexing can sometimes seem less straightforward at first. However, it is not difficult and (I am alsmost certain) you will be glad that you did.
0
Thank you for your very elaborate answer. I will put this to the test immediately.
As for your comment on indexing variables as matrices, I absolutely agree this normally makes programming clearer and faster to execute, but I ran across a problem that seems to prevent me from using matrices all the way through. I am using the "shiftr" command to generate series with different lags, which I will eventually use in an AR(p) model where I will let the lag length be determined on the basis of the minimum AIC. My amateur code that I used before (really inefficient) looks as follows:
p=12
l = seqa(1,1,p);
m_garch = ones(p,1).*.yd_garch';
m_sv = ones(p,1).*.yd_sv';
m_stv = ones(p,1).*.yd_stv';
m_rsv = ones(p,1).*.yd_rsv';
lags_yd_garch = shiftr(m_garch,l,0)';
lags_yd_sv = shiftr(m_sv,l,0)';
lags_yd_stv = shiftr(m_stv,l,0)';
lags_yd_rsv = shiftr(m_rsv,l,0)';
The rest of the code is not relevant, as this shows what I eventually want to achieve. So I have to do OLS for 4 different series. I have now started combining all the different series (yd_...) in a matrix "yd". Using the kronecker product on this matrix will change the usefulnedd of the shiftr command I think. The ideal solution would be:
for i(1,4,1);
m_(i) = ones(p,1).*.yd[.,i]'; //(i) is obviously not a valid command//
lags_yd_(i) = shiftr(m_(i),1,0)';
endfor;
So I hope I can do something like this using the code you proposed above.
Any further suggestions are always welcome of course!
Thanks again
0
I do have a couple of suggestions for you. The GAUSS function, lagn, will create multiple lags for you in one call. For example, you could change these two lines:
```p=12
l = seqa(1,1,p);
m_garch = ones(p,1).*.yd_garch';
lags_yd_garch = shiftr(m_garch,l,0)';
```
to this:
```p = 12;
l = seqa(1, 1, p);
lags_yd_garch = lagn(yd_garch, p);
```
The one difference between the lagn procedure and yours is that lagn fills in the empty matrix elements with GAUSS missing values, while yours fills them in with zeros. For example:
``` lagn output your method
1.1 . . 1.1 0 0
3.5 1.1 . 3.5 1.1 0
2.8 3.5 1.1 2.8 3.5 1.1
3.1 2.8 3.5 3.1 2.8 3.5
```
What is nice about filling the missing elements with GAUSS missing values is that you can easily remove all of the rows that contain a missing value with the GAUSS packr function. However, if you want to use zeros instead, you can modify the lagn code to do that.
My second suggestion would be to make a procedure do most of the work and then call that 4 times instead of doing all the work for all 4 vectors. Instead of doing something like this:
```p=12
l = seqa(1,1,p);
lags_yd_garch = lagn(yd_garch, l);
lags_yd_sv = lagn(yd_sv, l);
lags_yd_stv = lagn(yd_stv, l);
lags_yd_rsv = lagn(yd_rsv, l);
//remove missing value rows for ols regression
lags_yd_garch = packr(lags_yd_garch);
lags_yd_sv = packr(lags_yd_sv);
lags_yd_rsv = packr(lags_yd_rsv);
lags_yd_stv = packr(lags_yd_stv);
//calculate OLS parameter estimates
b_hat_yd_garch = lags_yd_garch[.,1]/lags_yd_garch[.,2:cols(lags_yd_garch)];
b_hat_yd_sv = lags_yd_sv[.,1]/lags_yd_sv[.,2:cols(lags_yd_sv)];
b_hat_yd_rsv = lags_yd_rsv[.,1]/lags_yd_rsv[.,2:cols(lags_yd_rsv)];
b_hat_yd_stv = lags_yd_stv[.,1]/lags_yd_stv[.,2:cols(lags_yd_stv)];
```
```p=12
bhat_yd_garch = olsLagN(yd_garch, p);
bhat_yd_sv = olsLagN(yd_sv, p);
bhat_yd_rsv = olsLagN(yd_rsv, p);
bhat_yd_stv = olsLagN(yd_stv, p);
proc (1) = olsLagN(y, p);
local l, lags_y, b_hat;
l = seqa(1,1,p);
lags_y = lagn(y, l);
//remove missing value rows for ols regression
lags_y = packr(lags_y);
//calculate OLS parameter estimates
b_hat = lags_y[.,1]/lags_y[.,2:cols(lags_y)];
retp(b_hat);
endp;
```
This version is a bit shorter and easier to read. It will also scale to more vectors much easier. I am not certain exactly what you are trying to do, so maybe it will not be exactly as simple to apply as this, but it is something to keep in mind at least.
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http://joe-mccullough.com/physics/Physics/Force_and_Newtons_Laws.html | 1,675,338,998,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500017.27/warc/CC-MAIN-20230202101933-20230202131933-00222.warc.gz | 24,970,341 | 7,194 | # Forces and Newton's Laws
Student Learning Objectives
Lessons / Lecture Notes
Important Equations
Example Problems
Applets and Animations
Student Learning Objectives
• To recognize what does and does not constitute a force.
• To understand and use Newton's 1st, 2nd, and 3rd laws.
• To identify the specific forces acting on an object and draw an accurate free-body diagram.
• To identify action/reaction pairs of forces.
• To begin the process of understanding the connection between force and motion.
• To distinguish mass, weight, and gravity.
• To learn and use simple models of friction and drag.
• To recognize and solve simple equilibrium problems.
• To apply the full strategy for force and motion problems to problems in single-particle dynamics.
• To learn how two objects interact.
Lessons / Lecture Notes
The Physics Classroom
(conceptual)
PY105 Notes from Boston University (algebra-based):
Physics 2A notes from Dr. Bobby W.S. Lau (algebra-based)
HyperPhysics (calculus-based)
Important Equations
(for algebra-based Physics)
Example Problems
Example Problems for algebra-based physics (from College Physics 2nd Edition by Knight, Jones, and Field):
Example Problems (Forces and Newton's Laws) | Example Problems (Applying Newton's Laws)
Solutions to Example Problems (Forces and Newton's Laws) | Solutions to Example Problems (Applying Newton's Laws)
Example Problems for calculus-based physics (from Physics for Scientists and Engineers 4th Edition by Knight):
Example Problems (Forces and Dynamics I) | Example Problems (Newton's Third Law) | Example Problems (Dynamics II)
Solutions to Example Problems (Forces and Dynamics I) | Solutions to Example Problems (Newton's Third Law) | Solutions to Example Problems (Dynamics II)
Applets and Animations | 384 | 1,781 | {"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-2023-06 | latest | en | 0.874938 |
http://classicwallclocks.com/wall-clock-xl-wall-clock-amazon.html | 1,569,108,120,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574710.66/warc/CC-MAIN-20190921231814-20190922013814-00198.warc.gz | 41,873,311 | 9,325 | Who doesn't love an alarm clock that verbally lets you know it's time to get up and start your day? Step aside, Siri, because "Reminder Rosie" is where it's at. You can easily record up to 25 different alarm-reminders at 6 seconds each. So why not wake up to a recording of your kids saying, "Good morning, Mom!"? (Or maybe you don't want that.) We love how you can turn off the alarm by quickly pressing a button or just saying out loud, "Reminder Off."
“This was a gift for my elderly, cognitively impaired mother. She was very often confused as to which day it was, the time of day, etc. She would show up at appointments on the wrong day, or think it was morning when it was evening. This clock is fantastic! It has made such a difference. Now she always knows what day and time it is. The clock is also very nice-looking. This was an excellent purchase and highly recommended. I can’t say enough good things about it.”
##### The next development in accuracy occurred after 1656 with the invention of the pendulum clock. Galileo had the idea to use a swinging bob to regulate the motion of a time-telling device earlier in the 17th century. Christiaan Huygens, however, is usually credited as the inventor. He determined the mathematical formula that related pendulum length to time (about 99.4 cm or 39.1 inches for the one second movement) and had the first pendulum-driven clock made. The first model clock was built in 1657 in the Hague, but it was in England that the idea was taken up.[38] The longcase clock (also known as the grandfather clock) was created to house the pendulum and works by the English clockmaker William Clement in 1670 or 1671. It was also at this time that clock cases began to be made of wood and clock faces to utilize enamel as well as hand-painted ceramics.
The most accurate mechanical timekeeper is the Shortt pendulum clock; it makes use of the movement described above for electric master clock systems. The Shortt pendulum clock consists of two separate clocks, one of which synchronizes the other. The timekeeping element is a pendulum that swings freely, except that once every half minute it receives an impulse from a gently falling lever. This lever is released by an electrical signal transmitted from its slave clock. After the impulse has been sent, a synchronizing signal is transmitted back to the slave clock that ensures that the impulse to the free pendulum will be released exactly a half minute later than the previous impulse. The pendulum swings in a sealed box in which the air is kept at a constant, low pressure. Shortt clocks in observatories are kept in a room, usually a basement, where the temperature remains nearly constant, and under these conditions they can maintain the correct time to within a few thousandths of a second per day.
For some scientific work timing of the utmost accuracy is essential. It is also necessary to have a standard of the maximum accuracy against which working clocks can be calibrated. An ideal clock would give the time to unlimited accuracy, but this is not realisable. Many physical processes, in particular including some transitions between atomic energy levels, occur at exceedingly stable frequency; counting cycles of such a process can give a very accurate and consistent time—clocks which work this way are usually called atomic clocks. Such clocks are typically large, very expensive, require a controlled environment, and are far more accurate than required for most purposes; they are typically used in a standards laboratory.
A captivating timepiece, this 29-inch wall clock stylishly pays homage to French country style with elegance. Especially fitting for an empty wall space, this wooden piece effortlessly draws the eye with its bold Roman numerals and matching clock hands that hover over a pale aqua and ivory-colored face. Its fleur-de-lis and floral details are a harmonious match for a romantic and feminine aesthetic in the living room, which is easily complemented by other weathered décor and Parisian...
Howard Miller 625-323 Alton wall clock. Matte black case with shatter-resistant acrylic crystal. White dial and large, easy-to-read black Arabic numerals. Auto Daylight-Savings movement automatically adjusts for Daylight Savings Time. The hands of the clock follow an LCD display on the back of the clock and make corrections to keep the correct time including Daylight Saving Time corrections.
Many alarm clocks have radio receivers that can be set to start playing at specified times, and are known as clock radios. Some alarm clocks can set multiple alarms, a useful feature for couples who have different waking up schedules. A progressive alarm clock, still new in the market, can have different alarms for different times (see Next-Generation Alarms) and even play music of your choice. Most modern televisions, mobile phones and digital watches have alarm clock functions to turn on or make sounds at a specific time.
Strong, durable, super large time display makes the model perfect for those who either want a big number clock, or for visually impaired individuals who need extra-large numerals, The red LED display on a black background, makes the contrast ideal for easy viewing. The electric power with battery backup lets you sleep with peace of mind and the extra-large snooze bar keeps nap-time within reach. This clock is a solid purchase.
In the 13th century, Al-Jazari, an engineer from Mesopotamia (lived 1136–1206) who worked for Artuqid king of Diyar-Bakr, Nasir al-Din, made numerous clocks of all shapes and sizes. A book on his work described 50 mechanical devices in 6 categories, including water clocks. The most reputed clocks included the Elephant, Scribe and Castle clocks, all of which have been successfully reconstructed. As well as telling the time, these grand clocks were symbols of status, grandeur and wealth of the Urtuq State.[citation needed]
Clockmakers developed their art in various ways. Building smaller clocks was a technical challenge, as was improving accuracy and reliability. Clocks could be impressive showpieces to demonstrate skilled craftsmanship, or less expensive, mass-produced items for domestic use. The escapement in particular was an important factor affecting the clock's accuracy, so many different mechanisms were tried.
```Now, RCA Digital Alarm Clock is pretty basic, but for some of us, that’s all we need, right? With a bright, beaming red LED display, you’re certain to rub your eyes and wonder what weird light is pulsating from your nightstand. There’s an alarm indicator and the classic AM/PM light that we all remember from early-2000 models. You can do something different from your late 1990s-inspired clock, and that is change the brightness levels so you don’t completely burn out your retinas.
```
Remember a time before everything was digital? Yeah… we don’t either. But a few excellent time-tested products pop up out of the woodwork from time to time, and a traditional dual bell alarm clock does the trick. The hands move silently while you sleep, and buzz like a mother when it’s time to rise and shine. If you’re like us, you’re sometimes bothered by LED flashing you in the face all night. With Peakeep Twin Bell Stereoscopic Clock, you simply touch the button to activate the stereoscopic dial backlight, check the time, and they cop-out for another fifteen minutes. Don’t worry – Peakeep will wake you up.
Alarm clocks are perfect for your nightstand or bedside table to make sure that you wake up on time. There are many different kinds of alarm clocks, ranging from digital clocks, analog clocks to projection clocks. Originally, clocks had difficulty with precision because they used a set of imprecise internal gears. Now, atomic alarm clocks are the perfect solution if you are looking for a clock that will get you to work on time. The best alarm clock depends on your needs and personal style. You should consider your purpose behind purchasing an alarm clock before making your selection.
This Oversized Rustic Wood 24" Wall Clock is one of the best sellers and for good reason - this wonderfully shabby elegance wall clocks really give that country home decor or rustic decor character that you cannot find anywhere else! The simple beauty of it is amazing it a part for that shabby elegance decor or rustic home decor charm! Gorgeous in kitchens or anywhere you want to complete that French country decor look as well!
Use your wall space as the board to create your scalable clock. The Blink wall clock by Umbra is a set of black metal indicators and 12 number that mount to the wall with adhesives. Comes with a template to set the scale of the clock on a wall. Clock mechanism mounts to the wall with a screw or nail operate on one AA battery (not included). Designed by Edward Lee and Adrienna Matzeg Umbra original, modern, casual, functional and affordable design for the home.
The returned instants from Clock work on a time-scale that ignores leap seconds, as described in Instant. If the implementation wraps a source that provides leap second information, then a mechanism should be used to "smooth" the leap second. The Java Time-Scale mandates the use of UTC-SLS, however clock implementations may choose how accurate they are with the time-scale so long as they document how they work. Implementations are therefore not required to actually perform the UTC-SLS slew or to otherwise be aware of leap seconds.
Amazon buyers love it as well: with more than 4,600 reviews, the clock has an average of 4.4 stars. Not only do most buyers comment that it easily awakens them even though they are heavy sleepers, but several hearing-impaired buyers also mentioned that even without the sound, the vibration and flashing lights were enough to wake them up. One customer summed it up well, "If you can sleep through this, then you may not be alive."
Some predecessors to the modern clock may be considered as "clocks" that are based on movement in nature: A sundial shows the time by displaying the position of a shadow on a flat surface. There is a range of duration timers, a well-known example being the hourglass. Water clocks, along with the sundials, are possibly the oldest time-measuring instruments. A major advance occurred with the invention of the verge escapement, which made possible the first mechanical clocks around 1300 in Europe, which kept time with oscillating timekeepers like balance wheels.[1][2][3][4]
The Howard Miller 625-542 Brass Works is an oversized, large wall clock featuring a metal outer frame finished in antique brass and accented by four decorative antique brass finished screws applied to the dial. The dial features antique brass-finished gears viewed through an open center. The gear in the middle of the clock turns as a second hand would rotate.. Quartz, battery-operated movement requires 1 AA battery. One year warranty and Free Shipping.
In mechanical clocks, the power source is typically either a weight suspended from a cord or chain wrapped around a pulley, sprocket or drum; or a spiral spring called a mainspring. Mechanical clocks must be wound periodically, usually by turning a knob or key or by pulling on the free end of the chain, to store energy in the weight or spring to keep the clock running.
No, that’s not a statement - it’s an app title. This little handy dandy game makes you solve puzzles or finish writing exercises before it’ll let you off the hook. If you try to snooze it too many times, it’s going to remember, and put you through hell. It’s like getting interest on the fact that you slept in an extra five minutes the day before. For the love of gear, just solve the puzzle, and muddle through the morning - you’ll be glad you did.
The naysayers might note that your cell phone alarm can already be set to vibrate, so why not just slip your smartphone under your pillowcase? All sanitary arguments aside, the SmartShaker 2 vibrates much more powerfully than a standard smartphone (three times more to be exact). The SmartShaker 2 also connects to your smartphone via Bluetooth and incorporates a functional, easy-to-use app that’s available for both Android and iOS devices. You can even set up to 10 different alarms and choose from an array of options, just in case you want an audio alarm as well as a vibration. After all, sometimes the snooze struggle is all too real.
The Howard Miller Angelina 625-636 Wall Clock is part of Howard Millers New 91st Anniversary Edition of clocks. This clock has an aged look that is achieved through spatter marks, the use of a rasp, and well placed dents. The Antique White finish features brown undertones on select hardwood and veneers. The Angelina clocks that are made in 2017 will have the 91st Anniversary Edition inscription on the dial. The dial has an aged look with black Arabic numerals that look worn and charcoal grey hands behind a glass crystal that is convex. One Year Warranty and Free Shipping.
You also need an alarm clock when you have a job or you're in school. It's important for you to wake up on time and avoid being tardy. Arriving late to work is an offense that could get you fired from your job, especially if happens more than once. Missing class because you slept in will also slow your educational progress and waste your money. When you put your alarm clock on your dresser instead of putting it right by your bed, you'll have to get up to turn it off, which helps you get going for the day.
Get up one minute earlier each and every day. We have biological clocks; it’s not just some weird term we use. Time doesn’t exist. I’m not going to get all existential on you and make you contemplate your place in the universe, because you already know what that is—being the version of you that gets up earlier. Your body needs time to adjust, and if you can wean into it like this, in a month, you’ll be waking up a half-hour early.
Tired of the same old clock hanging on your wall? Is the incessant ticking of the time a constant reminder you need to change it out? Purchase a new wall clock from our great selection. All of Zazzle's wall clock designs are printed in full color, so the design you choose will look vibrant for a long time. Not only can you choose from our great designs, you can also pick whether you want a round, square, or aqua clock to help keep track of time. Browse through the collection to find a clock you'll love every second and every minute!
Never squint to see what time it is again, especially when you're waking up 20 times in the middle of the night to, you know, see exactly what time it is. We love the projection feature that displays the time on your wall or ceiling for easy viewing. Plus, it'll even toss up the temperature — so you'll already know what you should wear to work before you even get out of bed.
A clock that looks good and runs smoothly instantly adds charm and functionality to your living space. For a fresh take on the traditional wall clock, consider this eye-catching design. Introducing Ribbon Wall Clock by Umbra Add a modern, minimalist twist to telling time. With Ribbon, you get a stylish piece of wall décor and a functional wall clock all-in-one. Made with high-quality metal that’s been shaped to resemble overlapping ribbons, this unique clock features contrasting hands...
```This Digital Bluetooth AM/FM Dual Alarm Radio Tabletop Clock from Jensen makes getting up in the morning a more appealing proposition. With the option to wake to radio or alarm, you can start your day off right. It even has an aux input so you can connect a smartphone or music player, letting you play whatever tunes you like right from your nightstand.
``` | 3,304 | 15,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-39 | latest | en | 0.981156 |
http://openstudy.com/updates/50801b63e4b0b5696005420c | 1,448,722,933,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453553.36/warc/CC-MAIN-20151124205413-00355-ip-10-71-132-137.ec2.internal.warc.gz | 175,259,914 | 10,691 | ## Biancao9o3o12 3 years ago Solve the inequality Write the solution in set builder notation -2x-13≥-25 ? HELP PLEASE !
1. Bhagyashree
what is a set builder notation? anyways i can help you solve the inequality it works mostly like if you were to solve an equation so -2x-13 (>or=) -25 sorry i cant type symbols add 13 to both sides
2. Bhagyashree
then you get -2x (> or =) -12
3. Bhagyashree
now to get x on its own you divide by -2 from both sides
4. Kamille
$-2x \ge-25+13$ add 25+13 and then divide by (-2). Don't forget to change sign into another (from $from \ge \to l$
5. Kamille
$from \ge$ to <(or=)
6. Biancao9o3o12
I just need the answer because I don't really get nothing of this.
7. Bhagyashree
now you are not done yet whenever you have to divide or multiply by a negetive number you have to change the sign so the answer is NOT: x (> or =) 6 but it is: x (< or =) 6
8. Bhagyashree
did you get it now @Biancao9o3o12 ?
9. Biancao9o3o12
Thank you Very much Kamille and Bhagyashree
10. Bhagyashree
you welcome! | 333 | 1,040 | {"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} | 3.59375 | 4 | CC-MAIN-2015-48 | longest | en | 0.930716 |
https://onlybooks.org/vector-optimization-lecture-notes-in-economics-and-mathematical-systems | 1,701,532,951,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00680.warc.gz | 504,083,578 | 44,864 | # Vector Optimization (lecture Notes In Economics And Mathematical Systems) by Guang-ya Chen / 2005 / English / PDF
Vector optimization model has found many important applications in decision making problems such as those in economics theory, management science, and engineering design (since the introduction of the Pareto optimal solu tion in 1896). Typical examples of vector optimization model include maxi mization/minimization of the objective pairs (time, cost), (benefit, cost), and (mean, variance) etc. Many practical equilibrium problems can be formulated as variational in equality problems, rather than optimization problems, unless further assump tions are imposed. The vector variational inequality was introduced by Gi- nessi (1980). Extensive research on its relations with vector optimization, the existence of a solution and duality theory has been pursued. The fundamental idea of the Ekeland's variational principle is to assign an optimization problem a slightly perturbed one having a unique solution which is at the same time an approximate solution of the original problem. This principle has been an important tool for nonlinear analysis and optimization theory. Along with the development of vector optimization and set-valued optimization, the vector variational principle introduced by Nemeth (1980) has been an interesting topic in the last decade. Fan Ky's minimax theorems and minimax inequalities for real-valued func tions have played a key role in optimization theory, game theory and math ematical economics. An extension was proposed to vector payoffs was intro duced by Blackwell (1955).
views: 348 | 328 | 1,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.936524 |
http://jakoblodwick.com/nova-scotia/how-to-make-a-balloon-car-go-5-meters.php | 1,558,891,652,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259327.59/warc/CC-MAIN-20190526165427-20190526191427-00012.warc.gz | 107,500,851 | 8,916 | ## How To Make A Balloon Car Go 5 Meters
#### Balloon Car
31/10/2010 · Best Answer: Maybe with elastic band powered engine would be possible,.
#### Balloon Car xkcd
• The target is 5 meters from the start line, in the center of the 2 meter wide track • Kits are not allowed! • The standard mousetrap must be mounted to the chassis and must not be painted or decoratedThe standard mousetrap must be mounted to the chassis and must not be painted or decorated
#### David and Alexander's Amazing Balloon Car Project
• The target is 5 meters from the start line, in the center of the 2 meter wide track • Kits are not allowed! • The standard mousetrap must be mounted to the chassis and must not be painted or decoratedThe standard mousetrap must be mounted to the chassis and must not be painted or decorated
#### Balloon Car xkcd
the car has to be able to travel on an L shaped course basically it has to travel 3 meters and at the end of the three meters, it has to make a 90 degree turn and continue for another2 meters. i just have no idea as how to construct such a car, and the car is due in two weeks!!! so plsss everyone
How to make a balloon car go 5 meters
#### How to make a balloon powered car which has to make a u
The problem is we need to avoid an obstacle 1 meter away and should get back to the line of distance from where my car starts. The longest distance the car traveled from the obstacle is the winner. The vehicle must start from behind a start-line and then clear an obstacle as it travels the longest possible distance in the longitudinal direction from the start line. The device must come to a
#### How to make a balloon powered car which has to make a u
How to Make a Mousetrap Car Go 10 Meters. 2013-05-30 Views:37. Body; Advertisement . A mousetrap car is a small homemade car that is propelled by the force of the spring in a mousetrap. It is a common science and physics experiment that teaches the laws of mechanical advantage and the basic automotive mechanics of axles and wheels. B. A mousetrap car is a small homemade car that is …
#### peer actvity and balloon car Amlenge's Blog
the car has to be able to travel on an L shaped course basically it has to travel 3 meters and at the end of the three meters, it has to make a 90 degree turn and continue for another2 meters. i just have no idea as how to construct such a car, and the car is due in two weeks!!! so plsss everyone
#### Balloon Car
Slide the straw into the balloon about 2” (5 cm) and use tape to make it an airtight connection. The tape needs to touch both the balloon and the straw for a couple of tight, spiral wraps. 5. Turn the base so the straws are on the bottom and decide which end will be the front and which will be the rear of your car. You want to place the straw toward the rear of the car and the balloon toward
#### Balloon Car
6/04/2010 · I did the balloon car as one of my previous lessons in the class, and it worked out about the same way as the whole class balloon car lesson did. It is hard to see what is happening that is making a car not work and go the full 5 meters. It was hard for some people to do both the balloon car on the same day that we did the peer activity, but in all fairness they were supposed to be done the
#### How to make a balloon powered car which has to make a u
I blew the balloon up and let the car go it went at a good fast pace but since the balloon was over weighted with air the balloon flopped over the nosel making air impossible to flow out. It was a devastation because i worked on this project for about 4 hours straight before the trial that Sunday evening. Our car ended with 1.9 meters, and did not achieve our goal of 10 meters. I believe this
#### Balloon Car xkcd
The problem is we need to avoid an obstacle 1 meter away and should get back to the line of distance from where my car starts. The longest distance the car traveled from the obstacle is the winner. The vehicle must start from behind a start-line and then clear an obstacle as it travels the longest possible distance in the longitudinal direction from the start line. The device must come to a
#### Balloon Car
With one hand holding the balloon and the other with crossed fingers, I let go of the balloon. The car traveled a distance of 4.76 meters in a time of 7 seconds. Therefore its average speed was 0.7 m/s. It did not pass the 5 meter line therefore failing to complete the run. I think that the axles were not formed properly and the balloon was too big therefore making the car …
#### Balloon Car xkcd
Balloon Car Rocket Project Erin B. Trevor S. Emily B. Hour 2 Introduction • Distance Traveled: 2.5 meters • Mass: 39.7 grams • Special Features: The car has a 3 straw nozzle for enhanced acceleration.
### How to make a balloon car go 5 meters - Balloon Car Rocket Project ppt PLHS - SlideGur.com
#### how to publish to the play store
Before uploading your app to the Play Store you need to prepare a few things. One of the questions you need to answer during the publishing process is about the maturity rating of your app. Google has four rating levels: Everyone, Low maturity, Medium maturity and High maturity.
#### how to prepare for skydiving
There’s a host of skydiving styles to live up to, from a solo skydive to accelerated free fall and tandem skydiving. Each and every type of experience is designed …
#### how to make rubber tires for hot wheels
Lay a pair of good tires on the wheels face down on this sheet of plastic, spaced 1/2" apart. An old set of AJ's silicone tires work very well for this. An old set of AJ's silicone tires work very well for this.
#### how to make miniature people out of paddle pop sticks
Bean Bag Animal Bed Crafts Project - Make a bed for your bean bag animal, Miniature figures, or other small stuffed animal with crafts sticks or popsicle sticks. Birds of a Feather Bookmark Crafts Ideas - Instructions to make a bookmark that looks like a bird with crafts sticks / popsicle sticks.
#### how to run adwcleaner without mouse
17/01/2018 · Mouse Without Borders is a Microsoft Garage project by Truong Do. Garage projects are side projects that Microsoft employees like Truong build for fun on their nights and weekends.
Pepole wondring where to start if want to know how to make money online. Adsense for me is a great way to make a passive income with little niche websites.
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Ontario: Port Dover ON, Kenabeek ON, Newcastle ON, Capreol, English Line ON, Cramahe ON, Middleville ON, ON Canada, M7A 9L5 | 2,163 | 8,496 | {"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-2019-22 | longest | en | 0.947195 |
https://gomathanswerkey.com/mcgraw-hill-math-grade-6-lesson-14-1-answer-key/ | 1,713,083,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00857.warc.gz | 263,055,936 | 52,363 | # McGraw Hill Math Grade 6 Lesson 14.1 Answer Key Understanding Percent
Practice questions available in McGraw Hill Math Grade 6 Answer Key PDF Lesson 14.1 Understanding Percent will engage students and is a great way of informal assessment.
Exercises
WRITE DECIMALS AND FRACTIONS
Question 1.
45% =
45% in the decimal form can be written as 0.45
45% in the fraction form can be written as 45/100
Question 2.
35% =
35% in the decimal form can be written as 0.35
35% in the fraction form can be written as 35/100
Question 3.
43% =
43% in the decimal form can be written as 0.43
43% in the fraction form can be written as 43/100
Question 4.
10% =
10% in the decimal form can be written as 0.10
10% in the fraction form can be written as 10/100
Question 5.
87% =
87% in the decimal form can be written as 0.87
87% in the fraction form can be written as 87/100
Question 6.
2% =
2% in the decimal form can be written as 0.2
2% in the fraction form can be written as 20/100
Question 7.
59% =
59% in the decimal form can be written as 0.59
59% in the fraction form can be written as 59/100
Question 8.
.1% =
0.1% in the decimal form can be written as 0.001
0.1% in the fraction form can be written as 1/1000
Question 9.
.45% =
0.45% in the decimal form can be written as 0.0045
0.45% in the fraction form can be written as 45/10000
Question 10.
Out of 100 questions, Yoshi answered 76 questions correct on his math exam. What percentage of questions did Yoshi answer correctly?
Given,
Out of 100 questions, Yoshi answered 76 questions correct on his math exam.
76/100 = 76% | 465 | 1,578 | {"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} | 4.09375 | 4 | CC-MAIN-2024-18 | latest | en | 0.954197 |
https://elinux.org/index.php?title=Sparkfun:_Flex_Sensor&diff=176288&oldid=176282 | 1,508,235,504,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00034.warc.gz | 683,106,676 | 8,989 | Difference between revisions of "Sparkfun: Flex Sensor"
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```Overview: 2
Wiring: 2
Code: 2
git: 2
Demo: 2
Total: 10/10
Comments: Thanks for the quick fix. Both on this and your other page.
```
Overview
Inputs and Outputs for FlexSensor
The product. The datasheet.
This is a simple flex sensor that measures 4.5" in length. As the sensor is flexed, the resistance across the sensor increases. What you do with this resistance is totally up to whatever you can come up with.
Inputs and Outputs
As for inputs and outputs, the sensor is basically a resistor and thus should be read from the analog input pins. To work with the analog input, you must put 1kΩ resistor from analog ground to the analog input you choose then connect the sensor from the analog VDD to the analog input you choose. The input voltage can then be read by the processor to determine how far the flex sensor is being bent (flexed).
Bone Usage
Wiring example for the Flex Sensor on a Bone
To interface this with the BeagleBone you just need to connect the flex resistor from the analog VCC to an analog input and another resistor (I used 1.1kΩ ) from analog input to analog ground(on P9). From there you can read the voltage input and manipulate it using code however desired. In my case I basically took the number that the analog input gave me and manipulated it to average about 1000 samples per second, and display the averaged value. When you bend the resistor the value increases, and when you relax the resistor the value decreases. You can see this in the sample code provided. There is also a picture included that will aid in wiring the sensor to the bone.
Reading and Interpreting Analog Input Data
Reading the analog input data is pretty simple. Basically, with the way I described hooking the component up, as you bend the flex senor the value of its resistance increases. So, you receive a lower value for the voltage read from the ain pin. The resistance of the flex sensor is approximately 10kΩ when flat, and increases to approximately 110kΩ whe bent as far as possible.
Sample C Code
The following code can be found through github.
You can do a git clone of this repository which includes MiniProject02 folder. Using the make command will make the two c files.
```
/****************************************************************
* James Popenhagen
* MiniProject02_FlexResistor.c
* 9/23/12
****************************************************************/
#include <errno.h>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <poll.h>
#include <signal.h>
#include <unistd.h>
/****************************************************************
* Constants
****************************************************************/
#define MAX_BUF 64
/****************************************************************
* Global Variables
****************************************************************/
int keepgoing = 1;
void signal_handler(int sig)
{
printf( "Ctrl-C pressed, cleaning up and exiting..\n" );
keepgoing = 0;
}
/****************************************************************
* read_ain
****************************************************************/
int read_ain(char* ain){
FILE *fp;
char ainPath[MAX_BUF];
char ainRead[MAX_BUF];
snprintf(ainPath, sizeof ainPath, "/sys/devices/platform/omap/tsc/%s", ain);
if((fp = fopen(ainPath, "r")) == NULL){
printf("Cannot open specified ain pin, %s\n", ain);
return 1;
}
if(fgets(ainRead, MAX_BUF, fp) == NULL){
printf("Cannot read specified ain pin, %s\n", ain);
}
fclose(fp);
return atoi(ainRead);
}
/****************************************************************
* Main
****************************************************************/
int main(int argc, char **argv){
int ainPin;
char ain[MAX_BUF];
float duty_cycle = 0;
float avgDutyCycle = 0;
int avgCount = 1;
if (argc < 2){
printf("Usage: ./MiniProject02 <ainpin>");
exit(-1);
}
signal(SIGINT, signal_handler);
ainPin = atoi(argv[1]);
snprintf(ain, sizeof ain, "ain%d", ainPin);
while (keepgoing) {
usleep(1000);
duty_cycle = read_ain(ain);
duty_cycle = duty_cycle/4095 * 100;
duty_cycle = 30 - duty_cycle;
duty_cycle = duty_cycle*4;
duty_cycle = duty_cycle - 28.6;
if(duty_cycle <= 6) duty_cycle = 0;
avgDutyCycle += duty_cycle;
if (avgCount == 250) {
printf("Calculating. \r");
fflush(stdout);
}
if (avgCount == 500) {
printf("Calculating . \r");
fflush(stdout);
}
if (avgCount == 750) {
printf("Calculating .\r");
fflush(stdout);
}
if(avgCount >= 1000){
avgDutyCycle = avgDutyCycle/avgCount;
avgDutyCycle = avgDutyCycle * 100;
printf(" \r");
printf("Calculating %d\r",(int)avgDutyCycle);
if(avgDutyCycle >= 9000) printf("Calculating It's over 9000!!!!\r");
fflush(stdout);
avgDutyCycle = 0;
avgCount = 0;
}
avgCount++;
}
fflush(stdout);
return 0;
}
```
You must specify the ain pin number as the first argument when executing this code after it is compiled. | 1,169 | 5,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-43 | latest | en | 0.891237 |
https://metanumbers.com/13376 | 1,675,175,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00377.warc.gz | 427,491,784 | 7,646 | 13376 (number)
13,376 (thirteen thousand three hundred seventy-six) is an even five-digits composite number following 13375 and preceding 13377. In scientific notation, it is written as 1.3376 × 104. The sum of its digits is 20. It has a total of 8 prime factors and 28 positive divisors. There are 5,760 positive integers (up to 13376) that are relatively prime to 13376.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 20
• Digital Root 2
Name
Short name 13 thousand 376 thirteen thousand three hundred seventy-six
Notation
Scientific notation 1.3376 × 104 13.376 × 103
Prime Factorization of 13376
Prime Factorization 26 × 11 × 19
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 8 Total number of prime factors rad(n) 418 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 13,376 is 26 × 11 × 19. Since it has a total of 8 prime factors, 13,376 is a composite number.
Divisors of 13376
1, 2, 4, 8, 11, 16, 19, 22, 32, 38, 44, 64, 76, 88, 152, 176, 209, 304, 352, 418, 608, 704, 836, 1216, 1672, 3344, 6688, 13376
28 divisors
Even divisors 24 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 28 Total number of the positive divisors of n σ(n) 30480 Sum of all the positive divisors of n s(n) 17104 Sum of the proper positive divisors of n A(n) 1088.57 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 115.655 Returns the nth root of the product of n divisors H(n) 12.2877 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 13,376 can be divided by 28 positive divisors (out of which 24 are even, and 4 are odd). The sum of these divisors (counting 13,376) is 30,480, the average is 10,88.,571.
Other Arithmetic Functions (n = 13376)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 5760 Total number of positive integers not greater than n that are coprime to n λ(n) 1440 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1588 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 5,760 positive integers (less than 13,376) that are coprime with 13,376. And there are approximately 1,588 prime numbers less than or equal to 13,376.
Divisibility of 13376
m n mod m 2 3 4 5 6 7 8 9 0 2 0 1 2 6 0 2
The number 13,376 is divisible by 2, 4 and 8.
Classification of 13376
• Abundant
Expressible via specific sums
• Polite
• Practical
• Non-hypotenuse
Base conversion (13376)
Base System Value
2 Binary 11010001000000
3 Ternary 200100102
4 Quaternary 3101000
5 Quinary 412001
6 Senary 141532
8 Octal 32100
10 Decimal 13376
12 Duodecimal 78a8
20 Vigesimal 1d8g
36 Base36 abk
Basic calculations (n = 13376)
Multiplication
n×y
n×2 26752 40128 53504 66880
Division
n÷y
n÷2 6688 4458.67 3344 2675.2
Exponentiation
ny
n2 178917376 2393198821376 32011427434725376 428184853366886629376
Nth Root
y√n
2√n 115.655 23.7379 10.7543 6.68752
13376 as geometric shapes
Circle
Diameter 26752 84043.9 5.62086e+08
Sphere
Volume 1.00246e+13 2.24834e+09 84043.9
Square
Length = n
Perimeter 53504 1.78917e+08 18916.5
Cube
Length = n
Surface area 1.0735e+09 2.3932e+12 23167.9
Equilateral Triangle
Length = n
Perimeter 40128 7.74735e+07 11584
Triangular Pyramid
Length = n
Surface area 3.09894e+08 2.82041e+11 10921.5
Cryptographic Hash Functions
md5 b3aa19fe9dc706a3b4cdaa8ddb37d852 489046eacfb9541c6a7bdbabfd2c66581b927f9d 8b7ca0db55864b0684fe3f42a929360544d59492ee48f659a6bf51f9f8ba79e2 b4689721db23712491dee1261166f1c70438aae57e2a1a3a3033c984875527602b22a31ef0f9958859afc6cb7620d4fd2ffd3a5e842b506ecac94e739614e025 0115758fb9525711470c38217a0ee2de02b9c2f2 | 1,503 | 4,154 | {"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} | 3.78125 | 4 | CC-MAIN-2023-06 | latest | en | 0.798686 |
https://www.physicsforums.com/threads/rotational-kinematics-challenge.229396/ | 1,575,869,356,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00466.warc.gz | 848,954,460 | 17,447 | # Rotational Kinematics, Challenge
Rotational Kinematics, Challenge!!!!!
1. Homework Statement
A satellite follows an elliptical orbit. The only force on the satellite is the gravity atraction from the planet. The satellites speed at point A is 8000m/s, and it is 6000km away, Point B is 24000km east of the planet, and Point C is component vector 9000ikm+12000jkm away, if the you take the planet as the origin.
a. Is there any torque acting on the satellite?
b. satellite speed at Point B
c, satellite speed at Point C
2. Homework Equations
We have just started rotational kinematics, and have not yet done space physics like newtons constant of gravity and that stuff. So just Lf=Li, conservation of momentum.
3. The Attempt at a Solution
a. i supect there is no torque because there is no external force acting on the satellite, as the gravity act as the moment arm.
b and c. i used La=Lb=Lc, therefore RaVa=RbRb=RcVc,
therefore giving Vb=(RaVa)/Rb and Vc=(RaVa)/Rc, and that Rc=sqrt(9000^2+12000^2)=15000, so my answers i get are 2000m/s at point B and 3200m/s at Point C, but the answers in the book say 2000m/s and 4000m/s..... so can someone please tell me where i have gone wrong??? TY
Related Advanced Physics Homework Help News on Phys.org
Dick
Homework Helper
Let's just start with a). Your answer is completely vague and wrong. What's an actual expression that will let you compute the torque and using that, say why you think it would be constant.
ok, as torque=rFsin(theta), since there is no Force, then there is no torque, is that better?
ok, i have thought about it a bit more, and is my answer wrong because i have ignored the angle? therefore RaVaSin(90)=RbVbSin(90)=RcVcSin(12000/9000) this gives the answer in the book, but is it correct, in the way it meant to be done?
fredrick08,
Yes, you have the correct answers in your 3rd post. Note that $$sin(90) = sin(\pi/2)= 1$$.
To answer why you were incorrect for the first part - recall that the earth is curved and Newton's First Law tells us that the satellite should fly into space except that there is the force of gravity from the earth. Since the earth diverges from it's tangent, there exists a small angle between the curved path and the satellites path, this is the angle you missed. Hence, $$rFsin(\theta)\neq 0$$ which means there does exist some amount of torque.
Last edited:
ok, sweet TY, but are you saying that for a) there is torque??? because the answer says theres not. and when i look at every diagram i have for toque, there is always a Force acting at an angle, to rotate an object, but in this case there is not.....???
i understand what you mean, and im not saying your wrong, but its only 1st year physics, so i think the book ignores it..... but im not sure....
Dick | 721 | 2,776 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-51 | latest | en | 0.927759 |
https://kerodon.net/tag/0005 | 1,653,791,359,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00574.warc.gz | 400,299,955 | 4,004 | # Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Example 1.1.0.1 (Singular Homology). For any topological space $X$, the singular homology groups $\operatorname{ \mathrm{H} }_{\ast }(X; \operatorname{\mathbf{Z}})$ are defined as the homology groups of a chain complex
$\cdots \xrightarrow {\partial } \operatorname{\mathbf{Z}}[ \operatorname{Sing}_{2}(X) ] \xrightarrow { \partial } \operatorname{\mathbf{Z}}[ \operatorname{Sing}_1(X) ] \xrightarrow {\partial } \operatorname{\mathbf{Z}}[ \operatorname{Sing}_0(X) ],$
where $\operatorname{\mathbf{Z}}[ \operatorname{Sing}_ n(X) ]$ denotes the free abelian group generated by the set $\operatorname{Sing}_ n(X)$ and the differential is given on generators by the formula
$\partial (\sigma ) = \sum _{i = 0}^{n} (-1)^{i} d_ i \sigma .$ | 261 | 813 | {"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": 2, "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-2022-21 | latest | en | 0.451118 |
https://electronics.stackexchange.com/questions/695450/zero-vs-pole-of-a-circuit | 1,718,203,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00669.warc.gz | 203,195,420 | 39,134 | # Zero vs pole of a circuit
Intuitively, what makes a resistor/capacitor combination become a pole or zero in around an amplifier system?
My intuition always says that a zero is formed when there is a direct path through a capacitor from input to output. And pole is formed when output has a load that causes output response to diminish in high frequency. However, I feel this is just experience from arbitrary examples.
Is there a more definitive intuition without the mathematics of which is based on calculating transfer function to be 0 at zero and $$\\infty\$$ at pole to get their locations?
• Poles and zeros are mathematical constructs so, to ask for a non-mathematical definition seems to miss the point. Dec 26, 2023 at 16:35
• scc28, probably start with this answer here on this site and then consider buying and reading Christophe P. Basso's "Linear Circuit Transfer Functions: An Introduction to Fast Analytical Techniques" book. He discusses the work of Dr. Middlebrook (now deceased.) I used to subscribe and read Dr. Middlebrook's regular emails. His wife did what she could after he died for those who followed his writing. But Dr. Basso has done the best of any to keep his flame alive. And it's probably the better answer you can get here. Start there. Dec 27, 2023 at 7:31 | 293 | 1,296 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.968284 |
http://www.freep.com/story/entertainment/nightlife/2014/11/06/mark-pilarski-casinos-blackjack/18421265/ | 1,508,287,712,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822625.57/warc/CC-MAIN-20171017234801-20171018014801-00142.warc.gz | 443,162,658 | 28,738 | QUESTION: We have a casino a little north of Toronto that has added a blackjack table with some variations. If you get a hard nine, 10 or 11, the house will double down for you. Also, if you get a split opportunity, the house will double down for you. If the dealer total is 22, it’s a push. A blackjack still pays 3:2. My question is: If you play basic strategy,s the house advantage lessened on this type of blackjack? — Bert W.
ANSWER: The variant of blackjack that you’re describing is called Free Bet Blackjack. It’s free to the extent that with your initial wager, the casino forks over the money needed for your double-downs and split wagers.
On this modified version, standard blackjack rules and table minimums and maximums apply and blackjacks pay 3:2 on a six-deck shoe. On a single deck game(not advised), blackjacks pay only 6:5.
There is no such thing as something for nothing. So there is a trade-off. If the dealer gets a 22, all wagers left standing push instead of it being a win for the player.
■ Free double-downs on a hard 9, 10 or 11.
■ Free splits on all pairs except 10s and fours.
■ You get up to four free re-splits, including aces.
■ You get a free double-down after your free splits.
■ Finally, with your own money, you can double down on other hands, including after a split, even if it’s a free split.
Here’s an example of what “free” means: When you reach a hard total of 9, 10 or an 11 (including drawing cards in some casinos), you can double down on the house’s dime and your regular wager is matched by the casino. The dealer will place a distinct button on the layout, and if your hand wins, that button is paid the same amount as your original wager.
As for some simplified basic strategy, you will want to double down for free on 9, 10 or l1 against any dealer up-card. Likewise, you’ll want to split — free — on any pair except on fives (you would double instead), and on 10s, where you would just stand.
With a much simpler basic strategy than your typical blackjack game, the house edge on this game is approximately 1%. That’s not too shabby for the added fun this game seems to provide. I’m giving it two thumbs up.
Q: In regard to your column about wide-area progressive machines, you stated: “As for long-term paybacks, they are consistent, as Casino A would not allow Casino B to offer a higher long-term payback.” Am I to assume that also means I have the same chance of hitting the progressive jackpot in Casino A as in Casino B? — Susan L.
A: Each machine on the linked network has the same likelihood of hitting those super-sized jackpots. However, if a large casino has more machines linked on the wide-area progressive, it is more likely that the progressive will hit in the larger casino. This is strictly because of the number of connected machines the larger casino has on its floor.
Mark Pilarski is a contributing editor for numerous gambling publications. E-mail questions to pilarski@markpilarski.com. | 696 | 2,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-43 | latest | en | 0.920992 |
https://mathchat.me/2014/02/19/how-to-find-the-gcf-for-three-or-more-numbers/?shared=email&msg=fail | 1,656,162,204,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103035636.10/warc/CC-MAIN-20220625125944-20220625155944-00592.warc.gz | 459,059,587 | 32,277 | ## Kiss those Math Headaches GOODBYE!
### How to Find the GCF for Three or More Numbers
To find the GCF for three or more numbers, follow these steps:
1) Determine which of the given numbers is smallest, then find the smallest difference between any pair of numbers.
2) See what is smaller: the smallest number, or the smallest difference. Whichever one is smallest, that number is the GPGCF (Greatest Possible GCF). That means that this is the biggest number that the GCF could possibly be. Or, more formally we would say: The GCF, if it exists, must be less than or equal to the GPGCF.
3) Check if the GPGCF itself goes into all of the given numbers. If so, then it is the GCF. If not, list the factors of the GPGCF from largest to the smallest and test them until you find the largest one that does divide evenly into the given numbers. The first factor (i.e., the largest factor) that divides evenly into the given numbers is, by definition, the GCF.
EXAMPLE:
Problem: Find the GCF for 18, 30, 54.
1) Note that the smallest number is 18, and the smallest difference between the pairs is 12 [54 – 30 = 24; 54 – 18 = 36; 30 – 18 = 12] .
2) Of those four quantities (the smallest number and the three differences), 12 is the least. This means that the
GPGCF = 12.
3) Check if 12 divides evenly into the three given numbers: 18, 30 and 54. In fact, 12 doesn’t divide evenly into ANY of these numbers. Next we check the factors of 12, in order from largest to smallest. Those factors are: 6, 4, 3 and 2. The first of those that divides evenly into all three numbers is 6. [18 ÷ 6 = 3; 30 ÷ 6 = 5; 54 ÷ 6 = 9]. So the GCF = 6. And we are done.
MORE CHALLENGING PROBLEM:
Find the GCF for 24, 148, 200.
1) Note that the smallest number is 24, and that the smallest difference between the pairs is 52 [200 – 148 = 52; 200 – 24 = 176; 148 – 24 = 124] .
2) Of those four quantities (the smallest number and the three differences), 24 is the least. This means that for this problem, the GPGCF = 24.
3) Check if 24 divides evenly into the three given numbers: 24, 148 and 200. While 24 does divide evenly into 24, it does not divide evenly into 148 or 200. So next we check the factors of 24, in order from largest to smallest. Those factors are: 12, 8, 6, 4, 3 and 2. The first of those that divides evenly into the three given numbers is 4. [24 ÷ 4 = 6; 148 ÷ 4 = 37; 200 ÷ 4 = 50]. So the GCF = 4. And, once again, we are done.
The process may seem a bit long, but once you get used to it and start doing it in your mind, not on paper, you should find that it actually is quite fast. And you’ll find yourself figuring out the GCF for three or more numbers all in your mind — with no need for pencil and paper — while everyone around you will be making prime factor trees or using calculators. And surely that is a good feeling.
Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com Just click the links in the sidebar for more information! | 922 | 3,201 | {"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} | 4.90625 | 5 | CC-MAIN-2022-27 | latest | en | 0.865098 |
https://focustutoring.com/gmat-number-properties-switching-digits/ | 1,685,898,603,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650201.19/warc/CC-MAIN-20230604161111-20230604191111-00072.warc.gz | 302,486,388 | 46,714 | Select Page
Welcome back to our series on number properties. In the last article, we learned about place-value problems. This article will cover the related topic of switching digits in two-digit integers. Here are some examples of what that means:
## Place-value problems
75 <-> 57
39 <-> 93
62 <-> 26
Some GMAT quant problems ask about the difference (in terms of subtraction) between two such numbers, or about the difference between the individual digits used to build these numbers.
### Here’s an official GMAT problem:
If a two-digit integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
These problems can be baffling the first time you see one, but there is a straightforward rule that makes them very easy.
The difference between the two-digit integers xy and yx is 9 times the difference between the digits x and y.
Given this rule, all you need to do on the problem above is take 27/9 = 3, and the correct answer choice is A.
Here are some examples of digit switching, to be followed by an explanation of the rule.
• 75 <-> 57
75 – 57 = 18
7 – 5 = 2
18 / 2 = 9
• 39 <-> 93
93 – 39 = 54
9 – 3 = 6
54 / 6 = 9
• 62 <-> 26
62 – 26 = 36
6 – 2 = 4
36 / 4 = 9
Why does it work this way? The answer has everything to do with the way our number system is built. In any number, the tens digit “counts” ten times as much as the units digit. When switching digits of a two-digit number, each digit changes by the same amount but in opposite directions. If the tens digit increases by 2, the units digit decreases by 2. If the tens digit decreases by 5, the units digit increases by 5. But the change for the tens digit “counts” ten times as much as the change for the units digit.
75 <-> 57
Here the digits differ by 2. If the 7 in the tens place becomes a 5, the value decreases by 20. Meanwhile, the 5 in the units place becomes a 7, and the value increases by 2. So when the digits switch places, the change in the units place always “undoes” 1/10 of the change in the tens place, resulting in a “net” change that is always a multiple of 9.
You can practice switching digits with any two-digit number to watch this property at work. And whenever you see it on GMAT quant, you’ll be ready.
In the next article, we’ll consider the “spacing” of multiples of a given integer and how this spacing relates to series of consecutive integers.
If you are in the middle of studying for the GMAT and are looking for a private GMAT tutor, our elite tutors have all scored over 770 on the GMAT and have years of professional experience with tutoring. You can meet with us for a 30-minute complimentary consultation call.
Contributor: Elijah Mize (Apex GMAT Instructor) | 710 | 2,795 | {"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} | 4.78125 | 5 | CC-MAIN-2023-23 | latest | en | 0.906938 |
https://www.reference.com/science/use-conversion-unit-table-496bf0597d338578 | 1,566,703,259,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00132.warc.gz | 952,551,855 | 18,879 | # How Do You Use a Conversion Unit Table?
Use a conversion table to determine the amount you have to multiply or divide the unit by when converting between units of measurement. Divide when moving up to a greater unit and multiply when moving down to a lesser one.
In general, conversions within the metric system are easier as units move up and down by multiples of ten. From small to large, milli-, centi- and deci- come before the unit name when indicating units smaller than the standard unit. For instance, a centimeter is 1/100 of a meter. Prefixes for larger units are deka-, hecto- and kilo-. Convert units up by dividing by 10, and down by multiplying by 10 for each step.
Using a unit conversion table, you can determine that you must multiply the amount you have in kilometers by 1000 to see its equivalent in meters. For instance, a 20 kilometer long path equals 20,000 meters.
Unlike the metric system, the imperial system does not follow a pattern. For instance, a foot equals .33 yards and a mile equals 1760 yards. Use the rates on a conversion table to convert imperial units to each other. For instance, to see the inch equivalent of a 5-yard long item, multiply the length by 36. By doing so, you find out that its length equals 180 inches.
You can also use unit conversion tables to convert the units of two systems to each other. The method is the exact same as moving up and down within either system. For instance, a mile equals 1.6 kilometers. As such, you multiply the amount in miles by 1.6 to see its equivalent in kilometers. A 300-mile long trip equals a 480-kilometer trip.
Similar Articles | 368 | 1,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-35 | latest | en | 0.892796 |
http://www.howeverythingworks.org/print1.php?QNum=755 | 1,531,849,221,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589757.30/warc/CC-MAIN-20180717164437-20180717184437-00302.warc.gz | 471,885,284 | 2,129 | MLA Citation: Bloomfield, Louis A. "Question 755"How Everything Works 17 Jul 2018. 17 Jul 2018 .
755. How does a refrigerator work? - SK
A refrigerator uses a material called a "working fluid" to transfer heat from the food inside the refrigerator to the air around the refrigerator. This working fluid moves through the refrigerator's three main components—the compressor, the condenser, and the evaporator—over and over again, in a continuous cycle. I'll begin as the fluid enters the refrigerator's compressor, which is usually located on the bottom of the refrigerator where it's exposed to the room air. The working fluid enters the compressor as a low-pressure gas at roughly room temperature. The compressor squeezes the molecules of that gas closer together, increasing the gas's density and pressure. Since squeezing a gas involves physical work (a force exerted on an object as that object moves in the direction of the force), the compressor transfers energy to the working fluid and that fluid becomes hotter as a result.. The working fluid leaves the compressor as a high-pressure gas that's well above room temperature. The working fluid then enters the condenser, which is typically a snake-like pipe on the back of the refrigerator. Since the fluid is hotter than the room air, heat flows out of the fluid and into the room air. The fluid then begins to condense into a liquid and it gives up additional thermal energy as it condenses. This additional thermal energy also flows as heat into the room air.
The working fluid leaves the condenser as a high-pressure liquid at roughly room temperature. It then flows into the refrigerator, then through a narrowing in the pipe, and then into the evaporator, which is another snake-like pipe that's wrapped around the freezing compartment (in a non-frostfree refrigerator) or hidden in the back of the food compartment (in a frostfree refrigerator). When the fluid goes through the narrowing in the pipe, it's pressure drops and it enters the evaporator as a low-pressure liquid at roughly room temperature. It immediately begins to evaporate and expands into a gas. In doing so, it uses its thermal energy to separate its molecules from one another and it becomes very cold. Heat flows from the food to this cold gas. The working fluid leaves the evaporator as a low-pressure gas a little below room temperature and heads off toward the compressor to begin the cycle again. Overall, heat has been extracted from the food and delivered to the room air. The compressor consumed electric energy during this process and that energy has become thermal energy in the room air. | 528 | 2,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-30 | latest | en | 0.947399 |
https://brainmass.com/math/discrete-math/basic-proof-contradiction-zero-product-property-1995 | 1,709,461,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00685.warc.gz | 137,983,771 | 7,205 | Purchase Solution
# Basic Proof by Contradiction of the Zero Product Property
Not what you're looking for?
Prove the Zero Producty Property in real numbers that:
If ab=0 then a=0 or b=0
(Question is repeated in attachment)
##### Solution Summary
This solution shows a proof that if ab=0, then a=0 or b=0. Gives an introduction to proving things by contradiction.
##### Solution Preview
I'm going to call our assumption (ab=0) P, and the thing we want to prove (a=0 or b=0) Q. So we want to show that P => Q (P implies Q) and we can do this by contradiction: assume P and ~Q (the negation of Q) and show that there is a contradiction with this assumption.
I know what P is (ab=0) but what is ~Q? It is ~(a=0 or b=0). Now ...
##### Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.
##### Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts
This quiz test you on how well you are familiar with solving quadratic inequalities.
##### Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form. | 334 | 1,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.90356 |
https://forum.allaboutcircuits.com/threads/sideband-rejection.18237/ | 1,674,805,514,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00151.warc.gz | 268,379,513 | 24,483 | # Sideband Rejection
#### msseng
Joined Jan 19, 2009
11
Hi,
I have a problem to reject the lower side band of a signal that exists very close to the carrier!
I have a 10MHz signal with a few hertz deviation. I mix it with a 10MHz signal and do some work on remained deviation signal.
But when I mix it again to the 1MHz I have an image that is undesirable.
because of very close distance between 10MHz and that deviation, I can't use any filter to reject the lower signal.
#### Wendy
Joined Mar 24, 2008
23,062
Lets see, you are mixing 10Mhz to 10Mhz. Why, for a reference (as in calibration)? Two 10Mhz signals = a few hertz + 20Mhz (assuming the two base frequencies are surpressed, a bad assumtion). I think we'll need a bit more background before we can be of much help.
If this is a calibration then why not use a Lessajous curve setup?
#### msseng
Joined Jan 19, 2009
11
Lets see, you are mixing 10Mhz to 10Mhz. Why, for a reference (as in calibration)? Two 10Mhz signals = a few hertz + 20Mhz (assuming the two base frequencies are surpressed, a bad assumtion). I think we'll need a bit more background before we can be of much help.
If this is a calibration then why not use a Lessajous curve setup?
I am mixing a 10+delat MHz signal to 10MHz,to get "delta". because only delta must be changed (up or downconvert). and then I mix it with 10MHz to have a 10+(delta)' signal without any image (it means that 10-(delta)' is undesirable)
because of small delta, I can't use any filter to filter undesirale signal.
I hope to say my purpose!
#### msseng
Joined Jan 19, 2009
11
You can build a crystal filter with a quite narrow passband.
Have a look here:
Dear
I can't use any filter.because the delta is about a few hertz.
and the filter that can filter this, must have a very very high Q (about 10^6)
if you have the other solution, I would be greateful to hear it!
thanks.
#### bertus
Joined Apr 5, 2008
22,118
Hello,
When you mix 10 Mhz and 10 Mhz + few hz, you will get a few hz and 20 + few hz.
This principe looks like a direct conversion reciever.
A direct conversion reciever uses a double balanced mixer like a SL440 or NE602.
To get the few hz they use a lowpass filter after the mixer (in your case to get rid of the 20 + few hz).
Greetings,
Bertus
#### KL7AJ
Joined Nov 4, 2008
2,229
You want to use the phasing method. You can get the bandpass right down to 0 Hz that way. It takes some careful adjustment, but it can be done.
Another variation is the Weaver modulator. Both are referenced fully on the internet .
eric
#### msseng
Joined Jan 19, 2009
11
thanks a lot
but I had visited that link previously, and there is a question: Is there any mixer can mix a 10 MHz with a few Hz signal?!
#### Wendy
Joined Mar 24, 2008
23,062
How about the old fashioned diode mixer? I have to step out, but it is a common enough amature radio circuit.
#### Wendy
Joined Mar 24, 2008
23,062
thanks a lot
but I had visited that link previously, and there is a question: Is there any mixer can mix a 10 MHz with a few Hz signal?!
Sure, that is basic AM. If I understand you correctly you want to reject either the Upper Side Band or Lower Side Band.
#### bertus
Joined Apr 5, 2008
22,118
Hello,
Did you also scroll down the page ?
There a schematic can be found for SSB.
Here phase canceling is used to suppress the sideband.
Greetings,
Bertus
#### Attachments
• 20.2 KB Views: 97
#### Wendy
Joined Mar 24, 2008
23,062
OK, I studied balanced mixers in school, but this is new stuff for me. How do you get a consistant 45° shift across the band? Answers some questions I used to have though about SSB.
#### bertus
Joined Apr 5, 2008
22,118
#### msseng
Joined Jan 19, 2009
11
dears,
I report my problem again:
i have a 10MHz carrier with a 1-10Hz message.
I need a 10MHz+(1-10)Hz signal, and so know I can use a thing such as SSB Modulator.
But my question is does it exist any SSB Modulator Module or Mixer that can mix 2 mentioned signal?!
please help me if you can, but please don't propose some thing such as AM and etc. if you don't understand my problem completely!
thanks
#### bertus
Joined Apr 5, 2008
22,118
Hello,
The main problem is the low LF frequencies.
You can use the schematic in post 16.
For the phase shifter circuit use 300 times higher capacitors to use it from 1 to 10 hz in stead of 300 to 3000 hz.
Greetings,
Bertus
#### KL7AJ
Joined Nov 4, 2008
2,229
Hello,
Did you also scroll down the page ?
There a schematic can be found for SSB.
Here phase canceling is used to suppress the sideband.
Greetings,
Bertus
HE HE! Eat your heart out, Bertus! I have a Central Electronics 100V...the first full power PHASING sideband rig ever commercially made. A real collector's item.
(I picked mine up at the estate of a Silent Key....I've seen them on Ebay for as much as \$6600! Still not selling mine, though!)
Eric the avaricious.
#### msseng
Joined Jan 19, 2009
11
Hello,
The main problem is the low LF frequencies.
You can use the schematic in post 16.
For the phase shifter circuit use 300 times higher capacitors to use it from 1 to 10 hz in stead of 300 to 3000 hz.
Greetings,
Bertus
Thanks dear! | 1,445 | 5,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-06 | latest | en | 0.943948 |
https://www.aqua-calc.com/calculate/food-weight-to-volume/substance/dark-blank-red-blank-kidney-blank-beans-coma-and-blank-upc-column--blank-035826082848 | 1,582,794,253,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146665.7/warc/CC-MAIN-20200227063824-20200227093824-00415.warc.gz | 598,064,464 | 7,899 | # Volume of DARK RED KIDNEY BEANS, UPC: 035826082848
## food weight to volume conversions
### calculate volume of generic and branded foods per weight
#### Volume, i.e. how many spoons, cups,gallons or liters in 100 gram of DARKRED KIDNEY BEANS, UPC: 035826082848
centimeter³ 91 milliliter 91 foot³ 0 US cup 0.38 Imperial gallon 0.02 US dessertspoon 12.31 inch³ 5.55 US fluid ounce 3.08 liter 0.09 US gallon 0.02 meter³ 9.1 × 10-5 US pint 0.19 metric cup 0.36 US quart 0.1 metric dessertspoon 9.1 US tablespoon 6.15 metric tablespoon 6.07 US teaspoon 18.46 metric teaspoon 18.2
#### Weight
gram 100 ounce 3.53 kilogram 0.1 pound 0.22 milligram 100 000
Nutrient (find foodsrich in nutrients) Unit Value /100 g BasicAdvancedAll Proximates Energy kcal 85 Protein g 5.38 Total lipid (fat) g 0 Carbohydrate,bydifference g 14.62 Fiber,totaldietary g 4.6 Sugars, total g 1.54 Minerals Calcium, Ca mg 31 Iron, Fe mg 1.38 Sodium, Na mg 92 Vitamins Vitamin C,totalascorbic acid mg 0 Vitamin A, IU IU 0 Lipids Fatty acids,totalsaturated g 0 Cholesterol mg 0
#### See how many calories in0.1 kg (0.22 lbs) of DARKRED KIDNEY BEANS, UPC:035826082848
From kilocalories(kcal) kilojoule(kJ) Carbohydrate 0 0 Fat 0 0 Protein 0 0 Other 85 355.64 Total 85 355.64
• About DARK RED KIDNEY BEANS, UPC: 035826082848
• 274.73893 grams [g] of DARK RED KIDNEY BEANS, UPC: 035826082848 fill 1 metric cup
• 9.17123 ounces [oz] of DARK RED KIDNEY BEANS, UPC: 035826082848 fill 1 US cup
• DARK RED KIDNEY BEANS, UPC: 035826082848 weigh(s) 274.74 gram per (metric cup) or 9.17 ounce per (US cup), and contain(s) 85 calories per 100 grams or ≈3.527 ounces [ weight to volume | volume to weight | price | density ]
• Ingredients: PREPARED KIDNEY BEANS, WATER, SUGAR, SALT, AND CALCIUM CHLORIDE. DISODIUM EDTA ADDED AS A PRESERVATIVE.
• Manufacturer: Food Town Stores Inc.
• A few foods with a name containing, like or similar to DARK RED KIDNEY BEANS, UPC: 035826082848:
• For instance, compute how many cups or spoons a pound or kilogram of “DARK RED KIDNEY BEANS, UPC: 035826082848” fills. Volume of the selected food item is calculated based on the food density and its given weight. Visit our food calculations forum for more details.
• Reference (ID: 10147)
• USDA National Nutrient Database for Standard Reference; National Agricultural Library; United States Department of Agriculture (USDA); 1400 Independence Ave., S.W.; Washington, DC 20250 USA.
#### Foods, Nutrients and Calories
ANTIPASTI PANINO, UPC: 017869804472 contain(s) 353 calories per 100 grams or ≈3.527 ounces [ price ]
CASHEWS, UPC: 833302008029 contain(s) 536 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Marine, CORALine, Volcano Beach weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
silver impregnated coconut shell carbon weighs 490 kg/m³ (30.5897 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-401B, liquid (R401B) with temperature in the range of -51.12°C (-60.016°F) to 68.34°C (155.012°F)
#### Weights and Measurements
A Byte is a unit of capacity that necessary to store eight binary digits
Scientifically, the weight of a body is the gravitational force acting on the body, and directed towards the center of gravity with magnitude of the force depending on the mass of the body.
Gm/s² to fur/s² conversion table, Gm/s² to fur/s² unit converter or convert between all units of acceleration measurement.
#### Calculators
Oils, fuels, refrigerants: compute volume by weight and temperature | 1,141 | 3,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-10 | latest | en | 0.4927 |
https://www.studentlance.com/solution/manova-spss-statistics-solution | 1,519,170,479,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813109.8/warc/CC-MAIN-20180220224819-20180221004819-00319.warc.gz | 982,465,803 | 6,366 | # MANOVA SPSS Statistics Solution - 18690
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## Cirara
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Request Description
Main Task: Activity #8 You will submit one Word document for this activity. You will create this Word document by cutting and pasting SPSS output into Word.
Part A. SPSS Activity In this exercise, you are playing the role of a researcher that is testing new medication designed to improve cholesterol levels. When examining cholesterol in clinical settings, we look at two numbers: low-density lipoprotein (LDL) and high-density lipoprotein (HDL). You may have heard these called “good” (HDL) and “bad” (LDL) cholesterol. For LDL, lower numbers are better (below 100 is considered optimal). For HDL, 60 or higher is optimal.
In this experiment, you will be testing three different versions of the new medication. In data file “Activity 8.sav” you will find the following variables: group (0=control, 1=Drug A, 2=Drug B, 3=Drug C), LDL, and HDL (cholesterol numbers of participants after 12 weeks).
Using a MANOVA, try to ascertain which version of the drug (A, B or C) shows the most promise. Perform the following analyses and paste the SPSS output into your Word document. 1. Exploratory Data Analysis. a. Perform exploratory data analysis on the relevant variables in the dataset. When possible, include appropriate graphs to help illustrate the dataset. b. Compose a one to two paragraph write up of the data. c. Create an APA style table that presents descriptive statistics for the sample.
2. Perform a MANOVA. Using the “Activity 8.sav” data set, perform a MANOVA. “Group” is your fixed factor and LDL and HDL are your dependent variables. Be sure to include simple contrasts to distinguish between the drugs (group variable). In the same analysis, include descriptive statistics and parameter estimates. Finally, be certain to inform SPSS that you want a post-hoc test to help you determine which drug works best. a. Is there any statistically significant difference in how the drugs perform? If so, explain the effect. Use the post hoc tests as needed. b. Write up the results using APA style and interpret them.
Solution Description
Solution
Attachments | 545 | 2,332 | {"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-2018-09 | latest | en | 0.841898 |
https://rdrr.io/github/OnofriAndreaPG/aomisc/man/CVA.html | 1,712,955,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816070.70/warc/CC-MAIN-20240412194614-20240412224614-00460.warc.gz | 441,336,890 | 7,707 | CVA: Canonical variate analysis for multienvironment and... In OnofriAndreaPG/aomisc: Statistical methods for the agricultural sciences
CVA R Documentation
Canonical variate analysis for multienvironment and multitrait genotype experiments
Description
This function performs canonical variate analysis as a descriptive visualisation tool. It is close to the 'lda()' function in the MASS package but it is not meant to be used for discriminant analyses.
Usage
``````CVA(dataset, groups, scale = TRUE, constraint = 3)
``````
Arguments
`dataset` dataset is a multidimensional matrix of observations `groups` groups is a vector coding for groupings `scale` whether the data needs to be standardised prior to analysis. Defaults to TRUE `constraint` It is the type of scaling for eigenvectors, so that canonical variates have: 1 = unit within-group standard deviations (most common); 2 = unit total standard deviations; 3 = unit within group norms; 4 = unit total norms. It defaults to 3
Details
More detail can be found in a blog page, at 'https://www.statforbiology.com/2023/stat_multivar_cva/'. Please, note that preliminary data transformations (e.g.: standardisation) are left to the user and must be performed prior to analyses (see example below).
Value
`TOT` matrix of total variances-covariances `B` matrix of 'between-groups' variances-covariances `W` matrix of 'within-group' variances-covariances `B/W` matrix of W^-1 B `eigenvalues` vector of eigenvalues `eigenvectors` matrix of eigenvectors `proportion` a vector containing the proportion of total discriminating ability captured by each canonical variate `correlation` vector of canonical correlations `squared.canonical.correlation` vector of squared canonical correlations `coefficients` matrix of canonical coefficients `scores` matrix of canonical scores `centroids` matrix of scores for centroids `total.structure` matrix of total canonical structure `between.structure` matrix of between-groups canonical structure `within.structure` matrix of within-groups canonical structure `class.fun` matrix of classifications functions `class.val` matrix of classification values `within.structure` matrix of within-groups canonical structure `class` vector of predicted classes
Andrea Onofri
References
https://www.statforbiology.com/2023/stat_multivar_cva/
Examples
``````fileName <- "https://www.casaonofri.it/_datasets/WheatQuality4years.csv"
dataset\$Year <- factor(dataset\$Year)
# Standardise the data
groups <- dataset\$Genotype
Z <- apply(dataset[,3:6], 2, scale, center = TRUE, scale = TRUE) | 578 | 2,578 | {"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-2024-18 | latest | en | 0.768801 |
http://www.docstoc.com/docs/93060898/Thinking-Mathematically-by-Robert-Blitzer | 1,405,075,615,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776426734.39/warc/CC-MAIN-20140707234026-00087-ip-10-180-212-248.ec2.internal.warc.gz | 229,484,003 | 19,880 | # Thinking Mathematically by Robert Blitzer
Document Sample
``` Thinking
Mathematically
Chapter 3
Logic
Thinking
Mathematically
Section 1
Statements, Negations, and
Quantified Statements
Statements
A statement is a sentence that is either true or false,
but not both at the same time.
Examples
“Miami is a city in Florida” is a true statement.
“Two plus two equals five” is a false statement.
“Today is Friday” is a statement that is
sometimes true and sometimes false, but never
both true and false at the same time.
“Go to the grocery store” is not a statement. It
is a command and is neither true nor false.
Negation of Statements
The negation of a statement is another statement that
has the opposite truth value. That is, when a
statement is true its negation is false and when the
statement is false its negation is true.
Examples
“Miami is not a city in Florida” is the negation
of the statement “Miami is a city in Florida”.
“Two plus two is not equal to five” is the negation
of the statement “Two plus two equals five”.
Be careful! “Two plus two equals four” is a true
statement but it is not the negation of “Two plus two
equals five”.
Symbolic Statements
Just as x can be used as a name for a number, a
symbol such as p can be used as a name for a
statement.
When p is used as a name for a statement the symbols
~p are used as a name for the negation of p.
Examples
Let p stand for “Miami is a city in Florida.”
Then ~p is the statement “Miami is not a city in
Florida.”
Quantified Statements
A quantified statement is one that says something
about all, some, or none of the objects in a collection.
Examples
“All students in the college are taking history.”
“Some students are taking mathematics.”
“No students are taking both mathematics and
history.”
Equivalent Statements
In any language there are many ways to say the same
thing. The different linguistic constructions of a
statement are considered equivalent.
Example
“All students in the college are taking history.”
“Every student in the college is taking history.”
Example
“Some students are taking mathematics.”
“At least one student is taking mathematics.”
Negating Quantified Statements
The negation of a statement about all objects is not
all. Not all can often be expressed by some are not.
Examples
p : All students in the college are taking history.
~p : Some students in the college are not taking
history.
Negating Quantified Statements
The negation of a statement about some objects is not
some. Not some can often be expressed by none or not
any.
Examples
p : Some students are taking mathematics.
~p : None of the students are taking mathematics.
Negating Quantified Statements
The negation of a statement about all objects is not
all. Not all can often be expressed by some are not.
The negation of a statement about some objects is not
some. Not some can often be expressed by none or not
any.
Statement Negation of statement
All A are B No A are B
Some A are B Some A are not B
Thinking
Mathematically
Section 2
Compound Statements and
Connectives
Compound Statements
Simple statements can be connected with and,
either … or, if … then, or if and only if. These more
complicated statements are called compound statements.
Examples
“Miami is a city in Florida” is a true statement.
“Atlanta is a city in Florida” is a false statement.
“Either Miami is a city in Florida or Atlanta is a city
in Florida” is a compound statement that is true.
“Miami is a city in Florida and Atlanta is a city in
Florida” is a compound statement that is false.
And Statements
When two statements are represented by p and q
the compound and statement is p q.
p: Harvard is a college.
q: Disney World is a college.
p q: Harvard is a college and Disney World is a
college.
p ~q: Harvard is a college and Disney World is
not a college.
Either ... or Statements
When two statements are represented by p and q
the compound either ... or statement is p q.
p: The bill receives majority approval.
q: The bill becomes a law.
p q: The bill receives majority approval or the
bill becomes a law.
p ~q: The bill receives majority approval or the
bill does not become a law.
The Two Meanings of Or
• Compound statements involving the word or
are a bit harder to understand.
• There are two meanings:
• 1. Exclusive or: At a Chinese Restaurant, the
lunch menu says: "Choose one from column A
or one from column B."
• You can only choose one item, so you are not
allowed to choose from both columns.
The Two Meanings of Or
• Compound statements involving the word or
are a bit harder to understand.
• There are two meanings:
• 2. Inclusive or: "A good teacher should be
either smart or funny." Here, you are implying
that it's possible for a good teacher to be both
smart AND funny, but at least one of these
characteristics should be true.
The Two Meanings of Or
In this class, we will use the word or to be the
inclusive or: one or the other or both!
If ... then Statements
When two statements are represented by p and q
the compound if ... then statement is: p q.
p: Ed is a poet.
q: Ed is a writer.
p q: If Ed is a poet, then Ed is a writer.
q p: If Ed is a writer, then Ed is a poet.
~q ~p: If Ed is not a writer, then Ed is not a
Poet
If and only if Statements
When two statements are represented by p and q the
compound if and only if statement is: p q.
p: The word I'm thinking of is "set".
q: The word has 464 meanings.
p q: The word I'm thinking of is "set" if and only if
the word has 464 meanings.
~q ~p: The word I'm thinking of does not have 464
meanings if and only if the word is not "set".
Translating sentences into
symbolic expressions
• Words such as and, but, or, either,
neither, if, then, and so on are used in
ordinary sentences and have meanings in
logic.
• Example:
– He went to the store but didn't buy anything.
Translating sentences into
symbolic expressions
He went to the store but didn't buy anything.
He went to the store and didn't buy anything.
He went to the store yet he didn't buy anything.
Although he went to the store,
He went to the store ; nevertheless,
Translating symbolic
expressions into sentences
• If he goes home, it is not true that
he missed his curfew or that
he will be late for work tomorrow.
– p: he goes home
– q: he missed his curfew
– r: he will be late for work tomorrow.
p ~( q ∨ r )
Symbolic Logic
Statements of Logic
Name Symbolic Form Meaning
Negation ~p not
Conjunction pq and
Disjunction pq or
Conditional pq if..then
Biconditional pq is the same as
Thinking
Mathematically
Section 3
Truth Tables for Negation,
Conjunction, and Disjunction
Truth Tables - Negation
If a statement is true then its negation is false
and if the statement is false then its negation
is true. This can be represented in the form
of a table called a truth table.
p ~p
T F
F T
Truth Tables - Conjunction
The conjunction of two statements is true
only when both of them are true. This can
be represented in the form of a truth table
in the following way.
p q pq
T T T
T F F
F T F
F F F
Truth Tables - Disjunction
The disjunction of two statements is false
only when both of them are false. This can
be represented in the form of a truth table
in the following way.
p q pq
T T T
T F T
F T T
F F F
Constructing a Truth Table
Construct a truth table for (p ~q) q.
p q ~q p ~q (p ~q) q
T T F F T
T F T T T
F T F F T
F F T F F
does this look familiar?
Constructing a Truth Table
Construct a truth table for ( p ~q) (~p q )
p q ~p ~q (p ~q) (~p q) (p ~q) (~p q)
T T F F F F F
T F F T T F T
F T T F F T T
F F T T F F F
Section Quiz
Construct a truth table for (p q) (~p ~q)
p q ~p ~q (p q) (~p ~q) (p q) (~p ~q)
T T F F T F T
T F F T F F F
F T T F F F F
F F T T F T T
Section Quiz
Construct a truth table for (p q) ~r
p q r ~r (p q) (p q) ~r
T T T F T T
T T F T T T
T F T F F F
T F F T F T
F T T F F F
F T F T F T
F F T F F F
F F F T F T
Section Quiz
• For each of the following 10 expressions,
write down the value of the expression:
• For example
2. F ∨F F ∧
3. ~(~F)F F
10. ∨ ∨F ∨
1. T TF∧∨T F T
•
5.
4.
9.
6.
8.
7.
• =T
Solutions
1. T ∧ F = F
2. F ∨ F = F
3. T ∧ T ∧ T = T
4. F ∨ T =T
5. T ∧ F =F
6. ~(~F) = F
7. F ∧ F =F
8. T∨F∨F =T
9. T ∧ T =T
10. F ∨ F =F
Section Quiz
• Construct a truth table for ~p ∨ q
p q ~q ~p ∨ q p q ~q ~p ∨ q p q ~q ~p ∨ q
T T F T T T T T T T F T
T F F T T F F T T F T T
F T T F F T T T F T F F
F F T T F F F F F F T T
p q ~q ~p ∨ q
T T F T
T F F T
F T T F
F F T T
Thinking
Mathematically
Section 4
Truth Tables for the Conditional
and Biconditional
Conditional Statements
The truth table for implication (→) is
p q p→ q
T T T
T F F
F T T
F F T
This one is difficult to understand.
Let's try anyway.
Conditional Statements
The truth table for implication (→) is
p q p→ q
T T T
T F F
F T T
F F T
Suppose I say, "If it rains tomorrow, I'll
be carrying an umbrella."
p: it is raining tomorrow
q: I'm carrying an umbrella
Conditional Statements
The truth table for implication (→) is
p q p→ q
T T T No. I told
T F F the truth
F T T
F F T
The next day, it's raining. You look out
your window and there I am, keeping dry
under an umbrella.
Did I lie?
Conditional Statements
The truth table for implication (→) is
p q p→ q
T T T I sure did!
T F F
F T T
F F T
The next day, it's raining. You look out
your window and there I am, soaking wet
without an umbrella.
Did I lie?
Conditional Statements
The truth table for implication (→) is
p q p→ q
T T T Not really.
T F F
F T T I never said
F F T what I'd do if it
didn't rain.
The next day, it's sunny. You look out your
window and there I am, holding an
umbrella.
Did I lie?
Conditional Statements
The truth table for implication (→) is
p q p→ q
T T T Not really.
T F F
F T T I never said
F F T what I'd do if it
didn't rain.
The next day, it's sunny. You look out your
window and there I am, without an
umbrella.
Did I lie?
Conditional Statements
The truth table for implication (→) is
p q p→ q
T T T
T F F
F T T
F F T
p is called the precedent (or hypothesis)
q is called the consequent (or consequence)
An implication is false if and only if the precedent is
true and the consequent is false.
Biconditional Statements
The truth table for the biconditional, "if
and only if", (↔) is
p q p↔q
T T T
T F F
F T F
F F T
This operator can be viewed in two ways:
- A two-way implication: p → q and q → p
Biconditional Statements
The truth table for the biconditional, "if
and only if", (↔) is
p q p↔q
T T T
T F F
F T F
F F T
This operator can be viewed in two ways:
- The concept of "has exactly the same
truth table as" or "is equivalent to"
Biconditional Statements
• Look at the truth tables below:
p q q↔p p q pq qp pq /\ qp
T T T T T T T T
T F F T F F T F
F T F F T T F F
F F T
F F T T T
Section Quiz
• For each of the following 10 expressions,
write down the value of the expression:
• For example
1. T TT TTF
10. ∨ ∧F
2. F ⟷ ~F
• ⟷
F
3. ~(~T)FF
5.
6.
9.
8.
4.
7.
• =T
Solutions
1. T F = F
2. F ⟷ F = T
3. T ∧ T = T
4. F T = T
5. T ∧ ~F = T
6. ~(~T) = T
7. F F =T
8. T⟷F = F
9. T ∧ T =T
10. T F = F
Thinking
Mathematically
Section 5
Equivalent Statements, Variations
of Conditional Statements, and
DeMorgan’s Laws
Equivalent Statements
We saw at the end of a previous lecture
that the truth table for (p /\ ~q) \/ q was the
same as the one for p \/ q .
Whenever two statements have the same
truth table values, we say they are
equivalent.
Equivalent Statements
Two statements are equivalent if they have
the same truth values.
p q pq p q ~p ~pq
T T T T T F T
T F T T F F T
F T T F T T T
F F F F F T F
So the statements pq and ~pq are equivalent.
They mean the same thing!
Equivalent Statements
Whoa!!!! The statements p q and
~p q mean the same thing??????????
Equivalent Statements
The statements p q and ~p q are equivalent.
They mean do the same thing!
•Why?
• In order for p q to be true, either p or q or
maybe both have to true. That means that if p is
false, we would need for q to be true.
• If p is false, then q is true.
• If ~p is true, then q is true.
• If not p then q.
• ~p q
The Contrapositive
The statements p q and ~q ~p are
equivalent. They are contrapositives.
p q pq p q ~p ~q
~q ~q ~p
~q ~p
T T T T T F F T
T F F T F F TT F
F T T F T T F T
F F T F
F F
F T
T TT T
For example, "If it rains tomorrow, I'll take my
umbrella" means exactly the same thing as "If I don't
take my umbrella tomorrow, it's because it isn't
raining."
The Contrapositive
The statements p q and ~q ~p are
equivalent. They are contrapositives.
For example, "If it rains tomorrow, I'll take my umbrella"
means exactly the same thing as "If I don't take my
umbrella tomorrow, it's because it isn't raining."
Think about it. The original statement says that on the
condition that it's raining, I'll definitely take an umbrella.
So if you see me not carrying an umbrella, then it can't be
raining!
Converse and Inverse
q p is the converse of p q.
~p ~q is the inverse of p q.
The converse and inverse are contrapositives
of each other and are equivalent.
p q qp p q ~p ~q ~p~q
T T T T T F F T
T F T T F F T T
F T F F T T F F
F F T
F F T T T
Conditional Statements
Let p and q be statements.
Name Symbolic Form
Conditional pq
Converse qp
Inverse ~p ~q
Contrapositive ~q ~p
Conditional Statements
Conditional: If it rains tomorrow, I'll take my umbrella.
Converse: If I have my umbrella tomorrow, it will be raining.
Inverse: If it doesn't rain tomorrow, I won't take my umbrella.
Contrapositive: If I don't have my umbrella tomorrow, it
won't be raining.
DeMorgan's Laws
• The statements ~(p ∧ q) and ~p ∨ ~q are
equivalent. This is one of two DeMorgan's Laws.
• ~(p ∧ q) says "It is false that p and q are true"
• ~p ∨ ~q says "Either p is not true or q is not true"
• These are the same because if it is false that p and
q are both true, then either p has to be false or q
has to be false.
DeMorgan's Laws
• The statements ~(p ∧ q) and ~p ∨ ~q are
equivalent. This is one of two
DeMorgan's Laws.
p q pq ~(pq) p q ~p ~q ~p~q
T T T F T T F F F
T F F T T F F T T
F T F T F T T F T
F F F T F F T T T
DeMorgan's Laws
• The statement ~(p\/q) and ~p/\~q are
equivalent. This is the other
DeMorgan's Law.
p q pq ~(pq) p q ~p ~q ~p~q
T T T F T T F F F
T F T F T F F T F
F T T F F T T F F
F F F T F F T T T
Another equivalence
• The statements p→q and ~p q are
equivalent.
p q p→q p q ~p ~p\/q
T T T T T F T
T F F T F F F
F T T F T T T
F F T T
F F T
Another equivalence
• The statements p→q and ~p q are
equivalent.
• If p is true (p), that means that ~p is
false.
• So, in order for ~p q to be true, if p
is true (in other words, ~p is false),
the other half of ~p q, namely q,
must be true. If p then q.
Thinking
Mathematically
Section 6
Arguments and Truth Tables
Arguments
There are two parts to an argument.
The first part consists of a collection of
statements called the premises.
The second part is a statement called the
conclusion.
Symbolically the argument is the conditional
statement whose if part is the conjunction of
all the premises and whose then part is the
conclusion.
An Example of an Argument
p q
pq If I get an A on the final I will pass the course.
p I got an A on the final.
q I will pass the course
The argument is “If I get an A on the final I will
pass the course and I got an A on the final.
Therefore I will pass the course.”
[(p q) /\ p] q
An Example of an Argument
pq If I get an A on the final I will pass the course.
p I got an A on the final.
q I will pass the course
The argument says that getting an A on the final will assure
me of passing the course. Since I did get an A, the professor
has to give me a passing grade (or else he lied to me.)
We can show that this argument makes sense by showing that
the corresponding truth table is a tautology.
A tautology is a logical statement that is always true; its
truth table has T on all lines.
Valid Arguments
Valid arguments are tautologies. That is, they
are always true.
pq If I get an A on the final I will pass the course.
p I got an A on the final.
q I will pass the course
[(p q) /\ p] q
p q pq (pq)p q [(pq)p]q tautology !!!
T T T T T T
this is a valid
T F F F F T argument !!!
F T T F T T
F F T F F T
Another Example of an Argument
pq If I get an A on the final I will pass the course.
q I passed the course !!!
p I must have gotten an A on the final
[(p q) /\ q] p
p q pq (pq)q p [(pq)q]p not a
tautology !!!
T T T T T T
T F F F T T not a valid
F T T T F F argument !!!
F F T F F T
Evaluating Real Arguments
• Step 1: Assign variables (p, q, r, …) to each basic
• Step 2: For each premise in the argument, write the
corresponding logical expression.
• Step 3: Write the corresponding logical expression
for the conclusion.
• Step 4: Connect all the premise expressions with .
Connect that big expression with the conclusion
expression with .
• Step 5: Construct the truth table for the overall
expression.
• Step 6: If the expression is a tautology, it is a valid
argument.
Another Example:
• Here's a real-world argument. Is it a valid
argument?
– If I go out, I'll go to the store.
– If I go to the store, I will buy some chips.
– I'm not going out.
– Therefore, I won't be buying chips.
Another Example:
p q
• If I go out, I'll go to the store.
• If I go to the store, I will buy some chips.
• I'm not going out. r
• Therefore, I won't be buying chips.
Step 1: Assign variables (p, q, r, …) to each basic statement
Another Example:
• If I go out, I'll go to the store. q
p
• If I go to the store, I will buy some chips.
qr
• I'm not going out. ~p
• Therefore, I won't be buying chips. ~r
[(p q) (q r) ~p] ~r
For each premise in the argument,
Step 2:3: Write the corresponding logical write the
Step Connect all the premise expressions with .
Step 4:
corresponding logical expression.
expression big expression with
Connect that for the conclusion. the conclusion
expression with .
Step 6: If the expression is atable for the overall expression.
Step 5: Construct the truth tautology, it is a valid argument.
p q r p q q r ~p [(p q) ~r [(p q) (q r) ~p] ~r
(q r)
~p
T T T T T F F F T
T T F T F F F T T
T F T F T F F F T
T F F F T F F T T
F T T T T T T F F
F T F T F T F T
F F T T T T T F
F F F T T T T T
not a tautology. not valid
Another Example:
• If it is not sunny, it is either overcast or it's
the evening.
• If it's the evening, it can't be overcast.
• If it's overcast, I will wear a raincoat.
• It's not the evening.
• Therefore, I am wearing a raincoat.
Another Example:
• p: it is sunny.
• q: it is overcast.
• r: it is the evening.
• s: I am wearing a raincoat.
Another Example:
• If it is not sunny, it is either overcast or it's
the evening. ~p(qr)
• If it's the evening, it can't be overcast. r~q
• If it's overcast, I will wear a raincoat. qs
• It's not the evening. ~r
• Therefore, I am wearing a raincoat. s
• [(~p(qr)) (r~q) (qs) ~r] s
First half of truth table. Other half is when p is false
p q r s ~p(qr) r~q qs ~r [(~p(qr)) (r~q) [(~p(qr))
(qs) ~r] (r~q)
(qs) ~r] s
T T T T T F T F F T
T T T F T F F F F T
T T F T T T T T T T
T T F F T T F T F T
T F T T T T T F F T
T F T F T T T F F T
T F F T T T T T T T
T F F F T T T T T F
not a tautology. not valid
~p is F
Section Quiz
• I'll either get my hair cut or I won't go
outside.
• I'm not getting my hair cut
• I won't go outside.
• Is this a valid argument?
p: get haircut
Section Quiz q: go outside
• I'll either get my hair cut or I won't go
outside. p ~q
• I'm not getting my hair cut. ~p
• Therefore, I won't go outside. ~q
[(p ~q) ~p] ~q
• Is this a valid argument?
Section Quiz
Is this a valid argument?
yes
~q
p q ~p ~q p ~q (p ~q)~p [(p~q)~p]~q
T T F F T F F T
T F F T T F T T
F T T F F F F T
F F T T T T T T
Section Quiz
• Which of the following is a tautology?
– a) p ~p
– b) p ~p
– c) p → ~p
– d) p ↔ ~p
```
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# lecture9 - Lecture 9 Directional Derivatives and the...
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Lecture 9: Directional Derivatives and the Gradient May 19, 2009 Lecture 9: Directional Derivatives and the Gradient
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Objectives 1 Compute the directional derivative of a function. 2 Interpret geometrically the directional derivative. 3 Construct the gradient vector of a multivariable function 4 Use the gradient to maximize the directional derivative. 5 Use the gradient in three dimensions to find tangent planes and normal lines to level surfaces. Lecture 9: Directional Derivatives and the Gradient
Partial derivatives f x ( x ) = lim h 0 f ( x + h , y ) - f ( x , y ) h f y ( x ) = lim h 0 f ( x , y + h ) - f ( x , y ) h Derivatives in the directions of ˆ i and ˆ j Lecture 9: Directional Derivatives and the Gradient
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Directional derivative Suppose that u = h a , b i is a unit vector. We define the directional derivative D u f ( x 0 , y o ) = lim h 0 f ( x 0 + ah , y 0 + bh ) - f ( x 0 , y 0 ) h Lecture 9: Directional Derivatives and the Gradient
Directional derivative Suppose that u = h a , b i is a unit vector. We define the directional derivative D u f ( x 0 , y o ) = lim h 0 f ( x 0 + ah , y 0 + bh ) - f ( x 0 , y 0 ) h We compute the directional derivative as follows: D u f ( x 0 , y o ) = f x ( x 0 , y 0 ) a + f y ( x 0 , y 0 ) b Lecture 9: Directional Derivatives and the Gradient
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Directional derivative Suppose that u = h a , b i is a unit vector. We define the directional derivative D u f ( x 0 , y o ) = lim h 0 f ( x 0 + ah , y 0 + bh ) - f ( x 0 , y 0 ) h We compute the directional derivative as follows: D u f ( x 0 , y o ) = f x ( x 0 , y 0 ) a + f y ( x 0 , y 0 ) b IMPORTANT: u must be a unit vector Lecture 9: Directional Derivatives and the Gradient
Gradient vector If f is a function of two variables, the gradient of f is defined as f ( x , y ) = h f x ( x , y ) , f y ( x , y ) i = f x ˆ i + f y ˆ j Lecture 9: Directional Derivatives and the Gradient
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Gradient vector If f is a function of two variables, the gradient of f is defined as f ( x , y ) = h f x ( x , y ) , f y ( x , y ) i = f x ˆ i + f y ˆ j We can then write the directional derivative as D u f ( x 0 , y o ) = f ( x , y ) · u Lecture 9: Directional Derivatives and the Gradient
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http://forum.allaboutcircuits.com/threads/confusion-about-vertical-polarization-and-horizontal-polarization.71484/ | 1,485,242,427,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284352.26/warc/CC-MAIN-20170116095124-00489-ip-10-171-10-70.ec2.internal.warc.gz | 105,141,065 | 15,593 | # confusion about vertical polarization and horizontal polarization
Discussion in 'Wireless & RF Design' started by donut, Jun 22, 2012.
1. ### donut Thread Starter Member
May 23, 2012
51
0
Could you please help me to understand some basic concepts about antenna polarization. The books I have are really confusing and do not have examples that I can learn from. This is really frustrating because books are biased to people who have experience and not have any experience like me.
Please look at the attached E plane radiation pattern image I attached (figure 13).
The books image (figure 13) tells me that the attached image is a E plane radiation pattern which means that the E field is perpendicular to the earth (so vertically polarized).
The books image (figure 13) tells me that the attached image dipole is located on the line from 90 degrees to 270 degrees.
How can you have a vertically polarized antenna with a dipole positioned horizontally (90 degrees to 270 degrees)?
What am I missing here? Im sure the book is accurate I just dont know how to read the radiation patter. Please help
Figure 19 (attached) clearly correlates to horizontal polarized dipole but figure 13 says that it is a vertical polarized dipole.
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2. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
783
Presumably 0° / 180° is the ground plane angle. So the dipole is offset from the ground plane by 90° and lies along the 90° / 270° axis perpendicular to the ground. Rotate the page by 90° and that should represent "reality" as you would perceive it standing on the ground and looking at the dipole.
3. ### vk6zgo Active Member
Jul 21, 2012
677
85
Yes,you are confused!
Fig 13 is ,if you could imagine it,looking down on a horizontally polarised
dipole.
In other words,the diagram is showing you the radiation pattern in azimuth ,not elevation.
My understanding is that polarisation is not determined by the orientation of the E-field.
If this is the book I think it is,this is covered earlier
Actually,I was confused too,the E field orientation does determine the antenna's polarisation.
I got E & H fields mixed up,but the other stuff in this posting is correct.
Fig 19,in my opinion,not a very good diagram,simply shows that a vertically mounted dipole will have vertical polarisation,while a horizontally mounted dipole will have horizontal polarisation.
Unless I miss my guess,the book is the ARRL Antenna Handbook.
It's a great book,but it does have a fairly messy layout,so you may find diagrams that seem related,just happen to appear together,or the Fig referred to on one page may be on the next or previous page.
The "Digest" type format,where a lot of the stuff is slightly modified articles from "QST" tends to lead to these problems.
PS: Looking at the dipole diagram from "end on",the radiation pattern looks like a circle,instead of a figure 8,& looking at it from an angle,it looks like a doughnut.
The following website has a PDF which includes (fig3.1) a graphical representation of this doughnut shape.
http://www.ictregulationtoolkit.org/en/Document.3643.pdf
Last edited: Aug 10, 2012
4. ### PRFGADGET Active Member
Aug 8, 2011
51
7
A "Quick & Dirty" way to look at the "Difference" between Vertical and Horizontal Polarization is to say that there is a 90 degree "PHASE" shift or "Differential" between the two.
The above statement can open a large can of worms in many circles but like I said it's "QUICK & DIRTY".
Something else you might want to do a bit of research on is "RADIO HORIZON" and I won't even get into that here.
Another "Q&D", Vertical polarization tends to be "LINE OF SITE" for the most part (this is somewhat Frequency / Band Dependent and dependent on ionospheric conditions) , where "HORIZONTAL" polarization tends to follow the curvature of the earth's surface which tends to allow for greater range in distance.
Another term you may want to do some research on is "DOPPLER SHIFT" .
As "ZGO" pointed out earlier, the ARRL book's are good but can be quite confusing , you might do better looking for book's by William Orr or Walter Maxwell . | 991 | 4,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-04 | latest | en | 0.944588 |
https://b-davies.github.io/env170/Week4/Week4Lab/02_LookingForPatterns.html | 1,723,755,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00615.warc.gz | 87,953,122 | 8,027 | # 2Looking for Patterns
``````library(modeldata)
library(tidyverse)``````
``````── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
✔ dplyr 1.1.2 ✔ readr 2.1.4
✔ forcats 1.0.0 ✔ stringr 1.5.0
✔ ggplot2 3.4.3 ✔ tibble 3.2.1
✔ lubridate 1.9.2 ✔ tidyr 1.3.0
✔ purrr 1.0.2
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors``````
When we start looking for patterning in the data, we’
### 2.0.1 Univariate patterns
#### 2.0.1.1 Counts
--bar charts
What if the counts are already in the data?
``````hairEye<-as_tibble(HairEyeColor)
ggplot(data=hairEye,aes(x=Hair,y=n,fill=Eye)) +
geom_col()``````
#### 2.0.1.3 Distributions
-Center
--Mean
--Median
--Range
--Standard Deviation
-Shape
--Normal
--Bimodal, multimodal
---Density plots
TRY IT YOURSELF: Look at X dataset
### 2.0.2 Bivariate
#### 2.0.2.2
``````hairEye<-as_tibble(HairEyeColor)
ggplot(data=hairEye,aes(x=Hair,y=n,fill=Eye)) +
geom_col()``````
#### 2.0.2.3 Categorical and numerical
Like histograms, boxplots show the center, spread, and shape of distributions, but these are most useful when comparing more than one group.
Median: the central line of the boxplot is
Interquartile range: The “box” part of the box is an indication of the interquartile range. This is a
Have a look at the Sacramento dataset:
``Sacramento``
``````# A tibble: 932 × 9
city zip beds baths sqft type price latitude longitude
<fct> <fct> <int> <dbl> <int> <fct> <int> <dbl> <dbl>
1 SACRAMENTO z95838 2 1 836 Residential 59222 38.6 -121.
2 SACRAMENTO z95823 3 1 1167 Residential 68212 38.5 -121.
3 SACRAMENTO z95815 2 1 796 Residential 68880 38.6 -121.
4 SACRAMENTO z95815 2 1 852 Residential 69307 38.6 -121.
5 SACRAMENTO z95824 2 1 797 Residential 81900 38.5 -121.
6 SACRAMENTO z95841 3 1 1122 Condo 89921 38.7 -121.
7 SACRAMENTO z95842 3 2 1104 Residential 90895 38.7 -121.
8 SACRAMENTO z95820 3 1 1177 Residential 91002 38.5 -121.
9 RANCHO_CORDOVA z95670 2 2 941 Condo 94905 38.6 -121.
10 RIO_LINDA z95673 3 2 1146 Residential 98937 38.7 -121.
# ℹ 922 more rows``````
``````ggplot(data=Sacramento,aes(x=type,y=price)) +
geom_boxplot()``````
#### 2.0.2.4 Two numerical
-Scatterplots
Form: Is
Direction: If increasing values on one axis correspond with increasing values on the other, then the direction is positive; if increasing values on
Strength:
Try it yourself!
### 2.0.3 Multivariate
With more than two variables,
Does my data show signs of clustering?
Does a pattern exist for some categories by not others? | 1,008 | 2,956 | {"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-2024-33 | latest | en | 0.502437 |
https://www.studypool.com/discuss/1099281/multi-step-word-problems-involving-multplication-of-fractions?free | 1,481,004,217,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541883.3/warc/CC-MAIN-20161202170901-00300-ip-10-31-129-80.ec2.internal.warc.gz | 953,835,075 | 14,558 | ##### Multi step word problems involving multplication of fractions
Algebra Tutor: None Selected Time limit: 1 Day
There were
runners to start a race. In the first half of the race, 1/3 of them dropped out. In the second half of the race, 2/3 of the remaining runners dropped out. How many runners finished the race?
Aug 1st, 2015
Total runners = 72
Runners dropped in the first half= (1/3) of 72 =(1/3)*72= 24
Runners remaining for second half = 72 - 24 = 48
Runners dropped in the second half = (2/3) of 48= (2/3)*48= 32
Therefore,
Runners who finished the race = 48 - 32 = 16
Aug 1st, 2015
...
Aug 1st, 2015
...
Aug 1st, 2015
Dec 6th, 2016
check_circle | 221 | 667 | {"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-2016-50 | longest | en | 0.961076 |
http://forum.arduino.cc/index.php?topic=72119.msg539915 | 1,475,280,289,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662430.97/warc/CC-MAIN-20160924173742-00087-ip-10-143-35-109.ec2.internal.warc.gz | 99,672,183 | 12,899 | Go Down
### Topic: Multiplexed LEDs without resistor? (Read 2359 times)previous topic - next topic
#### magagna
##### Sep 12, 2011, 09:07 pm
I'm planning a 24x16 LED matrix using these:
http://www.us.kingbright.com/images/catalog/SPEC/APT3216SYCK.pdf
The sheet says forward voltage is 2v and peak forward current is 175mA @ 10% duty cycle. The LED resistor calculator says I'll need 18 ohm resistors for this.
If my duty cycle will be 6.25% (1/16th instead of 1/10th), is it a bad idea if I don't use any resistor to limit current on the LEDs?
Thanks
http://en.wiktionary.org/wiki/magagna <-- My last name. Pretty apt.
#### Noorman
#1
##### Sep 12, 2011, 09:13 pm
I would not take a chance on that one.
A program hickup can ruin your project, and that for a 1 cent component.
Use the components as calculations indicate and do not gamble with that.
2B || !2B
... bonding electrons and bits!
#2
##### Sep 12, 2011, 09:53 pm
Peak Forward Current: 175 mA, 1/10 Duty Cycle, >>> 0.1ms Pulse Width<<<
How wide of a pulse were you planning?
Were you planning to run from an ATMega's 40mA ABSOLUTE MAX ( and not intended for long term use either) output pin, or something else?
Designing & building electrical circuits for over 25 years. Screw Shield for Mega/Due/Uno, Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at my website.
#### magagna
#3
##### Sep 12, 2011, 11:24 pm
Thanks, I hadn't considered that. I want to use pwm but if I'm doing my math right I'll need > 625 hertz so I don't exceed 0.1ms / pulse. I'll have to check that out.
If I can get a refresh rate > 625 hz will it be OK to skip the resistors?
As far as driving the LEDs, I'm using a modified verison of Evil Mad Scientist's Peggy2: the rows will be driven by 4 ATMega pins demultiplexed to 16 lines via a pair of 74hc138s 3:8 decoders, which in turn connect to PNP transistors to supply enough current. The columns will be driven by 3 74hc595 shift registers, which will drive 3 ULN2803a Darlington arrays as the 595's can't supply enough power to continuously light 8 LEDs. (EMS's Peggy2 uses a STP16DP05 in place of the 595 + ULN2803, but I can't find that chip cheap enough to justify using it over the two chip solution).
http://en.wiktionary.org/wiki/magagna <-- My last name. Pretty apt.
#### Grumpy_Mike
#4
##### Sep 12, 2011, 11:28 pm
Quote
If I can get a refresh rate > 625 hz will it be OK to skip the resistors?
No, it is NEVER ok to not use a resistor, see:-
http://www.thebox.myzen.co.uk/Tutorial/LEDs.html
Quote
The LED resistor calculator says I'll need 18 ohm resistors for this.
No it doesn't you are putting the wrong numbers into it.
#### magagna
#5
##### Sep 12, 2011, 11:39 pmLast Edit: Sep 12, 2011, 11:41 pm by Chris Magagna Reason: 1
Hi Mike, I use this calculator:
http://ledz.com/?p=zz.led.resistor.calculator
Supply voltage = 5v, voltage drop = 2v, desired current = 175mA, it says resistor = 18 ohm.
Is this wrong?
As far as needing the resistors -- I was pretty sure I do based on that web page (and some postings from you) but I've seen several multiplexed LED designs that don't use them so I wanted to check.
Thanks
http://en.wiktionary.org/wiki/magagna <-- My last name. Pretty apt.
#### Grumpy_Mike
#6
##### Sep 12, 2011, 11:46 pm
While the figure might be right using such a low resistance value is meaningless. This is because the forward voltage drop is not that stable and repeatable from device to device. And it changes over time. The resistor bit is the linear part of the circuit and once you get below 50R or so it becomes an increasingly tiny part with the variations in forward voltage drop making drastic changes to the current. Try calculating the current with a resistor 5% lower than 18R, I assume you will use 5% resistors. Now calculate the current at the maximum and minimum forward volts drop given in the data sheet. Once you start trying to go over 100mA with an LED it is time to think about constant current drive.
#### magagna
#7
##### Sep 12, 2011, 11:48 pm
Thank you for the information and advice Mike.
http://en.wiktionary.org/wiki/magagna <-- My last name. Pretty apt.
#### magagna
#8
##### Sep 13, 2011, 04:33 am
Mike et al,
I've been researching your advice on constant current drivers and think I can use 3 TLC5916's
(http://focus.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=TLC5916&fileType=pdf)
in place of the shift registers, Darilington arrays, and resistors.
Since the chip can handle 960mA each LED should get 120mA if I use a 160 ohm resistor (if I did the math right).
Please let me know if I'm missing anything; thanks again for your help.
http://en.wiktionary.org/wiki/magagna <-- My last name. Pretty apt.
#### Grumpy_Mike
#9
##### Sep 26, 2011, 05:02 am
The resistor value looks right. However running a chip at the maximum current stated in the data sheet rarely works because what you also have to consider is the power dissipation in the chip. Often the maximum current figure is not obtainable because the chip overheats first. You can calculate the maximum current based on the maximum chip temperature you want to use.
See this for detailed:-
http://www.thebox.myzen.co.uk/Tutorial/Power.html
#### magagna
#10
##### Sep 27, 2011, 02:33 am
Hi Mike,
Thanks again for all your help on my projects. I've been reading your links (and some of the other great documents you've written) and now (hopefully) have a better understanding of how everything fits together.
Chris
http://en.wiktionary.org/wiki/magagna <-- My last name. Pretty apt.
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Please enter a valid email to subscribe | 1,564 | 5,627 | {"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-2016-40 | longest | en | 0.874235 |
https://blogs.mathworks.com/student-lounge/2018/02/01/hydraulic-decoupled-suspension-part1/?from=en | 1,719,331,821,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00435.warc.gz | 116,563,021 | 26,935 | Hydraulic decoupled suspension – the journey starts with a bold idea
Great ideas don’t always stand the test of time. It may be due to some drawbacks that knock out the great advantage, or it simply is too expensive or difficult to implement. The following article will share a contradicting example. Enjoy this engineering success story of originality and rock-solid engineering skills.
Timothy Novotny and Alex Hönger, from AMZ Racing of ETH Zurich, presented their concept of a hydraulic decoupled suspension at the Formula Student Germany Workshop in October 2017 [video link]. AMZ Racing is one of the teams where model-based design is part of their DNA for racecar development. They use MathWorks products for a multitude of applications such as lap-time simulation, control design, optimize trajectory planning for their driverless car and many more.
For that specific project, the team accepted significant risk by allowing Timothy and Alex to pursue their bold idea from early stage concept until functioning in a winning racecar. The result is a proof what bright minds in conjunction with a powerful team and its partners can achieve.
Some Basics
The primary goals of a racecar suspension can be summed up briefly. It should maximize overall grip while minimizing wheel load variations. Furthermore, it should reliably exhibit predictable and driver-friendly behavior.
Movements of an acting suspension can be described by 4 modes: Pitch, Heave, Roll and Warp. Let’s use our imagination and assume a street bump just ahead of the front left wheel. The reaction force on the suspension will try to lift the whole chassis (heave), lift the front axle (pitch), lift both left wheels (roll), and twist the front axle relative to the rear axle (warp) – all at the same time.
1. Pitch is induced mainly through braking and acceleration. Ideally, pitch stiffness would show an approximately linear force-displacement behavior.
2. Heave results primarily from aerodynamics loads. The race engineer would want it to act progressively, i.e. increasing spring stiffness with increasing load.
3. Roll typically happens while cornering. For the sake of balancing, rolling stiffness should be linear and happen symmetrically between left and right.
4. Warp results from maneuvering and track irregularities (see illustration below). Ideally, this mode should be unsprung. This means that movements should be allowed to happen freely, without a spring or damper balancing the deformation. Of course, the tires are not rigid bodies ?.
Concept Idea
The key idea that Timothy and Alex had was to fully decouple modes. They imagined a system with 4 degrees of freedom, one for each mode, where stiffness and damping can be adjusted independently. This would simplify set-up and lower the single wheel stiffness by at least 25%. The team also could better leverage the potential of their already implemented magneto rheological fluid (MRF) dampers and integrate active suspension components.
Let’s revisit their thought process for an early stage concept. A main drawback of common suspension concepts is that decoupling of heave from pitch and roll from warp cannot be done without adding a connection between front and rear axle. Adding further rods was dismissed because of its potential added weight, the negative impact on the package, and the large number of attachment points it would require.
Timothy and Alex thought about various hybrid mechanical-hydraulic systems before coming up with the idea of a fully hydraulic suspension. See the illustration below and follow this link for an animated GIF.
Each wheel would have its own hydraulic system (see colors in the figure above). All elements that share a color would be connected to each other. Consequently, there are no connections between elements of different colors. Certainly, there would be a central unit comprised of four elements: one for each mode. Admittedly, in the figure above, there is a three-part central unit. This results from the combination of warp and roll elements through a lever arm. In an earlier concept, the central unit had different elements for roll and warp as can be seen below.
The warp element is a bit special. There is no spring because warp should be unsprung.
There are many added benefits of the lever arm configuration. The system is presumably lighter and allows simpler tuning of the roll balance by adjusting the length of the left and right lever arms. Roll balance can be described as a connection between roll and warp. Roll movements actuate warp and thus increase tire load on the wheels of one diagonal.
I hope you click back to the racing lounge for the second part of this story. You will read about detailed design, manufacturing and testing … and see some awesome videos of the system in action.
| | 940 | 4,820 | {"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} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.942721 |
http://slidegur.com/doc/23272/graphics-processing-unit--gpu--architecture-and-programming | 1,481,339,221,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542938.92/warc/CC-MAIN-20161202170902-00327-ip-10-31-129-80.ec2.internal.warc.gz | 260,972,331 | 10,211 | Graphics Processing Unit (GPU) Architecture and Programming
```Graphics Processing Unit (GPU)
Architecture and Programming
TU/e 5kk73
/ʤɛnju:/ /jɛ/
Zhenyu Ye
Henk Corporaal
2011-11-15
System Architecture
GPU Architecture
NVIDIA Fermi, 512 Processing Elements (PEs)
What Can It Do?
Render triangles.
NVIDIA GTX480 can render 1.6
billion triangles per second!
ref: "How GPUs Work", http://dx.doi.org/10.1109/MC.2007.59
Single-Chip GPU v.s. Fastest Super Computers
ref: http://www.llnl.gov/str/JanFeb05/Seager.html
GPUs Are In Top Supercomputers
The Top500 supersomputer ranking in June 2011.
ref: http://top500.org
GPUs Are Also Green
The Green500 supersomputer ranking in June 2011.
ref: http://www.green500.org
The Gap Between CPU and GPU
Note: This is from the perspective of NVIDIA.
ref: Tesla GPU Computing Brochure
The Gap Between CPU and GPU
•
Application performance benchmarked by Intel.
ref: "Debunking the 100X GPU vs. CPU myth", http://dx.doi.org/10.1145/1815961.1816021
In This Lecture, We Will Find Out...
•
•
What is the archticture in GPUs?
How to program GPUs?
Don't worry, we will start from C and RISC!
int A[2][4];
for(i=0;i<2;i++){
for(j=0;j<4;j++){
A[i][j]++;
}
}
Assembly
code of
inner-loop
lw r0, 4(r1)
sw r0, 4(r1)
Programmer's
view of RISC
Most CPUs Have Vector SIMD Units
Programmer's view of a vector SIMD, e.g. SSE.
Let's Program the Vector SIMD
Unroll inner-loop to vector operation.
int A[2][4];
int A[2][4];
for(i=0;i<2;i++){
for(i=0;i<2;i++){
for(j=0;j<4;j++){
movups xmm0, [ &A[i][0] ] // load
A[i][j]++;
}
movups [ &A[i][0] ], xmm0 // store
}
}
int A[2][4];
for(i=0;i<2;i++){
for(j=0;j<4;j++){
A[i][j]++;
}
}
Looks like the previous example,
but SSE instructions execute on 4 ALUs.
Assembly
code of
inner-loop
lw r0, 4(r1)
sw r0, 4(r1)
How Do Vector Programs Run?
int A[2][4];
for(i=0;i<2;i++){
movups xmm0, [ &A[i][0] ] // load
movups [ &A[i][0] ], xmm0 // store
}
CUDA Programmer's View of GPUs
A GPU contains multiple SIMD Units.
CUDA Programmer's View of GPUs
A GPU contains multiple SIMD Units. All of them can access global memory.
What Are the Differences?
SSE
2. The "Shared Memory" spaces
GPU
Grid
contains
contains
Let's Start Again from C
int A[2][4];
for(i=0;i<2;i++){
for(j=0;j<4;j++){
A[i][j]++;
}
convert into CUDA
}
int A[2][4];
__device__ kernelF(A){ // all threads run same kernel
i = blockIdx.x; // each thread block has its id
A[i][j]++; // each thread has a different i and j
}
thread 3 of block 1 operates
on element A[1][3]
int A[2][4];
__device__ kernelF(A){ // all threads run same kernel
i = blockIdx.x; // each thread block has its id
A[i][j]++; // each thread has a different i and j
}
int A[2][4];
kernelF<<<(2,1),(4,1)>>>(A);
__device__ kernelF(A){
i = blockIdx.x;
A[i][j]++;
}
mv.u32 %r0, %ctaid.x
// r0 = i = blockIdx.x
mv.u32 %r1, %ntid.x
// r2 = j = threadIdx.x
mv.u32 %r2, %tid.x
mad.u32 %r3, %r2, %r1, %r0 // r3 = i * "threads-per-block" + j
ld.global.s32 %r4, [%r3] // r4 = A[i][j]
// r4 = r4 + 1
st.global.s32 [%r3], %r4 // A[i][j] = r4
Utilizing Memory Hierarchy
Example: Average Filters
Average over a
3x3 window for
a 16x16 array
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
tmp = (A[i-1][j-1]
+ A[i-1][j]
...
+ A[i+1][i+1] ) / 9;
A[i][j] = tmp;
}
Utilizing the Shared Memory
Average over a
3x3 window for
a 16x16 array
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
}
Utilizing the Shared Memory
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
i = threadIdx.y; allocate shared mem
smem[i][j] = A[i][j]; // load to smem
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
}
However, the Program Is Incorrect
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
}
Let's See What's Wrong
scheduled on 8 PEs.
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
}
Let's See What's Wrong
scheduled on 8 PEs.
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
} Threads starts window operation as soon as it
Let's See What's Wrong
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
Threads starts window operation as soon as it
Some elements in the window are not}
scheduled on 8 PEs.
How To Solve It?
scheduled on 8 PEs.
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
}
Use a "SYNC" barrier!
scheduled on 8 PEs.
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
__sync(); // threads wait at barrier
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
+ smem[i+1][i+1] ) / 9;
}
Use a "SYNC" barrier!
scheduled on 8 PEs.
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
__sync(); // threads wait at barrier
A[i][j] = ( smem[i-1][j-1]
hit barrier.
...
+ smem[i+1][i+1] ) / 9;
}
Use a "SYNC" barrier!
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
__shared__ smem[16][16];
smem[i][j] = A[i][j]; // load to smem
__sync(); // threads wait at barrier
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
All elements in the window are loaded
+ smem[i+1][i+1] ) / 9;
}
scheduled on 8 PEs.
Review What We Have Learned
1. Single Instruction Multiple Thread (SIMT)
2. Shared memory
Vector SIMD can also have shared memory.
For Example, the CELL architecture.
Q: What are the fundamental differences between
the SIMT and vector SIMD programming models?
Take the Same Example Again
Average over a
3x3 window for
a 16x16 array
Assume vector SIMD and SIMT
both have shared memory.
What is the difference?
Vector SIMD v.s. SIMT
int A[16][16]; // global memory
__shared__ int B[16][16]; // shared mem
kernelF<<<(1,1),(16,16)>>>(A);
for(i=0;i<16;i++){
__device__
kernelF(A){
__shared__ smem[16][16];
for(j=0;i<4;j+=4){
movups xmm0, [ &A[i][j] ]
movups [ &B[i][j] ], xmm0 }}
smem[i][j] = A[i][j]; // load to smem
for(i=0;i<16;i++){
__sync(); // threads wait at barrier
for(j=0;i<4;j+=4){
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j]
...
...
divps xmm1, 9 }}
+ smem[i+1][i+1] ) / 9;
for(i=0;i<16;i++){
for(j=0;i<4;j+=4){
addps [ &A[i][j] ], xmm1 }}
}
Vector SIMD v.s. SIMT
int A[16][16];
__shared__ int B[16][16];
kernelF<<<(1,1),(16,16)>>>(A);
__device__ kernelF(A){
Programmers schedule __shared__ smem[16][16];
for(j=0;i<4;j+=4){
operations on PEs.
movups xmm0, [ &A[i][j] ]
# of PEs in HW is
movups [ &B[i][j] ], xmm0 }}
transparent to
smem[i][j]
=
A[i][j];
for(i=0;i<16;i++){
programmers.
You need to know how __sync(); // threads wait at barrier
for(j=0;i<4;j+=4){ many PEs are in HW.
A[i][j] = ( smem[i-1][j-1]
+ smem[i-1][j] Programmers
give up exec.
...
...
ordering to HW.
Each inst. is executed by
+ smem[i+1][i+1] ) / 9;
divps xmm1, 9 }} all PEs in locked step.
}
for(i=0;i<16;i++){
for(i=0;i<16;i++){
for(j=0;i<4;j+=4){
addps [ &A[i][j] ], xmm1 }}
CUDA programmers let the SIMT
hardware schedule operations on PEs.
Review What We Have Learned
Programmers convert data level parallelism (DLP) into thread
level parallelism (TLP).
Example of Implementation
Note: NVIDIA may use a more
complicated implementation.
Example
0x0008: sub r3, r4, r5
Assume warp 0 and
warp 1 are scheduled
for execution.
0x0008: sub r3, r4, r5
r1 for warp 0
r4 for warp 1
Buffer Src Op
0x0008: sub r3, r4, r5
Push ops to op collector:
r1 for warp 0
r4 for warp 1
0x0008: sub r3, r4, r5
r2 for warp 0
r5 for warp 1
Buffer Src Op
0x0008: sub r3, r4, r5
Push ops to op collector:
r2 for warp 0
r5 for warp 1
Execute
0x0008: sub r3, r4, r5
Compute the first 16
Execute
0x0008: sub r3, r4, r5
Compute the last 16
Write back
0x0008: sub r3, r4, r5
Write back:
r0 for warp 0
r3 for warp 1
A Brief Recap of SIMT Architecture
• Threads in the same warp are scheduled
•
together to execute the same instruction.
A warp of 32 threads can be executed on 16
(8) PEs in 2 (4) cycles by time-multiplexing.
Summary
•
•
The CUDA programming model.
The SIMT architecture.
Reference
•
•
•
•
NVIDIA Tesla: A Unified Graphics and Computing Architecture, IEEE | 3,310 | 8,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-50 | longest | en | 0.576935 |
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