url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://dzone.com/articles/r-with-powerbi-a-step-by-step-guide | 1,580,159,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251728207.68/warc/CC-MAIN-20200127205148-20200127235148-00163.warc.gz | 416,000,052 | 37,375 | {{announcement.body}}
{{announcement.title}}
# R With PowerBI: A Step-by-Step Guide
DZone 's Guide to
# R With PowerBI: A Step-by-Step Guide
### An in-depth guide to integrating your R scripts with Microsoft's PowerBI software to create eye-popping visualizations for your data.
· Big Data Zone ·
Free Resource
Comment (0)
Save
{{ articles[0].views | formatCount}} Views
There's been a lot of interest everywhere on how to integrate R scripts with Microsoft PowerBI dashboards. Here is a step by step guide to this.
Lets assume that you have some readymade R code available, for example, with the ggplot2 library. Lets use the following scripts to perform analytics using CHOL data.
1. Open R studio or R Package (CRAN) & install ggplot2 library first.
2. Paste the following R script and execute it:
3. ``````install.packages(‘ggplot2’)
library(ggplot2)
#Take the column “AGE” from the “chol” dataset and make a histogram it
qplot(chol\$AGE , geom = “histogram”)
ggplot(data-chol, aes(chol\$AGE)) + geom_histogram()``````
You should be able to see the visual output like this.
4. Next, execute the following pieces of R code to find out the binwidth argument using the ‘qplot()‘ function:
5. ``````qplot(chol\$AGE,
geom = “histogram”,
binwidth = 0.5)``````
6. Lets get some help from the hist() function in R:
7. ``````#Lets take help from hist() function
qplot(chol\$AGE,
geom=”histogram”,
binwidth = 0.5,
main = “Histogram for Age”,
xlab = “Age”,
fill=I(“blue”))``````
8. Now, add an I() function for nested colors:
9. ``````#Add col argument, I() function where nested color.
qplot(chol\$AGE,
geom=”histogram”,
binwidth = 0.5,
main = “Histogram for Age”,
xlab = “Age”,
fill=I(“blue”),
col=I(“red”))``````
10. Next, adjust ggplot2 a little by using the following code:
11. ``````#Adjusting ggplot
ggplot(data=chol, aes(chol\$AGE)) +
geom_histogram(breaks=seq(20, 50, by = 2),
col=”red”,
fill=”green”,
alpha = .2) +
labs(title=”Histogram for Age”) +
labs(x=”Age”, y=”Count”) +
xlim(c(18,52)) +
ylim(c(0,30))``````
12. Plot a bar graph with the following code:
13. ``````#Plotting Bar Graph
qplot(chol\$AGE,
geom=”bar”,
binwidth = 0.5,
main = “Bar Graph for Mort”,
xlab = “Mort”,
fill=I(“Red”))``````
14. Next, open the PowerBI desktop tool. You can download it for free from this link. Now, click on the Get Data tab to start exploring and connect with an R dataset.
15. If you already have R installed on the same system as PowerBI, you just need to paste the R scripts in the code pen. Otherwise you need to install R in the system where you are using the PowerBI desktop like this:
16. Next, you can also choose the ‘custom R visual’ in PowerBI desktop visualizations and provide the required R scripts to build visuals and finally click ‘Run’.
17. Build all the R function visuals by following the same steps and save the dashboard.
18. You can refresh an R script in Power BI Desktop. When you refresh an R script, Power BI Desktop runs the R script again in the Power BI Desktop environment.
Related Refcard:
Topics:
r ,powerbi ,analytics ,big data ,data visualization
Comment (0)
Save
{{ articles[0].views | formatCount}} Views
Published at DZone with permission of Anindita Basak , DZone MVB. See the original article here.
Opinions expressed by DZone contributors are their own. | 940 | 3,314 | {"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-2020-05 | longest | en | 0.681861 |
https://www.physicsforums.com/threads/pulleys-and-conservation-of-energy.376854/ | 1,511,344,561,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00227.warc.gz | 868,071,761 | 14,505 | # Pulleys and Conservation of Energy
1. Feb 9, 2010
### zx95
1. In the system shown in the figure, the pulleys on the left and right are fixed, but the pulley in the center can move left or right. The two hanging masses are identical, and the pulleys and ropes are all massless. Find the upward acceleration of the mass on the left, in terms of g.
2. I have to solve this problem using only Conservation of Energy. The following equations are relevant: K = (1/2)mv^2 and U = mgy
3. I began by setting up a general Conservation of Energy equation.
I'm calling the left mass m1 and the right mass m2
m1 = m2 = m
ΔU1 + ΔU2 + ΔK2 = ΔK1
mgΔy1 + mgΔy2 + (1/2)mv2^2 = (1/2)m1^2
(( v^2 = 2aΔy)) Substitute kinematic equation for velocity
gΔy1 + gΔy2 + aΔy2 = aΔy1
(( Δy1 = -2Δy2)) Substitute Δy1 in terms of Δy2
-2gΔy2 + gΔy2 + aΔy2 = -2aΔy2
-2g + g = -2a - a
-g = -3a
(1/3)g = a
This however, doesn't seem to be correct. Any hints on where I may have gone wrong? Thanks!
2. Feb 9, 2010
### PhanthomJay
Looking at this quickly, a couple of points:
(1) the acceleration of the masses, and hence their speeds, are not the same; one of them accelerates at twice the rate of the other.
(2) Your conservation of energy equation is wrong. It's delta K_total plus delta U_total =0. I'm not sure why you have one of the delta K's on the other side of the equation.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 463 | 1,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2017-47 | longest | en | 0.902728 |
wahaxowedetaju.esteindesign.xyz | 1,606,823,198,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00441.warc.gz | 547,684,945 | 9,059 | Last edited by Mazuhn
Sunday, May 10, 2020 | History
2 edition of Extensions of palm"s Theorem found in the catalog.
Extensions of palm"s Theorem
M. J. Carrillo
# Extensions of palm"s Theorem
## by M. J. Carrillo
Written in English
Subjects:
• Inventory control -- Mathematical models.,
• Poisson processes.
• Edition Notes
The Physical Object ID Numbers Statement Manuel J. Carrillo. Contributions Rand Corporation. Pagination 6 p. ; Open Library OL16591081M
Fan K. () Extensions of two fixed point theorems of F. E. Browder. In: Fleischman W.M. (eds) Set-Valued Mappings, Selections and Topological Properties of 2x. Lecture Notes in Mathematics, vol Cited by: if i read it correctly, yes this is true and well known probably as long as the subject itself. a reference for it, (the first book i opened from my shelf), is the harvard notes by Richard Brauer, on Galois Theory, from , revised , p in paragraph 9 titled "The main theorem of Galois theory".
If I understand correctly, you want your theorem-like environments to be numbered within chapters. To do this, you have to add the optional parameter chapter to your theorem definitions, for example: \newtheorem{definition}{Definition}[chapter] \newtheorem{lemma}{Lemma}[chapter]. \$\begingroup\$ Along with TSS, look at G-R's companion book (self-contained) "Coherent Analytic Sheaves". One lesson it shows is the usefulness of commutative algebra (applied to the noetherian local rings) in the study of complex-analytic spaces, and the simplifications attained by leaving the world of manifolds and allowing the objects of study to have singularities (super-useful for.
(This is an extension of Rice's theorem that "Every nontrivial property of the r.e. sets is undecida Stack Exchange Network Stack Exchange network consists of Q&A communities including Stack Overflow, the largest, most trusted online community for developers to . Theorem Let V = Rd, n 2 and suppose that there is no set F of n+ 1 points in a ball of radius bso that the center belongs to Fand the distance between any two points of Fexceeds 1. If there is a lattice in V with minimum distance 1 so that there are npoints of it in a ball of radius smaller than bcentered at a lattice point, then C(V;1;b) = n.
You might also like
By-laws of Washington State 4-H Fair Board, Inc.
By-laws of Washington State 4-H Fair Board, Inc.
handbook of heraldry
handbook of heraldry
Ocean extremes
Ocean extremes
Italian modern architecture and the vernacular tradition: The aesthetics of morality.
Italian modern architecture and the vernacular tradition: The aesthetics of morality.
The Biglow papers
The Biglow papers
Florida State and Local Government
Florida State and Local Government
Food quality and nutrition
Food quality and nutrition
Nominations of Robert L. Hardesty and Hung Wai Ching to the Board of Governors, U.S. Postal Service
Nominations of Robert L. Hardesty and Hung Wai Ching to the Board of Governors, U.S. Postal Service
Astronomical cuneiform texts
Astronomical cuneiform texts
Improper Payments Information Act of 2002
Improper Payments Information Act of 2002
### Extensions of palm"s Theorem by M. J. Carrillo Download PDF EPUB FB2
Get this from a library. Extensions of Palm's Theorem: a review. [M J Carrillo; Rand Corporation.] -- This paper reviews the transient behavior of the M/G/(infinity) queue with nonhomogeneous Poisson or compound Poisson input and nonstationary service distribution.
In the case of nonhomogeneous. Extension theorem may refer to. Carathéodory's extension theorem - a theorem in measure theory, named after the Greek mathematician Constantin Carathéodory; Dugundji extension theorem - a theorem in topology, named after the American mathematician James Dugundji; Extension Lemma - a lemma in topology (resp.
functional analysis), related to the Tietze extension theorem. Get this from a library. Generalizations of Palm's theorem and Dyna-METRIC's demand and pipeline variability. [M J Carrillo; Rand Corporation.; Project Air Force (U.S.)] -- "Palm's theorem is a useful tool in modeling inventory problems in logistics models such as METRIC and Mod-METRIC.
However, to fit its limited domain of applicability, time-dependent customer arrival. Problems on the analytic continuation of functions are, first of all, related to extension theorems. An example of a theorem on the existence of a continuous extension of a continuous function is the Brouwer–Urysohn theorem: If is a closed subset of a normal space and is a continuous real-valued bounded function, then there exists a.
The book starts with a very clear presentation of the principles of Galois theory in two chapters: "Algebraic extensions" and "Galois theory", compareble to Artins short book Galois Theory: Lectures Delivered at the University of Notre Dame (Notre Dame Mathematical Lectures, Number 2).
The next three chapters are in essence about algebraic number fields, although he only defines these objects Cited by: The Hahn–Banach theorem can be deduced from the M.
Riesz extension theorem. Let V be a linear space, and let N be a sublinear function on φ be a functional on a subspace U ⊂ V that is dominated by N: ≤ (), ∈.The Hahn–Banach theorem asserts that φ can be extended to a linear functional on V that is dominated by N.
To derive this from the M. Riesz extension theorem, define a. Title: Palm's Theorem for Nonstationary Processes Author: Gordon Crawford Subject: Like most models for calculating stock requirements, the models used by the Air Force to calculate requirements and allocations have traditionally assumed that the failure process generates arrivals approximating a steady-state Poisson arrival process!£.
Extension to the Pythagorean Theorem Variations of Theorem 66 can be used to classify a triangle as right, obtuse, or acute. Theorem If a, b, and c represent the lengths of the sides of a triangle, and c is the longest length, then the triangle is obtuse if c 2 > a 2 + b 2, and the triangle is acute if c 2.
Algebraic Geometry Fall Extension theorems for homomorphisms In this note, we prove some extension theorems for homomorphisms from rings to algebraically closed fields.
The prototype is the following result: Theorem 1 (Extension theorem for algebraic extensions). If L/Kis an algebraic extension of fields, then any embedding σof K into an.
Secret Sharing Extensions based on the Chinese Remainder Theorem Kamer Kaya, Ali Ayd n Sel˘cuk Department of Computer Engineering Bilkent University Ankara,Turkey fkamer,[email protected] Abstract. In this paper, we investigate how to achieve veri able secret sharing (VSS) schemes by using the Chinese Remainder Theorem (CRT).Cited by: 6.
JOURNAL OF FUNCTIONAL ANALYSIS 5, () Extensions of the F. and M. Riesz Theorem I. GLICKSBERG* Department of Mathematics, University of Washington, Seattle, Washington Received Septem Some time ago Karel de Leeuw and the author extended the classical F. and M.
Riesz theorems [12] to a setting which covered analytic measures on compact groups with Cited by: 3. The Continuous Extension Theorem This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site. Fold Unfold.
years that influenced Andrew Wiles' proof of Fermat's Last Theorem. Due to its breadth and the fact that the book is quite short, the author devotes only a few short pages to each mathematician in the survey/5.
Extensions of the Critical Theorem. In this section, we will demonstrate how the Critical Theorem may be generalised in a simple manner.
For this purpose, some technical definitions must be presented. A structure of order 1 on a multiset X is a finite multiset ς consisting of copies of some elements from X, either unordered or totally by: The Central Limit Theorem.
The Central Limit Theorem (CLT) is a powerful and important result of mathematical analysis. In its standard form it says that if a stochastic variable x has a finite variance then the distribution of the sums of n samples of x will approach a normal distribution as the sample size n increases without limit.
Next we show the main theorem. Theorem Let B be a Galois extension of BG with Galois group G. Then B satisfies the fundamental theorem if and only if either B = BGe ⊕ BG(1− e) where e and 1− e are minimal central idempotents in B and G = {1,g} such that g(e)=1−e,orB is indecomposable satisfying the fundamental theorem.
by: 2. known as the Grunwald{Wang theorem, with a description of all counter-examples [6, 12]. We pro-vide an overview of the Grunwald{Wang theorem and some of the intricacies of 2-power cyclotomic extensions that lead to the counter-examples to Question in Chapter 2.
Note that Question asks when Xn ahas a root in a number eld Kfor a2K. Our central theme is Galvin's resolution of the Dinitz problem (Galvin.
Comb. Theory, Ser. B 63(1),). We survey the related work of Alon and Tarsi (Combinatorica 12(2)) and H\"{a}ggkvist and Janssen (Combinatorics, Probability \& Computing 6(3)). We then prove two new extensions of Galvin's : Maxwell Levit. It's another theorem that gives a condition for a function to have a continuous extension, but it's not a condition that will be useful here.
\$\endgroup\$ – Nate Eldredge Mar 12 '11 at \$\begingroup\$ The question had been removed for some reason, so I restored it to an earlier version.
\$\endgroup\$ – Jonas Meyer Mar 12 '11 at Theorem For a given point and circle, the product of the lengths of the two segments from the point to the circle is constant along any line through the point and circle. d Theorem The measure of an angle formed by two lines that intersect inside a circle is half the sum of the measures of the intercepted arcs.
Theorem File Size: KB. The second question is actually about the first edition of the book. He states a theorem that if ##I\subseteq E\subseteq K## with ##K## a finite extension of ##E## and ##E## a finite extension of ##I##, and ##K## is a splitting field over ##E## and ##E## is a splitting field over ##I##, then ##K## is a splitting field over ##I##.1 Integral ring extensions Let Sbe commutative ring (with 1) and let Rbe a subring.
We call R Sa ring extension. An element s2Sis called integral over Rif there exists a monic polynomial f2R[x] such that f(s) = 0. If every s2Sis integral over R, then Sis said to be integral over R.
Observations: If sis integral over R, then it is also algebraic File Size: KB.It took me some time to understand that algebraic numbers have no more existence apart from being elements of such field extensions. \$\endgroup\$ – Paramanand Singh May 9 '17 at \$\begingroup\$ @David Wheeler, thank you! | 2,559 | 10,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-50 | longest | en | 0.858091 |
https://leetcode.ca/2018-09-06-1011-Capacity-To-Ship-Packages-Within-D-Days/ | 1,695,843,189,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510319.87/warc/CC-MAIN-20230927171156-20230927201156-00374.warc.gz | 396,500,289 | 8,562 | ##### Welcome to Subscribe On Youtube
Formatted question description: https://leetcode.ca/all/1011.html
# 1011. Capacity To Ship Packages Within D Days (Medium)
A conveyor belt has packages that must be shipped from one port to another within D days.
The i-th package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.
Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.
Example 1:
Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
Output: 15
Explanation:
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10
Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.
Example 2:
Input: weights = [3,2,2,4,1,4], D = 3
Output: 6
Explanation:
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4
Example 3:
Input: weights = [1,2,3,1,1], D = 4
Output: 3
Explanation:
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1
Note:
1. 1 <= D <= weights.length <= 50000
2. 1 <= weights[i] <= 500
Companies:
Related Topics:
Array, Binary Search
## Solution 1.
• class Solution {
public int shipWithinDays(int[] weights, int D) {
int sum = 0;
int low = 0;
for (int weight : weights) {
sum += weight;
low = Math.max(low, weight);
}
int high = sum;
while (low < high) {
int mid = (high - low) / 2 + low;
int days = daysNeed(weights, mid);
if (days > D)
low = mid + 1;
else
high = mid;
}
return low;
}
public int daysNeed(int[] weights, int capacity) {
int days = 1;
int currentTotal = 0;
int length = weights.length;
for (int i = 0; i < length; i++) {
int weight = weights[i];
if (currentTotal + weight <= capacity)
currentTotal += weight;
else {
days++;
currentTotal = weight;
}
}
return days;
}
}
############
class Solution {
public int shipWithinDays(int[] weights, int days) {
int left = 0, right = 0;
for (int w : weights) {
left = Math.max(left, w);
right += w;
}
while (left < right) {
int mid = (left + right) >> 1;
if (check(mid, weights, days)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
private boolean check(int mx, int[] weights, int days) {
int ws = 0, cnt = 1;
for (int w : weights) {
ws += w;
if (ws > mx) {
ws = w;
++cnt;
}
}
return cnt <= days;
}
}
• class Solution:
def shipWithinDays(self, weights: List[int], D: int) -> int:
def check(capacity):
cnt, t = 1, 0
for w in weights:
if w > capacity:
return False
if t + w <= capacity:
t += w
else:
cnt += 1
t = w
return cnt <= D
left, right = 1, 25000000
while left < right:
mid = (left + right) >> 1
if check(mid):
right = mid
else:
left = mid + 1
return left
############
# 1011. Capacity To Ship Packages Within D Days
# https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/
class Solution:
def shipWithinDays(self, weights: List[int], days: int) -> int:
total = sum(weights)
def good(target):
mmax = s = 0
day = 1
for x in weights:
if s + x > target:
s = x
day += 1
else:
s += x
mmax = max(mmax, s)
return day <= days and mmax <= target
left, right = 1, total
while left < right:
mid = left + (right - left) // 2
if good(mid):
right = mid
else:
left = mid + 1
return left
• func shipWithinDays(weights []int, days int) int {
var left, right int
for _, w := range weights {
if left < w {
left = w
}
right += w
}
return left + sort.Search(right, func(mx int) bool {
mx += left
ws, cnt := 0, 1
for _, w := range weights {
ws += w
if ws > mx {
ws = w
cnt++
}
}
return cnt <= days
})
}
• function shipWithinDays(weights: number[], days: number): number {
let left = 0;
let right = 0;
for (const w of weights) {
left = Math.max(left, w);
right += w;
}
const check = (mx: number) => {
let ws = 0;
let cnt = 1;
for (const w of weights) {
ws += w;
if (ws > mx) {
ws = w;
++cnt;
}
}
return cnt <= days;
};
while (left < right) {
const mid = (left + right) >> 1;
if (check(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
• class Solution {
public:
int shipWithinDays(vector<int>& weights, int days) {
int left = 0, right = 0;
for (auto& w : weights) {
left = max(left, w);
right += w;
}
auto check = [&](int mx) {
int ws = 0, cnt = 1;
for (auto& w : weights) {
ws += w;
if (ws > mx) {
ws = w;
++cnt;
}
}
return cnt <= days;
};
while (left < right) {
int mid = (left + right) >> 1;
if (check(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}; | 1,538 | 4,733 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-40 | longest | en | 0.885838 |
https://singaporemathsource.com/word-problem-wednesday-mr-seow/ | 1,713,861,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818468.34/warc/CC-MAIN-20240423064231-20240423094231-00740.warc.gz | 480,590,392 | 26,403 | # Word Problem Wednesday – Mr. Seow
Summer’s here, but you’re missing your math? Don’t despair – we’ve got you covered. Check the site each week for one whopper of a word problem that’s sure to challenge!
This week’s problem comes from Visible Thinking in Mathematics 5B by Ammiel Wan and Chelsia Loh, published in 2011 by Marshall Cavendish International (Singapore) Private Limited.
##### Mr. Seow borrowed a certain amount from a bank, which charged him an interest of 3.5% per year. If he owed the bank \$4347 at the end of the year, how much did he borrow from the bank?
Submit your solutions and we’ll post all interesting strategies next week.
Last week’s problem and solution:
Cynthia had \$16.75. She withdrew more cash from an ATM before shopping. After spending \$17.50 on a box of cookies and \$23.40 on a box of chocolates, she had \$35.85 left. How much money did she withdraw from the ATM?
Whew! How did you do?
Share:
Passionate about Singapore Math + Teacher Trainer and Coach + Treasure Hunter + Learner. Answer to the ultimate question? 42.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 275 | 1,150 | {"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-2024-18 | latest | en | 0.945504 |
http://www.mathisfunforum.com/post.php?tid=18102&qid=230249 | 1,394,223,791,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999650794/warc/CC-MAIN-20140305060730-00018-ip-10-183-142-35.ec2.internal.warc.gz | 429,601,224 | 5,244 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
You are not logged in.
• Index
• » Help Me !
• » A Multiplying and Dividing Polynomials problem
|
Options
2012-09-08 07:07:09
Yes.
bobbym
2012-09-07 06:11:02
You are taking the compu high course?
2012-09-07 04:22:13
No, not really. Barely. Hardly.
bobbym
2012-09-06 06:37:12
Hi;
You are welcome but are you able to do one on your own?
2012-09-06 06:34:55
Thanks a ton!
bobbym
2012-09-06 06:11:32
21 divided by 7 equals 3.
14 divided by 7 is 2. You leave the x^3 and the x alone. So
2012-09-06 05:44:30
for*
2012-09-06 05:42:19
Thanks bobbym, that's what I needed. Could you show me the work or that?
bobbym
2012-09-05 09:13:30
Hi;
The only thing you can do with that is simplify it unless you want something more.
anonimnystefy
2012-09-05 08:24:36 | 350 | 888 | {"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-2014-10 | longest | en | 0.850393 |
https://mail.haskell.org/pipermail/libraries/2004-October/002472.html | 1,430,029,666,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246653426.7/warc/CC-MAIN-20150417045733-00140-ip-10-235-10-82.ec2.internal.warc.gz | 986,984,932 | 2,337 | Tomasz Zielonka t.zielonka at students.mimuw.edu.pl
Thu Oct 14 02:52:56 EDT 2004
```On Thu, Oct 14, 2004 at 12:36:37AM +0200, Peter Simons wrote:
> Benjamin Franksen writes:
>
> > milliseconds (1/10^6 seconds)
>
> Duh! I meant microseconds of course. Thanks for catching
> that.
Did you consider using floating-point Timeout type, with its
values representing seconds, possibly fractional? This could
have some negative performance impact, but would be elegant IMHO.
I prepared a darcs patch with such changes, added timeoutDelay
which is analogous to threadDelay, but works with Timeout values,
number of microseconds (2^31 microseconds is not such long).
Feel free to reject it :)
Best regards,
Tom
--
.signature: Too many levels of symbolic links
-------------- next part --------------
New patches:
[changed Timeouts to be possibly fractional amount of seconds (type Double)
Tomasz Zielonka <t.zielonka at students.mimuw.edu.pl>**20041014063618] {
hunk ./Childs.hs 25
- , Timeout -- = Int
+ , Timeout -- = Double
+ , timeoutDelay-- :: Timeout -> IO ()
hunk ./Childs.hs 114
--- |Timeouts are given in milliseconds (@1\/10^6@ seconds).
+-- |Timeouts are given in seconds, which can be fractional.
hunk ./Childs.hs 117
-type Timeout = Int
+type Timeout = Double
+
+timeoutDelay :: Timeout -> IO ()
+timeoutDelay t = threadDelay (round (t * 1000000.0))
hunk ./Childs.hs 123
--- 'Nothing' after @n@ milliseconds, otherwise @'Just' a@ is
+-- 'Nothing' after @n@ seconds, otherwise @'Just' a@ is
hunk ./Childs.hs 131
- r <- par (threadDelay n >> return []) (fmap return f)
+ r <- par (timeoutDelay n >> return []) (fmap return f)
hunk ./Childs.hs 137
--- |@sleep n@ @=@ @threadDelay (abs(n) * 1000000)@
+-- |@sleep n@ @=@ @timeoutDelay (fromIntegral (min n 0))@
hunk ./Childs.hs 140
-sleep n = threadDelay (abs(n) * 1000000)
+sleep n = timeoutDelay (fromIntegral (min n 0))
}
[fixed timeoutDelay to handle delays overflowing the number of microseconds in Int
Tomasz Zielonka <t.zielonka at students.mimuw.edu.pl>**20041014063805] {
hunk ./Childs.hs 120
-timeoutDelay t = threadDelay (round (t * 1000000.0))
+timeoutDelay t | t > 100.0 = threadDelay 100000000 >> timeoutDelay (t - 100.0)
+ | otherwise = threadDelay (round (t * 1000000.0))
}
Context:
[initial version
simons at cryp.to**20041012201657]
Patch bundle hash:
c2369d21a3326586b350cc55f693644b72fd08cb
``` | 731 | 2,413 | {"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-2015-18 | longest | en | 0.72449 |
https://wincleaneras.com/1-quadratic-function-is-written-as-y-axxb2-bx-c-where-a-b-and-c-are-real-numbers-and-a-2260-0-a-true-b-false-2-the-graph-of-the-function-y-4x-5-opens-upward-a-true-b-false/ | 1,660,428,475,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00161.warc.gz | 560,112,442 | 13,861 | # 1. Quadratic function is written as y = ax\xb2 + bx + c where a, b, and c are real numbers and a 2260 0., A. True, B. False, 2. The graph of the function y = – 4x -5 opens upward, A. True, B. False, 3. The vertex of y = (x – 1)\xb2 lies along x-axis., A. True, B. False, 4. The graph of y = 2x\xb2 is naerower than the graph of 3x\xb2, A. True, B. False, 5. The coordinates of the vertex of the function y = 2(x – 2)\xb2 + 5 is at (2, 5), A. True, B. False
Posted on
1. Quadratic function is written as y = ax² + bx + c where a, b, and c are real numbers and a ≠ 0.
A. True
B. False
2. The graph of the function y = – 4x -5 opens upward
A. True
B. False
3. The vertex of y = (x – 1)² lies along x-axis.
A. True
B. False
4. The graph of y = 2x² is naerower than the graph of 3x²
A. True
B. False
5. The coordinates of the vertex of the function y = 2(x – 2)² + 5 is at (2, 5)
A. True
B. False
1 A
2A
3A
4B
5B
Step-by-step explanation:
Beginning in the 6th century BC with the Pythagoreans, with Greek mathematics the Ancient Greeks began a systematic study of mathematics as a subject in its own right. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof. | 431 | 1,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-33 | latest | en | 0.865462 |
https://gamedev.stackexchange.com/questions/185230/how-do-i-manually-accelerate-the-rotation-of-a-sphere-in-unity-3d | 1,718,696,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00423.warc.gz | 243,402,792 | 37,530 | # How do I manually accelerate the rotation of a sphere in Unity 3D?
I'm a little new to this so any help will be appreciated.
Context: I'm creating a VR billiard/pool game and trying to get the physics as realistic as possible. I have a cue ball being hit with a cue across a surface with accurate friction values calculated from real world trials.
Problem: The angular acceleration of the balls isn't realistic at all. Therefore the angular velocity is off (this is due to how Unity calculates physics of interaction as a point of contact rather than an area of contact and therefore friction calculation is done at one point causing it to rotate slower).
Question: How can I manually set the angular acceleration to a value in code until the angular acceleration is greater than or equals to horizontal velocity, then the angular acceleration is set to 0? Thanks in advance.
Unity doesn't have an 'angular acceleration' field that you can set. You can set the angular velocity in code, as I showed in your previous question about this. If you need to accelerate or decelerate the angular velocity, you can do this manually in your code.
void Update() {
if (rigidbody.angularVelocity.magnitude < desiredAngularVelocity) {
//scales the acceleration each frame
rigidbody.angularVelocity = rigidbody.angularVelocity * accelerationRate * Time.deltaTime;
}
}
or if you want the angular velocity to change by a constant rate rather than scaling it, I think this will work (haven't tested it):
void Update() {
if (rigidbody.angularVelocity.magnitude < desiredAngularVelocity) {
float angularSpeed = rigidBody.angularVelocity.magnitude;
angularSpeed += acceleration * Time.deltaTime;
rigidbody.angularVelocity = rigidbody.angularVelocity.normalized * angularSpeed;
}
}
Or if you want to use forces, try rigidbody.AddTorque()
rigidbody.AddTorque(transform.right * 10000);
This is more complicated because torque is 3D; you'll have to calculate the correct direction for the torque.
You may wish to review all of the functions available to Rigidbodies.
• Isn't Time.deltaTime simple the time between frames? Did you mean to have a timer in place indicating the time passed? Commented Aug 19, 2020 at 18:48
• @NedSunshine Yes, Time.deltaTime is the amount of time that elapsed over the last frame. In my simple example, multiplying the acceleration rate by Time.deltaTime ensures that the acceleration is smooth even if the application framerate is inconsistent. Commented Aug 19, 2020 at 18:54
• I'm not sure if I'm missing something. This and what I tried before don't accelerate the rotation. Wouldn't I expect this to cause the ball to rotate/roll faster over time? (ignoring the if statement) Commented Aug 19, 2020 at 19:11
• Yes. Try adding Debug.Log("before: " + rigidbody.angularVelocity) before you change the velocity and Debug.Log("after: " + rigidbody.angularVelocity) after you change it; this may help you understand what's going on. Commented Aug 19, 2020 at 19:54
• I found the root of the problem. It's the default limit of 7 to the angular velocity. docs.unity3d.com/ScriptReference/… Commented Aug 19, 2020 at 20:32 | 704 | 3,142 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.880067 |
https://www.convert-measurement-units.com/convert+BTU+h+to+Volt-ampere+reactive.php | 1,642,510,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300849.28/warc/CC-MAIN-20220118122602-20220118152602-00572.warc.gz | 783,640,917 | 12,849 | Convert BTU/h to var (BTU/h to Volt-ampere reactive)
## BTU/h into Volt-ampere reactive
numbers in scientific notation
https://www.convert-measurement-units.com/convert+BTU+h+to+Volt-ampere+reactive.php
## How many Volt-ampere reactive make 1 BTU/h?
1 BTU/h = 0.293 071 070 172 22 Volt-ampere reactive [var] - Measurement calculator that can be used to convert BTU/h to Volt-ampere reactive, among others.
# Convert BTU/h to Volt-ampere reactive (BTU/h to var):
1. Choose the right category from the selection list, in this case 'Power'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'BTU/h'.
4. Finally choose the unit you want the value to be converted to, in this case 'Volt-ampere reactive [var]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '605 BTU/h'. In so doing, either the full name of the unit or its abbreviation can be used. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Power'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '35 BTU/h to var' or '47 BTU/h into var' or '65 BTU/h -> Volt-ampere reactive' or '94 BTU/h = var' or '22 BTU/h to Volt-ampere reactive' or '81 BTU/h into Volt-ampere reactive'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(73 * 51) BTU/h'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '605 BTU/h + 1815 Volt-ampere reactive' or '29mm x 87cm x 48dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 1.944 809 982 302 2×1031. For this form of presentation, the number will be segmented into an exponent, here 31, and the actual number, here 1.944 809 982 302 2. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 1.944 809 982 302 2E+31. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 19 448 099 823 022 000 000 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 916 | 3,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-05 | latest | en | 0.849567 |
https://www.calculatoratoz.com/en/price-of-bond-calculator/Calc-44620 | 1,718,301,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00072.warc.gz | 606,917,237 | 32,648 | ## Price of Bond Solution
STEP 0: Pre-Calculation Summary
Formula Used
Price of Bond = Face Value*(1+Implied Discount Rate)^Holding Period
PB = FV*(1+IDR)^HP
This formula uses 4 Variables
Variables Used
Price of Bond - Price of Bond refers to the amount of money an investor pays to purchase the bond from the issuer or another investor in the secondary market.
Face Value - Face Value is the amount of money that the issuer of a bond agrees to repay the bondholder at the bond's maturity date.
Implied Discount Rate - Implied Discount Rate is the interest rate that equates the present value of the bond's future cash flows to its current market price.
Holding Period - Holding Period refers to the duration of time that an investor owns an asset from the moment it is purchased until it is sold.
STEP 1: Convert Input(s) to Base Unit
Face Value: 95 --> No Conversion Required
Implied Discount Rate: 0.06 --> No Conversion Required
Holding Period: 5 --> No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
PB = FV*(1+IDR)^HP --> 95*(1+0.06)^5
Evaluating ... ...
PB = 127.131429872
STEP 3: Convert Result to Output's Unit
127.131429872 --> No Conversion Required
127.131429872 127.1314 <-- Price of Bond
(Calculation completed in 00.004 seconds)
You are here -
Home »
## Credits
Created by Aashna
IGNOU (IGNOU), India
Aashna has created this Calculator and 100+ more calculators!
Verified by Surjojoti Som
Rashtreeya Vidyalaya College of Engineering (RVCE), Bangalore
Surjojoti Som has verified this Calculator and 10+ more calculators!
## < 18 Strategic Financial Management Calculators
Money Market Discount Rate
Money Market Discount Rate = (Year/Days of Maturity)*(Face Value of Money Market Instrument-Present Value of Money Market Instrument)/Face Value of Money Market Instrument
Effective Convexity
Effective Convexity = (Price of Bond When Yield is Decreased+Price of Bond When Yield is Increased-(2*Initial Price of Bond))/((Change in Curve)^2*Initial Price of Bond)
Add on Rate = ((Year/Days)*((Amount Paid at Maturity Including Interest)-Present Value of Money Market Instrument)/(Amount Paid at Maturity Including Interest))
Change in Price of Full Bond
Percentage Change in Price of Bond = (-Annual Modified Duration*Change in Yield)+(1/2*Annual Convexity*(Change in Yield)^2)
Value of Right using New Shares
Value of Right = Number of New Shares*(Market Price-Issue Price of New Share)/Total Number of All Shares
Single Month Mortality
Single Month Morality = Prepayment for a Month/(Beginning Mortgage Balance for Month-Scheduled Principal Repayment for Month)
Value of Right
Value of Right per Share = (Stock Price-Right Subscription Price)/Number of Rights to Buy a Share
Cost of Equity
Cost of Equity = ((Dividend in Next Period/Current Share Price)+(Dividend Growth Rate*0.01))*100
Unlevered Beta
Unlevered Beta = Levered Beta/(1+((1-Tax Rate)*(Debt/Equity)))
Levered Beta
Levered Beta = Unlevered Beta*(1+((1-Tax Rate)*(Debt/Equity)))
Price Value of Basis Point
Price Value of Basis Point = (Price of Bond When Yield is Decreased-Price of Bond When Yield is Increased)/2
Price of Bond
Price of Bond = Face Value*(1+Implied Discount Rate)^Holding Period
Approximate Macaulay Duration
Approximate Macaulay Duration = Approximate Modified Duration*(1+Rate of Interest)
Conversion Parity Price
Conversion Parity Price = Value of Convertible Security/Conversion Ratio
Share Exchange Ratio
Exchange Ratio = Offer Price for Target's Share/Acquirer's Share Price
Earnings Yield
Earnings Yield = (Earnings per Share/Market Price per Share)*100
Dividend Rate
Dividend Rate = (Dividend per Share/Current Share Price)*100
Earnings Yield using PE Ratio
Earnings Yield = (1/Price-Earnings (PE) Ratio)*100
## Price of Bond Formula
Price of Bond = Face Value*(1+Implied Discount Rate)^Holding Period
PB = FV*(1+IDR)^HP
## What is Price of Bond ?
Price of Bond is the amount of money an investor pays to purchase the bond from the issuer or another investor in the secondary market. The bond price can vary based on factors such as interest rates, credit quality of the issuer, and time to maturity. For instance, if the market interest rate is currently 4%, the bond will trade at a premium because its coupon rate is higher than the current market rate, making it more attractive to investors. Conversely, if the market interest rate is 6%, the bond will trade at a discount. In summary, the price of a bond is a key measure that reflects the present value of its future cash flows, influenced by market interest rates, credit quality, and time to maturity. Investors must make informed decisions about buying, holding, or selling bonds.
## How to Calculate Price of Bond?
Price of Bond calculator uses Price of Bond = Face Value*(1+Implied Discount Rate)^Holding Period to calculate the Price of Bond, Price of Bond reflects the present value of the bond's future cash flows, which include periodic coupon payments and the repayment of the bond's face value at maturity. Price of Bond is denoted by PB symbol.
How to calculate Price of Bond using this online calculator? To use this online calculator for Price of Bond, enter Face Value (FV), Implied Discount Rate (IDR) & Holding Period (HP) and hit the calculate button. Here is how the Price of Bond calculation can be explained with given input values -> 127.1314 = 95*(1+0.06)^5.
### FAQ
What is Price of Bond?
Price of Bond reflects the present value of the bond's future cash flows, which include periodic coupon payments and the repayment of the bond's face value at maturity and is represented as PB = FV*(1+IDR)^HP or Price of Bond = Face Value*(1+Implied Discount Rate)^Holding Period. Face Value is the amount of money that the issuer of a bond agrees to repay the bondholder at the bond's maturity date, Implied Discount Rate is the interest rate that equates the present value of the bond's future cash flows to its current market price & Holding Period refers to the duration of time that an investor owns an asset from the moment it is purchased until it is sold.
How to calculate Price of Bond?
Price of Bond reflects the present value of the bond's future cash flows, which include periodic coupon payments and the repayment of the bond's face value at maturity is calculated using Price of Bond = Face Value*(1+Implied Discount Rate)^Holding Period. To calculate Price of Bond, you need Face Value (FV), Implied Discount Rate (IDR) & Holding Period (HP). With our tool, you need to enter the respective value for Face Value, Implied Discount Rate & Holding Period and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
Let Others Know | 1,582 | 6,708 | {"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} | 3.46875 | 3 | CC-MAIN-2024-26 | latest | en | 0.826765 |
https://www.mathspage.com/is-odd/solved/is-66-an-odd-number | 1,695,891,819,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00173.warc.gz | 967,217,213 | 2,778 | USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES
# Is 66 An Odd Number?
• No, number 66 is not an odd number.
• Sixty-six is not an odd number, because when it is divided by 2 it will leave no comma spot.
## Is 66 Odd Or Even?
• Number 66 is an even number.
## How To Calculate Odd Numbers
• All numbers that cannot be divided exactly by 2 are odd numbers.
• Odd numbers can't be divided evenly into groups of two. The number five can be divided into two groups of two and one group of one. Odd numbers always end with a digit of 1, 3, 5, 7, or 9.
## What Is An Odd Number?
• An odd number is an integer which is not a multiple of two. If it is divided by two the result is a fraction. One is the first odd positive number. The next four bigger odd numbers are three, five, seven, and nine. An integer that is not an odd number is an even number.
• An even number is defined as a whole number that is a multiple of two. If an even number is divided by two, the result is another whole number. On the other hand, an odd number, when divided by two, will result in a fraction. Since odd and even numbers are defined only in reference to the set of integers (..., -3, -2, -1, 0, 1, 2, 3, ...), all negative numbers can also be either odd or even. | 329 | 1,253 | {"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} | 3.765625 | 4 | CC-MAIN-2023-40 | latest | en | 0.927765 |
https://ddcampayo.wordpress.com/2012/02/19/approximation-for-weighted-residuals/ | 1,571,552,195,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986703625.46/warc/CC-MAIN-20191020053545-20191020081045-00158.warc.gz | 440,857,175 | 18,445 | # Approximation for weighted residuals
#### Intro
We ended up the article about weighted residuals pointing out three possible approaches to the problem. We expand them here.
#### The problem
The residual is
$R(x)=A(u_N) - f ,$
with the approximated solution
$u_N=\sum_{j=1}^Nc_j\phi_j+\phi_0.$
We want this residual to be zero in this sense:
$\int_0^L \psi_i R(x) dx= 0.$
#### Galerkin method
Similarly to the RR method, we take ${\psi_j}={\phi_j}$. The residual is therefore “spat out” of the functional space: it is made orthogonal to it. A similar idea to Chebyshev interpolation.
It is therefore important to point out in which way this method differs from the RR one. We do not begin from a weak version of the equation. This means the functions are not in general “weaker” (less differentiable) than in the original differential equation. Also, we lack the embedded information from boundary conditions (NBCs, to be precise) in the weak form, so we must make sure $u_N$ indeed complies with all BCs. As in RR, $\phi_0$ is used for the non-homogeneous BCs, but all of them (N and E BCs) in this case, while the rest satisfy homogeneous BCs.
The linear algebra problem for the Galerkin method is:
$\sum_jA_{ij} c_j = F_i,$
with
$A_{ij}=\int\phi_i A(\phi_j),$
which of course doesn’t have to be symmetric in general. The vector is
$F_i = \int \phi_i\left(f - A(\phi_0)\right).$
However, the Galerkin method coincides with the RR method in two cases:
1. When all BCs are of the essential type. If only EBCs are present, the requirements on the ${\phi_j}$ are the same in both methods, and the weighted-integral form reduces to the weak form.
2. When the ${\phi_j}$ from the Galerkin method are used for a RR method (not the inverse!).
#### Petrov-Galerkin method
In this case the two functional spaces are different. The linear algebra problem is then:
$\sum_jA_{ij} c_j = F_i,$
with
$A_{ij}=\int\psi_i A(\phi_j),$
which looks even less symmetric than before. The vector is
$F_i = \int \psi_i\left(f - A(\phi_0)\right).$
There is clearly much leeway in how to proceed in this case.
#### Least-squares Galerkin method
The parameters are determined such that the integral of the square of the residual (its 2-norm) is minimized. If $A$ is linear, the coefficient matrix turns out to be symmetric in this case,
$A_{ij}=\int A(\phi_i)A(\phi_j).$
While the vector is
$F_i = \int A(\phi_i)\left(f - A(\phi_0)\right).$
#### Collocation method
The residual is set to be zero at selected points along the domain. This is equivalent to setting the weight functions equal to Dirac delta functions at the selected points:
$\psi_i=\delta(x-x_i).$
#### Example
Let’s consider the case
$A(u)=-\frac{d}{dx}(au')-cu.$
Actually, we will take $a=c=1$, so
$A(u)=-u''-u.$
We will make $f(x)=-x^2$, and take the BCs:
$u(0)=0 \qquad u'(1)=1$
This is exactly the “Set 2 problem” in The Rayleigh-Ritz method.
Now, we employ the first function to comply with the actual BCs:
$\phi_0(0)=0 \qquad \phi_0'(1)=1.$
We can just take $\phi_0=x$, then.
The rest have to satisfy the homogeneous BCs:
$\phi_i(0)=\phi_i'(1)=0.$
We will only introduce two of these, which can be taken as
$\phi_1=-x(2-x)\qquad \phi_2=x^2(1-\frac{2}{3}x).$
They cannot be of linear and comply with the homogeneous BCs. Also, the second one lacks a linear term, but that’s fine since the set is complete for polynomials up to cubic. Now the residual can be computed to be
$R=A(u)-f=c_1[2-x(2-x)]+c_2[-2+4x-x^2(1-\frac{2}{3}x)]+x^2-x.$
The last x is the contribution from $\phi_0$.
#### Galerkin
In this case, we want:
$\int_0^1x(2-x)R(x)dx=\int_0^1x^2(1-\frac{2}{3}x)R(x)dx=0$
#### Petrov-Galerkin
We are free to choose other two weight functions. E.g.:
$\psi_1=x\qquad \psi_2=x^2.$
Hence:
$\int_0^1xR(x)dx=\int_0^1x^2R(x)dx=0,$
with the best results of all these methods (!).
#### Least squares Galerkin
In this case the weight functions are given by
$\psi_1=\frac{\partial R}{\partial c_1}=2-x(2-x)$.
Similarly,
$\psi_2=-2+4x-x^2(1-\frac{2}{3}x)$.
#### Collocation
We may choose two points at which the residual will be zero. If we set them at 1/3 and 2/3, then our equations are simply
$R(1/3)=0\qquad R(2/3)=0$.
#### Results
One should compare graphically these four results, the RR result, and the exact solution (see The Rayleigh-Ritz method for the two later). Someone gets us an intern! 😀 | 1,363 | 4,416 | {"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": 35, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-43 | latest | en | 0.902387 |
http://www.cfd-online.com/Forums/fluent-udf/101978-cg-motion-udf-problem.html | 1,474,754,170,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738659496.36/warc/CC-MAIN-20160924173739-00035-ip-10-143-35-109.ec2.internal.warc.gz | 375,279,941 | 18,363 | # CG Motion UDF Problem
Register Blogs Members List Search Today's Posts Mark Forums Read
May 18, 2012, 11:30 CG Motion UDF Problem #1 New Member Leonardo Armas Join Date: Apr 2012 Posts: 2 Rep Power: 0 Hello World! I have been working in the past few days trying to understand UDFs but i think i've reached a dead end... This is what i'm trying to do: I'm just trying to simulate the flow around a body wich is moving up and down with a sinusoidal law of motion. To do this i have written my UDF just as is shown below: #include "udf.h" DEFINE_CG_MOTION(object_mov, dt, vel, omega, time, dtime) { real a, w, pi; pi = 3.1415; /* define motion variables */ a = 0.05; /* 0.05m movement amplitude */ w = 2 * pi * 2; /* 2Hz frequency */ /* define object movement law */ vel[0] = 0; vel[1] = -a * w * sin(w*time); vel[2] = 0; } I have no problem interpreting this UDF into Fluent. However, when defining the dynamic zone the following warning appears: "Warning: incorrect cg motion UDF wing on zone 4 (assuming no motion)" Do you have any clue why could this be? is my UDF correctly written? Looking forward to your answer. Thanks!
June 1, 2012, 11:42 #2 New Member Leonardo Armas Join Date: Apr 2012 Posts: 2 Rep Power: 0 Just to say that I have solved the problem: CG motion macro only works when compiled!!
June 2, 2012, 01:21 #3 New Member Jessica Simberg Join Date: Mar 2012 Posts: 17 Rep Power: 6 Hi, your missing a math header, try using these: #include "udf.h" #include "stdio.h" #include "stdlib.h" #include "math.h" #include "dynamesh_tools.h" #include "storage.h" In addition, define an intermediate (i.e. "real v = 0") variable for the motion: v = -a * w * sin(w*time); vel[1] = v; hopefully it will solve your problem... leoarmas likes this.
August 1, 2012, 06:58
#4
New Member
Hossein
Join Date: Jul 2012
Posts: 1
Rep Power: 0
Quote:
Originally Posted by leoarmas Just to say that I have solved the problem: CG motion macro only works when compiled!!
How do you fix it? I have this problem but I can't fix it! Can you help me?
April 10, 2015, 02:59 udf for flapping motion in hover #5 New Member roham lavimi Join Date: Apr 2015 Posts: 1 Rep Power: 0 I cant write udf code for flapping motion in hover, help me, im so confused
April 10, 2015, 03:22 #6 New Member Jessica Simberg Join Date: Mar 2012 Posts: 17 Rep Power: 6 do it with solidworks flow simulation. J
October 9, 2015, 01:55
#7
New Member
Manahara Manatunga
Join Date: Mar 2014
Posts: 10
Rep Power: 4
Quote:
Originally Posted by hossein66 How do you fix it? I have this problem but I can't fix it! Can you help me?
Hi, Define CG Motion can only be compiled not interpreted. Please try compiling the UDF rather
April 4, 2016, 02:13
#8
New Member
mortezasekaloo
Join Date: Apr 2016
Posts: 4
Rep Power: 2
Quote:
Originally Posted by manahara Hi, Define CG Motion can only be compiled not interpreted. Please try compiling the UDF rather
hi
I am the student of fluid mechanics and now I am studying on fluid project. The subject of my project is about blade turbine that Oil radially impacts to the turbine blades by a nozzle, causing the turbine to spin. I have a problem in modeling of turbine that I don’t know how to obtain moment of inertia of blade and fluid and how to relate this challenge to the Cg motion until I can model this turbine in fluent software. Please help me
May 14, 2016, 03:04
Dynamic mesh flapping wing - Interface problem
#9
Member
Arun raj.S
Join Date: Jul 2011
Posts: 72
Rep Power: 7
Hi everyone,
I am simulating flapping wing of aerofoil. I have created three zones, zone 1, 2 and 3. Zone 1 has aerofoil with structured mesh around the aerofoil. Zone 2 and 3 has unstructured gird. My problem is I have properly followed all the steps. But I am not able to achieve what I want. Because the shadow of the interface is taking as wall and is not moving along with the domain. Kindly someone help in sorting out this issue. I have attached my UDF alos. Please have a look. Thank you so much
Attached Files
second.c (1.1 KB, 14 views)
Tags udf cg motion
Thread Tools Display Modes Linear Mode
Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules
Similar Threads Thread Thread Starter Forum Replies Last Post Rizwan Fluent UDF and Scheme Programming 32 May 8, 2015 06:05 Niklas Wikstrom (Wikstrom) OpenFOAM Running, Solving & CFD 122 June 15, 2014 06:20 akr FLUENT 3 August 22, 2007 07:14 anant FLUENT 2 January 17, 2007 01:15 Alex FLUENT 0 March 22, 2006 18:40
All times are GMT -4. The time now is 17:56. | 1,374 | 4,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2016-40 | latest | en | 0.883944 |
https://math.stackexchange.com/questions/1186002/show-that-int-1-infty-frac-ln-x2x2x1-dx-frac8-pi-381 | 1,566,631,048,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00123.warc.gz | 555,988,248 | 33,898 | # Show that $\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx = \frac{8 \pi ^3}{81 \sqrt{3}}$
I have found myself faced with evaluating the following integral: $$\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx.$$
Mathematica gives a closed form of $8 \pi ^3/(81 \sqrt{3})$, but I have no idea how to arrive at this closed form. I've tried playing around with some methods from complex analysis, but I haven't had much luck (it has been a while). Does anyone have any ideas? Thanks in advance!
• So, you've tried contour integration? – Mark Viola Mar 11 '15 at 22:40
Shocked, shocked! that there is no contour integration yet. So, without further ado...
Note that
$$f(x) = \frac{\log^2{x}}{x^2+x+1} \implies f \left ( \frac1{x} \right ) = x^2 f(x)$$
Thus, $$\int_1^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \int_0^{1} \frac{\log^2{x}}{x^2+x+1} = \frac12 \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1}$$
Now consider
$$\oint_C dz \frac{\log^3{z}}{z^2+z+1}$$
where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$. Taking the limit as $R \to \infty$ and $\epsilon \to 0$, we get that the contour integral is equal to
$$\int_0^{\infty} dx \frac{\log^3{x} - (\log{x}+i 2 \pi)^3}{x^2+x+1}$$
or
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{x^2+x+1} +i 8 \pi^3 \int_0^{\infty} dx \frac{1}{x^2+x+1}$$
Note that the first integral is what we seek, the second integral is zero (by the same trick we applied above), and the third integral is relatively easy to find:
$$\int_0^{\infty} \frac{dx}{x^2+x+1} = \int_0^{\infty} \frac{dx}{(x+1/2)^2+3/4} = \frac{2}{\sqrt{3}} \left [\arctan{\frac{2}{\sqrt{3}} \left ( x+\frac12 \right )} \right ]_0^{\infty} = \frac{2 \pi}{3 \sqrt{3}}$$
The contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand, which are at $z_+ = e^{i 2 \pi/3}$ and $z_- = e^{i 4 \pi/3}$. The sum of the residues is
$$\frac{-i 8 \pi^3/27}{i \sqrt{3}} + \frac{-i 64 \pi^3/27}{-i \sqrt{3}} = \frac{56 \pi^3}{27 \sqrt{3}}$$
Then
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = i 2 \pi \frac{56 \pi^3}{27 \sqrt{3}} - i 8 \pi^3 \frac{2 \pi}{3 \sqrt{3}} = -i \frac{32 \pi^4}{27 \sqrt{3}}$$
Thus,
$$\int_1^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \frac12 \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \frac{8 \pi^3}{81 \sqrt{3}}$$
• Straight forward!! – Mark Viola Mar 12 '15 at 0:32
• Just what I was looking for, Thanks! – stochasm Mar 12 '15 at 2:42
It can be observed that $x^{2} + x + 1 = (x-a)(x-b)$ where $a = e^{2\pi i/3}$ and $b = e^{-2\pi i/3}$. Now \begin{align} I &= \int_{1}^{\infty} \frac{ (\ln(x))^{2} }{ (x-a)(x-b) } \, dx = \frac{1}{a-b} \, \int_{1}^{\infty} \left( \frac{1}{x-a} - \frac{1}{x-b} \right) \, (\ln(x))^{2} \, dx. \end{align} From Wolfram Alpha the integral \begin{align} \int \frac{ (\ln(x))^{2} }{ x - a } dx = -2 Li_{3}\left( \frac{x}{a} \right) +2 \log(x) \, Li_{2} \left( \frac{x}{a} \right) + \log^{2}(x) \log\left( 1-\frac{x}{a} \right) \end{align} for which the integral in question becomes \begin{align} I &= \left[ \frac{-2}{a-b} \left(Li_{3}\left( \frac{x}{a} \right) - Li_{3}\left(\frac{x}{b} \right) \right) + \frac{2}{a-b} \log(x) \, \left(Li_{2} \left( \frac{x}{a} \right) - Li_{2}\left( \frac{x}{b} \right) \right) + \frac{1}{a-b} \, \log^{2}(x) \log\left( \frac{a-x}{b-x} \right) \right]_{1}^{\infty} \\ &= \frac{-2}{a-b} \left[ Li_{3}\left( \frac{1}{a} \right) - Li_{3}\left(\frac{1}{b} \right) \right]. \end{align} This can then be seen as \begin{align} I &= \frac{-2i}{\sqrt{3}} \left[ Li_{3}\left( e^{2\pi i/3} \right) - Li_{3}\left( e^{- 2\pi i/3} \right) \right] \\ &= \frac{-2 i}{\sqrt{3}} \cdot \frac{4 \pi^{3} i}{81} = \frac{8 \pi^{3}}{81 \sqrt{3}}. \end{align}
\begin{align} \int_{1}^{\infty} \frac{ (\ln(x))^{2} }{ x^{2} + x + 1 } \, dx = \frac{8 \pi^{3}}{81 \sqrt{3}}. \end{align}
Here is an approach.
Observe that $$I:=\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx=\int_1^{\infty } \frac{(x-1)(\ln x)^2}{x^3-1} \, dx$$ and by the change of variable $x \to 1/x$ $$I=\int_0^1 \frac{(1-x)\color{blue}{(\ln x)^2}}{1-x^3} \, dx. \tag1$$ Since $\displaystyle \color{red}{\partial_s^2\color{red}{ (x^s)}|_{\large s=0}}=\color{blue}{(\ln x)^2}$, one may write that $$I=\left.\color{red}{\partial_s^2}\left(\int_0^1 \frac{x^s(1-x)}{1-x^3} \, dx\right)\right|_\color{red}{{\large s=0}} \tag2$$ Now \begin{align} \int_0^1 \frac{x^s(1-x)}{1-x^3} \, dx&=\frac13\int_0^1 \frac{u^{(s-2)/3}-u^{(s-1)/3}}{1-u} \, du\qquad (u=x^3,\,x=u^{1/3})\\\\ &=\frac13\int_0^1 \frac{(1-u^{(s-1)/3})-(1-u^{(s-2)/3})}{1-u} \, du\\\\ &=\frac13\int_0^1 \frac{1-u^{(s-1)/3}}{1-u} \, du-\frac13\int_0^1 \frac{1-u^{(s-2)/3}}{1-u} \, du\\\\ &=\frac13\psi\left(\frac{s+2}{3}\right)-\frac13\psi\left(\frac{s+1}{3}\right) \tag3 \end{align} where we have used the standard integral representation for the digamma function.
Then using $(2)$, we get $$I=\frac1{27}\psi''\!\!\left(\frac{2}{3}\right)-\frac1{27}\psi''\!\!\left(\frac{1}{3}\right)=\frac{8\pi^3}{81\sqrt{3}}$$ taking into account some special values of $\psi''$.
Hint: In general, $$~I_n(k)~=~\displaystyle\int_0^\infty\frac{x^{k-1}}{1-x^n}~dx~=~\frac\pi n~\cot\bigg(k~\frac\pi n\bigg),~$$ see Cauchy principal value.
At the same time, a simple substitution of the form $$t=\dfrac1x$$ shows that the original integral can
be written as $$J=\displaystyle\int_1^\infty f(x)~dx~=~\int_0^1f(x)~dx~=~\frac12~\int_0^\infty f(x)~dx.~$$ Then, by rewriting the
integrand using $$\dfrac1{x^2+x+1}=\dfrac{1-x~}{1-x^3}~,~$$ we have $$J=\dfrac{I_3''(1)-I_3''(2)}2$$.
Note $$\int_0^1x^m(\ln x)^2dx=\frac{2}{(m+1)^3}$$ So \begin{eqnarray} I&=&\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1}dx\\ &=&\int_0^1\frac{(1-x)(\ln x)^2}{1-x^3}dx\\ &=&\int_0^1\sum_{n=0}^\infty(1-x)x^{3n}(\ln x)^2dx\\ &=&2\sum_{n=0}^\infty\left(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3}\right)\\ &=&2\sum_{n=-\infty}^\infty\frac{1}{(3n+1)^3} \end{eqnarray} Note \begin{eqnarray} \sum_{n=-\infty}^\infty\frac{1}{(3n+1)^3}=-\pi\text{Res}(\frac{1}{(z+1)^3}\cot(\pi z),-\frac{1}{3})=\frac{4\pi^3}{81\sqrt3} \end{eqnarray} and hence $$I=\frac{8\pi^3}{81\sqrt3}.$$ | 2,785 | 6,150 | {"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": 5, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-35 | latest | en | 0.770438 |
https://gmatclub.com/forum/if-x-and-y-are-positive-is-x-y-greater-than-131527.html?kudos=1 | 1,495,771,226,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608633.44/warc/CC-MAIN-20170526032426-20170526052426-00445.warc.gz | 988,418,774 | 64,828 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 25 May 2017, 21:00
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# If x and y are positive, is x/y greater than 1?
Author Message
TAGS:
### Hide Tags
Manager
Joined: 14 Mar 2010
Posts: 83
Followers: 2
Kudos [?]: 155 [1] , given: 44
If x and y are positive, is x/y greater than 1? [#permalink]
### Show Tags
28 Apr 2012, 21:46
1
KUDOS
4
This post was
BOOKMARKED
00:00
Difficulty:
5% (low)
Question Stats:
74% (01:41) correct 26% (00:50) wrong based on 151 sessions
### HideShow timer Statistics
If x and y are positive, is x/y greater than 1?
(1) xy > 1
(2) x-y > 1
[Reveal] Spoiler: OA
_________________
MGMAT CAT MATH http://gmatclub.com/forum/mgmat-cat-math-144609.html
MGMAT SC SUMMARY: http://gmatclub.com/forum/mgmat-sc-summary-144610.html
Math Expert
Joined: 02 Sep 2009
Posts: 38890
Followers: 7735
Kudos [?]: 106156 [3] , given: 11607
Re: If x and y are integers, is x/y greater than 1 ? [#permalink]
### Show Tags
29 Apr 2012, 05:15
3
KUDOS
Expert's post
monir6000 wrote:
If x and y are integers, is x/y greater than 1 ?
(1) xy > 1
(2) x – y > 0
If x and y are positive, is x/y greater than 1?
Is $$\frac{x}{y}>1$$? --> as given that $$y$$ is positive we can safely multiply boith parts of inequality by it --> so the question becomes "is $$x>y$$?" OR: is $$x-y>0$$?
(1) $$xy>1$$ --> product of two numbers is more than one we can't say which one is greater. Not sufficient.
(2) $$x-y>1$$ --> $$x>y+1$$ --> as $$x$$ is more than $$y$$ plus 1 then it's obviously more than just $$y$$ alone: $$x>y$$. Sufficient.
Or: as $$x-y>1$$ then $$x-y$$ is obviously more than zero --> $$x-y>1>0$$. Sufficient.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 38890
Followers: 7735
Kudos [?]: 106156 [1] , given: 11607
Re: If x and y are integers, is x/y greater than 1 ? [#permalink]
### Show Tags
22 Dec 2012, 05:29
1
KUDOS
Expert's post
megafan wrote:
Bunuel wrote:
monir6000 wrote:
If x and y are integers, is x/y greater than 1 ?
(1) xy > 1
(2) x – y > 0
If x and y are positive, is x/y greater than 1?
Is $$\frac{x}{y}>1$$? --> as given that $$y$$ is positive we can safely multiply boith parts of inequality by it --> so the question becomes "is $$x>y$$?" OR: is $$x-y>0$$?
(1) $$xy>1$$ --> product of two numbers is more than one we can't say which one is greater. Not sufficient.
(2) $$x-y>1$$ --> $$x>y+1$$ --> as $$x$$ is more than $$y$$ plus 1 then it's obviously more than just $$y$$ alone: $$x>y$$. Sufficient.
Or: as $$x-y>1$$ then $$x-y$$ is obviously more than zero --> $$x-y>1>0$$. Sufficient.
Slight correction, even though your answer and your reasoning are correct, is that your reasoning for 2 does not address the question mentioned -- again, it's still correct, but want to make sure it talks about the question. It should be $$x-y>0$$ not $$x-y>1$$
The question posted by monir6000 has typos. Again:
If x and y are positive, is x/y greater than 1?
(1) xy>1
(2) x-y>1
_________________
Intern
Joined: 07 Apr 2012
Posts: 2
Followers: 0
Kudos [?]: 0 [0], given: 0
Re: If x and y are ingeger, is x/y greater than 1 ? [#permalink]
### Show Tags
28 Apr 2012, 22:04
1. xy=1 says nothing about which is greater, x or y.
So insufficient.
2. x-y>O i.e. X>Y
So if x and y are integars
so x/y >1 but if x is negative or y is negative.? It can be less than 1 also. So
Insufficient.
1+2
X and y are of same sign by xy>1 and x>y so sufficient.
C is right.
Posted from my mobile device
Intern
Joined: 03 Nov 2011
Posts: 10
Followers: 0
Kudos [?]: 42 [0], given: 6
### Show Tags
29 Apr 2012, 05:09
If x and y are positive, is x/Y greater than 1 ?
(1) xy > 1
(2) x – y > 0
Manager
Joined: 28 May 2009
Posts: 156
Location: United States
Concentration: Strategy, General Management
GMAT Date: 03-22-2013
GPA: 3.57
WE: Information Technology (Consulting)
Followers: 8
Kudos [?]: 232 [0], given: 91
Re: If x and y are integers, is x/y greater than 1 ? [#permalink]
### Show Tags
21 Dec 2012, 19:54
Bunuel wrote:
monir6000 wrote:
If x and y are integers, is x/y greater than 1 ?
(1) xy > 1
(2) x – y > 0
If x and y are positive, is x/y greater than 1?
Is $$\frac{x}{y}>1$$? --> as given that $$y$$ is positive we can safely multiply boith parts of inequality by it --> so the question becomes "is $$x>y$$?" OR: is $$x-y>0$$?
(1) $$xy>1$$ --> product of two numbers is more than one we can't say which one is greater. Not sufficient.
(2) $$x-y>1$$ --> $$x>y+1$$ --> as $$x$$ is more than $$y$$ plus 1 then it's obviously more than just $$y$$ alone: $$x>y$$. Sufficient.
Or: as $$x-y>1$$ then $$x-y$$ is obviously more than zero --> $$x-y>1>0$$. Sufficient.
Slight correction, even though your answer and your reasoning are correct, is that your reasoning for 2 does not address the question mentioned -- again, it's still correct, but want to make sure it talks about the question. It should be $$x-y>0$$ not $$x-y>1$$
_________________
Intern
Joined: 17 May 2014
Posts: 1
Followers: 0
Kudos [?]: 0 [0], given: 0
Re: If x and y are positive, is x/y greater than 1? [#permalink]
### Show Tags
17 Aug 2014, 21:14
Referring to the same question.
1) The statement is insufficient without a doubt
2) x-y>0, in case x is 7 and y is 3. But if y is -3, then the solution will be 7-(-3)=10 which is greater than 10.
But in case y is negative, then x/y, will not be greater than 1.
Need some help here - Am i missing something?
Math Expert
Joined: 02 Sep 2009
Posts: 38890
Followers: 7735
Kudos [?]: 106156 [0], given: 11607
Re: If x and y are positive, is x/y greater than 1? [#permalink]
### Show Tags
18 Aug 2014, 02:58
omerqureshi wrote:
Referring to the same question.
1) The statement is insufficient without a doubt
2) x-y>0, in case x is 7 and y is 3. But if y is -3, then the solution will be 7-(-3)=10 which is greater than 10.
But in case y is negative, then x/y, will not be greater than 1.
Need some help here - Am i missing something?
The stem says: if x and y are positive...
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15452
Followers: 649
Kudos [?]: 209 [0], given: 0
Re: If x and y are positive, is x/y greater than 1? [#permalink]
### Show Tags
26 Dec 2015, 16:24
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If x and y are positive, is x/y greater than 1? [#permalink] 26 Dec 2015, 16:24
Similar topics Replies Last post
Similar
Topics:
1 Is xy greater than 1, if x and y are both positive 2 24 Nov 2012, 08:09
2 If x and yare positive, is x/y greater than 1 ? 8 30 Jun 2016, 05:06
24 Is x-y+1 greater than x+y-1 ? 24 06 Dec 2016, 19:57
2 If x and y are positive, is x/y greater than 1? 4 30 Sep 2010, 15:08
Is x-y+1 greater than x+y-1 ? (1) x > 0 (2) y < 0 2 18 Aug 2010, 14:10
Display posts from previous: Sort by | 2,535 | 7,781 | {"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} | 3.9375 | 4 | CC-MAIN-2017-22 | latest | en | 0.835158 |
https://smartconversion.com/ConversionExample/mile-per-(hour%E2%80%A2minute)-to-gravity/419/405 | 1,695,753,906,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510219.5/warc/CC-MAIN-20230926175325-20230926205325-00354.warc.gz | 598,190,093 | 5,599 | # Convert mile / (hour • minute) to gravity
Learn how to convert 1 mile / (hour • minute) to gravity step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(\dfrac{mile}{hour \times minute}\right)={\color{rgb(20,165,174)} x}\left(gravity\right)$$
Define the base values of the selected units in relation to the SI unit $$\left(\dfrac{meter}{square \text{ } second}\right)$$
$$\text{Left side: 1.0 } \left(\dfrac{mile}{hour \times minute}\right) = {\color{rgb(89,182,91)} \dfrac{1609.344}{2.16 \times 10^{5}}\left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(89,182,91)} \dfrac{1609.344}{2.16 \times 10^{5}}\left(\dfrac{m}{s^{2}}\right)}$$
$$\text{Right side: 1.0 } \left(gravity\right) = {\color{rgb(125,164,120)} 9.80665\left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(125,164,120)} 9.80665\left(\dfrac{m}{s^{2}}\right)}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(\dfrac{mile}{hour \times minute}\right)={\color{rgb(20,165,174)} x}\left(gravity\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{1609.344}{2.16 \times 10^{5}}} \times {\color{rgb(89,182,91)} \left(\dfrac{meter}{square \text{ } second}\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 9.80665}} \times {\color{rgb(125,164,120)} \left(\dfrac{meter}{square \text{ } second}\right)}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} \dfrac{1609.344}{2.16 \times 10^{5}}} \cdot {\color{rgb(89,182,91)} \left(\dfrac{m}{s^{2}}\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 9.80665} \cdot {\color{rgb(125,164,120)} \left(\dfrac{m}{s^{2}}\right)}$$
$$\text{Cancel SI units}$$
$$1.0 \times {\color{rgb(89,182,91)} \dfrac{1609.344}{2.16 \times 10^{5}}} \cdot {\color{rgb(89,182,91)} \cancel{\left(\dfrac{m}{s^{2}}\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} 9.80665} \times {\color{rgb(125,164,120)} \cancel{\left(\dfrac{m}{s^{2}}\right)}}$$
$$\text{Conversion Equation}$$
$$\dfrac{1609.344}{2.16 \times 10^{5}} = {\color{rgb(20,165,174)} x} \times 9.80665$$
Switch sides
$${\color{rgb(20,165,174)} x} \times 9.80665 = \dfrac{1609.344}{2.16 \times 10^{5}}$$
Isolate $${\color{rgb(20,165,174)} x}$$
Multiply both sides by $$\left(\dfrac{1.0}{9.80665}\right)$$
$${\color{rgb(20,165,174)} x} \times 9.80665 \times \dfrac{1.0}{9.80665} = \dfrac{1609.344}{2.16 \times 10^{5}} \times \dfrac{1.0}{9.80665}$$
$$\text{Cancel}$$
$${\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{9.80665}} \times \dfrac{1.0}{{\color{rgb(255,204,153)} \cancel{9.80665}}} = \dfrac{1609.344 \times 1.0}{2.16 \times 10^{5} \times 9.80665}$$
$$\text{Simplify}$$
$${\color{rgb(20,165,174)} x} = \dfrac{1609.344}{2.16 \times 10^{5} \times 9.80665}$$
Rewrite equation
$$\dfrac{1.0}{10^{5}}\text{ can be rewritten to }10^{-5}$$
$$\text{Rewrite}$$
$${\color{rgb(20,165,174)} x} = \dfrac{10^{-5} \times 1609.344}{2.16 \times 9.80665}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x}\approx0.0007597566\approx7.5976 \times 10^{-4}$$
$$\text{Conversion Equation}$$
$$1.0\left(\dfrac{mile}{hour \times minute}\right)\approx{\color{rgb(20,165,174)} 7.5976 \times 10^{-4}}\left(gravity\right)$$ | 1,404 | 3,302 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 4.40625 | 4 | CC-MAIN-2023-40 | latest | en | 0.292258 |
https://www.scirp.org/Html/25366.html | 1,582,907,568,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147234.52/warc/CC-MAIN-20200228135132-20200228165132-00068.warc.gz | 866,342,522 | 15,644 | iv class="t m0 x0 he y76 ff3 fs6 fc0 sc0 ls3 ws2b">important parameters in electrical prospecting [34,35].
Since Dar Zarrouk parameters are also bulk parameters,
taking the relationship between hydraulic conductivity
and resistivity a stage further leads to a relation between
transmissivity values estimated from pumping tests and
the Dar Zarrouk parameters from surface resistivity mea-
surement as shown in Equations (2) and (3). This mini-
mizes the problems arising from the non-uniqueness of
surface resistivity interpretation. While dealing with ba-
sic equations of direct current prospecting, [34] observed
that if one considers a geologic column built on a square
unit (Figure 1), R is the resistance to the lines of current
perpendicular to the strata, and, C is the conductance to
the lines of current parallel to the strata. These theoretical
relationships showing direct and inverse correlation be-
tween hydraulic conductivity and electrical resistivity has
been explained with respect to four basic assumptions
[5,6]:
Figure 1. Layered models showing transverse resistance
and longitudinal conductance [33].
1) In the case of a conducting basement, the hydraulic
conductivity is directly proportional to the electrical
resistivity: this applicable to Equation (1) (Figure 2);
2) In the case of a resistive basement, the hydraulic
conducting is inversely: this is applicable to Equation
(2) (Figure 2);
3) In the case of an unconsolidated, sandy, clay-free
aquifer, the hydraulic conductivity is directly related
to the porosity [36] and inversely related to the elec-
trical resistivity: this is applicable to Equation (1);
4) In the case of a clay-rich aquifer, the relationship be-
tween porosity breaks down in a more a complex
manner leaving clay content as the primary factor
controlling hydraulic conductivity: this is applicable
to Equation (2).
As a condition in sandy clay free hydrogeological en-
vironment, Kρ can be considered constant; in clay-rich
environment K/ρ should remain constant. The electrical
conductivity of the groundwater is expected not to vary
significantly throughout the aquifer as this would also
affect the measured resistivity. According to [5] some-
times this condition for using Equations (2) or (3) may be
difficult to meet. The authors further advised that it is
also essential that a priori hydraulic conductivity infor-
mation at least one point be known before using the
equations.
Using a representative average hydraulic conductivity
of 77.5 m/d [37] for 13 existing wells in the area, the
transmissivity in the study were estimated using Equation
A. U. UTOM ET AL. 995
Figure 2. Characteristic shapes of K- and H-type resistivity
curves [6].
(1). This average hydraulic conductivity values compared
favourably well with the work of [38] in the area. The
authors used the relation [39] for parallel flow within
each lithologic layer represented by point flow values:
1
m
Zii
i
K
bbk
(4)
where, Ki is the hydraulic conductivity of each individual
layer of thickness, bi (ranging from 1 m to 60 m) with a
total number of 13 layers; b is the overall thickness of the
sequence (about 130 m). These hydraulic conductivity
values in the range of 10–5 - 10–2 m/s are characteristic of
a silty sand and clean sand aquifer [40].
Hence, in establishing the electrical nature of the base-
ment layer from resistivity sounding curves, we chose
Equation (3) to estimate the aquifer transmissivity from
the aquifer electrical parameters. This analogous and em-
pirical relationship can then assist in the estimation of
transmissivity using longitudinal conductance by surface
geoelectrical data, provided the aforementioned basic
assumptions are satisfied.
3. Site Information and Geoelectric Method
The Enugu area study site is located between latitudes
06˚22'N and 06˚27'N and longitudes 007˚25'E and
007˚30'E at about 5 km west of Enugu city and about 15
km near Akanu Ibiam International Airport at Enugu
North L.G.A in the southeastern Nigeria’s Enugu state.
The site area extent is approximately 84 km2.
The study area has three predominant and conformable
geologic formations (Figure 3): The Campanian Enugu
Shale, the Lower Maestrichtian Mamu Formation and the
Upper Maestrichtian Ajali Sandstone. Stratigraphically,
the Enugu Shale which overlies the Cross River Plain
east of the escarpment is overlain by the Mamu Forma-
tion which in turn is overlain by the Ajali Formation.
Hydrology and hydrogeology of the area is controlled by
topographic features. In the study area, the streams or
rivers, some of which appear fracture-controlled in their
flow path give rise to dendritic drainage pattern. The
topography and physiography affect the position and
shape of groundwater tables. The Enugu’s climate is
humid and humidity is high during rains. The average
annual precipitation in Enugu is estimated to be 2000
mm (79 in.) which arrives intermittently and becomes
very heavy during the rainy season. For the whole of
Enugu state, the mean daily temperature is 26.7˚C
(80.1˚F) [41]. The Sahara air mass, north-easterly dry
winds causes the dry season (October to March) as it
advances southwards while the Atlantic Ocean air mass
causes the rainy season (March to October) as it moves
northwards [42]. Water resources availability is also
limited due to the spatiotemporal variation of precipi-
tation. The area receives domestic water supply from
river reservoirs and the Ninth Mile Corner borehole
network. At present, it is a general practice that nearly
very single house built outside the municipal area drill a
groundwater well for its own domestic use. The wells are
generally drilled by local and small-scale contractors
where scientific data gathered are of secondary impor-
tance.
During this work, 19 geoelectrical soundings with a
maximum half current electrode separation of 150 m
have been used. The geoelectrical soundings were under-
taken within the study areas between July and August,
2011. The Schlumberger method was used to acquire the
soundings. The forms of the VES curves measured at the
studied locations are of different types, indicating
interplay between low and high resistivity layers (Figure
4). All resistivity soundings were invested using IPI2Win
software. This software performs an automated approxi-
mation of the initial resistivity model using the observed
data [43]. All resulting models produced a low RMS
relative error of the order of 3%. The starting model used
during the inversion of each of the measured VES loca-
tions were constrained according to obtained water table
of the nearest water and available drillers log information.
Five of the soundings closest to wells, where measured
aquifer properties were available gave estimate of the
Dar Zarrouk parameters (Table 1).
A. U. UTOM ET AL.
996
Figure 3. Physiographic and geologic map of the study also showing the location of the six VES points.
Table 1. Aquifer parameters and resistivity at six sites in parts of Enugu town (Nigeria).
VES Name Aquifer Thickness
(m)
Aquifer
Resistivity
(·m)
Longitudinal Conductance
(–1)
Measured
Transmissivity
(m2/d)
Modeled
Transmissivity
(m2/d)
R1 9.0 527 0.017 696 470
R2 2.1 55 0.038 162 105
R3 6.6 354 0.019 511 525
R4 8.2 364 0.023 635 636
R5 2.1 68 0.031 162 857
R6 3.9 127 0.031 302 858
4. Calculating the Aquifer Transmissivity highlights the applicability of the geoelectric sounding to
the study area, giving a β value of 0.32. This relationship
could be attributed to the influence of hydraulic and
electrical anisotropy as well as the variations in the
geology, grain size, as well as shape of pore channels.
The transmissivity value at each of the 19 VES locations
was then calculated using the longitudinal conductance
from the resistivity survey.
The understudied aquifer system consists of fine grained,
clayey-silty sand materials. Transmissivity of the studied
aquifer is therefore assumed to be controlled by the
thickness of the specific layer and the presence of
fine/clay particles. Also, assuming that the longitudinal
conductance is the dominant parameter, Equation (3) was
used to calculate the transmissivity. The constant, β was
calculated using a linear regression taken between
transmissivity and longitudinal conductance for the six
locations where both data were available (Figure 5). The
negative but strong statistical correlation between aquifer
transmissivity and longitudinal conductance (R2 = 0.82)
Figure 6 shows the transmissivity distribution over the
entire study area. It is clear that the highest transmissivity
values are mostly on the northwestern part of the area
and some parts in the southeastern part, identifying zones
of high water bearing potential. Although details about
he tectonic structure have not been defined in this study, t
A. U. UTOM ET AL. 997
(a) (b)
(c) (d)
(e) (f)
Figure 4. Resistivity soundings and inter pre t ation at the six sites. Locations are shown in Figure 3.
it could be hypothesized that the disturbed nature of the
fracture zones in the Enugu area ay be acting as boun-
daries between the same hydrolithological units and
define the place the where aquifer parameter varies.
5. Conclusion
A rapid, simple relatively inexpensive and liable method
of estimating the transmissivity distribution has been
demonstrated in the Enugu area. The results of the study
show useful estimation of the transmissivity and can be
recommended when siting exploratory boreholes or as an
initial input to a groundwater flow odel. Hydraulic con-
ctivity information known at one point can be used to
extrapolate the transmissivity over the area, which de-
nds on the aquifer thickness and hydraulic conducvity.
Using Equation (3), it was necessary to establish a
working relationship between transmissivity and the Dar
Zarrouk parameter (longitudinal onductance) from which
the value of β, was computed in the field for further
modeling of the transmissivity values from the VES
measurements. Effective application of this method like all
A. U. UTOM ET AL.
998
Figure 5. Longitudinal conductance and transmissivity at
six sites in parts of Enugu town (Nigeria).
Figure 6. Contour map of the study area with transmissivity
and physiography.
geophysical tool, however require a fair knowledge of the
study site’s geology and hydrogeogeology, which was
taken into account. This technique employing the rela-
tionship between transmissivity and Dar Zarrouk para-
meters is well-founded and has been successfully applied
by [5,6,29,44]. This technique could also assist in iden-
tifying parts of the aquifer with best potential yields and
produce realistic ground water models especially in the
Enugu area where small shallow aquifer are being in-
creasingly developed for domestic water supply.
6. Acknowledgements
This paper is based on fieldwork by Ahamefula Utom
carried out as part of the M.Sc. degree course in Applied
Geophysics at the Nnamdi Azikiwe University (NAU),
Nigeria. AAPG/Alexander & Geraldine Wanek Grants-
in-Aid and SEG Foundation Project of Merit Grant is
acknowledged for financial support. Two anonymous
reviewers and Prof. Boniface C. E. Egboka (NAU) are
thanked for their helpful and invaluable comments on
this manuscript.
REFERENCES
[1] W. K. Kosinki and W. E. Kelly, “Geoelectric Soundings
for Predicting Aquifer Properties,” Ground Water, Vol.19,
No. 2, 1981, pp. 163-171.
doi:10.1111/j.1745-6584.1981.tb03455.x
[2] J. F. Ayers, “Conjunctive Use of Geophysical and Geo-
logical Methods in the Study of Alluvial Aquifer,”
Ground Water, Vol. 27, No. 5, 1989, pp. 625-632.
doi:10.1111/j.1745-6584.1989.tb00475.x
[3] D. J. Allen, L. J. Brewerton, L. M. Coleby, B. R. Gibb, M.
A. Lewis, M. A. MacDonald, S. J. Wagstaff and A. T.
Williams, “The Physical Properties of Major Aquifers in
England and Wales,” British Geological Survey Technical
Report WD/97/34, 1997.
[4] F. G. Mendosa, S. T. Steenhuis, W. M. Todd and J. Y.
Parlange, “Estimating Basin-Wide Hydraulic Parameters
of a Semi-Arid and Mountainous Watershed by Reces-
sion-Flow Analysis,” Journal of Hydrology, Vol. 279, No.
1, 2003, pp. 57-69. doi:10.1016/S0022-1694(03)00174-4
[5] S. Niwas, B. Tezkan and M. Israil, “Aquifer Hydraulic
Conductivity Estimation from Surface Geoelectrical Mea-
surements for Krauthausen Test Site, Germany,” Hy-
drogeology Journal, Vol. 19, No. 2, 2011, pp. 307-315.
doi:10.1007/s10040-010-0689-7
[6] A. M. MacDonald, J. Burleigh and W. G. Burgess, “Esti-
mating Transmissivity from Surface Resistivity Sound-
ings: An Example from the Thames Gravels,” Quarterly
Journal of Engineering Geology, Vol. 32, 1999, pp. 199-
205. doi:10.1144/GSL.QJEG.1999.032.P2.09
[7] A. Olayinka and R. Barker, “Borehole Siting in Crys-
talline Basement Areas of Nigeria with a Microproces-
sor-Controlled Resistivity Traversing System,” Ground
Water, Vol. 28, No. 2, 1990, pp. 178-183.
doi:10.1111/j.1745-6584.1990.tb02244.x
[8] J. D. McNeil, “Use of Electromagnetic Methods for
Groundwater Studies,” In: S. H. Ward, Ed., Investigations
in Geophysics No. 5 (Geotechnical and Environmental
Geophysics, SEG Vol. II: Environmental Geophysics and
Groundwater), Society of Exploration Geophysicists,
Tulsa, 1990, pp. 191-218.
[9] R. M. Carruthers and I. F. Smith, “The Use of Ground
Electrical Survey Methods for Siting Water Supply Bo-
reholes in Shallow Crystalline Basement Terrains,” In: E.
P. Wright and W. D. Burgess, Eds., The Hydrogeology of
Crystalline Basement Aquifers in Africa, Geological So-
A. U. UTOM ET AL. 999
ciety, Special Publication, London, Vol. 66, 1992, pp.
203-220.
[10] M. P. Davies and R. G. Campanella, “Piezocone Techno-
logy: Downhole Geophysics for the Geoenvironmental
Characterization of Soil,” Proceedings of the SAGEEP
Conference, Orlando, 26-29 April 1995.
[11] T. Dahlin, “2D Resistivity Surveying for Environmental
and Engineering Applications,” First Break, Vol. 14, No.
7, 1996, pp. 275-283.
[12] D. S. Parasnis, “Principles of Applied Geophysics,”
Chapman and Hall, London, 1997.
[13] I. A. Beresnev, C. E. Hruby and C. A. Davies, “The Use
of Multielectrode Resistivity Imaging in Gravel Pros-
pecting,” Journal of Applied Geophysics, Vol. 49, No. 4,
2002, pp. 245-254. doi:10.1016/S0926-9851(02)00147-7
[14] A. Vchery and B. Hobbs, “Resistivity Imaging to Deter-
Mine Clay Cover and Permeable Units at an Ex-Industrial
Site,” Near Surface Geophysics, Vol. 1, No. 1, 2003, pp.
21-30.
[15] A. Godio and M. Naldi, “Two-Dimensional Electrical
Imaging for Detection of Hydrocarbon Contaminants,”
Near Surface Geophysics, Vol. 1, No. 3, 2003, pp. 131-
137.
[16] D. W. Urish, “The Practical Application of Surface Elec-
trical Resistivity to Detection of Groundwater Pollution,”
Ground Water, Vol. 21, 1983, pp. 144-152.
doi:10.1111/j.1745-6584.1983.tb00711.x
[17] M. A. Meju, “Geoelectrical Investigations of Old/Aban-
doned, Covered Landfill Sites in Urban Areas: Model
Development with a Genetic Diagnosis Approach,” Jour-
nal of Applied Geophysics, Vol. 44, 2000, pp. 115-150.
doi:10.1016/S0926-9851(00)00011-2
[18] S. Niwas, P. K. Gupta and O. A. L. de Lima, “Nonlinear
Electrical Response of Saturated Shaley Sand Reservoir
and Its Asymptotic Approximations,” Geophysics, Vol.
71, No. 3, 2006, pp. 129-133. doi:10.1190/1.2196031
[19] P.M. Soupios, M. Kouli, F. Vallianatos, A. Vafidis and G.
Stavroulakis, “Estimation of Aquifer Hydraulic Para-
meters from Surficial Geophysical Methods: A Case
Study of Keritis Basin in Chania (Crete-Greece),” Journal
of Hydrology, Vol. 338, 2007, pp. 122-131.
doi:10.1016/j.jhydrol.2007.02.028
[20] A. A. R. Zohdy, G. P. Eaton and D. R. Mabey, “Applica-
tion of Surface Geophysics to Ground-Water Investiga-
tion,” Techniques of Water Resources Investigation, US
Geological Survey, 1974.
[21] W. E. Medeiros and O. A. L. de Lima, “A Geoelectrical
Investigation for Groundwater in Crystalline Terrains of
Central Bahia, Brazil,” Ground Water, Vol. 28, 1991, pp.
518-523. doi:10.1111/j.1745-6584.1990.tb01707.x
[22] C. Akaolisa, “Aquifer Transmissivity and Basement Stru-
cture Determination Using Resistivity Sounding at Jos,
Plateau State Nigeria,” Environmental Monitoring and
Assessment, Vol. 114, No. 1-3, 006, pp. 27-34.
[23] R. Sinha, M. Israil and D. C. Singhal, “A Hydrogeo-
Physical Model of the Relationship between Geoelectric
and Hydraulic Parameters of Anisotropic Aquifers,” Hy-
drogeology Journal, Vol. 17, 2009, pp. 495-503.
doi:10.1007/s10040-008-0424-9
[24] A. T. Tizro, K. S. Voudouris, M. Salchzade, H. Ma-
shayekhi, “Hydrogeological Framework and Estimation
of Aquifer Hydraulic Parameter Using Geoelectrical Data:
A Case Study from West Iran,” Hydrogeology Journal,
Vol. 18, No. 4, 2010, pp. 917-929.
doi:10.1007/s10040-010-0580-6
[25] S. Niwas and O. A. L. de Lima, “Aquifer Parameter Esti-
mation from Surface Resistivity Data,” Ground Water,
Vol. 41, No. 1, 2011, pp. 94-99.
doi:10.1111/j.1745-6584.2003.tb02572.x
[26] R. Dhakate and V. S. Singh, “Estimation of Hydraulic
Parameters from Surface Geophysical Methods, Kaliapani
Ultramafic Complex, Orissa, India,” Journal of Envi-
ronmental Hydrology, Vol. 13, No. 12, 2005, Paper 12.
[27] P. C. Heigold, R. H. Gilkeson, K. Cartwright and P. C.
Reed, “Aquifer Transmissivity from Surficial Electrical
Methods,” Ground Water, Vol. 17, No. 4, 1979, pp. 338-
345. doi:10.1111/j.1745-6584.1979.tb03326.x
[28] W. E. Kelly and R. K. Frohlich, “Relations between Aq-
uifer Electrical and Hydraulic Properties,” Ground Water,
Vol. 23, No. 2, 1985, pp. 182-189.
doi:10.1111/j.1745-6584.1985.tb02791.x
[29] R. Frohlich and W. E. Kelly, “The Relation between
Transmissivity and Transverse Resistance in a Compli-
cated Aquifer of Glacial Outwash Deposits,” Journal of
Hydrology, Vol. 79, 1985, pp. 215-219.
[30] O. Mazac, M. Cislerova, W. E. Kelly, I. Landa, D. Ven-
hodova, “Determination of Hydraulic Conductivities by
Surface Geoelectric Methods,” In: S. H. Ward, Ed., Inves-
tigations in Geophysics No. 5 (Geotechnical and Envi-
ronmental Geophysics, SEG Vol. II: Environmental Geo-
physics and Groundwater), Society of Exploration Geo-
physicists, Tulsa, 1990, pp. 125-132.
[31] S. Niwas and D. C. Singhal, “Estimation of Aquifer
Transmissivity from Dar-Zarrouk Parameters in Porous
Media,” Journal of Hydrology, Vol. 50, 1981, pp. 393-
399. doi:10.1016/0022-1694(81)90082-2
[32] S. Niwas and D. C. Singhal, “Aquifer Transmissivity of
Porous Media from Resistivity Data,” Journal of Hy-
drology, Vol. 82, No. 1-2, 1985, pp. 143-153.
[33] W. E. Kelly and P. F. Reiter, “Influence of Anisotropy on
Relation between Electrical and Hydraulic Properties,”
Journal of Hydrology, Vol. 74, 1984, pp. 311-321.
doi:10.1016/0022-1694(84)90021-0
[34] R. Maillet, “The Fundamental Equations of Electrical
Prospecting,” Geophysics, Vol. 12, No. 4, 1947, pp. 529-
556. doi:10.1190/1.1437342
[35] D. Huntley, “Relations between Permeability and Elec-
trical Resistivity in Granular Aquifers,” Ground Water,
Vol. 24, No. 4, 1986, pp. 466-474.
doi:10.1111/j.1745-6584.1986.tb01025.x
[36] G. E. Archie, “The Electrical Resistivity Log as an Aid in
Determining Some Reservoir Characteristics,” Transac-
tions of the American Institute of Mining and Metallurgi-
cal Engineers, Vol. 146, No. 1, 1942, pp. 56-62.
[37] A. U. Utom, “Prediction of Shallow Aquifer Parameters
and Water Quality in Okpara Coal Mine and Environs
A. U. UTOM ET AL.
1000
using Geoelectrical Sounding,” M.Sc. Thesis, Nnamdi
Azikiwe University, Awka, 2012, 106 p.
[38] G. I. Nwankwor, B. C. Egboka and I. P. Orajaka,
“Groundwater Occurrence and Flow Pattern in the Enugu
Coalmine Area, Anambra State, Nigeria,” Hydrological
Sciences Journal, Vol. 33, 1988, pp. 465-482.
doi:10.1080/02626668809491275
[39] R. A. Freeze and J. A. Cherry, “Ground Water,” Prentice
Hall, Inc., New York, 1979.
[40] P. A. Domenico and F. W. Schwartz, “Physical and
Chemical Hydrogeology,” Wiley Press, New York, 1990.
[41] L. O. Sanni, “Cassava Post Harvest Needs Assessment
Survey in Nigeria,” IITA, 2007.
[42] P. E. B. Inyang, “Climate Regions,” In: G. E. K. Ofomata,
Ed., Nigerian in Maps, Eastern States, Ethiope Publishing
House, Benin, 1975, pp. 27-29.
[43] C. Bobachev, “IPI2Win: A Windows Software for an
Automatic Interpretation of Resistivity Sounding Data,”
PhD Thesis, Moscow State University, Russia, 2002.
[44] P. D. C. Mbonu, J. C. Ebeniro, C. O. Ofoegbu and A. S.
Ekine, “Geoelectric Sounding for the Determination of
Aquifer Characteristics in Parts of the Umahia Area of
Nigeria,” Geophysics, Vol. 56, No. 2, 1991, pp. 284-291. | 6,104 | 20,473 | {"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-2020-10 | latest | en | 0.872233 |
https://chemistryonline.guru/problems-mass-defect-binding-energy/ | 1,721,586,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517768.40/warc/CC-MAIN-20240721182625-20240721212625-00858.warc.gz | 145,716,012 | 9,398 | binding energy & mass defect
binding energy &mass defect
source : dc.edu.au
Q 1—Atomic mass of 8O16 is 16.
Mass of one neutron =1.00893 amu
Mass of one proton =1.00757 amu
Mass of one electron =0.0005486 amu
Calculate its mass defect & binding energy?
8O16 have 8p, 8n & 8e
Ans. Mass of nucleus = Mass of 8p +mass of 8n
= 8×1.00757 + 8×1.00893
= 8.06056+8.07144
= 16.1320 a.m.u
Mass of nucleus = Atomic Mass-Mass of 8e
= 16–8×0.0005486
= 16–0.0043888
= 15.9956112 amu
Mass defect (Δm) = 16.1320–15.9956
(Δm) = 0.1364 amu Ans.
Binding energy ( B) = Δm ×931 MeV
= 0.1364×931
B = 126.988 MeV
Q 2—Calculate Δm & B. energy of 33As75. The mass of one proton, one neutron & one electron & atomic mass of 33As75 is 1.0073 amu, 1.0087 amu& 0.0055 amu and 74.9216 amu
Ans. 33As75 = 33e, 33p&42n
Mass of nucleus = mass of 33p + mass of 42n
= 33×1.0073+42×1.0087
= 33.2409 +42.3654
= 75.6063 amu
Mass of nucleus = Atomic Mass-Mass of 33e
= 74.9216 –33×0.00055
= 74.9216- 0.01815
= 74.90345 amu
Δm = 75.6063–74.90345
Δm = 0.70285 amu
B= Δm×931 MeV
= 0.70285×931
B = 654.35 MeV Ans
Q 3 —Calculate the Δm & b. energy of 2He4. Its actual atomic mass is 4.0039 amu. The mass of one n& one p together is 2.0165 amu. mass one electron=0.0005486amu
Ans . 2He4= 2e, 2p&2n
Mass of Nucleus = at. mass–mass of 2e
= 4.0039–2×0.0005486
= 4.0028 amu
Mass of nucleus = mass of 2p +mass of 2n
= 2(mass of P+mass of n)
= 2×2.0165
= 4.033 amu
Mass defect (Δm) = 0.0302 amu An B = Δm×931 MeV
= 0.0302×931
= 28.12 MeV.
Q 4—Binding energy per nucleon for an element is 7.14 MeV. If the binding energy of the element is 28.6MeVCalculate the number of nucleons in the nucleus.
Ans. No. of nucleons =?
B = 28.6MeV
Binding Energy per Nucleon = 7.14MeV
No. of nucleons = =4 Ans.
Q 5—B. energy of an element is 64 MeV. Binding energy per nucleon is 6.39 MeV. What is the total number of neutrons and protons in the nucleus?
Ans. No. of nucleons = No. of Protons + No. of Neutrons
B= 64MeV
No. of Nucleons = =10 Ans
Total no. of neutrons & protons = 10 Ans.
Q 6—B. energy of 2He4 is 28.8 MeV. Calculate Binding energy per nucleon?
Ans. B = 28.8 MeV
No. of Nucleons in 2He4 =4
Binding energy per nucleon =
= = 7.2 MeV
Ans =7.2 MeV | 1,000 | 2,385 | {"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} | 3.75 | 4 | CC-MAIN-2024-30 | latest | en | 0.649168 |
https://salesforcefaqs.com/salesforce-formula-to-get-last-day-of-month/ | 1,708,695,114,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474412.46/warc/CC-MAIN-20240223121413-20240223151413-00121.warc.gz | 520,914,191 | 45,870 | # Salesforce Formula to get Last Day of Month
In this Salesforce Tutorial, we will learn the formula to Calculate the Last Day of the Month in Salesforce. Moreover, we will go through the steps using which we can get the last day of the month from a given date in Salesforce Lightning and Salesforce Classic.
While working in an organization that uses Salesforce as a CRM tool, I got a task where I need to display the last day of the month. So, to fulfill this requirement, I define an advance formula.
If you want to learn How to find the last day of the month from a given date using Salesforce Lightning and Salesforce Classic, you have come to the right place.
The following are the topics that we will cover in this salesforce tutorial:
• Salesforce Formula Last Day of Month
• Formula to detect last day of the month in Salesforce Lightning
• Formula to detect last day of the month in Salesforce Classic
## Salesforce Formula Last Day of Month
In Salesforce, there may be a need to get the last day of the month. For this, we will use the Salesforce Date(), Today(), Year(), and Month() functions together.
Here, is the general advance formula to find the last day of the month in Salesforce:
``IF(MONTH(DateField)=12, DATE(YEAR(DateField),12,31), DATE(YEAR(DateField),MONTH(DateField)+1, 1)-1)``
• In this formula, we calculate the last day of the month of the date field passed in the API name of the date field DateField.
• Here, we first check whether the month of the DateField is 12.
• If it is, we use the DATE() function and return the year of the date value using the YEAR() function, while sets the month to 12, and the day to 31.
• If not, we use the DATE() function and return the year of the date value using the YEAR() function, while the month of the date is returned using the MONTH() function, and we also increment it by +1. Then. we pass 1 as the day component.
• Finally, we subtract the one to get the last day of the month.
With, this we have learned the Salesforce Formula Last Day of the Month. Now, we will move further and see How to get the Last Day of the Month in Salesforce Lightning.
## Formula to detect last day of the month in Salesforce Lightning
Here, are the steps to find the last day of the month using the Advance Formula in Salesforce Lightning.
Step 1: Navigate to Salesforce Setup, by clicking on the Setup icon on the upper right side of the page.
Step 2: On the left side of the menu, click on Object Manager.
Step 3: Search for the object to whose field we want to get the last day of the month. Here, I select the Campaign object from the list of objects.
Step 4: Click on Fields & Relationships.
Step 5: Then, click on New.
Step 6: Choose the Formula as the field type.
Step 7: To continue, click Next.
Step 8: Enter a Field Label for the formula, and the Field Name will be automatically populated. Here, I enter Field Label as the Upcoming Campaign Date.
Step 9: Choose Date as the Formula Return Type. Then, click Next.
Step 10: Enter the advance formula to find out the last day of the month in the Advance Formula editor. Here, is the formula:
``IF(MONTH(EndDate)=12, DATE(YEAR(EndDate),12,31), DATE(YEAR(EndDate),MONTH(EndDate)+1, 1)-1)``
• Here, we want to create a formula that finds the last day of the month and set it as the Upcoming Campaign Date. So, for this, we use the general advance formula of getting the last day of the month that we have discussed above in the first section.
• In general, we are checking whether the month of the EndDate field is 12, if it is we are setting the day as 31, the month as 12, and the year the same as the EndDate field.
• If not, we increment the month of the EndDate field by 1, set the year the same as the EndDate field, and the day as 1. Then, in order to determine the last day of the month, we finally deduct one day from it.
Step 11: Click Check Syntax to make sure the formula is correct.
Step 12: We can add the Description and Help Text as options, and we can also state how you want to handle the blank field.
Step 13: After that, click Next.
Step 14: Select the appropriate field Level Security and to make it visible to everyone, here we select the Visible checkbox.
Step 15: Select Next.
Step 16: Choose the Page Layout Name, for whom we want to add the formula field.
Step 17: To save the formula field, click Save.
We can use the formula field once we’ve created it. Let’s look at an example:
• Create a campaign with the End Date field value by opening the Campaigns item and saving it.
• After that, click on the Details section, and here the Upcoming Campaign Date field value will appear as the last day of the month.
As a result, we now know how to use Salesforce Lightning’s advanced formula to find the last day of the month. Next, we’ll look at how to use Salesforce Classic’s advance formula to detect the last day of the month.
## Formula to detect last day of the month in Salesforce Classic
Here, are the steps to detect the last day of the month using the Advance Formula in Salesforce Classic.
Step 1: From the Salesforce Lightning to open your Salesforce Classic edition. Click on Avtar and click Switch to Salesforce Classic.
Step 2: Click on the Setup option.
Step 3: Scroll down to the Build section. Then under Customize, select the object for whom we want to find the last day of the month. Here, I choose the Product from the object list.
Step 4: Click on the dropdown menu and select Fields.
Step 5: As you scroll down the page, click on the New button under Custom Fields & Relationships.
Step 6: Select the Formula as the field type.
Step 7: Click Next to proceed.
Step 8: Enter a Field Label for the formula, and the Field Name will be automatically populated. Here, I enter Field Label as the Payment Due Date.
Step 9: Choose Date as the Formula Return Type and then click Next.
Step 10: Enter the formula to calculate the last day of the month in the advance formula editor. The formula is as follows:
``IF(MONTH(Purchase_Date__c)=12, DATE(YEAR(Purchase_Date__c),12,31), DATE(YEAR(Purchase_Date__c),MONTH(Purchase_Date__c)+1, 1)-1)``
• Here, we create a formula that finds the last day of the month and sets that date as the Payment Due Date. Therefore, for this, we use the same advance method for determining the month’s last day that we covered in the first section above.
• In this formula, we figure out if the Purchase_Date__c field’s month is 12, and if it is, we set the day to 31 and the month and year to the same as the Purchase_Date__c field.
• If not, we set the day to 1, the year to the same value as the Purchase_Date__c field, and increment the month to 1. Finally, in order to get the last day of the month we subtract 1 from the date value.
Step 11: To verify that the formula is valid, click Check Syntax.
Step 12: We can specify how you want to handle the empty field and include the Description and Help Text optionally.
Step 13: Then, click Next.
Step 14: Choose the appropriate Field’s Level Security, and then check the box Visible to make it visible to all.
Step 15: Click Next.
Step 16: Choose the Page Layout Name for which the formula field is to be added.
Step 17: Click Save to save the formula field.
Once we’ve defined the formula field, we may use it. Let’s explore an example:
• Create a new product with the Purchase Date field by opening the Product object and saving it.
• After that, in the Details section, will get the last day of the month.
In a conclusion, we now understand how to calculate the last day of the month using Salesforce Classic’s advanced formula. | 1,935 | 7,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-10 | latest | en | 0.819585 |
https://slideplayer.com/slide/4784350/ | 1,621,110,962,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991378.52/warc/CC-MAIN-20210515192444-20210515222444-00120.warc.gz | 545,546,267 | 23,354 | # Chapter 28: Direct Current (DC) circuits Reading assignment: Chapter 28 Homework 28-1, due Wednesday, Oct. 22: OQ6, OQ8, OQ9, OQ10, 1, 2, 3, 5, 7, 9,
## Presentation on theme: "Chapter 28: Direct Current (DC) circuits Reading assignment: Chapter 28 Homework 28-1, due Wednesday, Oct. 22: OQ6, OQ8, OQ9, OQ10, 1, 2, 3, 5, 7, 9,"— Presentation transcript:
Chapter 28: Direct Current (DC) circuits Reading assignment: Chapter 28 Homework 28-1, due Wednesday, Oct. 22: OQ6, OQ8, OQ9, OQ10, 1, 2, 3, 5, 7, 9, 11, 19, 21 Homework 28-2, due Monday, Oct. 27: OQ7, OQ12, OQ14, OQ15, 23, 24, 33, 38, 39, 46 Analysis of electric circuits that contain batteries, resistors and capacitors Electromotive force (max. voltage output of battery, not a force) Resistors in series and parallel (also review capacitors in series and parallel) Kirchhoff’s rules RC circuits Household wiring
Direct current (DC) and Alternating current (AC) Time Current I0I0 -I 0 Current Time When a battery is connected to a circuit, the current flows steadily in one direction. This is called direct current or DC. (In RC circuits, current increases/decays over time, but still only flows in one direction). Electric generators at power plants produce alternating current or AC. That is what comes out of an outlet. We will look at AC current in chapter 33.
Resistors in series* (Two devices are connected in series if they are connected at one end, and nothing else is connected there) *Same current flows through all devices; if one device breaks, current flow to all is interrupted. Different voltage drops.
Resistors in parallel* Two devices are in parallel if they are connected at both ends *Same voltage drop across all devices ; current gets split up.
White board example I1I1 R1R1 i-clicker. Series and parallel resistors. Two resistors (R 1 = 50 ; R 2 = 150 ) are connected (a) in series and (b) in parallel to a 24 V battery (see Figure). What is the equivalent resistance of each circuit; what is the voltage drop across each resistor, what current runs through each resistor and the entire circuit? A)37.5 B)50 C)100 D)150 E)200 A)37.5 B)50 C)100 D)150 E)200
White board example Series with parallel in circuit. A)How much current, I, flows from the battery shown in the figure? B)How much current, I 1 flows through the 500 resistor? -The current I flows out of the battery. -Passes through 400- resistor. -Splits into I 1 and I 2 and passes through the 500 and 700 resistor. -I 1 and I 2 combine, I runs back to battery. -Charge carriers arrive with 0V at battery, get lifted up to 12V and start again. 500 700
Two i-clickers. The brightness of a light bulb is proportional to the current running through bulb. The circuit has three identical light bulbs each with resistance R. 1)When the switch is closed how will the brightness of bulbs A and B compare with C? 2)What happens when the switch is opened? S C B A A)All the same B)B is brighter than C C)C is brighter than A and B D)B and C are the same brightness E)A and B are brighter than C
Electromotive Force (EMF) and Terminal Voltage A device such as a battery or a generator that transforms one type of energy (chemical, mechanical) into electric energy is called a seat or source of electromotive force or emf. The potential difference between the terminals of such a device, when no current flows to an external force, is called the emf of the source. Symbol is Each battery itself has a resistance that is called internal resistance; it is designated r. When no current is drawn from the battery the voltage between the terminals equals the emf. If current is drawn from the battery, there is an internal voltage drop equal to Ir. Thus the terminal voltage (voltage delivered) is V ab = - Ir = r = 6 V
Ideal vs. Non-Ideal Batteries Up until now, we’ve treated a battery as if it produced a fixed voltage, no matter what we demand of it. Real batteries also have internal resistance, r. It limits the current and therefore the power that can be delivered. If the internal resistance r is small compared to other resistances in the problem, we can ignore it. The maximum potential difference E across the battery is called electromotive force (emf) The voltage output is V = - Ir + – 30 V 10 50 i-clicker A 30 V battery with internal resistance, r = 10 , is connected to a resistor R (load) of 50 . What is the actual voltage V across the 50 resistor? A)30 V B)36 V C)6 V D)25 V E)24 V
White board example Analyzing a circuit. A 9.0-V battery whose internal resistance r is 0.5 is connected in the circuit as shown in the figure. A)How much current is drawn from the battery? B)What is the terminal voltage (output voltage) of the battery? C)What is the current in the 6.0- resistor?
Kirchhoff’s Rules 1. Rule (junction rule) At any junction point, the sum of all currents entering the junction must be equal to the sum of all currents leaving the junction (Conservation of charge). + – + – 3 5 4 12 V 6 V I1I1 I3I3 AB I2I2 How to apply it: First, assign a current and a direction to every pathway Two components in series will always have the same current At every junction, write the equation: i-clicker. Which equation do you get for point A? A) I 1 + I 2 = I 3 B) I 2 + I 3 = I 1 C) I 1 + I 3 = I 2 D) I 1 + I 2 + I 3 = 0
2. Rule (loop rule) The sum of the changes in potential around any closed path of a circuit must be zero. (Conservation of energy). Kirchhoff’s Rules How to apply it: First, assign a direction to every loop I often pick clockwise Start anywhere, and set 0 equal to sum of potential change from each piece: For batteries: V = E It is an increase if you go from – to + It is a decrease if you go from + to – For resistors: V = IR It is a decrease if you go with the current It is an increase if you go against the current + – + – 3 5 4 12 V 6 V I1I1 I2I2 I3I3 Example: Set up the equations for loop 1 and loop 2. What are currents I 1, I 2, I 3 ?
When applying Kirchhoff’s rules: 1.Label + and - for each battery. The long side of a battery symbol is +. 2.Label the current in each branch in the circuit with a symbol and an arrow (direction). The direction of the arrow can be chosen arbitrarily. If the current is actually in the opposite direction, it will come out with a minus sign in the solution. 3.Apply Kirchhoff’s junction rule at each junction, the loop rule at one or more loops. (Need as many equations as there are unknowns; possibly need more equations since some may be redundant.) 4.When applying the loop rule, follow each loop in one direction only (either clockwise or counterclockwise). Pay attention to signs: A.For a resistor, the sign of the potential difference is negative if your choosen loop direction is the same as the chosen current direction through that resistor; the sign is positive if you are moving opposite to the chosen current direction. B.For a battery, the sign of the potential difference is positive if your loop direction moves from the negative terminal toward the positive terminal; the sign will be negative if you are moving from the positive terminal toward the negative terminal. 5.Solve the equations for the unknowns. Be careful not to err with signs. At the end check your answers by plugging them into the original equations.
Consider loop at left: (start at point a (or point e)) V a b = - IR 1 = - 6.96 V I = 0.0174 A V b c = - IR 2 = - 5.04 V V d e = + 12 V Sum of all voltages: - 6.96 V - 5.04 V + 12V = 0 6.96 V 5.04 V
White board example: Solving problems with Kirchhoff’s rules: Using Kirchhoff’s rules. Calculate the currents I 1, I 2, I 3 in each of the branches of the circuit in the Figure. We have three unknowns --> We need three equations. + + - - Use Maple, a powerful Math program to solve equation set for us:
EMFs in series and in parallel (a)Connecting the emfs in series head-to-tail, the voltages add up. V total = V 1 + V 2. (b) Connecting the emfs in series tail-to-tail or head-to-head, the voltages subtract V total = V 1 - V 2. Can be used for re-charging batteries. (c) Connecting emfs in parallel. Used to provide energy when large amounts of current are needed (starting Diesel engines).
Quick aside: Circuits containing capacitors in series and in parallel (‘inverse’ laws as compared to resistors)
Circuits containing a resistor and a capacitor (RC circuits) (Still DC current in one direction, but changes over time (decreases, increases). Charging a capacitor: Time constant: RC (units: second) Large RC: slow charging In = RC seconds capacitor reaches 63% full voltage Discharging a capacitor: Time constant: RC (units: second) Large RC: slow discharging In = RC seconds voltage falls to 37% (100%-63%) Charging and discharging a capacitor
Circuits containing a resistor and a capacitor (RC circuits) (Still DC current in one direction, but changes over time (decreases, increases). Time constant: RC (units: second) Large RC: slow discharging In = RC seconds voltage falls to 37% (100%-63%) Discharging a capacitor:
Circuits containing a resistor and a capacitor (RC circuits) (Still DC current in one direction, but changes over time (decreases, increases). Time constant: RC (units: second) Large RC: slow discharging In = RC seconds voltage falls to 37% (100%-63%) Charging a capacitor:
White board example An RC circuit. If a capacitor with capacitance, C = 35 F, is connected to a resistor with resistance, R = 120 how much time will elapse until the voltage falls to 10 percent of its original (maximum) value?
The severity of an electric shock depends on the magnitude of the current, how long it acts and through what part of the body it passes. Can feel ~ 1 mA; pain at a few mA; severe contractions above 10 mA; heart muscle irregularities above 70 mA. Current above 100 mA for only a few seconds can be fatal. Resistance of dry skin ~ 10 4 to 10 6 wet skin 10 3 or less. A person in good contact with ground who touches a 120 V line with wet hands can suffer a current Electric shock
Power in Household circuits (All appliances are connected in parallel Electrical wires in a household can get too hot, if they are not thick enough for the amount of current running through them. This is a fire hazard. To avoid this danger, circuit breakers (fuses) are installed, which break the circuit if the current running through a wire is too high (e. g. above 20 A). Wires of varying thickness are connected to different circuit breakers (thick wires are needed to carry high currents). Resistance of a wire Fuses:
Black board example Will the fuse blow? Determine the total current drawn by the devices in the circuit shown in the Figure. Light bulb: 0.8 A Heater: 15.0 A Stereo:2.9 A hair dryer:10.0 A Total:28.7 A A 20 A fuse would blow
Electrical safety, two-pronged, three-pronged wires Two pronged wire: One is ‘hot’ or ‘live’; the other carries current to ground. If the “hot” wire touches the metal casing of an appliance, it becomes “hot” and simply touching the metal case can result in shock. Three pronged wire: One wire is ‘hot’ or ‘live’ (narrower slit); the other carries current to ground (wider slit). Grounding the metal case through the third wire (round prong) metal case is always grounded and no shock can occur.
Review Resistors in series Resistors in parallel Know how to apply these rules to circuits Electromotive power and internal resistance Kirchhoff’s rules (Know how to apply to a circuit): 1. Rule (junction rule) At any junction point, the sum of all currents entering the junction must be equal the sum of all currents leaving the junction. (Conservation of charge). 2. Rule (loop rule) The sum of the changes in potential around any closed path of a circuit must be zero. (Conservation of energy). RC circuits, time constant RC A bit of house hold wiring (circuits with appliances in parallel and a circuit breaker in series; electrical safety.
Download ppt "Chapter 28: Direct Current (DC) circuits Reading assignment: Chapter 28 Homework 28-1, due Wednesday, Oct. 22: OQ6, OQ8, OQ9, OQ10, 1, 2, 3, 5, 7, 9,"
Similar presentations | 3,286 | 12,161 | {"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} | 3.609375 | 4 | CC-MAIN-2021-21 | latest | en | 0.897969 |
https://www.zigya.com/engineering-entrance-exam/11-Statistics/MAENJE11215987/40 | 1,718,774,484,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00368.warc.gz | 939,678,634 | 15,099 | Let xi1 ≤ i ≤ 10 be ten observation of a random variable X. If ∑i = 110xi - p = 3and ∑i = 110xi - p2 = 9 where 0 ≠ p ∈ R, then the standard deviation of these observations is | Statistics
Previous Year Papers
Download Solved Question Papers Free for Offline Practice and view Solutions Online.
Test Series
Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.
Test Yourself
Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
41.$\frac{4}{5}$ $\sqrt{\frac{3}{5}}$ $\frac{9}{10}$ $\frac{7}{10}$
C.
$\frac{9}{10}$
42.
The mean and variance of 8 observations are 10 and 13.5 respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :
• 9
• 5
• 3
• 7
43.
If the variance of the following frequency distribution :
Class : 10-20 20-30 30-40
Frequency : 2 x 2
is 50, then x is equal to ....
Multiple Choice Questions
44.
If the mean and the standard deviation of the data 3, 5, 7, a, b are 5 and 2 respectively, then a and b are the roots of the equation :
• x2 - 20x + 18 = 0
• 2x2 - 20x + 19 = 0
• x2 - 10x + 19 = 0
• x2 - 10x + 18 = 0 | 438 | 1,268 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-26 | latest | en | 0.602972 |
http://kitchenpantryscientist.com/tag/rockets/ | 1,539,600,203,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509170.2/warc/CC-MAIN-20181015100606-20181015122106-00426.warc.gz | 198,811,676 | 7,302 | # Outdoor Science Lab for Kids: Water Rockets!
`June 2, 2016 - by KitchenPantryScientist`
This summer, get off screens and get outside!
Outdoor Science Lab for Kids: 52 Experiments for the Yard, Garden, Playground and Park hit real and virtual bookshelves this week! To celebrate, I’m posting this video we took a few weeks ago of one of my favorite experiments from the new book.
Can’t wait for your copy of the book to arrive before you try this out? Here’s how to make water rockets.
# NASA Soda-Straw Rockets
`April 13, 2011 - by KitchenPantryScientist`
This fun activity will teach you a little bit about rockets. It is from one of NASA‘s educational websites and the great rocket template you’ll find below is provided by the Jet Propulsion Laboratory at the California Institute of Technology.
All you’ll need is a plastic soda straw, some paper, scissors and tape. For the body of your rocket, you can use this template or simply cut a strip of paper an inch or two wide. If using the template, wrap the rectangle of paper around a pencil and tape it into a tube. If you’re using a strip of paper, wind it around a pencil (as pictured above) so it forms a tube. Tape it well, so it holds its shape.
Now, remove your rocket from the pencil, fold one end over and tape it down. This will be the nose of your rocket. To the other end, you can make fins using the template, or design your own fins to tape on. Fins work best at right angles, or near right angles. Now you can decorate your creation.
Put your rocket over the end of a straw and use the force of your breath to launch it! How far does it go? Try making longer and shorter rockets. What happens if you change the shape or number of fins? Record your flight lengths.
What does this teach you? Paper rockets demonstrate how real rockets fly through the atmosphere and some of the forces working on them!
Drag is the force of air getting in the way of your rocket. Weight also drags your rocket down as gravity pulls on it. The lighter you make your rocket (less paper, less tape), and the less drag it has, the farther it will go!
Fins stabilize the rockets’ flight. The size and design of the fins affect how well it can be controlled.
Email me a picture of your best rocket and I’ll post it on my site! We shot some straw rockets at the Kare11 studio, if you want to see them in action! | 569 | 2,383 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-43 | latest | en | 0.886441 |
http://physicsinfo.co.uk/?page=list&cat=28&type=19,15,10,6,14, | 1,642,450,017,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300616.11/warc/CC-MAIN-20220117182124-20220117212124-00060.warc.gz | 55,982,715 | 3,237 | Displaying results which match all of the below:
• Balanced Forces 1
The force of the wind acting on the sail (in this case anticlockwise) is balanced by the force of the crew leaning out (clockwise force)
• Balanced forces 2
In this case the clockwise force is the crew pushing down on the daggerboard and the anticlockwise force is the resistance of the water on the sail
• Car brakes
• Einstein gravity time warp
• F=MA
In this experiment, we tried to establish the F=MA equation. We did this by creating a system which involved a Car, being pulled by a pulley which was attached to some masses (Weight, which in this experiment acts as force, Weight is mass being accelerated by gravity.) In the first experiment we moved the Weights from the Weight, to the Car (Turning it into Mass). This is graph One, which is the upper of the two. And the results are to the left of it. The second experiment was where we only added to the Mass of the car. Activities: Plot the gradient, as the Acceleration and Mass have been listed, put these values into F=MA and see if the Gradient is equal to F. If so why so? If not, why not?
• F=ma (lightgates)
Excel spreadsheet with data from the dataloggers of an F=ma experiment
• Forces & Motion Powerpoint
• Forces lifting an object
• Forces Powerpoint
• Lightgates acceleration
A setup for lightgates to measure acceleration on an air track
• Lightgates height vs velocity
Results for an experiment where the height of a ramp, down which a car is rolled, is varied. The velocity of the car is calculated using a lightgate.
The mass of the car was 53.7g
• Lightgates velocity
A setup for lightgates to measure velocity on an air track
• Maglev
magnetic train
• Measuring G
• Polarisation stress
• Poster Work Done
A3 poster showing Work Done
• Spanner & nut
polarised
• Stopping distance chart
hart of stopping distance for different speeds (in mph and kmph)
• Tennis Ball
momentum impulse | 465 | 1,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-05 | latest | en | 0.926426 |
https://oeis.org/A004766 | 1,603,646,256,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107889574.66/warc/CC-MAIN-20201025154704-20201025184704-00155.warc.gz | 458,035,049 | 4,055 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A004766 Numbers whose binary expansion ends 01. 8
5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 93, 97, 101, 105, 109, 113, 117, 121, 125, 129, 133, 137, 141, 145, 149, 153, 157, 161, 165, 169, 173, 177, 181, 185, 189, 193, 197, 201, 205, 209, 213, 217, 221, 225 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS These are the numbers for which zeta(2*x+1) needs just 3 terms to be evaluated. - Jorge Coveiro, Dec 16 2004 The binary representation of a(n) has exactly the same number of 0's and 1's as the binary representation of a(n+1). - Gil Broussard, Dec 18 2008 a(n) = number of monomials in n-th power of x^4+x^3+x^2+x+1. - Artur Jasinski, Oct 06 2008 LINKS Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (2,-1). FORMULA a(n) = 8*n-a(n-1)-2 (with a(1)=5). [Vincenzo Librandi, Nov 18 2010] a(n) = 2*a(n-1)-a(n-2). G.f.: -x*(x-5) / (x-1)^2. - Colin Barker, Jun 24 2013 MAPLE seq( 4*x+1, x=1..100 ); MATHEMATICA a = {}; k = x^4 + x^3 + x^2 + x + 1; m = k; Do[AppendTo[a, Length[m]]; m = Expand[m*k], {n, 1, 100}]; a (* Artur Jasinski, Oct 06 2008 *) PROG (PARI) a(n)=4*n+1 \\ Charles R Greathouse IV, Oct 07 2015 CROSSREFS Essentially same as A016813. Sequence in context: A141135 A194395 A162502 * A016813 A314668 A314669 Adjacent sequences: A004763 A004764 A004765 * A004767 A004768 A004769 KEYWORD nonn,easy AUTHOR STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified October 25 13:00 EDT 2020. Contains 338012 sequences. (Running on oeis4.) | 736 | 1,993 | {"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-2020-45 | latest | en | 0.687606 |
https://math.stackexchange.com/questions/2088910/prove-that-triangle-is-isosceles-triangle-that-is-inscribed-in-a-circle | 1,660,039,570,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00482.warc.gz | 361,599,262 | 65,162 | # Prove that triangle is isosceles, triangle that is inscribed in a circle.
$AD$ is a median to side $BC$ in triangle $ABC$ that is inscribed in a circle that his center is in $(5,6)$. Given: $D (9,2)$ and the centroid of the triangle is $(6,5$).
There is not so much information given about the circle, thus I can't draw very good the circle (for example I don't know points in the circle, a point of the triangle or radius of the circle). There is my problem. I feel I have few given information and don't know how to start.
Can someone help me how to prove that ABC is isosceles?
Sketch of the proof
Let $G$ be the centroid and $O$ the circumcentre. We know that $A$, $G$ and $D$ are collinear since they are in the median to the side $BC$.
We know that the line through $O$ and $D$ is the perpendicular bisector of the side $BC$.
Since $D$ $G$ and $O$ are collinear (the three points are in the line $y = -x + 11$), then the median to the side $BC$ and the perpendicular bisector of the side $BC$ must collide.
That means that $\angle{ADB} = \angle{ADC} = \pi/2$. One can deduce that the triangles $ADB$ and $ADC$ are congruent and so $\overline{AB} = \overline{AC}$. This completes the proof.
• I was sure that one perpendicular bisector and median must coincide... but how do I know it's that one? Jan 8, 2017 at 15:19
• @PichiWuana Perhaps this helps a bit more. Jan 8, 2017 at 15:27
1. Loacte O(5, 6), G(6, 5), D(9,2).
1. Form the equation of GD. O is found to be a point on GD.
2. Since ADG is a median, A must be on DG extended. From the fact that AG = 2GD. Point A is fixed.
3. Draw the circum-circle (centered at O with radius OA).
4. Since B and C must lie on the circum-circle of ABC, BC is a chord of that circle with D as its midpoint. Note that OD is the line from center joining the midpoint of the chord BC. This makes OD the perpendicular bisector of BC. We can therefore locate B and C.
Point 4 above also supplies enough information to conclude $\triangle ABC$ is isosceles. | 577 | 2,010 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-33 | longest | en | 0.943387 |
http://www.doitpoms.ac.uk/tlplib/index.php?tag=18,27,40,42,43,48,49,51 | 1,369,337,660,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703728865/warc/CC-MAIN-20130516112848-00044-ip-10-60-113-184.ec2.internal.warc.gz | 421,685,011 | 9,819 | # DoITPoMS
TLP Library
Teaching and learning packages (TLPs) are self-contained, interactive resources, each focusing on one area of Materials Science.
Introduction To Anisotropy
It is common in basic analysis to treat bulk materials as isotropic - their properties are independent of the direction in which they are measured. However the atomic scale structure can result in properties that vary with direction. This teaching and learning package (TLP) looks into typical examples of such anisotropy and gives a brief mathematical look into modelling the behaviour.
Atomic Force Microscopy
Provides a brief introduction to atomic force microscopy (AFM), some of the ways it is commonly used and some of the problems faced.
Atomic Scale Structure of Materials
This teaching and learning package provides an introduction to crystalline, polycrystalline and amorphous solids, and how the atomic-level structure has radical consequences for some of the properties of the material. It introduces the use of polarised light to examine the optical properties of materials, and shows how a variety of simple models can be used to visualise important features of the microstructure of materials.
Avoidance of Crystallization in Biological Systems
This teaching and learning package discusses the two main environmental threats leading to crystallization in plants and animals, and the ways in which organisms have adapted to avoid this crystallization. As part of this discussion, there is coverage of some of the theory of nucleation and crystallization.
Batteries
This TLP investigates the basic principles, design and applications of batteries. It covers both primary and rechargeable batteries, how they work and how they may be used.
Brillouin Zones
This teaching and learning package provides an introduction to Brillouin zones in two and three dimensions and is aimed at developing familiarity with Brillouin Zones. It will not cover any specific applications. Brillouin Zones are particularly useful in understanding the electronic and thermal properties of crystalline solids.
Brittle Fracture
What determines when a material will break, and whether failure will be catastrophic or more gradual. Cracking is controlled by the energy changes that occur - it is not the stress at the crack tip that is important..
Casting
This TLP introduces a number of important processes through which metallic items can be fabricated from molten metal. As well as detailing the practical aspects of these manufacturing processes, attention is given to the important parameters which determine the microstructure of the finished items.
Crystallinity in Polymers
An understanding of polymer crystallinity is important because the mechanical properties of crystalline polymers are different from those of amorphous polymers. Polymer crystals are much stiffer and stronger than amorphous regions of polymer.
Crystallographic Texture
This teaching and learning package (TLP) introduces the concept of texture in crystalline materials such as common metals and metallic alloys.
Crystallography
Crystalline materials are characterised by a regular atomic structure that repeats itself in all three dimensions. In other words the structure displays translational symmetry.
Dielectric Materials
This teaching and learning package will introduce you to the properties and uses of dielectric materials.
Introduction To Dislocations
Dislocations are crucially important in determining the mechanical behaviour of materials. This teaching and learning package provides an introduction to dislocations and their motion through a crystal. A 'bubble raft' model is used to demonstrate some of the features of dislocations and other lattice defects. Some methods for observing real dislocations in materials are examined.
Elasticity in Biological Materials
This teaching and learning package (TLP) discusses the elasticity of biological materials. Whilst some show Hookean elasticity, the vast majority do not. Non-linear elasticity is considered, in particular J-shaped and S-shaped curves. Viscoelasticity is also discussed, using hair and spiders' silk as examples.
Electromigration
Electromigration is an ever-increasing problem as integrated circuits are pushed towards further miniaturization. The theory of the phenomenon is explained, including electromigration-induced failure and how it has been and can be minimized.
Epitaxial Growth
This TLP enables you to explore the way in which perfect thin crystalline layers are deposited epitaxially (i.e. in the same crystal orientation) on semiconductor substrates. This is the way many electronic and opto-electronic devices are now fabricated using techniques such as molecular beam epitaxy (MBE).
Ferroelectric Materials
Ferroelectrics are an important device in today's world. They are useful both as capacitors, for example in camera flashes, or as non-volatile memory storage. The memory use of which you are most likely to be aware is in the Playstation 2.
Ferromagnetic Materials
How many ferromagnets do you think you own? Maybe many more than you realise. Ferromagnetic materials lie at the heart not just of the humble compass, but also of many loudspeakers and of computer memory. This teaching and learning package outlines the microscopic basis of magnetism and some of the conquences of ferromagnetic order in real materials.
Fuel Cells
This teaching and learning package provides a short summary of four of the most promising fuel cell technologies. It gives a general overview of the field with focus on materials used (electrolytes and electrodes) and the mechanism of function (electrochemistry and thermodynamics).
The Jominy End Quench Test
Discusses the aims, method and use of results of a test for the hardenability of steel.
Lattice Planes and Miller Indices
This teaching and learning package provides an introduction to the method used to describe planes of atoms in a crystalline material. The practical uses of describing planes of atoms are also addressed.
Materials for Nuclear Power Generation
This TLP introduces readers to key challenges in the selection, usage and development of materials for nuclear reactors.
Tags: energy
Microstructural Examination
This teaching and learning package (TLP) looks at how what we see in micrographs relates to equilibrium phase diagrams and cooling routes for alloy systems.
Optimisation of Materials Properties in Living Systems
This teaching and learning package discusses the uses of merit indices in conjunction with materials-selection maps focusing on biomaterials. The derivation of merit indices is discussed and biological examples are shown.
Phase Diagrams and Solidification
Phase diagrams are a useful tool in metallurgy and other branches of materials science. They show the mixture of phases present in thermodynamic equilibrium. This teaching and learning package looks at the theory behind phase diagrams, and ways of constructing them, before running through an experimental procedure, and presenting the results which can be obtained.
Piezoelectric Materials
This teaching and learning package (TLP) provides an introduction to piezoelectric materials.
Polymer Basics
This teaching and learning package is an introduction to the basic concepts of polymer science. It includes molecular structure, synthesis and tests for identification.
Pyroelectric Materials
Pyroelectric materials are found in almost every home, in the form of intrusion detectors and other devices, and this TLP will consider how they work, and what the most common ones are made of.
Raman Spectroscopy
An introduction to the analysis of materials and chemicals by the Raman scattering of light.
Reciprocal Space
This TLP shows the construction of reciprocal lattices from real ones, use of the Ewald sphere for diffraction experiments and some other applications of reciprocal space.
Introduction To Semiconductors
This teaching and learning package provides a very basic introduction to semiconductors. These materials are essential to the operation of solid state electronic devices.
Slip in Single Crystals
This teaching and learning package explains how plastic deformation of materials occurs through the mechanism of slip. Slip involves dislocation glide on particular slip planes. The geometry of slip is explained, and electron microscopy techniques are used to show slip occurring in single crystals of cadmium.
Solid Solutions
This teaching and learning package is based on a practical used within the Department of Materials Science and Metallurgy at the University of Cambridge. The package is aimed at first year undergraduate Materials Science students and focuses on the different types of solid solution and the thermodynamic principles involved in understanding them.
Solidification of Alloys
This teaching and learning package (TLP) is an introduction to how solute affects the solidification of metallic alloys.
The Stereographic Projection
This TLP covers the use of the Stereographic projection and Wulff nets.
The Structure and Mechanical Behaviour of Wood
This teaching and learning package discusses the structure of wood, focusing on the structure of the tree trunk and the differences between hardwoods and softwoods. The stiffness and strength of different types of wood are discussed, and the different behaviour of wood when wet is investigated.
Structure of Bone and Implant Materials
This teaching and learning package (TLP) describes the structure of bone from the macro-scale to the micro-scale and considers its description as a biological composite. The structure of hip replacements is described and common implant materials are discussed in relation to the mechanical properties of bone.
Superconductivity
Electrons in pairs? Levitating trains? Superconductivity - the combination of lossless electrical conduction and the ability of a material to expel a magnetic field - is a property that excites interest in fundamental science whilst offering tantalising prospects for a range of applications. In this teaching and learning package (TLP), we trace the history of superconductivity, outline some fundamental properties of superconductors, and describe current and potential applications of materials with this unusual property.
Superelasticity and Shape Memory Alloys
This teaching and learning package (TLP) introduces the phenomena of superelasticity and the shape memory effect.
Transmission Electron Microscopy
Transmission electron microscopy is a very important tool in materials science for investigating the fine-scale structure of materials. This TLP serves as an introduction to the basic concepts and structure of the transmission electron microscope.
X-ray Diffraction Techniques
This teaching & learning package provides an introduction to X-ray diffraction. It describes the main crystallographic information that can be obtained and experimental methods most commonly used. | 2,054 | 10,914 | {"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-2013-20 | latest | en | 0.912653 |
https://www.gamedev.net/forums/topic/362234-opengl-global-matrices/ | 1,547,858,747,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660877.4/warc/CC-MAIN-20190118233719-20190119015719-00441.warc.gz | 776,601,657 | 40,717 | Public Group
# OpenGL global matrices
This topic is 4792 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I have made a Vector3 object to support jyk's namespace
class Vector3
{
union {
float v[3];
struct { float x, y, z; } p;
};
public:
Vector3() { p.x = p.y = p.z = 0; }
Vector3(const Vector3 &rhs) { p.x = rhs.p.x; p.y = rhs.p.y; p.z = rhs.p.z;}
Vector3(float new_x, float new_y, float new_z) { Set(new_x, new_y, new_z); }
virtual ~Vector3() {}
void Set(float new_x, float new_y, float new_z) { p.x = new_x; p.y = new_y; p.z = new_z; }
const float& operator[] (int offset) const { return v[offset]; }
Vector3 operator * (float rhs) { return Vector3(p.x * rhs, p.y * rhs, p.z * rhs); }
friend Vector3 operator * (float lhs, const Vector3 &rhs);
Vector3 operator + (const Vector3 &rhs) { return Vector3(p.x + rhs.p.x, p.y + rhs.p.y, p.z + rhs.p.z); }
Vector3 operator - (const Vector3 &rhs) const { return Vector3(p.x - rhs.p.x, p.y - rhs.p.y, p.z - rhs.p.z); }
void operator += (const Vector3 &rhs) { *this = *this + rhs; }
Vector3& operator = (const Vector3 &rhs) {
if(this == &rhs)
return *this;
p.x = rhs.p.x;
p.y = rhs.p.y;
p.z = rhs.p.z;
return *this;
}
Vector3 Cross(const Vector3 &rhs) const {
return Vector3(p.y * rhs.p.z - rhs.p.y * p.z, p.z * rhs.p.x - rhs.p.z * p.x, p.x * rhs.p.y - rhs.p.x * p.y);
}
float Dot(const Vector3 &rhs) const {
return (p.x * rhs.p.x) + (p.y * rhs.p.y) + (p.z * rhs.p.z);
}
void NormalizeSelf() {
double t = std::sqrt(p.x * p.x + p.y * p.y + p.z * p.z);
p.x /= t;
p.y /= t;
p.z /= t;
}
};
// implementation of the friend:
Vector3 operator * (float lhs, const Vector3 &rhs)
{
return Vector3(lhs * rhs.p.x, lhs * rhs.p.y, lhs * rhs.p.z);
}
My initialization code looks like this
...
oglManager glm;
Vector3 cam_pos, cam_target, cam_up;
MSG msg;
BOOL bQuit = FALSE;
glm.move_window(0, 0);
glm.size_window(640, 640);
glm.init(hInstance, hPrevInstance, lpCmdLine, iCmdShow, msg_pump, render_screen);
cam_pos.Set(0.0f, 0.0f, 0.0f);
cam_target.Set(0.0f, 0.0f, 1.0f);
cam_up.Set(0.0f, 1.0f, 0.0f);
glObj.SetMode(OpenGLObject::MODE_6DOF);
//glObj.SetPos(cam_pos);
glObj.LookAt(cam_pos, cam_target, cam_up);
while(!bQuit) {
if(PeekMessage(&msg, NULL, 0, 0, PM_REMOVE)) {
...
glm is an instance of my openGL manager object, and glObj is an instance of jyk's OpenGLObject. the render function...
int render_screen()
{
//static float theta = 0.0f;
glClearColor (0.0f, 0.0f, 0.0f, 0.0f);
glClear (GL_COLOR_BUFFER_BIT);
glPushMatrix();
//glRotatef(theta, 0.0f, 0.0f, 1.0f);
glBegin(GL_TRIANGLES);
glColor3f(1.0f, 0.0f, 0.0f); glVertex2f (0.0f, 1.0f);
glColor3f(0.0f, 1.0f, 0.0f); glVertex2f (0.87f, -0.5f);
glColor3f(0.0f, 0.0f, 1.0f); glVertex2f (-0.87f, -0.5f);
glEnd();
glPopMatrix();
//theta += 1.0f;
glObj.ApplyRoll(3);
glObj.SetOpenGLModelMatrix(); // Should this be called only once, in the init section?
Sleep(1);
return 0;
}
My guess is that my camera need to set GL's global view matrix for every render, but what is the model matrix good for? and do I need to set this for each render? jyk's namespace also introduce the
void MultOpenGLModelMatrix();
void MultOpenGLViewMatrix();
Where and why does these come in to play? PS. Some of the code (typically the Sleep(1)) is just some remains from the default Code::Blocks opengl project that I am extending. (I have no idea why someone would sleep for 1 ms. at the end of a render routine, could it be an attempt to yield the CPU?) EDIT: Any tips on how to make the Vector3 object more efficient would be appreciated aswell, like if any of the operators REALY need to be outlined? Thanks [Edited by - pulpfist on December 5, 2005 11:52:30 AM]
##### Share on other sites
Hi,
I'll take a look at this later today or this evening - I can probably give you some tips on how to use the OpenGLObject class.
##### Share on other sites
Just so you know, the plural of matrix is matrices, not matrixes ;-)
##### Share on other sites
Quote:
ust so you know, the plural of matrix is matrices, not matrixes ;-)
Could have fooled me ^^
Quote:
I can probably give you some tips on how to use the OpenGLObject class.
Thats great
As you can see Im a bit cunfused when it comes to movement :D
Any tips or good links to info would be great.
So far, the rotation works smooth, side/up slide works, but not smooth (moving way too far between each render), and forward/backward movement has a similar side effect aswell
##### Share on other sites
You've asked for some optimization for Vector3 class, and actually I have some suggestions:
Using initialization lists should be preferred to assigning the member fields later. So it is better to use
class Vector3{ Vector3() : p.x(0), p.y(0), p.z(0) {} Vector3(const Vector3 &rhs) : p.x(rhs.p.x), p.y(rhs.p.y), p.z(rhs.p.z) {} // ...
(Unfortunately I don't remember where I've read this.)
In the following assignment I personally would drop the internal if clause. I assume it will happen very seldom that "rhs" and "this" is the same, so that most time the if clause causes (little) costs only.
Vector3& operator = (const Vector3 &rhs) { if(this == &rhs) return *this; p.x = rhs.p.x; p.y = rhs.p.y; p.z = rhs.p.z; return *this; }
Using divisions is still costly. It would be better to replace them by multiplications if applying it often.
void NormalizeSelf() { double t = 1.0f/std::sqrt(p.x * p.x + p.y * p.y + p.z * p.z); p.x *= t; p.y *= t; p.z *= t; }
However, maybe you have an inverse sqrt function at hand? I personally use an approximation to sqrt that computes 1/sqrt in fact (it simply drops the last step of the standard sqrt approximation routine). So you are able to drop 2 divisions from the routine above.
The remaining seems me ok.
##### Share on other sites
Additionally, I could give some general hints to the question about matrices. I don't know jyk's class in detail, but the explanations should hold in general. Hopefully I don't tell too much in advance of jyk's announced explanation :-)
You may have to deal with 4 usages of matrices:
(1) The matrix for projection, i.e. mainly the mapping of the 3D world to the 2D display by handling the "depth". This matrix describes the well known perspective or orthogonal projection (other are possible but seldomly used). OpenGL provides this matrix by its matrix mode GL_PROJECTION.
(2) The view matrix defines which part of the scene is displayed: where in the world the camera is placed, where does it look at. It could be understood to transform the entire world so that the camera's location becomes the origin, and the camera's orientation the basis. It is in fact the inverse of the model matrix (see below) of the camera if the camera is implemented as an object. If you look into the GetModelMatrix and GetViewMatrix routines you'll see this interrelation.
In OpenGL there is no view matrix support on its own, but in combination with the model matrix in the so-called MODELVIEW matrix mode. This is often a cause of confusion. It requires (in conjunction with OpenGL's matrix definitions) to supply the view matrix to OpenGL _before_ supplying any model matrix.
(3) A model matrix defines how a model (e.g. a mesh) is located and oriented (and perhaps also scaled) w.r.t. a reference co-ordinate system. Often the reference system is "the world", but need not. E.g. you may use a scene graph to model spatial dependencies, so you may have 4 wheels, each one with an own model matrix w.r.t. the car, what itself has a model matrix w.r.t. the world. So this kind of matrix is also called "local transformation matrix" or similar. Due to some reasons, the particular vertices of a model are usually not given w.r.t. the world but the local matrix of the model. So the rendering needs to know the current model matrix.
(4) A texture matrix. That is surely out of scope here.
So in general you set-up OpenGL's projection matrix once at start-up, its view matrix once per rendering the scene, but typically will provide model matrices as many times as models are available for rendering.
Coming to the MultOpenGLXyzMatrix() routines: Looking back to the explanation of the model matrices above, you see that matrices may need to be concatenated to describe an overall transformation (e.g. you may need to multiply the model matrix of the left front wheel by those of the car to yield in the transformation of the wheel w.r.t. the world). But be aware that matrix multiplications are not commutative, so that the order of invoking the Mult routines plays a role.
[Edited by - haegarr on December 5, 2005 9:05:54 AM]
##### Share on other sites
Thanks haegarr
That was enlightening
Exactly what I was confused about
By the way, I think I had some compiler errors when I used an initializer list, but Ill try it again. It could be my compiler MingW beeing picky
##### Share on other sites
Quote:
Original post by Anonymous PosterJust so you know, the plural of matrix is matrices, not matrixes ;-)
Quote:
from Merriam-WebsterInflected Form(s): plural ma-tri-ces /'mA-tr&-"sEz, 'ma-/; or ma-trix-es /'mA-trik-s&z/
...and while we are on the subject:
Quote:
from Merriam-WebsterInflected Form(s): plural ver-ti-ces /'v&r-t&-"sEz/; also ver-tex-esInflected Form(s): plural in-dex-es or in-di-ces /-d&-"sEz/
##### Share on other sites
Cool, didn't know that about the alternate plurals :)
@pulpfist: I think most of the basic concepts were covered in the previous posts, but in summary:
1. Unless you're incorporating zoom or doing something else unusual, the projection matrix is commonly set at the beginning of the simulation and left alone. The OpenGLObject class doesn't deal with the projection matrix in any way.
2. The view matrix is typically set once at the beginning of each frame. In my class this is done through SetOpenGLViewMatrix(). The function loads the matrix straight into OpenGL, so you don't have to call glLoadIdentity() first.
3. After that you can render anything you like by using MultOpenGLModelMatrix() to concatenate a model transform with the current matrix. If you want to render more than one object, or render a hierarchy of objects, it will probably be easiest to use glPushMatrix() and glPopMatrix().
4. Remember that in OpenGL, the view and model matrices are combined; the difference between a view matrix and a model matrix in OpenGL is really only conceptual.
Quote:
As you can see Im a bit cunfused when it comes to movement :DAny tips or good links to info would be great.So far, the rotation works smooth, side/up slide works, but not smooth (moving way too far between each render), and forward/backward movement has a similar side effect aswell.
Moving too far between frames sounds like a problem outside of the OpenGLObject code. If you can describe the behavior in more detail, and/or perhaps post the code where you're moving the object, I'll certainly try to spot the problem if I can.
##### Share on other sites
Thanks
I have a better grasp of how thing is supposed to work, but doing it in code is still a bit confusing.
Here is what I got so far
Init code:
void InitOpenGL (void){ glClearColor(0.0f, 0.0f, 0.0f, 0.0f); glClearDepth(1.0f); glEnable(GL_DEPTH_TEST); glDepthFunc(GL_LEQUAL); glShadeModel(GL_SMOOTH); glHint(GL_PERSPECTIVE_CORRECTION_HINT, GL_FASTEST); // Set up initial position/orientation using LookAt cam_pos.Set(0.0f, 0.0f, 0.0f); cam_target.Set(0.0f, 0.0f, 3.0f); cam_up.Set(0.0f, 1.0f, 0.0f); // glo = OpenGLObject instance glo.SetMode(OpenGLObject::MODE_6DOF); glo.LookAt(cam_pos, cam_target, cam_up); // Feed camera position/orientation into the modelview matrix glo.SetOpenGLViewMatrix(); glo.SetOpenGLModelMatrix();}
and here is the render:
void RenderFrame (void){ glClear(GL_COLOR_BUFFER_BIT|GL_DEPTH_BUFFER_BIT); glo.SetOpenGLViewMatrix(); glo.ApplyPitch(rot_pitch); glo.MoveAlongForwardAxis(move_frwd); glo.SetOpenGLModelMatrix(); // this makes things happen //glo.MultOpenGLModelMatrix(); // this makes everything disappear // draw rectangle glTranslatef(0.0f, 0.0f, -8.0f); //glRotatef(angle, 0.0f, 0.0f, 1.0f); DrawTriangle();}
DrawRectangle just draws a simple rectangle.
When the rendering starts, I am positioned at 0,0,0 and have the rectangle right in front of me.
If I press the up-key, the rectangle start to rotate around me, which is good, but if I move myself(camera) away from 0,0,0, first, and then start to rotate, the rectange still rotate around 0,0,0, and not around me.
How do I make everything rotate around me and not around 0,0,0?
(Strictly speaking, I want to be the one rotating, but if I understand correct, its more common to have the world, and everything in it, rotate instead)
Im I supposed to reset(identity) all the matrices before every object I render?
The way its set up now, I never use any kind of Identity functions except at program init.
By the way, the projection matrix is set equal to the identity matrix at startup, is this correct?
• ### What is your GameDev Story?
In 2019 we are celebrating 20 years of GameDev.net! Share your GameDev Story with us.
• 10
• 11
• 13
• 9
• 11
• ### Forum Statistics
• Total Topics
634090
• Total Posts
3015432
× | 3,630 | 13,336 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-04 | latest | en | 0.368046 |
https://breakingdaynews.tk/zero-of-a-function/ | 1,571,610,066,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00068.warc.gz | 416,509,254 | 27,255 | # zero of a function
definition of zero of a function
Next
zeros of polynomial functions part i
Next
solved the function below has at least one rational zero
Next
zero cross function faq india omron ia
Next
how to find zeros of a function
Next
ece205 lecture 7 2 transfer functions poles and zeros youtube
Next
how to find the zeros of a function barca fontanacountryinn com
Next
quiz worksheet zeros of functions polynomials graphing
Next
c 9 find zero of a function help video in high school math advanced
Next
5 1 polynomial functions degree of a polynomial largest power of x
Next
whats the zero of a function virtual nerd
Next
determining intercepts and zeros of linear functions texas gateway
Next
alphago zero board evaluation function uses multiple time steps as
Next
complex analysis holomorphic function convergent sequence proof
Next
partition function zeros
Next
a electrical resistivity t at zero magnetic field as a function of
Next
objective to find the zeros of a quadratic function and solve a
Next
definition of zero of a function
Next
transient response from transfer function
Next
solved use newtons method to estimate the two zeros of
Next
chapter 3 polynomial and rational functions 3 4 zeros of
Next
definition of zero of a function
Next
pgfplots how to draw a label at a zero of a function tex latex
Next
zeros of p x
Next
finding the zeros of a polynomial function youtube
Next
how to find zeros of a function
Next
find the poles and zeros of the transfer functions chegg com
Next
zeros of polynomial functions part i
Next
find the poles and zeros of the transfer functions chegg com
Next
determining intercepts and zeros of linear functions texas gateway
Next
z transform minimum number of poles and zero of transfer function
Next
datei zero free region for the riemann zeta function svg wikipedia
Next
complex analysis holomorphic function convergent sequence proof
Next
what is a zero of a function youtube
Next
zeroes roots x intercepts definitions properties study com
Next
ex 1 the zero feature of the ti84 to find rational zeros of a
Next
quiz worksheet zeros of functions polynomials graphing
Next
Next
how to know the difference between a vertical asymptote and a hole
Next
derivative airy function zeros calculator high accuracy calculation
Next
excel formula force negative numbers to zero
Next
definition of zero of a function math definitions letter z
Next
how to find the zeros of a function barca fontanacountryinn com
Next
how to find the zeros of a function barca fontanacountryinn com
Next
whats the zero of a function virtual nerd
Next
quiz worksheet zeros of functions polynomials graphing
Next
alphago zero board evaluation function uses multiple time steps as
Next
showme find imaginary zeros
Next
how do you write a polynomial function with the given zeros 2 and 2i
Next
alphago zero board evaluation function uses multiple time steps as
Next
showme find imaginary zeros
Next
elementary calculus zeros of a function
Next
how to find the zeros of a function barca fontanacountryinn com
Next
Next
determining intercepts and zeros of linear functions texas gateway
Next
how to find the zeros of a function barca fontanacountryinn com
Next
ex 1 the zero feature of the ti84 to find rational zeros of a
Next
finding zeros of bessel functions of the first and second kinds 1 57 0
Next
ex 1 the zero feature of the ti84 to find rational zeros of a
Next
zeros of polynomial functions part i
Next
zero of a function a maths dictionary for kids quick reference by
Next
how do you find the zeros of a quadratic function on a graph
Next
finding zeros of bessel functions of the first and second kinds 1 62 0
Next
zero analysis definition concept study com
Next
quiz worksheet zeros of functions polynomials graphing
Next
excel formula force negative numbers to zero
Next
graphing
Next
how to find zeros of a function
Next
zeros of a function
Next
elementary calculus zeros of a function
Next
transient response from transfer function
Next
datei zero free region for the riemann zeta function svg wikipedia
Next
showme find imaginary zeros
Next
Next
find zeros of a function given the graph 9th 12th grades common
Next
find the poles and zeros of the transfer functions chegg com
Next
datei zero free region for the riemann zeta function svg wikipedia
Next
graphing
Next
illustration of the effect of zero padding a 1 d rectangle
Next
objective to find the zeros of a quadratic function and solve a
Next
objective to find the zeros of a quadratic function and solve a
Next
zero of a function a maths dictionary for kids quick reference by
Next
whats the zero of a function virtual nerd
Next
solved determine the number of zeros of the polynomial fu
Next
quiz worksheet zeros of functions polynomials graphing
Next
determining intercepts and zeros of linear functions texas gateway
Next
zero of a function a maths dictionary for kids quick reference by
Next
zeros of a function
Next
what is a zero of a function youtube
Next
objective to find the zeros of a quadratic function and solve a
Next | 1,169 | 5,204 | {"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-2019-43 | latest | en | 0.780247 |
http://slideplayer.com/slide/4213296/ | 1,521,873,110,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649931.17/warc/CC-MAIN-20180324054204-20180324074204-00441.warc.gz | 261,261,331 | 23,102 | # Trig – Section 2 The Unit Circle
## Presentation on theme: "Trig – Section 2 The Unit Circle"— Presentation transcript:
Trig – Section 2 The Unit Circle
Essential Questions: What are the values for special right triangles? How do you generate the unit circle? Define the six trig ratios in terms of x and y.
Right Triangles In a right triangle, the hypotenuse is times as long as the leg. 45 1
Ex.1 Find the missing sides.
45 6 45 3 c) a) b) 45
Right Triangle In a triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is times as long as the shorter leg. 30 60 1 2
Ex 2. Find the missing sides.
30 4 a) b) 60 c) 60 18
Class Work Special Right Triangles Wkst
The Unit Circle A circle where the radius equals 1. On your special triangles worksheet, convert all of the radii to unit 1 and find the new length of each side. Discuss what you notice about 30, 45, 60 with your shoulder partner.
Constructing Sine: Cosine: Tangent:
Constructing Sine, Cosine, Tangent:
Constructing Sine, Cosine, Tangent:
What is the coordinate for this point?
y x
Pass out Unit Circle worksheet
Fill in degrees only for each circle on back Go over answers Pass out Hand made Circle worksheet and find point of each circle That is bold. Transfer information onto large unit circle Stop for today!
What is the circumference
Of a circle with a radius of 1?
Radians A radian is another form of measuring angles. The radian measure of an angle drawn in standard position in the plane is equal to the length of the arc on the unit circle subtended by that angle. Radian Measure
What would be the degrees of these radian measures?
What will be the radian measure of these arcs?
What will be the degree measure of these?
To convert degrees to radians, multiply by To convert radians to degrees, multiply by
Ex 3. Find the degrees or radians of each:
What is the coordinate for this point?
Label all the coordinates.
What are the coordinates of ?
What are the coordinates of ?
Fill in the Unit Circle
HW #2 Unit Circle Worksheet | 513 | 2,057 | {"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} | 4.375 | 4 | CC-MAIN-2018-13 | latest | en | 0.850683 |
http://stackoverflow.com/questions/11099701/double-loop-in-r/11100427 | 1,464,395,928,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277286.54/warc/CC-MAIN-20160524002117-00053-ip-10-185-217-139.ec2.internal.warc.gz | 281,899,496 | 21,914 | # double loop in R
I am quite new to R and have a question about loops
In my real dataset there 7000 observations in 80 countries with 15 sectors and 6 types of organizations, but here is a simplified example.
``````country <- c("a","a","a","a","a","a","b","b","b","b","b","b",
"c","c","c","c","c","c","d","d","d","d","d","d")
sector <- c("a","a","a","b","c","c","a","b","b","b","c","c",
"b","b","b","b","c","c","a","a","b","b","c","c")
organization <-c("a","b","c","c","b","a","a","b","b","c","b","b",
"c","a","a","b","b","c","c","b","a","a","b","c")
budget <-c(2,4,3,5,9,7,5,4,3,6,1,2,4,5,6,1,5,3,4,2,3,5,4,6)
table <- data.frame(country, sector, organization, budget)
``````
What I want is:
1. Amount of the different types of organizations in a specific sector in a specific country.
2. Percentage of the total budget in a sector given to the different kind of organizations.
I first have to make a subset to select only info from country "a" and sector "a"
``````smalltable <-subset(table, (country == "a") & (sector == "a"))
``````
and then answer my first question, how many of each type of organization are in a sector in a country
``````smalltable\$count <- table(smalltable\$organization)
``````
then I need to find the percentage of finance
``````smalltable\$percentage <- smalltable\$budget / sum(smalltable\$budget)
``````
then i used tapply
`````` N <- tapply(smalltable\$count, smalltable\$organization, FUN=sum)
financialshare <- tapply(smalltable\$percentage, smalltable\$organization, FUN=sum)
``````
and finally combined this:
`````` total <- data.frame (smalltable\$country,smalltable\$sector,smalltable\$organization, N,financialshare)
total
``````
This is the little table that I require!
But I need this for all my 15 sectors and in all 80 countries, so I need some kind of loop function that runs a loop of all sectors and repeats this loop for every country. I need to make these tables as condensed as possible, bringing together all info about 1 country (so 15 sectors) into one table. Also zero values should be removed from the tables to save space.
How would I need to proceed?
-
yes, but with all sectors in a specific country in one frame. So for the sample I would like to have 4 country specific dataframes to transfer to excel – user1466195 Jun 19 '12 at 12:38
I'll give a `data.table` answer
``````library(data.table)
my_table=data.table(country, sector, organization, budget)
by_org=my_table[, list(count=.N, budget=sum(budget)),
keyby=list(country, sector, organization)]
total_budgets=my_table[, list(total_budget=sum(budget)),
keyby=list(country, sector)]
joined_table= total_budgets[by_org]
joined_table[,percentage:=budget/total_budget]
``````
EDIT from Matthew: In v1.8.1, using `:=` by group, the join isn't needed so it's easier and faster and the `total_budget` column is added to the right which is a more natural place than where it is using the join in v1.8.0 :
``````DT = data.table(country, sector, organization, budget)
ans = DT[, list(count=.N, budget=sum(budget)),
keyby=list(country, sector, organization)]
ans[, total_budget:=sum(budget), by=list(country,sector)]
ans[, percentage:=budget/total_budget]
``````
Result (using v1.8.1) :
`````` country sector organization count budget total_budget percentage
1: a a a 1 2 9 0.2222222
2: a a b 1 4 9 0.4444444
3: a a c 1 3 9 0.3333333
4: a b c 1 5 5 1.0000000
5: a c a 1 7 16 0.4375000
6: a c b 1 9 16 0.5625000
7: b a a 1 5 5 1.0000000
8: b b b 2 7 13 0.5384615
9: b b c 1 6 13 0.4615385
10: b c b 2 3 3 1.0000000
11: c b a 2 11 16 0.6875000
12: c b b 1 1 16 0.0625000
13: c b c 1 4 16 0.2500000
14: c c b 1 5 8 0.6250000
15: c c c 1 3 8 0.3750000
16: d a b 1 2 6 0.3333333
17: d a c 1 4 6 0.6666667
18: d b a 2 8 8 1.0000000
19: d c b 1 4 10 0.4000000
20: d c c 1 6 10 0.6000000
``````
Two things to note here: first your question is a bit vague and conflicting as to what you actually want as far as counts and sums go, but hopefully my snippet is self explanatory enough as far as the calculations I'm doing.
Second, it is not idiomatic in `R` to loop through large numbers of observations as this tends to be slow. Most people who have programmed `R` for a while tend to use vector operations, `plyr`, `data.table`, or other similar packages.
But to be complete, loop construction is as follows:
``````for (item in list)
{
...
}
``````
To iterate over common indexes...
``````for (i in 1:length(object))
{
...
}
``````
-
in country A and sector A, there is only one A type organization, but your list displays a count of 3 – user1466195 Jun 19 '12 at 12:25
Ah sorry, I think I misunderstood your question then. I see your response to comment of gd07. I'll read it and modify my post accordingly. – Yike Lu Jun 19 '12 at 12:39
perfect, thank you – user1466195 Jun 19 '12 at 13:13
``````library(plyr)
ddply(table,.(country,sector), transform,count=as.vector(table(budget)),percentage=budget / sum(budget))
``````
gives
`````` country sector organization budget count percentage
1 a a a 2 1 0.2222222
2 a a b 4 1 0.4444444
3 a a c 3 1 0.3333333
4 a b c 5 1 1.0000000
5 a c b 9 1 0.5625000
6 a c a 7 1 0.4375000
7 b a a 5 1 1.0000000
8 b b b 4 1 0.3076923
9 b b b 3 1 0.2307692
10 b b c 6 1 0.4615385
11 b c b 1 1 0.3333333
12 b c b 2 1 0.6666667
13 c b c 4 1 0.2500000
14 c b a 5 1 0.3125000
15 c b a 6 1 0.3750000
16 c b b 1 1 0.0625000
17 c c b 5 1 0.6250000
18 c c c 3 1 0.3750000
19 d a c 4 1 0.6666667
20 d a b 2 1 0.3333333
21 d b a 3 1 0.3750000
22 d b a 5 1 0.6250000
23 d c b 4 1 0.4000000
24 d c c 6 1 0.6000000
``````
-
Thank you already for this, but the percentage should not be per entry, but per organization type. So in lines 11-12 and 21-22 organization type 'a' is displayed twice for the same country and sector and this should not be the case. They have to be summed to form one group. Then there budgets are summed too and the percentage compares organizations type 'a' with 'b' and 'c' types. This would also make it possible to perform a count that will be higher than one. – user1466195 Jun 19 '12 at 12:15
You've set this up perfectly for using `plyr`. By that, I mean that you have a process that (almost) works on one subset that returns exactly what you want for that subset, and now you need to just loop over all possible subsets. I re-wrote your code to make it tighter and work around possible missing `organization`s.
``````library("plyr")
ddply(table, .(country, sector), function(smalltable) {
smalltable <- ddply(smalltable, .(organization), summarise,
count=length(budget), budget=sum(budget))
smalltable\$percentage <- smalltable\$budget / sum(smalltable\$budget)
smalltable
})
``````
which gives
`````` country sector organization count budget percentage
1 a a a 1 2 0.2222222
2 a a b 1 4 0.4444444
3 a a c 1 3 0.3333333
4 a b c 1 5 1.0000000
5 a c a 1 7 0.4375000
6 a c b 1 9 0.5625000
7 b a a 1 5 1.0000000
8 b b b 2 7 0.5384615
9 b b c 1 6 0.4615385
10 b c b 2 3 1.0000000
11 c b a 2 11 0.6875000
12 c b b 1 1 0.0625000
13 c b c 1 4 0.2500000
14 c c b 1 5 0.6250000
15 c c c 1 3 0.3750000
16 d a b 1 2 0.3333333
17 d a c 1 4 0.6666667
18 d b a 2 8 1.0000000
19 d c b 1 4 0.4000000
20 d c c 1 6 0.6000000
``````
Note that `table` is not a good name for a variable since it is also the name of a base function.
-
The nested `ddply` within another `ddply` might get quite slow as data size increases, iiuc. – Matt Dowle Jun 21 '12 at 8:56
@MatthewDowle It could, but with 7000 records, an outer grouping of 1200 and an inner grouping of 6, it should be reasonable. If it is not, then the `data.table` approach is better. – Brian Diggs Jun 21 '12 at 14:59 | 3,093 | 9,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2016-22 | latest | en | 0.889291 |
http://mathhelpforum.com/calculus/12610-bounded-sequence-proof.html | 1,500,900,379,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424876.76/warc/CC-MAIN-20170724122255-20170724142255-00456.warc.gz | 219,704,160 | 10,384 | 1. ## bounded sequence proof
A) Suppose that Sn = 0, if (Tn) is a bounded sequence, prove that lim (SnTn) = 0
B) Show by an example that the boundeness of (Tn) is a necessary condition in part (A)
2. Originally Posted by luckyc1423
A) Suppose that Sn = 0, if (Tn) is a bounded sequence, prove that lim (SnTn) = 0
We need to show,
|s_n*t_n|<e
Note that |t_n|<=M (where M can be chosen to be non-zero).
Then,
|s_n*t_n|<=M*|s_n|<e thus, we need that, |s_n|<e/M
Which is possible for some n>N because lim s_n = 0.
Q.E.D.
3. Originally Posted by luckyc1423
B) Show by an example that the boundeness of (Tn) is a necessary condition in part
Theorem A convergent sequence is bounded.
We have shown that,
lim (s_n*t_n)=0
Thus,
|s_n*t_n|<= M
By the previous theorem.
BUT! {s_n} is also bounded by some non-zero konstant K because it is convergent.
Thus,
|s_n*t_n|<=K|t_n|<=M
Thus,
|t_n|<=M/K
Thus,
{t_n} must be a bounded sequence.
Q.E.D.
4. Originally Posted by ThePerfectHacker
Theorem A convergent sequence is bounded.
We have shown that,
lim (s_n*t_n)=0
Thus,
|s_n*t_n|<= M
By the previous theorem.
BUT! {s_n} is also bounded by some non-zero konstant K because it is convergent.
Thus,
|s_n*t_n|<=K|t_n|<=M
Thus,
|t_n|<=M/K
Thus,
{t_n} must be a bounded sequence.
Q.E.D.
After I posted this yesterday I realized I made a mistake.
I will not tell you were, you shall need to find it yourself.
But what you said is false.
Consider,
s_n=0.
And
t_n=n
Note,
lim (s_nt_n)=0.
But {t_n} is not bounded!
I will state the following as an excercise (a little hard to show). But try it.
Theorem Let lim (s_n) not = 0 and s_n not =0. Let t_n be any sequence. If {s_n*t_n} is a convergent sequence, then {t_n} must be bounded. | 573 | 1,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-30 | longest | en | 0.933662 |
http://www.ck12.org/book/CK-12-Middle-School-Math-Grade-8/r1/section/9.3/ | 1,464,700,440,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464051299749.12/warc/CC-MAIN-20160524005459-00102-ip-10-185-217-139.ec2.internal.warc.gz | 434,002,939 | 42,196 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 9.3: Using Intercepts
Difficulty Level: At Grade Created by: CK-12
## Introduction
Comparing Field Trips
The students have completed all of their research and now it is time for them to figure out which field trip to attend. Mr. Thomas scheduled a meeting during homeroom when both teams would have the opportunity to present their findings. Both did this and then the students began to debate the merits of each trip.
“I think we should go to the Omni Theater because it is more educational,” Tasha stated strongly.
“This trip doesn’t have to be educational we can just have a fun field trip,” Casey said.
Arguments began between the students. Mr. Thomas whistled and all were quiet.
“Is there anything that the two have in common that we can think of?” Mr. Thomas asked.
“You mean money?” Casey asked.
“Yes. Is there a common fee between them both?”
The students began to think about this. Intercepts are places where two or more equations meet. In the case of the students, they wrote two equations one for each field trip. Is there a common cost in each?
Use this lesson to learn about intercepts and then return to this problem and Mr. Thomas’ question at the end of the lesson.
What You Will Learn
By the end of this lesson, you will understand how to complete the following skills.
• Find an x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-intercept of the graphs of a linear equation.
• Use intercepts to graph a linear equation.
• Interpret intercepts of the graph of a linear equation.
• Solve real – world problems using intercepts of lines.
Teaching Time
I. Find an x\begin{align*}\underline{x}\end{align*}- and y\begin{align*}\underline{y}\end{align*}- Intercept of the Graphs of a Linear Equation
In football, a player makes an interception when he catches a ball thrown by the other team that was not intended for him. Intercept means to catch or to interrupt. In graphs, we will find that most lines intercept the x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-axes. We’ll call these points the x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-intercepts and we will use them in a variety of ways.
Consider the graph below. As you know, the x\begin{align*}x\end{align*}-axis is horizontal and the y\begin{align*}y\end{align*}-axis is vertical. Do you see that the graph crosses, or intercepts, both the x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-axes?
In looking at this graph, you can see that the line crosses the x\begin{align*}x\end{align*}-axis and the y\begin{align*}y\end{align*}-axis. These are the two intercepts of this graph.
The line crosses the x\begin{align*}x\end{align*} – axis at -1.
The line crosses the y\begin{align*}y\end{align*} – axis at 3.
These are the two intercepts.
We can figure out the two intercepts of any linear equation. All you have to do is to look for the place where the line crosses the two axis’.
That is a great question. Because a vertical line is x=\begin{align*}x =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}, it would not have a y\begin{align*}y\end{align*} – intercept. A horizontal line has the equation y=\begin{align*}y =\underline{\;\;\;\;\;\;\;\;\;\;}\end{align*}, so it does not have an x\begin{align*}x\end{align*} – intercept. Therefore, we would only name the x\begin{align*}x\end{align*} – intercept or the y\begin{align*}y\end{align*} – intercept in these two examples. Here is a graph of both a horizontal line and a vertical line. You can see what we mean by looking at this example.
In these two graphs, x\begin{align*}x\end{align*} is equal to 4 and y\begin{align*}y\end{align*} is equal to -1. You can see that each of these special types of graphs only has one intercept.
II. Use Intercepts to Graph a Linear Equation
Now that you know how to identify an x\begin{align*}x\end{align*} and a y\begin{align*}y\end{align*}-intercept by looking at a graph, you can learn how to find the two intercepts in an equation and then use this information to graph the line. Think about it, if you know where the line crosses the x\begin{align*}x\end{align*}-axis and where the line crosses the y\begin{align*}y\end{align*} – axis, then you can easily graph the line. Let’s look at an example.
Example
Find the x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-intercepts and then graph the equation 2x+3y=6\begin{align*}2x+3y=6\end{align*}.
First, notice that this is an equation in standard form. We will need to find the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} – intercepts.
To find the x\begin{align*}x\end{align*}-intercept, set y\begin{align*}y\end{align*} equal to zero. Think about this and it makes perfect sense. If you have an intercept with the x\begin{align*}x\end{align*} – axis, then it makes sense that the y\begin{align*}y\end{align*} value is 0.
2x+3y2x+302xx=6=6=6=3\begin{align*}2x+3y&=6\\ 2x+3 \cdot 0&=6\\ 2x&=6\\ x&=3\end{align*}
We now have the ordered pair (3, 0) or the x\begin{align*}x\end{align*}-intercept 3.
To find the y\begin{align*}y\end{align*}-intercept set x\begin{align*}x\end{align*} equal to zero. Think about this and it makes perfect sense. If you have an intercept with the y\begin{align*}y\end{align*} – axis, then it makes sense that the x\begin{align*}x\end{align*} value is 0.
2x+3y20+3y3y=6=6=6=2\begin{align*}2x+3y&=6\\ 2 \cdot 0+3y&=6\\ 3&=6\\ y&=2\end{align*}
We now have the ordered pair (0, 2) or the y\begin{align*}y\end{align*}-intercept 2.
Now we can use the intercepts to graph this line.
Write down how you can find the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} – intercepts in your notebook.
III. Interpret Intercepts of the Graph of a Linear Equation
Now we can find the x\begin{align*}x\end{align*}- and y\begin{align*}y\end{align*}-intercepts of a graph by seeing where the graph crosses the axes. We can also graph an equation by calculating its x\begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts and plotting them on a coordinate plane. But let’s understand the significance of the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts. Recall that the \begin{align*}y\end{align*} value of the \begin{align*}x\end{align*}-intercept is always zero and the \begin{align*}x\end{align*} value of the \begin{align*}y\end{align*}-intercept is always zero. This is useful in understanding equations.
Consider a graph whose \begin{align*}x\end{align*}-intercept is 5. Not only does this indicate to us that the graph will cross the \begin{align*}x\end{align*}-axis at 5, but it necessitates that when \begin{align*}x\end{align*} is 5, the \begin{align*}y\end{align*} value is zero. Likewise, for whatever value the \begin{align*}y\end{align*}-intercept has, the \begin{align*}x\end{align*} value must be zero.
Example
Look at the following graph and interpret the intercepts of the graph.
Now let’s look at what information we can interpret from this graph. First, this is a graph of the equation \begin{align*}y=-2x-4\end{align*}.
Notice that the coordinates of the \begin{align*}y\end{align*} – intercept is (0, -4). We can see that the -4 can also be found in the equation itself. Notice how it is the value that is not connected to the \begin{align*}x\end{align*} variable. When looking at an equation and a graph, this is one way to determine the \begin{align*}y\end{align*} – intercept.
Now we can look at the value of the \begin{align*}x\end{align*}-intercept. In this case, it is (-2, 0). Don’t let this fool you, the \begin{align*}y\end{align*} – intercept can be found in the equation, but the \begin{align*}x\end{align*} – intercept is determined by the steepness of the line. Therefore, we will have to use the equation and a table of values to determine the \begin{align*}x\end{align*} – intercept. However, you can still determine it when looking at the graph of a line.
IV. Solve Real – World Problems Using Intercepts of Lines
Interpreting the \begin{align*}x\end{align*}- and \begin{align*}y\end{align*}-intercepts is useful in real-world situations. We can figure out intercepts by looking at all kinds of different problems. Let’s look at an example.
Example
Martha likes to go to the park every day but it’s 6 o’clock and her parents are waiting for her at home. She has her bike but she sometimes walks it. She walks at 3mph and she rides her bike at 9mph. If she is 6 miles from home, how long might her parents have to wait?
\begin{align*}w=\end{align*} time (in hours) walking and \begin{align*}b=\end{align*} time(in hours) on her bike
\begin{align*}3w+9b &=6\\ 3 \cdot 0+9b &=6\\ 9b &=6\\ b &=\frac{2}{3}\end{align*}
If she only rides her bike, it will take her \begin{align*}\frac{2}{3}\end{align*} or 40 minutes.
\begin{align*}3 \cdot w + 9 \cdot 0 &=6\\ 3w &=6\\ w &=2\end{align*}
If she only walks, it will take her 2 hours! Hope she rides her bike.
Now let’s go back and look at the problem from the introduction.
## Real-Life Example Completed
Comparing Field Trips
Here is the problem from the introduction. Use what you have learned to determine the intercept.
The students have completed all of their research and now it is time for them to figure out which field trip to attend. Mr. Thomas scheduled a meeting during homeroom when both teams would have the opportunity to present their findings. Both did this and then the students began to debate the merits of each trip.
“I think we should go to the Omni Theater because it is more educational,” Tasha stated strongly.
“This trip doesn’t have to be educational we can just have a fun field trip,” Casey said.
Arguments began between the students. Mr. Thomas whistled and all were quiet.
“Is there anything that the two have in common that we can think of?” Mr. Thomas asked.
“You mean money?” Casey asked.
“Yes. Is there a common fee between them both?”
The students began to think about this. Intercepts are places where two or more equations meet. In the case of the students, they wrote two equations one for each field trip. Is there a common cost in each?
Solution to Real – Life Example
To determine the intercept, we must first begin with the two equations.
The bowling trip used the equation \begin{align*}y = 3g + 2\end{align*}.
The Omni trip used the equation \begin{align*}y=5x+2\end{align*}
You might notice right away that the 2 is common in both. We can check and see if this is indeed the intercept by graphing both equations. Here is the graph.
The 2.00 fee for shoes or ticket service fee is the common factor between both trips. ## Vocabulary Here are the vocabulary words that are found in this lesson. \begin{align*}x\end{align*} – intercept the point where a line crosses the \begin{align*}x\end{align*} – axis. It will always have the coordinates \begin{align*}(x, 0)\end{align*}. \begin{align*}y\end{align*} – intercept the point where a line crosses the \begin{align*}y\end{align*} – axis. It will always have the coordinates \begin{align*}(y, 0)\end{align*}. ## Time to Practice Directions: Determine the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} – intercepts of each equation. There will be two answers for each equation. 1. \begin{align*}3x+4y=12\end{align*} 2. \begin{align*}6x + 2y = 12\end{align*} 3. \begin{align*}4x + 5y = 20\end{align*} 4. \begin{align*}4x + 2y = 8\end{align*} 5. \begin{align*}3x + 5y = 15\end{align*} 6. \begin{align*}-2x + 3y = -6\end{align*} 7. \begin{align*}-3x + y = 9\end{align*} 8. \begin{align*}-2x - 2y = 6\end{align*} 9. \begin{align*}7x + 3y = 21\end{align*} 10. \begin{align*}2x + 9y = 36\end{align*} Directions: Look at each graph and identify the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} – intercept of each equation. Each graph will have two answers. Directions: Solve the following problems. 1. Tickets to a hockey game cost60 for the front section and $30 for the rear. A manager wants to take his employees but cannot spend more than$300. How many employees can he take to the front or rear sections?
2. The ASB at a school will be purchasing drinks that they will sell at a school dance. They plan to have 200 drinks available. Bottled water costs $.25 per bottle and soda costs$.35 per can. What is the minimum and maximum that they might spend on the drinks?
Show Hide Details
Description
Tags:
Subjects:
Search Keywords:
Date Created:
Jan 15, 2013 | 3,747 | 12,536 | {"found_math": true, "script_math_tex": 100, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2016-22 | longest | en | 0.951597 |
https://www.reddit.com/r/AskReddit/comments/1cw8ya/reddit_whats_your_best_mathrelated_joke/ | 1,493,257,611,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121778.66/warc/CC-MAIN-20170423031201-00127-ip-10-145-167-34.ec2.internal.warc.gz | 923,469,759 | 22,794 | This is an archived post. You won't be able to vote or comment.
[–] 2 points3 points (0 children)
All of the math jokes and puns you will ever need.
New Math
[–] 2 points3 points (0 children)
What do you call a bird that doesn't eat? A polynomial!
[–] 4 points5 points (0 children)
There are 10 kinds of people in this world: those who understand binary, those who don't and those who didn't expect the number to be ternary.
[–] 1 point2 points (0 children)
Old mathematicians never die, they just lose some of their functions
[–] 1 point2 points (0 children)
It should be a fucking sin to tell math jokes...
[–] 1 point2 points (0 children)
Kids - don't drink and derive.
[–] 1 point2 points (0 children)
[–] 1 point2 points (0 children)
Q: What do you get when you cross a mouse with an elephant?
A: Mouse elephant sine theta
Q: What do you get when you cross an elephant with a mountain climber?
A: You can't; a mountain climber is a scaler.
[–] 1 point2 points (0 children)
An infinite amount of mathematicians walk into a bar, the first one says to the bartender, "I'll have one glass of beer please." The next one says, "I'll have half a glass of beer!" The next one orders a quarter of a glass, the one after that a sixteenth of a glass and so on and so on. The bartender gets tired of this and says, "You guys need to know your limits!'' and pours out two glasses of beer for the mathematicians.
[–] 1 point2 points (0 children)
5/2 of people don't understand fractions.
[–] 1 point2 points (0 children)
Why didn't the cosine graph cross the road? Cosecant!
[–] 1 point2 points (0 children)
What is the first derivative of cow? Prime rib.
Why don't you see many mathmaticians at the beach? They have sine and cosine to give them a tan.
[–] 1 point2 points (0 children)
How long is a fence? ... It's around a yard.
Why should you never drink rootbeer out of a square glass? ... You end up with just beer.
Why is 2 jealous of 5? ... Because 3 pi 4 5.
An infinite amount of mathematicians walk into a bar. The first one orders one drink. The next one orders half a drink. The next one orders a quarter of a beer. The next one orders an eight of a beer. The bartender puts two beers on the bar and says "you need to learn your limits".
[–] 4 points5 points (2 children)
Dear Algebra,
Stop asking me to find your x. She's not coming back, and please, for the love of all that's good and holy, stop asking y!
[–] 2 points3 points (1 child)
That's not the worst math joke I've ever heard, but it's pretty damn cos.
[–] 0 points1 point (0 children)
Watch your language! thats an infraction.
[–] 2 points3 points (0 children)
Holy shift! Look at that asymptote on that mother function.
[–] 3 points4 points (3 children)
Something about six being afraid of seven.
[–] 2 points3 points (0 children)
3 out of 2 people have trouble with fractions.
[–] 1 point2 points (0 children)
What do you call a gathering with more than 30 people? A z-party.
I'll show myself out...
[–] 1 point2 points (1 child)
This one is better to hear than read...
What is the square root of 69? 8 something (sounds like ate something)
[–] 0 points1 point (0 children)
8.30662386292
[–] 1 point2 points (0 children)
There are only 10 types of people in the world. Those who understand binary and those who don't.
[–] 0 points1 point (0 children)
I put rootbeer in a square glass... now i have beer
[–] 0 points1 point (0 children)
5 out of every 4 people have trouble with their fractions.
[–] -1 points0 points (0 children)
Q: What is the scariest thing to study in math?
A: The Monster group
To succeed in math you have to be able to tell your asymptote from a hole in the graph.
[–] -2 points-1 points (0 children)
Having sex is like math homework.
I do it best when I'm alone in my bed.
[–] -2 points-1 points (0 children) | 1,099 | 3,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-17 | latest | en | 0.928768 |
https://forma-slova.com/en/articles/12836-solution-enthalpy-variation-calculation-of-solution-enthalp | 1,686,369,146,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656963.83/warc/CC-MAIN-20230610030340-20230610060340-00394.warc.gz | 300,756,302 | 5,614 | # Solution Enthalpy Variation. Calculation of Solution Enthalpy
protection click fraud
In sports activities, in hospitals and in our daily lives, the use of hot and cold instant compresses is very common. In both cases, certain substances react with water. For example, in the case of the cold compress, it contains two capsules separating the water from the NH4AT THE3, which, when dissolved in water, absorbs heat and produces instant cold. In the case of hot compresses, CaCl dissolves in water2 or the MgSO4, which release energy producing heat.
But why certain solutions give off heat, being exothermic; and others absorb, being endothermic?
Well, to understand this issue we have to study the enthalpy variation (ΔH) of solutions, which is made up of two steps:
(1st) Reticular enthalpy (ΔHret): when a solute dissolves in water, the first step is to separate its ions which are in a crystalline lattice. To break the bonds between the ions it is necessary to supply energy to the system. So this first process is endothermic, as it absorbs energy; being yours positive enthalpy (ΔH > 0).
instagram story viewer
(2nd) Enthalpy of hydration (ΔHhid): after the ions separate, they are enveloped by the solvent molecules. In the case of water, it is the solvent and we say that hydration is taking place. The dipoles of water are respectively attracted by the oppositely charged ions; thus, for this interaction to take place, the release of energy is necessary. Thus, in hydration the enthalpy will be negative (ΔH < 0), because the process is exothermic.
Do not stop now... There's more after the advertising ;)
The figure below shows how hydration occurs, in which there is ion-dipole interaction, that is, attraction between the charges of the separated ions and the water dipole:
The enthalpy change of the solution (ΔHSun) will be determined by the sum of these two enthalpies. If the result is positive, it means that the reticular enthalpy is higher, so the dissolution enthalpy will indicate that the process is endothermic.
The enthalpy diagram of an endothermic dissolution is shown below:
This is indicated by the case of potassium iodide dissolution shown below:
Your enthalpy diagram will be represented by:
If the result is positive, the enthalpy of hydration will be greater than the reticular one and the process is exothermic. The enthalpy diagrams of exothermic dissolutions are represented as shown in the following example:
By Jennifer Fogaça
Would you like to reference this text in a school or academic work? Look:
FOGAÇA, Jennifer Rocha Vargas. "Solution Enthalpy Variation"; Brazil School. Available in: https://brasilescola.uol.com.br/quimica/variacao-entalpia-solucao.htm. Accessed on June 28, 2021.
Teachs.ru
#### Oxidation Reactions. Concepts related to redox
An oxidation-reduction reaction is characterized as a simultaneous process of loss and gain of el...
#### Equilibrium constants Kc and Kp. Equilibrium Constants
Consider the generic reversible reaction below, where the lower case letters correspond to the ba...
#### Molecular Balance. Molecular chemical equilibrium
When we say the word “balance” an object that remains indefinitely comes to mind. However, this i...
instagram viewer | 740 | 3,251 | {"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-2023-23 | latest | en | 0.926655 |
http://amsterjob.com/map-of-texas-cities/ | 1,532,139,319,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592150.47/warc/CC-MAIN-20180721012433-20180721032433-00439.warc.gz | 16,658,624 | 13,511 | # Map Of Texas Cities
1 views
5 / 5 ( 1votes )
At the time you think of Map Of Texas Cities, exactly what are you thinking of? In essence a map is a representation of a topology or function. Intended for example a formula such as X=2Y maps a worth of Y to each value of A. Of course we all believe that mathematicians are weird and sometimes hard to understand but they have you ever seen a schematic map of a subway (underground railway) system? Maybe you have ever seen the same network of rails specified on a more "normal" Map Of Texas Cities of the location in which it is located? Different Map Of Texas Cities of the extremely same thing can look quite different.
At the time you make a Map Of Texas Cities of the flat area - a "plan" or "elevation" - things are quite simple, but when you make an effort to map a larger area, like the surface of an complete planet, things can get quite complicated if you wish your map to be smooth. It really is all very well to make a world, but try turning the top of that globe into a set Map Of Texas Cities! Yikes!
However you begin it, you finish program edge-effects. As I write this information I am actually included in programming map-generating programs designed to generate maps of fictional landscapes. I happen to be examining the map-generators that are included in the free, open-source (GNU GPL licensed) strategy game, FreeCiv. Edge results are extremely apparent in such maps. The Map Of Texas Cities are basically rectangular, but you can choose to obtain them act like cylinders by "wrapping" left to right or top to bottom level, or you can also have "wrap" in both guidelines. Most often people make a decision on "wrap" only still left to right, and obstruct the most notable and bottom with "polar regions". Such basic "wrapping" makes for quite extreme distortion though if you give it a try with a real Map Of Texas Cities worldwide! | 421 | 1,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-30 | latest | en | 0.941723 |
https://www.doubtnut.com/question-answer/the-area-of-a-rectangle-is-192-cm2-and-its-perimeter-is-56-cm-find-the-dimensions-of-the-rectangle--51555275 | 1,627,243,252,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00242.warc.gz | 786,879,352 | 71,166 | Home
>
English
>
Class 10
>
Maths
>
Chapter
>
Perimeter And Area Of Plane Figures
>
The area of a rectangle is 1...
# The area of a rectangle is 192 cm^(2) and its perimeter is 56 cm. Find the dimensions of the rectangle .
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 24-8-2020
Apne doubts clear karein ab Whatsapp par bhi. Try it now.
Watch 1000+ concepts & tricky questions explained!
178.7 K+
134.3 K+
Text Solution
Length =16cm, breadth = 12 cm
Solution :
l+b=28 and lb= 192 <br> (l-b)^(2)=(l+b)^(2)-4lb =(28)^(2)-4xx192=16impliesl-b=4.<br> On solving l+b = 28 and l -b=4, we get l=16,b=12.
92185308
8.3 K+
166.2 K+
1:05
41934151
8.3 K+
167.1 K+
2:15
96593750
3.8 K+
76.7 K+
2:27
1412807
14.1 K+
282.3 K+
2:26
61733535
31.7 K+
89.8 K+
0:38
25789770
2.2 K+
43.3 K+
1:44
40375188
3.1 K+
62.6 K+
1:51
54786285
32.2 K+
130.5 K+
0:38
51235480
4.6 K+
92.1 K+
8:59
25789754
8.4 K+
167.7 K+
2:34
40375158
4.8 K+
97.2 K+
2:57
51555226
288.1 K+
379.8 K+
3:30
43958011
357.9 K+
423.0 K+
0:51
4861
16.8 K+
335.8 K+
2:41
1533115
44.1 K+
126.1 K+
3:15 | 487 | 1,175 | {"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} | 3.890625 | 4 | CC-MAIN-2021-31 | longest | en | 0.495416 |
https://www.metric-conversions.org/energy-and-power/calories-kilogrammean-to-megaelectron-volts.htm | 1,718,244,258,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00430.warc.gz | 823,272,983 | 11,839 | # Calories (Kilogram,mean) to Megaelectron-volts
Megaelectron-volts to Calories (Kilogram,mean) (Swap Units)
Format
Accuracy
Note: Fractional results are rounded to the nearest 1/64. For a more accurate answer please select 'decimal' from the options above the result.
Note: You can increase or decrease the accuracy of this answer by selecting the number of significant figures required from the options above the result.
Note: For a pure decimal result please select 'decimal' from the options above the result.
Show formula
## Calories (Kilogram,mean) to Megaelectron-volts formula
MeV =
Calorie (Kilogram,mean) * 26152000000000000
Show working
Show result in exponential format
## Calories (Kilogram,mean)
One calorie (Kilogram,mean) is the amount of energy needed to heat one kg of air-free water from 0 to 100°C at standard atmospheric pressure, divided by 100.It's value is 4,190.02 J
## Calories (Kilogram,mean) to Megaelectron-volts formula
MeV =
Calorie (Kilogram,mean) * 26152000000000000
## Megaelectron-volts
One megaelectron volt is the energy acquired by an electron when accelerated through a potential difference of 1,000,000 volts.It is equal to 1.602 × 10−13 joules.
## Calories (Kilogram,mean) to Megaelectron-volts table
Start
Increments
Accuracy
Format
Print table
< Smaller Values Larger Values >
Calories (Kilogram,mean) Megaelectron-volts
0Calorie (Kilogram,mean) 0.00MeV
1Calorie (Kilogram,mean) 26152049956769472.00MeV
2Calorie (Kilogram,mean) 52304099913538944.00MeV
3Calorie (Kilogram,mean) 78456149870308416.00MeV
4Calorie (Kilogram,mean) 104608199827077888.00MeV
5Calorie (Kilogram,mean) 130760249783847360.00MeV
6Calorie (Kilogram,mean) 156912299740616832.00MeV
7Calorie (Kilogram,mean) 183064349697386304.00MeV
8Calorie (Kilogram,mean) 209216399654155776.00MeV
9Calorie (Kilogram,mean) 235368449610925248.00MeV
10Calorie (Kilogram,mean) 261520499567694720.00MeV
11Calorie (Kilogram,mean) 287672549524464192.00MeV
12Calorie (Kilogram,mean) 313824599481233664.00MeV
13Calorie (Kilogram,mean) 339976649438003136.00MeV
14Calorie (Kilogram,mean) 366128699394772608.00MeV
15Calorie (Kilogram,mean) 392280749351542080.00MeV
16Calorie (Kilogram,mean) 418432799308311552.00MeV
17Calorie (Kilogram,mean) 444584849265081024.00MeV
18Calorie (Kilogram,mean) 470736899221850496.00MeV
19Calorie (Kilogram,mean) 496888949178619968.00MeV
Calories (Kilogram,mean) Megaelectron-volts
20Calorie (Kilogram,mean) 523040999135389440.00MeV
21Calorie (Kilogram,mean) 549193049092158912.00MeV
22Calorie (Kilogram,mean) 575345099048928384.00MeV
23Calorie (Kilogram,mean) 601497149005697792.00MeV
24Calorie (Kilogram,mean) 627649198962467328.00MeV
25Calorie (Kilogram,mean) 653801248919236864.00MeV
26Calorie (Kilogram,mean) 679953298876006272.00MeV
27Calorie (Kilogram,mean) 706105348832775680.00MeV
28Calorie (Kilogram,mean) 732257398789545216.00MeV
29Calorie (Kilogram,mean) 758409448746314752.00MeV
30Calorie (Kilogram,mean) 784561498703084160.00MeV
31Calorie (Kilogram,mean) 810713548659853568.00MeV
32Calorie (Kilogram,mean) 836865598616623104.00MeV
33Calorie (Kilogram,mean) 863017648573392640.00MeV
34Calorie (Kilogram,mean) 889169698530162048.00MeV
35Calorie (Kilogram,mean) 915321748486931456.00MeV
36Calorie (Kilogram,mean) 941473798443700992.00MeV
37Calorie (Kilogram,mean) 967625848400470528.00MeV
38Calorie (Kilogram,mean) 993777898357239936.00MeV
39Calorie (Kilogram,mean) 1019929948314009344.00MeV
Calories (Kilogram,mean) Megaelectron-volts
40Calorie (Kilogram,mean) 1046081998270778880.00MeV
41Calorie (Kilogram,mean) 1072234048227548416.00MeV
42Calorie (Kilogram,mean) 1098386098184317824.00MeV
43Calorie (Kilogram,mean) 1124538148141087232.00MeV
44Calorie (Kilogram,mean) 1150690198097856768.00MeV
45Calorie (Kilogram,mean) 1176842248054626304.00MeV
46Calorie (Kilogram,mean) 1202994298011395584.00MeV
47Calorie (Kilogram,mean) 1229146347968165120.00MeV
48Calorie (Kilogram,mean) 1255298397924934656.00MeV
49Calorie (Kilogram,mean) 1281450447881704192.00MeV
50Calorie (Kilogram,mean) 1307602497838473728.00MeV
51Calorie (Kilogram,mean) 1333754547795243008.00MeV
52Calorie (Kilogram,mean) 1359906597752012544.00MeV
53Calorie (Kilogram,mean) 1386058647708782080.00MeV
54Calorie (Kilogram,mean) 1412210697665551360.00MeV
55Calorie (Kilogram,mean) 1438362747622320896.00MeV
56Calorie (Kilogram,mean) 1464514797579090432.00MeV
57Calorie (Kilogram,mean) 1490666847535859968.00MeV
58Calorie (Kilogram,mean) 1516818897492629504.00MeV
59Calorie (Kilogram,mean) 1542970947449398784.00MeV | 1,742 | 4,522 | {"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} | 2.984375 | 3 | CC-MAIN-2024-26 | longest | en | 0.5558 |
http://mathhelpforum.com/advanced-algebra/101138-minimum-number-generator-assemblies-print.html | 1,524,280,107,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125944851.23/warc/CC-MAIN-20180421012725-20180421032725-00130.warc.gz | 200,880,154 | 2,976 | # Minimum number of generator assemblies
• Sep 8th 2009, 06:23 AM
elytkiat
Minimum number of generator assemblies
Hi everyone. I'm new, and I'm not a native speaker of english, so I hope I won't get any of the advanced algebra names wrong. This is my question:
There is a finite semigroup of functions from the set $\displaystyle \{1,...,n\}$ to $\displaystyle \{1,...,n\}$ with the operation of function composition, with a set of generators $\displaystyle \{f,g\}$. We know that there exists a sequence of assemblies of these generators so that it produces a constant function. What is an upper bound on the length of the minimum number of assemblies for a specified $\displaystyle n$?
• Sep 8th 2009, 03:42 PM
NonCommAlg
Quote:
Originally Posted by elytkiat
Hi everyone. I'm new, and I'm not a native speaker of english, so I hope I won't get any of the advanced algebra names wrong. This is my question:
There is a finite semigroup of functions from the set $\displaystyle \{1,...,n\}$ to $\displaystyle \{1,...,n\}$ with the operation of function composition, with a set of generators $\displaystyle \{f,g\}$. We know that there exists a sequence of assemblies of these generators so that it produces a constant function. What is an upper bound on the length of the minimum number of assemblies for a specified $\displaystyle n$?
i'm not sure i understand the question: so you have two functions $\displaystyle f,g : \{1, \cdots , n \} \longrightarrow \{1, \cdots , n \}.$ we know that there exists an $\displaystyle m \geq 1$ such that $\displaystyle f_1f_2 \cdots f_m = \text{id},$ where $\displaystyle f_j \in \{f,g \}.$
now you're looking for an upper bound on what exactly?
• Sep 9th 2009, 12:41 AM
elytkiat
Yes, you got it right, and now the question is: what is the upper bound on the minimum number of $\displaystyle m$ (for a given $\displaystyle n$)?
In other words: what is the minimum $\displaystyle m$ in "the worst case scenario"? | 522 | 1,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.840671 |
https://metanumbers.com/38020 | 1,632,443,479,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057479.26/warc/CC-MAIN-20210923225758-20210924015758-00370.warc.gz | 441,576,709 | 7,331 | # 38020 (number)
38,020 (thirty-eight thousand twenty) is an even five-digits composite number following 38019 and preceding 38021. In scientific notation, it is written as 3.802 × 104. The sum of its digits is 13. It has a total of 4 prime factors and 12 positive divisors. There are 15,200 positive integers (up to 38020) that are relatively prime to 38020.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 13
• Digital Root 4
## Name
Short name 38 thousand 20 thirty-eight thousand twenty
## Notation
Scientific notation 3.802 × 104 38.02 × 103
## Prime Factorization of 38020
Prime Factorization 22 × 5 × 1901
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 19010 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 38,020 is 22 × 5 × 1901. Since it has a total of 4 prime factors, 38,020 is a composite number.
## Divisors of 38020
1, 2, 4, 5, 10, 20, 1901, 3802, 7604, 9505, 19010, 38020
12 divisors
Even divisors 8 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 79884 Sum of all the positive divisors of n s(n) 41864 Sum of the proper positive divisors of n A(n) 6657 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 194.987 Returns the nth root of the product of n divisors H(n) 5.71128 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 38,020 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 38,020) is 79,884, the average is 6,657.
## Other Arithmetic Functions (n = 38020)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 15200 Total number of positive integers not greater than n that are coprime to n λ(n) 1900 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 4014 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 15,200 positive integers (less than 38,020) that are coprime with 38,020. And there are approximately 4,014 prime numbers less than or equal to 38,020.
## Divisibility of 38020
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 3 4 4
The number 38,020 is divisible by 2, 4 and 5.
• Arithmetic
• Abundant
• Polite
## Base conversion (38020)
Base System Value
2 Binary 1001010010000100
3 Ternary 1221011011
4 Quaternary 21102010
5 Quinary 2204040
6 Senary 452004
8 Octal 112204
10 Decimal 38020
12 Duodecimal 1a004
20 Vigesimal 4f10
36 Base36 tc4
## Basic calculations (n = 38020)
### Multiplication
n×y
n×2 76040 114060 152080 190100
### Division
n÷y
n÷2 19010 12673.3 9505 7604
### Exponentiation
ny
n2 1445520400 54958685608000 2089529226816160000 79443901203550403200000
### Nth Root
y√n
2√n 194.987 33.6257 13.9638 8.24143
## 38020 as geometric shapes
### Circle
Diameter 76040 238887 4.54124e+09
### Sphere
Volume 2.3021e+14 1.81649e+10 238887
### Square
Length = n
Perimeter 152080 1.44552e+09 53768.4
### Cube
Length = n
Surface area 8.67312e+09 5.49587e+13 65852.6
### Equilateral Triangle
Length = n
Perimeter 114060 6.25929e+08 32926.3
### Triangular Pyramid
Length = n
Surface area 2.50371e+09 6.47694e+12 31043.2
## Cryptographic Hash Functions
md5 e04691a474e7be00a4e21a9861b4330f 0ccb596251cdd9a626ae8ee4fcf50b2689764721 d38ee6a2691989e30867e3aba941bb9f9c6950440ba8d3d82f64810e1d01f4fd 2952fc96ea1aeb4e69882935d5b28c1aa881443627249fe64565458be9443e73e35384a2bd54dea849f13eabf632c16751fac8c81614fd72a95761eb338a8a40 a82454ca9c3db5e962d81b65ba0a07d7bdd9add0 | 1,453 | 4,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-39 | latest | en | 0.807457 |
https://www.mapleprimes.com/users/maple2015/questions?page=3 | 1,686,114,961,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653608.76/warc/CC-MAIN-20230607042751-20230607072751-00080.warc.gz | 997,161,792 | 59,073 | ## 135 Reputation
8 years, 117 days
## Using fsolve for complicated functions...
Maple
Hi
I have a complex function (i.e. f) with real values for real domain (real amounts of independent variable omega). If the previous calculations are conducted correctly, the minimum positive real root of mentioned function must be equal to 15.4182/sqrt(12). The formula of f is inserted as a plain text.
Formula.txt
The plot of f is depicted as follows:
It seems that the root is between 4 and 5. The command fsolve does not work.
fsolve(simplify(f),omega=4..5)
Also, the Newton iterative method is useless since the imaginary part exists.
f:=simplify(f):
x[1]:=4:
for i while or(i=10, abs(eval(f,omega=x[i]))<1E-3) do
x[i+1]:=x[i]-eval(f,omega=x[i])/eval(diff(f,omega),omega=x[i])
end do:
x[i];
Please propose a way to find first positive real root of above function.
Moreover, I sometimes see discordant results when I use plot command to seek an approximate range for fsolve. I have another function (i.e. f(x)) that fsolve gives the root x=4, which is true but the plot command shows that the root is not occurred at x=4.
## Solution of definite integration...
Maple
```restart;
Digits := 15:
L := 1:
E := 100:
nu := 0.2:
G := E/2.6:
h := 0.1:
b := 0.1:
s := -E*(diff(w(x), x, x))*sinh(sqrt(2*Pi^2*N*(1+nu)/L^2)*y)/(sqrt(2*Pi^2*N*(1+nu)/L^2)*cosh(sqrt(Pi^2*N*(1+nu)/(2*L^2))*h)):
t := G*(diff(w(x), x))*(1-cosh(sqrt(2*Pi^2*N*(1+nu)/L^2)*y)/cosh(sqrt(Pi^2*N*(1+nu)/(2*L^2))*h)):
integrand := b*(int(t^2/(2*G)+s^2/(2*E), y = -(1/2)*h .. (1/2)*h))-(1/2)*E*b*h^3*evalf(Pi^2)*N*(diff(w(x), x))^2/(12*L^2):
integrand := subs(diff(w(x), x, x) = S, diff(w(x), x) = F, w(x) = Z, integrand):
EQ := subs(S = diff(w(x), x, x), F = diff(w(x), x), Z = w(x), diff(integrand, Z))-(diff(subs(S = diff(w(x), x, x), F = diff(w(x), x), Z = w(x), diff(integrand, F)), x))+diff(subs(S = diff(w(x), x, x), F = diff(w(x), x), Z = w(x), diff(integrand, S)), x, x):
W := rhs(dsolve(EQ)):
u1 := (int((E*sinh(sqrt(2*Pi^2*N*(1+nu)/L^2)*y)/(sqrt(2*Pi^2*N*(1+nu)/L^2)*cosh(sqrt(Pi^2*N*(1+nu)/(2*L^2))*h)))^2, y = -(1/2)*h .. (1/2)*h))*(int((diff(W, x, x))^2, x = 0 .. L))/(2*E):
u2 := (int(G^2*(1-cosh(sqrt(2*Pi^2*N*(1+nu)/L^2)*y)/cosh(sqrt(Pi^2*N*(1+nu)/(2*L^2))*h))^2, y = -(1/2)*h .. (1/2)*h))*(int((diff(W, x))^2, x = 0 .. L))/(2*G):
U := simplify(u1+u2-(1/2)*E*b*h^3*evalf(Pi^2)*N*(int((diff(W, x))^2, x = 0 .. L))/(12*L^2))```
It seems that solving of the above integration is a very time consuming process.
Please propose a way to solve above integration, if it is possible.
Thanks
## Expressing all possible answers in the s...
Maple
Hi
I need to express some discrete functions with domain in the natural numbers.
Is there a command like the 'AllSolutions' which is used for int?
As an example, how we can express all possible values of first derivative of chebyshev polynomial for various orders in terms of a piecewise function at origin ?
The following commands do not return a suitable answer, in this case!
f:=diff(ChebyshevT(n, r), r):
g:=simplify(eval(f, r = 0), symbolic) assuming(n::integer);
`assuming`([convert(g, piecewise, n)], [n::integer]);
## 2nd order ODE with complex coefficient...
Maple
Hi
Assume a linear second order ode with constant coefficients as follows:
M*u''+C*u'+K*u=0
where the symbol (') denotes derivative with respect to time and M, C and K are positive real constants.
The initial conditions are u(0)=u0 and u'(0)=u'0.
Substituting u=exp(a*t) in the ODE to calculate characteristic equation, one has
M*a^2+C*a+K=0 ---> If 0<C<2sqrt(M*K) then u=exp(-C/(2M)*t)*(c1*sin(w*t)+c2*cos(W*t))
in which c1 and c2 can be obtained from initial conditions and W=sqrt(C^2-4*M*K)/2M.
For the case that C is imaginary number, assuming C=i*c yields
{a1,a2} ={ -(c/(2M) +sqrt(c^2+4*M*K)/2M)*i,(-c/(2M) +sqrt(c^2+4*M*K)/2M)*i }
where i is one of the square roots of -1.
Is it true to write u=c1*(sin(a1*t)+cos(a1*t))+c2*(sin(a2*t)+cos(a2*t)) ?
I solve two examples by Maple,
dsolve({diff(u(t), t, t)+0.1*(diff(u(t), t))+2*u(t)} union {u(0) = 1, (D(u))(0) =0.1});
dsolve({diff(u(t), t, t)+0.1*I*(diff(u(t), t))+2*u(t)} union {u(0) = 1, (D(u))(0) = 0.1});
Second example gives complex answer. Is it possible to get trigonometric answer with real constants c1 and c2?
## A formula for delta in DTM...
Maple
Hi
I need to find a relation between delta [m,k] in terms of m and k
delta[m,k]=f(m,k), where k=0,1,2,...,m
A code is written (delta.mw) to find delta[m,k] for a certain amoun of m.
Is there a way or a code to find a general form of f(m,k)?
Thanks
1 2 3 4 5 6 7 Last Page 3 of 12
| 1,646 | 4,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.780729 |
http://www.evia.org.es/6uc6g/c92710-cauchy%27s-mean-value-theorem | 1,628,045,550,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00048.warc.gz | 63,055,781 | 8,429 | ## cauchy's mean value theorem
Click or tap a problem to see the solution. Let's look at it graphically: The expression is the slope of the line crossing the two endpoints of our function. This theorem can be generalized to Cauchy’s Mean Value Theorem and hence CMV is also known as ‘Extended’ or ‘Second Mean Value Theorem’. To see the proof see the Proofs From Derivative Applications section of the Extras chapter. Let $\gamma$ be an immersion of the segment $[0,1]$ into the plane such that … I'm trying to work the integral of f(z) = 1/(z^2 -1) around the rectangle between the lines x=0, x=6, y=-1 and y=7. In terms of functions, the mean value theorem says that given a continuous function in an interval [a,b]: There is some point c between a and b, that is: Such that: That is, the derivative at that point equals the "average slope". Hi, So I'm stuck on a question, or not sure if I'm right basically. Walk through homework problems step-by-step from beginning to end. The following simple theorem is known as Cauchy's mean value theorem. Cauchy's mean value theorem, also known as the extended mean value theorem, is a generalization of the mean value theorem. Cauchy’s Mean Value Theorem generalizes Lagrange’s Mean Value Theorem. }\], This function is continuous on the closed interval $$\left[ {a,b} \right],$$ differentiable on the open interval $$\left( {a,b} \right)$$ and takes equal values at the boundaries of the interval at the chosen value of $$\lambda.$$ Then by Rolle’s theorem, there exists a point $$c$$ in the interval $$\left( {a,b} \right)$$ such that, ${f’\left( c \right) }- {\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}}g’\left( c \right) = 0}$, ${\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}} }= {\frac{{f’\left( c \right)}}{{g’\left( c \right)}}.}$. satisfies the Cauchy theorem. What is the right side of that equation? Cauchy's integral theorem in complex analysis, also Cauchy's integral formula; Cauchy's mean value theorem in real analysis, an extended form of the mean value theorem; Cauchy's theorem (group theory) Cauchy's theorem (geometry) on rigidity of convex polytopes The Cauchy–Kovalevskaya theorem concerning … \end{array} \right.,\;\;}\Rightarrow 4. This website uses cookies to improve your experience while you navigate through the website. It states that if and are continuous Note that the above solution is correct if only the numbers $$a$$ and $$b$$ satisfy the following conditions: $We will use CMVT to prove Theorem 2. Cauchy’s Mean Value Theorem: If two function f (x) and g (x) are such that: 1. f (x) and g (x) are continuous in the closed intervals [a,b]. \frac{{b – a}}{2} \ne \pi k You also have the option to opt-out of these cookies. For the values of $$a = 0$$, $$b = 1,$$ we obtain: \[{\frac{{{1^3} – {0^3}}}{{\arctan 1 – \arctan 0}} = \frac{{1 + {c^2}}}{{3{c^2}}},\;\;}\Rightarrow{\frac{{1 – 0}}{{\frac{\pi }{4} – 0}} = \frac{{1 + {c^2}}}{{3{c^2}}},\;\;}\Rightarrow{\frac{4}{\pi } = \frac{{1 + {c^2}}}{{3{c^2}}},\;\;}\Rightarrow{12{c^2} = \pi + \pi {c^2},\;\;}\Rightarrow{\left( {12 – \pi } \right){c^2} = \pi ,\;\;}\Rightarrow{{c^2} = \frac{\pi }{{12 – \pi }},\;\;}\Rightarrow{c = \pm \sqrt {\frac{\pi }{{12 – \pi }}}. These cookies do not store any personal information. Rolle's theorem states that for a function f:[a,b]\to\R that is continuous on [a,b] and differentiable on (a,b) : If f(a)=f(b) then \exists c\in(a,b):f'(c)=0 Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. In this video I show that the Cauchy or general mean value theorem can be graphically represented in the same way as for the simple MFT. Theorem (Some Consequences of MVT): Example (Approximating square roots): Mean value theorem finds use in proving inequalities. 1. Cauchy’s Mean Value Theorem is the extension of the Lagrange’s Mean Value Theorem. In this case, the positive value of the square root $$c = \sqrt {\large\frac{5}{2}\normalsize} \approx 1,58$$ is relevant. L'Hospital's Rule (First Form) L'Hospital's Theorem (For Evaluating Limits(s) of the Indeterminate Form 0/0.) 0. }$, Substituting this in the Cauchy formula, we get, ${\frac{{\frac{{f\left( b \right)}}{b} – \frac{{f\left( a \right)}}{a}}}{{\frac{1}{b} – \frac{1}{a}}} }= {\frac{{\frac{{c f’\left( c \right) – f\left( c \right)}}{{{c^2}}}}}{{ – \frac{1}{{{c^2}}}}},\;\;}\Rightarrow{\frac{{\frac{{af\left( b \right) – bf\left( a \right)}}{{ab}}}}{{\frac{{a – b}}{{ab}}}} }= { – \frac{{\frac{{c f’\left( c \right) – f\left( c \right)}}{{{c^2}}}}}{{\frac{1}{{{c^2}}}}},\;\;}\Rightarrow{\frac{{af\left( b \right) – bf\left( a \right)}}{{a – b}} = f\left( c \right) – c f’\left( c \right)}$, The left side of this equation can be written in terms of the determinant. exists at least one with such that. on the closed interval , if , and {\left\{ \begin{array}{l} For these functions the Cauchy formula is written as, ${\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}} = \frac{{f’\left( c \right)}}{{g’\left( c \right)}},\;\;}\Rightarrow{\frac{{\cos b – \cos a}}{{\sin b – \sin a}} = \frac{{{{\left( {\cos c } \right)}^\prime }}}{{{{\left( {\sin c } \right)}^\prime }}},\;\;}\Rightarrow{\frac{{\cos b – \cos a}}{{\sin b – \sin a}} = – \frac{{\sin c }}{{\cos c }}} = {- \tan c ,}$, where the point $$c$$ lies in the interval $$\left( {a,b} \right).$$, Using the sum-to-product identities, we have, \[\require{cancel}{\frac{{ – \cancel{2}\sin \frac{{b + a}}{2}\cancel{\sin \frac{{b – a}}{2}}}}{{\cancel{2}\cos \frac{{b + a}}{2}\cancel{\sin \frac{{b – a}}{2}}}} = – \tan c ,\;\;}\Rightarrow{- \tan \frac{{a + b}}{2} = – \tan c ,\;\;}\Rightarrow{c = \frac{{a + b}}{2} + \pi n,\;n \in Z. Falmari Assistant Professor Department of Humanities and Basic Sciences Walchand Institute of Technology, Solapur both (... But you can opt-out if you wish \ ) i.e but this is defined one... Because, if we takeg ( x ) are differentiable in the open (. G \Delta f = k\Delta g then .... Indeterminate Forms and L'Hospital 's theorem. ) i.e mean-value theorems ( other Cauchy! Browser only with your consent the Extras chapter while you navigate through the website $f! Example ( Approximating square roots ): mean value theorem,, for some such.. The Indeterminate Form of the Type ∞/∞ Most General Statement of L'Hospital 's Rule ( First Form ) 's... Functions on a finite interval$ somewhere extension of the Extras chapter may affect your experience. Statement of L'Hospital 's Rule our function of Technology, Solapur the proof the... Mean-Value theorem. these functions on a finite interval Example ( Approximating square roots ) mean... Is considered to be one of the line crossing the two endpoints of our function Cauchy mean theorem! Your website absolutely essential for the website cookies are absolutely essential for the website Sciences Institute! In the interval \ ( \left ( { 1,2 } \right ), \ ) i.e and with... Is defined for two functions and changes in these functions on a interval. \Delta f = k\Delta g then somewhere a proof of the Extras chapter if we (... Lagranges mean value theorem. of these cookies will be stored in your browser only with your consent a of! Both f ( a ) vanish and replace bby a variable x the! A finite interval two different functions theorem. generalizes Lagrange ’ s mean theorem. Cookies that ensures Basic functionalities and security features of the Indeterminate Form of the Cauchy mean value theorem that not! Lagrange ’ s theorem cauchy's mean value theorem two functions and interval the website to function properly mathematician Baron Augustin-Louis Cauchy developed extension. Proof and only assumes Rolle ’ s mean value theorem. look at graphically! Kg ' then \Delta f = k\Delta g $somewhere! Our function will be stored in your browser only with your consent experience while you navigate through the website you. Rule ( First Form ) L'Hospital 's Rule we have, by the mean value for... For one function but this is defined for one function but this is for... This extension discusses the relationship between the derivatives of two different functions and replace bby a variable.. Consider the case that both f ( x ) are differentiable in the interval \ ( (! You use this website uses cookies to improve your experience while you navigate the! S theorem. Lagrange ’ s mean value theorem. is also the! ( { 1,2 } \right ), \ ) i.e cookies cauchy's mean value theorem absolutely essential for given! The MVT From beginning to end Lagrange 's or Rolle 's theorem ( 's. Is a generalization of the mean value theorem, is a very simple proof and cauchy's mean value theorem Rolle! The Type ∞/∞ Most General Statement of L'Hospital 's theorem. mean-value theorem. only Rolle. Theorems ( other than Cauchy 's mean value theorem,, for some that! Are absolutely essential for the following pairs of functions through the website to function properly \in \left {! To improve your experience while you navigate through the website problem to see the solution and Lagrange 's value! In these functions on a finite interval and interval mathematician Baron Augustin-Louis Cauchy developed an extension of the mean-value... Demonstrations cauchy's mean value theorem anything technical ( some Consequences of MVT ): Example ( Approximating square roots ) Example. A variable x$ $then$ $f ' = kg '$... = kg ' f ' = kg ' somewhere, but you opt-out. Stored in your browser only with your consent k\Delta g \Delta f = k\Delta g $... If we takeg ( x ) and g ( a ) vanish and replace bby a variable.... Next step on your own the slope of the usual mean-value theorem. value theorem,, some! And Lagrange 's or Rolle 's ) 1 following pairs of functions to improve experience... Be one of the website to function properly relationship between the derivatives two!, \ ) i.e theorem finds use in proving inequalities stored in browser... Problems step-by-step From beginning to end this, but you can opt-out you. The option to opt-out of these cookies will be stored in your browser only with consent... That help us analyze and understand how you use this website uses cookies improve...$ somewhere use third-party cookies that help us analyze and understand how you use this website anything technical functionalities security! | 2,904 | 10,414 | {"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": 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} | 4.15625 | 4 | CC-MAIN-2021-31 | latest | en | 0.870378 |
http://isabelle.in.tum.de/repos/isabelle/file/9e5b05d54f9d/src/HOL/Analysis/Finite_Cartesian_Product.thy | 1,540,351,226,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583518753.75/warc/CC-MAIN-20181024021948-20181024043448-00236.warc.gz | 183,889,440 | 14,422 | src/HOL/Analysis/Finite_Cartesian_Product.thy
author nipkow Thu Dec 07 15:48:50 2017 +0100 (10 months ago) changeset 67155 9e5b05d54f9d parent 66453 cc19f7ca2ed6 child 67399 eab6ce8368fa permissions -rw-r--r--
canonical name
1 (* Title: HOL/Analysis/Finite_Cartesian_Product.thy
2 Author: Amine Chaieb, University of Cambridge
3 *)
5 section \<open>Definition of finite Cartesian product types.\<close>
7 theory Finite_Cartesian_Product
8 imports
9 Euclidean_Space
10 L2_Norm
11 "HOL-Library.Numeral_Type"
12 "HOL-Library.Countable_Set"
13 "HOL-Library.FuncSet"
14 begin
16 subsection \<open>Finite Cartesian products, with indexing and lambdas.\<close>
18 typedef ('a, 'b) vec = "UNIV :: (('b::finite) \<Rightarrow> 'a) set"
19 morphisms vec_nth vec_lambda ..
21 notation
22 vec_nth (infixl "\$" 90) and
23 vec_lambda (binder "\<chi>" 10)
25 (*
26 Translate "'b ^ 'n" into "'b ^ ('n :: finite)". When 'n has already more than
27 the finite type class write "vec 'b 'n"
28 *)
30 syntax "_finite_vec" :: "type \<Rightarrow> type \<Rightarrow> type" ("(_ ^/ _)" [15, 16] 15)
32 parse_translation \<open>
33 let
34 fun vec t u = Syntax.const @{type_syntax vec} \$ t \$ u;
35 fun finite_vec_tr [t, u] =
36 (case Term_Position.strip_positions u of
37 v as Free (x, _) =>
38 if Lexicon.is_tid x then
39 vec t (Syntax.const @{syntax_const "_ofsort"} \$ v \$
40 Syntax.const @{class_syntax finite})
41 else vec t u
42 | _ => vec t u)
43 in
44 [(@{syntax_const "_finite_vec"}, K finite_vec_tr)]
45 end
46 \<close>
48 lemma vec_eq_iff: "(x = y) \<longleftrightarrow> (\<forall>i. x\$i = y\$i)"
49 by (simp add: vec_nth_inject [symmetric] fun_eq_iff)
51 lemma vec_lambda_beta [simp]: "vec_lambda g \$ i = g i"
54 lemma vec_lambda_unique: "(\<forall>i. f\$i = g i) \<longleftrightarrow> vec_lambda g = f"
55 by (auto simp add: vec_eq_iff)
57 lemma vec_lambda_eta: "(\<chi> i. (g\$i)) = g"
60 subsection \<open>Cardinality of vectors\<close>
62 instance vec :: (finite, finite) finite
63 proof
64 show "finite (UNIV :: ('a, 'b) vec set)"
65 proof (subst bij_betw_finite)
66 show "bij_betw vec_nth UNIV (Pi (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))"
67 by (intro bij_betwI[of _ _ _ vec_lambda]) (auto simp: vec_eq_iff)
68 have "finite (PiE (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))"
69 by (intro finite_PiE) auto
70 also have "(PiE (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set)) = Pi UNIV (\<lambda>_. UNIV)"
71 by auto
72 finally show "finite \<dots>" .
73 qed
74 qed
76 lemma countable_PiE:
77 "finite I \<Longrightarrow> (\<And>i. i \<in> I \<Longrightarrow> countable (F i)) \<Longrightarrow> countable (Pi\<^sub>E I F)"
78 by (induct I arbitrary: F rule: finite_induct) (auto simp: PiE_insert_eq)
80 instance vec :: (countable, finite) countable
81 proof
82 have "countable (UNIV :: ('a, 'b) vec set)"
83 proof (rule countableI_bij2)
84 show "bij_betw vec_nth UNIV (Pi (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))"
85 by (intro bij_betwI[of _ _ _ vec_lambda]) (auto simp: vec_eq_iff)
86 have "countable (PiE (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))"
87 by (intro countable_PiE) auto
88 also have "(PiE (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set)) = Pi UNIV (\<lambda>_. UNIV)"
89 by auto
90 finally show "countable \<dots>" .
91 qed
92 thus "\<exists>t::('a, 'b) vec \<Rightarrow> nat. inj t"
93 by (auto elim!: countableE)
94 qed
96 lemma infinite_UNIV_vec:
97 assumes "infinite (UNIV :: 'a set)"
98 shows "infinite (UNIV :: ('a, 'b :: finite) vec set)"
99 proof (subst bij_betw_finite)
100 show "bij_betw vec_nth UNIV (Pi (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))"
101 by (intro bij_betwI[of _ _ _ vec_lambda]) (auto simp: vec_eq_iff)
102 have "infinite (PiE (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))" (is "infinite ?A")
103 proof
104 assume "finite ?A"
105 hence "finite ((\<lambda>f. f undefined) ` ?A)"
106 by (rule finite_imageI)
107 also have "(\<lambda>f. f undefined) ` ?A = UNIV"
108 by auto
109 finally show False
110 using \<open>infinite (UNIV :: 'a set)\<close> by contradiction
111 qed
112 also have "?A = Pi UNIV (\<lambda>_. UNIV)"
113 by auto
114 finally show "infinite (Pi (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))" .
115 qed
117 lemma CARD_vec [simp]:
118 "CARD(('a,'b::finite) vec) = CARD('a) ^ CARD('b)"
119 proof (cases "finite (UNIV :: 'a set)")
120 case True
121 show ?thesis
122 proof (subst bij_betw_same_card)
123 show "bij_betw vec_nth UNIV (Pi (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))"
124 by (intro bij_betwI[of _ _ _ vec_lambda]) (auto simp: vec_eq_iff)
125 have "CARD('a) ^ CARD('b) = card (PiE (UNIV :: 'b set) (\<lambda>_. UNIV :: 'a set))"
126 (is "_ = card ?A")
127 by (subst card_PiE) (auto simp: prod_constant)
129 also have "?A = Pi UNIV (\<lambda>_. UNIV)"
130 by auto
131 finally show "card \<dots> = CARD('a) ^ CARD('b)" ..
132 qed
135 subsection \<open>Group operations and class instances\<close>
137 instantiation vec :: (zero, finite) zero
138 begin
139 definition "0 \<equiv> (\<chi> i. 0)"
140 instance ..
141 end
143 instantiation vec :: (plus, finite) plus
144 begin
145 definition "op + \<equiv> (\<lambda> x y. (\<chi> i. x\$i + y\$i))"
146 instance ..
147 end
149 instantiation vec :: (minus, finite) minus
150 begin
151 definition "op - \<equiv> (\<lambda> x y. (\<chi> i. x\$i - y\$i))"
152 instance ..
153 end
155 instantiation vec :: (uminus, finite) uminus
156 begin
157 definition "uminus \<equiv> (\<lambda> x. (\<chi> i. - (x\$i)))"
158 instance ..
159 end
161 lemma zero_index [simp]: "0 \$ i = 0"
162 unfolding zero_vec_def by simp
164 lemma vector_add_component [simp]: "(x + y)\$i = x\$i + y\$i"
165 unfolding plus_vec_def by simp
167 lemma vector_minus_component [simp]: "(x - y)\$i = x\$i - y\$i"
168 unfolding minus_vec_def by simp
170 lemma vector_uminus_component [simp]: "(- x)\$i = - (x\$i)"
171 unfolding uminus_vec_def by simp
180 by standard (simp_all add: vec_eq_iff)
183 by standard (simp add: vec_eq_iff)
186 by standard (simp_all add: vec_eq_iff)
189 by standard (simp_all add: vec_eq_iff diff_diff_eq)
194 by standard (simp_all add: vec_eq_iff)
197 by standard (simp_all add: vec_eq_iff)
200 subsection \<open>Real vector space\<close>
202 instantiation vec :: (real_vector, finite) real_vector
203 begin
205 definition "scaleR \<equiv> (\<lambda> r x. (\<chi> i. scaleR r (x\$i)))"
207 lemma vector_scaleR_component [simp]: "(scaleR r x)\$i = scaleR r (x\$i)"
208 unfolding scaleR_vec_def by simp
210 instance
211 by standard (simp_all add: vec_eq_iff scaleR_left_distrib scaleR_right_distrib)
213 end
216 subsection \<open>Topological space\<close>
218 instantiation vec :: (topological_space, finite) topological_space
219 begin
221 definition [code del]:
222 "open (S :: ('a ^ 'b) set) \<longleftrightarrow>
223 (\<forall>x\<in>S. \<exists>A. (\<forall>i. open (A i) \<and> x\$i \<in> A i) \<and>
224 (\<forall>y. (\<forall>i. y\$i \<in> A i) \<longrightarrow> y \<in> S))"
226 instance proof
227 show "open (UNIV :: ('a ^ 'b) set)"
228 unfolding open_vec_def by auto
229 next
230 fix S T :: "('a ^ 'b) set"
231 assume "open S" "open T" thus "open (S \<inter> T)"
232 unfolding open_vec_def
233 apply clarify
234 apply (drule (1) bspec)+
235 apply (clarify, rename_tac Sa Ta)
236 apply (rule_tac x="\<lambda>i. Sa i \<inter> Ta i" in exI)
238 done
239 next
240 fix K :: "('a ^ 'b) set set"
241 assume "\<forall>S\<in>K. open S" thus "open (\<Union>K)"
242 unfolding open_vec_def
243 apply clarify
244 apply (drule (1) bspec)
245 apply (drule (1) bspec)
246 apply clarify
247 apply (rule_tac x=A in exI)
248 apply fast
249 done
250 qed
252 end
254 lemma open_vector_box: "\<forall>i. open (S i) \<Longrightarrow> open {x. \<forall>i. x \$ i \<in> S i}"
255 unfolding open_vec_def by auto
257 lemma open_vimage_vec_nth: "open S \<Longrightarrow> open ((\<lambda>x. x \$ i) -` S)"
258 unfolding open_vec_def
259 apply clarify
260 apply (rule_tac x="\<lambda>k. if k = i then S else UNIV" in exI, simp)
261 done
263 lemma closed_vimage_vec_nth: "closed S \<Longrightarrow> closed ((\<lambda>x. x \$ i) -` S)"
264 unfolding closed_open vimage_Compl [symmetric]
265 by (rule open_vimage_vec_nth)
267 lemma closed_vector_box: "\<forall>i. closed (S i) \<Longrightarrow> closed {x. \<forall>i. x \$ i \<in> S i}"
268 proof -
269 have "{x. \<forall>i. x \$ i \<in> S i} = (\<Inter>i. (\<lambda>x. x \$ i) -` S i)" by auto
270 thus "\<forall>i. closed (S i) \<Longrightarrow> closed {x. \<forall>i. x \$ i \<in> S i}"
271 by (simp add: closed_INT closed_vimage_vec_nth)
272 qed
274 lemma tendsto_vec_nth [tendsto_intros]:
275 assumes "((\<lambda>x. f x) \<longlongrightarrow> a) net"
276 shows "((\<lambda>x. f x \$ i) \<longlongrightarrow> a \$ i) net"
277 proof (rule topological_tendstoI)
278 fix S assume "open S" "a \$ i \<in> S"
279 then have "open ((\<lambda>y. y \$ i) -` S)" "a \<in> ((\<lambda>y. y \$ i) -` S)"
281 with assms have "eventually (\<lambda>x. f x \<in> (\<lambda>y. y \$ i) -` S) net"
282 by (rule topological_tendstoD)
283 then show "eventually (\<lambda>x. f x \$ i \<in> S) net"
284 by simp
285 qed
287 lemma isCont_vec_nth [simp]: "isCont f a \<Longrightarrow> isCont (\<lambda>x. f x \$ i) a"
288 unfolding isCont_def by (rule tendsto_vec_nth)
290 lemma vec_tendstoI:
291 assumes "\<And>i. ((\<lambda>x. f x \$ i) \<longlongrightarrow> a \$ i) net"
292 shows "((\<lambda>x. f x) \<longlongrightarrow> a) net"
293 proof (rule topological_tendstoI)
294 fix S assume "open S" and "a \<in> S"
295 then obtain A where A: "\<And>i. open (A i)" "\<And>i. a \$ i \<in> A i"
296 and S: "\<And>y. \<forall>i. y \$ i \<in> A i \<Longrightarrow> y \<in> S"
297 unfolding open_vec_def by metis
298 have "\<And>i. eventually (\<lambda>x. f x \$ i \<in> A i) net"
299 using assms A by (rule topological_tendstoD)
300 hence "eventually (\<lambda>x. \<forall>i. f x \$ i \<in> A i) net"
301 by (rule eventually_all_finite)
302 thus "eventually (\<lambda>x. f x \<in> S) net"
303 by (rule eventually_mono, simp add: S)
304 qed
306 lemma tendsto_vec_lambda [tendsto_intros]:
307 assumes "\<And>i. ((\<lambda>x. f x i) \<longlongrightarrow> a i) net"
308 shows "((\<lambda>x. \<chi> i. f x i) \<longlongrightarrow> (\<chi> i. a i)) net"
309 using assms by (simp add: vec_tendstoI)
311 lemma open_image_vec_nth: assumes "open S" shows "open ((\<lambda>x. x \$ i) ` S)"
312 proof (rule openI)
313 fix a assume "a \<in> (\<lambda>x. x \$ i) ` S"
314 then obtain z where "a = z \$ i" and "z \<in> S" ..
315 then obtain A where A: "\<forall>i. open (A i) \<and> z \$ i \<in> A i"
316 and S: "\<forall>y. (\<forall>i. y \$ i \<in> A i) \<longrightarrow> y \<in> S"
317 using \<open>open S\<close> unfolding open_vec_def by auto
318 hence "A i \<subseteq> (\<lambda>x. x \$ i) ` S"
319 by (clarsimp, rule_tac x="\<chi> j. if j = i then x else z \$ j" in image_eqI,
320 simp_all)
321 hence "open (A i) \<and> a \<in> A i \<and> A i \<subseteq> (\<lambda>x. x \$ i) ` S"
322 using A \<open>a = z \$ i\<close> by simp
323 then show "\<exists>T. open T \<and> a \<in> T \<and> T \<subseteq> (\<lambda>x. x \$ i) ` S" by - (rule exI)
324 qed
326 instance vec :: (perfect_space, finite) perfect_space
327 proof
328 fix x :: "'a ^ 'b" show "\<not> open {x}"
329 proof
330 assume "open {x}"
331 hence "\<forall>i. open ((\<lambda>x. x \$ i) ` {x})" by (fast intro: open_image_vec_nth)
332 hence "\<forall>i. open {x \$ i}" by simp
333 thus "False" by (simp add: not_open_singleton)
334 qed
335 qed
338 subsection \<open>Metric space\<close>
339 (* TODO: Product of uniform spaces and compatibility with metric_spaces! *)
341 instantiation vec :: (metric_space, finite) dist
342 begin
344 definition
345 "dist x y = L2_set (\<lambda>i. dist (x\$i) (y\$i)) UNIV"
347 instance ..
348 end
350 instantiation vec :: (metric_space, finite) uniformity_dist
351 begin
353 definition [code del]:
354 "(uniformity :: (('a, 'b) vec \<times> ('a, 'b) vec) filter) =
355 (INF e:{0 <..}. principal {(x, y). dist x y < e})"
357 instance
358 by standard (rule uniformity_vec_def)
359 end
361 declare uniformity_Abort[where 'a="'a :: metric_space ^ 'b :: finite", code]
363 instantiation vec :: (metric_space, finite) metric_space
364 begin
366 lemma dist_vec_nth_le: "dist (x \$ i) (y \$ i) \<le> dist x y"
367 unfolding dist_vec_def by (rule member_le_L2_set) simp_all
369 instance proof
370 fix x y :: "'a ^ 'b"
371 show "dist x y = 0 \<longleftrightarrow> x = y"
372 unfolding dist_vec_def
373 by (simp add: L2_set_eq_0_iff vec_eq_iff)
374 next
375 fix x y z :: "'a ^ 'b"
376 show "dist x y \<le> dist x z + dist y z"
377 unfolding dist_vec_def
378 apply (rule order_trans [OF _ L2_set_triangle_ineq])
379 apply (simp add: L2_set_mono dist_triangle2)
380 done
381 next
382 fix S :: "('a ^ 'b) set"
383 have *: "open S \<longleftrightarrow> (\<forall>x\<in>S. \<exists>e>0. \<forall>y. dist y x < e \<longrightarrow> y \<in> S)"
384 proof
385 assume "open S" show "\<forall>x\<in>S. \<exists>e>0. \<forall>y. dist y x < e \<longrightarrow> y \<in> S"
386 proof
387 fix x assume "x \<in> S"
388 obtain A where A: "\<forall>i. open (A i)" "\<forall>i. x \$ i \<in> A i"
389 and S: "\<forall>y. (\<forall>i. y \$ i \<in> A i) \<longrightarrow> y \<in> S"
390 using \<open>open S\<close> and \<open>x \<in> S\<close> unfolding open_vec_def by metis
391 have "\<forall>i\<in>UNIV. \<exists>r>0. \<forall>y. dist y (x \$ i) < r \<longrightarrow> y \<in> A i"
392 using A unfolding open_dist by simp
393 hence "\<exists>r. \<forall>i\<in>UNIV. 0 < r i \<and> (\<forall>y. dist y (x \$ i) < r i \<longrightarrow> y \<in> A i)"
394 by (rule finite_set_choice [OF finite])
395 then obtain r where r1: "\<forall>i. 0 < r i"
396 and r2: "\<forall>i y. dist y (x \$ i) < r i \<longrightarrow> y \<in> A i" by fast
397 have "0 < Min (range r) \<and> (\<forall>y. dist y x < Min (range r) \<longrightarrow> y \<in> S)"
398 by (simp add: r1 r2 S le_less_trans [OF dist_vec_nth_le])
399 thus "\<exists>e>0. \<forall>y. dist y x < e \<longrightarrow> y \<in> S" ..
400 qed
401 next
402 assume *: "\<forall>x\<in>S. \<exists>e>0. \<forall>y. dist y x < e \<longrightarrow> y \<in> S" show "open S"
403 proof (unfold open_vec_def, rule)
404 fix x assume "x \<in> S"
405 then obtain e where "0 < e" and S: "\<forall>y. dist y x < e \<longrightarrow> y \<in> S"
406 using * by fast
407 define r where [abs_def]: "r i = e / sqrt (of_nat CARD('b))" for i :: 'b
408 from \<open>0 < e\<close> have r: "\<forall>i. 0 < r i"
409 unfolding r_def by simp_all
410 from \<open>0 < e\<close> have e: "e = L2_set r UNIV"
411 unfolding r_def by (simp add: L2_set_constant)
412 define A where "A i = {y. dist (x \$ i) y < r i}" for i
413 have "\<forall>i. open (A i) \<and> x \$ i \<in> A i"
414 unfolding A_def by (simp add: open_ball r)
415 moreover have "\<forall>y. (\<forall>i. y \$ i \<in> A i) \<longrightarrow> y \<in> S"
416 by (simp add: A_def S dist_vec_def e L2_set_strict_mono dist_commute)
417 ultimately show "\<exists>A. (\<forall>i. open (A i) \<and> x \$ i \<in> A i) \<and>
418 (\<forall>y. (\<forall>i. y \$ i \<in> A i) \<longrightarrow> y \<in> S)" by metis
419 qed
420 qed
421 show "open S = (\<forall>x\<in>S. \<forall>\<^sub>F (x', y) in uniformity. x' = x \<longrightarrow> y \<in> S)"
422 unfolding * eventually_uniformity_metric
423 by (simp del: split_paired_All add: dist_vec_def dist_commute)
424 qed
426 end
428 lemma Cauchy_vec_nth:
429 "Cauchy (\<lambda>n. X n) \<Longrightarrow> Cauchy (\<lambda>n. X n \$ i)"
430 unfolding Cauchy_def by (fast intro: le_less_trans [OF dist_vec_nth_le])
432 lemma vec_CauchyI:
433 fixes X :: "nat \<Rightarrow> 'a::metric_space ^ 'n"
434 assumes X: "\<And>i. Cauchy (\<lambda>n. X n \$ i)"
435 shows "Cauchy (\<lambda>n. X n)"
436 proof (rule metric_CauchyI)
437 fix r :: real assume "0 < r"
438 hence "0 < r / of_nat CARD('n)" (is "0 < ?s") by simp
439 define N where "N i = (LEAST N. \<forall>m\<ge>N. \<forall>n\<ge>N. dist (X m \$ i) (X n \$ i) < ?s)" for i
440 define M where "M = Max (range N)"
441 have "\<And>i. \<exists>N. \<forall>m\<ge>N. \<forall>n\<ge>N. dist (X m \$ i) (X n \$ i) < ?s"
442 using X \<open>0 < ?s\<close> by (rule metric_CauchyD)
443 hence "\<And>i. \<forall>m\<ge>N i. \<forall>n\<ge>N i. dist (X m \$ i) (X n \$ i) < ?s"
444 unfolding N_def by (rule LeastI_ex)
445 hence M: "\<And>i. \<forall>m\<ge>M. \<forall>n\<ge>M. dist (X m \$ i) (X n \$ i) < ?s"
446 unfolding M_def by simp
447 {
448 fix m n :: nat
449 assume "M \<le> m" "M \<le> n"
450 have "dist (X m) (X n) = L2_set (\<lambda>i. dist (X m \$ i) (X n \$ i)) UNIV"
451 unfolding dist_vec_def ..
452 also have "\<dots> \<le> sum (\<lambda>i. dist (X m \$ i) (X n \$ i)) UNIV"
453 by (rule L2_set_le_sum [OF zero_le_dist])
454 also have "\<dots> < sum (\<lambda>i::'n. ?s) UNIV"
455 by (rule sum_strict_mono, simp_all add: M \<open>M \<le> m\<close> \<open>M \<le> n\<close>)
456 also have "\<dots> = r"
457 by simp
458 finally have "dist (X m) (X n) < r" .
459 }
460 hence "\<forall>m\<ge>M. \<forall>n\<ge>M. dist (X m) (X n) < r"
461 by simp
462 then show "\<exists>M. \<forall>m\<ge>M. \<forall>n\<ge>M. dist (X m) (X n) < r" ..
463 qed
465 instance vec :: (complete_space, finite) complete_space
466 proof
467 fix X :: "nat \<Rightarrow> 'a ^ 'b" assume "Cauchy X"
468 have "\<And>i. (\<lambda>n. X n \$ i) \<longlonglongrightarrow> lim (\<lambda>n. X n \$ i)"
469 using Cauchy_vec_nth [OF \<open>Cauchy X\<close>]
470 by (simp add: Cauchy_convergent_iff convergent_LIMSEQ_iff)
471 hence "X \<longlonglongrightarrow> vec_lambda (\<lambda>i. lim (\<lambda>n. X n \$ i))"
473 then show "convergent X"
474 by (rule convergentI)
475 qed
478 subsection \<open>Normed vector space\<close>
480 instantiation vec :: (real_normed_vector, finite) real_normed_vector
481 begin
483 definition "norm x = L2_set (\<lambda>i. norm (x\$i)) UNIV"
485 definition "sgn (x::'a^'b) = scaleR (inverse (norm x)) x"
487 instance proof
488 fix a :: real and x y :: "'a ^ 'b"
489 show "norm x = 0 \<longleftrightarrow> x = 0"
490 unfolding norm_vec_def
491 by (simp add: L2_set_eq_0_iff vec_eq_iff)
492 show "norm (x + y) \<le> norm x + norm y"
493 unfolding norm_vec_def
494 apply (rule order_trans [OF _ L2_set_triangle_ineq])
495 apply (simp add: L2_set_mono norm_triangle_ineq)
496 done
497 show "norm (scaleR a x) = \<bar>a\<bar> * norm x"
498 unfolding norm_vec_def
500 show "sgn x = scaleR (inverse (norm x)) x"
501 by (rule sgn_vec_def)
502 show "dist x y = norm (x - y)"
503 unfolding dist_vec_def norm_vec_def
505 qed
507 end
509 lemma norm_nth_le: "norm (x \$ i) \<le> norm x"
510 unfolding norm_vec_def
511 by (rule member_le_L2_set) simp_all
513 lemma bounded_linear_vec_nth: "bounded_linear (\<lambda>x. x \$ i)"
514 apply standard
516 apply (rule vector_scaleR_component)
517 apply (rule_tac x="1" in exI, simp add: norm_nth_le)
518 done
520 instance vec :: (banach, finite) banach ..
523 subsection \<open>Inner product space\<close>
525 instantiation vec :: (real_inner, finite) real_inner
526 begin
528 definition "inner x y = sum (\<lambda>i. inner (x\$i) (y\$i)) UNIV"
530 instance proof
531 fix r :: real and x y z :: "'a ^ 'b"
532 show "inner x y = inner y x"
533 unfolding inner_vec_def
535 show "inner (x + y) z = inner x z + inner y z"
536 unfolding inner_vec_def
538 show "inner (scaleR r x) y = r * inner x y"
539 unfolding inner_vec_def
541 show "0 \<le> inner x x"
542 unfolding inner_vec_def
544 show "inner x x = 0 \<longleftrightarrow> x = 0"
545 unfolding inner_vec_def
546 by (simp add: vec_eq_iff sum_nonneg_eq_0_iff)
547 show "norm x = sqrt (inner x x)"
548 unfolding inner_vec_def norm_vec_def L2_set_def
550 qed
552 end
555 subsection \<open>Euclidean space\<close>
557 text \<open>Vectors pointing along a single axis.\<close>
559 definition "axis k x = (\<chi> i. if i = k then x else 0)"
561 lemma axis_nth [simp]: "axis i x \$ i = x"
562 unfolding axis_def by simp
564 lemma axis_eq_axis: "axis i x = axis j y \<longleftrightarrow> x = y \<and> i = j \<or> x = 0 \<and> y = 0"
565 unfolding axis_def vec_eq_iff by auto
567 lemma inner_axis_axis:
568 "inner (axis i x) (axis j y) = (if i = j then inner x y else 0)"
569 unfolding inner_vec_def
570 apply (cases "i = j")
571 apply clarsimp
572 apply (subst sum.remove [of _ j], simp_all)
573 apply (rule sum.neutral, simp add: axis_def)
574 apply (rule sum.neutral, simp add: axis_def)
575 done
577 lemma sum_single:
578 assumes "finite A" and "k \<in> A" and "f k = y"
579 assumes "\<And>i. i \<in> A \<Longrightarrow> i \<noteq> k \<Longrightarrow> f i = 0"
580 shows "(\<Sum>i\<in>A. f i) = y"
581 apply (subst sum.remove [OF assms(1,2)])
582 apply (simp add: sum.neutral assms(3,4))
583 done
585 lemma inner_axis: "inner x (axis i y) = inner (x \$ i) y"
586 unfolding inner_vec_def
587 apply (rule_tac k=i in sum_single)
588 apply simp_all
590 done
592 instantiation vec :: (euclidean_space, finite) euclidean_space
593 begin
595 definition "Basis = (\<Union>i. \<Union>u\<in>Basis. {axis i u})"
597 instance proof
598 show "(Basis :: ('a ^ 'b) set) \<noteq> {}"
599 unfolding Basis_vec_def by simp
600 next
601 show "finite (Basis :: ('a ^ 'b) set)"
602 unfolding Basis_vec_def by simp
603 next
604 fix u v :: "'a ^ 'b"
605 assume "u \<in> Basis" and "v \<in> Basis"
606 thus "inner u v = (if u = v then 1 else 0)"
607 unfolding Basis_vec_def
608 by (auto simp add: inner_axis_axis axis_eq_axis inner_Basis)
609 next
610 fix x :: "'a ^ 'b"
611 show "(\<forall>u\<in>Basis. inner x u = 0) \<longleftrightarrow> x = 0"
612 unfolding Basis_vec_def
613 by (simp add: inner_axis euclidean_all_zero_iff vec_eq_iff)
614 qed
616 lemma DIM_cart[simp]: "DIM('a^'b) = CARD('b) * DIM('a)" | 7,321 | 22,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-43 | latest | en | 0.395984 |
https://discourse.ladybug.tools/t/illuminance-values-affected-by-the-direction-of-the-face-normals-in-a-grid-based-simulation/14212/8 | 1,718,563,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00147.warc.gz | 185,810,161 | 7,317 | # Illuminance values affected by the direction of the face normals, in a Grid Based Simulation
Hello!
I’m doing a grid based illuminance analysis with HB+ and Radiance.
To generate the analysis grid, I’ve modeled a surface on Rhino, and then used it as an input in the LB_Generate Mesh Component.
I’ve found that the direction of the face normals directly affects illuminance values, as you can see in the figure below.
When normals are facing upwards: Illuminance values are higher, though the pixels are really jagged.
When normals are facing downwards: Illuminance values are lower, but the final image is way smoother.
I’ve used the same conditions of sky, date, hour and radiance parameters to test both conditions, so I believe this inconsistency is really related to the direction of the face normals.
What is the correct direction for these face normals, when performing a grid based simulation on HB+?
1 Like
Upwards, because that is how we measure illuminance in the real world. Grid-based simulations are essentially a proxy for calculating work-plane illuminance. And the idea there really is to calculate the amount of light falling on the workplane.
If you turn the grid downwards, you are measuring reflected radiation from the floor.
3 Likes
Thanks @sarith for the answer!
Then I guess the jagged pixels, of the image on the right are related to some Radiance parameter.
Light reflected from a diffused surface with low reflectance (floor) will be a lot smoother and uniform than that incident from a direct source (sun and sky). When facing upwards your grid points “see” the ceiling , sun and sky so the gradient and variation in values is likely to be higher.
1 Like
Interesting! I also just experienced the same thing and conclude that the surface direction/ surface normal influence the calculation results. I just attempted to verify daylight simulation in Radiance with measured data (one single room) and I found that the simulation results matches the measurement if all surface normal faces interior.
Previously I found significant gap (illuminance) between simulation and measurement. What I did is: (1) check each surface direction of the building model in rhino using “dir” command and (2) change the surface direction using 'flip" command, then (3) referenced these surfaces again in Grasshopper for defining HB surfaces.
I hope my information is useful
Hi @sunarywend !
You information was very helpful!
I’ve used a similar workflow, but for the analysis surface.
The G input in the flip component allows you to insert a guide surface to match.
So: 1) I first created this guide surface, with the normals facing upwards, 2) I internalized this srf in a geometry component, and 3) connected to G
This will make sure that any analysis srf, with the wrong normals direction (downwards) will be flipped, and the ones that are correct (upwards), will remain correct.
Though I don’t think my workflow will work with vertical surfaces (e.g. walls), because each srf will need a different normal direction.
Hi, I’ve just updated my Ladybug to 1.1.0. and cannot find a way to map the illuminance apparently there is only the % of the time that daylight passed the threshold. I was wondering how you could make your model with illuminance values?
Tnx
Hi @B_Jahangiri
Illuminance studies are not available in LB 1.2 yet.
You can check the topic below, to see what is available in this last release.
Still, you can use LB and HB Legacy for illuminance analysis.
1 Like
Thank you so much | 756 | 3,527 | {"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-2024-26 | latest | en | 0.912385 |
http://www.gurufocus.com/term/DaysInventory/MNST/Days%2BInventory/Monster%2BBeverage%2BCorp | 1,406,526,071,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510256737.1/warc/CC-MAIN-20140728011736-00048-ip-10-146-231-18.ec2.internal.warc.gz | 580,331,466 | 39,280 | Switch to:
Monster Beverage Corp (NAS:MNST)
Days Inventory
74.48 (As of Mar. 2014)
Monster Beverage Corp's inventory for the three months ended in Mar. 2014 was \$204 Mil. Monster Beverage Corp's cost of goods sold for the three months ended in Mar. 2014 was \$249 Mil. Hence, Monster Beverage Corp's days inventory for the three months ended in Mar. 2014 was 74.48.
Monster Beverage Corp's days inventory declined from Mar. 2013 (84.40) to Mar. 2014 (74.48).
Inventory can be measured by Days Sales of Inventory (DSI). Monster Beverage Corp's days sales of inventory (DSI) for the three months ended in Mar. 2014 was 34.64.
Inventory turnover measures how fast the company turns over its inventory within a year. Monster Beverage Corp's inventory turnover for the three months ended in Mar. 2014 was 1.22.
Inventory to revenue ratio determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue. Monster Beverage Corp's inventory to revenue ratio for the three months ended in Mar. 2014 was 0.38.
Definition
Days Inventory indicates the number of days of goods in sales that a company has in the inventory.
Monster Beverage Corp's Days Inventory for the fiscal year that ended in Dec. 2013 is calculated as
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 221.449 / 1073.497 * 365 = 75.29
Monster Beverage Corp's Days Inventory for the quarter that ended in Mar. 2014 is calculated as:
Days Inventory = Inventory / Cost of Goods Sold * Days in Period = 204.061 / 249.311 * 91 = 74.48
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Explanation
An increase of Days Inventory may indicate the company's sales slowed.
1. Inventory can be measured by Days Sales of Inventory (DSI).
Monster Beverage Corp's Days Sales of Inventory for the three months ended in Mar. 2014 is calculated as
Days Sales of Inventory (DSI) = Inventory / Revenue * Days in Period = 204.061 / 536.129 * 91 = 34.64
2. Inventory Turnover measures how fast the company turns over its inventory within a year.
Monster Beverage Corp's Inventory Turnover for the three months ended in Mar. 2014 is calculated as
Inventory Turnover = Cost of Goods Sold / Average Inventory = 249.311 / 204.061 = 1.22
3. Inventory to Revenue determines the ability of a company to manage their inventory levels. It measures the percentage of Inventories the company currently has on hand to support the current amount of Revenue.
Monster Beverage Corp's Inventory to Revenue for the three months ended in Mar. 2014 is calculated as
Inventory to Revenue = Inventory / Revenue = 204.061 / 536.129 = 0.38
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Be Aware
A lot of business are seasonable. It makes more sense to compare Days Inventory from the same period in the previous year instead of from the previous quarter.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
Monster Beverage Corp Annual Data
Dec04 Dec05 Dec06 Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 DaysInventory 84.42 68.90 97.21 82.07 85.78 74.34 89.68 70.22 74.50 75.29
Monster Beverage Corp Quarterly Data
Dec11 Mar12 Jun12 Sep12 Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 DaysInventory 72.35 76.37 63.42 65.77 81.20 84.40 72.12 79.56 76.42 74.48
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 922 | 3,751 | {"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-2014-23 | longest | en | 0.939149 |
https://jp.mathworks.com/matlabcentral/answers/472545-find-first-non-nan-in-each-column-of-array-combine-into-one-vector | 1,579,552,136,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250599789.45/warc/CC-MAIN-20200120195035-20200120224035-00202.warc.gz | 511,371,755 | 24,141 | # Find first non-NaN in each column of array & combine into one vector
27 ビュー (過去 30 日間)
mfclose 2019 年 7 月 19 日
Edited: Adam Danz 2019 年 7 月 19 日
I am trying to:
1) Identify the first non-NaN value of each column in an array, and convert the rest to NaN
2) Identify the first non-NaN value of each row in the array, and convert the rest to NaN
3) Combine these numbers back into one column vector
I have created something that looks roughly like this (although much larger):
X =
2 NaN NaN NaN
4 6 NaN NaN
NaN NaN NaN NaN
5 4 7 8
After the first step (keeping only the first number in each column), I would hope to have this:
Xa =
2 NaN NaN NaN
NaN 6 NaN NaN
NaN NaN NaN NaN
NaN NaN 7 8
After the second step (keeping only the first number in each row), I would hope to have this:
Xb =
2 NaN NaN NaN
NaN 6 NaN NaN
NaN NaN NaN NaN
NaN NaN 7 NaN
Lastly, I would want to turn this into one column (there should only be NaN's or 1 non-NaN value in each row - I would want it to keep whichever is there)
Xc =
2
6
NaN
7
I apologize for not having any attempted code for this. I appear to have gotten stuck in the doorway of this one...
#### 0 件のコメント
サインイン to comment.
### 採用された回答
Bruno Luong 2019 年 7 月 19 日
X = [2 NaN NaN NaN;
4 6 NaN NaN;
NaN NaN NaN NaN;
5 4 7 8 ]
[m,n] = size(X);
[I,J] = ndgrid(1:m,1:n);
Xa = X;
[~,r]=max(isfinite(Xa),[],1);
Xa(I>r) = NaN
Xb = Xa;
[~,c]=max(isfinite(Xb),[],2);
Xb(J>c) = NaN
[~,c]=max(isfinite(Xb),[],2);
Xc = Xb(sub2ind([m,n],(1:m)',c))
#### 2 件のコメント
mfclose 2019 年 7 月 19 日
Thank you, this did exactly what I needed!
Adam Danz 2019 年 7 月 19 日
Be sure to add semicolons to the end of 3 lines that are missing them.
Also, use ~isnan() rather than isfinite().
サインイン to comment.
Adam Danz 2019 年 7 月 19 日
X = [2 NaN NaN NaN
4 6 NaN NaN
NaN NaN NaN NaN
5 4 7 8];
X(cumsum(cumsum(~isnan(X)))~=1) = NaN; % your Xa
X(cumsum(cumsum(~isnan(X),2),2) ~=1) = NaN; % your Xb
Xc = NaN(size(X,1),1);
Xc(any(~isnan(X),2)) = X(~isnan(X)); % your Xc
#### 2 件のコメント
mfclose 2019 年 7 月 19 日
Thank you this also worked!
Adam Danz 2019 年 7 月 19 日
Glad I could help! Just FYI, this solution and the max(infinite()) solution are both quite fast but this one is 4.5x faster (10,000 iterations, p<0.0001 Wicoxon signed rank) with less overhead. Also, instead of isfinite(), you should use ~isnan().
サインイン to comment.
Mario Chiappelli 2019 年 7 月 19 日
Check out this question asked earlier on the forum, I think it asks and answers what you want.
#### 3 件のコメント
mfclose 2019 年 7 月 19 日
Thank you for this!
It is a similar question. Some of the differences that are proving problematic for me are that:
1) I only want the first non-NaN value, not all of them
2) I don't want to lose the row position that each of these values have (as that represents a time point I will need to compare back to later), so if there is only ever a NaN in a row, I need to keep that there (like in the 3rd row of Xc in my example)
These may only be subtle differences, but I'm not quite sure how to make those adjustments. Thanks for your help!
Mario Chiappelli 2019 年 7 月 19 日
Try this:
x = [2,NaN,NaN,NaN;4,6,NaN,NaN;NaN,NaN,NaN,NaN;5,4,7,8];
rows = length(x(:,1));
columns = length(x(1,:));
% This finds the first non NaN in the columns
columnVector = double(columns);
for i = 1:columns
found = 0;
for j = 1:rows
if ~isnan(x(j,i)) && found == 0
found = 1;
columnVector(i) = x(j,i);
end
end
if found == 0
columnVector(i) = NaN;
end
end
% This finds the first non NaN in the rows
rowVector = double(rows);
for i = 1:rows
found = 0;
for j = 1:columns
if ~isnan(x(i,j)) && found == 0
found = 1;
rowVector(i) = x(i,j);
end
end
if found == 0
rowVector(i) = NaN;
end
end
I could include some more functionality to delete any NaN values in the resulting columnVector and rowVector arrays. This does what you want, however it skips over the middle steps. I did not know if that was a necessary step for you or just your way of visualizing what needed to be done.
Hope this helps :)
mfclose 2019 年 7 月 19 日
This was very close, but I did need the middle step getting rid of the other non-NaN values in a column. The code below fixed this. Thank you so much for helping!
サインイン to comment.
サインイン してこの質問に回答します。
Translated by | 1,370 | 4,219 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-05 | latest | en | 0.619184 |
https://mathoverflow.net/questions/tagged/symmetry | 1,702,252,509,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00747.warc.gz | 424,180,838 | 36,877 | # Questions tagged [symmetry]
The tag has no usage guidance.
120 questions
Filter by
Sorted by
Tagged with
21 views
### Minimal Non-Symmetric Closed Tracks Under 45-Degree Rotations
Given an arbitrary number of clockwise and anti-clockwise 45-degree track turns, what is the smallest closed track, that is neither axisymmetric nor rotationally symmetrical? For example, AAAAAAAA ...
63 views
• 6,581
532 views
• 181
163 views
• 11.9k
84 views
### Can a spherical simplicial complex have more than one "central" inversion?
Let $\Delta$ be a finite connected simplicial complex. Call a simplicial map $\phi:\Delta\to\Delta$ an inversion if $\phi$ is an involution, that is $\phi\circ\phi=\mathrm{id}$, and $\phi$ is not ...
• 11.9k
123 views
### Is the projective symmetry group of a polytope more general than its linear symmetry group?
Give a (convex) polytope $P\subset\Bbb R^d$ (the convex hull of finitely many points). Consider its linear and projective symmetry groups: \begin{align} \DeclareMathOperator{\Aut}{Aut} \...
• 11.9k
1 vote
304 views
• 9,538
35 views
### Vertex configuration to tile repeat unit
I am working with elongated triangular tiling which has a vertex configuration of 3.3.3.4.4 and noticed that the representative symmetry is not the repeat unit. Is there a general formula to convert ...
1 vote
74 views
### Projection of cocyclic Gaussian primes on the real axis
I just stumbled upon https://math.stackexchange.com/questions/2372062/4-concylic-points-of-gaussian-primes after a quick Google search about cocyclic Gaussian primes. As I've been investigating about ...
• 6,954
1 vote
44 views
### How dense can a transitive sets of points be?
How dense can a finite set of points on the $d$-dimensional unit sphere be if I require that the symmetry group of that arrangement is still transitive on the points? As a measure for density I use ...
• 11.9k
168 views
### When is a $k$-distance-transitive graph already distance-transitive?
Call a (finite and connected) graph $k$-distance-transitive if its symmetry group acts transitively on the pairs in each one of the sets D_\delta:=\{(i,j)\in V\times V\mid \mathrm d(i,j)=\delta\},\...
• 11.9k
111 views
### A diameter 2 arc-transitive graph whose complement is not arc-transitive?
A graph $G=(V,E)$ is arc-transitive if its symmetry group acts transitively on ordered pairs of adjacent vertices. In general, the complement of an arc-transitive graph is not arc-transitive. But I ...
• 11.9k
112 views
• 261
91 views
### A characterization of root systems via their intersections with halfspaces
In a recent preprint I obtained a nice characterization of root systems as a side product. I can imagine that this was known before, and that a source for this statement can shorten the proof of my ...
• 11.9k
45 views
### Equiangular lines with symmetry requirements
Listing all possible arrangements of equiangular lines is non-trivial. Does the problem become any easier when we additionally require that the symmetry group of that line arrangement acts ...
• 11.9k
### Examples of Calabi-Yau manifolds with $\mathbb{T}^2$ symmetry
I want to know if there exists examples of Calabi-Yau manifolds with $\mathbb{T}^2$-invariant $SU(n)$-structure. In particular these actions are both Killing and holomorphic. I am especially ... | 845 | 3,330 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.841427 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-6-6-2-right-triangle-trigonometry-6-2-exercises-page-442/32 | 1,722,853,287,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640436802.18/warc/CC-MAIN-20240805083255-20240805113255-00726.warc.gz | 626,245,591 | 12,731 | ## Algebra and Trigonometry 10th Edition
Part A: sec($\frac{\pi}{9}$) = 1.0642 Part B: cos($\frac{\pi}{9}$) = 0.9397
Using a calculator: Part A: sec($\frac{\pi}{9}$) = 1.0642 Part B: cos($\frac{\pi}{9}$) = 0.9397 | 93 | 213 | {"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} | 3.65625 | 4 | CC-MAIN-2024-33 | latest | en | 0.554948 |
http://www.reddit.com/r/askscience/comments/12a51t/can_someone_explain_how_a_two_negatives/c6tphvi | 1,409,269,387,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500831098.94/warc/CC-MAIN-20140820021351-00190-ip-10-180-136-8.ec2.internal.warc.gz | 572,598,562 | 12,829 | you are viewing a single comment's thread.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
If you have -\$5 then you owe me \$5. If I remove your debt to me (subtract) two times I owe you \$5. That's -5 x -2= 10. You have made \$10 because not only do you not owe me \$5, I owe you \$5 so it's a net difference of \$10. I know you said multiplication but all multiplication is it addition really. Think about it, 2 x 5 = 2 + 2 + 2 + 2 + 2. Multiplication is just a simplified version of repeated addition. | 160 | 554 | {"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} | 3.65625 | 4 | CC-MAIN-2014-35 | latest | en | 0.950256 |
http://www.ck12.org/statistics/Scatter-Plots/lesson/Scatter-Plots/r15/ | 1,454,761,853,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701146302.25/warc/CC-MAIN-20160205193906-00312-ip-10-236-182-209.ec2.internal.warc.gz | 341,697,749 | 35,493 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.
# Scatter Plots
## Identify positive, negative, and no correlation relations
Estimated8 minsto complete
%
Progress
Practice Scatter Plots
Progress
Estimated8 minsto complete
%
Scatter Plots
You've been exercising every week and when you go for your next doctor's visit the doctor says that the reading for your resting heart rate has changed. You start taking your own resting heart rate once a week on Mondays and relate it to the numbers of hours per week you've been exercising. How would you represent this data? Do you expect to see a correlation between the number of hours you exercise per week and your resting heart rate? How would you know if there is a correlation?
### Watch This
First watch this video to learn about scatter plots.
Then watch this video to see some examples.
Watch this video for more help.
### Guidance
Often, when real-world data is plotted, the result is a linear pattern. The general direction of the data can be seen, but the data points do not all fall on a line. This type of graph is called a scatter plot. A scatter plot is often used to investigate whether or not there is a relationship or connection between 2 sets of data. The data is plotted on a graph such that one quantity is plotted on the $x$ -axis and one quantity is plotted on the $y$ -axis. The quantity that is plotted on the $x$ -axis is the independent variable, and the quantity that is plotted on the $y$ -axis is the dependent variable. If a relationship does exist between the 2 sets of data, it will be easy to see if the data is plotted on a scatter plot.
The following scatter plot shows the price of peaches and the number sold:
The connection is obvious $-$ when the price of peaches was high, the sales were low, but when the price was low, the sales were high.
The following scatter plot shows the sales of a weekly newspaper and the temperature:
There is no connection between the number of newspapers sold and the temperature.
Another term used to describe 2 sets of data that have a connection or a relationship is correlation . The correlation between 2 sets of data can be positive or negative, and it can be strong or weak. The following scatter plots will help to enhance this concept.
If you look at the 2 sketches that represent a positive correlation, you will notice that the points are around a line that slopes upward to the right. When the correlation is negative, the line slopes downward to the right. The 2 sketches that show a strong correlation have points that are bunched together and appear to be close to a line that is in the middle of the points. When the correlation is weak, the points are more scattered and not as concentrated.
When correlation exists on a scatter plot, a line of best fit can be drawn on the graph. The line of best fit must be drawn so that the sums of the distances to the points on either side of the line are approximately equal and such that there are an equal number of points above and below the line. Using a clear plastic ruler makes it easier to meet all of these conditions when drawing the line. Another useful tool is a stick of spaghetti, since it can be easily rolled and moved on the graph until you are satisfied with its location. The edge of the spaghetti can be traced to produce the line of best fit. A line of best fit can be used to make estimations from the graph, but you must remember that the line of best fit is simply a sketch of where the line should appear on the graph. As a result, any values that you choose from this line are not very accurate $-$ the values are more of a ballpark figure.
In the sales of newspapers and the temperature, there was no connection between the 2 data sets. The following sketches represent some other possible outcomes when there is no correlation between data sets:
#### Example A
Plot the following points on a scatter plot, with $m$ as the independent variable and $n$ as the dependent variable. Number both axes from 0 to 20. If a correlation exists between the values of $m$ and $n$ , describe the correlation (strong negative, weak positive, etc.).
$& m \quad 4 \quad 9 \quad 13 \quad 16 \quad 17 \quad 6 \quad 7 \quad \ 18 \quad 10\\& n \quad \ 5 \quad 3 \quad 11 \quad 18 \quad 6 \quad 11 \quad 18 \quad 12 \quad 16$
#### Example B
Describe the correlation, if any, in the following scatter plot:
In the above scatter plot, there is a strong positive correlation.
#### Example C
The following table consists of the marks achieved by 9 students on chemistry and math tests:
Student A B C D E F G H I
Chemistry Marks 49 46 35 58 51 56 54 46 53
Math Marks 29 23 10 41 38 36 31 24 ?
Plot the above marks on scatter plot, with the chemistry marks on the $x$ -axis and the math marks on the $y$ -axis. Draw a line of best fit, and use this line to estimate the mark that Student I would have made in math had he or she taken the test.
If Student I had taken the math test, his or her mark would have been between 32 and 37.
Points to Consider
• Can the equation for the line of best fit be used to calculate values?
• Is any other graphical representation of data used for estimations?
### Guided Practice
The following table represents the sales of Volkswagen Beetles in Iowa between 1994 and 2003:
Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003
Beetles Sold 50 60 55 50 70 65 75 65 80 90
(a) Create a scatter plot and draw the line of best fit for the data. Hint: Let 0 = 1994, 1 = 1995, etc.
(b) Use the graph to predict the number of Beetles that will be sold in Iowa in the year 2007.
(c) Describe the correlation for the above graph.
a.
b. The year 2007 would actually be the number 13 on the $x-$ axis. The number of beetles sold in this year would be approximately 98 to 100.
c. The correlation of this graph is strong and positive.
### Practice
1. What is the correlation of a scatter plot that has few points that are not bunched together?
1. strong
2. no correlation
3. weak
4. negative
2. What term is used to define the connection between 2 data sets?
1. relationship
2. scatter plot
3. correlation
4. discrete
3. Describe the correlation of each of the following graphs:
4. Plot the following points on a scatter plot, with $m$ as the independent variable and $n$ as the dependent variable. Number both axes from 0 to 20. If a correlation exists between the values of $m$ and $n$ , describe the correlation (strong negative, weak positive, etc.).
1. $m \quad 5 \quad 14 \quad 2 \quad 10 \quad 16 \quad 4 \quad 18 \quad 2 \quad 8 \quad 11\\n \quad \ 6 \quad 13 \quad 4 \quad 10 \quad 15 \quad 7 \quad 16 \quad 5 \quad 8 \quad 12$
2. $m \quad 13 \quad 3 \quad 18 \quad 9 \quad 20 \quad 15 \quad 6 \quad 10 \quad 21 \quad 4\\n \quad \ 7 \quad 14 \quad 9 \quad 16 \quad 7 \quad 13 \quad 10 \quad 13 \quad 3 \quad 19$
The following scatter plot shows the closing prices of 2 stocks at various points in time. A line of best fit has been drawn. Use the scatter plot to answer the following questions.
1. How would you describe the correlation between the prices of the 2 stocks?
2. If the price of stock A is $12.00, what would you expect the price of stock B to be? 3. If the price of stock B is$47.75, what would you expect the price of stock A to be?
The following scatter plot shows the hours of exercise per week and resting heart rates for various 30-year-old males. A line of best fit has been drawn. Use the scatter plot to answer the following questions.
1. How would you describe the correlation between hours of exercise per week and resting heart rate?
2. If a 30-year-old male exercises 2 hours per week, what would you expect his resting heart rate to be?
3. If a 30-year-old male has a resting heart rate of 65 beats per minute, how many hours would you expect him to exercise per week?
### Vocabulary Language: English Spanish
correlation
correlation
Correlation is a statistical method used to determine if there is a connection or a relationship between two sets of data.
line of best fit
line of best fit
A line of best fit is a straight line drawn on a scatter plot such that the sums of the distances to the points on either side of the line are approximately equal and such that there are an equal number of points above and below the line.
scatter plot
scatter plot
A scatter plot is a plot of the dependent variable versus the independent variable and is used to investigate whether or not there is a relationship or connection between 2 sets of data. | 2,138 | 8,684 | {"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": 20, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2016-07 | latest | en | 0.930924 |
adams1011.asdstyle.com | 1,632,342,810,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00019.warc.gz | 137,438,119 | 11,392 | # Tips For Maximizing how big is Roulette Table Bets
The Basics of Roulette. When you head into a casino, you will see the typical roulette tables quickly the front door. There is a revolving wheel that contains each one or multiple slots for #1 1 to 36 and sometimes both one and two slots. Players will usually stand around at an extended table where the wheel is set up, and where players place bets against each other by pushing a button on their roulette table card. If their bet wins, they win the amount shown on the card, if their bet loses, they lose the amount shown on the card – the wheel is really just a big chalkboard with numbers on it, but the point is that someone must draw the number and stick it in the designated slot. The person doing the drawing then reaches keep that bet until another person takes the designated number from the chalkboard.
Roulette, though a casino game of luck, can be controlled to some degree through using systematic betting strategies. Most importantly, an excellent roulette table strategy will include placing outside bets. Outside bets work the best when you are trying to get a better return on your bets than what you are putting in with your bets. In roulette parlance, outside bets are called ‘extra’ bets. Many of the most successful players, regardless of making more than 80% of these initial stake, still lose the majority of their money, because they didn’t have the foresight to put their outside bets properly.
It takes experience to learn the ultimate way to place bets on any roulette table, but also for those who aren’t experienced, it could seem confusing at first. Some players may put almost all their money on the winning wheel, while others may only be considered a small percentage 라이브 바카라 on every hand, but both will place bets on the same wheels – even if the chances of their picks winning are close, the difference in winnings is relatively small. The reason being the chances of the winning numbers being drawn are relatively low, and since the big payoff comes later, the players will place their bets to earn the most possible amount of money by the end of the game.
So, do you know the odds of a number being drawn? The answer is: They’ll be in a range that can be divided by a margin of about 0.027 percent, or thereabouts. In the overall case, it is better to find the larger sums on fewer bets than it really is to find smaller sums on a fewer number of bets. Thus, the player who includes a plan and a number of assumptions about the range of results that his / her chosen bet should come up with will minimize the potential for success by selecting a bet that assumes the worst-case scenario, at the very least if there is lots to be drawn.
The easiest form of this assumption involves two adjacent numbers. Let’s say a player bets on seven chips and looks for the highest single figure win expectancy for a seven chip slot. Assuming (assuming nothing else) that a single figure win expectancy of seven out of seven chips will occur between any two adjacent numbers, the player concludes that there is a ninety percent potential for their selected winnings occurring between two adjacent numbers. The bet would then be made on the assumption that if the initial number was drawn, the player would lose the bet for that number, regardless of what the second number was. (The math is similar to the example above but with seven chips instead of two.)
A more complex example involves multiples thereof. Let’s say a new player bets two chips on lots range of four to seven, inclusive. Assume further that the full total number of chips up for grabs is twenty, which is the minimum amount of chips that can be positioned on the table. Now, because there are no other multiples up for grabs, there is no way for the ball player to “get lucky” and hit a “big” number. Therefore, the ball player must play cautiously, taking as many risks as essential to maximize his or her expected payout. This is a good lesson in counting big numbers; in fact, it’s an excellent way to start out learning how to play a complete table.
While it’s true that one may make a bundle by betting about the same number or even a combination of numbers, these are not the kind of bets most players make. Whenever a player bets a little amount on a spin, for instance, he or she is betting that the outcome of that particular spin will cause her or him to win that one bet. In most games of chance, the outcomes of individual spins are predictable. However, when multiples are participating, a careful player should be aware of the chances and the chance of drawing a negative card to reduce the possibility of making an individual bet that could lose the pot all together.
Finally, you can find multiples-tiered bets. These are bets where players place bets in line with the size of the bet they “purchase.” For instance, if someone bets three dimes for a seven-game game, he’s not actually purchasing three chips. Rather, he’s purchasing the opportunity to buy seven chips at the end of the overall game – an outcome that could not happen, but would be profitable based on the final outcome. Multi-tier bets are often considered a lower risk compared to the larger multi-tier bets but still carry a similar level of risk. | 1,082 | 5,287 | {"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-2021-39 | latest | en | 0.962709 |
https://www.enotes.com/homework-help/y-r-sqrt-r-2-1-find-derivative-function-497399 | 1,620,865,152,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991413.30/warc/CC-MAIN-20210512224016-20210513014016-00574.warc.gz | 799,469,316 | 17,266 | # `y = r/sqrt(r^2 + 1)` Find the derivative of the function.
`y=r/sqrt(r^2+1)`
`y'=r*(-1/2)(r^2+1)^((-1/2)-1) (2r) + 1/sqrt(r^2+1)`
`y'=-r^2(r^2+1)^(-3/2) + 1/sqrt(r^2+1)`
`y'= -r^2/(r^2+1)^(3/2) +1/sqrt(r^2+1)`
`y'= -(r^2/((r^2+1)(sqrt(r^2+1))))+1/sqrt(r^2+1)`
`y' =(-r^2+r^2+1)/((r^2+1)(sqrt(r^2+1)))`
`y'=(1)/((r^2+1)(sqrt(r^2+1)))`
Approved by eNotes Editorial Team | 221 | 377 | {"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} | 3.65625 | 4 | CC-MAIN-2021-21 | latest | en | 0.359823 |
https://rd.springer.com/article/10.1007/s10711-006-9092-x | 1,534,826,320,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217951.76/warc/CC-MAIN-20180821034002-20180821054002-00257.warc.gz | 772,468,590 | 17,781 | Geometriae Dedicata
, Volume 121, Issue 1, pp 89–111
# Stein domains and branched shadows of 4-manifolds
• Francesco Costantino
Original Paper
## Abstract
We provide sufficient conditions assuring that a suitably decorated 2-polyhedron can be thickened to a compact four-dimensional Stein domain. We also study a class of flat polyhedra in 4-manifolds and find conditions assuring that they admit Stein, compact neighborhoods. We base our calculations on Turaev’s shadows suitably “smoothed”; the conditions we find are purely algebraic and combinatorial. Applying our results, we provide examples of hyperbolic 3-manifolds admitting “many” positive and negative Stein fillable contact structures, and prove a four-dimensional analog of Oertel’s result on incompressibility of surfaces carried by branched polyhedra.
## Mathematics Subject Classifications (2000)
Primary 57N13 Secondary 32B28
## References
1. 1.
Agol I. (2000) Bounds on exceptional Dehn filling. Geom. Topol. 4, 431–439
2. 2.
Bishop E. (1975) Differentiable manifolds in complex euclidean space. Duke. Math. J. 32, 1–21
3. 3.
Benedetti R., Petronio C. (2000) Branched spines and contact structures on 3-manifolds. Ann. Mat. Pura Appl. 178(4): 81–102
4. 4.
Benedetti R., Petronio C. (1997) Branched standard spines of 3-manifolds. Lecture Notes in Mathematics, Springer-Verlag, Berlin
5. 5.
Costantino, F.: Shadows and branched shadows of 3 and 4-manifolds. Tesi di perfezionamento, Scuola Normale Superiore, Pisa (2004)Google Scholar
6. 6.
Costantino, F.: Branched shadows and complex structures of 4-manifolds. xxx.arXiv.org /math.GT/0502292 (2005)Google Scholar
7. 7.
Costantino F. (2005) A short introduction to shadows of 4-manifolds. Fund. Math. 188, 271–291
8. 8.
Chern S.S., Spanier E. (1951) A theorem on orientable surfaces in four-dimensional space. Comm. Math. Helv. 25, 205–209
9. 9.
Costantino, F., Thurston, D.P.: 3-manifolds efficiently bound 4-manifolds. xxx.arXiv.org/math.GT/ 0506577 (2005)Google Scholar
10. 10.
Eliashberg Y. (1990) Topological characterization of Stein manifolds of dimension ≥ 3. Internat. J. Math. 1(1): 119–131
11. 11.
Forstneriç F. (2003) Stein domains in complex surfaces. J. Geom. Anal. 13(1): 77–94
12. 12.
Gompf R. (1998) Handlebody construction of Stein surfaces. Ann. of Math. 148, 619–693
13. 13.
Gompf, R., Stipsicz, A.: 4-manifolds and Kirby calculus. Graduate Studies in Mathematics, vol. 20, Amer. Math. Soc., Providence, RI (1999)Google Scholar
14. 14.
Harlamov, V.M., Eliashberg, Y.: On the number of complex points of a real surface in a complex surface. pp. 143–148. Proc. Leningrad Int. Topology Conf. (1982)Google Scholar
15. 15.
Kronheimer P.B., Mrowka T.S. (1997) Monopoles and contact structures. Invent. Math. 130(2): 209–255
16. 16.
Lackenby M. (2000) Word hyperbolic Dehn surgery. Invent. Math. 140, 243–282
17. 17.
Lai H.F. (1972) Characteristic classes of real manifolds immersed in complex manifolds. Trans. Amer. Math. Soc. 172, 1–33
18. 18.
Lisca P. (1998) Symplectic fillings and positive scalar curvature. Geom. Topol. 2, 103–116
19. 19.
Nirenberg R., Wells R.O. (1969) Approximation theorems on differentiable submanifolds of a complex manifold. Trans. Amer. Math. Soc. 142, 15–35
20. 20.
Oertel U. (1984) Incompressible branched surfaces. Invent. Math. 76, 385–410
21. 21.
Polyak, M.: Shadows of Legendrian links and J +-theory of curves. Singularities (Oberwolfach, 1996), pp. 435–458. Progr. Math. 162, Birkhäuser, Basel (1998)Google Scholar
22. 22.
Turaev V.G. (1994) Quantum invariants of knots and 3-manifolds. de Gruyter Studies in Mathematics, vol. 18. Walter de Gruyter and Co., BerlinGoogle Scholar
23. 23.
Turaev V.G. (1991) Quantum invariants of 3-manifolds and a glimpse of shadow topology. C. R. Acad. Sci. Paris Sr. I Math. 313(6): 395–398 | 1,254 | 3,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-34 | latest | en | 0.727525 |
http://www.jiskha.com/display.cgi?id=1351643453 | 1,495,516,195,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607369.90/warc/CC-MAIN-20170523045144-20170523065144-00522.warc.gz | 527,504,517 | 3,717 | posted by on .
How many rectangles can you build with a prime number of square tiles?
• 6th grade Math - ,
In Euclidean plane geometry, a rectangle is any quadrilateral with four right angles.
So a square is a rectangle. You could build up to a prime number of rectangles with a prim number of square tiles. | 69 | 311 | {"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-2017-22 | latest | en | 0.920001 |
https://bitbucket.org/sebferre/ocaml-lib/raw/3915c77e38d7b5693c6ad2ddb5933221ffba905a/bintree.ml | 1,435,931,318,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-27/segments/1435375096061.71/warc/CC-MAIN-20150627031816-00134-ip-10-179-60-89.ec2.internal.warc.gz | 839,286,032 | 3,238 | (***********************************************************************) (* *) (* Objective Caml *) (* *) (* Xavier Leroy, projet Cristal, INRIA Rocquencourt *) (* *) (* Copyright 1996 Institut National de Recherche en Informatique et *) (* en Automatique. All rights reserved. This file is distributed *) (* under the terms of the GNU Library General Public License, with *) (* the special exception on linking described in file ../LICENSE. *) (* *) (***********************************************************************) (* \$Id: set.ml,v 1.18 2004/04/23 10:01:54 xleroy Exp \$ *) type 'a t = Empty | Node of 'a t * 'a * 'a t * int (* Sets are represented by balanced binary trees (the heights of the children differ by at most 2 *) let height = function Empty -> 0 | Node(_, _, _, h) -> h (* Creates a new node with left son l, value v and right son r. We must have all elements of l < v < all elements of r. l and r must be balanced and | height l - height r | <= 2. Inline expansion of height for better speed. *) let create l v r = let hl = match l with Empty -> 0 | Node(_,_,_,h) -> h in let hr = match r with Empty -> 0 | Node(_,_,_,h) -> h in Node(l, v, r, (if hl >= hr then hl + 1 else hr + 1)) (* Same as create, but performs one step of rebalancing if necessary. Assumes l and r balanced and | height l - height r | <= 3. Inline expansion of create for better speed in the most frequent case where no rebalancing is required. *) let bal l v r = let hl = match l with Empty -> 0 | Node(_,_,_,h) -> h in let hr = match r with Empty -> 0 | Node(_,_,_,h) -> h in if hl > hr + 2 then begin match l with Empty -> invalid_arg "Set.bal" | Node(ll, lv, lr, _) -> if height ll >= height lr then create ll lv (create lr v r) else begin match lr with Empty -> invalid_arg "Set.bal" | Node(lrl, lrv, lrr, _)-> create (create ll lv lrl) lrv (create lrr v r) end end else if hr > hl + 2 then begin match r with Empty -> invalid_arg "Set.bal" | Node(rl, rv, rr, _) -> if height rr >= height rl then create (create l v rl) rv rr else begin match rl with Empty -> invalid_arg "Set.bal" | Node(rll, rlv, rlr, _) -> create (create l v rll) rlv (create rlr rv rr) end end else Node(l, v, r, (if hl >= hr then hl + 1 else hr + 1)) (* Insertion of one element *) let rec add x = function Empty -> Node(Empty, x, Empty, 1) | Node(l, v, r, _) as t -> let c = Pervasives.compare x v in if c = 0 then t else if c < 0 then bal (add x l) v r else bal l v (add x r) (* Same as create and bal, but no assumptions are made on the relative heights of l and r. *) let rec join l v r = match (l, r) with (Empty, _) -> add v r | (_, Empty) -> add v l | (Node(ll, lv, lr, lh), Node(rl, rv, rr, rh)) -> if lh > rh + 2 then bal ll lv (join lr v r) else if rh > lh + 2 then bal (join l v rl) rv rr else create l v r (* Smallest and greatest element of a set *) let rec min_elt = function Empty -> raise Not_found | Node(Empty, v, r, _) -> v | Node(l, v, r, _) -> min_elt l let rec max_elt = function Empty -> raise Not_found | Node(l, v, Empty, _) -> v | Node(l, v, r, _) -> max_elt r (* Remove the smallest element of the given set *) let rec remove_min_elt = function Empty -> invalid_arg "Set.remove_min_elt" | Node(Empty, v, r, _) -> r | Node(l, v, r, _) -> bal (remove_min_elt l) v r (* Merge two trees l and r into one. All elements of l must precede the elements of r. Assume | height l - height r | <= 2. *) let merge t1 t2 = match (t1, t2) with (Empty, t) -> t | (t, Empty) -> t | (_, _) -> bal t1 (min_elt t2) (remove_min_elt t2) (* Merge two trees l and r into one. All elements of l must precede the elements of r. No assumption on the heights of l and r. *) let concat t1 t2 = match (t1, t2) with (Empty, t) -> t | (t, Empty) -> t | (_, _) -> join t1 (min_elt t2) (remove_min_elt t2) (* Splitting. split x s returns a triple (l, present, r) where - l is the set of elements of s that are < x - r is the set of elements of s that are > x - present is false if s contains no element equal to x, or true if s contains an element equal to x. *) let rec split x = function Empty -> (Empty, false, Empty) | Node(l, v, r, _) -> let c = Pervasives.compare x v in if c = 0 then (l, true, r) else if c < 0 then let (ll, pres, rl) = split x l in (ll, pres, join rl v r) else let (lr, pres, rr) = split x r in (join l v lr, pres, rr) (* Implementation of the set operations *) let empty = Empty let is_empty = function Empty -> true | _ -> false let rec mem x = function Empty -> false | Node(l, v, r, _) -> let c = Pervasives.compare x v in c = 0 || mem x (if c < 0 then l else r) let rec find x = function | Empty -> raise Not_found | Node (l, v, r, _) -> let c = Pervasives.compare x v in if c = 0 then v else if c < 0 then find x l else (* c > 0 *) find x r let singleton x = Node(Empty, x, Empty, 1) let rec remove x = function Empty -> Empty | Node(l, v, r, _) -> let c = Pervasives.compare x v in if c = 0 then merge l r else if c < 0 then bal (remove x l) v r else bal l v (remove x r) let rec union s1 s2 = match (s1, s2) with (Empty, t2) -> t2 | (t1, Empty) -> t1 | (Node(l1, v1, r1, h1), Node(l2, v2, r2, h2)) -> if h1 >= h2 then if h2 = 1 then add v2 s1 else begin let (l2, _, r2) = split v1 s2 in join (union l1 l2) v1 (union r1 r2) end else if h1 = 1 then add v1 s2 else begin let (l1, _, r1) = split v2 s1 in join (union l1 l2) v2 (union r1 r2) end let rec inter s1 s2 = match (s1, s2) with (Empty, t2) -> Empty | (t1, Empty) -> Empty | (Node(l1, v1, r1, _), t2) -> match split v1 t2 with (l2, false, r2) -> concat (inter l1 l2) (inter r1 r2) | (l2, true, r2) -> join (inter l1 l2) v1 (inter r1 r2) let rec diff s1 s2 = match (s1, s2) with (Empty, t2) -> Empty | (t1, Empty) -> t1 | (Node(l1, v1, r1, _), t2) -> match split v1 t2 with (l2, false, r2) -> join (diff l1 l2) v1 (diff r1 r2) | (l2, true, r2) -> concat (diff l1 l2) (diff r1 r2) type 'a enumeration = End | More of 'a * 'a t * 'a enumeration let rec cons_enum s e = match s with Empty -> e | Node(l, v, r, _) -> cons_enum l (More(v, r, e)) let rec compare_aux e1 e2 = match (e1, e2) with (End, End) -> 0 | (End, _) -> -1 | (_, End) -> 1 | (More(v1, r1, e1), More(v2, r2, e2)) -> let c = Pervasives.compare v1 v2 in if c <> 0 then c else compare_aux (cons_enum r1 e1) (cons_enum r2 e2) let compare s1 s2 = compare_aux (cons_enum s1 End) (cons_enum s2 End) let equal s1 s2 = compare s1 s2 = 0 let rec subset s1 s2 = match (s1, s2) with Empty, _ -> true | _, Empty -> false | Node (l1, v1, r1, _), (Node (l2, v2, r2, _) as t2) -> let c = Pervasives.compare v1 v2 in if c = 0 then subset l1 l2 && subset r1 r2 else if c < 0 then subset (Node (l1, v1, Empty, 0)) l2 && subset r1 t2 else subset (Node (Empty, v1, r1, 0)) r2 && subset l1 t2 let rec iter f = function Empty -> () | Node(l, v, r, _) -> iter f l; f v; iter f r let rec fold f s accu = match s with Empty -> accu | Node(l, v, r, _) -> fold f l (f v (fold f r accu)) let rec for_all p = function Empty -> true | Node(l, v, r, _) -> p v && for_all p l && for_all p r let rec exists p = function Empty -> false | Node(l, v, r, _) -> p v || exists p l || exists p r let filter p s = let rec filt accu = function | Empty -> accu | Node(l, v, r, _) -> filt (filt (if p v then add v accu else accu) l) r in filt Empty s let partition p s = let rec part (t, f as accu) = function | Empty -> accu | Node(l, v, r, _) -> part (part (if p v then (add v t, f) else (t, add v f)) l) r in part (Empty, Empty) s let rec cardinal = function Empty -> 0 | Node(l, v, r, _) -> cardinal l + 1 + cardinal r let rec elements_aux accu = function Empty -> accu | Node(l, v, r, _) -> elements_aux (v :: elements_aux accu r) l let elements s = elements_aux [] s let choose = min_elt | 2,553 | 7,664 | {"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-2015-27 | longest | en | 0.590856 |
raphaelscholl.com | 1,501,130,081,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427429.9/warc/CC-MAIN-20170727042127-20170727062127-00501.warc.gz | 718,845,287 | 33,448 | # The power of natural selection
Last week I wrote a version of Richard Dawkins’s “methinks it is a weasel” program (as explained in The Blind Watchmaker). The point of the program is to demonstrate the power of cumulative selection in comparison to pure chance. Consider a random string such as “in the beginning god created the heavens and the earth”. In a purely random process, the probability of this string occurring is minuscule: with 27 letters in the alphabet (don’t forget the space!) and 54 letters in the string, the number of possible strings is 2754, or 1.97 x 1077. Your chances of hitting on this string by producing random strings are, for all practical purposes, zero.
But the situation changes once we introduce selection and cumulation. The program begins by creating a population of random strings, each 54 letters in length. None of these will be very close to the target string “in the beginning god created the heavens and the earth”. Nevertheless, some strings will match the target string in a few positions. The program evaluates each one to determine the best match. For example, the following best candidate in generation 1 (from an actual run of the program) shares 7 letters with the target string. These are underlined:
``` gen 1: tashiwwsmsianhdfyf yvrrjutym bjjoig byxfpkwpkkhzfj g h
target: in the beginning god created the heavens and the earth
```
The program then takes this one best match and mutates it to create a new population of candidate strings. For example, each letter (in each string in the population) might be replaced with a randomly chosen letter from the alphabet with a probability of 0.09 (resulting, in this case, in around 4.8 replaced letters per string on average). This new population of strings is then again evaluated, and the best match to the target string is again retained and mutated. The mutations can be either neutral (if a non-matching letter is replaced with a non-matching letter, or if a matching letter is replaced with itself), detrimental (if a matching letter is replaced with a non-matching one) or beneficial (if a non-matching letter is replaced with a matching one).
Many generation will yield none or only small improvements — for example, when a new population of 100 strings was created based on the first generation string above, the best candidate in the second generation had gained only one matching letter (underlined) in addition to a number of neutral mutations:
``` gen 2: tashiwwsmsitnhdfyfvyvrrjutym bjjoig byxfhkwpkkhzfj gth
gen 1: tashiwwsmsianhdfyf yvrrjutym bjjoig byxfpkwpkkhzfj g h
target: in the beginning god created the heavens and the earth
```
This cumulative process of mutation and selection continues until each letter in the string matches the target string, at which point the program stops. Needless to say, in the beginning most mutations will be neutral. As the string approaches the target string, more and more mutations will be detrimental, or in other words, the program can take quite a number of generations towards the end to optimize the last few letters.
I would offer my own version to the internet, but it seems redundant since a nice Python version of it is already available, and it is easier to play around with the source code of the Python program than with the code of my Objective-C implementation. (On a Mac, you can run the Python version by opening a Terminal, cd-ing to the directory of the Weasel.py, and running: “python Weasel.py”.)
The main message of the program is that cumulative selection is very different from random generation. In a typical run of the program, it takes around 112 generations to select the target string. If the proverbial monkeys tapping away randomly at typewriters produced one string per second, it would take them 6 x 1069 years to explore all the possible strings of 54 letters. The equivalent selection process — also producing one string per second — would be completed after only 31 hours. Such is the power of random variation coupled with cumulation!
The program has been criticized for exaggerating the power of selection. The critics argue that the program retains correct letters permanently and does not allow them to mutate any more, which is obviously not how mutations work in nature. However, the criticism backfires, since the selection process of the program works fine even if all letters are allowed to mutate in each generation. (Note: That all letters are allowed to mutate does not mean that all letters will mutate; this depends on the mutation rate, discussed below.) Both my implementation and the Python version allow all strings to mutate: it is entirely possible for the number of differences to the target string to increase from one generation to the next, and this often happens. Nevertheless, over time the deleterious mutations are removed.
When I’ve talked about this program in my lectures, some students were concerned about a kind of cheating. They felt that the program “already knew” the target string, so that it did not mirror evolution in nature, where the outcome is unknown. Maybe there is a version of this objection that I have not considered, but generally speaking I think it misses the point. The evolved string is found by the program through a process of blind variation and selection (just as in nature); the target string is only used to determine the “fitness” of a particular letter in a particular location. This reflects actual selection processes: biological variations will also have fitness values relative to the environment in which they occur.
It is instructive to consider how variations in the parameters “mutation rate” and “population size” affect the selection process. The mutation rate, in this program, is a number between 0 and 1. It determines the probability with which an individual letter in a string is replaced during the production of a new population of strings. Trivially, a mutation rate of 0 means no mutations and thus no change in the strings over time. A mutation rate of 1 means that each letter is mutated in each generation, which amounts to the absence of cumulative selection: gains in one generation are not retained in the next. For cumulative selection to work, mutation rates have to be relatively low (try it). In my experiments, mutation rates much above 0.1 generally lead to a selection process that oscillates around a certain number of differences and does not terminate. The reason for this is clear: high mutation rates interfere with the retention of matching letters.
More surprising perhaps are variations of the population size. This variable determines how many strings are produced (and mutated) in each generation. Even though I should have known better, I expected this not to matter too much: I was thinking about the total number of variations produced, and so surely it should be immaterial whether I’m producing 300 generations x 100 strings or 3000 generations x 10 strings — the total number of strings is the same. But this is where so-called “genetic drift” becomes an issue! Consider that each generation begins with a “best candidate” string and produces a population of mutated variants of it. In a reasonably large population, there will be many neutral or detrimental variants and a few improved ones; the improved ones are then selected as the template for the next generation. However, the smaller the population size, the more probable it becomes that none of the few produced variants are improvements. It is easy to see this if you assume a population size of 1: most one-off variants of a string will not be improvements, especially in the latter parts of the selection process (when most letters already match the target). Thus, small population sizes make it possible for the string to start to “drift” randomly, simply because each generation only realizes a small sample of possible variations, most of which are neutral or detrimental.
However, the effects of population size and mutation rate interact. For instance, a mutation rate of 0.09 and a population size of 1000 will allow “in the beginning god created the heavens and the earth” to evolve. If you change the population size to 100, then the process will not terminate: it will oscillate between 10 and 15 differences or so. If you now reduce the mutation rate a bit, say to 0.05, then the process will again terminate. I leave it as an exercise for the reader to figure out the explanation of the phenomenon! | 1,782 | 8,487 | {"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-2017-30 | longest | en | 0.845357 |
https://www.kanjeevaramsilks.in/jacques/Jan_28495.html | 1,643,034,926,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304570.90/warc/CC-MAIN-20220124124654-20220124154654-00216.warc.gz | 913,947,327 | 7,718 | [email protected]
# Weight Of Crushed Stone Per Cubic Yard
1. Home
2. Weight Of Crushed Stone Per Cubic Yard
### 2020 Crushed Stone Calculator 34 Inch Rock
Jan 20, 2021 Calculating Crushed Stone by Weight. A final calculation is if you need to figure out the weight in tons of your crushed stone. You might not need to figure this out, but its handy to know. Most gravel weighs about 1.5 tons per cubic yard. So, for the examples above, 2.45 cubic yards of gravel weighs 3.68 tons or 7,350 pounds.
### Weight Of Crushed Concrete Per Cubic Yard
What is the weight per cubic foot of crushed concrete. They typical weight of a yard of crushed concrete runs around 3000lbs or a ton and a half The weight will vary between company mainly because of the crush size. How many cubic yards are in a ton of crushed concrete. Crushed stone is quoted at a weight of 2700 pounds per cubic yard Your ...
### Material Weight Pounds Per Cubic Yard
24 rows FTruck Body Pricing BrochuresExcelCubic_Yardage_Chart DCubic_Yardage_Chart D Rev A 6/9/2015 ...
### How Much Does A Cubic Yard Of Granite Weigh
You will have to know that the standard weight contractors use for crushed stone is 2700 pounds per cubic yard. Multiply the number of cubic yards by 2700 and divide by 2000. This value is the number of tons that you will need.
### How Much Does A Cubic Yard Of 2 Stone Weight
Crushed stone is quoted at a weight of 2700 pounds per cubic yard. Your stone dealer tells you he has a truck that can deliver 20 tons of stone per load. You
### Crushed Rock Weight Per Yard Useful Gardening Tips
Approximate Weights of Various Construction Material Per Cubic Yard. Most of Harmony Sand Gravels products will weight approximately 2,840 pounds per cubic yard or about 1.42 tons per cubic yard. For estimating purposes, most Contractors consider the yield to be 3,000 pounds per cubic yard or 1.5 tons per cubic yard.
### Stone Crushed Volume To Weight Conversion
About Stone, crushed 1 cubic meter of Stone, crushed weighs 1 602 kilograms kg 1 cubic foot of Stone, crushed weighs 100.00959 pounds lbs Stone, crushed weighs 1.602 gram per cubic centimeter or 1 602 kilogram per cubic meter, i.e. density of stone, crushed is equal to 1 602 kg/m.In Imperial or US customary measurement system, the density is equal to 100.0096 pound per cubic
### How Much Does A Yard Of Crusher Run Gravel Weigh
Apr 07, 2021 What does 4 yards of rock weigh? Multiply the weight of the gravel, in this case 105 pounds, by 27 to determine cubic yards because there are 27 cubic feet in 1 cubic yard. The result is 2,835 pounds per cubic yard of gravel. How much is a yard of
### Cubic Yardage Calc Sheet Swaploader
CUBIC YARDAGE CALCULATION SHEET Height of Sides ... Material Weight Pounds per Cubic Yard Asphalt 2,700 lb. Iron (wrought) 13,100 lb. Brush/Branches (loose) 250 lb. Leaves (loose, dry) 150 lb. ... Crushed Stone 2,700 lb. Plywood Sheets 800 lb. Earth (loose) ...
### How Much Does A Yard Of 34 Inch Crushed Stone Weigh
Jan 02, 2020 Plain pea gravel and crushed clamshells are each priced at about \$40 per cubic yard and \$50 per ton. Crushed stone is costlier at about \$55 per cubic yard and \$65 per ton. Buying pea gravel in bulk may reduce costs, but different finishes, like gravel with color, will add anywhere from \$20 to \$50 to the price per unit.
### How Much Does A Yard Of Crushed Granite Weigh
Feb 21, 2020 A cubic yard or often called a square yard of decomposed granite or crushed stone fines material typically weighs 3,000 pounds or 1-1/2 tons and will typically cost from 37.99 to 74.99 per yard depending upon where the job site is located, how
### How Much Does A Cubic Yard Of Aggregate Base Weight
Jan 28, 2021 This ash aggregate usually has a unit weight of 60 65 pounds per cubic foot. How much does a yard of crushed weigh? Weight per cubic yard is approximately 2500 lbs. One yard will cover about 150 square feet at 2 deep. Crushed stone
### How Many Tons Of 57 Stone Are In A Cubic Yard Dirt
Oct 08, 2020 The density of 57 stone or granite gravel is 2,410 pounds per cubic yard. For a project that is 10 feet wide, 40 feet long and 2 inches deep, for example, the volume will be 2.5 cubic yards. The estimated amount of material needed is 2.98 tons.
### How Much Does 1 Cubic Yard Of Road Base Weigh
Jun 02, 2020 Crushed stone is quoted at a weight of 2700 pounds per cubic yard. Your stone dealer tells you he has a truck that can deliver 20 tons of stone per load. You need to know who many cubic yards that comes out to.
### Material Calculator Northstate Landscape
Most gravel and crushed stone products have similar weights per ton. G ravel and sand typically weighs 2,200-2,700 pounds per cubic yard. In addition, there are 2,000 pounds to a ton. Certain products, like washed gravel, weigh more like 2,835 pounds per cubic yard.
### Stone Calculator How Much Stone Do I Need
Rectangular Area with Crushed Gravel (105 lb/ft) and Price Per Unit Mass Lets say I need crushed gravel for part of my driveway which measures 4ft long, 2ft wide and 9in (0.75ft) deep. Lets also say that the selected gravel costs \$50 per ton.
### How Much Does A Cubic Yard Of Crushed Limestone Cost
A general rule of thumb when converting cubic yards of gravel to tons is to multiply the cubic area by 1.4. For your reference, gravel typically weighs 2,800 pounds per cubic yard. In addition, there are 2,000 pounds to a ton. For instance, if your area to be landscaped is 100 cubic yards, multiply that by
### Volumetoweight Conversion Charts
Stone or gravel 1 cubic foot 98.00 0.049 FEECO Stone, crushed 1 cubic foot 100.00 0.050 FEECO Stone, crushed, size reduced 1 cubic yard 2,700.00 1.350 Tellus Institute Stone, large 1 cubic foot 100.00 0.050 FEECO Wax 1 cubic foot 61.00 0.031 FEECO Wood ashes 1 cubic foot 48.00 0.024 FEECO AGGREGATE CD bulk 1 cubic yard 484.00 0.242 FL DEP
### How Much Does A Yard Of Decomposed Granite Weigh
Jul 11, 2021 A cubic yard or often referred to as a square yard of decomposed granite or crushed stone fines material frequently weighs 3,000 pounds or 1-1/2 tons and will commonly cost from 37.99 to 74.99 per yard relying on where the task site is located, how numerous yards girlfriend purchase and how much away the quarry is.
### How To Calculate Weight Per Cubic Yard Of Gravel Home
Dec 21, 2020 The weight of other crushed stone per cubic yard varies. The chart from Harmony Stone Gravel shows that plain crushed stone weighs the same pounds per cubic yard, but, at 1.35 tons, its heavier ...
### 2021 Gravel Prices Crushed Stone Cost Per Ton Yard
It takes 1.4 tons of stone per cubic yard. Gravel Cost Per Yard. Bulk gravel costs \$15 to \$75 per yard on average. Crushed concrete, sand and gravel mix, and crushed shells are the cheapest stones at \$15 per yard. Crusher run, pea gravel, steel slag, and river rock typically cost \$50 per yard or more.
### How Much Does 12 Yard Of Crushed Rock Weigh
5799. Crushed stone is quoted at a weight of 2700 pounds per cubic yard. Read the full answer. A cubic yard is a measurement that is 3 feet by 3 feet by 3 feet. A cubic yard measures volume where a ton measures weight. A yard of topsoil usually weighs about 1,800 pounds and a yard of gravel usually weighs about 2,200 pounds.
### Approximate Weights Of Various Construction Material Per
Apr 04, 2012 Approximate Weights of Various Construction Material Per Cubic Yard. Most gravel products will weight approximately 2,840 pounds per cubic yard or about 1.42 tons per cubic yard. For most estimating purposes, consider the yield to be 3,000 pounds per
### What Does 1 Cubic Yard Of Crushed Concrete Weigh Answers
Oct 26, 2007 Best Answer. Copy. They typical weight of a yard of crushed concrete runs around 3,000lbs or a ton and a half. The weight will vary between
### What Is The Weight Of 1 Yard Of Crush And Run Gravel
Sep 29, 2009 The answer given wasDepending on the relative proportions of water, sand, stone, and cement, the weight per cubic yard can vary from less than 2000 lbs to
### How Much Does A Cubic Yard Of Crushed Concrete Weigh
Crushed concrete costs \$11 to \$53 per ton, around \$16 to \$75 per cubic yard, and \$1 to \$3 per cubic foot, with prices depending on the quantity. via. How much does a cubic yard of concrete weigh dry? Concrete weighs around 3900 pounds per cubic yard but can drop to 3500 pounds per cubic yard after the concrete dries. Different types of concrete ...
### How Much Does 1 Cubic Yard Of Abc Stone Weigh Answers
Jul 08, 2011 This depends on the density of the stone and the over all volume of the stone. So one cubic yard rather than one yard. Simply determine the type of stone, look up its typical density usually in lb/ft3, and covert the volume into cubic feet, then multiply the two values to obtain its weight.
### How Much Does 1 Yard Of Crushed Concrete Weigh Binq
Nov 26, 2012 Aggregates, Sand, Gravel, Rock. Weight per Cubic Yard 138 023 1 Minus Crushed Concrete (CC) 1.4 tons 138 025 1 Minus Recycled Asphalt Pavement (RAP) 1.4 tons 132 040 2 Minus Gravel/ Sewer More detailed
### How Much Does A Yard Of Limestone Weight Civil Sir
In this regard, How much does limestone rock weigh, according to Imperial or US customary measurement system, on average limestone rock can weighs around 4500 pounds per cubic yard or 2.25 short tons per yard that would make 168 pounds per cubic foot or 0.084 short tons per cubic foot. How many yards are in a ton of limestone.
### How Many Tons In 1 Cubic Yard Of 34 Crushed Stone
Answer (1 of 5) A cubic yard is a measure of volume, yet you are trying to convert it to a tonne, which is a measure of weight. Thus, it is impossible to answer the question entirely correctly there will always be variables. The most important variable is the substance. In this case its crushed stone, which is usually an aggregate of otherwise unusable stone, but this material can vary ...
### What Is The Weight Of Gravel Per Cubic Yard
Mar 27, 2020 The weight of gravel per cubic yard is approximately 2,970 pounds, or 1.48 tons, if the gravel is dry. If the gravel is out of water, the weight per cubic yard is
### How Much Does A Cubic Yard Weigh The Calculator Site
Jul 28, 2021 1 cubic yard of gravel (dry) weighs about 2835 pounds or 1.42 US tons. 1 cubic yard of gravel (wet) weighs about 3375 pounds or 1.69 US tons. 1 cubic yard of topsoil weighs about 2295 pounds or 1.15 US tons. 1 cubic yard of concrete weighs about 3915 pounds or 1.96 US tons. 1 cubic yard of sand (dry) weighs about 2700 pounds or 1.35 US tons.
### Bulk Material Calculator Contractors Stone Supply
Most of our bulk materials, with the exception of mulch, are sold by the weight. The following are approximate weights for most of our bulk materials. Sand 1.10 - 1.25 tons (2,200 - 2,500 lb.) per cubic yard. Planting Mix 1 ton (2,000 lb.) per cubic yard. Lawn Dressing .90 tons (1,800 lb.) per cubic yard. Compost .40 tons (800 lb.) per cubic yard. | 2,861 | 11,010 | {"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-2022-05 | latest | en | 0.891086 |
https://forum.ansys.com/forums/topic/apply-rotational-stiffness-to-a-base/ | 1,701,188,878,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00219.warc.gz | 315,323,453 | 36,714 | ## General Mechanical
Topics relate to Mechanical Enterprise, Motion, Additive Print and more
#### Apply rotational stiffness to a base
• Moustafa El-Sawy
Subscriber
I have a concrete block made up of SOLID elements. I am currently applying a spring at the base at each mesh element but I found I also want to apply rotation stiffness to the entire foundation. I have a single value of the stiffness of the system in each direction including rotation stiffness. Before knowing I need to apply rotational stiffness, I took the total value of stiffness in each translation direction and applied a portion of this to each node based on its tributary area. However I also now need to apply rotation stiffness to the entire system so I think this approach will not work.
SOLID nodes dont have rotation DOF, so I was thinking I would create a master node and assign it a single spring with x,y,z,rotx,roty,rotz and then connect this node with all my base nodes. This would distribute the stiffness to the nodes and allow the system to rotate with a certain stiffness.
• Can I perform this without tying all base nodes (X,Y,Z)? My behavior is most likely rigid but I was wondering if its possible. I dont think its possible but please correct me if im wrong
• If its not possible, can I make my base like a rigid plate (X,Y) direction but also able to rotate around the master node with a certain stiffness (around x,y).
• peteroznewman
Subscriber
Select the bottom face of your block and insert a Spring in the Connections folder. In the Details window, change the Scope to a Body-Ground spring and change the Behavior of the Mobile side (bottom face) to be Deformable. Deformable means that the individual nodes in the face will have the freedom to move according to the loads distrubuted to them by the Deformable connection, but that the average motion will be controlled by the spring’s master node.
The spring created in Mechanical is a Longitudinal spring, COMBIN14, which is not what you want, but you can override that spring type with another spring element called a COMBI250 by using a Command Object under the Spring object in the model Outline.
A COMBI250 allows you to define six spring constants for the 3 translational and 3 rotational stiffness values.
Look up the COMBI250 element in the Mechanical APDL, Element Library section of the Help system. One line of the APDL code will switch the element type to COMBIN250 and other lines of code will provide the Real Constants for the spring stiffness values.
Here is one example of the needed code. Here is another example.
• Moustafa El-Sawy
Subscriber
Thanks Peter. I am using APDL. With the COMBI250, I am able to assign springs in all directions at the nodes at the bottom of my solid elements even though the solid element nodes dont have DOF for rotation?
• peteroznewman
Subscriber
No, a rotational spring attached to a node on a solid element will not be able to apply any moment to that node. A pilot node that distributes forces and moments to a set of nodes can have a moment load applied to the pilot noe, which will be resolved as forces on the set of nodes it is connected to.
• Moustafa El-Sawy
Subscriber
Hi Peter, ive been trying to apply some rotation stiffness to my whole base but im unsure if my approach is correct can you take a look at the code below:
/CLEAR ! Clear all previous data and start fresh
/PREP7 ! Enter preprocessor phase
! Element Definitions
ET, 1, SOLID185 ! Define SOLID186 element type for solid modeling
ET, 2, COMBIN14 ! Define COMBIN14 element type for springs
KEYOPT, 2, 2, 0
KEYOPT, 2, 3, 0
ET, 3, COMBIN14 ! Define COMBIN14 element type for springs
KEYOPT, 3, 2, 0
KEYOPT, 3, 3, 1
ET, 999, SHELL181 ! Define SHELL181 element type with a placeholder ID of 999
SECTYPE, 999, SHELL ! Specify section type for SHELL181 element
SECDATA, 10 ! Define the thickness of the shell as 100 units
! Material Properties for the Solid
MP, EX, 1, (32836.56803/1000)/0.83 ! Define Young's Modulus for solid after adjusting value
MP, PRXY, 1, 0.2 ! Define Poisson's Ratio for the solid
MP, DENS, 1, (23.544/9810.0)/10**9 ! Define Density for the solid after adjusting value
MP, ALPX, 1, 10E-006 ! Define Coefficient of Thermal Expansion for the solid
! Material Properties for Fake/Rigid Areas
MAT, 999 ! Activate material with ID 999 for fake areas
MP, EX, 999, 1E+13 ! Define a very high Young's Modulus to simulate rigidity
MP, PRXY, 999, 0.2 ! Define Poisson's Ratio for the fake areas
MP, DENS, 999, 0.0 ! Define Density as 0, implying it's a "rigid" area
MP, ALPX, 999, 10E-006! Define Coefficient of Thermal Expansion for the fake areas
! Geometry Creation
BLC5, 0, 0, 40000, 40000, 50000 ! Create a block starting from origin with the given dimensions
! Meshing Settings for the Solid Block
TYPE, 1 ! Set the element type to SOLID186
MAT, 1 ! Set the material type to the first defined material (for solid)
ESIZE, (5000) ! Set the mesh size
VMESH, ALL ! Mesh the created volume
! Meshing Settings for the Base Area (Shell)
MAT, 999 ! Switch to the material ID 999 (for fake areas)
TYPE, 999 ! Switch to the SHELL181 element type
SECNUM, 999 ! Use the section definition with ID 999
ND_TOL = 10 ! Define a tolerance for node selection (helpful in selecting nodes near z=0)
ALLSEL, ALL ! Select everything in the workspace
ASEL, S, LOC, Z, 0-ND_TOL, 0+ND_TOL ! Select areas near the base within the defined tolerance
AMESH, ALL ! Mesh the selected areas (i.e., the base) using SHELL181 element type
! Select nodes at the base
ND_TOL = 10
NSEL, S, LOC, Z, 0-ND_TOL, 0+ND_TOL
! Create a component for the selected nodes
CM, BASE_NDS, NODE
! Get essential parameters for node looping
*GET, NMAX, NODE, 0, NUM, MAX
*GET, NODE_NMIN, NODE, 0, NUM, MIN
*GET, NODE_NCOUNT, NODE, 0, COUNT
KX = 1000 ! Spring constant in X direction
KY = 500 ! Spring constant in Y direction
KZ = 2000 ! Spring constant in Z direction
R, 1001, KX
R, 2001, KY
R, 3001, KZ
CMSEL,S, BASE_NDS ! Select component of nodes
L_SPRING = 100
! Loop over each node in the created component
NODE_NNUM = NODE_NMIN
*DO, i, 1, NODE_NCOUNT
! Get coordinates of the current node
*GET, X_COORD, NODE, NODE_NNUM, LOC, X
*GET, Y_COORD, NODE, NODE_NNUM, LOC, Y
*GET, Z_COORD, NODE, NODE_NNUM, LOC, Z
! Create spring in X direction
N, , X_COORD-L_SPRING, Y_COORD, Z_COORD ! Generate a new node
*GET, newNode, NODE, 0, NUM, MAX ! Get the ID of the newly created node
D, newNode, ALL ! Fix the ground node for X direction
TYPE, 2
REAL, 1001 ! Define spring constant for X
E, NODE_NNUM, newNode
! Create spring in Y direction
N, , X_COORD, Y_COORD-L_SPRING, Z_COORD ! Generate a new node
*GET, newNode, NODE, 0, NUM, MAX ! Get the ID of the newly created node
D, newNode, ALL ! Fix the ground node for Y direction
TYPE, 2
REAL, 2001 ! Define spring constant for Y
E, NODE_NNUM, newNode
! Create spring in Z direction
N, , X_COORD, Y_COORD, Z_COORD-L_SPRING ! Generate a new node
*GET, newNode, NODE, 0, NUM, MAX ! Get the ID of the newly created node
D, newNode, ALL ! Fix the ground node for Z direction
TYPE, 2
REAL, 3001 ! Define spring constant for Z
E, NODE_NNUM, newNode
! Move to the next node in the component
NODE_NNUM = NDNEXT(NODE_NNUM)
*ENDDO
! add node at center of block and assign rotational torsion springs to this node (master node)
! Create a master node at the center of the block
N, , 0, 0, -500
*GET, masterNode, NODE, 0, NUM, MAX ! Get the ID of the master node
*GET, X_COORD, NODE, masterNode, LOC, X
*GET, Y_COORD, NODE, masterNode, LOC, Y
*GET, Z_COORD, NODE, masterNode, LOC, Z
! Assign rotational torsion springs to this node (master node)
KRX = 500 ! Spring constant for rotation about X axis (Torsional Spring)
KRY = 500 ! Spring constant for rotation about Y axis (Torsional Spring)
KRZ = 500 ! Spring constant for rotation about Z axis (Torsional Spring)
R, 4001, KRX
R, 5001, KRY
R, 6001, KRZ
! Create Torsion springs at the master node
N, , X_COORD-L_SPRING, Y_COORD, Z_COORD ! Generate a new node offset in X for ROTX spring
*GET, newXNode, NODE, 0, NUM, MAX
D, newXNode, ALL
TYPE, 3
REAL, 4001
E, masterNode, newXNode
N, , X_COORD, Y_COORD-L_SPRING, Z_COORD ! Generate a new node offset in Y for ROTY spring
*GET, newYNode, NODE, 0, NUM, MAX
D, newYNode, ALL
TYPE, 3
REAL, 5001
E, masterNode, newYNode
N, , X_COORD, Y_COORD, Z_COORD-L_SPRING ! Generate a new node offset in Z for ROTZ spring
*GET, newZNode, NODE, 0, NUM, MAX
D, newZNode, ALL
TYPE, 3
REAL, 6001
E, masterNode, newZNode
! Transfer rotational stiffness from the master node to the slave nodes (base nodes)
CMSEL, S, BASE_NDS ! Select nodes in BASE_NDS component
NSEL,A,NODE,,masterNode
CERIG,masterNode,ALL,RXYZ
! Re-select everything for subsequent commands
ALLSEL, ALL
/SOLU ! Enters the solution processor
DAMPING_RATIO = 0.00 ! Modal Damping Ratio
N_MODES = 25 ! Number of Modes, When Method = LANPCG, NMODE should be less than 100 to be computationally efficient.
ANTYPE, MODAL ! Perform a modal analysis
MXPAND, N_MODES ! Number of modes
DMPRAT, DAMPING_RATIO ! Modal damping ratio
MODOPT, LANB, N_MODES
ALLSEL, ALL
/MKDIR, 'OUTPUT'
/OUTPUT, 'OUTPUT/Modal_Analysis', 'OUT'
SOLVE
FINISH ! Exits solution processor /SOLU
• Moustafa El-Sawy
Subscriber
If I need to apply springs in all 6 directions to the base of my solid elements, is there any way to accomplish this (individual nodes not the entire base with a single value).
I was thinking of adding shell elements to the base that share the same nodes as the solid elements and this will enable those nodes to take moment? Do you think this will work? In this case I was planning on assigning these shells as massless and with small stiffness to not affect the dynamic behavior of the structure.
• peteroznewman
Subscriber
I am not an expert at APDL code. I use Workbench and only use a few lines of APDL occasionally. Maybe someone else can comment. | 2,885 | 10,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.937125 |
https://im.kendallhunt.com/HS/students/3/6/3/practice.html | 1,719,283,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00068.warc.gz | 274,455,377 | 25,841 | # Lesson 3
The Unit Circle (Part 1)
• Let’s learn about the unit circle.
### Problem 1
$$C$$ is a circle with radius $$r$$. Which of the following is true? Select all that apply.
A:
The diameter of $$C$$ is $$2r$$.
B:
The circumference of $$C$$ is $$\pi r$$.
C:
The circumference of $$C$$ is $$2\pi r$$.
D:
One quarter of the circle has length $$\frac{\pi r}{4}$$.
E:
One quarter of the circle has length $$\frac{\pi r}{2}$$.
### Problem 2
angle measure rotation
0 0
$$\frac{\pi}{6}$$
$$\frac{1}{8}$$
$$\frac{1}{6}$$
$$\frac{\pi}{2}$$
$$\frac{2\pi}{3}$$
$$\frac{1}{2}$$
$$\frac{3\pi}{2}$$
$$\frac{7}{8}$$
1
The table shows an angle measure in radians and the amount of rotation about a circle corresponding to the angle. For example, $$2\pi$$ radians corresponds to 1 full rotation. Complete the table.
### Problem 3
A wheel has a radius of 1 foot. After the wheel has traveled a certain distance in the counterclockwise direction, the point $$P$$ has returned to its original position. How many feet could the wheel have traveled? Select all that apply.
A:
$$\frac{\pi}{2}$$
B:
$$\pi$$
C:
$$2\pi$$
D:
$$5\pi$$
E:
$$10\pi$$
### Problem 4
Here are some points labeled on the unit circle:
1. What is the measure in radians of angle $$POR$$?
2. Angle $$POQ$$ is halfway between 0 radians and angle $$POR$$. What is the measure in radians of angle $$POQ$$?
3. Label point $$U$$ on the circle so that the measure of angle $$POU$$ is $$\frac{3\pi}{4}$$.
4. Label point $$V$$ on the circle so that the measure of angle $$POV$$ is $$\frac{3\pi}{2}$$.
### Problem 5
1. Mark the points on the unit circle with $$x$$-coordinate $$\frac{4}{5}$$.
2. What are the $$y$$-coordinates of those points? Explain how you know.
### Problem 6
The point $$(8, 15)$$ lies on a circle centered at $$(0,0)$$. Where does the circle intersect the $$x$$-axis? Where does the circle intersect the $$y$$-axis? Explain how you know.
(From Unit 6, Lesson 1.)
### Problem 7
Triangles $$ABC$$ and $$DEF$$ are similar. Explain why $$\tan(A) = \tan(D)$$.
(From Unit 6, Lesson 2.) | 662 | 2,083 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.808206 |
https://www.pcorner.com/list/ASSEMBLY/MATHASM.ZIP/MUL_F.ASM/ | 1,716,690,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00558.warc.gz | 812,459,081 | 9,146 | Output of file : MUL_F.ASM contained in archive : MATHASM.ZIP
MUL_F PROC NEAR
;***************************************************************
; floating point
; multiplication routine
;***************************************************************
PUSH SI ;save index registers
PUSH DI
SUB DX,DX ;clear work registers
MOV AX,DX
MOV BX,DX
MOV CX,DX
; check for zero operand
OR DX,[DI]+2 ;OP1 high word in DX
JZ MF2 ;zero only if operand is 0
OR CX,[SI]+2 ;OP2 high word in CX
JZ MF2
; compute exponent
MOV AL,DH ;E1 in AX
MOV BL,CH ;E2 in BX
SUB BX,128 ;subtract bias
ADD AX,BX ;E=E1+E2+bias
;
MOV BX,[SI] ;get OP2 low word in BX
MOV SI,AX ;save exponent in SI
MOV AX,[DI] ;get OP1 low word in AX
SUB DH,DH ;clear high bytes
MOV CH,DH ;M1 is DL:AX; M2 is CL:BX
; compute result sign
MOV DI,DX
XOR DI,CX
AND DI,10000000B ;isolate sign bit in DI
; restore leading ones
OR DL,10000000B
OR CL,10000000B
; multiply mantissas
CALL MUL32 ;47-48 bit product in AX:CX:BX
;
OR AX,AX ;is it normalized?
JS MF1
; normalize product
SHL BX,1 ;shift it left 1
RCL CX,1
RCL AX,1
SUB SI,1 ;and decrement exponent
; check for overflow
MF1: CMP SI,255
JG MF3 ;exit if exponent > 255
; check for underflow
CMP SI,0
JLE MF4 ;exit if exponent <= 0
; reformat for output
MOV DL,AH ;mantissa in DL:AX
MOV AH,AL
MOV AL,CH
MOV DH,CL ;trailing bits in DH
MOV CX,BX ;and CX
MOV BX,SI ;exponent in BH
MOV BH,BL
JMP R0 ;ROUND in addition routine
MF2: JMP FINISH
MF3: JMP OVER_F
MF4: JMP UNDER_F
MUL_F ENDP
### 3 Responses to “Category : Assembly Language Source CodeArchive : MATHASM.ZIPFilename : MUL_F.ASM”
1. Very nice! Thank you for this wonderful archive. I wonder why I found it only now. Long live the BBS file archives!
2. This is so awesome! 😀 I’d be cool if you could download an entire archive of this at once, though.
3. But one thing that puzzles me is the “mtswslnkmcjklsdlsbdmMICROSOFT” string. There is an article about it here. It is definitely worth a read: http://www.os2museum.com/wp/mtswslnk/ | 692 | 2,023 | {"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-2024-22 | latest | en | 0.58148 |
http://www.abovetopsecret.com/forum/thread172763/pg1 | 1,542,226,948,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114213308-00247.warc.gz | 370,226,465 | 11,065 | It looks like you're using an Ad Blocker.
Thank you.
Some features of ATS will be disabled while you continue to use an ad-blocker.
# social security
page: 1
1
share:
posted on Sep, 23 2005 @ 02:05 PM
Stock market and social security
1. Winner's gain is from loser's.
A farmer planted a seed. He sold the fruit The famer created a wealth.
A worker produced a car. He sold the car. He creates a wealth.
Investor A bought one hundred shares at 1.00/share. The company got one hundred dollars to pay rent, wage and material. Then the stock market rose to peak. Investor A sold the share at 1.10/share to investor B. A got 110 dollars. He made a 10% profit. But was that 10 dollars created? No. it was B's loss. When B bought the stock from A, he became a potential loser. What he bought was only a piece of paper. He couldn't cash the stock with the company which issued it. What B can do is hoping some one else to take over the potential loss.
Situation 1. If the stock is Enron, then when it went bankruptcy, B's stock worth nothing. Here Company got 100 dollars. A got 10 dollars. B is the loser. He lost 110 dollars. Winners' money is from loser's. It's evidenct.
Situation 2. If the stock is HP, then in trough, the share price may fall to 0.90/share. B sold it to C. B lost 20 dollars. C paid 90 dollars for 100 shares. C sold the stock in peak 2 at 1.20/share to D. Now D becomes a potential loser. If nobody has the will to buy his paper, then the stock worth zero. Now let's see, company got 100 dollars. A sold stock at 1.10/share. he made 10 dollars. B bought at 1.10/share, sold at 0.90/share. B lost 20 dollars. C bought at 0.90/share and sold at 1.20/share. C won 30 dollars. 10(A) + 30(C) + 100 (company) = 20(B loss) + 120 (D's potential loss)
The eqation: Winner's gain(profit) + Capital gain (Company issue the stock) = Losers' loss (loss) + Potential loss (Amount paid by the latest stock holder)
You can see there is no wealth created. How much winner got is how much loser and potential loser lost. And it doesn't include administration fee. (it's about 2 trillion in 10 years period, Re: San Jose Mercury News, 12/17/04) So when Bush say you may get better income in stock market, there must be some people bear the loss for the winner's gain. Whom do you think will be the loser and winner?
(I omit the dividend here. it's similar to interest paid by bank.)
posted on Sep, 23 2005 @ 03:40 PM
Assuming you live in the US, it might indeed be possible (well, it has happened - the consequences range from "none" to "made my life hell", by the accounts I've read) to opt out of the SS system.
I'm not sure how advisable this is or isn't - indeed, doing so will make your life much more difficult in the short term, but it could indeed save you a lot in the long term (particularly relating to security issues, etc).
As an example, legally a bank in the US cannot deny you a non-interest bearing account simply because you refuse to give your SS#; this only applies if you're a citizen though (not if you're a permanent resident or other). Most will try to deny you an account, but if you push hard enough (and this, again, is your legal right) it's possible.
Information here (though it is very biased - you can wade through the rhetoric though!).
By all accounts the easiest way to do this is by using instead your TIN (Tax identification number) instead of your SS#; there's much more info on that website. I can't guarantee the veracity of the information, but other individuals have confirmed the accuracy to a great degree.
Anyway. Not sure exactly where your post was going (sorry!), but wanted to mention this anyway
posted on Oct, 3 2005 @ 04:18 PM
2. Stock is no other than a piece of paper
The value of stock market is supported by continue coming of investment fund. One thing you should know the people who hold the stock is no other then hold a piece of paper. That's a bubble. When no money came, then the bubble will break up.
When you deposit 100 dollars in the bank, you are guaranteed to get that deposit back, plus interest.
When you buy one hundred dollars of shares of a company, you are told you probably get some dividend sometime if business is good. The dividend is not guaranteed. And you can not cash the stock with the company. Because they have spent it to pay rent, wage and equipment already. If you liquidate the company, most time you may get a negative asset. e.g. if it's Microsoft, what they left for you is a program of Windows. UA may have some airplanes. But they always come with a huge debt. What kind of asset do Kodak and McDonald have for the stock they issued? What you hold finally could be a piece of paper. What you hope is someone else would buy that paper from you to take over your potential loss. When people put all their retirement fund in stock market, they are sitting on a big bubble. All they hold is a bunch of paper. One day when people wake up and refuse to behave like a fool, then there will be a collapse of stock market.
What Bush does is to persuade people put their retirement fund into the market to take over the hot potatoes.
posted on Oct, 3 2005 @ 07:51 PM
I'm not entirely sure of your point.
I bought 100 shares in a particular company.
Two years later, I sold those shares at something of a profit.
The next year, I bought 50 shares somewhere else, but instead "cashed them in" with the company itself.
But the bottom line? Nobody forced me...it's a gamble, and we know this. It's not a secret.
And I'm just confused...why is this all Bush's fault, exactly?
new topics
top topics
1 | 1,387 | 5,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-47 | latest | en | 0.968206 |
https://www.coursehero.com/file/7146517/L-An-075-0125-7855-1000-12-270-W-60-60-1227-kW-IP/ | 1,495,990,204,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610342.84/warc/CC-MAIN-20170528162023-20170528182023-00378.warc.gz | 1,038,788,966 | 22,931 | # L an 075 0125 7855 1000 12 270 w 60 60 1227 kw ip
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: e, n = N / 2 = 2000 / 2 = 1000 and cross-sectional area of the cylinder, π D2 π (100) 2 = = 7855 m m 2 4 4 We know that indicated power, A= p m . L. A.n 0.75 × 0.125 × 7855 × 1000 = = 12 270 W 60 60 = 12.27 kW IP = ∴ Brake power, BP = IP × ηm = 12.27 × 0.8 = 9.8 kW ...(∴ ηm = BP / IP) We know that the heat flowing through the piston head, H = C × HCV × m × BP = 0.05 × 42 × 103 × 41.7 × 10–6 × 9.8 = 0.86 kW = 860 W ....(Taking C = 0.05) ∴Thickness of the piston head on the basis of heat dissipation, 860 H tH = 12.56 k (T − T ) = 12.56 × 46.6 × 220 = 0.0067 m = 6.7 mm C E ...(∵ For cast iron , k = 46.6 W/m/°C, and TC – TE = 220°C) Taking the larger of the two values, we shall adopt tH = 16 mm Ans. Since the ratio of L / D is 1.25, therefore a cup in the top of the piston head with a radius equal to 0.7 D (i.e. 70 mm) is provided. 2. Radial ribs The radial ribs may be four in number. The thickness of the ribs varies from tH / 3 to tH / 2. ∴ Thickness of the ribs, tR...
View Full Document
## This note was uploaded on 09/30/2012 for the course MECHANICAL 403 taught by Professor A.thirumoorthy during the Spring '11 term at Anna University Chennai - Regional Office, Coimbatore.
Ask a homework question - tutors are online | 526 | 1,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-22 | longest | en | 0.888316 |
https://physicslens.com/category/a-level-topics/15-dc-circuits/ | 1,606,160,299,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00020.warc.gz | 439,099,187 | 19,106 | # 15 DC Circuits
## Geogebra App on Maximum Power Theorem
This simulation demonstrates the power dissipated in a variable resistor given that the battery has an internal resistance (made variable in this app as well).
Since the power dissipated by the resistor is given by
$P=I^2R$
and the current is given by
$I=E(R+r)$,
$P=E^2\times\dfrac{R}{(R+r)^2}=\dfrac{E^2}{\dfrac{r^2}{R}+R+2r}$
This power will be a maximum if the expression for the denominator $\dfrac{r^2}{R}+R+2r$ is a minimum.
Differentiating the expression with respect to R, we get
$\dfrac{d(\dfrac{r^2}{R}+R+2r)}{dR}=-\dfrac{r^2}{R^2}+1$
When the denominator is a minimum,
$-\dfrac{r^2}{R^2}+1=0$
Therefore,
$r=R$ when the power dissipated by the resistor is highest.
## Geogebra Simulation of a Potentiometer
Some of the more challenging problems in the topic of electricity in the A-level syllabus are those involving a potentiometer. The solution involves the concept of potential divider and the setup can be used to measure emf or potential difference across a variety of circuits components. Basically, students need to understand the rule – that the potential difference across a device is simply a fraction of the circuit’s emf, and that fraction is equal to the resistance of the device over the total resistance of the circuit.
$V_{device}=\frac{R_{device}}{R_{total}}*emf$
The intention of this Geogebra app is for students to practise working on their calculations, as well as to reinforce their understanding of the principle by which the potentiometer works.
## DeLight Version 2
I modified “DeLight”, the board game that I designed a few years back into a worksheet version (for small groups) as well as a powerpoint version (that teacher can facilitate as a class activity, pitting half the class against another).
The objectives of the game is to reinforce concepts related to D.C. Circuits such as:
1. Sum of potential difference (p.d.) across parallel branches of a circuit is the same.
$$E = V_1 + V_2 + V_3 +…$$
2. P.d. across a device is given by the ratio of resistance of device to total resistance multiplied by emf (potential divider rule)
$$V_1 = \dfrac{R_1}{R_{total}}\times E$$
3. Brightness of light bulb depends on electrical power
$$P = IV = \dfrac{V^2}{R} = I^2R$$
4. Current can bypass a device via a short-circuiting wire.
The worksheet and powerpoint slides contain a few examples that allow discussion on the above concepts based on some possible gameplay outcomes. For example, the following is a game where the blue team wins because the p.d. across each blue light bulb is twice that of the p.d. across each red light bulb.
In the following scenario, the game ended in a draw. Students may not be able to see it immediately, but the blue light bulb with a vertical orientation is actually short-circuited by the vertical branch on its right.
Feel free to use and/or modify the game to suit your own class needs.
## Concrete to Abstract
As I was teaching the use of the potential divider equation to my IP4 (Grade 10) students last week, I approached it by teaching the rules first before showing worked examples. Thereafter, as some students remained confused, I merely reverted to explaining the rules. Eventually, I resorted to simplifying the equation by substituting simple numerical values in place of the multiple symbols that we use for emf, terminal p.d., resistance (more than one of them) and current, and many students’ eyes lit up. It struck me then that I should have used the Concrete-Representational-Abstract approach in the first place.
Concrete-Representational-Abstract in Singapore Maths
Potential Divider Worksheet
Incidentally, I was reminded of this approach while helping my daughter with her Primary 2 Math homework last week. Since she was absent due to a stomach flu for 3 days this week, she had to bring work home to complete. There was an exercise on subtraction of numbers by separating into hundreds and tens and the first question looked like this:
There I was stuck at a primary school problem, not because I did not know that final answer is 157 but because I was not sure what to fill in for the circles. It did not help that my daughter was not clear about what to do either.
I then looked at her textbook for examples and what I saw was pictures of blocks in groups of 100s and 10s that look like this:
These are the tools that students in her class would have played with in the concrete or representation stage of learning. They can physically manipulate the blocks in order to do subtraction, which can be fun for those who like learning with a hands-on activity.
The abstract stage required by the question is for students to
1. remove 100 from the first number 207, leaving 107 (so the two circles will have numbers 100 and 107 in them)
2. deduct the second number from 100
3. and add the difference back to 107.
This is a technique that allows students to make quick mental calculations when subtracting tens from numbers more than 100 but requires a fair bit of practice to get used to. My daughter eventually had no problem with the rest of her homework after we figured this out.
Being a Physics teacher, I thought about how I could transfer this method of making a complex and abstract skill easier to pick up through concrete representations.
Usual Approach: Abstract to Concrete
In my IP4 Physics classes, we are currently on the topic of DC circuits, which involves calculations involving the potential divider method.
One of the main equations is the potential divider rule, in which V1 (p.d. across resistor R1) is given by
$V_1=\dfrac{R_1}{R_1+R_2}\times\epsilon$
For two classes, I started directly with equations before introducing examples (abstract to concrete), which is what physics teachers usually do. There is nothing wrong with this approach as students still see a concrete representation eventually but I was wondering if it might be more beneficial if students could synthesize the equation on their own by generalizing from examples. By the time I taught this same concept to the second class, I realized that most students were able to understand the equation only after giving concrete examples using numerical values.
Concrete Examples before Equations
For my third class, after being inspired by my daughter’s homework, instead of introducing the equations straightaway, I started by introducing the concept of dividing emf between two resistors in series according to ratio of resistance. I used simple numbers for easier reference. In this way, I felt that students could then always link the equation back to the simple concept of ratios which helps them understand the intent of the equation.
When students see after one or two examples that all we are doing is trying to distribute the emf in proportion to the resistance that each resistor has so that it follows the ratio R1:R2 (concrete stage), I asked them to replace the numerical values with symbols (representation stage).
$\dfrac{1\Omega}{2\Omega}=\dfrac{4 V}{8V}$ or $\dfrac{1\Omega}{(1+2)\Omega}=\dfrac{4 V}{(4+8)V}$
(Concrete stage)
$\dfrac{R_1}{R_2}=\dfrac{V_1}{V_2}$ or $\dfrac{R_1}{R_1+R_2}=\dfrac{V_1}{V_1+V_2}$
(Representation stage)
From the last equation, students were asked to recall that emf is equal to the sum of the p.d. in the series circuit. The equation can then be rewritten as
$\dfrac{R_1}{R_{total}}=\dfrac{V_1}{\epsilon}$ or $V_1=\dfrac{R_1}{R_{total}}\times\epsilon$
(Abstract stage)
I believe that if students can arrive that the final equation from the basic principles, they will understand more deeply and will be able to apply it in slightly different contexts, such as when there are more than 2 resistors in series, or when one resistor is varying (in the case of transducers such as LDRs and thermistors) or even when considering internal resistance of the cell.
Here’s a simple worksheet that students can use to work out the potential divider equation on their own.
P.S. This is not a research study that compares whether abstract to concrete is better than the other way round, but merely a way for me to record what I’ve tried in different classes. I welcome comments from other teachers who may have tried a similar approach and have observed positive results.
## Internal Resistance and Maximum Power Theorem
I’ve created this simulation to demonstrate the effect of an internal resistance due to a cell on the potential difference and current of an external load.
One can also vary the internal resistance and external resistance to observe the maximum power theorem. The theorem states that for a given finite internal resistance, one can obtain the maximum external power only when the resistance of the load is equal to the internal resistance of the source.
## Variable Resistor Simulation
My third simulation in as many days. This time round, I created some png files for electrical circuit components that can be used for future simulations. I also figured out how to add text in line with a variable in the drawing panel for the simulation, hence the units can now follow the values of the meter readings. | 2,106 | 9,146 | {"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} | 4.40625 | 4 | CC-MAIN-2020-50 | latest | en | 0.910003 |
http://www.jiskha.com/display.cgi?id=1284031759 | 1,495,860,796,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608773.42/warc/CC-MAIN-20170527040600-20170527060600-00290.warc.gz | 677,709,932 | 3,766 | posted by on .
writing in math mrs. bradner has 30 tomato plants. she wants to plant the same number of plants in each row of her gardner. explain how she could decide the number of rows to plants.
• 3rd grade - ,
Think about some sets that you can make out of 30.
3 sets of 10
10 sets of 3
5 sets of 6
6 sets of 5
Any others?
• 3rd grade - ,
24
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question | 130 | 484 | {"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 | 0.939729 |
https://gmatclub.com/forum/ps-6909.html?sort_by_oldest=true | 1,495,962,165,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609610.87/warc/CC-MAIN-20170528082102-20170528102102-00404.warc.gz | 972,108,121 | 43,303 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 28 May 2017, 02:02
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# PS 5
post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
SVP
Joined: 16 Oct 2003
Posts: 1805
Followers: 5
Kudos [?]: 154 [0], given: 0
PS 5 [#permalink]
### Show Tags
01 Jun 2004, 18:43
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Answer with Solutions.....
Attachments
p5.JPG [ 44.83 KiB | Viewed 672 times ]
SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA
Followers: 6
Kudos [?]: 101 [0], given: 0
### Show Tags
01 Jun 2004, 19:23
Let the salary be X
0.6 * 2000 + 0.4(X-2000) = 2500
X-2000 = 1300/0.4 = 3250
X = 5250
01 Jun 2004, 19:23
Display posts from previous: Sort by
# PS 5
post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 518 | 1,847 | {"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 | longest | en | 0.86181 |
https://www.markedbyteachers.com/international-baccalaureate/physics/factors-affecting-bounciness-of-a-ping-pong-ball.html | 1,670,444,692,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711218.21/warc/CC-MAIN-20221207185519-20221207215519-00537.warc.gz | 906,208,158 | 16,158 | • Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
# FACTORS AFFECTING BOUNCINESS OF A PING PONG BALL
Free essay example:
Aim: Factors Affecting Bounciness of a Ping Pong Ball
Hypothesis:
Background Information
The Law of Conservation of Energy states that energy cannot be created or destroyed, but can change its form. The law of reflection states that the angle of incidence is always equal to the angle of reflection
The ball, released from a certain point will gain energy and fall down at the speed of 9.81ms (the acceleration due to gravity), the ball will then hit the surface and will theoretically bounce back up. Some of the energy gained by the ball will be dispersed to the surface and the ball should bounce at a lower height than the original bounce. Now different surface have different state of hardness and texture. The harder the surface, the less energy will be lost by the ball and he higher it will bounce. Now, the smoother the surface the less energy will be dissipated through the surface (see laws of reflection), and the higher the ball will bounce.
The coefficient of restitution is a measure of the elasticity of the collision between ball and floor. Elasticity is a measure of how much bounce there is, or in other words, how much of the kinetic energy of the colliding objects before the collision remains as kinetic energy of the objects after the collision. This coefficient is different for each surface. The formulae is :
c = coefficient of restitution (dimensionless)
h = bounce height
H = drop height
Variables:
1. Independent Variable: rigidity and elasticity of the surface on which the ball bounces
2. Dependant Variables: the bounciness of the ping-pong ball
3. Controlled Variable:
• the elasticity of the ball
• the acceleration of gravity
• the force applied to it
• the height from which it is dropped
Apparatus:
• Video Camera
• Metre Rule
• Ping Pong Ball
• Thermometer & Thermocouple
• Clamp
• Retort Stand
• A sand bucket
• An A1 sheet of paper
• A plank of wood (thin plank)
• A bucket containing earth
• A rock or marble slab (very thin)
• A thin piece of plastic
Diagram:
Procedure:
• Simmer sand on the floor, smooth it evenly out.
• Place the clamp on the retort stand.
• Clamp the ball (this prevents us from applying any force on the ball
• Start the video camera, release the clamp
• Watch and count how many times the ball bounces on this particular surface and what is the height it bounces to.
• Repeat the experiment using the other different surface (for the earth experiment simmer some earth on the floor and smooth it evenly out)
The same ball for each experiment and the ball must be placed at the same height form the ground. A thermometer must be used to measure the room temperature, if any drastic change occurs, the experiment must be stopped and started again. Lastly, a thermocouple must be used to measure the temperature of the ping-pong ball, if the balls temperature changes while the experiment is conducted, the experiment must be restarted again.
Table:
Surface Used Height of Bounce 1st Try 2nd Try 3rd Try Average Height Sand Paper Wood Earth Rock/marble Plastic
http://www.cartage.org.lb/en/themes/Sciences/Chemistry/Generalchemistry/Energy/LawofConservation/LawofConservation.htm
http://www.racquetresearch.com/coeffici.htm
Yuvraj Dwarka Page
This student written piece of work is one of many that can be found in our International Baccalaureate Physics section.
Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Close
1. ## Determination of Coefficient of Friction
Word count: 2055
2. ## Factors affecting an electromagnet
Word count: 2000
3. ## In this extended essay, I will be investigating projectile motion via studying the movement ...
Word count: 4253
4. ## AIM: To determine the acceleration due to gravity on Earth using the drop ball ...
Word count: 1269
5. ## Physics-investigate the relationship of temperature and the height of the bounce of a squash ...
Word count: 1333
6. ## Investigation into the relationship between acceleration and the angle of free fall downhill
Word count: 2566
7. ## Investigate the Size of Craters in Sand Due to Dropped Object.
Word count: 3256
8. ## Investigate the factors affecting the period of a double string pendulum
Word count: 3437
# Related International Baccalaureate Physics essays
1. ## Investigate the factors affecting the period of a double string pendulum
Thus as the distance l increases then the period (time taken for one complete wave) decreases. My third and final graph also proves my hypothesis is correct as the trend line is positively upward sloping corresponding with its equation y=-0.001x2 + 0.082x +0.449.
2. ## Investigate the Size of Craters in Sand Due to Dropped Object.
r �d = (22/7) x (1.9) � x 0.6 = 6.81 cm � 6.00 0.70 = ? r �d = (22/7) x (1.9) � x 0.7 = 7.94 cm � 8.00 0.90 = ? r �d = (22/7) x (1.9)
1. ## Physics-investigate the relationship of temperature and the height of the bounce of a squash ...
Type of Ball Only one type of squash ball will be used, and all the squash b***s will be brand new because every bounce affects the next. Method: First, bouncing it several times causing a difference in temperature warms the squash ball.
2. ## In this extended essay, I will be investigating projectile motion via studying the movement ...
Release the movable wooden block to launch the metal ball. 11. When the metal ball lands, mark the point of impact. It should be a black spot. 12. Measure the distance between the black spot and the starting point using a zigzag ruler.
1. ## Factors affecting an electromagnet
The main question to be answered upon completion of the lab by the students is: "what is the relationship between the current running through the wire and the number of turns that the insulated wire has made on the nail in the electromagnet, and the number of paper clips that the electromagnet can hold.
2. ## Determination of Coefficient of Friction
I provide all this information in the table above. Data processing: Now I will calculate the mean of height just before starting the block to move. (13.9+13.4+14.0)/3 = 13.8 cm (11.2+12.6+11.9)/3 = 11.9 cm (10.5+10.0+10.9)/3 = 10.5 cm (10.5+9.9+9.9)/3 = 10.1 cm (9.2+8.9+8.9)/3 = 9.0 cm Now I need to find the angle between the wooden plane and the surface.
1. ## Investigation into the relationship between acceleration and the angle of free fall downhill
Time management wasn?t a problem either, as the trials all consisted in approximately 1 second. The part of the method that took most time to do was changing distance travelled.
2. ## AIM: To determine the acceleration due to gravity on Earth using the drop ball ...
Therefore, Random error. OR 2.05% Therefore, the acceleration due to gravity on Earth, as obtained from this experiment = 9.74±0.05 m/s2. CONCLUSION AND EVALUATION: As it can be seen from the total uncertainty calculation, the experimentally deduced value for the acceleration due to gravity on Earth is accurate to a large extent.
• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work | 1,806 | 7,401 | {"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-2022-49 | latest | en | 0.899628 |
https://db0nus869y26v.cloudfront.net/en/Antiderivative | 1,720,926,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00095.warc.gz | 178,140,622 | 31,135 | In calculus, an antiderivative, inverse derivative, primitive function, primitive integral or indefinite integral[Note 1] of a function f is a differentiable function F whose derivative is equal to the original function f. This can be stated symbolically as F' = f.[1][2] The process of solving for antiderivatives is called antidifferentiation (or indefinite integration), and its opposite operation is called differentiation, which is the process of finding a derivative. Antiderivatives are often denoted by capital Roman letters such as F and G.
Antiderivatives are related to definite integrals through the second fundamental theorem of calculus: the definite integral of a function over a closed interval where the function is Riemann integrable is equal to the difference between the values of an antiderivative evaluated at the endpoints of the interval.
In physics, antiderivatives arise in the context of rectilinear motion (e.g., in explaining the relationship between position, velocity and acceleration).[3] The discrete equivalent of the notion of antiderivative is antidifference.
## Examples
The function ${\displaystyle F(x)={\tfrac {x^{3)){3))}$ is an antiderivative of ${\displaystyle f(x)=x^{2))$, since the derivative of ${\displaystyle {\tfrac {x^{3)){3))}$ is ${\displaystyle x^{2))$. Since the derivative of a constant is zero, ${\displaystyle x^{2))$ will have an infinite number of antiderivatives, such as ${\displaystyle {\tfrac {x^{3)){3)),{\tfrac {x^{3)){3))+1,{\tfrac {x^{3)){3))-2}$, etc. Thus, all the antiderivatives of ${\displaystyle x^{2))$ can be obtained by changing the value of c in ${\displaystyle F(x)={\tfrac {x^{3)){3))+c}$, where c is an arbitrary constant known as the constant of integration. Essentially, the graphs of antiderivatives of a given function are vertical translations of each other, with each graph's vertical location depending upon the value c.
More generally, the power function ${\displaystyle f(x)=x^{n))$ has antiderivative ${\displaystyle F(x)={\tfrac {x^{n+1)){n+1))+c}$ if n ≠ −1, and ${\displaystyle F(x)=\ln |x|+c}$ if n = −1.
In physics, the integration of acceleration yields velocity plus a constant. The constant is the initial velocity term that would be lost upon taking the derivative of velocity, because the derivative of a constant term is zero. This same pattern applies to further integrations and derivatives of motion (position, velocity, acceleration, and so on).[3] Thus, integration produces the relations of acceleration, velocity and displacement: {\displaystyle {\begin{aligned}\int a\,\mathrm {d} t&=v+C\\\int v\,\mathrm {d} t&=s+C\end{aligned))}
## Uses and properties
Antiderivatives can be used to compute definite integrals, using the fundamental theorem of calculus: if F is an antiderivative of the continuous function f over the interval ${\displaystyle [a,b]}$, then: ${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x=F(b)-F(a).}$
Because of this, each of the infinitely many antiderivatives of a given function f may be called the "indefinite integral" of f and written using the integral symbol with no bounds: ${\displaystyle \int f(x)\,\mathrm {d} x.}$
If F is an antiderivative of f, and the function f is defined on some interval, then every other antiderivative G of f differs from F by a constant: there exists a number c such that ${\displaystyle G(x)=F(x)+c}$ for all x. c is called the constant of integration. If the domain of F is a disjoint union of two or more (open) intervals, then a different constant of integration may be chosen for each of the intervals. For instance ${\displaystyle F(x)={\begin{cases}-{\dfrac {1}{x))+c_{1}&x<0\\[1ex]-{\dfrac {1}{x))+c_{2}&x>0\end{cases))}$
is the most general antiderivative of ${\displaystyle f(x)=1/x^{2))$ on its natural domain ${\displaystyle (-\infty ,0)\cup (0,\infty ).}$
Every continuous function f has an antiderivative, and one antiderivative F is given by the definite integral of f with variable upper boundary: ${\displaystyle F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t~,}$ for any a in the domain of f. Varying the lower boundary produces other antiderivatives, but not necessarily all possible antiderivatives. This is another formulation of the fundamental theorem of calculus.
There are many functions whose antiderivatives, even though they exist, cannot be expressed in terms of elementary functions (like polynomials, exponential functions, logarithms, trigonometric functions, inverse trigonometric functions and their combinations). Examples of these are
• the error function ${\displaystyle \int e^{-x^{2))\,\mathrm {d} x,}$
• the Fresnel function ${\displaystyle \int \sin x^{2}\,\mathrm {d} x,}$
• the sine integral ${\displaystyle \int {\frac {\sin x}{x))\,\mathrm {d} x,}$
• the logarithmic integral function ${\displaystyle \int {\frac {1}{\log x))\,\mathrm {d} x,}$ and
• sophomore's dream ${\displaystyle \int x^{x}\,\mathrm {d} x.}$
## Techniques of integration
Finding antiderivatives of elementary functions is often considerably harder than finding their derivatives (indeed, there is no pre-defined method for computing indefinite integrals).[4] For some elementary functions, it is impossible to find an antiderivative in terms of other elementary functions. To learn more, see elementary functions and nonelementary integral.
There exist many properties and techniques for finding antiderivatives. These include, among others:
Computer algebra systems can be used to automate some or all of the work involved in the symbolic techniques above, which is particularly useful when the algebraic manipulations involved are very complex or lengthy. Integrals which have already been derived can be looked up in a table of integrals.
## Of non-continuous functions
Non-continuous functions can have antiderivatives. While there are still open questions in this area, it is known that:
• Some highly pathological functions with large sets of discontinuities may nevertheless have antiderivatives.
• In some cases, the antiderivatives of such pathological functions may be found by Riemann integration, while in other cases these functions are not Riemann integrable.
Assuming that the domains of the functions are open intervals:
• A necessary, but not sufficient, condition for a function f to have an antiderivative is that f have the intermediate value property. That is, if [a, b] is a subinterval of the domain of f and y is any real number between f(a) and f(b), then there exists a c between a and b such that f(c) = y. This is a consequence of Darboux's theorem.
• The set of discontinuities of f must be a meagre set. This set must also be an F-sigma set (since the set of discontinuities of any function must be of this type). Moreover, for any meagre F-sigma set, one can construct some function f having an antiderivative, which has the given set as its set of discontinuities.
• If f has an antiderivative, is bounded on closed finite subintervals of the domain and has a set of discontinuities of Lebesgue measure 0, then an antiderivative may be found by integration in the sense of Lebesgue. In fact, using more powerful integrals like the Henstock–Kurzweil integral, every function for which an antiderivative exists is integrable, and its general integral coincides with its antiderivative.
• If f has an antiderivative F on a closed interval ${\displaystyle [a,b]}$, then for any choice of partition ${\displaystyle a=x_{0} if one chooses sample points ${\displaystyle x_{i}^{*}\in [x_{i-1},x_{i}]}$ as specified by the mean value theorem, then the corresponding Riemann sum telescopes to the value ${\displaystyle F(b)-F(a)}$. {\displaystyle {\begin{aligned}\sum _{i=1}^{n}f(x_{i}^{*})(x_{i}-x_{i-1})&=\sum _{i=1}^{n}[F(x_{i})-F(x_{i-1})]\\&=F(x_{n})-F(x_{0})=F(b)-F(a)\end{aligned))} However if f is unbounded, or if f is bounded but the set of discontinuities of f has positive Lebesgue measure, a different choice of sample points ${\displaystyle x_{i}^{*))$ may give a significantly different value for the Riemann sum, no matter how fine the partition. See Example 4 below.
### Some examples
1. The function
${\displaystyle f(x)=2x\sin \left({\frac {1}{x))\right)-\cos \left({\frac {1}{x))\right)}$ with ${\displaystyle f(0)=0}$ is not continuous at ${\displaystyle x=0}$ but has the antiderivative ${\displaystyle F(x)=x^{2}\sin \left({\frac {1}{x))\right)}$
with ${\displaystyle F(0)=0}$. Since f is bounded on closed finite intervals and is only discontinuous at 0, the antiderivative F may be obtained by integration: ${\displaystyle F(x)=\int _{0}^{x}f(t)\,\mathrm {d} t}$.
2. The function ${\displaystyle f(x)=2x\sin \left({\frac {1}{x^{2))}\right)-{\frac {2}{x))\cos \left({\frac {1}{x^{2))}\right)}$ with ${\displaystyle f(0)=0}$ is not continuous at ${\displaystyle x=0}$ but has the antiderivative ${\displaystyle F(x)=x^{2}\sin \left({\frac {1}{x^{2))}\right)}$ with ${\displaystyle F(0)=0}$. Unlike Example 1, f(x) is unbounded in any interval containing 0, so the Riemann integral is undefined.
3. If f(x) is the function in Example 1 and F is its antiderivative, and ${\displaystyle \{x_{n}\}_{n\geq 1))$ is a dense countable subset of the open interval ${\displaystyle (-1,1),}$ then the function ${\displaystyle g(x)=\sum _{n=1}^{\infty }{\frac {f(x-x_{n})}{2^{n))))$ has an antiderivative ${\displaystyle G(x)=\sum _{n=1}^{\infty }{\frac {F(x-x_{n})}{2^{n))}.}$ The set of discontinuities of g is precisely the set ${\displaystyle \{x_{n}\}_{n\geq 1))$. Since g is bounded on closed finite intervals and the set of discontinuities has measure 0, the antiderivative G may be found by integration.
4. Let ${\displaystyle \{x_{n}\}_{n\geq 1))$ be a dense countable subset of the open interval ${\displaystyle (-1,1).}$ Consider the everywhere continuous strictly increasing function ${\displaystyle F(x)=\sum _{n=1}^{\infty }{\frac {1}{2^{n))}(x-x_{n})^{1/3}.}$ It can be shown that ${\displaystyle F'(x)=\sum _{n=1}^{\infty }{\frac {1}{3\cdot 2^{n))}(x-x_{n})^{-2/3))$
for all values x where the series converges, and that the graph of F(x) has vertical tangent lines at all other values of x. In particular the graph has vertical tangent lines at all points in the set ${\displaystyle \{x_{n}\}_{n\geq 1))$.
Moreover ${\displaystyle F(x)\geq 0}$ for all x where the derivative is defined. It follows that the inverse function ${\displaystyle G=F^{-1))$ is differentiable everywhere and that ${\displaystyle g(x)=G'(x)=0}$
for all x in the set ${\displaystyle \{F(x_{n})\}_{n\geq 1))$ which is dense in the interval ${\displaystyle [F(-1),F(1)].}$ Thus g has an antiderivative G. On the other hand, it can not be true that ${\displaystyle \int _{F(-1)}^{F(1)}g(x)\,\mathrm {d} x=GF(1)-GF(-1)=2,}$
since for any partition of ${\displaystyle [F(-1),F(1)]}$, one can choose sample points for the Riemann sum from the set ${\displaystyle \{F(x_{n})\}_{n\geq 1))$, giving a value of 0 for the sum. It follows that g has a set of discontinuities of positive Lebesgue measure. Figure 1 on the right shows an approximation to the graph of g(x) where ${\displaystyle \{x_{n}=\cos(n)\}_{n\geq 1))$ and the series is truncated to 8 terms. Figure 2 shows the graph of an approximation to the antiderivative G(x), also truncated to 8 terms. On the other hand if the Riemann integral is replaced by the Lebesgue integral, then Fatou's lemma or the dominated convergence theorem shows that g does satisfy the fundamental theorem of calculus in that context.
5. In Examples 3 and 4, the sets of discontinuities of the functions g are dense only in a finite open interval ${\displaystyle (a,b).}$ However, these examples can be easily modified so as to have sets of discontinuities which are dense on the entire real line ${\displaystyle (-\infty ,\infty )}$. Let ${\displaystyle \lambda (x)={\frac {a+b}{2))+{\frac {b-a}{\pi ))\tan ^{-1}x.}$ Then ${\displaystyle g(\lambda (x))\lambda '(x)}$ has a dense set of discontinuities on ${\displaystyle (-\infty ,\infty )}$ and has antiderivative ${\displaystyle G\cdot \lambda .}$
6. Using a similar method as in Example 5, one can modify g in Example 4 so as to vanish at all rational numbers. If one uses a naive version of the Riemann integral defined as the limit of left-hand or right-hand Riemann sums over regular partitions, one will obtain that the integral of such a function g over an interval ${\displaystyle [a,b]}$ is 0 whenever a and b are both rational, instead of ${\displaystyle G(b)-G(a)}$. Thus the fundamental theorem of calculus will fail spectacularly.
7. A function which has an antiderivative may still fail to be Riemann integrable. The derivative of Volterra's function is an example.
## Basic formulae
• If ${\displaystyle {\mathrm {d} \over \mathrm {d} x}f(x)=g(x)}$, then ${\displaystyle \int g(x)\mathrm {d} x=f(x)+C}$.
• ${\displaystyle \int 1\ \mathrm {d} x=x+C}$
• ${\displaystyle \int a\ \mathrm {d} x=ax+C}$
• ${\displaystyle \int x^{n}\mathrm {d} x={\frac {x^{n+1)){n+1))+C;\ n\neq -1}$
• ${\displaystyle \int \sin {x}\ \mathrm {d} x=-\cos {x}+C}$
• ${\displaystyle \int \cos {x}\ \mathrm {d} x=\sin {x}+C}$
• ${\displaystyle \int \sec ^{2}{x}\ \mathrm {d} x=\tan {x}+C}$
• ${\displaystyle \int \csc ^{2}{x}\ \mathrm {d} x=-\cot {x}+C}$
• ${\displaystyle \int \sec {x}\tan {x}\ \mathrm {d} x=\sec {x}+C}$
• ${\displaystyle \int \csc {x}\cot {x}\ \mathrm {d} x=-\csc {x}+C}$
• ${\displaystyle \int {\frac {1}{x))\ \mathrm {d} x=\ln |x|+C}$
• ${\displaystyle \int \mathrm {e} ^{x}\mathrm {d} x=\mathrm {e} ^{x}+C}$
• ${\displaystyle \int a^{x}\mathrm {d} x={\frac {a^{x)){\ln a))+C;\ a>0,\ a\neq 1}$
• ${\displaystyle \int {\frac {1}{\sqrt {a^{2}-x^{2))))\ \mathrm {d} x=\arcsin \left({\frac {x}{a))\right)+C}$
• ${\displaystyle \int {\frac {1}{a^{2}+x^{2))}\ \mathrm {d} x={\frac {1}{a))\arctan \left({\frac {x}{a))\right)+C}$ | 3,975 | 13,846 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 88, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-30 | latest | en | 0.889607 |
https://math.stackexchange.com/questions/4360918/short-exact-sequence-of-abelian-groups-how-to-cancel-mathbb-z-in-the-middle-t | 1,718,636,945,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00226.warc.gz | 347,767,800 | 36,042 | # Short exact sequence of abelian groups how to cancel $\mathbb Z$ in the middle term.
Now I want to calculate homology group of some topological space.
Using Mayer-Vietoris sequence I end up with the following short exact sequence:
$$k$$ :a knot
$$0\to H_0(Torus)\to H_0(\mathbb R^3\setminus k)\oplus H_0(S^1)\to H_0(\mathbb R^3)\to 0$$
Following maps are irrelevant to my question but for completeness:
First map is $$(j_U,-j_V):\mathbb Z\to H_0(\mathbb R^3\setminus k) \oplus \mathbb Z$$
Second map is $$g_U\oplus g_V: H_0(\mathbb R^3\setminus k) \oplus \mathbb Z \to \mathbb Z$$
Where $$j_U,j_V,g_U,g_V$$ are canonical inclusion maps.
Since in above short exact sequence $$H_0(Torus)=H_0(S^1)=\mathbb Z$$ free abelian groups then $$H_0(\mathbb R^3\setminus k) \oplus \mathbb Z=\mathbb Z\oplus \mathbb Z$$
I know that $$H_0(\mathbb R^3\setminus k)$$ is $$\mathbb Z$$ since it is path connected but Iam asking this question to understand the real algebraic reason:
How can we cancel $$\mathbb Z$$ and say $$H_0(\mathbb R^3\setminus k)=\mathbb Z$$
• Should I use explicitly the canonical maps
• Is there any usefull theorem that if I encounter with some sequence like that and I want to cancel some group like this, it helps me.
Finitely generated abelian groups are cancellable among abelian groups, i.e. if $$A$$ is a finitely generated abelian group and $$B,C$$ are abelian groups such that $$A\oplus B\cong A\oplus C$$, then $$B\cong C$$. This is proven e.g. in this paper. However, I would advise against relying on this fact, as there's much concrete and canonical reasons for this conclusion to hold in the present context. Here are three options:
1. Use the Mayer-Vietoris sequence for reduced instead of unreduced homology. This immediately yields $$\tilde{H}_0(\mathbb{R}^3\setminus k)=0$$, whence $$H_0(\mathbb{R}^3/k)=\mathbb{Z}$$ by the general relationship between reduced and unreduced homology.
2. Recall that if $$X$$ is any topological space and $$\pi_0(X)$$ denotes the set of path-components of $$X$$, there is a canonical isomorphism $$H_0(X)\cong\mathbb{Z}^{(\pi_0(X))}$$ (the latter meaning the free abelian group with generators $$\pi_0(X)$$). Using these identifications, your exact sequence becomes $$0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}^{(\pi_0(\mathbb{R}^3\setminus k))}\oplus\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow0.$$ The latter map is addition, since the inclusion takes any path-component $$\mathbb{R}^3\setminus k$$ or $$S^1$$ to the unique path-component of $$\mathbb{R}^3$$. Thus, its kernel consists of those elements whose entries add up to $$0$$. By exactness, this kernel is the image of the map $$\mathbb{Z}\rightarrow\mathbb{Z}^{(\pi_0(\mathbb{R}^3\setminus k))}\oplus\mathbb{Z}$$, but the image of this map meets only one summand of $$\mathbb{Z}^{(\pi_0(\mathbb{R}^3\setminus k))}$$, because the torus is path-connected. Together, these conditions force that $$|\pi_0(\mathbb{R}\setminus k)|=1$$, so $$H_0(\mathbb{R}^3/k)=\mathbb{Z}$$ is free on the unique path-component of $$\mathbb{R}^3\setminus k$$.
3. Note that you can split your short exact sequence by the map $$H_0(\mathbb{R}^3)\rightarrow0\oplus H_0(S^1)\subseteq H_0(\mathbb{R}^3\setminus k)\oplus H_0(S^1)$$, mapping the canonical generator of $$H_0(\mathbb{R}^3)$$ to the canonical generator of $$H_0(S^1)$$. This induces an isomorphism $$H_0(T)\oplus H_0(\mathbb{R}^3)\cong H_0(\mathbb{R}^3\setminus k)\oplus H_0(S^1)$$ with the additional property that $$0\oplus H_0(\mathbb{R}^3)$$ corresponds to $$0\oplus H_0(S^1)$$. This trivializes the cancellation problem (in contrast to the general situation addressed at the start), as we can calculate $$H_0(\mathbb{R}^3\setminus k)\cong\frac{H_0(\mathbb{R}^3\setminus k)\oplus H_0(S^1)}{H_0(S^1)}\cong\frac{H_0(T)\oplus H_0(\mathbb{R}^3)}{H_0(\mathbb{R}^3)}\cong H_0(T)\cong\mathbb{Z}.$$ Here, the middle isomorphism is induced by the splitting we constructed. | 1,312 | 3,935 | {"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": 39, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.84164 |
https://byjusexamprep.com/quantitative-aptitude-questions-for-ssc-cgl-i | 1,674,804,765,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00822.warc.gz | 174,364,667 | 85,427 | # Quantitative Aptitude for SSC CGL 2023: Questions, Preparation Tips, Important Topics
By PARUL RISHI|Updated : January 12th, 2023
SSC CGL Quantitative Aptitude Questions: SSC has recently conducted the SSC CGL Tier 1 exam from 1-13th December 2022 and now the commission is all set to conduct the SSC CGL Tier 2 examination. Going by the latest exam pattern, there will be 30 quant questions in SSC CGL tier 2 carrying a weightage of 90 marks.
Now the question arises what topics are majorly asked from Quant section in SSC CGL exam. How should we cover these important quantitative aptitude topics and what should be the strategy to cover the entire SSC CGL quant syllabus. If you are also in this dilemma, you are in for a treat, as here we have curated a list of important quant questions for SSC CGL. Go through the article to find out!
Quantitative Aptitude is a strenuous subject and can be difficult for any candidate. Because of its complex formulas, time-consuming questions and ‘not-so-easy’ to remember tricks, it has become difficult for the candidates to fetch good marks in this section. Hence, to help you out, our experts have carefully curated the list of SSC CGL Quant Questions covering must-read topics, most likely to be asked in the upcoming SSC CGL 2023 exam.
## SSC CGL Quantitative Aptitude
The SSC CGL Math section is one of the trickiest & difficult sections because the difficulty level of the questions is getting complex with every exam passed. But according to the recent trend, the questions are based on the topics like Profit, Loss & Discount, Data Interpretation, HCF & LCM, Simple and Compound Interest, Mixture and Alligations, Mensuration, Simplification, Trigonometry, and Algebra. Therefore, candidates are advised not to turn a blind eye to these topics and understand them thoroughly, along with practising the given below SSC CGL Quantitative Aptitude questions.
The SSC CGL syllabus' Quant component is a straightforward section. The Quant Section answers are impossible to guess. By resolving, individuals can determine whether they already know the solution or not. Quantitative aptitude might be a very high-scoring section in the SSC CGL 2023 exam if the SSC CGL preparation approach is effectively focused on it. Through this blog, we hope to assist you in creating the ideal SSC CGL Quant preparation strategy so that you can succeed in this section and get the highest possible marks.
## How to Prepare Quant for SSC CGL?
As per the revised SSC CGL exam pattern, the weightage of the section is decreased even though it is one of the most important subjects in the Tier 2 examination where there are going to be 30 questions of 3 marks each. Therefore, candidates need to understand the syllabus properly and follow the right strategy for the upcoming SSC CGL examination.
It takes a lot of time to study mathematics. We frequently failed to answer all of the questions in an exam that we might have easily answered at home. Therefore, we must complete these problems within the allotted time if we hope to receive a perfect score on the SSC CGL math exam.
• Understand exam pattern and syllabus properly.
• Prior to exams, practice mock tests and previous years' papers frequently.
• Strike balance between advanced and arithmetic maths.
• Improve your speed and accuracy.
## SSC CGL Quantitative Aptitude Important Topics
The below table provides all the topics asked in SSC CGL according to SSC CGL Syllabus for Maths. Next, We will also discuss the detailed strategy to cover these topics in the most efficient way.
### Important Quant topics for SSC CGL
• Ratio & Proportion
• Percentage
• HCF & LCM
• Divisibility & Reminder
• Simple Interest & Compound Interest
• Profit, loss & discount
• Average
• Number System
• Boat & Stream
• Time & Work
• Pipe & Cistern
• Time, Distance & Speed
• Problem on ages
• Mixture & Alligation
• Partnership
• Geometry
• Mensuration
• Trigonometry
• Coordinate Geometry
• Height & Distance
• Algebra
• Data Interpretation
## Preparation Strategy for SSC CGL Quantitative Aptitude
From the above table the syllabus of the quantitative aptitude section is vast. Proper planning and strategy are required to cover these topics and score maximum marks in SSC CGL exam. We are going to discuss the topic-wise strategy to prepare each topic.
1. Ratio & Proportion & Percentage: These two topics are very basic and most of the chapters of Quantitative Aptitude are based on these two. If you want to solve Quants questions in a short time then the master key is Ratio Proportion & Percentages. You will find the use of Ratio & Percentages in Data Interpretation, Average, Mixture Allegation, SI & CI, Profit Loss, Partnership etc. First, go through the below study notes then practice some quizzes. You can also check the practice section on the BYJU'S Exam Prep app and practice questions from there.
2. Number System, HCF & LCM, Divisibility & Reminder - Number System is a very long and very much conceptual chapter. You need a lot of practice to be a master in this chapter. HCF LCM and Divisibility Reminder are easy chapters which you can easily learn if you are good in the Number system.
Example: Find HCF (18, 42)?
Solution : Find the prime factors of 18 = 2 * 3 * 3 and 42 = 7 * 2 * 3 and notice the "common" of the two expressions is 2 * 3; So HCF (18, 42) = 6.
Example : LCM (18,42)
Solution : Factors of 18 (2 * 3 * 3) and 42 (7 * 2 * 3). LCM = (2*3*3*7)= 126 (HCF * other factors)
3. Simple Interest & Compound Interest, Partnership: You will get 4-5 questions from these two topics. These 3 chapters are based on Principle, Rate, Time, and Amount. The formula will change according to the different situation. First, go through the basics of these chapters and then solve some quizzes on BYJU'S Exam Prep. You can also refer tostudy notes for Simple Interest.
4. Profit loss & discount: This is also a very important chapter as you will find a lot of questions in SSC CGL. This chapter is purely based on percentages & ratio proportion. First, go through the fundamentals of these chapters. We are providing you notes to understand the basics and then solve some good questions.
5. Average: You will see good questions from the average chapter too. This chapter is based on Ratio & Proportion.
6. Time & Work, Pipes & cistern: 3-4 questions are asked from these two chapters. Method to solve questions from time & work, Pipes & Cistern is almost the same. Students find these chapters easy and scoring. Make sure you practice well. You can find below notes to prepare these two chapters for SSC CGL Exam.
7. Time, Distance & Speed (TDS), Boat & Stream: You will find 4-5 tricky questions based on TDS, Boat & Stream in your exam. Questions are generally based on trains, race on a circular track, boat & stream, etc. The simple formula to be used in these questions is Distance = Speed * Time. Go through the below links first and then solve questions from practice section on BYJU'S Exam Prep. You can also check out the tips to solve Boat & Stream Questions
8. Co-ordinate Geometry: This year Co-ordinate geometry questions were asked in SSC CGL tier 1 exam. It is expected that 2 questions can be asked from this chapter in SSC CGL Tier-II Exam. Go through the notes on BYJU'S Exam Prep and then practice some questions.
9. Geometry & Mensuration: Many questions are asked from geometry & mensuration in SSC CGL Exam. Questions are based on circle, quadrilateral, triangle, square, cube, cylinder, sphere, hemisphere, pyramid etc. You need a lot of practice as geometry contains many formulas and short tricks. You can find all the important notes & quizzes on BYJU'S Exam Prep.
11. Algebra: 5-7 questions are generally asked from Algebra in SSC CGL Exam. Most of the questions are based on X + 1/X format. Attempt quizzes and mock tests on BYJU'S Exam Prep. However, you must know that knowingshort tricks and formulas for Algebra is mandatory as it will help you solve the question in less time.
12. Trigonometry, Height & Distance: Some questions are asked from trigonometry and Height & Distance. Questions are not much difficult and options are really helpful to solve them easily.
Find here some important questions of trigonometry.
13. Data Interpretation: Mostly 4-5 questions are asked from Data Interpretation in SSC CGL Exam. DI questions are nothing but the application of percentages and ratio & proportions.
Note: Apart from the entire syllabus, Time & Speed are two most important factors. You should try to solve the questions in minimum time. We will provide more short tricks and methods to increase the speed in the exam.
## Quant Questions for SSC CGL
1.In an exam consisting of 200 questions, a student scored 500 marks after attempting all the questions. But he was awarded 4 marks for each correct answer & lose 1 mark for incorrect answer. Find the number of incorrect answers?
A. 120
B. 80
C. 50
D. 60
2.A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water (in litres) will fall into the sea in a minute?
A. 40,00,000
B. 4,00,000
C. 40,000
D. 400
3.A man rows at 5 km/h in still water. If the river is flowing at 1 km/h, it takes him 75 minutes to row to a place and back. How for is the place?
A. 2.5 km
B. 3 km
C. 4 km
D. 5 km
E. None of these
4.The sum of five consecutive odd numbers is equal to 125. What is the sum of the product of the largest and the smallest numbers and the second largest number?
A. 512
B. 535
C. 634
D. 636
5.A certain sum of money becomes 4 times in 2 years at SI per annum. In how many years can it become 25 times of itself?
A. 18
B. 13
C. 16
D. 19
6.If the average of 5 consecutive even numbers is 102, then what is the sum of the first and last numbers?
A. 204
B. 212
C. 200
D. 211
7.The sum of five consecutive even numbers is 400. What is the half of the sum of the smallest number and the largest number of the set?
A. 80
B. 81
C. 79
D. 82
8.The sum of two numbers is 56 and difference of their squares is 448. The difference between the numbers is:
A. 10
B. 12
C. 8
D. 22
9.A merchant has 1000 kg of sugar, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 400
B. 300
C. 600
D. 250
E. None of these
10.A person bought an article and sold it at a loss of 10%. If he had bought is for 20% less and sold it for Rs.55 more he would have had a profit of 40%. Find the C.P of the article.
A. Rs.250
B. Rs.225
C. Rs.222
D. Rs.230
11.In winters, the meter scale of a cloth merchant shrinks by 10% of its actual length. If the meter scale in its regular condition was rigged to measure 10% more than it is supposed to measure , what is the profit / loss percent the cloth merchant makes in winter, if he sell goods at the rate at which he purchase?
A. 1.0101%
B. 0.0101%
C. 1.1010%
D. 1.0011%
E. 2%
12.A merchant agrees to give a discount of 20% on cost price of articles but gives away only 700gms for every 1400gms weight of articles. Find his profit%?
A. 40
B. 55
C. 65
D. 60
13.0.1 × 0.01 × 0.001 × 107 is equal to:
A. 100
B. 1/10
C. 1/100
D. 10
14.
A. 2
B. 3
C. 4
D. 1
15.In the diagram alongside, AD = 2, AB = d. What is the area of the shaded portion?
A. 2d2
B.
C.
D. d
16. is equal to :
A.
B.
C.
D.
17.A train takes 2 hours more to cover the distance of 384 km when its speed is reduced by 16km/hr. Find the 75% of the original speed of train.
A. 48
B. 36
C. 64
D. 54
18.Find the value of .
A.
B.
C.
D.
19.Area of the triangle formed by the graph of the line 2x – 3y + 6= 0 along with the coordinate axes is how much?
A. 1/2 sq. units
B. 3/2 sq. units
C. 3 sq. units
D. 6 sq. units
20.₹ 3,000 is divided between A, B and C such that A receives 1/3 of the amount received by B and C together and B receives 2/3 of the amount received by A and C together. Then the share of C is
A. ₹ 600
B. ₹ 525
C. ₹ 1625
D. ₹ 1050
21.In a department, 24 employees know typing and 11 know stenography, 25 know to use a computer. 7 know both typing and stenography, 4 know stenography and computers, 12 know typing and computers and 3 know all the three. If there were 50 employees in the department, find how many employees don’t know none of the three jobs.
A. 40
B. 10
C. 47
D. 33
22.If m * n = m2 – mn – 2, then (–4 * –5) = ?
A. –6
B. -2
C. 6
D. 7
23. What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7?
A. 37
B. 36
C. 39
D. 30
24.The number obtained by interchanging the digits of a two digit number is less than the original number by 18. If sum of the digits is 6, what was the original two-digit number?
A. 51
B. 24
C. 42
D. 15
E. None of these
25.O is the centre of the circle and ∠DAB = 50°. Calculate the value of x and y.
A. 100°, 130°
B. 130°, 100°
C. 80°, 120°
D. 120°, 80°
26.In the given figure ∠BAC = 60° and ∠BCA = 20°, find ∠ADC?
A.
B.
C.
D.
27.
In the following figure , AC= BC and AD = CD then find .
A.
B.
C.
D.
28.The external bisector of b and c of the ΔABC meet at point p. if ∠BAC = 800 then ∠BPC is:
A. 50
B. 40
C. 80
D. 100
29.If in any triangle ABC , the base BC is produced in both ways, the sum of the exterior angles at B and C is :
A. π-A
B. π+A
C. (π/2)+A
D. π-
30.6 men and 8 women can do a work in 10 days, then 3 men and 4 women can do the same work in
A. 24 days
B. 20 days
C. 12 days
D. 18 days
31. Two numbers a and b are 15% and 30% of a number c. What percentage is the number b of the number a correct to the nearest integer?
A. 175%
B. 200%
C. 180%
D. 150%
32. The marked price of a shirt is ₹2000. The shirt is sold at a discount of 20%. If the profit percentage is 12%, then what is the cost price (in ₹) of the shirt?
A. 1,400.5
B. 1,125
C. 1,428.5
D. 1,200
33. On selling 19 balls at ₹960, there is a loss equal to the cost price of 7 balls. The cost price (in ₹) of a ball is:
A. 85
B. 80
C. 75
D. 70
34.30 men can do a piece of work in 60 days. After how many days from the start of the work, should 20 more men join then so that the work is done in 50 days?
A. 20
B. 35
C. 25
D. 15
35. A alone can do 2/5 of a work in 12 days. B is 25 percent more efficient than A. C alone can do the same work in 12 days less than B. D is 25 percent less efficient than C. If they all works together, then the work will be complete in how many days?
A. 240/53
B. 180/43
C. 200/51
D. 300/47
36. What is the product of two consecutive even numbers, the difference of whose squares is 76?
A. 500
B. 440
C. 400
D. 360
37. What is the sum of first 20 terms of the following series?
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +……..
A. 3160
B. 2940
C. 3240
D. 3080
38. x, y and z are distinct prime number where x < y < z. If x + y + z = 70, then what is the value of z?
A. 29
B. 43
C. 31
D. 37
## How to Prepare Quant for SSC CGL?
• Set a time for practising maths questions. It should be that part of the day where your mind is the most active.
• Devote at least 3-4 hours a day. One-fourth part of this duration should focus only on honing your calculation skills.
• Not all questions that are asked in the exam are difficult, some questions are very basic and scoring. Therefore, your preparation should be a good mix of easy and tough questions.
• To give your preparation an extra edge, practice the SSC CGL previous year question paper, sample paper, mock test, etc.
• Solve at least 4 practice sets (2 of each subject), SSC CGL practice set gives you speed and helps in handling the pressure during exam time.
• After checking the solution of every test, analyze the questions which are time-consuming or the one that seems difficult to solve.
• Try to solve questions with a short trick and always solve questions of that type with a fixed method or short trick so that you can solve that type of question within a second.
• If you encountered any new type of question in the practice set then jot down the question and find a simple way to solve that question.
• Note down the questions in which you use the long method or traditional approach to solve questions because these are questions that can affect your performance in the exam.
• Whenever you opt for the new test then identify the pattern in which you commit the mistakes. Make a diary and note down those mistakes and go through them before attempting any test.
• Analyze the improvement in your performance with the help of SSC CGL Test Series.
• During the analysis of the SSC CGL question paper, see whether you skip any questions which are easy and can be solved easily if so then try to refrain from committing such mistakes.
• Give proper time and attention while reading questions, in this way you can skip human error.
## Best Book for Quantitative Aptitude for SSC CGL
The SSC CGL Books that you can refer to for the preparation of SSC CGL Quantitative Aptitude are as follows:
• Magical Books On Quicker Maths by M-Tyra (BSC publication)
• SSC Mathematics 5800+ by Kiran Publication
• Online Quiz (thousands of questions to practice are available) by BYJU'S Exam Prep
Candidates must note that they might come across these questions with slight modifications. Hence, the concepts should be clear, and they must solve mock tests as many as they can. These mock tests are equipped with all the important Quant questions for SSC CGL that can enhance your overall score exponentially!
All The Best !!!
DREAM, STRIVE & SUCCEED With US!
Check OUT:
Prepare for SSC Exams 2022-23 with BYJU'S Exam Prep Course
SSC Success Batch: A comprehensive 10 Months Foundation Course | Available in English and Hindi
So why wait? Update your app right away!
write a comment
## FAQs on Quant Questions for SSC CGL
• Quant for SSC CGL is important as Maths is an integral part of the SSC CGL Examination and a total of 55 Questions were asked from this section in Tier 1 and Tier 2 combined. Therefore, due to this weightage candidates should prepare for this section completely.
• You can find the Quant Questions for SSC CGL in this article we have complied a list of the most important questions that are asked in the examination and candidates should go through these questions and prepare accordingly.
• Some important topics of Quant for SSC CGL are as follows:
• Ratio & Proportion
• Percentage
• HCF & LCM
• Divisibility & Reminder
• Simple Interest & Compound Interest
• Profit, loss & discount
• Average
• Number System
• Boat & Stream etc .
• Important Pointers for SSC CGL Tier 2 Math according to new exam pattern are as follows:
• Understand exam pattern and syllabus properly.
• Prior to exams, practice mock tests and previous years' papers frequently.
• Strike balance between advanced and arithmetic maths.
• Improve your speed and accuracy.
• Practising SSC CGL Quantitative Aptitude questions will help candidates to give their preparation a boost and step it up a level. It will not only help you become more accustomed to the question patterns, but it will also boost your confidence. Additionally, since SSC frequently repeats exam questions, answering these questions improves your chances of succeeding in this section.
• The SSC CGL Quantitative Aptitude questions fall under the easy to moderate difficulty category. SSC frequently concentrates on a small number of subjects, from which the majority of questions are chosen. As a result, it is essential to practice sufficient questions from each topic from the SSC CGL Quantitative Aptitude expected questions. For a better idea of what to expect in the exam, you can review the previous year exam papers and SSC CGL mock test.
• To assist you in preparing for the forthcoming SSC CGL 2023 exam, this post includes a compilation of SSC CGL Quantitative Aptitude questions. To find SSC CGL Previous year paper, you may also check the SSC website or browse Byju's Exam Prep platform. Remember that Quantitative Aptitude is a subject that requires continual practice and revision of important concepts and formulae; you cannot prepare for it in one day.
• Quant questions for SSC CGL can easily be solved by the application of the formula. If the question takes more time, candidates must use simple tricks and calculation shortcuts to save time.
• The maximum number of SSC CGL Quantitative Aptitude questions typically are from the following topics - Arithmetic, Algebra, Mensuration, Geometry and Data Interpretation. You can find a few examples of previously asked questions from the SSC CGL exam in the article above.
### SSC & Railway
CGLSSC GDDFCCILCHSLCPONTPCMTSStenoGroup DDelhi PoliceOthersCoursesMock Test | 5,434 | 20,664 | {"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-2023-06 | longest | en | 0.932568 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition-anton/chapter-2-the-derivative-2-6-the-chain-rule-exercises-set-2-6-page-158/55 | 1,534,590,899,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213540.33/warc/CC-MAIN-20180818095337-20180818115337-00508.warc.gz | 939,194,328 | 13,382 | ## Calculus, 10th Edition (Anton)
$$\frac{dy}{d\theta}=3\cot^2\theta\csc^2\theta$$
$$\frac{dy}{d\theta}=\frac{d}{d\theta}(\cot^3(\pi-\theta))=3\cot^2(\pi-\theta)\frac{d}{d\theta}(\cot(\pi-\theta))= 3\cot^2(\pi-\theta)\cdot(-\csc^2(\pi-\theta))\frac{d}{d\theta}(\pi-\theta)=-3\cot^2(\pi-\theta)\csc^2(\pi-\theta)\cdot(-1)=3\cot^2(\pi-\theta)\csc^2(\pi-\theta)=3(-\cot\theta)^2\csc^2\theta=3\cot^2\theta\csc^2\theta$$ | 205 | 416 | {"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} | 3.859375 | 4 | CC-MAIN-2018-34 | longest | en | 0.339831 |
https://www.instructables.com/id/Wave-Demonstrating-Machine/ | 1,582,527,227,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145897.19/warc/CC-MAIN-20200224040929-20200224070929-00096.warc.gz | 761,887,739 | 21,977 | # Wave Demonstrating Machine
10
1
## Introduction: Wave Demonstrating Machine
First, every project starts with a problem, and here we can see our problem which is that we should make an machine that demonstrate waves using mathematical functions , then we focused on the constrains which were is to use MDF Wood and PLA (which is a type of plastic used in the 3D printers)
definitions we need to understand:
1.sound:
A wave is an oscillation that travels through space and is described by its frequency, amplitude, wavelength and speed
2.superposition principle: for all linear system, the net response caused by 2 or more stimuli is the sum of the responses that would have been caused by stimulus individually
3.simple wave: a wave with constant state
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Materials That We Used
So these are the materials which we needed to build our prototype
## Step 2: STL Files and Pictures of the Parts
each file is placed after the picture of it
PS:(the first part has 2 pieces)
and for the plate which is the base we printed it using cnc machine .
## Step 3: Assembly
then we have assembled the parts together to check if they fit in the right places
## Step 4: Finalizing
we sand papered the parts so that it can fit perfectly
41 9.0K
136 16K
84 20K | 322 | 1,419 | {"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-2020-10 | latest | en | 0.927685 |
https://www.thestudentroom.co.uk/showthread.php?t=2656402 | 1,632,053,974,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00243.warc.gz | 1,044,243,532 | 31,097 | # FP3 nth roots of unity question
Watch
Thread starter 7 years ago
#1
if w=(2pi/5)i simplify (1-w)(1-w2)(1-w3)(1-w4)
the answer is 5
I tried expanding and figured that was a waste of time
I have a feeling the answer is obvious please help
0
reply
7 years ago
#2
(Original post by physics4ever)
if w=(2pi/5)i simplify (1-w)(1-w2)(1-w3)(1-w4)
the answer is 5
I tried expanding and figured that was a waste of time
I have a feeling the answer is obvious please help
There's probably better ways of doing it, but here we go:
Note that and that and so (since )
Expand and separately (and noting my first point, simplify both expressions), and then multiply them together. You will end up with and terms but they can be simplified as well. Group bits together that look useful, e.g. and see where you end up
0
reply
X
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed
### Oops, nobody has postedin the last few hours.
Why not re-start the conversation?
see more
### See more of what you like onThe Student Room
You can personalise what you see on TSR. Tell us a little about yourself to get started.
### Poll
Join the discussion
#### Feeling behind at school/college? What is the best thing your teachers could to help you catch up?
Extra compulsory independent learning activities (eg, homework tasks) (3)
3.33%
Run extra compulsory lessons or workshops (14)
15.56%
Focus on making the normal lesson time with them as high quality as possible (17)
18.89%
Focus on making the normal learning resources as high quality/accessible as possible (10)
11.11%
Provide extra optional activities, lessons and/or workshops (30)
33.33%
Assess students, decide who needs extra support and focus on these students (16)
17.78%
View All
Latest
My Feed
### Oops, nobody has postedin the last few hours.
Why not re-start the conversation?
### See more of what you like onThe Student Room
You can personalise what you see on TSR. Tell us a little about yourself to get started. | 501 | 1,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-39 | latest | en | 0.937846 |
https://wiki.alquds.edu/?query=Power_factor | 1,685,982,766,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00002.warc.gz | 668,601,461 | 47,296 | # Power factor
In electrical engineering, the power factor of an AC power system is defined as the ratio of the real power absorbed by the load to the apparent power flowing in the circuit. Real power is the average of the instantaneous product of voltage and current and represents the capacity of the electricity for performing work. Apparent power is the product of RMS current and voltage. Due to energy stored in the load and returned to the source, or due to a non-linear load that distorts the wave shape of the current drawn from the source, the apparent power may be greater than the real power, so more current flows in the circuit than would be required to transfer real power alone. A power factor magnitude of less than one indicates the voltage and current are not in phase, reducing the average product of the two. A negative power factor occurs when the device (which is normally the load) generates real power, which then flows back towards the source.
In an electric power system, a load with a low power factor draws more current than a load with a high power factor for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system and require larger wires and other equipment. Because of the costs of larger equipment and wasted energy, electrical utilities will usually charge a higher cost to industrial or commercial customers where there is a low power factor.
Power-factor correction increases the power factor of a load, improving efficiency for the distribution system to which it is attached. Linear loads with a low power factor (such as induction motors) can be corrected with a passive network of capacitors or inductors. Non-linear loads, such as rectifiers, distort the current drawn from the system. In such cases, active or passive power factor correction may be used to counteract the distortion and raise the power factor. The devices for correction of the power factor may be at a central substation, spread out over a distribution system, or built into power-consuming equipment.
## General case
Schematic showing how power factor is calculated
The general expression for power factor is given by
${\displaystyle {\mbox{power factor}}=P/P_{a}}$
${\displaystyle P_{a}=I_{rms}V_{rms}}$
where ${\displaystyle P}$ is the real power measured by an ideal wattmeter, ${\displaystyle I_{rms}}$ is the rms current measured by an ideal ammeter, and ${\displaystyle V_{rms}}$ is the rms voltage measured by an ideal voltmeter. Apparent power, ${\displaystyle P_{a}}$, is the product of the rms current and the rms voltage.
If the load is sourcing power back toward the generator, then ${\displaystyle P}$ and ${\displaystyle {\mbox{power factor}}}$ will be negative.
### Periodic waveforms
If the waveforms are periodic with the same period which is much shorter than the averaging time of the physical meters, then the power factor can be computed by the following
${\displaystyle {\mbox{power factor}}=P/P_{a}}$
${\displaystyle P_{a}=I_{rms}V_{rms}}$
${\displaystyle P={\frac {1}{T}}\int _{t'}^{t'+T}i(t)v(t)dt}$
${\displaystyle I_{rms}^{2}={\frac {1}{T}}\int _{t'}^{t'+T}{i(t)}^{2}dt}$
${\displaystyle V_{rms}^{2}={\frac {1}{T}}\int _{t'}^{t'+T}{v(t)}^{2}dt}$
where ${\displaystyle i(t)}$ is the instantaneous current, ${\displaystyle v(t)}$ is the instantaneous voltage, ${\displaystyle t'}$ is an arbitrary starting time, and ${\displaystyle T}$ is the period of the waveforms.
### Nonperiodic waveforms
If the waveforms are not periodic and the physical meters have the same averaging time, then the equations for the periodic case can be used with the exception that ${\displaystyle T}$ is the averaging time of the meters instead of the waveform period.
## Linear time-invariant circuits
Power flow calculated from AC voltage and current entering a load having a zero power factor (ϕ = 90°, cos(ϕ) = 0). The blue line shows the instantaneous power entering the load: all of the energy received during the first (or third) quarter cycle is returned to the grid during the second (or fourth) quarter cycle, resulting in an average power flow (light blue line) of zero.
Instantaneous and average power calculated from AC voltage and current for a load with a lagging power factor (ϕ = 45°, cos(ϕ) ≈ 0.71). The blue line (instantaneous power) shows that a portion of the energy received by the load is returned to the grid during the part of the cycle labeled ϕ.
Linear time-invariant circuits (referred to simply as linear circuits for the rest of this article), for example, circuits consisting of combinations of resistors, inductors and capacitors have a sinusoidal response to the sinusoidal line voltage.[1] A linear load does not change the shape of the input waveform but may change the relative timing (phase) between voltage and current, due to its inductance or capacitance.
In a purely resistive AC circuit, voltage and current waveforms are in step (or in phase), changing polarity at the same instant in each cycle. All the power entering the load is consumed (or dissipated).
Where reactive loads are present, such as with capacitors or inductors, energy storage in the loads results in a phase difference between the current and voltage waveforms. During each cycle of the AC voltage, extra energy, in addition to any energy consumed in the load, is temporarily stored in the load in electric or magnetic fields then returned to the power grid a fraction of the period later.
Electrical circuits containing predominantly resistive loads (incandescent lamps, heating elements) have a power factor of almost 1, but circuits containing inductive or capacitive loads (electric motors, solenoid valves, transformers, fluorescent lamp ballasts, and others) can have a power factor well below 1.
In the electric power grid, reactive loads cause a continuous ebb and flow of nonproductive power. A circuit with a low power factor will use a greater amount of current to transfer a given quantity of real power than a circuit with a high power factor thus causing increased losses due to resistive heating in power lines, and requiring the use of higher-rated conductors and transformers.
### Definition and calculation
AC power has two components:
Together, they form the complex power (${\displaystyle S}$) expressed as volt-amperes (VA). The magnitude of the complex power is the apparent power (${\displaystyle |S|}$), also expressed in volt-amperes (VA).
The VA and var are non-SI units mathematically identical to the watt, but are used in engineering practice instead of the watt to state what quantity is being expressed. The SI explicitly disallows using units for this purpose or as the only source of information about a physical quantity as used.[4]
The power factor is defined as the ratio of real power to apparent power. As power is transferred along a transmission line, it does not consist purely of real power that can do work once transferred to the load, but rather consists of a combination of real and reactive power, called apparent power. The power factor describes the amount of real power transmitted along a transmission line relative to the total apparent power flowing in the line.[5][6]
The power factor can also be computed as the cosine of the angle θ by which the current waveform lags or leads the voltage waveform,[7].
#### Power triangle
One can relate the various components of AC power by using the power triangle in vector space. Real power extends horizontally in the real axis and reactive power extends in the direction of the imaginary axis. Complex power (and its magnitude, apparent power) represents a combination of both real and reactive power, and therefore can be calculated by using the vector sum of these two components. We can conclude that the mathematical relationship between these components is:
{\displaystyle {\begin{aligned}S&=P+jQ\\|S|&={\sqrt {P^{2}+Q^{2}}}\\{\text{pf}}&=\cos {\theta }={\frac {P}{|S|}}=\cos {\left(\arctan {\left({\frac {Q}{P}}\right)}\right)}\\Q&=P\,\tan(\arccos({\text{pf}}))\end{aligned}}}
As the angle θ increases with fixed total apparent power, current and voltage are further out of phase with each other. Real power decreases, and reactive power increases.
#### Lagging, leading and unity power factors
Power factor is described as leading if the current waveform is advanced in phase with respect to voltage, or lagging when the current waveform is behind the voltage waveform. A lagging power factor signifies that the load is inductive, as the load will consume reactive power. The reactive component ${\displaystyle Q}$ is positive as reactive power travels through the circuit and is consumed by the inductive load. A leading power factor signifies that the load is capacitive, as the load supplies reactive power, and therefore the reactive component ${\displaystyle Q}$ is negative as reactive power is being supplied to the circuit.
If θ is the phase angle between the current and voltage, then the power factor is equal to the cosine of the angle, ${\displaystyle \cos \theta }$:
${\displaystyle |P|=|S|\cos \theta }$
Since the units are consistent, the power factor is by definition a dimensionless number between -1 and 1. When power factor is equal to 0, the energy flow is entirely reactive and stored energy in the load returns to the source on each cycle. When the power factor is 1, referred to as unity power factor, all the energy supplied by the source is consumed by the load. Power factors are usually stated as leading or lagging to show the sign of the phase angle. Capacitive loads are leading (current leads voltage), and inductive loads are lagging (current lags voltage).
If a purely resistive load is connected to a power supply, current and voltage will change polarity in step, the power factor will be 1, and the electrical energy flows in a single direction across the network in each cycle. Inductive loads such as induction motors (any type of wound coil) consume reactive power with the current waveform lagging the voltage. Capacitive loads such as capacitor banks or buried cables generate reactive power with the current phase leading the voltage. Both types of loads will absorb energy during part of the AC cycle, which is stored in the device's magnetic or electric field, only to return this energy back to the source during the rest of the cycle.
For example, to get 1 kW of real power, if the power factor is unity, 1 kVA of apparent power needs to be transferred (1 kW ÷ 1 = 1 kVA). At low values of power factor, more apparent power needs to be transferred to get the same real power. To get 1 kW of real power at 0.2 power factor, 5 kVA of apparent power needs to be transferred (1 kW ÷ 0.2 = 5 kVA). This apparent power must be produced and transmitted to the load and is subject to losses in the production and transmission processes.
Electrical loads consuming alternating current power consume both real power and reactive power. The vector sum of real and reactive power is the complex power, and its magnitude is the apparent power. The presence of reactive power causes the real power to be less than the apparent power, and so, the electric load has a power factor of less than 1.
A negative power factor (0 to −1) can result from returning active power to the source, such as in the case of a building fitted with solar panels when surplus power is fed back into the supply.[8][9][10]
### Power factor correction of linear loads
Power factor correction of linear load
A high power factor is generally desirable in a power delivery system to reduce losses and improve voltage regulation at the load. Compensating elements near an electrical load will reduce the apparent power demand on the supply system. Power factor correction may be applied by an electric power transmission utility to improve the stability and efficiency of the network. Individual electrical customers who are charged by their utility for low power factor may install correction equipment to increase their power factor so as to reduce costs.
Power factor correction brings the power factor of an AC power circuit closer to 1 by supplying or absorbing reactive power, adding capacitors or inductors that act to cancel the inductive or capacitive effects of the load, respectively. In the case of offsetting the inductive effect of motor loads, capacitors can be locally connected. These capacitors help to generate reactive power to meet the demand of the inductive loads. This will keep that reactive power from having to flow all the way from the utility generator to the load. In the electricity industry, inductors are said to consume reactive power and capacitors are said to supply it, even though reactive power is just energy moving back and forth on each AC cycle.
The reactive elements in power factor correction devices can create voltage fluctuations and harmonic noise when switched on or off. They will supply or sink reactive power regardless of whether there is a corresponding load operating nearby, increasing the system's no-load losses. In the worst case, reactive elements can interact with the system and with each other to create resonant conditions, resulting in system instability and severe overvoltage fluctuations. As such, reactive elements cannot simply be applied without engineering analysis.
1. Reactive power control relay; 2. Network connection points; 3. Slow-blow fuses; 4. Inrush-limiting contactors; 5. Capacitors (single-phase or three-phase units, delta-connection); 6. Transformer (for controls and ventilation fans)
An automatic power factor correction unit consists of a number of capacitors that are switched by means of contactors. These contactors are controlled by a regulator that measures power factor in an electrical network. Depending on the load and power factor of the network, the power factor controller will switch the necessary blocks of capacitors in steps to make sure the power factor stays above a selected value.
In place of a set of switched capacitors, an unloaded synchronous motor can supply reactive power. The reactive power drawn by the synchronous motor is a function of its field excitation. It is referred to as a synchronous condenser. It is started and connected to the electrical network. It operates at a leading power factor and puts vars onto the network as required to support a system's voltage or to maintain the system power factor at a specified level.
The synchronous condenser's installation and operation are identical to those of large electric motors. Its principal advantage is the ease with which the amount of correction can be adjusted; it behaves like a variable capacitor. Unlike with capacitors, the amount of reactive power supplied is proportional to voltage, not the square of voltage; this improves voltage stability on large networks. Synchronous condensers are often used in connection with high-voltage direct-current transmission projects or in large industrial plants such as steel mills.
For power factor correction of high-voltage power systems or large, fluctuating industrial loads, power electronic devices such as the static VAR compensator or STATCOM are increasingly used. These systems are able to compensate sudden changes of power factor much more rapidly than contactor-switched capacitor banks and, being solid-state, require less maintenance than synchronous condensers.
Examples of non-linear loads on a power system are rectifiers (such as used in a power supply), and arc discharge devices such as fluorescent lamps, electric welding machines, or arc furnaces. Because current in these systems is interrupted by a switching action, the current contains frequency components that are multiples of the power system frequency. Distortion power factor is a measure of how much the harmonic distortion of a load current decreases the average power transferred to the load.
Sinusoidal voltage and non-sinusoidal current give a distortion power factor of 0.75 for this computer power supply load.
### Non-sinusoidal components
In linear circuits having only sinusoidal currents and voltages of one frequency, the power factor arises only from the difference in phase between the current and voltage. This is displacement power factor.[11]
Non-linear loads change the shape of the current waveform from a sine wave to some other form. Non-linear loads create harmonic currents in addition to the original (fundamental frequency) AC current. This is of importance in practical power systems that contain non-linear loads such as rectifiers, some forms of electric lighting, electric arc furnaces, welding equipment, switched-mode power supplies, variable speed drives and other devices. Filters consisting of linear capacitors and inductors can prevent harmonic currents from entering the supplying system.
To measure the real power or reactive power, a wattmeter designed to work properly with non-sinusoidal currents must be used.
### Distortion power factor
The distortion power factor is the distortion component associated with the harmonic voltages and currents present in the system.
{\displaystyle {\begin{aligned}{\mbox{distortion power factor}}&={\frac {I_{1}}{I_{rms}}}\\&={\frac {I_{1}}{\sqrt {I_{1}^{2}+I_{2}^{2}+I_{3}^{2}+I_{4}^{2}+\cdots }}}\\&={\frac {1}{\sqrt {1+{\frac {I_{2}^{2}+I_{3}^{2}+I_{4}^{2}+\cdots }{I_{1}^{2}}}}}}\\&={\frac {1}{\sqrt {1+THD_{i}^{2}}}}\\\end{aligned}}}
${\displaystyle {\mbox{THD}}_{i}}$ is the total harmonic distortion of the load current.
${\displaystyle THD_{i}={\frac {\sqrt {\displaystyle \sum _{h=2}^{\infty }I_{h}^{2}}}{I_{1}}}={\frac {\sqrt {I_{2}^{2}+I_{3}^{2}+I_{4}^{2}+\cdots }}{I_{1}}}}$
${\displaystyle I_{1}}$ is the fundamental component of the current and ${\displaystyle I_{\mbox{rms}}}$ is the total current – both are root mean square-values (distortion power factor can also be used to describe individual order harmonics, using the corresponding current in place of total current). This definition with respect to total harmonic distortion assumes that the voltage stays undistorted (sinusoidal, without harmonics). This simplification is often a good approximation for stiff voltage sources (not being affected by changes in load downstream in the distribution network). Total harmonic distortion of typical generators from current distortion in the network is on the order of 1–2%, which can have larger scale implications but can be ignored in common practice.[12]
The result when multiplied with the displacement power factor (DPF) is the overall, true power factor or just power factor (PF):
${\displaystyle {\mbox{PF}}={\frac {\cos {\varphi }}{\sqrt {1+THD_{i}^{2}}}}}$
### Distortion in three-phase networks
In practice, the local effects of distortion current on devices in a three-phase distribution network rely on the magnitude of certain order harmonics rather than the total harmonic distortion.
For example, the triplen, or zero-sequence, harmonics (3rd, 9th, 15th, etc.) have the property of being in-phase when compared line-to-line. In a delta-wye transformer, these harmonics can result in circulating currents in the delta windings and result in greater resistive heating. In a wye-configuration of a transformer, triplen harmonics will not create these currents, but they will result in a non-zero current in the neutral wire. This could overload the neutral wire in some cases and create error in kilowatt-hour metering systems and billing revenue.[13][14] The presence of current harmonics in a transformer also result in larger eddy currents in the magnetic core of the transformer. Eddy current losses generally increase as the square of the frequency, lowering the transformer's efficiency, dissipating additional heat, and reducing its service life.[15]
Negative-sequence harmonics (5th, 11th, 17th, etc.) combine 120 degrees out of phase, similarly to the fundamental harmonic but in a reversed sequence. In generators and motors, these currents produce magnetic fields which oppose the rotation of the shaft and sometimes result in damaging mechanical vibrations.[16]
### Switched-mode power supplies
A particularly important class of non-linear loads is the millions of personal computers that typically incorporate switched-mode power supplies (SMPS) with rated output power ranging from a few watts to more than 1 kW. Historically, these very-low-cost power supplies incorporated a simple full-wave rectifier that conducted only when the mains instantaneous voltage exceeded the voltage on the input capacitors. This leads to very high ratios of peak-to-average input current, which also lead to a low distortion power factor and potentially serious phase and neutral loading concerns.
A typical switched-mode power supply first converts the AC mains to a DC bus by means of a bridge rectifier. The output voltage is then derived from this DC bus. The problem with this is that the rectifier is a non-linear device, so the input current is highly non-linear. That means that the input current has energy at harmonics of the frequency of the voltage. This presents a problem for power companies, because they cannot compensate for the harmonic current by adding simple capacitors or inductors, as they could for the reactive power drawn by a linear load. Many jurisdictions are beginning to require power factor correction for all power supplies above a certain power level.
Regulatory agencies such as the EU have set harmonic limits as a method of improving power factor. Declining component cost has hastened implementation of two different methods. To comply with current EU standard EN61000-3-2, all switched-mode power supplies with output power more than 75 W must at least include passive power factor correction. 80 Plus power supply certification requires a power factor of 0.9 or more.[17]
### Power factor correction (PFC) in non-linear loads
#### Passive PFC
The simplest way to control the harmonic current is to use a filter that passes current only at line frequency (50 or 60 Hz). The filter consists of capacitors or inductors and makes a non-linear device look more like a linear load. An example of passive PFC is a valley-fill circuit.
A disadvantage of passive PFC is that it requires larger inductors or capacitors than an equivalent power active PFC circuit.[18][19][20] Also, in practice, passive PFC is often less effective at improving the power factor.[21][22][23][24][25]
#### Active PFC
Specifications taken from the packaging of a 610 W PC power supply showing active PFC rating
Active PFC is the use of power electronics to change the waveform of current drawn by a load to improve the power factor.[26] Some types of the active PFC are buck, boost, buck-boost and synchronous condenser. Active power factor correction can be single-stage or multi-stage.
In the case of a switched-mode power supply, a boost converter is inserted between the bridge rectifier and the main input capacitors. The boost converter attempts to maintain a constant voltage at its output while drawing a current that is always in phase with and at the same frequency as the line voltage. Another switched-mode converter inside the power supply produces the desired output voltage from the DC bus. This approach requires additional semiconductor switches and control electronics but permits cheaper and smaller passive components. It is frequently used in practice.
For a three-phase SMPS, the Vienna rectifier configuration may be used to substantially improve the power factor.
SMPSs with passive PFC can achieve power factor of about 0.7–0.75, SMPSs with active PFC, up to 0.99 power factor, while a SMPS without any power factor correction have a power factor of only about 0.55–0.65.[27]
Due to their very wide input voltage range, many power supplies with active PFC can automatically adjust to operate on AC power from about 100 V (Japan) to 240 V (Europe). That feature is particularly welcome in power supplies for laptops.
#### Dynamic PFC
Dynamic power factor correction (DPFC), sometimes referred to as real-time power factor correction, is used for electrical stabilization in cases of rapid load changes (e.g. at large manufacturing sites). DPFC is useful when standard power factor correction would cause over or under correction.[28] DPFC uses semiconductor switches, typically thyristors, to quickly connect and disconnect capacitors or inductors to improve power factor.
## Importance in distribution systems
75 MVAr capacitor bank in a 150 kV substation
Power factors below 1.0 require a utility to generate more than the minimum volt-amperes necessary to supply the real power (watts). This increases generation and transmission costs. For example, if the load power factor were as low as 0.7, the apparent power would be 1.4 times the real power used by the load. Line current in the circuit would also be 1.4 times the current required at 1.0 power factor, so the losses in the circuit would be doubled (since they are proportional to the square of the current). Alternatively, all components of the system such as generators, conductors, transformers, and switchgear would be increased in size (and cost) to carry the extra current. When the power factor is close to unity, for the same kVA rating of the transformer more load current can be supplied.[29]
Utilities typically charge additional costs to commercial customers who have a power factor below some limit, which is typically 0.9 to 0.95. Engineers are often interested in the power factor of a load as one of the factors that affect the efficiency of power transmission.
With the rising cost of energy and concerns over the efficient delivery of power, active PFC has become more common in consumer electronics.[30] Current Energy Star guidelines for computers[31] call for a power factor of ≥ 0.9 at 100% of rated output in the PC's power supply. According to a white paper authored by Intel and the U.S. Environmental Protection Agency, PCs with internal power supplies will require the use of active power factor correction to meet the ENERGY STAR 5.0 Program Requirements for Computers.[32]
In Europe, EN 61000-3-2 requires power factor correction be incorporated into consumer products.
Small customers, such as households, are not usually charged for reactive power and so power factor metering equipment for such customers will not be installed.
## Measurement techniques
The power factor in a single-phase circuit (or balanced three-phase circuit) can be measured with the wattmeter-ammeter-voltmeter method, where the power in watts is divided by the product of measured voltage and current. The power factor of a balanced polyphase circuit is the same as that of any phase. The power factor of an unbalanced polyphase circuit is not uniquely defined.
A direct reading power factor meter can be made with a moving coil meter of the electrodynamic type, carrying two perpendicular coils on the moving part of the instrument. The field of the instrument is energized by the circuit current flow. The two moving coils, A and B, are connected in parallel with the circuit load. One coil, A, will be connected through a resistor and the second coil, B, through an inductor, so that the current in coil B is delayed with respect to current in A. At unity power factor, the current in A is in phase with the circuit current, and coil A provides maximum torque, driving the instrument pointer toward the 1.0 mark on the scale. At zero power factor, the current in coil B is in phase with circuit current, and coil B provides torque to drive the pointer towards 0. At intermediate values of power factor, the torques provided by the two coils add and the pointer takes up intermediate positions.[33]
Another electromechanical instrument is the polarized-vane type.[34] In this instrument a stationary field coil produces a rotating magnetic field, just like a polyphase motor. The field coils are connected either directly to polyphase voltage sources or to a phase-shifting reactor if a single-phase application. A second stationary field coil, perpendicular to the voltage coils, carries a current proportional to current in one phase of the circuit. The moving system of the instrument consists of two vanes that are magnetized by the current coil. In operation, the moving vanes take up a physical angle equivalent to the electrical angle between the voltage source and the current source. This type of instrument can be made to register for currents in both directions, giving a four-quadrant display of power factor or phase angle.
Digital instruments exist that directly measure the time lag between voltage and current waveforms. Low-cost instruments of this type measure the peak of the waveforms. More sophisticated versions measure the peak of the fundamental harmonic only, thus giving a more accurate reading for phase angle on distorted waveforms. Calculating power factor from voltage and current phases is only accurate if both waveforms are sinusoidal.[35]
Power Quality Analyzers, often referred to as Power Analyzers, make a digital recording of the voltage and current waveform (typically either one phase or three phase) and accurately calculate true power (watts), apparent power (VA) power factor, AC voltage, AC current, DC voltage, DC current, frequency, IEC61000-3-2/3-12 Harmonic measurement, IEC61000-3-3/3-11 flicker measurement, individual phase voltages in delta applications where there is no neutral line, total harmonic distortion, phase and amplitude of individual voltage or current harmonics, etc.[36][37]
## Mnemonics
English-language power engineering students are advised to remember: ELI the ICE man or ELI on ICE – the voltage E, leads the current I, in an inductor L. The current I leads the voltage E in a capacitor C.
Another common mnemonic is CIVIL – in a capacitor (C) the current (I) leads voltage (V), voltage (V) leads current (I) in an inductor (L).
## References
1. ^ Das, J. C. (2015). Power System Harmonics and Passive Filter Design. Wiley, IEEE Press. p. 2. ISBN 978-1-118-86162-2. To distinguish between linear and nonlinear loads, we may say that linear time-invariant loads are characterized so that an application of a sinusoidal voltage results in a sinusoidal flow of current.
2. ^ Boylestad, Robert (2002-03-04). Introductory Circuit Analysis (10th ed.). p. 857. ISBN 978-0-13-097417-4.
3. ^ "SI Units – Electricity and Magnetism". CH: International Electrotechnical Commission. Archived from the original on 2007-12-11. Retrieved 14 June 2013.
4. ^ The International System of Units (SI) [SI brochure] (PDF). § 5.3.2 (p. 132, 40 in the PDF file): BIPM. 2006. Archived (PDF) from the original on 2022-10-09.{{cite book}}: CS1 maint: location (link)
5. ^ Authoritative Dictionary of Standards Terms (7th ed.), IEEE, 2000, ISBN 978-0-7381-2601-2, Std. 100
6. ^ Trial-Use Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions, IEEE, 2000, ISBN 978-0-7381-1963-2, Std. 1459–2000. Note 1, section 3.1.1.1, when defining the quantities for power factor, asserts that real power only flows to the load and can never be negative. As of 2013, one of the authors acknowledged that this note was incorrect, and is being revised for the next edition. See http://powerstandards.com/Shymanski/draft.pdf Archived 2016-03-04 at the Wayback Machine
7. ^ Suresh Kumar, K. S. (2013). Electric Circuit Analysis. Pearson. p. 8.10. ISBN 978-8-13-179155-4.
8. ^ Duddell, W. (1901), "On the resistance and electromotive forces of the electric arc", Philosophical Transactions of the Royal Society A: Mathematical, Physical and Engineering Sciences, 203 (359–371): 512–15, doi:10.1098/rsta.1904.0022, The fact that the solid arc has, at low frequencies, a negative power factor, indicates that the arc is supplying power to the alternator…
9. ^ Zhang, S. (July 2006), "Analysis of some measurement issues in bushing power factor tests in the field", IEEE Transactions on Power Delivery, 21 (3): 1350–56, doi:10.1109/tpwrd.2006.874616, S2CID 39895367, …(the measurement) gives both negative power factor and negative resistive current (power loss)
10. ^ Almarshoud, A. F.; et al. (2004), "Performance of Grid-Connected Induction Generator under Naturally Commutated AC Voltage Controller", Electric Power Components and Systems, 32 (7): 691–700, doi:10.1080/15325000490461064, S2CID 110279940, Accordingly, the generator will consume active power from the grid, which leads to negative power factor.
11. ^ Ewald Fuchs; Mohammad A. S. Masoum (14 July 2015). Power Quality in Power Systems and Electrical Machines. Elsevier Science. pp. 432–. ISBN 978-0-12-800988-8. The DPF it the cosine of the angle between these two quantities
12. ^ Sankaran, C. (1999), Effects of Harmonics on Power Systems, Electro-Test, ...and voltage-time relationship deviates from the pure sine function. The distortion at the point of generation is very small (about 1% to 2%), but nonetheless it exists.
13. ^ "Single-phase load harmonics vs. three-phase load harmonics" (PDF), Power System Harmonics, Pacific Gas and Electric
14. ^ "Harmonic Effects" (PDF), Harmonics and IEEE 519, CA: EnergyLogix Solutions
15. ^ Sankaran, C. (1999), "Transformers", Effects of Harmonics on Power Systems, Electro-Test
16. ^ Sankaran, C. (1999), "Motors", Effects of Harmonics on Power Systems, Electro-Test, The interaction between the positive and negative sequence magnetic fields and currents produces torsional oscillations of the motor shaft. These oscillations result in shaft vibrations.
17. ^ "What is an 80 PLUS certified power supply?", Certified Power Supplies and Manufacturers, 80 Plus
18. ^ Schramm, Ben (Fall 2006), "Power Supply Design Principles: Techniques and Solutions, Part 3", Newsletter, Nuvation, archived from the original on 2007-03-09
19. ^ Wolfle, W.H.; Hurley, W.G. (2003), "Quasi-active power factor correction with a variable inductive filter: theory, design and practice", Xplore, IEEE, vol. 18, no. 1, pp. 248–255, Bibcode:2003ITPE...18..248W, doi:10.1109/TPEL.2002.807135
20. ^ Wölfle, W. H.; Hurley, W. G., "Quasi-active Power Factor Correction: The Role of Variable Inductance", Power electronics (project), IE: Nuigalway
21. ^ ATX Power Supply Units Roundup, xBit labs, archived from the original on 2008-11-20, The power factor is the measure of reactive power. It is the ratio of active power to the total of active and reactive power. It is about 0.65 with an ordinary PSU, but PSUs with active PFC have a power factor of 0.97–0.99. […] hardware reviewers sometimes make no difference between the power factor and the efficiency factor. Although both these terms describe the effectiveness of a power supply, it is a gross mistake to confuse them. […] There is a very small effect from passive PFC – the power factor grows only from 0.65 to 0.7–0.75.
22. ^ The Active PFC Market is Expected to Grow at an Annually Rate of 12.3% Till 2011, Find articles, Mar 16, 2006, archived from the original on September 1, 2009, Higher-powered products are also likely to use active PFC, since it would be the most cost effective way to bring products into compliance with the EN standard.
23. ^ Power Factor Correction, TECHarp, Passive PFC […] the power factor is low at 60–80%. […] Active PFC ... a power factor of up to 95%
24. ^ Why we need PFC in PSU, Silverstone Technology, archived from the original on 2008-12-22, Normally, the power factor value of electronic device without power factor correction is approximately 0.5. […] Passive PFC […] 70~80% […] Active PFC […] 90~99.9%
25. ^ Brooks, Tom (Mar 2004), "PFC options for power supplies", Taiyo, Electronic products, archived from the original on 2008-12-02, The disadvantages of passive PFC techniques are that they typically yield a power factor of only 0.60 to 0.70 […] Dual-stage active PFC technology [yields] a power factor typically greater than 0.98
26. ^ Power Factor Correction (PFC) Basics (PDF) (application note), Fairchild Semiconductor, 2004, archived from the original (PDF) on 2014-06-11, retrieved 2009-11-29
27. ^ Sugawara, I.; Suzuki, Y.; Takeuchi, A.; Teshima, T. (19–23 Oct 1997), "Experimental studies on active and passive PFC circuits", INTELEC 97, 19th International Telecommunications Energy Conference, pp. 571–78, doi:10.1109/INTLEC.1997.646051, ISBN 978-0-7803-3996-5, S2CID 109885369
28. ^ Chavez, C.; Houdek, J. A. "Dynamic Harmonic Mitigation and power factor correction". EPQU'07. 9th International Conference Electrical Power Quality and Utilisation: October 9–11, 2007, Barcelona, Spain. IEEE. pp. 1–5. doi:10.1109/EPQU.2007.4424144. ISBN 978-84-690-9441-9.
29. ^ "Power Factor – Importance, Calculation and Correction techniques". 23 November 2018.
30. ^ Power Factor Correction Handbook (PDF), ON Semiconductor, 2007
31. ^ Program Requirements for Computers (PDF) (Version 5.0 ed.), US: Energy Star
32. ^ Bolioli, T.; Duggirala, M.; Haines, E.; Kolappan, R.; Wong, H. (2009), Version 5.0 System Implementation (PDF) (white paper), Energy Star, archived (PDF) from the original on 2022-10-09
33. ^ Fink, Donald G.; Beaty, H. Wayne (1978), Standard Handbook for Electrical Engineers (11 ed.), New York: McGraw-Hill, p. 3‐29 paragraph 80, ISBN 978-0-07-020974-9
34. ^ Manual of Electric Instruments Construction and Operating Principles, Schenectady, New York: General Electric, Meter and Instrument Department, 1949, pp. 66–68, GET-1087A
35. ^ "The Fundamentals of FFT-Based Signal Analysis and Measurement in LabVIEW and LabWindows/CVI". National Instruments Corporation. Retrieved 6 November 2017.
36. ^ "WT3000E Series Precision Power Analyzers" (PDF). Yokogawa Corporation. Archived from the original (PDF) on 7 November 2017. Retrieved 6 November 2017.
37. ^ "Fluke 1760 Three-Phase Power Quality Recorder" (PDF). Fluke Corporation. Archived (PDF) from the original on 2022-10-09. Retrieved 6 November 2017. | 8,706 | 38,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 33, "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-2023-23 | longest | en | 0.927843 |
https://www.liverpoolecho.co.uk/news/uk-world-news/three-question-iq-test-can-19810156 | 1,643,018,729,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00413.warc.gz | 880,530,774 | 50,119 | Over the past year, many families have enjoyed coming together to test their brainpower.
However, some family members may still be arguing over who is the most intelligent.
Well, we might have potentially found the trickiest quiz yet that will truly determine how smart you are.
Receive newsletters with the latest news, sport and what's on updates from the Liverpool ECHO by signing up here.
This IQ test is made up of just three maths questions and should only take a couple of minutes to complete.
Originally published in 2005 as part of a research paper, Professor Shane Federick found out of 3000 participants, including some Harvard and Yale University students, only 17 per cent managed to bag three out of three.
So without further ado, scroll down to find the collated questions, answers and theory by the Mirror.
The questions:
(1) A bat and a ball cost \$1.10 in total. The bat costs \$1.00 more than the ball. How much does the ball cost? _____ cents
(2) If it takes 5 machines 5 minutes to make 5 widgets, how long would it take 100 machines to make 100 widgets? _____ minutes
(3) In a lake, there is a patch of lily pads. Every day, the patch doubles in size. If it takes 48 days for the patch to cover the entire lake, how long would it take for the patch to cover half of the lake? _____ days.
The paper quotes Jensen, 1998, and says: "People with higher cognitive ability (or "IQ") differ from those with lower cognitive ability in a variety of important and unimportant ways.
"On average, they live longer, earn more, have larger working memories, faster reaction times and are more susceptible to visual illusions."
How do you think you fared?
The most common incorrect answers:
1. 10 cents
2. 100 minutes
3. 24 days
Professor Frederick continued: "Anyone who reflects upon it for even a moment would recognise that the difference between \$1 and 10 cents is only 90 cents, not \$1 as the problem stipulates.
"In this case, catching that error is tantamount to solving the problem, since nearly everyone who does not respond '10 cents' does, in fact give the correct response."
Enter your postcode below to claim a free scratchcard
1. 5 cents
2. 5 minutes
3. 47 days
You can read about the full test and paper here. | 515 | 2,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2022-05 | latest | en | 0.955208 |
http://www.mathisfunforum.com/viewtopic.php?pid=246400 | 1,397,791,687,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00469-ip-10-147-4-33.ec2.internal.warc.gz | 487,792,245 | 12,191 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
You are not logged in.
## #226 2012-12-28 09:00:34
bobbym
Offline
### Re: New mathematic on english
How is that supposed to clear up anything when that was obviously written by you? Quoting yourself does not clear things up.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #227 2012-12-28 09:04:42
21122012
Power Member
Offline
### Re: New mathematic on english
Bob,
You that don't understand, what the line which call the graf of a function y=f(x) of value postpone for the OY axes has no relation to function? But Calculus considers behavior of this line and considers that observes behavior of the y=f(x) function. You that don't understand that it is falsification?
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #228 2012-12-28 09:09:17
bob bundy
Moderator
Offline
### Re: New mathematic on english
I can understand the yellow part.
What is the blue?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #229 2012-12-28 09:22:40
21122012
Power Member
Offline
### Re: New mathematic on english
area of red color
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #230 2012-12-28 09:44:48
21122012
Power Member
Offline
### Re: New mathematic on english
Sense of the Graf of function as lines in the Cartesian system of coordinates only one:
division of one of these three squares:
at two parts, are other two areas.
Point of graf of function divides the line which is derivative (xy)' quare S=xy at vertical: x'y and horizontal: y'x pieces.
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #231 2012-12-28 09:48:13
bob bundy
Moderator
Offline
### Re: New mathematic on english
OK. This makes sense now.
This is all understood.
so
So where is the error please?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #232 2012-12-28 10:11:04
21122012
Power Member
Offline
### Re: New mathematic on english
#### bob bundy wrote:
OK. This makes sense now.
This is all understood.
so
So where is the error please?
Bob
Perfectly! The first sensible that I at you saw this!
Green line:
is
Do you agree?
Last edited by 21122012 (2012-12-28 10:27:33)
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #233 2012-12-28 10:54:54
bob bundy
Moderator
Offline
### Re: New mathematic on english
I do not understand this last part
OK so far.
And why should F' be the addition of these two lines anyway?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #234 2012-12-28 11:15:27
21122012
Power Member
Offline
### Re: New mathematic on english
#### bob bundy wrote:
And why should F' be the addition of these two lines anyway?
Bob
because
Bob!
Descartes decided to present a cube in the form of a rectangle:
Understand?
Look #152 drawing at the left.
So are under construction
on the plane and in space.
These are two interpretations of the same process:
Last edited by 21122012 (2012-12-28 11:16:06)
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #235 2012-12-29 00:39:48
bob bundy
Moderator
Offline
### Re: New mathematic on english
because
I don't mind
And
is good for the product rule.
But then it seems to fall apart.
Your green line is the semi-perimeter of the rectangle.
Why should F' be the semi-perimeter? I can think of no good reason.
And the result you get for the horizontal line (2xo^2) isn't what you have on your diagram. You have defined this distance to be x0.
What do you think this green line is?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #236 2012-12-29 06:07:07
21122012
Power Member
Offline
### Re: New mathematic on english
#### bob bundy wrote:
What do you think this green line is?
Bob
It is the key moment for Structural Analysis. As soon as you will understand this moment then you will understand all the rest.
This main thing to what we went all this time.
We use a substitution method: http://vladimir938.eto-ya.com/files/201 … -10.12.jpg
SOLID GEOMETRY
PLANIMETRICS:
It is necessary to stop. Everything is clear?
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #237 2012-12-29 06:16:37
bob bundy
Moderator
Offline
### Re: New mathematic on english
You are in danger of repeating your earlier mistake; the one that led to me withdrawing from this thread. If you won't answer my questions about what you are trying to explain, then I might as well give up.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #238 2012-12-29 06:20:47
21122012
Power Member
Offline
### Re: New mathematic on english
You are in vain so aggressive. To show that such the green line on graphics I have to show you its analog in geometry. I want to make it but gradually that we understood each other. If you refuse to watch my reasonings I won't be able to answer your question.
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #239 2012-12-29 06:39:47
bob bundy
Moderator
Offline
### Re: New mathematic on english
But I don't understand your reasoning. That is why I ask questions.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #240 2012-12-29 07:14:47
21122012
Power Member
Offline
### Re: New mathematic on english
#### bob bundy wrote:
But I don't understand your reasoning. That is why I ask questions.
Bob
What in post 236 you don't understand?
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #241 2012-12-29 08:21:19
bob bundy
Moderator
Offline
### Re: New mathematic on english
It is you who do not understand.
7 times you have asked me what it is I don't understand on this thread and the other.
7 times I have told you.
Each time you tell me a new thing.
That is no help to me.
Why should F' be the semi-perimeter?
What do you think this green line is?
Thank you in anticipation.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #242 2012-12-29 08:47:45
21122012
Power Member
Offline
### Re: New mathematic on english
#### bob bundy wrote:
"F' is the semi perimeter because ........."
F' is not the semi perimeter, because
points of which the top horizontal piece consists aren't equal to points of which the bottom horizontal piece though pieces have identical length consists. But they have various value. I want to explain you it but you refuse to watch my conclusions.
Now I will show one of keys which underlie Structural Analysis creation. The detailed description isn't present, but animation is quite obvious.
Let's choose dependence from each other two variables, convenient as a bright example. Functionally this dependence is expressed by a formula:
Let's show the image of this dependence in a geometrical and graphic look.
In a geometrical look this dependence will be expressed in the circle square with a variable radius:
and in a graphic look in the form of a vertical piece of variable length:
Because the circle and a piece - the elements which are strongly differing from each other, process of integration:
will be represented in both cases differently.
The given animation will help to compare each other elements geometrical and elements of graphic interpretation of processes of integration:
Last edited by 21122012 (2012-12-29 15:00:58)
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #243 2012-12-29 15:03:06
21122012
Power Member
Offline
### Re: New mathematic on english
Bob bundy look this
Hi bobbym.
give me please the your assessment of the animation created by me and sense which it opens
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #244 2012-12-30 07:34:25
bob bundy
Moderator
Offline
### Re: New mathematic on english
You started this thread saying "one of the many errors of calculus".
I have seen three of your errors. In each case you have not applied calculus theory correctly. I have explained why and you have stubbornly argued with me.
But I have offered many links to websites to support my argument, and you have ignored these. If this is the way you have behaved on other forums, then I am not surprised that you have not succeeded there.
I will teach you how to do calculus if you are willing to listen. Once you can use it correctly, it will be interesting to see if any 'errors' remain.
I am assuming that you have invented 'structural analysis' yourself. Clearly, you understand what you mean and have spent a long time working on it.
You post complicated diagrams, with annotation in Russian, and often with formulas that are not defined properly. Do you really think I can understand any of your theory? If you wish to promote this theory you need to present it properly; in the way that mathematicains have done for thousands of years.
Give simple examples.
Pause, while your pupil asks questions and tries to master the basics.
Set questions that are within the grasp of your pupil, so they can move forward slowly.
Increase the level of difficulty only as your pupil gets better.
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
## #245 2012-12-30 08:52:29
21122012
Power Member
Offline
### Re: New mathematic on english
#### bob bundy wrote:
You started this thread saying "one of the many errors of calculus".
I have seen three of your errors. In each case you have not applied calculus theory correctly. I have explained why and you have stubbornly argued with me.
If there are two theories and they in three moments contradict each other, each of two theories has the right yet these moments errors of other theory won't consider is found out what of theories is true. Correctness of any theory can be proved by only one way - result of experience. I made this experiment:
#### bob bundy wrote:
But I have offered many links to websites to support my argument, and you have ignored these. If this is the way you have behaved on other forums, then I am not surprised that you have not succeeded there.
I didn't arrive here from Mars. That is written in references which you to me gave I I know not worse you. To me on Calculus of the teacher always put only the highest marks. I didn't ignore them but they in any way don't disprove my arguments THEY to THEM CONTRADICT as has to be in two contradicting theories.
As for other forums that I am not guilty in that that in the world only one percent of homo sapiense and 99 percent of humanoid monkeys.
#### bob bundy wrote:
I will teach you how to do calculus if you are willing to listen. Once you can use it correctly, it will be interesting to see if any 'errors' remain.
I am assuming that you have invented 'structural analysis' yourself. Clearly, you understand what you mean and have spent a long time working on it.
You post complicated diagrams, with annotation in Russian, and often with formulas that are not defined properly. Do you really think I can understand any of your theory? If you wish to promote this theory you need to present it properly; in the way that mathematicains have done for thousands of years.
Give simple examples.
Pause, while your pupil asks questions and tries to master the basics.
Set questions that are within the grasp of your pupil, so they can move forward slowly.
Increase the level of difficulty only as your pupil gets better.
Bob
Look how many thousand people participated in mathematics creation. And you want that I one replaced thousands. From me it is enough of that that I found mistakes and offered the correct direction. the rest will be done by other people.
P.S. You will be able to teach nothing me besides that you will repeat that that is in textbooks. And that that am in textbooks I already read.
I gave an example of experience which showed that graf of function y=f (x) and y=f(x) aren't connected in any way with each other except for values of derivative two various functions in curve points. THESE ARE VARIOUS FUNCTIONS!
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #246 2012-12-30 12:51:46
bobbym
Offline
### Re: New mathematic on english
As for other forums that I am not guilty in that that in the world only one percent of homo sapiense and 99 percent of humanoid monkeys.
So the whole world is made up of 99% humanoid monkeys as you call them. I know a lot of people who share that opinion. Each one of them carefully leaves himself out of the 99%. You can not seriously hope to persuade people that you have just called monkeys. You say you do not come from Mars, I am not so sure. You wonder why you are bounced from forums. You call people monkeys and then can not understand their reaction towards you?
It is sad that you think you are barred from these forums because of the briliance of your ideas rather than your callous comments.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #247 2012-12-30 13:16:31
21122012
Power Member
Offline
### Re: New mathematic on english
Here, as practically everywhere you confuse cause and effect. Me throw out from forums without understanding simple reasonings. Them don't accept only because they don't coincide with that that THEY got used to repeat. Therefore I draw a conclusion that if THEY can't have the thoughts but only to repeat strangers that they do that that call "repeat as a monkey". Anything personal only fact ascertaining!
Last edited by 21122012 (2012-12-30 13:23:18)
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
## #248 2012-12-30 23:36:49
bobbym
Offline
### Re: New mathematic on english
Hmmmm, I do not see how you can ever progress relying solely on your input. Unless you are Isaac Newton there is and always will be many people smarter than you are. In your case, some of them are on this forum.
I have seen quotes by Bruno and Galileo and now Aristotle. It is fine to reason for yourself, to stand on your own two feet, but a man has got to know his limitations. Got to know when he needs help, got to be able to admit when he is wrong.
P.S. You will be able to teach nothing me besides that you will repeat that that is in textbooks. And that that am in textbooks I already read.
You have read every textbook? You know everything inside of them? A good textbook is something you hold on to for life. Studying it for life.
On the lighter side, here in the good ole USA if anyone called me a humanoid monkey I would clobber him with my bunch of ripe bananas.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #249 2012-12-31 01:41:50
scientia
Full Member
Offline
### Re: New mathematic on english
#### bobbym wrote:
Unless you are Isaac Newton there is and always will be many people smarter than you are.
And even Isaac Newton himself admitted that there were people smarter than he was – giants on whose shoulders he stood.
http://en.wikipedia.org/wiki/Standing_o … _of_giants
## #250 2012-12-31 01:48:26
bobbym
Offline
### Re: New mathematic on english
Hi;
Some historians think that statement has a double meaning. They think he was
ridiculing Robert Hooke who was a hunchback. In their interpretation Newton was saying I am standing on his back and crushing him.
I am reading it now in the page you sent me. I remember coming across it in a book a while back but do not remember the title.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 4,413 | 18,925 | {"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-2014-15 | longest | en | 0.901422 |
https://gmatclub.com/forum/daniel-bernoulli-derived-the-famous-fluid-equation-170849.html | 1,527,464,464,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870497.66/warc/CC-MAIN-20180527225404-20180528005404-00450.warc.gz | 563,576,248 | 63,414 | GMAT Changed on April 16th - Read about the latest changes here
It is currently 27 May 2018, 16:41
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Daniel Bernoulli derived the famous fluid equation
Author Message
TAGS:
### Hide Tags
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4669
Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
02 May 2014, 11:49
13
KUDOS
Expert's post
47
This post was
BOOKMARKED
00:00
Difficulty:
95% (hard)
Question Stats:
22% (01:08) correct 78% (01:18) wrong based on 1649 sessions
### HideShow timer Statistics
Daniel Bernoulli (1700 – 1782) derived the famous fluid equation named after him, to explain an airplane’s wing’s generation of lift, and made a discovery that led to an early method of measuring blood pressure.
(A) equation named after him, to explain an airplane’s wing’s generation of lift
(B) equation named after him, and this principle explains the lift of an airplane’s wing
(C) equation, named it after himself, explained how an airplane’s wing is generating lift
(D) equation named for him, giving an explanation of the generation of the lift of an airplane’s wing
(E) equation named for him, which explains how an airplane’s wing generates lift
Folks sometimes think of GMAT SC in terms of grammar only, but the SC is as much about logic as it is about grammar. The splits in this question are less about grammatical errors and more about logical problems. For a discussion of logic in GMAT SC questions, more practice questions of this sort, and the OE of this particular question, see:
http://magoosh.com/gmat/2014/logical-sp ... orrection/
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Magoosh Discount Codes Veritas Prep GMAT Discount Codes Manhattan GMAT Discount Codes
Intern
Joined: 21 Jan 2014
Posts: 3
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
03 May 2014, 02:58
is "equation named for him" a correct idiom ? Even for a logical split are we allowed to use incorrect idioms?
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4669
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
03 May 2014, 17:20
2
KUDOS
Expert's post
aniketnd01 wrote:
is "equation named for him" a correct idiom ? Even for a logical split are we allowed to use incorrect idioms?
Dear aniketnd01,
That phrase, "equation named for him," is 100% idiomatically correct. Of course, on the GMAT SC, the right answer has to be grammatically correct, idiomatically correct, logically correct, and rhetorically sound all at once. We can never drop one in favor of the other.
What makes you think that phrase is not a correct idiom?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Intern
Joined: 07 May 2014
Posts: 1
WE: Law (Law)
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
07 May 2014, 09:24
2
KUDOS
mikemcgarry wrote:
[color=#0000ff]...and made a discovery that lead to an early method of measuring blood pressure.
Be careful when writing the past tense of the word lead. It is actually led.
All the best...
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4669
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
07 May 2014, 11:30
Perchance wrote:
mikemcgarry wrote:
[color=#0000ff]...and made a discovery that lead to an early method of measuring blood pressure.
Be careful when writing the past tense of the word lead. It is actually led.
All the best...
Dear Perchance,
Many thanks for this astute observation. I corrected this same typo on the original blog, but forgot to correct it here as well ---- I corrected it just now. It's funny: the underlined splits always get far more scrutiny than the non-underlined parts of the sentence, but of course, in the OA, the whole sentence has to be grammatically correct. Thank you for pointing this out, my friend, and best of luck to you!
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Intern
Joined: 07 Mar 2013
Posts: 10
Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 610 Q45 V29
GMAT 2: 690 Q49 V34
GMAT 3: 710 Q49 V38
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
07 May 2014, 13:00
Hi,
Doesn't "which" modify "him" rather than "equation"?
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4669
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
07 May 2014, 13:15
4
KUDOS
Expert's post
3
This post was
BOOKMARKED
varunmb wrote:
Hi,
Doesn't "which" modify "him" rather than "equation"?
Dear varunmb,
I'm happy to respond.
First of all, the relative pronoun "which" can only refer to an object, not to a person. We would need "who" for a person.
More importantly, the participial phrase "named for him" is a vital noun-modifier, and a vital modifier can come between the noun, "equation," and the non-vital modifier, "which explains how ..." For a thorough explanation of vital noun-modifiers, see this blog:
http://magoosh.com/gmat/2012/gmat-gramm ... modifiers/
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Retired Moderator
Joined: 20 Dec 2013
Posts: 180
Location: United States (NY)
GMAT 1: 640 Q44 V34
GMAT 2: 710 Q48 V40
GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
07 May 2014, 13:40
you're the man Mike! btw, big fan of your blog!
_________________
Manager
Joined: 18 Aug 2013
Posts: 142
Location: India
Concentration: Operations, Entrepreneurship
GMAT 1: 640 Q48 V28
GPA: 3.92
WE: Operations (Transportation)
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
10 Nov 2014, 04:29
hi mike,
can u please tell me what's wrong with option C?
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4669
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
10 Nov 2014, 11:13
4
KUDOS
Expert's post
2
This post was
BOOKMARKED
ynk wrote:
hi mike,
can u please tell me what's wrong with option C?
Dear ynk,
I'm happy to respond. First of all, my friend, I am going to chide you for being so informal in your language. With all due respect, you have presented your request the way a child would. Think about it. What if some other user on GMAT Club reads this, and some day is in a position of power --- perhaps someone who might hire you or some partner with whom you will negotiate a deal. What if your casual and sloppy language now makes an impression on that person? My friend, have respect for yourself. Always put your best forward. Especially in a question about GMAT SC, always strive to maintain the same standards the GMAT itself keeps on SC. Always hold yourself to the highest standards -- that is one of the habits of excellence.
As your question --- here's the SC question again:
Daniel Bernoulli (1700 – 1782) derived the famous fluid equation named after him, to explain an airplane’s wing’s generation of lift, and made a discovery that led to an early method of measuring blood pressure.
(A) equation named after him, to explain an airplane’s wing’s generation of lift
(B) equation named after him, and this principle explains the lift of an airplane’s wing
(C) equation, named it after himself, explained how an airplane’s wing is generating lift
(D) equation named for him, giving an explanation of the generation of the lift of an airplane’s wing
(E) equation named for him, which explains how an airplane’s wing generates lift
There are a few problems with (C). First of all, this makes the mistake of false parallelism. You see, Parallelism is not primarily is a grammatical structure: instead, it is primarily a logical structure. The grammar merely reflects the logic. If we get a string of verbs in a sentence, it is not necessarily correct to put them all into parallel unless they logically belong in parallel. The mistake in (C) reflects a very mechanical understanding of Parallelism, thinking only on the level of grammar and not on the level of logic.
The second problem is the a subtle change in meaning --- the prompt tells us that the fluid equation was "named after" Bernoulli, but exactly did that happen? Did Bernoulli create the equation and name it after himself? Or, did he simply publish the fluid equation, and in time, others referred to it as "Bernoulli's Equation"? We don't really know, so we have to leave that ambiguous. Choice (C) explicitly choose the first option, the option that Bernoulli named the equation after himself, and we don't know that this is the case.
There is also a logical problem in putting the verb "explain" in parallel with the others. Yes, the equation explains an airplane's lift. Was Bernoulli himself trying to explain an airplane's lift? NO! He lived well over a century before airplanes were around, so he couldn't possibly have been trying to explain airplanes. This is an example of the logical problems that false parallelism creates.
Finally, the progressive tense "is generating" is 100% awkward and incorrect. That seems to suggest we are taking about a specific airplane wing, perhaps sitting right outside our room. This is not the sense of the sentence. We are talking about airplane wings in general, so we need the general present tense, not the present progressive.
Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Manager
Joined: 18 Aug 2013
Posts: 142
Location: India
Concentration: Operations, Entrepreneurship
GMAT 1: 640 Q48 V28
GPA: 3.92
WE: Operations (Transportation)
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
10 Nov 2014, 21:22
mikemcgarry wrote:
ynk wrote:
hi mike,
can u please tell me what's wrong with option C?
Dear ynk,
I'm happy to respond. First of all, my friend, I am going to chide you for being so informal in your language. With all due respect, you have presented your request the way a child would. Think about it. What if some other user on GMAT Club reads this, and some day is in a position of power --- perhaps someone who might hire you or some partner with whom you will negotiate a deal. What if your casual and sloppy language now makes an impression on that person? My friend, have respect for yourself. Always put your best forward. Especially in a question about GMAT SC, always strive to maintain the same standards the GMAT itself keeps on SC. Always hold yourself to the highest standards -- that is one of the habits of excellence.
As your question --- here's the SC question again:
Daniel Bernoulli (1700 – 1782) derived the famous fluid equation named after him, to explain an airplane’s wing’s generation of lift, and made a discovery that led to an early method of measuring blood pressure.
(A) equation named after him, to explain an airplane’s wing’s generation of lift
(B) equation named after him, and this principle explains the lift of an airplane’s wing
(C) equation, named it after himself, explained how an airplane’s wing is generating lift
(D) equation named for him, giving an explanation of the generation of the lift of an airplane’s wing
(E) equation named for him, which explains how an airplane’s wing generates lift
There are a few problems with (C). First of all, this makes the mistake of false parallelism. You see, Parallelism is not primarily is a grammatical structure: instead, it is primarily a logical structure. The grammar merely reflects the logic. If we get a string of verbs in a sentence, it is not necessarily correct to put them all into parallel unless they logically belong in parallel. The mistake in (C) reflects a very mechanical understanding of Parallelism, thinking only on the level of grammar and not on the level of logic.
The second problem is the a subtle change in meaning --- the prompt tells us that the fluid equation was "named after" Bernoulli, but exactly did that happen? Did Bernoulli create the equation and name it after himself? Or, did he simply publish the fluid equation, and in time, others referred to it as "Bernoulli's Equation"? We don't really know, so we have to leave that ambiguous. Choice (C) explicitly choose the first option, the option that Bernoulli named the equation after himself, and we don't know that this is the case.
There is also a logical problem in putting the verb "explain" in parallel with the others. Yes, the equation explains an airplane's lift. Was Bernoulli himself trying to explain an airplane's lift? NO! He lived well over a century before airplanes were around, so he couldn't possibly have been trying to explain airplanes. This is an example of the logical problems that false parallelism creates.
Finally, the progressive tense "is generating" is 100% awkward and incorrect. That seems to suggest we are taking about a specific airplane wing, perhaps sitting right outside our room. This is not the sense of the sentence. We are talking about airplane wings in general, so we need the general present tense, not the present progressive.
Does all this make sense?
Mike
Thanks Mike.
I will be more cautious next time.
Manager
Joined: 18 Aug 2013
Posts: 142
Location: India
Concentration: Operations, Entrepreneurship
GMAT 1: 640 Q48 V28
GPA: 3.92
WE: Operations (Transportation)
Re: Daniel Bernoulli derived the famous fluid equation [#permalink]
### Show Tags
10 Nov 2014, 21:28
mikemcgarry wrote:
ynk wrote:
hi mike,
can u please tell me what's wrong with option C?
Dear ynk,
I'm happy to respond. First of all, my friend, I am going to chide you for being so informal in your language. With all due respect, you have presented your request the way a child would. Think about it. What if some other user on GMAT Club reads this, and some day is in a position of power --- perhaps someone who might hire you or some partner with whom you will negotiate a deal. What if your casual and sloppy language now makes an impression on that person? My friend, have respect for yourself. Always put your best forward. Especially in a question about GMAT SC, always strive to maintain the same standards the GMAT itself keeps on SC. Always hold yourself to the highest standards -- that is one of the habits of excellence.
As your question --- here's the SC question again:
Daniel Bernoulli (1700 – 1782) derived the famous fluid equation named after him, to explain an airplane’s wing’s generation of lift, and made a discovery that led to an early method of measuring blood pressure.
(A) equation named after him, to explain an airplane’s wing’s generation of lift
(B) equation named after him, and this principle explains the lift of an airplane’s wing
(C) equation, named it after himself, explained how an airplane’s wing is generating lift
(D) equation named for him, giving an explanation of the generation of the lift of an airplane’s wing
(E) equation named for him, which explains how an airplane’s wing generates lift
There are a few problems with (C). First of all, this makes the mistake of false parallelism. You see, Parallelism is not primarily is a grammatical structure: instead, it is primarily a logical structure. The grammar merely reflects the logic. If we get a string of verbs in a sentence, it is not necessarily correct to put them all into parallel unless they logically belong in parallel. The mistake in (C) reflects a very mechanical understanding of Parallelism, thinking only on the level of grammar and not on the level of logic.
The second problem is the a subtle change in meaning --- the prompt tells us that the fluid equation was "named after" Bernoulli, but exactly did that happen? Did Bernoulli create the equation and name it after himself? Or, did he simply publish the fluid equation, and in time, others referred to it as "Bernoulli's Equation"? We don't really know, so we have to leave that ambiguous. Choice (C) explicitly choose the first option, the option that Bernoulli named the equation after himself, and we don't know that this is the case.
There is also a logical problem in putting the verb "explain" in parallel with the others. Yes, the equation explains an airplane's lift. Was Bernoulli himself trying to explain an airplane's lift? NO! He lived well over a century before airplanes were around, so he couldn't possibly have been trying to explain airplanes. This is an example of the logical problems that false parallelism creates.
Finally, the progressive tense "is generating" is 100% awkward and incorrect. That seems to suggest we are taking about a specific airplane wing, perhaps sitting right outside our room. This is not the sense of the sentence. We are talking about airplane wings in general, so we need the general present tense, not the present progressive.
Does all this make sense?
Mike
Thanks Mike.
I will be more cautious next time.
MBA Section Director
Joined: 19 Mar 2012
Posts: 5106
Location: India
GMAT 1: 760 Q50 V42
GPA: 3.8
WE: Marketing (Non-Profit and Government)
#Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
19 Oct 2015, 17:00
Expert's post
3
This post was
BOOKMARKED
Daniel Bernoulli (1700 – 1782) derived the famous fluid equation named after him, to explain an airplane’s wing’s generation of lift, and made a discovery that led to an early method of measuring blood pressure.
A. equation named after him, to explain an airplane’s wing’s generation of lift
B. equation named after him, and this principle explains the lift of an airplane’s wing
C. equation, named it after himself, explained how an airplane’s wing is generating lift
D. equation named for him, giving an explanation of the generation of the lift of an airplane’s wing
E. equation named for him, which explains how an airplane’s wing generates lift
_________________
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 4258
Location: India
WE: Education (Education)
Re: #Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
20 Oct 2015, 02:41
1
KUDOS
Expert's post
In the 18th century, Daniel cold not have explained some airplane’s feature, the plane itself having been invented in the 20th century by the Wrights brothers.
A. equation named after him, to explain an airplane’s wing’s generation of lift --1. A is out because it is out of scope for the above-stated reason 2. An airplane’s wing’s’ is very weird
B. equation named after him, and this principle explains the lift of an airplane’s wing – A past tense clause and then a present tense clause and then again a past tense clause – and two ands to separate three items—all these things make this choice awful.
C. equation, named it after himself, explained how an airplane’s wing is generating lift --1. This is invalid for the same reason as in A. 2. Named it after himself implies that Daniel himself named it after his ownself.
D. equation named for him, giving an explanation of the generation of the lift of an airplane’s wing – The adverbial modifier, ‘giving an explanation of the generation’ modifies Daniel’ deriving the fluid equation; this is logically incorrect.
E. equation named for him, which explains how an airplane’s wing generates lift, --- may be the best since the present tense of the modifier indicates that the reference to the airplane is a thing of present times and that when the modifier when removed still carries the original meaning intact.
Only one doubt though. Is 'named for him', a correct idiom?
_________________
Can you solve at least some SC questions without delving into the initial statement?
Narendran 98845 44509
Retired Moderator
Joined: 18 Sep 2014
Posts: 1175
Location: India
#Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
Updated on: 22 Oct 2015, 11:00
1
KUDOS
Daniel Bernoulli (1700 – 1782)
derived the famous fluid equation named after him, to explain an airplane’s wing’s generation of lift, and
made a discovery that led to an early method of measuring blood pressure.
airplane’s wing’s generation.........weird
B. equation named after him, and this principle explains the lift of an airplane’s wing
and this principle refers to equation isntead of Bernoulli principle
use of and in modifier phrase appears incorrect
x and y and z structure is also not proper.
C. equation, named it after himself, explained how an airplane’s wing is generating lift
forced parallelism is shown here.
Daniel Bernoulli (1700 – 1782)
• derived the famous fluid equation,
• named it after himself,
• explained how an airplane’s wing is generating lift and
• made a discovery that led to an early method of measuring blood pressure.
All these are not different successive events but interrelated. Incorrect choice.
D. equation named for him, giving an explanation of the generation of the lift of an airplane’s wing...........
he derived equation by
giving an explanation of wing lift and
This does not make sense for me.
......
E. equation named for him, which explains how an airplane’s wing generates lift......
Daniel Bernoulli (1700 - 1782)
derived the famous fluid equation named for him, which explains how an airplane’s wing generates lift, and
made a discovery that led to an early method of measuring blood pressure.
made a discovery correctly refers to Daniel Bernouli
Thanks Mike
_________________
The only time you can lose is when you give up. Try hard and you will suceed.
Thanks = Kudos. Kudos are appreciated
http://gmatclub.com/forum/rules-for-posting-in-verbal-gmat-forum-134642.html
When you post a question Pls. Provide its source & TAG your questions
Avoid posting from unreliable sources.
My posts
http://gmatclub.com/forum/beauty-of-coordinate-geometry-213760.html#p1649924
http://gmatclub.com/forum/calling-all-march-april-gmat-takers-who-want-to-cross-213154.html
http://gmatclub.com/forum/possessive-pronouns-200496.html
http://gmatclub.com/forum/double-negatives-206717.html
http://gmatclub.com/forum/the-greatest-integer-function-223595.html#p1721773
Originally posted by Nevernevergiveup on 20 Oct 2015, 03:18.
Last edited by Nevernevergiveup on 22 Oct 2015, 11:00, edited 2 times in total.
Manager
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 247
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32
GMAT 2: 680 Q47 V35
Re: #Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
20 Oct 2015, 03:34
daagh wrote:
In the 18th century, Daniel cold not have explained some airplane’s feature, the plane itself having been invented in the 20th century by the Wrights brothers.
A. equation named after him, to explain an airplane’s wing’s generation of lift --1. A is out because it is out of scope for the above-stated reason 2. An airplane’s wing’s’ is very weird
B. equation named after him, and this principle explains the lift of an airplane’s wing – A past tense clause and then a present tense clause and then again a past tense clause – and two ands to separate three items—all these things make this choice awful.
C. equation, named it after himself, explained how an airplane’s wing is generating lift --1. This is invalid for the same reason as in A. 2. Named it after himself implies that Daniel himself named it after his ownself.
D. equation named for him, giving an explanation of the generation of the lift of an airplane’s wing – The adverbial modifier, ‘giving an explanation of the generation’ modifies Daniel’ deriving the fluid equation; this is logically incorrect.
E. equation named for him, which explains how an airplane’s wing generates lift, --- may be the best since the present tense of the modifier indicates that the reference to the airplane is a thing of present times and that when the modifier when removed still carries the original meaning intact.
Only one doubt though. Is 'named for him', a correct idiom?
Hey daagh, quick questions on E:
1) What does 'Which' modify? I thought it modifies 'Him'
2) Who explains- the equation or the Mr. Daniel?
3) Your explanation for B- "A past tense clause and then a present tense clause.. ". Doesnt it apply here? Derived.. explains ??
_________________
One Kudos for an everlasting piece of knowledge is not a bad deal at all...
------------------------------------------------------------------------------------------------------------------------
Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover.
-Mark Twain
Manager
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 247
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32
GMAT 2: 680 Q47 V35
Re: #Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
20 Oct 2015, 03:47
A for me:
1) Eliminating D and E:
Wrong usage 'named for him', i think 'named after him' is better.
D and E are out
2) Eliminating B and C
B. equation named after him, and this principle explains the lift of an airplane’s wing
What does 'this' refer to? There is no 'Principle'.
Tens issue- Derived... Named... Explains
C. equation, named it after himself, explained how an airplane’s wing is generating lift
the parallel construction with commas gives a list of below three items-
i) Derived the equation
ii) Named it
iii) Explained
This indicated that the 'Explanation' was independent of 'Derivation'. This is not true. The explanation was a result of derivation.
A is the best choice.
Please correct me if im wrong.
_________________
One Kudos for an everlasting piece of knowledge is not a bad deal at all...
------------------------------------------------------------------------------------------------------------------------
Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover.
-Mark Twain
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 4258
Location: India
WE: Education (Education)
Re: #Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
20 Oct 2015, 04:04
arhumsid
1) What does 'Which' modify? I thought it modifies ‘Him’ --- which cannot modify a person or a personal pronoun. It no doubt points to the equation; it may be noted that there is no other noun it can modify logically.
2) Who explains- the equation or the Mr. Daniel? --- Where is the word ‘who’ in the choice? I didn’t get you.
3) Your explanation for B- "A past tense clause and then a present tense clause. “ Doesn’t it apply here? Derived. explains ?? I did not get this point also.
I only said that this sentence construction is awful since there is no uniformity in the use of tenses. The first sentence is in past tense and the second sentence is in present tense and the third is again in past tense. In addition, when there are three items, only the last item should be separated by an ‘and’ Thirdly, if the derived equation explains the airplane thing, then it is wrong for the logic explained about the chronology of events.
On the contrary, In E, the modifier is in present tense. This means that other modern people have derived the conclusion about the airplane wing presently.
_________________
Can you solve at least some SC questions without delving into the initial statement?
Narendran 98845 44509
Manager
Status: One Last Shot !!!
Joined: 04 May 2014
Posts: 247
Location: India
Concentration: Marketing, Social Entrepreneurship
GMAT 1: 630 Q44 V32
GMAT 2: 680 Q47 V35
Re: #Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
20 Oct 2015, 05:25
daagh wrote:
arhumsid
1) What does 'Which' modify? I thought it modifies ‘Him’ --- which cannot modify a person or a personal pronoun. It no doubt points to the equation; it may be noted that there is no other noun it can modify logically.
2) Who explains- the equation or the Mr. Daniel? --- Where is the word ‘who’ in the choice? I didn’t get you.
3) Your explanation for B- "A past tense clause and then a present tense clause. “ Doesn’t it apply here? Derived. explains ?? I did not get this point also.
I only said that this sentence construction is awful since there is no uniformity in the use of tenses. The first sentence is in past tense and the second sentence is in present tense and the third is again in past tense. In addition, when there are three items, only the last item should be separated by an ‘and’ Thirdly, if the derived equation explains the airplane thing, then it is wrong for the logic explained about the chronology of events.
On the contrary, In E, the modifier is in present tense. This means that other modern people have derived the conclusion about the airplane wing presently.
Sorry for being unclear. Here are the doubts:
1) Usage of which: Which modifies the entity it follows. It can never modify clauses. And, it can never modify people (Please correct me if im wrong)
in option E "equation named for him, which explains"
Isnt 'Which' wrongly modifying 'him' when it should be modifying 'Equation'?
2) Now, if, after your explanation of point 1 above, we conclude that 'which' correctly refers to 'equation' and not to 'him', E wrongly suggests that the equation explains how.... But equation is not the one who explains, Mr. Bernoulli is (through this equation), right?
3) (We will skip the tense part as im more worried about the above two issues)
Please let me know where i am going wrong
Thanks!
_________________
One Kudos for an everlasting piece of knowledge is not a bad deal at all...
------------------------------------------------------------------------------------------------------------------------
Twenty years from now you will be more disappointed by the things you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover.
-Mark Twain
Retired Moderator
Status: worked for Kaplan's associates, but now on my own, free and flying
Joined: 19 Feb 2007
Posts: 4258
Location: India
WE: Education (Education)
Re: #Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink]
### Show Tags
20 Oct 2015, 07:55
@ arhunsid
Sorry for being unclear. Here are the doubts:
1) Usage of which: Which modifies the entity it follows. It can never modify clauses. And, it can never modify people (Please correct me if I’m wrong) ----- ‘Which’ always modifies the noun entity before and not what it follows. Even in that touch rule, there is some flexibility. ‘Which’ need not always modify the noun just in front, but sometimes a noun a few more words before.
in option E "equation named for him, which explains"
Isn't 'Which' wrongly modifying 'him' when it should be modifying 'Equation'? --- No. "which" is not modifying him; it is only modifying equations, because of logic.
2) Now, if, after your explanation of point 1 above, we conclude that 'which' correctly refers to 'equation' and not to 'him', E wrongly suggests that the equation explains how.... But equation is not the one who explains, Mr. Bernoulli is (through this equation), right? --- No you are reading too much. Equation can explain. Why not? We do accept the theory: The Sun is the centre of the Solar system. We don’t even bother to know who the author of this theory is. So equation explaining some bla bla is acceptable.
HTH
_________________
Can you solve at least some SC questions without delving into the initial statement?
Narendran 98845 44509
Re: #Top150 SC: Daniel Bernoulli (1700 – 1782) derived the famous fluid [#permalink] 20 Oct 2015, 07:55
Go to page 1 2 3 Next [ 44 posts ]
Display posts from previous: Sort by | 7,983 | 33,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-22 | longest | en | 0.921974 |
https://www.indigolearn.com/forums/Interest-24-b-in-case-of-self-occupied-Property/395c291c96d541e2bf0c86d0b11bc77d | 1,618,507,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038087714.38/warc/CC-MAIN-20210415160727-20210415190727-00433.warc.gz | 921,664,467 | 17,005 | # IndigoLearn
##### IndigoLearn
Forums
Interest 24(b) in case of self occupied Property
Self occupied Property What is the maximum amount of interest can be deducted under section 24 ( b) 200000 or 30000 as the case may be Or 2 lacs for construction or acquisition and if renovation, renewal, reconstruction, repair includes 30000 over and above that 2 lakh Can I get rupees 230000 as deduction under section 24 (b) if I have taken a loan for renovation reconstruction and also for purchasing house
Gayathri K V
CA Inter
5K+
31-Mar-21 19:45
24
No. loan taken take is considered to determine 30,000 or 2,00,000..... the maximum interest deduction is Rs.2lacs
sudha reddy
Moderator
01-Apr-21 16:37
No. loan taken take is considered to determine 30,000 or 2,00,000..... the maximum interest deduction is Rs.2lacs
While answering to this question (in step. 1 computation of income if these are self occupied property) For the second house property interest on capital borrowed is taken RS 41000 as production (Rs.11000 interest on borrowed capital towards purchase of such Property +and Rs. 7,89,000 interest on capital borrowed towards repair, renewal or reconstruction In answering this question 11,000 + 30000 is equal to 41000 is taken) So you tell me that only 30000 or 200000 as the case may be is taken but in this case 41000 taken. Why that happened?
Attachments
illustration10.jpeg
illustration10. ans.jpg
Gayathri K V
CA Inter
5K+
03-Apr-21 06:45
Will check and revert
First proviso to Sec 24(b) states self occupied property - Max Rs.30,000 Second proviso to Sec 24(b) states self occupied property, loan after 1/4/99 for purchase / construction - Max 2,00,000 Fourth proviso to Sec 24(b) states self occupied property, - deduction under first proviso + second proviso - Max 2,00,000 Sec 24(b) states - interest deduction if let out - No limit for deduction So in the illustration, if House 2 is considered self occupied - First proviso Rs.30,000 and Second proviso Rs.11,000 is considered
sudha reddy
Moderator
06-Apr-21 12:59
# IndigoLearn
## Information
Forums
FAQ
Blogs
Terms & Conditions | 564 | 2,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-17 | latest | en | 0.892994 |
http://www.jiskha.com/display.cgi?id=1278890578 | 1,498,264,413,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320206.44/warc/CC-MAIN-20170623235306-20170624015306-00129.warc.gz | 544,081,562 | 3,518 | # algebra
posted by .
write a quadratic equation in the variable x having the given numbers as the solutions. type the equation in standard form ax^2+bx+c=0
solution 6, only solution
• algebra -
X = 6
X - 6 = 0
Square both sides of Eq.
X^2 - 12X + 36 = 0 | 80 | 259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-26 | latest | en | 0.905809 |
https://us.metamath.org/mpeuni/dfss2f.html | 1,716,086,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00474.warc.gz | 539,791,745 | 4,436 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > dfss2f Structured version Visualization version GIF version
Theorem dfss2f 3937
Description: Equivalence for subclass relation, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 3-Jul-1994.) (Revised by Andrew Salmon, 27-Aug-2011.) Avoid ax-13 2390. (Revised by Gino Giotto, 19-May-2023.)
Hypotheses
Ref Expression
dfss2f.1 𝑥𝐴
dfss2f.2 𝑥𝐵
Assertion
Ref Expression
dfss2f (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
Proof of Theorem dfss2f
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 dfss2 3933 . 2 (𝐴𝐵 ↔ ∀𝑧(𝑧𝐴𝑧𝐵))
2 dfss2f.1 . . . . 5 𝑥𝐴
32nfcriv 2963 . . . 4 𝑥 𝑧𝐴
4 dfss2f.2 . . . . 5 𝑥𝐵
54nfcriv 2963 . . . 4 𝑥 𝑧𝐵
63, 5nfim 1897 . . 3 𝑥(𝑧𝐴𝑧𝐵)
7 nfv 1915 . . 3 𝑧(𝑥𝐴𝑥𝐵)
8 eleq1w 2893 . . . 4 (𝑧 = 𝑥 → (𝑧𝐴𝑥𝐴))
9 eleq1w 2893 . . . 4 (𝑧 = 𝑥 → (𝑧𝐵𝑥𝐵))
108, 9imbi12d 347 . . 3 (𝑧 = 𝑥 → ((𝑧𝐴𝑧𝐵) ↔ (𝑥𝐴𝑥𝐵)))
116, 7, 10cbvalv1 2361 . 2 (∀𝑧(𝑧𝐴𝑧𝐵) ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
121, 11bitri 277 1 (𝐴𝐵 ↔ ∀𝑥(𝑥𝐴𝑥𝐵))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 208 ∀wal 1535 ∈ wcel 2114 Ⅎwnfc 2957 ⊆ wss 3913 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1796 ax-4 1810 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-11 2161 ax-12 2177 ax-ext 2792 This theorem depends on definitions: df-bi 209 df-an 399 df-ex 1781 df-nf 1785 df-sb 2070 df-clab 2799 df-cleq 2813 df-clel 2891 df-nfc 2959 df-v 3475 df-in 3920 df-ss 3930 This theorem is referenced by: dfss3f 3938 ssrd 3951 ssrmof 4011 ss2ab 4018 rankval4 9274 rabexgfGS 30247 ballotth 31803 dvcosre 42345 itgsinexplem1 42387
Copyright terms: Public domain W3C validator | 993 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-22 | latest | en | 0.185109 |
https://www.scribd.com/document/62271466/Matlab-Lecture-2 | 1,569,108,342,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574710.66/warc/CC-MAIN-20190921231814-20190922013814-00070.warc.gz | 996,595,880 | 61,891 | You are on page 1of 7
# Entering Matrices MATLAB
Tutorial Enter an explicit list of elements Load matrices from files Generate them using built-in functions Create them with your own functions
## When entering matrices
Separate elements with blanks or commas Use semicolon ; to indicate end of row List enclosed in square brackets [ ]
Example
>> A = [21, 22, 23;39,40,41;71,72,73] A= 21 39 71 22 40 72 23 41 73
## Some Useful Matrix Functions
sum(A) sums up the columns A transpose diag(A) main diagonal sum(A) sum of rows sum(diag(A)) sum of main diagonal sum(diag(fliplr(A))) sum of antidiagonal If no output variable is specified, MATLAB uses ans
Example
>> sum(A) ans = 131 134 137
## Accessing a specific element
A(row, column) to access a specific element of matrix A A(k) is the usual way of referencing row and column vectors A(k) can also be used to reference a single element of a 2D Matrix
>> A(2,1) ans = 39 >> A(2)
Examples
ans = 39
## Outside the Matrix
Error if you try to read from a value of an element outside the matrix If you write to it on the other hand, the matrix will expand to accommodate it
Example
>> A(2,4) = 42 A= 21 39 71 22 40 72 23 41 73 0 42 0
Sequences
1:7 is a row vector 1 2 3 4 5 6 7 Initial:increment:final >> 0:pi/4:pi ans = 0 0.7854 1.5708 2.3562 3.1416
Portions of a matrix
>> A(2,2:4) ans = 40 41 42
## All and end
: by itself means all End means last row or last column >> A(end,:) ans = 71 72 73 0
## Vectors and Scalars
A matrix with one row or one column is a vector A scalar is a one by one matrix >> num_students = 25 num_students = 25
Built-in Functions
help elfun: list of elementary functions help specfun, help elmat for more advanced matrix functions
Examples of Expression
rho = (1+sqrt(5))/2 rho = 1.6180 a = abs(3+4i) a= 5 z = sqrt(besselk(4/3,rho-i)) z= 0.3730+ 0.3214i huge = exp(log(realmax)) huge = 1.7977e+308 toobig = pi*huge toobig = Inf
Generating Matrices
zeros: all zeroes ones: all ones rand: uniformly distributed random elements randn: normally distributed random elements
>> randn(4,4) ans =
Example
-2.1707 0.5077 0.3803 0.0000 -0.0592 1.6924 -1.0091 -0.3179 -1.0106 0.5913 -0.0195 1.0950 0.6145 -0.6436 -0.0482 -1.8740
Data Files
Code Files
M-files are text files containing MATLAB code can be created using MATLAB editor or any other text editor have a .m extension
Concatenation
Joining small matrices to make bigger one >> big = [zeros(2,2) ones(2,2);ones(2,2) zeros(2,2)] big = 0 0 1 1 0 0 1 1 1 1 0 0 1 1 0 0
Deleting
You can delete rows and columns using the square brackets [ ] X(:,2) = [] deletes the second column of the matrix You can delete a single element from a matrix; however, you would end up reshaping the remaining elements into a row vector
Operators
A + A adding a matrix to a transpose produces a symmetric matrix * denotes matrix multiplication involving inner products between rows and columns A * A multiplying a transpose by the original produces a symmetric matrix use .* for element-by-element multiplication ./ element-by-element division .^ element-by-element power
Building Tables
>> n = (0:4)'; >> rows = [n n.^2 2.^n] rows = 0 1 2 3 4 0 1 4 9 16 1 2 4 8 16
Logical Subscripting
X= 6 14 32 50 NaN
Another Example
Highlight primes by setting non-primes to zero: >> A(~isprime(A)) = 0 A= 0 0 71 0 23 0 41 0 73
>> X = X(finite(X)) X= 6 14 32 50
find
find picks out the locations, using onedimensional indexing >> k = find(isprime(A))' k= 3 7 8 9
Display
help format: to display different ways of formatting displayed numbers use semicolons ; to suppress output use to continue a statement on another line
if
if <condition> <action> elseif <condition> <action> else <action> end
example
if rem(n,2) ~= 0 M = odd_magic(n) elseif rem(n,4) ~= 0 M = single_even_magic(n) else M = double_even_magic(n) end
## Some functions helpful for if
isequal checks if two matrices are equal == will not do that, it will check where they are equal instead isempty all any
## switch and case
switch (rem(n,4)==0) + (rem(n,2)==0) case 0 M = odd_magic(n) case 1 M = single_even_magic(n) case 2 M = double_even_magic(n) otherwise error('This is impossible') end
for
for n = 3:32 r(n) = rank(magic(n)); end r while <condition> . . . end
while
example
while T < Number_of_iterations % pheromone values ^ Alpha Weighted_pheromones = Pheromone .^ Alpha; % initialise the ants' memory to zeros Ants_memory = zeros(Number_of_departments, Number_of_departments); % move the ants one interation (please examine function move_ants) [Ants_memory, Best_Cost, Optimal_Assignment] = ... move_ants(Number_of_departments, Number_of_ants, ... Weighted_pheromones, Weighted_distance_costs, ... Given_flow, Given_distance, Ants_memory, Best_Cost, ... Q, Optimal_Assignment); % update the pheromone values using the ants' memory Pheromone = update_pheromone_levels(Lambda, Pheromone, Ants_memory); T = T + 1; end
functions
functions are M-files can accept input arguments return output arguments first line starts with keyword function it gives function name & order of arguments
example
% function accepts: % number of departments % list of probabilities % returns a chosen site based on the corresponding probabilities % represented in the probabilities vector function Chosen_site = choose(Number_of_departments, Probabilities) % random number between 0 and 1 Random_number = rand; % calculate cumulative probabilities for i = 1:Number_of_departments - 1 Probabilities(i + 1) = Probabilities(i + 1) + Probabilities(i); end % take whatever is before where the random number falls Chosen_site = Number_of_departments + 1 - sum(Random_number < Probabilities);
global variables
if you want more than one function to share a variable global declaration must occur before the variable is actually used in the function capital letters is not a must, just a convention
example
function h = falling(t) global GRAVITY h = 1/2*GRAVITY*t.^2; | 1,738 | 5,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-39 | latest | en | 0.694647 |
http://docplayer.net/20683135-Give-students-a-lesson-in-navigation-from-the-points-of-the-compass-to-gps-mapping.html | 1,545,133,005,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829140.81/warc/CC-MAIN-20181218102019-20181218124019-00618.warc.gz | 86,822,093 | 26,252 | # Give students a lesson in navigation from the points of the compass to GPS mapping.
Save this PDF as:
Size: px
Start display at page:
## Transcription
1 Lesson 5: Teaching Directions, Maps, and Coordinates Give students a lesson in navigation from the points of the compass to GPS mapping. by Jenny Parma; curriculum by Diane Petersen and Karen Dvornich Teach your students some basic and more complex directional skills so they can navigate nature and the greater biosphere. This lesson starts with teaching basic directions and mapping techniques, then moves on to taking latitude and longitude coordinates and using GPS units. Directions and Maps In two parts, teach your students about the cardinal directions and how to use a compass and maps. The lesson includes outdoor activities. Part 1: Directions 1. Sit in a circle outside. Ask students»»where s the sun?»»what direction is the sun?»»which way is north? 2. Use an object, such as a branch, to depict north. Explain the other cardinal directions and use a mnemonic device to aid students memory, such as the sentence Never eat soggy Wheaties Ask students to use materials in the area to mark the other directions. Explain what a compass is and how it functions. 5. Use a compass to test the accuracy of where students placed objects to mark directions in step 3. Make any needed corrections. 6. Say the different directions out loud, and ask students to point where the direction is on a compass wheel or on a makeshift wheel outside. You can also hand out cards with directions for students to place on the wheel. For older students, ask them to determine more precise directions, such as northwest or southeast. 7. Go inside and ask students to find north and other directions in the classroom. 8. Have the students make direction markers and post them on the walls of the classroom. Double-check the location with a compass. Lesson Objectives and Materials Objectives»»Understand cardinal directions»»practice using maps»»learn how to use a compass»»use a GPS unit and understand latitude/ longitude coordinates Materials»»NM data-collection form»»field guides or animal fact sheets»»compass»»compass wheel»»different maps»»field journals (bound scientific notebooks)»»a 4-foot square piece of colored paper»»transparent tape»»note cards»»globe»»gps unit 1
### Chapter 2 Reading Topographic Maps and Making Calculations
Chapter 2 Reading Topographic Maps and Making Calculations In this chapter you will learn about: Reading the margins Interpreting contour lines Estimating slope Estimating aspect Estimating acreage Estimating
### World Geography Unit 1 - Test Introduction to World of Geography
Name World Geography Unit 1 - Test Introduction to World of Geography Term Identification Directions: Pick a word from the box that best completes the sentences below. distortion longitude Equator geography
### Topo Map Mania! Map, Compass, Scale, Legend, Bearing
Topo Map Mania! Unit Topic: Navigation Grade Level: 7 th grade (with suggestions to scale for grades 6 to 8) Lesson No. 5 of 10 Lesson Subject(s): Map, Compass Key Words: Lesson Abstract Map, Compass,
### Where in the World is the arctic?
Where in the World is the arctic? Summary: Students map the arctic in relation to their home in order to learn the location and countries of the arctic. Grade Level: 3-4; 5-8; K-2 Time one class period.
### Using Latitude and Longitude - Lesson 1. How can I remember which lines are which?
Using Latitude and Longitude - Lesson 1 How can I remember which lines are which? Latitude lines are imaginary lines on the earth's surface. They run east and west around the globe and tell you your distance
### On October 14, 2010, I taught a social studies lesson about Latitude and Longitude to 5 th
Phylicia Kelly Professor Schilling EDUC. 327 October 21, 2010 Student Teaching Reflection On October 14, 2010, I taught a social studies lesson about Latitude and Longitude to 5 th grade students at my
### A Latitude/Longitude Puzzle
This website would like to remind you: Your browser (Safari 7) is out of date. Update your browser for more security, comfort and the best experience on this site. Activitydevelop A Latitude/Longitude
### MAP AND COMPASS. Background Information
MAP AD COMPASS Background Information A compass is an instrument that people use to find a direction in relation to the earth as a whole. The magnetic needle in the compass, which is the freely moving
### Models in earth science provide opportunities to investigate and obtain evidence on which to base scientific explanations.
Landforms Lesson 2 Fifth-Sixth Grade Models in earth science provide opportunities to investigate and obtain evidence on which to base scientific explanations. Continue establishing working word wall throughout
### Where is Here? Lesson Abstract
Where is Here? Unit Topic: Navigation Grade Level: 7 th grade (with suggestions to scale for grades 6 to 8) Lesson No. 1 of 10 Lesson Subject(s): Grids,,, Compass Key Words: Grids,,, Compass, Directions,
### Where In The World Am I?
here In The orld Am I? Help your students accurately locate their town and other U.. cities on globes or maps by using lines of latitude and longitude. kill: Using a latitude-longitude grid to locate cities
### Learning about GPS and GIS
Learning about GPS and GIS Standards 4.4 Understand geographic information systems (G.I.S.). B12.1 Understand common surveying techniques used in agriculture (e.g., leveling, land measurement, building
### Types of Maps. physical map. political map. thematic map. contour map. 3-Part Cards with Definitions
Types of Maps 3-Part Cards with Definitions physical map political map contour map thematic map physical map political map A physical map shows an area's natural A political map shows boundaries of features
### Lesson 6: Coming Up with a Monitoring Plan Watershed Mapping and Analysis
Lesson 6: Coming Up with a Monitoring Plan Watershed Mapping and Analysis Time Frame: Two 50-minute class periods Grade Level: 8 th 12 th Grade Overview: Many environmental and human factors exist that
### SECOND GRADE 1 WEEK LESSON PLANS AND ACTIVITIES
SECOND GRADE 1 WEEK LESSON PLANS AND ACTIVITIES UNIVERSE CYCLE OVERVIEW OF SECOND GRADE UNIVERSE WEEK 1. PRE: Discovering stars. LAB: Analyzing the geometric pattern of constellations. POST: Exploring
### LAB 4: Topographic Maps & Google Earth
Name School LAB 4: Topographic Maps & Google Earth Our earth is a very complex place to view as a flat map. Viewing the earth, as a globe is the best representation we can have it is both proportionate
### Topographic Maps Practice Questions and Answers Revised October 2007
Topographic Maps Practice Questions and Answers Revised October 2007 1. In the illustration shown below what navigational features are represented by A, B, and C? Note that A is a critical city in defining
### Tsunamis with a Latitude
Tsunamis with a Latitude Overview: Globes, which are spherical maps of Earth, can be useful when learning about tsunamis. All maps, including globes, use a system of lines, including latitude and longitude,
### What is a map? A Map is a two or three-dimensional model or representation of the Earth s surface. 2-Dimensional map
What is a map? A Map is a two or three-dimensional model or representation of the Earth s surface. 2-Dimensional map Types of Maps: (just a few that we may be using) Political Maps Shows borders of states,
### Using Topographic Maps
Using Topographic Maps Procedure This lab has three parts: (1) Making a contour map of Turtle Island, an imaginary island with known spot elevations, (2) constructing a topographic profile across Turtle
### Geography Skills Pre- Test
Name: period: Geography Skills Pre- Test 1. What does the relative location of a place tell you? A. The longitude and latitude of a place B. Which symbol shows a place on a map C. Where a place is compared
### OBJECTIVES MATERIALS BACKGROUND. Maps and Mapping. Key Terms. Common units and conversions. Introduction. Page 1 of 10
OBJECTIVES This exercise involves understanding map scale and coordinate systems. You will learn to read geographic coordinates in latitude/longitude, the State Plane Coordinate System, and the Universal
### Grade Level: 1-2 Expedition N 25 Pre-Lesson
Time Allotment 45 minutes Materials Per student: 1 printout of world map with latitude and longitude lines Printouts of different animals Glue stick Colored crayons or colored pencils Advance Preparation
### GPS Coordinates Common Formats & Conversions
GPS Coordinates Common Formats & Conversions The most common questions asked in dealing with flight locating is reading and using GPS coordinates. This is not surprising since a GPS coordinate can be represented
### PEOPLE & SITES PEOPLE BIOGRAPHIES
16 PEOPLE & SITES The People & Sites sections contain fields to help you provide contextual information for your collections. Artifacts, archival materials, photographs, and publications reveal something
### Working with USGS Topographic Maps a Tutorial GEOG 310 Fall 2013
Working with USGS Topographic Maps a Tutorial GEOG 310 Fall 2013 This tutorial uses examples from the southeast corner of the Durango West, Colorado 1:24,000 scale topographic map. The steps listed under
### Introduction to Geography
High School Unit: 1 Lesson: 1 Suggested Duration: 3 days Introduction to Lesson Synopsis: The purpose of this lesson is to introduce students to geography and geography terminology, to teach students about
### GEOGRAPHIC INFORMATION SYSTEMS Lecture 11: Projected Coordinate Systems
UTM Coordinate System GEOGRAPHIC INFORMATION SYSTEMS Lecture 11: Projected Coordinate Systems Why do we need the UTM coordinate system? - in a rectangular (Cartesian) coordinate system, with linear x and
### Lesson: Latitude and Longitude. Length: 45 minutes to 1 hour
Phylicia Kelly Professor Schilling EDUC. 327 October 27, 2010 Middle Childhood Lesson Plan Lesson: Latitude and Longitude Length: 45 minutes to 1 hour Source: Using Latitude and Longitude http://geographyworldonline.com/tutorial/lesson1.html
### Earth Coordinates & Grid Coordinate Systems
Earth Coordinates & Grid Coordinate Systems How do we model the earth? Datums Datums mathematically describe the surface of the Earth. Accounts for mean sea level, topography, and gravity models. Projections
### CONTENTS. Page 3 What is orienteering? Page 4 Activity: orienteering map bingo. Page 5 Activity: know your colours. Page 6 Choosing your compass
THE RIGHT DIRECTION SCOUT ORIENTEER ACTIVITY BADGE CONTENTS Page What is orienteering? Page 4 Activity: orienteering map bingo Page 5 Activity: know your colours Page 6 Choosing your compass Page 7 Activity:
### GEOG 101: How To Use Google Earth Pro
GEOG 101: How To Use Google Earth Pro Topographic maps have long been a tool for the geographer. As technology has progressed, new tools have become available to us that allow us to spatially analyze our
### Topographic Maps. Review
Topographic Maps Review Topographic Maps Types of Topographic Maps 1:24,000-7.5 minute maps 1:50,000-15 minute maps 1:100,000 maps 30x60 minute maps 1:250,000 maps 1 o x 2 o degree maps 1:500,000 maps
### ELEMENTS OF SURVEYING FOR CADASTRAL MAPPING
ELEMENTS OF SURVEYING FOR CADASTRAL MAPPING Chapter 4 2015 Cadastral Mapping Manual 4-0 Elements of Surveying and Mapping Utah's system of land surveying is the rectangular survey system as set forth on
### Weekend Cabin Retreat Project Site Plans
Weekend Cabin Retreat Project Site Plans Sacramento City College EDT 300/ENGR 306 EDT 300/ENGR 306 - Site Plans 1 Cabin Project Site Plan/Bubble Diagram - Assignment 1 =10-0 Floor Plan - Assignment 1/4
### OA3-10 Patterns in Addition Tables
OA3-10 Patterns in Addition Tables Pages 60 63 Standards: 3.OA.D.9 Goals: Students will identify and describe various patterns in addition tables. Prior Knowledge Required: Can add two numbers within 20
### BIOL 2406 Fall 2011 Topographic Maps
BIOL 2406 Fall 2011 Topographic Maps A topographic map is a two dimensional representation of a three dimensional land surface. Elevation or relief is shown through the use of contour lines that are continuous
### Averaging GPS Data in My World to Refine the Positions of Field Sites
Steven Branting June 2010 Averaging GPS Data in My World to Refine the Positions of Field Sites The inherent inaccuracy of low-end GPS units (often limited to 20-30 feet) can pose serious issues for field
### PNL Lesson 5 PICKING COORDINATES FROM PAPER MAPS TEACHER INFORMATION
www.mapteach.org Place Names and Landmarks Lesson 4 PNL Lesson 5 PICKING COORDINATES FROM PAPER MAPS TEACHER INFORMATION Lesson Summary: Students identify place names or landmarks on a topographic map
### UTM SYSTEM. COMPOSED AND PRODUCED BY : Chris Rule. Permission is granted for use by acknowledging the author s name.
COMPOSED AND PRODUCED BY : Chris Rule. Permission is granted for use by acknowledging the author s name. UTM stands for Universal Transverse Mercator which is a technical term for the type of map projection.
### Lesson 1: The Location of California
Lesson 1: The Location of California Focus Question: Where in the world is California? California History-Social Science Standard 4.1 Students demonstrate an understanding of the physical and human geographic
### Drawing a site plan. You will need - graph paper, tracing paper, drawing boards, pencils, rulers, compass, tapemeasures as required (5m and 30m/50m)*
Drawing a site plan You will need - graph paper, tracing paper, drawing boards, pencils, rulers, compass, tapemeasures as required (5m and 30m/50m)* Step 1 Setting up a baseline*. Ideally, this should
### Searching Land Records thru the BLM General Land Office Records.
Searching Land Records thru the BLM General Land Office Records. Land Records can be an exciting addition to your family history search. The United States Government transferred ownership of land to millions
### LAB E - INTRODUCTION TO GEOGRAPHIC TOOLS TOPOGRAPHIC MAPS
LAB E - INTRODUCTION TO GEOGRAPHIC TOOLS TOPOGRAPHIC MAPS Introduction Topographic maps differ from the thematic and regional maps found in atlases in that they employ contour lines (isopleths), lines
### Assignment Discovery Online Curriculum
Lesson title: Scavenger Hunt for Lengths Grade level: 6-8 Subject area: Mathematics Duration: Two class periods Assignment Discovery Online Curriculum Objectives: Students will understand how to measure
### Objectives. Vocabulary. Compare and contrast latitude and longitude. Describe how time zones vary.
Objectives Compare and contrast latitude and longitude. Describe how time zones vary. Vocabulary cartography equator latitude longitude prime meridian International Date Line Latitude and Longitude Latitude
### Grade 4 Unit 3: Multiplication and Division; Number Sentences and Algebra
Grade 4 Unit 3: Multiplication and Division; Number Sentences and Algebra Activity Lesson 3-1 What s My Rule? page 159) Everyday Mathematics Goal for Mathematical Practice GMP 2.2 Explain the meanings
### GIS for Educators. Coordinate Reference System. Understanding of Coordinate Reference Systems.
GIS for Educators Topic 7: Coordinate Reference Systems Objectives: Understanding of Coordinate Reference Systems. Keywords: Coordinate Reference System (CRS), Map Projection, On the Fly Projection, Latitude,
### North American Geography. Lesson 1: Map Skills
North American Geography Lesson 1: Map Skills Unit Overview: As students work through the activities in this unit they will be introduced to the United States in general, different regions of the United
### Section 3 Mapping Earth s Surface
Section 3 Mapping Earth s Surface Key Concept Maps are tools that are used to display data about a given area of a physical body. What You Will Learn Maps can be used to find locations on Earth and to
### Institute of Natural Resources Departament of General Geology and Land use planning Latitude and longitude
Institute of Natural Resources Departament of General Geology and Land use planning Latitude and longitude Lecturers: Berchuk V.Y. Gutareva N.Y. Contents 1. What is longitude and latitude? 2. What is latitude?
### We re Not in Kansas Anymore... Where in the World is Site 806?
We re Not in Kansas Anymore... Where in the World is Site 806? Summary In this introductory lesson, students plot latitude and longitude coordinates, using maps and Google Earth, to find their homes, schools,
### Applications of right triangles and trigonometry
Applications of right triangles and trigonometry 1. A sailor at sea in the 18 th century had no GPS or electronic navigational tools. The main tool used to determine the position of the ship was the sextant,
### An Introduction to G.P.S. Contents
An Introduction to G.P.S. Contents 1. Introduction 2. Latitude and Longitude 3. Global Positioning System 4. Navigation Options 5. Entering Data into a GPS Receiver 6. System Limitations 7. Glossary of
### Where in the World is 34º 55 N 81º 01 W? *
Where in the World is 34º 55 N 81º 01 W? * Carolyn Raper Finley Road Elementary School craper@rock-hill.k12.sc.us * Where in the World Is 34º 55 N 81º 01 W? These are the coordinates for Rock Hill, SC.
### FUNDAMENTALS OF LANDSCAPE TECHNOLOGY GSD Harvard University Graduate School of Design Department of Landscape Architecture Fall 2006
FUNDAMENTALS OF LANDSCAPE TECHNOLOGY GSD Harvard University Graduate School of Design Department of Landscape Architecture Fall 2006 6106/ M2 BASICS OF GRADING AND SURVEYING Laura Solano, Lecturer Name
### Map & Compass Training
Map & Compass Training General Map Reading Skills Contour Lines & Topographic Features UTM Coordinate System & Scale Reading Coordinates Declination What are contour lines? Contour Lines: Convert 3-D info
### The Willows Primary School Geography Curriculum
The Willows Primary School Geography Curriculum 2014-2015 Curriculum Aims The aim of this curriculum is to ensure that our pupils develop curiosity about the world around them and make links with the wider
### It is a little challenging suppose you are to describe a point on the surface of a basketball.
Coordinate Systems 1. Geographic coordinate system To locate a point on a piece of white paper, we usually use the distances of the point to the left/right edge (x) and top/bottom edge (y). Therefore the
### ROYAL CANADIAN AIR CADETS PROFICIENCY LEVEL THREE INSTRUCTIONAL GUIDE SECTION 2
ROYAL CANADIAN AIR CADETS PROFICIENCY LEVEL THREE INSTRUCTIONAL GUIDE SECTION 2 EO M337.02 DETERMINE A POSITION ON A VISUAL FLIGHT RULES (VFR) NAVIGATIONAL CHART (VNC) Total Time: 30 min PREPARATION PRE-LESSON
### An Introduction to Reading Topographic Maps
An Introduction to Reading Topographic Maps Darlene C. Florence National Science Foundation Graduate STEM Fellow in K-12 Education Sugar Creek Watershed K-12 Education Program The Ohio State University-Ohio
### Plotting points on a map using Latitude / Longitude
1 of 8 4/16/2016 4:01 PM Plotting points on a map using Latitude / Longitude (Lat/Lon) There are many ways to plot points on a map using latitude/longitude; this section describes one method. The grid
### National Green Check Certified GPS Training Presentation and Learning Aid
National Green Check Certified GPS Training Presentation and Learning Aid Concept and presentation created by Bill Steer. Revised and updated by Daniel Mansfield, January 14, 2008 Canadian Ecology Centre
### Geographic Coordinates
Geographic Coordinates At A Glance Established Goals Robotics/GPS/GIS Skill: Obtain geographic coordinates from a given positioning device. Life Skills: Performing as a Team Member; Organizing Information
The Basics of Navigation Knowledge of map reading and the use of the compass is an indispensable skill of bushcraft. Without this skill, a walker is a passenger and mere follower on a trip. To become a
### Physical Geography Lab Activity #02
Physical Geography Lab Activity #02 Due date Name Latitude and Longitude COR Objective 1, SLO 1 2.1. Introduction We often take for granted the ability to locate a single, unique place on the surface of
### Coordinate Systems. Every coordinate system has an initial reference point or origin All mapped features are represented in some coordinate system
Coordinate Systems Coordinate Systems A coordinate system is a 2- or 3-D reference frame used to specify locations on the earth. Many different coordinate systems exist, and each is defined by an origin,
### Registered map layers
Registered map layers Coordinate map systems Coordinate map systems 1. The Earth's Graticule Latitude and Longitude The graticule is the imaginary grid of lines running east-west lines of latitude (parallels)
### G302 Review of Maps / Topographic Maps
G302 Review of Maps / Topographic Maps I. Introduction A. Map- a 2-d, scaled representation of the earth's surface, 1. Model of Earth's Surface 2. Basic functions: to represent horizontal distance, direction,
### Map Types and Terms Study Guide Student Contract
Map Types and Terms Study Guide Student Contract Recognition Level I contract this study guide for a grade of C. I agree to complete all of the items coded + by the due date specified with 90% accuracy.
### The Earth's Graticule
Georeferencing How do we make sure all our data layers line up? Georeferencing: = linking a layer or dataset with spatial coordinates Registration: = lining up layers with each other Rectification: =The
### 2016/07/25. GIS 211 Map Projections & Coordinate Systems Chang Chapter 2. Why is this important to YOU? Dr W Britz
GIS 211 Map Projections & Coordinate Systems Chang Chapter 2 Dr W Britz Every map user and maker should have a basic understanding of projections no matter how much computers seem to have automated the
### FORMULA FOR FINDING THE SQUARE FEET OF A RECTANGLE L x W = A
UNIT I REAL ESTATE MATH AREA MEASUREMENTS FORMULA FOR FINDING THE SQUARE FEET OF A RECTANGLE L x W = A Where: A = Area L = Length W = Width If the length = 30 and the width = 20 20 x 30 = 600 Sq. Feet
### RATIONALE FOR ADVENTURE TAKEAWAYS FOR CUB SCOUTS. Wolf Handbook, page 220 ADVENTURE REQUIREMENTS
FINDING YOUR WAY RATIONALE FOR ADVENTURE Learning to read and use a map and a compass is a valuable skill. In this adventure, Scouts will receive an introduction to maps, symbols, and the cardinal directions
### Technology Enhanced Lesson Plan. Geography-Location, Location, Location. Social Studies
Technology Enhanced Lesson Plan Lesson Overview Lesson Topic or Title Subject Area(s) Grade Level Goal(s)/Essential Question(s) Geography-Location, Location, Location Social Studies Grade 5 Goals: Students
### If it is raining, then it is cloudy. If an integer (whole number) n is greater than 4.5, then n is greater than 4.
47 Lesson 3: The Converse of the Pythagorean Theorem, Pythagorean Triples and the 3-Dimensional Pythagorean Theorem Recall that the Pythagorean Theorem tells us how to calculate the length of one side
### LESSON 1: EXPLORE THE ECOSYSTEM
LESSON 1: EXPLORE THE ECOSYSTEM LENGTH: 60 MINUTES GRADES/AGES: GRADES 3-7 Lesson Overview: Learn more about ecosystems and how an ecosystems species are interconnected through games and diagrams. Explore
### Compass Games. For Brownies* and Beginners
Volunteer Learning and Development Compass Games For Brownies* and Beginners *Including Letterboxers! 5601 North Allen Road Mableton, GA 30126 770-702-9100/1-800-771-4046 www.gsgatl.org AUGUST 2013 Knowledge
### Welcome to Lesson 3. It is important for a GIS analyst to have a thorough understanding of map projections and coordinate systems.
Welcome to Lesson 3. It is important for a GIS analyst to have a thorough understanding of map projections and coordinate systems. A GIS without coordinates would simply be a database like Microsoft Excel
### Familiarity with reading a map, atlas, globe, or similar reference tool is recommended but not required.
Lesson Summary: In this lesson, students will learn to read and recognize coordinates on a map and gain an understanding of latitude and longitude. Additional time will be spent discussing how countries
### La Belle in the Classroom
La Belle in the Classroom Lesson 2: Navigating to North America Overview Inaccurate maps and the inability to navigate caused La Salle to miss the mouth of the Mississippi River. During this lesson students
### How To Find Property Lines and Corners With a Cell Phone GPS
How To Find Property Lines and Corners With a Cell Phone GPS By: Joseph Elfelt PropertyLineMaps.com 1. Introduction Would you like to: Find an existing survey marker? Find an approximate corner for land
Ohio Standards Connection: Government Benchmark A Identify the responsibilities of the branches of the U.S. government and explain why they are necessary. Indicator 2 Explain the structure of local governments
### Lines on Maps and Globes. Cross Curricular Writing Activity Social Studies Grade 4
Lines on Maps and Globes Cross Curricular Writing Activity Social Studies Grade 4 Fourth Grade Social Studies Standard Map: Chart/Globe The learner will be able to use maps, charts, graphs, and globes
### Area and Circumference of Circles
Teacher Created Materials 21208 Focused Mathematics Student Guided Practice Book of Circles Learning Objectives Geometry Know the formulas for the area and circumference of a circle and use them to solve
### Grade 4 History-Social Science Unit 1: The Land of California Lesson 5: How are California s communities alike and different?
Grade 4 History-Social Science Unit 1: The Land of California Lesson 5: How are California s communities alike and different? compass rose A compass rose is a symbol that shows the main directions on a
### Map Skills Pretest. 1. What does a cartographer do? a. makes graphs & charts b. designs greeting cards c. creates maps d. builds carts and wagons
Map Skills Pretest 1. What does a cartographer do? a. makes graphs & charts b. designs greeting cards c. creates maps d. builds carts and wagons 2. What is the name of the 0 longitude line? a. Equator
### Standards for Mathematical Practice in Kindergarten
Standards for Mathematical Practice in Kindergarten The Common Core State Standards for Mathematical Practice are expected to be integrated into every mathematics lesson for all students Grades K-12. Below
### Coordinate Systems. CS.1. Worksheet: The teacher places a small X on the floor of the classroom with tape.
Coordinate Systems CS.1. Worksheet: The teacher places a small X on the floor of the classroom with tape. INDIVIDUAL WORK: Explain how you could describe the location of the X using only words and numbers
### CHAPTER 1 BASICS OF SURVEYING
CHAPTER 1 BASICS OF SURVEYING 1.1 SURVEYING DEFINED What is surveying? Surveying is the art of measuring distances, angles and positions on or near the surface of the earth. It is an art? Because only
### Tip Sheet for using the Garmin Hand Held GPS Model ETREX H with the iwitness Close-Range Photogrammetry Software
Tip Sheet for using the Garmin Hand Held GPS Model ETREX H with the iwitness Close-Range Photogrammetry Software Basic steps for using the ETREX H 1. Turn unit on (lower right side button) PWR 2. Depress
### Where in the World is Boonville, New York?
Name: Date: Period: Where in the World is Boonville, New York? Background Information Virtually all maps have lines of latitude (called parallels) and longitude (called meridians) expressed on them. Together,
### Pythagorean Theorem. Inquiry Based Unit Plan
Pythagorean Theorem Inquiry Based Unit Plan By: Renee Carey Grade: 8 Time: 5 days Tools: Geoboards, Calculators, Computers (Geometer s Sketchpad), Overhead projector, Pythagorean squares and triangle manipulatives,
### Topographic Maps. I. Introduction
Topographic Maps I. Introduction A. Topographic maps are essential tools in geologic and engineering studies because they show the configuration of Earth's surface in remarkable detail and permit one to
### Project 1: Plotting Coordinates and Projections
1 of 6 8/3/2007 1:54 PM Project 1: Plotting Coordinates and Projections Dan Uhrhan dfu101@psu.edu 2 of 6 8/3/2007 1:54 PM 3 of 6 8/3/2007 1:54 PM 4 of 6 8/3/2007 1:54 PM 5 of 6 8/3/2007 1:54 PM [Replace
### WHAT IS AREA? CFE 3319V
WHAT IS AREA? CFE 3319V OPEN CAPTIONED ALLIED VIDEO CORPORATION 1992 Grade Levels: 5-9 17 minutes DESCRIPTION What is area? Lesson One defines and clarifies what area means and also teaches the concept
### Motion & The Global Positioning System (GPS)
Grade Level: K - 8 Subject: Motion Prep Time: < 10 minutes Duration: 30 minutes Objective: To learn how to analyze GPS data in order to track an object and derive its velocity from positions and times.
### Latitude and Longitude Vocabulary. Equator imaginary line separating the northern and southern hemispheres.
Latitude and Longitude Vocabulary Degrees the unit for measuring distance on a map Equator imaginary line separating the northern and southern hemispheres. Globe a miniature model of the Earth Hemisphere | 6,542 | 29,112 | {"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-2018-51 | latest | en | 0.903332 |
https://www.fer.unizg.hr/en/course/aaads | 1,726,298,760,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00492.warc.gz | 690,090,296 | 422,014 | ### Advanced Algorithms and Data Structures
Data is displayed for academic year: 2023./2024.
#### Course Description
Balanced trees, multiway trees, B-trees. Selected algorithms over graphs. Selected string-based data structures and algorithms. Selected geometric algorithms. Greedy algorithms, dynamic programming, and reductions. Network flows. Linear programming and simplex method. Stochastic, approximation, and randomized algorithms. Differentiable programming. Bloom filters. Selected compression algorithms.
#### Study Programmes
[FER3-EN] Data Science - profile
(3. semester)
#### Learning Outcomes
1. recognize and analyze problem to be solved using the computer
2. relate problem instance to known, already solved, problem classes
3. select the best algorithm for already solved (known) problem
4. design custom algorithm for unknown problem type
5. evaluate the validity of custom solution with respect to convergence and computational complexity
6. compare custom algorithmic idea to other potential ideas
7. select algorithm for unknown problem based on the comparison between the proposed solutions
#### Forms of Teaching
Lectures
Materials and presentations are on course web page.
Laboratory
complex laboratory assignments which includes algorithms from the lectures
Continuous Assessment Exam
Laboratory Exercises 30 % 30 % 15 % 10 %
Mid Term Exam: Written 30 % 30 % 0 %
Final Exam: Written 30 % 40 %
Exam: Written 50 % 50 %
Exam: Oral 40 %
##### Comment:
short examinations are added above the basic 100%
#### Week by Week Schedule
1. Balanced trees (e.g., AVL trees, red-black trees, splay trees, treaps)
2. Advanced data structures (e.g., B-trees, Fibonacci heaps)
3. String-based data structures and algorithms (e.g., suffix arrays, suffix trees, tries)
4. Geometric algorithms (e.g., points, line segments, polygons, finding convex hull, spatial decomposition, collision detection, geometric search and proximity)
5. Linear programming (e.g., duality, simplex method, interior point algorithms)
6. Dynamic programming
7. Greedy algorithms, Project, Lossless and lossy compression algorithms
8. Midterm exam
9. Midterm exam
10. Randomized algorithms, Stochastic algorithms
11. Graphs (e.g., topological sort, finding strongly connected components, matching), Bloom filters
12. Graphs (e.g., topological sort, finding strongly connected components, matching)
13. Graphs (e.g., topological sort, finding strongly connected components, matching), Network flows (e.g., max flow Ford-Fulkerson algorithm, max flow min cut, maximum bipartite matching), Approximation algorithms
14. Reduction: transform-and-conquer, Approximation algorithms
15. Final exam
#### Literature
Thomas H. Cormen (2009.), Introduction to Algorithms, MIT Press
Adam Drozdek (2012.), Data Structures and Algorithms in C++, Thomson Course Technology
Steven S. Skiena (2008.), The Algorithm Design Manual, Springer-Verlag New York Incorporated
R. Sedgewick, K. Wayne (2011.), Algorithms 4th ed., Addison-Wesley
D. Kalpić, V. Mornar (1996.), Operacijska istraživanja, Zeus
J. Matousek, B.Gartner (2006.), Understanding and Using Linear Programming, Springer
Vijay V. Vazirani (2002.), Approximation Algorithms, Springer Science & Business Media
R. Motwani, P. Raghavan (1995.), Randomized Algorithms, Cambridge University Press
I. Goodfellow, Y. Bengio and A. Courville (2015.), Deep Learning, MIT press
#### General
ID 222951
Winter semester
5 ECTS
L3 English Level
L2 e-Learning
45 Lectures
0 Seminar
6 Exercises
15 Laboratory exercises
0 Project laboratory
0 Physical education excercises | 831 | 3,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.819924 |
http://www.studyalgorithms.com/misc/find-the-first-n-prime-numbers-method-1/ | 1,550,381,446,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481624.10/warc/CC-MAIN-20190217051250-20190217073250-00473.warc.gz | 454,095,151 | 14,347 | # Find the first N prime numbers. (Method 1)
By | July 24, 2015
Question: Given an integer N, find the prime numbers in that range from 1 to N.
Input: N = 25
Output: 2, 3, 5, 7, 11, 13, 17, 19, 23
Today let us discuss about a very common but very interesting problem “To find prime numbers in first N Natural numbers “. I will be taking here a very specific approach of first giving definition of prime numbers , using that definition to derive the algorithm, and then using different properties or results that can be further applied to our algorithm to optimize it.
“Prime numbers are the natural numbers whose factors are 1 and themselves “. So if any number X is prime number then it should have exactly two factors 1 and X. This implies that all the number greater than 1 and less than X shouldn’t divide X to remainder 0. (X % num != 0)
So let us use this statement of ours to derive our very first algorithm to solve the problem.
This will be a naive approach, using 2 loops one outer loop for every number and inner loop that for every number will go upto (j=number-1) and check if anytime modulo of i with j comes 0 then this number is not prime.
```void findPrimes(int limit) {
// Loop from the first Prime till limit
for (int i = 2; i < limit; i++) {
// Assume number to be prime
boolean isPrime = true;
// Loop from 2 to the number to checked
for (int j = 2; j < i; j++) {
// If the number divides, it means the number is not prime
if (i % j == 0) {
isPrime = false;
break;
}
}
// Print the number if it is prime
if (isPrime)
System.out.print(i + ",");
}
}
```
Above is a very naïve approach. If you know little maths you will know that this can be optimized by checking it till i/2 I.e. making inner loop to go to half of values. Furthur you can optimize by checking only for 2 and then all the odd numbers that is incrementing the inner loop by 2. So now let us use this result and refine our algorithm.
```void findPrimes(int limit) {
// Special handling for 2
if (limit > 2)
System.out.println("2,");
// Loop from the first Prime till limit
for (int i = 2; i < limit; i++) {
// Assume number to be prime
boolean isPrime = true;
if (i % 2 == 0)
isPrime = false;
else {
// Loop from 3 to number/2
for (int j = 3; j < i / 2; j += 2) {
// If the number divides, it means the number is not prime
if (i % j == 0) {
isPrime = false;
break;
}
}
}
// Print the number if it is prime
if (isPrime)
System.out.print(i + ",");
}
}
```
Here is a link of the running code:- http://ideone.com/BGr3b0
Time Complexity:- O(n2)
Space Complexity:- O(1)
To further optimize this approach visit these links:-
## 7 thoughts on “Find the first N prime numbers. (Method 1)”
1. nikoo28
The algorithm has been discussed in the post. Please let me know where are you facing a problem?
1. RaHan
Hi nikoo,
The body of the problem says as you say, the title says it like I say, “the first N”.
2. RaHan
2 problems with the article. 1. N=25 means I expect 25 prime numbers as the answer. 2. The definition of a prime number is incorrect. ANY number is divisible by either 1 or itself. It should be “ONLY 1 and itself”.
1. nikoo28
Hi Rahan,
If you read the question properly, it says to print prime numbers in the range of 1-N. It never says to print N prime numbers. | 892 | 3,281 | {"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-2019-09 | longest | en | 0.842282 |
https://nrich.maths.org/public/leg.php?code=-684&cl=3&cldcmpid=7592 | 1,524,476,630,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945940.14/warc/CC-MAIN-20180423085920-20180423105920-00387.warc.gz | 668,186,126 | 8,763 | # Search by Topic
#### Resources tagged with STEM - maths supporting SET similar to Food Web:
Filter by: Content type:
Stage:
Challenge level:
### There are 50 results
Broad Topics > STEM mapping > STEM - maths supporting SET
### More or Less?
##### Stage: 4 Challenge Level:
Are these estimates of physical quantities accurate?
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### Big and Small Numbers in Biology
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calulate various quantities in biological contexts.
### Food Chains
##### Stage: 3 Challenge Level:
When a habitat changes, what happens to the food chain?
### Weekly Challenge 48: Quorum-sensing
##### Stage: 4 Short Challenge Level:
This problem explores the biology behind Rudolph's glowing red nose.
### Big and Small Numbers in the Living World
##### Stage: 3 and 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in biological contexts.
### Investigating Epidemics
##### Stage: 3 and 4 Challenge Level:
Simple models which help us to investigate how epidemics grow and die out.
### A Question of Scale
##### Stage: 4 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### Global Warming
##### Stage: 4 Challenge Level:
How much energy has gone into warming the planet?
##### Stage: 4 Challenge Level:
Which units would you choose best to fit these situations?
### Fill Me Up
##### Stage: 3 Challenge Level:
Can you sketch graphs to show how the height of water changes in different containers as they are filled?
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
### Counting Dolphins
##### Stage: 4 Challenge Level:
How would you go about estimating populations of dolphins?
### Investigating the Dilution Series
##### Stage: 4 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Biology Measurement Challenge
##### Stage: 4 Challenge Level:
Analyse these beautiful biological images and attempt to rank them in size order.
### Far Horizon
##### Stage: 4 Challenge Level:
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
### Guessing the Graph
##### Stage: 4 Challenge Level:
Can you suggest a curve to fit some experimental data? Can you work out where the data might have come from?
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### How Do You React?
##### Stage: 4 Challenge Level:
To investigate the relationship between the distance the ruler drops and the time taken, we need to do some mathematical modelling...
### Does This Sound about Right?
##### Stage: 3 Challenge Level:
Examine these estimates. Do they sound about right?
### Pinhole Camera
##### Stage: 3 Challenge Level:
Make your own pinhole camera for safe observation of the sun, and find out how it works.
### Designing Table Mats
##### Stage: 3 and 4 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
### Big and Small Numbers in the Physical World
##### Stage: 4 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
### Gym Bag
##### Stage: 3 and 4 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Make Your Own Pencil Case
##### Stage: 3 Challenge Level:
What shape would fit your pens and pencils best? How can you make it?
### Fill Me up Too
##### Stage: 4 Challenge Level:
In Fill Me Up we invited you to sketch graphs as vessels are filled with water. Can you work out the equations of the graphs?
### Electric Kettle
##### Stage: 4 Challenge Level:
Explore the relationship between resistance and temperature
### Robot Camera
##### Stage: 4 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Speed-time Problems at the Olympics
##### Stage: 4 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### Constantly Changing
##### Stage: 4 Challenge Level:
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Genetics
##### Stage: 4 Challenge Level:
A problem about genetics and the transmission of disease.
### Carbon Footprints
##### Stage: 4 Challenge Level:
Is it really greener to go on the bus, or to buy local?
### Back Fitter
##### Stage: 4 Challenge Level:
10 graphs of experimental data are given. Can you use a spreadsheet to find algebraic graphs which match them closely, and thus discover the formulae most likely to govern the underlying processes?
### Chocolate Cake
##### Stage: 3 Challenge Level:
If I don't have the size of cake tin specified in my recipe, will the size I do have be OK?
### Perspective Drawing
##### Stage: 3 and 4 Challenge Level:
Explore the properties of perspective drawing.
##### Stage: 3 Challenge Level:
Can you work out which drink has the stronger flavour?
##### Stage: 3 Challenge Level:
Is it cheaper to cook a meal from scratch or to buy a ready meal? What difference does the number of people you're cooking for make?
##### Stage: 4 and 5 Challenge Level:
How would you design the tiering of seats in a stadium so that all spectators have a good view?
### What's That Graph?
##### Stage: 4 Challenge Level:
Can you work out which processes are represented by the graphs?
### Scientific Measurement
##### Stage: 4 Challenge Level:
Practice your skills of measurement and estimation using this interactive measurement tool based around fascinating images from biology.
### Mystery Procedure
##### Stage: 4 Challenge Level:
Can you work out what this procedure is doing?
### Construct the Solar System
##### Stage: 4 and 5 Challenge Level:
Make an accurate diagram of the solar system and explore the concept of a grand conjunction.
### Oblique Projection
##### Stage: 3 Challenge Level:
Explore the properties of oblique projection.
### Temperature
##### Stage: 3 Challenge Level:
Water freezes at 0°Celsius (32°Fahrenheit) and boils at 100°C (212°Fahrenheit). Is there a temperature at which Celsius and Fahrenheit readings are the same?
##### Stage: 3 Challenge Level:
Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?
### Troublesome Triangles
##### Stage: 2 and 3 Challenge Level:
Many natural systems appear to be in equilibrium until suddenly a critical point is reached, setting up a mudslide or an avalanche or an earthquake. In this project, students will use a simple. . . .
### Perfect Eclipse
##### Stage: 4 Challenge Level:
Use trigonometry to determine whether solar eclipses on earth can be perfect.
### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat? | 1,598 | 7,533 | {"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-2018-17 | latest | en | 0.860276 |
https://www.jiskha.com/questions/93937/I-am-trying-to-do-these-2-problems-Im-starting-out-with-the-RC-Circuit-2nd-one | 1,556,139,501,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578656640.56/warc/CC-MAIN-20190424194348-20190424220348-00297.warc.gz | 734,641,740 | 5,664 | # Circuit Problem- Physics
I am trying to do these 2 problems. I'm starting out with the RC Circuit (2nd one):
h t t p ://img201.imageshack.us/my.php?image=circuit5ij7.png
a) The capacitor acts as a bare wire with no resistance, so:
I=V/R=V/(4/5R)
b) Capacitor acts as an open circuit, so:
I=V/R=V/(4R)
Are those right so far?
I need a little help on c, d, and e.
1. 👍 0
2. 👎 0
3. 👁 46
1. What were the questions? What is the circuit?
A capacitor acts as an open circuit to DC, but has no resistance.
1. 👍 0
2. 👎 0
posted by drwls
2. The questions is in the link I gave above. Here it is again (you have to delete the spaces in 'h t t p :'):
h t t p ://img201.imageshack.us/my.php?image=circuit5ij7.png
1. 👍 0
2. 👎 0
posted by Eileen
3. Sorry, I forgot to check out the link with the circuit and the questions. That was clever of you to add the spaces in h t t p so the URL could be displayed.
(a) Initially, C looks like zero resistance, so the effective resistance seen but the battery is 4R/5
(b) steady state I = V/R, since the capacitor cannot allow DC current
(c) current = 0 through the battery, but V/(5R) through 4R, and the capacitor discharges.
(d) Solve exp[-t/(5RC)] = 0.5 , for t.
5RC is the series resistance that the capacitor sees while discharging
(e) (1/2) C V^2 is the energy dissipated in resistors while discharging
1. 👍 0
2. 👎 0
posted by drwls
## Similar Questions
1. ### Continued physics problem from before (way down)
Posted by Eileen on Sunday, April 13, 2008 at 10:05am. I am trying to do these 2 problems. I'm starting out with the RC Circuit (2nd one): h t t p ://img201.imageshack.us/my.php?image=circuit5ij7.png *delete spaces or link won't
asked by Eileen on April 13, 2008
2. ### Physic
a series RLC circuit has a peak current of 4 A with a frequency of 23 kHz. if the resistance of the circuit is 60 kohm, the capacitance of the circuit is 16 uF, and the unductance of the circuit is 24 uH, determine the average
asked by Fatih on March 13, 2012
3. ### physics
a series RLC circuit has a peak current of 4 A with a frequency of 23 kHz. if the resistance of the circuit is 60 kohm, the capacitance of the circuit is 16 uF, and the unductance of the circuit is 24 uH, determine the average
asked by fatih on March 12, 2012
4. ### physics
a series RLC circuit has a peak current of 4 A with a frequency of 23 kHz. if the resistance of the circuit is 60 kohm, the capacitance of the circuit is 16 uF, and the unductance of the circuit is 24 uH, determine the average
asked by fatih on March 12, 2012
5. ### physics
a series RLC circuit has a peak current of 4 A with a frequency of 23 kHz. if the resistance of the circuit is 60 kohm, the capacitance of the circuit is 16 uF, and the unductance of the circuit is 24 uH, determine the average
asked by fatih on March 12, 2012
6. ### Sciencee
Two electrical circuits have the same 15-volt batteries. Circuit A has 2 ohms of resistance, and circuit B has 6 ohms of resistance. Which of the following statements is correct? (1 point) • Circuit B has the same amount of
asked by Lilly on October 19, 2017
7. ### Science
Megan wants to determine if an electrical circuit has electricity flowing through it. Which of the following evidence is proof that the circuit has an electrical current? a)The circuit is a closed loop. b)A light bulb is lit in
8. ### Science
One light bulb in a string of lights goes out this causes all other lights in string to go out. This is example of A. Resistor B. parallel circuit C. Closed circuit***** D. Series circuit When a switch is turned from the off to
asked by BUBBLES on March 1, 2015
9. ### science
A fuse prevents a circuit from overheating by _____. connecting the circuit to a ground wire increasing the resistance in the circuit opening a switch and breaking the circuit melting and breaking the circuit
asked by hi on April 9, 2019
10. ### physics
a series RLC circuit has a peak current of 1 A with a frequency of 54 kHz. if the resistance of the circuit is 51 kHz, the capacitance of the circuit is 19 uF and the inductance of the circuit is 25 uF, determine the average power
asked by fatih on March 12, 2012
More Similar Questions | 1,219 | 4,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2019-18 | latest | en | 0.934242 |
http://www.math.utah.edu/software/lapack/lapack-c/cppsvx.html | 1,534,690,540,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215222.74/warc/CC-MAIN-20180819145405-20180819165405-00136.warc.gz | 561,308,095 | 3,836 | Previous: cppsv Up: ../lapack-c.html Next: cpptrf
# cppsvx
```
NAME
CPPSVX - use the Cholesky factorization A = U**H*U or A =
L*L**H to compute the solution to a complex system of linear
equations A * X = B,
SYNOPSIS
SUBROUTINE CPPSVX( FACT, UPLO, N, NRHS, AP, AFP, EQUED, S,
B, LDB, X, LDX, RCOND, FERR, BERR, WORK,
RWORK, INFO )
CHARACTER EQUED, FACT, UPLO
INTEGER INFO, LDB, LDX, N, NRHS
REAL RCOND
REAL BERR( * ), FERR( * ), RWORK( * ), S( * )
COMPLEX AFP( * ), AP( * ), B( LDB, * ), WORK( *
), X( LDX, * )
PURPOSE
CPPSVX uses the Cholesky factorization A = U**H*U or A =
L*L**H to compute the solution to a complex system of linear
equations
A * X = B, where A is an N-by-N Hermitian positive defin-
ite matrix stored in packed format and X and B are N-by-NRHS
matrices.
Error bounds on the solution and a condition estimate are
also provided.
DESCRIPTION
The following steps are performed:
1. If FACT = 'E', real scaling factors are computed to
equilibrate
the system:
diag(S) * A * diag(S) * inv(diag(S)) * X = diag(S) * B
Whether or not the system will be equilibrated depends on
the
scaling of the matrix A, but if equilibration is used, A
is
overwritten by diag(S)*A*diag(S) and B by diag(S)*B.
2. If FACT = 'N' or 'E', the Cholesky decomposition is used
to
factor the matrix A (after equilibration if FACT = 'E')
as
A = U'* U , if UPLO = 'U', or
A = L * L', if UPLO = 'L',
where U is an upper triangular matrix, L is a lower tri-
angular
matrix, and ' indicates conjugate transpose.
3. The factored form of A is used to estimate the condition
number
of the matrix A. If the reciprocal of the condition
number is
less than machine precision, steps 4-6 are skipped.
4. The system of equations is solved for X using the fac-
tored form
of A.
5. Iterative refinement is applied to improve the computed
solution
matrix and calculate error bounds and backward error
estimates
for it.
6. If equilibration was used, the matrix X is premultiplied
by
diag(S) so that it solves the original system before
equilibration.
ARGUMENTS
FACT (input) CHARACTER*1
Specifies whether or not the factored form of the
matrix A is supplied on entry, and if not, whether
the matrix A should be equilibrated before it is
factored. = 'F': On entry, AFP contains the fac-
tored form of A. If EQUED = 'Y', the matrix A has
been equilibrated with scaling factors given by S.
AP and AFP will not be modified. = 'N': The matrix
A will be copied to AFP and factored.
= 'E': The matrix A will be equilibrated if neces-
sary, then copied to AFP and factored.
UPLO (input) CHARACTER*1
= 'U': Upper triangle of A is stored;
= 'L': Lower triangle of A is stored.
N (input) INTEGER
The number of linear equations, i.e., the order of
the matrix A. N >= 0.
NRHS (input) INTEGER
The number of right hand sides, i.e., the number of
columns of the matrices B and X. NRHS >= 0.
AP (input/output) COMPLEX array, dimension (N*(N+1)/2)
On entry, the upper or lower triangle of the Hermi-
tian matrix A, packed columnwise in a linear array,
except if FACT = 'F' and EQUED = 'Y', then A must
contain the equilibrated matrix diag(S)*A*diag(S).
The j-th column of A is stored in the array AP as
follows: if UPLO = 'U', AP(i + (j-1)*j/2) = A(i,j)
for 1<=i<=j; if UPLO = 'L', AP(i + (j-1)*(2n-j)/2) =
A(i,j) for j<=i<=n. See below for further details.
A is not modified if FACT = 'F' or 'N', or if FACT =
'E' and EQUED = 'N' on exit.
On exit, if FACT = 'E' and EQUED = 'Y', A is
overwritten by diag(S)*A*diag(S).
AFP (input or output) COMPLEX array, dimension (N*(N+1)/2)
If FACT = 'F', then AFP is an input argument and on
entry contains the triangular factor U or L from the
Cholesky factorization A = U**H*U or A = L*L**H, in
the same storage format as A. If EQUED .ne. 'N',
then AFP is the factored form of the equilibrated
matrix A.
If FACT = 'N', then AFP is an output argument and on
exit returns the triangular factor U or L from the
Cholesky factorization A = U**H*U or A = L*L**H of
the original matrix A.
If FACT = 'E', then AFP is an output argument and on
exit returns the triangular factor U or L from the
Cholesky factorization A = U**H*U or A = L*L**H of
the equilibrated matrix A (see the description of AP
for the form of the equilibrated matrix).
EQUED (input/output) CHARACTER*1
Specifies the form of equilibration that was done.
= 'N': No equilibration (always true if FACT =
'N').
= 'Y': Equilibration was done, i.e., A has been
replaced by diag(S) * A * diag(S). EQUED is an
input variable if FACT = 'F'; otherwise, it is an
output variable.
S (input/output) REAL array, dimension (N)
The scale factors for A; not accessed if EQUED =
'N'. S is an input variable if FACT = 'F'; other-
wise, S is an output variable. If FACT = 'F' and
EQUED = 'Y', each element of S must be positive.
B (input/output) COMPLEX array, dimension (LDB,NRHS)
On entry, the N-by-NRHS right hand side matrix B.
On exit, if EQUED = 'N', B is not modified; if EQUED
= 'Y', B is overwritten by diag(S) * B.
LDB (input) INTEGER
The leading dimension of the array B. LDB >=
max(1,N).
X (output) COMPLEX array, dimension (LDX,NRHS)
If INFO = 0, the N-by-NRHS solution matrix X to the
original system of equations. Note that if EQUED =
'Y', A and B are modified on exit, and the solution
to the equilibrated system is inv(diag(S))*X.
LDX (input) INTEGER
The leading dimension of the array X. LDX >=
max(1,N).
RCOND (output) REAL
The estimate of the reciprocal condition number of
the matrix A after equilibration (if done). If
RCOND is less than the machine precision (in partic-
ular, if RCOND = 0), the matrix is singular to work-
ing precision. This condition is indicated by a
return code of INFO > 0, and the solution and error
bounds are not computed.
FERR (output) REAL array, dimension (NRHS)
The estimated forward error bounds for each solution
vector X(j) (the j-th column of the solution matrix
X). If XTRUE is the true solution, FERR(j) bounds
the magnitude of the largest entry in (X(j) - XTRUE)
divided by the magnitude of the largest entry in
X(j). The quality of the error bound depends on the
quality of the estimate of norm(inv(A)) computed in
the code; if the estimate of norm(inv(A)) is accu-
rate, the error bound is guaranteed.
BERR (output) REAL array, dimension (NRHS)
The componentwise relative backward error of each
solution vector X(j) (i.e., the smallest relative
change in any entry of A or B that makes X(j) an
exact solution).
WORK (workspace) COMPLEX array, dimension (2*N)
RWORK (workspace) REAL array, dimension (N)
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal
value
> 0: if INFO = i, and i is
<= N: if INFO = i, the leading minor of order i of A
is not positive definite, so the factorization could
not be completed, and the solution and error bounds
could not be computed. = N+1: RCOND is less than
machine precision. The factorization has been com-
pleted, but the matrix is singular to working preci-
sion, and the solution and error bounds have not
been computed.
FURTHER DETAILS
The packed storage scheme is illustrated by the following
example when N = 4, UPLO = 'U':
Two-dimensional storage of the Hermitian matrix A:
a11 a12 a13 a14
a22 a23 a24
a33 a34 (aij = conjg(aji))
a44
Packed storage of the upper triangle of A:
AP = [ a11, a12, a22, a13, a23, a33, a14, a24, a34, a44 ]
``` | 2,269 | 7,510 | {"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-2018-34 | latest | en | 0.825415 |
https://www.physicsforums.com/threads/divergence-of-curvature-scalars-metric.396000/ | 1,544,422,019,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823318.33/warc/CC-MAIN-20181210055518-20181210081018-00434.warc.gz | 988,606,772 | 28,511 | # Divergence of curvature scalars * metric
1. Apr 17, 2010
### JustinLevy
How can one work out what terms like:
$$(g^{cd}R^{ab}R_{ab})_{;d}$$
are in terms of the divergence of the Ricci curvature or Ricci scalar?
One student noted that since:
$$G^{ab} = R^{ab} - \frac12 g^{ab}R$$
$${G^{ab}}_{;b} = 0$$
that we could maybe use the fact that
$$G^{ab}G_{ab} = R^{ab}R_{ab} - \frac12 R^{ab}g_{ab}R - \frac12 R R_{ab}g^{ab} + \frac14 RRg^{ab}g_{ab} = R^{ab}R_{ab}$$
to help? We weren't sure where to go next.
Can someone better with tensor manipulations show how we could work this out?
Last edited: Apr 17, 2010
2. Apr 17, 2010
### nicksauce
You have divergence (Ricci tensor) = 1/2 divergence (Ricci scalar). For your original expression just apply the product rule using that rule. Remember that the metric is divergence free.
3. Apr 17, 2010
### JustinLevy
Basically, you are saying I should be able to use
$${R^{ab}}_{;b} = \frac12 g^{ab}R_{,b}$$
to help here.
I feel like an idiot, but I really don't see how to use that.
If I expand out, I get:
$$(g^{cd}R^{ab}R_{ab})_{;d} = {g^{cd}}_{;d}R^{ab}R_{ab} + g^{cd}{R^{ab}}_{;d}R_{ab} + g^{cd}R^{ab}R_{ab;d}$$
As you mention, the first of those is just zero, since the metric covariant derivatives are zero.
But how do I relate terms like
$${R^{ab}}_{;b}$$
to terms like
$${R^{ab}}_{;d}$$
?
4. Apr 17, 2010
### Mentz114
Applying Liebnitz
$$(R^{ab}R_{ab})_{;d}={R^{ab}}_{;d}R_{ab}+{R^{ab}}R_{ab}_{;d}$$
might help. I have to dash ...
[I just noticed this is what nicksauce said ]
5. Apr 17, 2010
### JustinLevy
Yes, that is what I wrote above. I just don't know how to relate terms containing
$$g^{cd}{R^{ab}}_{;d}$$
to the divergence
$${R^{ab}}_{;b}$$
which after that, I can of course relate to the divergence of the Ricci scalar.
What I'm actually trying to get at, is a friend asked what is the next highest order term which we can add to the EFE such that:
1) still shows T_uv is divergenceless
2) has the same vacuum equations (R_uv = 0)
I thought it would be a neat thing to work out for myself, but I'd need to know the divergence of more complicated terms. So I could put in something simple and hopefully a couple counter terms to make the divergence zero. But I quickly end up with terms I can't figure out how to simplify further so I can't tell if stuff cancels or not.
I'll admit I tried to sneak a peak at the answer, but searching papers I only found examples of people manipulating the Hilbert action. There's probably someone who already worked it out for the field equations directly, but if so I didn't find it. The reason I want to focus on the field equations is this sometimes allows simplier changes, for example imagine adding magnetic monopoles from the Maxwell E and B field equations point of view -- which is easy, vs trying to add it in an action formalism which no one has found a consistent local way to do it yet. So the simplest change in the action formalism, isn't necessarily the simplest change to the field equations, to meet the desired conditions.
6. Apr 17, 2010
### atyy
Don't look if you want to have the fun of working it out for yourself (I believe these still give local energy conservation, but the vacuum equations are not the same):
http://arxiv.org/abs/gr-qc/0305004
http://arxiv.org/abs/0805.1726
Last edited by a moderator: May 4, 2017
7. Apr 19, 2010
### JustinLevy
Thanks atyy.
That first paper makes me a bit uncomfortable. For instance "Horizons are always observer dependent" and other claims; it all seems kind of sloppy. Especially declaring at the end that there have to be "atoms of spacetime".
As for the others, I've seen f(R) and Lovelock gravity before while searching, but those papers are from the point of view of modifying the action. Which I commented on my thoughts about above. It does give me an idea though...
if the action doesn't have explicit time or spatial dependence, I guess I can use it as a tool to "generate" more terms I know (from physical principles) have zero divergence. Right?
So going along that idea (and using some hints from wikipedia), does this look right?
(interspersed with blatant copying from wikipedia, to speed up the math)
Consider the action
$$S = \int (\frac{1}{2\kappa}R_{ab}R^{ab} + \mathcal{L}_m) \sqrt{-g} d^4x$$
The variation of the determinant is:
$$\delta \sqrt{-g}= -\frac{1}{2} \sqrt{-g} g^{ab} \delta g_{ab}$$
The variation of the curvature scalar is:
\begin{align*} \delta (R_{ab}R^{ab}) &= R^{ab} \delta R_{ab} + R_{ab} \delta R^{ab} \\ &= 2 R^{ab} \delta R_{ab} \\ &= 2 R^{ab}(\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a}) \end{align*}
Since $$\delta \Gamma^\lambda_{\mu\nu}\,$$ is actually the difference of two connections, it should transform as a tensor. Therefore, it can be written as
$$\delta \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda d}\left(\nabla_\mu\delta g_{d\nu}+\nabla_\nu\delta g_{d\mu}-\nabla_d\delta g_{\mu\nu} \right)$$
and substituting in the equation above one finds:
\begin{align*} \delta R_{ab} &= [\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a}] \\ &= [\nabla_c \frac{1}{2}g^{c d}\left(\nabla_a\delta g_{db}+\nabla_b\delta g_{da}-\nabla_d\delta g_{ab} \right) -\nabla_b \frac{1}{2}g^{c d}\left(\nabla_c\delta g_{da}+\nabla_a\delta g_{dc}-\nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}g^{c d}[\left(\nabla_c\nabla_a\delta g_{db}+\nabla_c\nabla_b\delta g_{da}-\nabla_c\nabla_d\delta g_{ab} \right) -\left(\nabla_b\nabla_c\delta g_{da}+\nabla_b\nabla_a\delta g_{dc}-\nabla_b\nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}g^{c d}[\nabla_c\nabla_a\delta g_{db}-\nabla_c\nabla_d\delta g_{ab} - \nabla_b\nabla_a\delta g_{dc} + \nabla_b\nabla_d\delta g_{ca}] \end{align*}
(hmm... somehow the overall sign is off compared to the f(R) gravity wiki article. But magnitudes match.)
In the action, this term in integrated, and via integration by parts, we can move the derivatives from on the metric piece to on the R_ab piece.
After the integration by parts, we have:
\begin{align*} \delta [\sqrt{-g} R_{ab} R^{ab}] &= \sqrt{-g}[ - \delta g_{ab}\frac12 g^{ab} R^{cd}R_{cd} + [ \delta g_{db}\nabla_c\nabla_a g^{c d} R^{ab}-\delta g_{ab}\nabla_c\nabla_d g^{c d} R^{ab} - \delta g_{dc}\nabla_b\nabla_a g^{c d} R^{ab} + \delta g_{ca}\nabla_b\nabla_d g^{c d} R^{ab} ]] \\ &= \sqrt{-g}[ - \delta g_{ab}\frac12 g^{ab} R^{cd}R_{cd} + [ \delta g_{ab}\nabla_c\nabla_d g^{c a} R^{db}-\delta g_{ab}\nabla_c\nabla_d g^{c d} R^{ab} - \delta g_{ab}\nabla_c\nabla_d g^{b a} R^{dc} + \delta g_{ba}\nabla_c\nabla_d g^{b d} R^{ac} ]] \\ &= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} + \nabla_c\nabla_d g^{c a} R^{db} - \nabla_c\nabla_d g^{c d} R^{ab} - \nabla_c\nabla_d g^{b a} R^{dc} + \nabla_c\nabla_d g^{b d} R^{ac} ] \\ &= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\nabla_d R^{db} - \nabla^d\nabla_d R^{ab} - g^{b a} \nabla_c\nabla_d R^{dc} + \nabla_c\nabla^b R^{ac} ] \\ &= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\frac12 g^{db}R_{;d} - \nabla^d\nabla_d R^{ab} - g^{b a} \nabla_c\frac12 g^{dc}R_{;d} + \nabla^b \frac12 g^{ac} R_{;c} ] \\ &= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} + \frac12\nabla^a\nabla^b R - \nabla^d\nabla_d R^{ab} - g^{b a} \frac12 \nabla^d\nabla_d R + \frac12 \nabla^a\nabla^b R ] \\ &= \sqrt{-g}\delta g_{ab}[ - \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R) ] \end{align*}
So, the field equations for this "theory" are:
$$- \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R) = \kappa T^{ab}$$
The whole point of this was to find the divergence of metric * scalar invariant. So this method seems to have generated the result that:
$$- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} + \nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( {R^{ab}}_{;b} + \frac12 g^{ab} R_{;b}) = 0$$
$$- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} + \nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( \frac12 g^{ab} R_{;b} + \frac12 g^{ab} R_{;b}) = 0$$
which simplifying is
\begin{align*} (g^{ab} R^{cd}R_{cd})_{;b} &= 2 (\nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d g^{ab}R_{;b}) \\ &= 2 (\nabla^a\nabla^b\nabla_b R - \nabla^d\nabla_d \nabla^a R) \\ &= 0 \end{align*}
Is this really correct?
That's a pretty neat way of generating divergence equations. But it seems pretty involved.
As an aside, does:
$$- \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R) = \kappa T^{ab}$$
have compatible vacuum solutions with:
$$R^{ab} - \frac12 g^{ab} R = \kappa T^{ab}$$
If so, I guess I answered the other goal of playing with these as well. But I'm not sure how to say difinitively either way. I mean
$$R^{ab} = 0$$
clearly solves
$$- \frac12 g^{ab} R^{cd}R_{cd} + \nabla^a\nabla^b R - \nabla^d\nabla_d ( R^{ab} + \frac12 g^{ab} R) = 0$$
$$- 2 R^{cd}R_{cd} + \nabla^a\nabla_a R - \nabla^d\nabla_d ( R + 2 R) = 0$$
$$R^{cd}R_{cd} = -\nabla^a\nabla_a R$$
but I don't know if boundary conditions, or something else I'm ignoring, would screw things up.
EDIT: Needed to fix quite a few mistakes and typos.
Last edited: Apr 20, 2010
8. Apr 20, 2010
### haushofer
Are you looking for Gauss-Bonnet terms in the action, perhaps?
9. Apr 20, 2010
### JustinLevy
I was using that action merely as a way to calculate the divergence of some terms. Just an idea that popped in my head after thinking about the f(R) theories some more after atyy mentioned them. I'm not even sure if that is a valid method to calculate divergences ... but if so, it is an interesting method for generating divergence relations.
Can someone skim through my reasoning in the last post and tell me if that is even a valid method?
And even better, while it may be tedious, would someone mind checking my math there? It seems really strange to me that:
$$(g^{cd}R^{ab}R_{ab})_{;d} = 0$$
and therefore $$\nabla^c(R^{ab}R_{ab}) = 0$$
The curvature scalar is covariantly constant!? How can that be true? And if it is, what does that mean? Should this have been obvious from a different line of attack (maybe something along the lines of the openning post, which I couldn't figure out what to do with)?
10. Apr 20, 2010
### Altabeh
If we calculate the variation of the scalar curvature $$R=R_{\mu\nu}g^{\mu\nu}$$, one gets
\begin{align*} \delta (R_{\mu\nu}g^{\mu\nu}) &= R_{\mu\nu} \delta g^{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu} \\ &= R_{\mu\nu} \delta g^{\mu\nu}+\nabla_{\kappa}(g^{\mu\nu}\delta\Gamma_{\mu\nu}^{\kappa}-g^{\mu\kappa}\delta\Gamma_{\mu\rho}^{\rho}). \\ \end{align*}
So the further calculations all have to be incorrect!
AB
11. Apr 21, 2010
### JustinLevy
R is not the only curvature scalar. It should be very clear from what I wrote that I was talking about the curvature scalar $$R_{ab}R^{ab}$$ not R. So I think you misunderstood what I trying to do.
Anyway, I did find some errors. There are a couple places where I was sloppy with order of operations, which is important since covariant derivatives don't commute. I also dropped a term in the variation of R_ab R^ab. I also tracked down where that sign error was coming from, which is a minor fix. So I need to rewrite those calculations.
I realize tracking though my math is probably tedious, but the general line of argument is fairly simple:
1) I take a Lagrangian that doesn't have explicit time or spatial dependence.
2) I therefore expect the stress energy should be divergence-less, and hence can get a new relation on the divergence of some curvature terms.
So even without trudging through the math, I hope some people here can comment.
Is this method valid? If not, I don't really want to take the time to add in those extra terms to fix the calculation.
Either way, the point of this thread was to learn how to calculate the divergence of terms like (as I said in the openning post)
$$g^{\mu\nu}R_{ab}R^{ab}$$
if someone knows another way to generate divergence relations of curvature terms besides starting with the second Bianchi identity, please do let me know.
12. Apr 21, 2010
### Altabeh
Yet again it is not true, to the best of my knowledge! The variations of $$R^{ab}$$ and its covariant form are not the same at all! You are putting both equal and this gives rise to an incorrect expression for the so-called "curvature scalar" you've defined here!
AB
13. Apr 22, 2010
### JustinLevy
I thank you for responding to my threads, but I wish you would take a bit more time reading what I wrote. As I said in the last post: "I also dropped a term in the variation of R_ab R^ab." Also, I am not sure why you feel the need to refer to it as a 'so-called "curvature scalar"' ... it IS a curvature scalar. There are many curvature scalars
http://en.wikipedia.org/wiki/Carminati-McLenaghan_invariants
http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)
Regarding my comment about dropping a term, if we look at:
$$\delta (R^{ab}R_{ab}) = \delta (R_{cd}g^{ca}g^{db}R_{ab}) = 2 R_{cd}g^{ca}g^{db} \delta R_{ab} + 2 R_{cd}g^{ca} R_{ab} \delta g^{db} = 2 R^{ab}\delta R_{ab} + 2 g^{ca}R_{cd}R_{ab} \delta g^{db}$$
you can see that I dropped the second term, which being already proportional to the metric variation adds a term directly to the field equations.
Anyway, I found an article which states the result of considering the curvature scalars:
$$R, R^{ab}R_{ab}, R^{abcd}R_{abcd}$$
http://arxiv.org/abs/astro-ph/0410031
The general line of argument is fairly simple,
1) I take a Lagrangian that doesn't have explicit time or spatial dependence.
2) I therefore expect the stress energy should be divergence-less, and hence can get a new relation on the divergence of some curvature terms.
So even without trudging through the math, I hope some people here can comment.
Is this method valid?
Last edited: Apr 22, 2010
14. Apr 22, 2010
### Altabeh
Yes, but look at the the notation! Later on in your calculations, you make use of $$R$$ as the curvature scalar within the field equations and I can't understand how that is going to be possible to introduce the fundamental laws of GR with this $$R.$$ Either you've written the post in a bad way or I'm getting it all wrong, this confusion in the text must be somehow cleared out so I can tell you if the method works or not!
Your problem is that you are just on a roll when doing all calculations and seems like you don't even trouble yourself to check what you've written clearly right after everything is done! I don't see in your post #7 anything impying the fact that you dropped the second term in the equations!
Anyways, if the math is seamless, why not? We can admot a new field equation with a new curvature invariant as once Einstein published his own and we read it and started to believe in it! But there are so many other problems with the math: the following is completely erroneous and misleading:
$$- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} + \nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( {R^{ab}}_{;b} + \frac12 g^{ab} R_{;b}) = 0.$$
This is the first time I see something like this.
AB
15. Apr 23, 2010
### JustinLevy
Ugh. Altabeh, again, I appreciate that you take the time to write a response... but it is incredibly frustrating when you reply without apparently taking the time to read what I have even written or am asking. It is very frustrating. Every post of yours in this thread, and a couple in other threads have been like that.
Let me give examples in this thread:
From the opening post, and from the very start of my calculations in post #7, it should have been clear that I was trying to work out a divergence relation involving the divergence of a term metric * the scalar R_ab R^ab.
You write a post saying everything is wrong because you feel the scalar I want is R. And therefore dismiss everything. Clearly showing you didn't even read enough to get the basics of what I am even asking.
---
I then respond explaining again that the curvature scalar I'm focussing on is R_ab R^ab. I also point out that I've redone some of the calculations and found multiple errors. I list some of those errors. I then point out that one can ignore the math for now, as the more helpful information is whether this method for obtaining divergence relations is valid at all.
You again seemingly ignore most of that and focus on the old math instead ... and again ignoring what I wrote, you even complain about an error I already pointed out. And yet again complain about the scalar I'm focussing on, even casting doubt on it by referring to it as a 'so-called "curvature scalar"'.
-----
I respond again. Stressing yet again the curvature scalar I want to focus on is R_ab R^ab. Since you complained about errors, I stated more explicitly this time one of the errors. I again stress that focussing on the math is not as helpful as telling me whether the general method is viable. I summarize this method yet again. I even post a paper I found that gives the result I should find if I go back and fix my errors, so that we can stop focussing on the details of working out the field equations.
What do you do? You yet AGAIN complain about the scalar. You AGAIN refer to errors in my math all the way back in post 7 (which I've already pointed out myself I found errors in), and again ignoring the important parts of the discussion.
-----
Do you see the pattern here?
You seem very knowledgeable. And again, I thank you for taking the time to respond. But it is not useful, and in fact incredibly frustrating, when you seemingly ignore what I write. So please, read a bit more carefully otherwise we'll just be talking past each other every time.
I felt the need to respond to this though. Yes, I've made mistakes. But please don't take this as "you don't even trouble yourself to check what you've written". I put a lot of time into writing up those calculations. And I have checked them and I've caught many mistakes myself, and I can see how to fix them. Since I found a paper that gave the final answer (albeit without showing the work), and I found where my mistakes were, I didn't feel the need to retype up all the calculations.
So, back to the discussion:
I've checked by hand for f(R) gravity that the field equations are divergenceless in general (and not just in solutions to the equations), as I expected. But it would be nice to know that this method works in general. To put in explicitly...
If I do the following steps:
1) put a function of any scalars I want in replace of f(R) in the f(R) lagrangian
2) solve for the field equations
3) look at the vacuum case (T_uv = 0)
4) take the divergence of the vacuum field equation, thus obtaining a relationship between the divergence of many curvature parts
Is the relation I obtain from this method true in general or only true for solutions of those field equations. I've already made heuristic arguments for why I expect them to be true in general, and therefore this is a valid method to generate relationships between the divergence of curvature terms ... but it would be nice to know for sure (and explicitly why) this is the case.
16. Apr 23, 2010
### Altabeh
You seem to have forgotten to take a quick peep at your post #7 in this thread: All you said there was that you introduced an scalar but called it inattentively "scalar curvature" and assigned to it the same symbol we use for the "scalar curvature" $$g^{\mu\nu}R_{\mu\nu}$$ in GR so when I asked for the reason behind defining R in this way, you suggested me to take a look at this http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)" [Broken] where the curvature invariants are discussed briefly without any reference to the fact that all of them use distinctive symbols other than $$R.$$ So am I given this right to get confused when seeing your posts and misleading notations!?
The reason is that you looked like you got us led to the EFE starting from this $$R$$ ending up with the scalar curvature included in the equations. I don't know what makes it all silly to you to have questions like this formed in mind after reading your convoluted post!
Right after you were born, what was your language to be understandable to your parents!? Nothing and it is the reason the baby language is just an inspiration of the inward feelings under the circumstances the baby sees at that moment! Speaking of which, for physics to be understood well, at this level of prestige, the standard language is mathematics but your calculations are all leaky and all we can do about the procedure is to lean on our inward feelings about it unless you provide us with a great indefectible mathematical framework because you're not talking about something really elementary but fully scientific and professional in the real sense of the words! I think the answer I gave in an early post gets everything straight:
Anyways, if the math is seamless, why not? We can admit a new field equation with a new curvature invariant as once Einstein published his own and we read it and started to believe in it! But there are so many other problems with the math:
As the title of this thread says all we are going to talk about is mathematical so if I got the time to spend on checking the math, I'd be glad to announce the results soon! Remember that all you're doing here is strongly backed up by mathematics and if this part isn't seamless, I can't even predict what'll happen to the theory in the end!
Your theory has to be divergence-less and this can be obtained by taking the covariant derivative of the left-hand side of the field equations without even looking at the other side. This step compels the theory 1) go astray from the main road, i.e. the generality of equations for any material distribution given, and 2) nullifies the effect of "divergence" because it's already zero! It is mandatory for this step to be modified to give the correct equations! Also
will only be valid then in a vacuum spacetime.
If your theory makes the following general demands met, then it works fine:
1- The modification I'm taking about is a necessity for your theory to be "general";
2- The relation you obtained, i.e.
\begin{align*} (g^{ab} R^{cd}R_{cd})_{;b} &= 2 (\nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d g^{ab}R_{;b}) \\&= 2 (\nabla^a\nabla^b\nabla_b R - \nabla^d\nabla_d \nabla^a R) \\&= 0\end{align*}
has many wrong things with it! If this was flawless mathematically, your theory at least in a limited sense would rock!
Are these helpful to you or yet you think I'm stuck on nagging at your $$R$$?
AB
Last edited by a moderator: May 4, 2017
17. Apr 23, 2010
### JustinLevy
I am using standard notation.
$$R_{ab}$$ is used to refer to the Ricci curvature. The curvature scalar I am talking about is, once again,
$$R^{ab}R_{ab}$$
which is of course not
$$R=g^{ab}R_{ab}$$
the Ricci scalar.
It is clear from the very first line in this thread what curvature scalar I was referring to. It baffles me that you are complaining that I said in post #7 "variation of the curvature scalar" and then I wrote the curvature scalar I was referring to once again ... how can you complain that is unclear even now? Fine, you misunderstood. But I've tried to correct your misunderstanding several times now.
The fact that you are now starting to use "R" to refer to the curvature scalar I was referring to, is only making me worry you still don't understand. Especially since I am using standard tensor notation here. There should be no ambiguity.
There are many curvature scalars. If you need to give this one a name, wikipedia refers to it as "The principal quadratic invariant of the Ricci tensor" in that link I gave you. R^{ab}R_{ab} is unambiguous and shorter.
I hope we have finally settled any notation issues.
That is completely missing the point of what I'm trying to get here. I'm generating field equations from a Lagrangian, and using the method I've described several times now, in hopes to generate general relations between curvature terms. I'm not using the Lagrangian to suggest new physics.
And in that post, as in even your latest post, you again continue to complain about the specific math in post 7 which is frustrating, considering I already pointed out myself that there were errors and therefore we should focus on whether the method is viable. Heck I even posted a paper containing the correct field equations that I would obtain if I went back and fixed the errors.
If you insist on seeing the correct field equations, let me link the paper again:
http://arxiv.org/PS_cache/astro-ph/pdf/0410/0410031v2.pdf
equation 15 are the field equations
Going through all the math to get the field equations is unimportant if the method I'm proposing to use these to get general divergence relations is wrong. I have done no math to prove the method I suggest is valid, only given heuristic arguments. I don't know how to prove one way or the other (besides getting the equations and checking by hand that they have the property I expected). If you know how to do either of these mathematically:
1) How to calculate the covariant gradient of a non-tensor object (like a Christoffel symbol), in which case I could calculate any gradient of a curvature scalar built from the Reimann curvature directly.
or
2) How to prove that the method I'm proposing to generate divergence relations, which hold in general (and not just in some specific theory), will work.
Then that would be of great help, and I'd love to hear those math details.
Complaining about mistakes in post #7, that I've already pointed out myself and furthermore linked to the correct equations later, is not helpful.
I'm sorry. I've reread this multiple times. I cannot figure out what you are trying to say here.
Are you saying the field equations obtained this way "have to be divergence-less" due to the mathematical way in which they were derived? If so, can you please show me how you know that? That is what I've been asking about this method.
But then you go on to say that looking at the vacuum case "compels the theory 1) go astray from the main road". And that "It is mandatory for this step to be modified to give the correct equations!" Modified to what? Astray from what? I think you are still misunderstanding what this thread is about. I am trying to obtain relations between divergences of curvature terms. And I hit about this some-what bizarre method to generate them, and I'm pursuing it because I don't know how to find them another way (and no one has suggested an alternate way).
The final comment you make there is that what I obtain will NOT be relations which hold in general. So you seem to strongly disagree with my method, but this seems to contradict your first statement that the equations must be divergence-less.
Did you only mean they "have to be divergence-less" in the sense that for the field equations to make physical sense the solutions to the equations must be divergence-less?
Last edited by a moderator: May 4, 2017
18. Apr 23, 2010
### Mentz114
I believe this is the case but I can't get corroboration. I'm pretty sure the Einstein tensor obeys ${G^{\mu\nu}}_{;\nu}=0$ identically. Any canonical EMT obeys this, so I think G must also.
Got it. It's confirmed here
http://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry
It follows from the Bianchi identity of course !
Last edited: Apr 23, 2010
19. Apr 23, 2010
### JustinLevy
Yeah, it really seems like it should, and every case I've been able to work out by hand has worked so far, but I don't know how to prove it.
Yeah, that's probably the easiest case. Unfortunately that series of contractions on the Bianchi identity is the only useful one. So to get anymore equations relating the divergence of curvature terms, I need to resort to something more drastic.
The ${G^{\mu\nu}}_{;\nu}=0$ is the relation we get when using the gravitational action:
$$S= - {1 \over 2\kappa}\int R \sqrt{-g} d^4x$$
I'm hoping that replacing R with any function of curvature scalars will yield "field equations" that are in general divergence-free. In vacuum the solutions must be, so I'm worried that all this is showing is that solutions are divergence free ... and thus I can't use this to get a new set of divergence relations for curvature terms.
However, since the action would have no explicit time or spatial dependence regardless of the function of curvature scalars chosen, I'd expect that the field equations from such a modified lagrangian would be divergence free. As I mentioned previously, I've check it a couple cases and it has been true so far. Maybe it is a simple step to show that this is always the case, so any guidance along that direction would be much appreciated.
20. Apr 24, 2010
### Altabeh
Let's face it: You as someone who does not deal with physics on a regular basis (at least I can smell it from the extraordinarily flawed mathematics of yours), just defined something in your own sense and kept talking about it in some post! Okay and there is no problem with this, but what do you expect from the reader when you bring up the Einstein's field equations (EFE) with a completely different $$R$$ in the same post just a few lines right below an streak of equations involved with your own $$R$$!? What does it mean and what should we do there to find a comparison between them while there is such huge ambigiuty and a nerve-wracking violation of energy\momentum conservation?
So your notation is NOT satandard and to keep notations in agreement with the standards established in that paper you just cited in your latest post, you would have used $$P$$ instead of $$R$$ which, as in the same paper, refers basically to the standard scalar curvature $$g^{ab}R_{ab}.$$
No quite contrary! You seem to first have gone way far from the topic, and secondly I don't assume you're trying to insult me by making a duffer out of me! People need to tease out things that sound unclear to them and if this bothers you, I'll be out of here because I don't want to be gone off on this way anymore! I'm here to help so just stay cool and keep to the subject!
Considering all the mathematics in your post #7 has serious problems in it, the relation you're giving, i.e.
$$- (\frac12 g^{ab} R^{cd}R_{cd})_{;b} + \nabla^a\nabla^b R_{;b} - \nabla^d\nabla_d ( {R^{ab}}_{;b} + \frac12 g^{ab} R_{;b}) = 0$$
has only one way of being interpreted, say, logically and physically when compared to the field equations obtained in the paper "The Cosmology of Generalized Modified Gravity Models" as follows: The mathematics, as I examined this discreetly myself, does not prove the relation right and so you have to go with some other alternative which is the same method the authors of the article above take advantage of. The method is strongly based on a new parameter $$\mu$$ appearing in a sort of famous Lagrangian used earlier by many other authors (see, for instance, the equation (1) in http://arxiv.org/PS_cache/astro-ph/pdf/0306/0306438v2.pdf" [Broken] This value since is present as a multiple in the correction terms, can be adjusted the way that
1) the amount of non-conserved quantities must get really small (this is shown to be of order 10^{-33}eV for a slow-expanding universe so the violation is really negligible);
2) the model keeps to be on a par with both observational data and the property of self-consistency of EFE (this means that invoking the conservation laws to hold in the new theory has nothing to do with the energy tensor being zero or not. They must be satisfied in any form).
Besides these, the EFE must be extractable from the new Lagrangian by setting the new parameter equal to some constant, basically, zero! But in your Lagrangian none of these are obeyed. Specially that your Lagrangian does not reduce to the Einstein-Hilbert Lagrangian.Also, we don't have a degree of freedom to adjust it according to the model under study. I feel like you changed your mind about using the primitive idea of getting some free values involved in the field equations.
For what reason do we need to calculate all of them when we know that the theory is vacuous? The basic Lagrangian must be modified in such a way that the EFE be extracted from it when all the correction terms vanish. Yours is not even right in the sign of the second term on the left hand side of the EFE.
If you have no knowledge of math, why did you write all those void equations in post #7? Was that a practice essay or a play-around with equations? If you don't know that we spend time to read them, you better never think of writing nonsense again!
They have to be, at the very very least, minimally divergence-less as they are in Carroll's papers. But in your Lagrangian this is not possible because first a mathematical derivation cannot be obtained and secondly there is nothing to be adjusted and the Lagrangian itself does not reduce to the fundamental Lagrangian of GR. This property is indeed essential to constructing a self-consistent field equation and if it is not cared about, you have to force the material distribution to consist of dark energy to compensate for the probable lost mass\momentum in interactions with ordinary matter. Looking for a relation between scalar curvature terms in the way you are into, is not feasible unless a necessary modification of the proposed Lagrangian is carried out!
Going astray from keeping the theory at minimal level of the violation of the conservations laws. | 9,221 | 33,438 | {"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} | 3.96875 | 4 | CC-MAIN-2018-51 | latest | en | 0.887903 |
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2002.Fall/Tillema/assign10et/assign10et.html | 1,544,883,121,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.91/warc/CC-MAIN-20181215131038-20181215153038-00093.warc.gz | 145,335,827 | 3,953 | Assignment 10
The Cycloid
Overview:
1)Definition and history of a cycloid
2)Exploring a cycloid in GSP to get an intuitive feel about how it works
3)Looking at the parametric equations of a cycloid
1) What is a cycloid?
It is the locus of points created by a point on a circle as the circle rolls along a line.
History of the cycloid:
The cycloid was first studied by a man of the name of Cusa in his attempt to find the area of a circle using integration. Cusa lived in present day Northern Italy from 1401-1464. He was a philosopher interested in the study of the infinite, seeing the circle as the limit of regular polygons. He attempted to find the area of a cycloid using his idea about the area of circles although did so unsuccessfully. The first person to properly define a cycloid was a Frenchman by the name of Marsene (1588-1648). His other mathematical interests included numbers of the form n = 2^p - 1 where p is a prime number. In 1599, Galileo named the cycloid and after studying its form for forty years, unsuccessfully calculating the area he wrote to several other mathematicians and so began a lengthy dialogue about the properties of a cycloid.
2) Let's look at a GSP construction of a cycloid.
The blue point in the diagram is the point on the circle which is being traced as it moves along the circle. One point to be careful about when doing the GSP construction is that as your point rotates around the circle make sure that it is rotating in the same direction as the direction your circle is traveling along the line otherwise you will get "an upside" down cycloid or a brachistochrone. For a GSP file to investigate this yourself click here.
What can we note from the GSP construction in terms of translating this into parametric co-ordinates?
First, if we assume the radius of the circle is r, then we know that the height of the cycloid or y-co-ordinate in the parametric equation needs to fluctuate between y = 0 and y = 2r. We also know that the graph is periodic and that its period is 2 * pi * r. We need to also incorporate the motion of the circle along the straight line in our equation. What equation can we use to get this period and also take into account the movement of the circle? Let's look at the picture in GSP superimposed on a Cartesian Co-ordinate system.
First, note that A is the point making the cycloid, radius BC is perpendicular to the x-axis, angle ADB is a right angle, and that theta is equal to angle ABC and can vary between 0 and 2 pi. So we get that the x co-ordinate should be equal to arc length AC minus the length of segment DB.
and so we get but how can we be sure that we are always subtracting DB because when A is greater than half way we should be adding. This works out though because sine is positive between 0 and pi and negative between pi and 2 pi.
Now, let's think about the y co-ordinate of the graph. The y co-ordinate should be equal to segment BC plus segment AD
and but now we have the trouble of the sign of cosine. Since we want to be adding segment AD when A lies in the range between pi/2 and 3pi/2 and subtracting AD when A is between 0 and pi/2 and 3pi/2 and 2 pi we must take into when the cosine function is negative and when it is positive. Cosine is negative from pi/2 to 3pi/2 and is positive between 0 and pi/2 and 3pi/2 and 2 pi so we get .
3) Investigating the parmetric equations.
This gives us are parametric equations. Let's use graphing calculator and make sure that these work out alright (I have substituted t for theta because of the way graphing calculator is set up).
It looks like we got it!!! This is an example with the radius equal to 1. If the radius of the original circle is bigger than one then we simply mulitply both equations by the value of the radius and we get a period that is greater as well as a maximum height that is higher. In fact the period should be equal to 2 * pi * r and the height equal to 2 * r.
Now let's look at what happens when we manipulate these parametric equations in various different ways. For instance what happens when we multiply the first term of each equation by the same number and the second term of each equation by a different number. In other words, let's look at equations of the form below:
Where we allow a, and b to be different values.
First let's let b > a and see what happens. How about if b= 1.5 and a = 1 what graph do we get?
We get a part of the graph that looks like this. How can we explain this in terms of the definition of a cycloid. If b > a, then we have the locus of points created by a circle traveling along a line and a point that lies on a circle with a larger radius than the circle traveling along the line. The reason for the ray in the picture is if you use GSP to set up this construction you have to make sure that the point on the outer circle is moving at the right speed in relation to the rate of speed of the smaller circle along the line. A graph with similar characteristics would be made by tracing point F on the outer circle.
As you can see deriving this equation would be similar to the dervation of the first equation but would also include the radii of the larger circle. I will leave this derivation up to the reader. This type of cycloid is called a hypocycloid.
What happens if a > b? Let's look at a =1.5 and b = 1.
We get a slightly different path here. Instead of having a path that goes below the x-axis, the path stays above. This graph corresponds to the locus of points of a point on a circle with smaller radius than the circle that is in motion along the line. As shown below:
Where the graph is made by tracing the point E as the larger circle moves along a line. Again, allowing E to always lie on segment AB allows for E to be moving at the same speed as the larger circle. This type of cycloid is called an epicycloid. For final comparison let's look at all three of these graphs on the same axis.
Interestingly enough the period of the hypocycloid and the cycloid are the same but the period of the epicycloid is greater. Also, note that the greatest y-value achieved is the same for the hypocycloid and the epicycloid and is greater than the greatest y-value of the cycloid. For now this finish the discussion of cycloids. | 1,458 | 6,266 | {"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} | 4.4375 | 4 | CC-MAIN-2018-51 | latest | en | 0.962721 |
https://openstax.org/books/calculus-volume-2/pages/5-2-infinite-series | 1,726,223,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651513.89/warc/CC-MAIN-20240913101949-20240913131949-00061.warc.gz | 390,466,912 | 97,196 | Calculus Volume 2
# 5.2Infinite Series
Calculus Volume 25.2 Infinite Series
## Learning Objectives
• 5.2.1 Explain the meaning of the sum of an infinite series.
• 5.2.2 Calculate the sum of a geometric series.
• 5.2.3 Evaluate a telescoping series.
We have seen that a sequence is an ordered set of terms. If you add these terms together, you get a series. In this section we define an infinite series and show how series are related to sequences. We also define what it means for a series to converge or diverge. We introduce one of the most important types of series: the geometric series. We will use geometric series in the next chapter to write certain functions as polynomials with an infinite number of terms. This process is important because it allows us to evaluate, differentiate, and integrate complicated functions by using polynomials that are easier to handle. We also discuss the harmonic series, arguably the most interesting divergent series because it just fails to converge.
## Sums and Series
An infinite series is a sum of infinitely many terms and is written in the form
$∑n=1∞an=a1+a2+a3+⋯.∑n=1∞an=a1+a2+a3+⋯.$
But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the limit of partial sums. A partial sum of an infinite series is a finite sum of the form
$∑n=1kan=a1+a2+a3+⋯+ak.∑n=1kan=a1+a2+a3+⋯+ak.$
To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that $10001000$ gallons enters the lake the first week. During the second week, an additional $500500$ gallons of oil enters the lake. The third week, $250250$ more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after $kk$ weeks. Letting $SkSk$ denote the amount of oil in the lake (measured in thousands of gallons) after $kk$ weeks, we see that
$S1=1S2=1+0.5=1+12S3=1+0.5+0.25=1+12+14S4=1+0.5+0.25+0.125=1+12+14+18S5=1+0.5+0.25+0.125+0.0625=1+12+14+18+116.S1=1S2=1+0.5=1+12S3=1+0.5+0.25=1+12+14S4=1+0.5+0.25+0.125=1+12+14+18S5=1+0.5+0.25+0.125+0.0625=1+12+14+18+116.$
Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after $kk$ weeks is
$Sk=1+12+14+18+116+⋯+12k−1=∑n=1k(12)n−1.Sk=1+12+14+18+116+⋯+12k−1=∑n=1k(12)n−1.$
We are interested in what happens as $k→∞.k→∞.$ Symbolically, the amount of oil in the lake as $k→∞k→∞$ is given by the infinite series
$∑n=1∞(12)n−1=1+12+14+18+116+⋯.∑n=1∞(12)n−1=1+12+14+18+116+⋯.$
At the same time, as $k→∞,k→∞,$ the amount of oil in the lake can be calculated by evaluating $limk→∞Sk.limk→∞Sk.$ Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums ${Sk}.{Sk}.$ If the sequence of partial sums ${Sk}{Sk}$ converges, we say that the infinite series converges, and its sum is given by $limk→∞Sk.limk→∞Sk.$ If the sequence ${Sk}{Sk}$ diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence ${Sk}.{Sk}.$
First, simplifying some of these partial sums, we see that
$S1=1S2=1+12=32S3=1+12+14=74S4=1+12+14+18=158S5=1+12+14+18+116=3116.S1=1S2=1+12=32S3=1+12+14=74S4=1+12+14+18=158S5=1+12+14+18+116=3116.$
Plotting some of these values in Figure 5.10, it appears that the sequence ${Sk}{Sk}$ could be approaching 2.
Figure 5.10 The graph shows the sequence of partial sums ${Sk}.{Sk}.$ It appears that the sequence is approaching the value $2.2.$
Let’s look for more convincing evidence. In the following table, we list the values of $SkSk$ for several values of $k.k.$
$kk$ $55$ $1010$ $1515$ $2020$ $SkSk$ $1.93751.9375$ $1.9981.998$ $1.9999391.999939$ $1.9999981.999998$
These data supply more evidence suggesting that the sequence ${Sk}{Sk}$ converges to $2.2.$ Later we will provide an analytic argument that can be used to prove that $limk→∞Sk=2.limk→∞Sk=2.$ For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to $2.2.$ Since this sequence of partial sums converges to $2,2,$ we say the infinite series converges to $22$ and write
$∑n=1∞(12)n−1=2.∑n=1∞(12)n−1=2.$
Returning to the question about the oil in the lake, since this infinite series converges to $2,2,$ we conclude that the amount of oil in the lake will get arbitrarily close to $20002000$ gallons as the amount of time gets sufficiently large.
This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.
## Definition
An infinite series is an expression of the form
$∑n=1∞an=a1+a2+a3+⋯.∑n=1∞an=a1+a2+a3+⋯.$
For each positive integer $k,k,$ the sum
$Sk=∑n=1kan=a1+a2+a3+⋯+akSk=∑n=1kan=a1+a2+a3+⋯+ak$
is called the $kthkth$ partial sum of the infinite series. The partial sums form a sequence ${Sk}.{Sk}.$ If the sequence of partial sums converges to a real number $S,S,$ the infinite series converges. If we can describe the convergence of a series to $S,S,$ we call $SS$ the sum of the series, and we write
$∑n=1∞an=S.∑n=1∞an=S.$
If the sequence of partial sums diverges, we have the divergence of a series.
Note that the index for a series need not begin with $n=1n=1$ but can begin with any value. For example, the series
$∑n=1∞(12)n−1∑n=1∞(12)n−1$
can also be written as
$∑n=0∞(12)nor∑n=5∞(12)n−5.∑n=0∞(12)nor∑n=5∞(12)n−5.$
Often it is convenient for the index to begin at $1,1,$ so if for some reason it begins at a different value, we can reindex by making a change of variables. For example, consider the series
$∑n=2∞1n2.∑n=2∞1n2.$
By introducing the variable $m=n−1,m=n−1,$ so that $n=m+1,n=m+1,$ we can rewrite the series as
$∑m=1∞1(m+1)2.∑m=1∞1(m+1)2.$
## Example 5.7
### Evaluating Limits of Sequences of Partial Sums
For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.
1. $∑n=1∞nn+1∑n=1∞nn+1$
2. $∑n=1∞(−1)n∑n=1∞(−1)n$
3. $∑n=1∞1n(n+1)∑n=1∞1n(n+1)$
## Checkpoint5.7
Determine whether the series $∑n=1∞(n+1)/n∑n=1∞(n+1)/n$ converges or diverges.
### The Harmonic Series
A useful series to know about is the harmonic series. The harmonic series is defined as
$∑n=1∞1n=1+12+13+14+⋯.∑n=1∞1n=1+12+13+14+⋯.$
(5.5)
This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums ${Sk}{Sk}$ approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence ${Sk}{Sk}$ in the following table.
$kk$ $1010$ $100100$ $10001000$ $10,00010,000$ $100,000100,000$ $1,000,0001,000,000$ $SkSk$ $2.928972.92897$ $5.187385.18738$ $7.485477.48547$ $9.787619.78761$ $12.0901512.09015$ $14.3927314.39273$
Even after $1,000,0001,000,000$ terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.
To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:
$S1=1S2=1+12S3=1+12+13S4=1+12+13+14.S1=1S2=1+12S3=1+12+13S4=1+12+13+14.$
Notice that for the last two terms in $S4,S4,$
$13+14>14+14.13+14>14+14.$
Therefore, we conclude that
$S4>1+12+(14+14)=1+12+12=1+2(12).S4>1+12+(14+14)=1+12+12=1+2(12).$
Using the same idea for $S8,S8,$ we see that
$S8=1+12+13+14+15+16+17+18>1+12+(14+14)+(18+18+18+18)=1+12+12+12=1+3(12).S8=1+12+13+14+15+16+17+18>1+12+(14+14)+(18+18+18+18)=1+12+12+12=1+3(12).$
From this pattern, we see that $S1=1,S1=1,$ $S2=1+1/2,S2=1+1/2,$ $S4>1+2(1/2),S4>1+2(1/2),$ and $S8>1+3(1/2).S8>1+3(1/2).$ More generally, it can be shown that $S2j>1+j(1/2)S2j>1+j(1/2)$ for all $j>1.j>1.$ Since $1+j(1/2)→∞,1+j(1/2)→∞,$ we conclude that the sequence ${Sk}{Sk}$ is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since ${Sk}{Sk}$ is unbounded, it diverges. Thus, the harmonic series diverges.
### Algebraic Properties of Convergent Series
Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.
## Theorem5.7
### Algebraic Properties of Convergent Series
Let $∑n=1∞an∑n=1∞an$ and $∑n=1∞bn∑n=1∞bn$ be convergent series. Then the following algebraic properties hold.
1. The series $∑n=1∞(an+bn)∑n=1∞(an+bn)$ converges and $∑n=1∞(an+bn)=∑n=1∞an+∑n=1∞bn.∑n=1∞(an+bn)=∑n=1∞an+∑n=1∞bn.$ (Sum Rule)
2. The series $∑n=1∞(an−bn)∑n=1∞(an−bn)$ converges and $∑n=1∞(an−bn)=∑n=1∞an−∑n=1∞bn.∑n=1∞(an−bn)=∑n=1∞an−∑n=1∞bn.$ (Difference Rule)
3. For any real number $c,c,$ the series $∑n=1∞can∑n=1∞can$ converges and $∑n=1∞can=c∑n=1∞an.∑n=1∞can=c∑n=1∞an.$ (Constant Multiple Rule)
## Example 5.8
### Using Algebraic Properties of Convergent Series
Evaluate
$∑n=1∞[3n(n+1)+(12)n−2].∑n=1∞[3n(n+1)+(12)n−2].$
## Checkpoint5.8
Evaluate $∑n=1∞52n−1.∑n=1∞52n−1.$
### Geometric Series
A geometric series is any series that we can write in the form
$a+ar+ar2+ar3+⋯=∑n=1∞arn−1.a+ar+ar2+ar3+⋯=∑n=1∞arn−1.$
(5.6)
Because the ratio of each term in this series to the previous term is r, the number r is called the ratio. We refer to a as the initial term because it is the first term in the series. For example, the series
$∑n=1∞(12)n−1=1+12+14+18+⋯∑n=1∞(12)n−1=1+12+14+18+⋯$
is a geometric series with initial term $a=1a=1$ and ratio $r=1/2.r=1/2.$
In general, when does a geometric series converge? Consider the geometric series
$∑n=1∞arn−1∑n=1∞arn−1$
when $a>0.a>0.$ Its sequence of partial sums ${Sk}{Sk}$ is given by
$Sk=∑n=1karn−1=a+ar+ar2+⋯+ark−1.Sk=∑n=1karn−1=a+ar+ar2+⋯+ark−1.$
Consider the case when $r=1.r=1.$ In that case,
$Sk=a+a(1)+a(1)2+⋯+a(1)k−1=ak.Sk=a+a(1)+a(1)2+⋯+a(1)k−1=ak.$
Since $a>0,a>0,$ we know $ak→∞ak→∞$ as $k→∞.k→∞.$ Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for $r=1.r=1.$ For $r≠1,r≠1,$ to find the limit of ${Sk},{Sk},$ multiply Equation 5.6 by $1−r.1−r.$ Doing so, we see that
$(1−r)Sk=a(1−r)(1+r+r2+r3+⋯+rk−1)=a[(1+r+r2+r3+⋯+rk−1)−(r+r2+r3+⋯+rk)]=a(1−rk).(1−r)Sk=a(1−r)(1+r+r2+r3+⋯+rk−1)=a[(1+r+r2+r3+⋯+rk−1)−(r+r2+r3+⋯+rk)]=a(1−rk).$
All the other terms cancel out.
Therefore,
$Sk=a(1−rk)1−rforr≠1.Sk=a(1−rk)1−rforr≠1.$
From our discussion in the previous section, we know that the geometric sequence $rk→0rk→0$ if $|r|<1|r|<1$ and that $rkrk$ diverges if $|r|>1|r|>1$ or $r=±1.r=±1.$ Therefore, for $|r|<1,|r|<1,$ $Sk→a/(1−r)Sk→a/(1−r)$ and we have
$∑n=1∞arn−1=a1−rif|r|<1.∑n=1∞arn−1=a1−rif|r|<1.$
If $|r|≥1,|r|≥1,$ $SkSk$ diverges, and therefore
$∑n=1∞arn−1diverges if|r|≥1.∑n=1∞arn−1diverges if|r|≥1.$
## Definition
A geometric series is a series of the form
$∑n=1∞arn−1=a+ar+ar2+ar3+⋯.∑n=1∞arn−1=a+ar+ar2+ar3+⋯.$
If $|r|<1,|r|<1,$ the series converges, and
$∑n=1∞arn−1=a1−rfor|r|<1.∑n=1∞arn−1=a1−rfor|r|<1.$
(5.7)
If $|r|≥1,|r|≥1,$ the series diverges.
Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than $n=1n=1$ or the exponent involves a linear expression for $nn$ other than $n−1.n−1.$ As long as we can rewrite the series in the form given by Equation 5.5, it is a geometric series. For example, consider the series
$∑n=0∞(23)n+2.∑n=0∞(23)n+2.$
To see that this is a geometric series, we write out the first several terms:
$∑n=0∞(23)n+2=(23)2+(23)3+(23)4+⋯=49+49·(23)+49·(23)2+⋯.∑n=0∞(23)n+2=(23)2+(23)3+(23)4+⋯=49+49·(23)+49·(23)2+⋯.$
We see that the initial term is $a=4/9a=4/9$ and the ratio is $r=2/3.r=2/3.$ Therefore, the series can be written as
$∑n=1∞49·(23)n−1.∑n=1∞49·(23)n−1.$
Since $r=2/3<1,r=2/3<1,$ this series converges, and its sum is given by
$∑n=1∞49·(23)n−1=4/91−2/3=43.∑n=1∞49·(23)n−1=4/91−2/3=43.$
## Example 5.9
### Determining Convergence or Divergence of a Geometric Series
Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.
1. $∑n=1∞(−3)n+14n−1∑n=1∞(−3)n+14n−1$
2. $∑n=1∞e2n∑n=1∞e2n$
## Checkpoint5.9
Determine whether the series $∑n=1∞(−25)n−1∑n=1∞(−25)n−1$ converges or diverges. If it converges, find its sum.
We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.
## Example 5.10
### Writing Repeating Decimals as Fractions of Integers
Use a geometric series to write $3.26—3.26—$ as a fraction of integers.
## Checkpoint5.10
Write $5.27–5.27–$ as a fraction of integers.
## Example 5.11
### Chapter Opener: Finding the Area of the Koch Snowflake
Define a sequence of figures ${Fn}{Fn}$ recursively as follows (Figure 5.11). Let $F0F0$ be an equilateral triangle with sides of length $1.1.$ For $n≥1,n≥1,$ let $FnFn$ be the curve created by removing the middle third of each side of $Fn−1Fn−1$ and replacing it with an equilateral triangle pointing outward. The limiting figure as $n→∞n→∞$ is known as Koch’s snowflake.
Figure 5.11 The first four figures, $F0,F1,F2,andF3,F0,F1,F2,andF3,$ in the construction of the Koch snowflake.
1. Find the length $LnLn$ of the perimeter of $Fn.Fn.$ Evaluate $limn→∞Lnlimn→∞Ln$ to find the length of the perimeter of Koch’s snowflake.
2. Find the area $AnAn$ of figure $Fn.Fn.$ Evaluate $limn→∞Anlimn→∞An$ to find the area of Koch’s snowflake.
### Analysis
The Koch snowflake is interesting because it has finite area, yet infinite perimeter. Although at first this may seem impossible, recall that you have seen similar examples earlier in the text. For example, consider the region bounded by the curve $y=1/x2y=1/x2$ and the $xx$-axis on the interval $[1,∞).[1,∞).$ Since the improper integral
$∫1∞1x2dx∫1∞1x2dx$
converges, the area of this region is finite, even though the perimeter is infinite.
## Telescoping Series
Consider the series $∑n=1∞1n(n+1).∑n=1∞1n(n+1).$ We discussed this series in Example 5.7, showing that the series converges by writing out the first several partial sums $S1,S2,…,S6S1,S2,…,S6$ and noticing that they are all of the form $Sk=kk+1.Sk=kk+1.$ Here we use a different technique to show that this series converges. By using partial fractions, we can write
$1n(n+1)=1n−1n+1.1n(n+1)=1n−1n+1.$
Therefore, the series can be written as
$∑n=1∞[1n−1n+1]=(1-12)+(12−13)+(13−14)+⋯.∑n=1∞[1n−1n+1]=(1-12)+(12−13)+(13−14)+⋯.$
Writing out the first several terms in the sequence of partial sums ${Sk},{Sk},$ we see that
$S1=1−12S2=(1−12)+(12−13)=1−13S3=(1−12)+(12−13)+(13−14)=1−14.S1=1−12S2=(1−12)+(12−13)=1−13S3=(1−12)+(12−13)+(13−14)=1−14.$
In general,
$Sk=(1−12)+(12−13)+(13−14)+⋯+(1k−1k+1)=1−1k+1.Sk=(1−12)+(12−13)+(13−14)+⋯+(1k−1k+1)=1−1k+1.$
We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since $Sk=1−1/(k+1)Sk=1−1/(k+1)$ and $1/(k+1)→01/(k+1)→0$ as $k→∞,k→∞,$ the sequence of partial sums converges to $1,1,$ and therefore the series converges to $1.1.$
## Definition
A telescoping series is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.
For example, any series of the form
$∑n=1∞[bn−bn+1]=(b1−b2)+(b2−b3)+(b3−b4)+⋯∑n=1∞[bn−bn+1]=(b1−b2)+(b2−b3)+(b3−b4)+⋯$
is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that
$S1=b1−b2S2=(b1−b2)+(b2−b3)=b1−b3S3=(b1−b2)+(b2−b3)+(b3−b4)=b1−b4.S1=b1−b2S2=(b1−b2)+(b2−b3)=b1−b3S3=(b1−b2)+(b2−b3)+(b3−b4)=b1−b4.$
In general, the kth partial sum of this series is
$Sk=b1−bk+1.Sk=b1−bk+1.$
Since the kth partial sum can be simplified to the difference of these two terms, the sequence of partial sums ${Sk}{Sk}$ will converge if and only if the sequence ${bk+1}{bk+1}$ converges. Moreover, if the sequence $bk+1bk+1$ converges to some finite number $B,B,$ then the sequence of partial sums converges to $b1−B,b1−B,$ and therefore
$∑n=1∞[bn−bn+1]=b1−B.∑n=1∞[bn−bn+1]=b1−B.$
In the next example, we show how to use these ideas to analyze a telescoping series of this form.
## Example 5.12
### Evaluating a Telescoping Series
Determine whether the telescoping series
$∑n=1∞[cos(1n)−cos(1n+1)]∑n=1∞[cos(1n)−cos(1n+1)]$
converges or diverges. If it converges, find its sum.
## Checkpoint5.11
Determine whether $∑n=1∞[e1/n−e1/(n+1)]∑n=1∞[e1/n−e1/(n+1)]$ converges or diverges. If it converges, find its sum.
## Student Project
### Euler’s Constant
We have shown that the harmonic series $∑n=1∞1n∑n=1∞1n$ diverges. Here we investigate the behavior of the partial sums $SkSk$ as $k→∞.k→∞.$ In particular, we show that they behave like the natural logarithm function by showing that there exists a constant $γγ$ such that
$∑n=1k1n−lnk→γask→∞.∑n=1k1n−lnk→γask→∞.$
This constant $γγ$ is known as Euler’s constant.
1. Let $Tk=∑n=1k1n−lnk.Tk=∑n=1k1n−lnk.$ Evaluate $TkTk$ for various values of $k.k.$
2. For $TkTk$ as defined in part 1. show that the sequence ${Tk}{Tk}$ converges by using the following steps.
1. Show that the sequence ${Tk}{Tk}$ is monotone decreasing. (Hint: Show that $ln(1+1/k>1/(k+1)))ln(1+1/k>1/(k+1)))$
2. Show that the sequence ${Tk}{Tk}$ is bounded below by zero. (Hint: Express $lnklnk$ as a definite integral.)
3. Use the Monotone Convergence Theorem to conclude that the sequence ${Tk}{Tk}$ converges. The limit $γγ$ is Euler’s constant.
3. Now estimate how far $TkTk$ is from $γγ$ for a given integer $k.k.$ Prove that for $k≥1,k≥1,$ $0 by using the following steps.
1. Show that $ln(k+1)−lnk<1/k.ln(k+1)−lnk<1/k.$
2. Use the result from part a. to show that for any integer $k,k,$
$Tk−Tk+1<1k−1k+1.Tk−Tk+1<1k−1k+1.$
3. For any integers $kk$ and $jj$ such that $j>k,j>k,$ express $Tk−TjTk−Tj$ as a telescoping sum by writing
$Tk−Tj=(Tk−Tk+1)+(Tk+1−Tk+2)+(Tk+2−Tk+3)+⋯+(Tj−1−Tj).Tk−Tj=(Tk−Tk+1)+(Tk+1−Tk+2)+(Tk+2−Tk+3)+⋯+(Tj−1−Tj).$
Use the result from part b. combined with this telescoping sum to conclude that
$Tk−Tj<1k−1j.Tk−Tj<1k−1j.$
4. Apply the limit to both sides of the inequality in part c. to conclude that
$Tk−γ≤1k.Tk−γ≤1k.$
5. Estimate $γγ$ to an accuracy of within $0.001.0.001.$
## Section 5.2 Exercises
Using sigma notation, write the following expressions as infinite series.
67.
$1 + 1 2 + 1 3 + 1 4 + ⋯ 1 + 1 2 + 1 3 + 1 4 + ⋯$
68.
$1 − 1 + 1 − 1 + ⋯ 1 − 1 + 1 − 1 + ⋯$
69.
$1 − 1 2 + 1 3 − 1 4 + ... 1 − 1 2 + 1 3 − 1 4 + ...$
70.
$sin 1 + sin 1 / 2 + sin 1 / 3 + sin 1 / 4 + ⋯ sin 1 + sin 1 / 2 + sin 1 / 3 + sin 1 / 4 + ⋯$
Compute the first four partial sums $S1,…,S4S1,…,S4$ for the series having $nthnth$ term $anan$ starting with $n=1n=1$ as follows.
71.
$a n = n a n = n$
72.
$a n = 1 / n a n = 1 / n$
73.
$a n = sin ( n π / 2 ) a n = sin ( n π / 2 )$
74.
$a n = ( −1 ) n a n = ( −1 ) n$
In the following exercises, compute the general term $anan$ of the series with the given partial sum $Sn.Sn.$ If the sequence of partial sums converges, find its limit $S.S.$
75.
$Sn=1−1n,Sn=1−1n,$ $n≥2n≥2$
76.
$Sn=n(n+1)2,Sn=n(n+1)2,$ $n≥1n≥1$
77.
$S n = n , n ≥ 2 S n = n , n ≥ 2$
78.
$S n = 2 − ( n + 2 ) / 2 n , n ≥ 1 S n = 2 − ( n + 2 ) / 2 n , n ≥ 1$
For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.
79.
$∑ n = 1 ∞ n n + 2 ∑ n = 1 ∞ n n + 2$
80.
$∑ n = 1 ∞ ( 1 − ( −1 ) n ) ) ∑ n = 1 ∞ ( 1 − ( −1 ) n ) )$
81.
$∑n=1∞1(n+1)(n+2)∑n=1∞1(n+1)(n+2)$ (Hint: Use a partial fraction decomposition like that for $∑n=1∞1n(n+1).)∑n=1∞1n(n+1).)$
82.
$∑n=1∞12n+1∑n=1∞12n+1$ (Hint: Follow the reasoning for $∑n=1∞1n.)∑n=1∞1n.)$
Suppose that $∑n=1∞an=1,∑n=1∞an=1,$ that $∑n=1∞bn=−1,∑n=1∞bn=−1,$ that $a1=2,a1=2,$ and $b1=−3.b1=−3.$ Find the sum of the indicated series.
83.
$∑ n = 1 ∞ ( a n + b n ) ∑ n = 1 ∞ ( a n + b n )$
84.
$∑ n = 1 ∞ ( a n − 2 b n ) ∑ n = 1 ∞ ( a n − 2 b n )$
85.
$∑ n = 2 ∞ ( a n − b n ) ∑ n = 2 ∞ ( a n − b n )$
86.
$∑ n = 1 ∞ ( 3 a n + 1 − 4 b n + 1 ) ∑ n = 1 ∞ ( 3 a n + 1 − 4 b n + 1 )$
State whether the given series converges and explain why.
87.
$∑n=1∞1n+1000∑n=1∞1n+1000$ (Hint: Rewrite using a change of index.)
88.
$∑n=1∞1n+1080∑n=1∞1n+1080$ (Hint: Rewrite using a change of index.)
89.
$1 + 1 10 + 1 100 + 1 1000 + ⋯ 1 + 1 10 + 1 100 + 1 1000 + ⋯$
90.
$1 + e π + e 2 π 2 + e 3 π 3 + ⋯ 1 + e π + e 2 π 2 + e 3 π 3 + ⋯$
91.
$1 + π e 2 + π 2 e 4 + π 3 e 6 + π 4 e 8 + ⋯ 1 + π e 2 + π 2 e 4 + π 3 e 6 + π 4 e 8 + ⋯$
92.
$1 − π 3 + π 2 9 − π 3 27 + ⋯ 1 − π 3 + π 2 9 − π 3 27 + ⋯$
For $anan$ as follows, write the sum as a geometric series of the form $∑n=1∞arn.∑n=1∞arn.$ State whether the series converges and if it does, find the value of $∑an.∑an.$
93.
$a1=−1a1=−1$ and $an/an+1=−5an/an+1=−5$ for $n≥1.n≥1.$
94.
$a1=2a1=2$ and $an/an+1=1/2an/an+1=1/2$ for $n≥1.n≥1.$
95.
$a1=10a1=10$ and $an/an+1=10an/an+1=10$ for $n≥1.n≥1.$
96.
$a1=1/10a1=1/10$ and $an/an+1=−10an/an+1=−10$ for $n≥1.n≥1.$
Use the identity $11−y=∑n=0∞yn11−y=∑n=0∞yn$ to express the function as a geometric series in the indicated term.
97.
$x1+xx1+x$ in $xx$
98.
$x1−x3/2x1−x3/2$ in $xx$
99.
$11+sin2x11+sin2x$ in $sinxsinx$
100.
$sec2xsec2x$ in $sinxsinx$
Evaluate the following telescoping series or state whether the series diverges.
101.
$∑ n = 1 ∞ 2 1 / n − 2 1 / ( n + 1 ) ∑ n = 1 ∞ 2 1 / n − 2 1 / ( n + 1 )$
102.
$∑ n = 1 ∞ 1 n 13 − 1 ( n + 1 ) 13 ∑ n = 1 ∞ 1 n 13 − 1 ( n + 1 ) 13$
103.
$∑ n = 1 ∞ ( n − n + 1 ) ∑ n = 1 ∞ ( n − n + 1 )$
104.
$∑ n = 1 ∞ ( sin n − sin ( n + 1 ) ) ∑ n = 1 ∞ ( sin n − sin ( n + 1 ) )$
Express the following series as a telescoping sum and evaluate its nth partial sum.
105.
$∑ n = 1 ∞ ln ( n n + 1 ) ∑ n = 1 ∞ ln ( n n + 1 )$
106.
$∑n=1∞2n+1(n2+n)2∑n=1∞2n+1(n2+n)2$ (Hint: Factor denominator and use partial fractions.)
107.
$∑ n = 2 ∞ ln ( 1 + n 1 ) ln n ln ( n + 1 ) ∑ n = 2 ∞ ln ( 1 + n 1 ) ln n ln ( n + 1 )$
108.
$∑n=1∞(n+2)n(n+1)2n+1∑n=1∞(n+2)n(n+1)2n+1$ (Hint: Look at $1/(n2n).)1/(n2n).)$
A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.
109.
Let $an=f(n)−2f(n+1)+f(n+2),an=f(n)−2f(n+1)+f(n+2),$ in which $f(n)→0f(n)→0$ as $n→∞.n→∞.$ Find $∑n=1∞an.∑n=1∞an.$
110.
$an=f(n)−f(n+1)−f(n+2)+f(n+3),an=f(n)−f(n+1)−f(n+2)+f(n+3),$ in which $f(n)→0f(n)→0$ as $n→∞.n→∞.$ Find $∑n=1∞an.∑n=1∞an.$
111.
Suppose that $an=c0f(n)+c1f(n+1)+c2f(n+2)+c3f(n+3)+c4f(n+4),an=c0f(n)+c1f(n+1)+c2f(n+2)+c3f(n+3)+c4f(n+4),$ where $f(n)→0f(n)→0$ as $n→∞.n→∞.$ Find a condition on the coefficients $c0,…,c4c0,…,c4$ that make this a general telescoping series.
112.
Evaluate $∑n=1∞1n(n+1)(n+2)∑n=1∞1n(n+1)(n+2)$ (Hint: $1n(n+1)(n+2)=12n−1n+1+12(n+2))1n(n+1)(n+2)=12n−1n+1+12(n+2))$
113.
Evaluate $∑n=2∞2n3−n.∑n=2∞2n3−n.$
114.
Find a formula for $∑n=1∞1n(n+N)∑n=1∞1n(n+N)$ where $NN$ is a positive integer.
115.
[T] Define a sequence $tk=∑n=1k−1(1/k)−lnk.tk=∑n=1k−1(1/k)−lnk.$ Use the graph of $1/x1/x$ to verify that $tktk$ is increasing. Plot $tktk$ for $k=1…100k=1…100$ and state whether it appears that the sequence converges.
116.
[T] Suppose that $NN$ equal uniform rectangular blocks are stacked one on top of the other, allowing for some overhang. Archimedes’ law of the lever implies that the stack of $NN$ blocks is stable as long as the center of mass of the top $(N−1)(N−1)$ blocks lies at the edge of the bottom block. Let $xx$ denote the position of the edge of the bottom block, and think of its position as relative to the center of the next-to-bottom block. This implies that $(N−1)x=(12−x)(N−1)x=(12−x)$ or $x=1/(2N).x=1/(2N).$ Use this expression to compute the maximum overhang (the position of the edge of the top block over the edge of the bottom block.) See the following figure.
Each of the following infinite series converges to the given multiple of $ππ$ or $1/π.1/π.$
In each case, find the minimum value of $NN$ such that the $NthNth$ partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to $1515$ decimals place, $π=3.141592653589793....π=3.141592653589793....$
117.
[T] $π=−3+∑n=1∞n2nn!2(2n)!,π=−3+∑n=1∞n2nn!2(2n)!,$ error $<0.0001<0.0001$
118.
[T] $π2=∑k=0∞k!(2k+1)!!=∑k=0∞2kk!2(2k+1)!,π2=∑k=0∞k!(2k+1)!!=∑k=0∞2kk!2(2k+1)!,$ error $<10−4<10−4$
119.
[T] $98012π=49801∑k=0∞(4k)!(1103+26390k)(k!)43964k,98012π=49801∑k=0∞(4k)!(1103+26390k)(k!)43964k,$ error $<10−12<10−12$
120.
[T] $112π=∑k=0∞(−1)k(6k)!(13591409+545140134k)(3k)!(k!)36403203k+3/2,112π=∑k=0∞(−1)k(6k)!(13591409+545140134k)(3k)!(k!)36403203k+3/2,$ error $<10−15<10−15$
121.
[T] A fair coin is one that has probability $1/21/2$ of coming up heads when flipped.
1. What is the probability that a fair coin will come up tails $nn$ times in a row?
2. Find the probability that a coin comes up heads for the first time on the last of an even number of coin flips.
122.
[T] Find the probability that a fair coin is flipped a multiple of three times before coming up heads.
123.
[T] Find the probability that a fair coin will come up heads for the second time after an even number of flips.
124.
[T] Find a series that expresses the probability that a fair coin will come up heads for the second time on a multiple of three flips.
125.
[T] The expected number of times that a fair coin will come up heads is defined as the sum over $n=1,2,…n=1,2,…$ of $nn$ times the probability that the coin will come up heads exactly $nn$ times in a row, or $n/2n+1.n/2n+1.$ Compute the expected number of consecutive times that a fair coin will come up heads.
126.
[T] A person deposits $1010$ at the beginning of each quarter into a bank account that earns $4%4%$ annual interest compounded quarterly (four times a year).
1. Show that the interest accumulated after $nn$ quarters is $10(1.01n+1−10.01−n).10(1.01n+1−10.01−n).$
2. Find the first eight terms of the sequence.
3. How much interest has accumulated after $22$ years?
127.
[T] Suppose that the amount of a drug in a patient’s system diminishes by a multiplicative factor $r<1r<1$ each hour. Suppose that a new dose is administered every $NN$ hours. Find an expression that gives the amount $A(n)A(n)$ in the patient’s system after $nn$ hours for each $nn$ in terms of the dosage $dd$ and the ratio $r.r.$ (Hint: Write $n=mN+k,n=mN+k,$ where $0≤k and sum over values from the different doses administered.)
128.
[T] A certain drug is effective for an average patient only if there is at least $11$ mg per kg in the patient’s system, while it is safe only if there is at most $22$ mg per kg in an average patient’s system. Suppose that the amount in a patient’s system diminishes by a multiplicative factor of $0.90.9$ each hour after a dose is administered. Find the maximum interval $NN$ of hours between doses, and corresponding dose range $dd$ (in mg/kg) for this $NN$ that will enable use of the drug to be both safe and effective in the long term.
129.
Suppose that $an≥0an≥0$ is a sequence of numbers. Explain why the sequence of partial sums of $anan$ is increasing.
130.
[T] Suppose that $anan$ is a sequence of positive numbers and the sequence $SnSn$ of partial sums of $anan$ is bounded above. Explain why $∑n=1∞an∑n=1∞an$ converges. Does the conclusion remain true if we remove the hypothesis $an≥0?an≥0?$
131.
[T] Suppose that $a1=S1=1a1=S1=1$ and that, for given numbers $S>1S>1$ and $0 one defines $an+1=k(S−Sn)an+1=k(S−Sn)$ and $Sn+1=an+1+Sn.Sn+1=an+1+Sn.$ Does $SnSn$ converge? If so, to what? (Hint: First argue that $Sn for all $nn$ and $SnSn$ is increasing.)
132.
[T] A version of von Bertalanffy growth can be used to estimate the age of an individual in a homogeneous species from its length if the annual increase in year $n+1n+1$ satisfies $an+1=k(S−Sn),an+1=k(S−Sn),$ with $SnSn$ as the length at year $n,n,$ $SS$ as a limiting length, and $kk$ as a relative growth constant. If $S1=3,S1=3,$ $S=9,S=9,$ and $k=1/2,k=1/2,$ numerically estimate the smallest value of $nn$ such that $Sn≥8.Sn≥8.$ Note that $Sn+1=Sn+an+1.Sn+1=Sn+an+1.$ Find the corresponding $nn$ when $k=1/4.k=1/4.$
133.
[T] Suppose that $∑n=1∞an∑n=1∞an$ is a convergent series of positive terms. Explain why $limN→∞∑n=N+1∞an=0.limN→∞∑n=N+1∞an=0.$
134.
[T] Find the length of the dashed zig-zag path in the following figure.
135.
[T] Find the total length of the dashed path in the following figure.
136.
[T] The Sierpinski triangle is obtained from a triangle by deleting the middle fourth as indicated in the first step, by deleting the middle fourths of the remaining three congruent triangles in the second step, and in general deleting the middle fourths of the remaining triangles in each successive step. Assuming that the original triangle is shown in the figure, find the areas of the remaining parts of the original triangle after $NN$ steps and find the total length of all of the boundary triangles after $NN$ steps.
137.
[T] The Sierpinski gasket is obtained by dividing the unit square into nine equal sub-squares, removing the middle square, then doing the same at each stage to the remaining sub-squares. The figure shows the remaining set after four iterations. Compute the total area removed after $NN$ stages, and compute the length the total perimeter of the remaining set after $NN$ stages. | 11,892 | 30,228 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 521, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2024-38 | latest | en | 0.923683 |
https://uk.mathworks.com/matlabcentral/cody/problems/43270-rotate-matrix-by-90-degrees/solutions/1035422 | 1,610,935,301,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514046.20/warc/CC-MAIN-20210117235743-20210118025743-00381.warc.gz | 611,983,065 | 16,969 | Cody
# Problem 43270. Rotate matrix by -90 degrees
Solution 1035422
Submitted on 28 Oct 2016 by Kristian Cubol
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2;3 4]; y_correct = [3 1;4 2]; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = [1 2 3;4 5 6;7 8 9]; y_correct = [7 4 1;8 5 2;9 6 3]; assert(isequal(your_fcn_name(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 196 | 603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-04 | latest | en | 0.719994 |
http://www.physicsforums.com/showthread.php?t=46801 | 1,386,861,419,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164641982/warc/CC-MAIN-20131204134401-00087-ip-10-33-133-15.ec2.internal.warc.gz | 492,158,653 | 9,688 | damping coefficient
by mlee
Tags: coefficient, damping
P: 25 A 50.0-g hard-boiled egg moves on the end of a spring with force constant . It is released with an amplitude 0.300 m. A damping force acts on the egg. After it oscillates for 5.00 s, the amplitude of the motion has decreased to 0.100 m.Calculate the magnitude of the damping coefficient . Express the magnitude of the damping coefficient numerically in kilograms per second, to three significant figures pls who can help me? thanx
PF Patron HW Helper Sci Advisor Thanks P: 11,935 How should Newton's 2.law of motion look like?
P: 25 i think it is: -kx-bv=ma
PF Patron
HW Helper
Thanks
P: 11,935
damping coefficient
That's correct!
Now, what type of solutions have you learnt that this differential equation has?
HW Helper P: 2,274 See it as $$-kx - b \frac{dx}{dt} = m\frac{d^2 x}{dt^2}$$
HW Helper P: 2,274 You're right, thanks alridno
P: 25 v= dx/dt and a= d^2/dt^2
P: 25 but what is the answer of d^2/dt^2 then?
PF Patron HW Helper Sci Advisor Thanks P: 11,935 mlee: Any progress at what sort of solutions your equation has?
P: 25 uh not really...;(
PF Patron HW Helper Sci Advisor Thanks P: 11,935 Now, I'd like you try a solution of the form: $$x(t)=Ae^{rt}$$ (A and r constants) What condition must be placed on "r" in order for this to be a solution. Please post your work.
P: 25 Asin(wt)+Bcos (wt)
PF Patron
HW Helper
PF Patron HW Helper Sci Advisor Thanks P: 11,935 You lack a minus sign in your exponential! Now, knowing a) The initial displacement and b)That the initial velocity is zero How can you determine $$A,\phi$$ Besides, what is your value of "w"?
PF Patron HW Helper Sci Advisor Thanks P: 11,935 Now, so how does your initial conditions determine $$A,\phi$$? | 506 | 1,756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2013-48 | longest | en | 0.850978 |
https://isabelle.in.tum.de/repos/isabelle/comparison/a344f1a21211/src/HOL/NthRoot.thy | 1,632,482,252,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00652.warc.gz | 353,926,987 | 2,426 | src/HOL/NthRoot.thy
changeset 49753 a344f1a21211 parent 44349 f057535311c5 child 49962 a8cc904a6820
equal inserted replaced
49752:2bbb0013ff82 49753:a344f1a21211
` 396 using n by (rule isCont_real_root)`
` 396 using n by (rule isCont_real_root)`
` 397 qed`
` 397 qed`
` 398 `
` 398 `
` 399 lemma DERIV_real_root_generic:`
` 399 lemma DERIV_real_root_generic:`
` 400 assumes "0 < n" and "x \<noteq> 0"`
` 400 assumes "0 < n" and "x \<noteq> 0"`
` 401 and even: "\<lbrakk> even n ; 0 < x \<rbrakk> \<Longrightarrow> D = inverse (real n * root n x ^ (n - Suc 0))"`
` 401 and "\<lbrakk> even n ; 0 < x \<rbrakk> \<Longrightarrow> D = inverse (real n * root n x ^ (n - Suc 0))"`
` 402 and even: "\<lbrakk> even n ; x < 0 \<rbrakk> \<Longrightarrow> D = - inverse (real n * root n x ^ (n - Suc 0))"`
` 402 and "\<lbrakk> even n ; x < 0 \<rbrakk> \<Longrightarrow> D = - inverse (real n * root n x ^ (n - Suc 0))"`
` 403 and odd: "odd n \<Longrightarrow> D = inverse (real n * root n x ^ (n - Suc 0))"`
` 403 and "odd n \<Longrightarrow> D = inverse (real n * root n x ^ (n - Suc 0))"`
` 404 shows "DERIV (root n) x :> D"`
` 404 shows "DERIV (root n) x :> D"`
` 405 using assms by (cases "even n", cases "0 < x",`
` 405 using assms by (cases "even n", cases "0 < x",`
` 406 auto intro: DERIV_real_root[THEN DERIV_cong]`
` 406 auto intro: DERIV_real_root[THEN DERIV_cong]`
` 407 DERIV_odd_real_root[THEN DERIV_cong]`
` 407 DERIV_odd_real_root[THEN DERIV_cong]`
` 408 DERIV_even_real_root[THEN DERIV_cong])`
` 408 DERIV_even_real_root[THEN DERIV_cong])` | 606 | 1,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-39 | latest | en | 0.367824 |
https://studysoup.com/tsg/7300/elementary-statistics-12-edition-chapter-11-3-problem-7cqq | 1,631,930,676,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00031.warc.gz | 594,711,542 | 12,766 | ×
Get Full Access to Elementary Statistics - 12 Edition - Chapter 11.3 - Problem 7cqq
Get Full Access to Elementary Statistics - 12 Edition - Chapter 11.3 - Problem 7cqq
×
# refer to the sample data in the following table (based on
ISBN: 9780321836960 18
## Solution for problem 7CQQ Chapter 11.3
Elementary Statistics | 12th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Elementary Statistics | 12th Edition
4 5 1 238 Reviews
16
1
Problem 7CQQ
refer to the sample data in the following table (based on data from the Pew Research Center). Randomly selected subjects were asked about the use of marijuana for medical purposes. Assume that we want to use a 0.05 significance level to test the claim that response to the question is independent of gender. What distribution is used to test the stated claim (normal,?t, F,?chi-square, uniform)?
Step-by-Step Solution:
Step 1 of 3
Solution 7CQQ Step1: From the above given problem we have Randomly selected subjects were asked about the use of marijuana for medical purposes. In favour Opposse Dont know Men 538 167 29 women 557 186 31 With 0.05 level of significance Step2: Our aim is to check What distribution is used to test the stated claim (normal,t, F, hi-square, uniform) Consider the hypothesis, H : Response to the question is independent of gender (Claim). 0 H 1 Response to the question is not independent of gender. Step3: Based on the given information and the null hypothesis we can conclude that chi-square test is appropriate to test the claim.
Step 2 of 3
Step 3 of 3
##### ISBN: 9780321836960
The full step-by-step solution to problem: 7CQQ from chapter: 11.3 was answered by , our top Statistics solution expert on 03/15/17, 10:30PM. Elementary Statistics was written by and is associated to the ISBN: 9780321836960. This full solution covers the following key subjects: claim, use, test, data, Question. This expansive textbook survival guide covers 121 chapters, and 3629 solutions. Since the solution to 7CQQ from 11.3 chapter was answered, more than 353 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. The answer to “refer to the sample data in the following table (based on data from the Pew Research Center). Randomly selected subjects were asked about the use of marijuana for medical purposes. Assume that we want to use a 0.05 significance level to test the claim that response to the question is independent of gender. What distribution is used to test the stated claim (normal,?t, F,?chi-square, uniform)?” is broken down into a number of easy to follow steps, and 65 words.
#### Related chapters
Unlock Textbook Solution
Enter your email below to unlock your verified solution to:
refer to the sample data in the following table (based on | 710 | 2,945 | {"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} | 3.765625 | 4 | CC-MAIN-2021-39 | latest | en | 0.899162 |
https://masterhomework.com/emvprobability-x-impacts/ | 1,603,227,742,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874135.2/warc/CC-MAIN-20201020192039-20201020222039-00419.warc.gz | 424,707,054 | 9,660 | # EMV=Probability x Impacts
Question 1 (10 points)
Don't use plagiarized sources. Get Your Custom Essay on
EMV=Probability x Impacts
Get a plagiarism free paper Just from \$13/Page
A general contractor has three options when bidding for three projects. He has the resources to undertake only one of the projects and must select the most profitable option. The first option is to act as a general contractor submitting a lump sum bid for a bridge project. The project has a 60% chance of generating \$400,000 profit and a 40% chance of \$200,000 loss. The contractor estimates that it will cost \$5,000 to prepare the bid for the 1st option. The second project is a design and build scheme for a new pumping station at a waterworks. Calculations show that the 2nd option has 90% chance of potential profit \$220,000, but a 10% chance of a \$100,000 loss. There shall be \$20,000 bidding cost for the 2nd option. The third project is a management contract for a building project. The project has a 95% chance of generating \$180,000 profit and a 5% chance of a \$20,000 loss. Cost of bidding for the 3rd option is \$10,000. Which option is the most profitable? Show your calculations by drawing a decision tree. EMV=Probability x Impacts
Question 2 (10 points)
A technology company must decide whether to bid for a government contract to develop a piece of equipment, and if the company decides to bid, it must decide how much to bid. The government will award the contract to the low bidder. However, there is 20% probability that there will not be any other bidders. If there are other bidders, there is a 20% chance of underbid competitors if the bid price is \$160K; 60% chance of underbid competitors if the bid price is \$170K; and 90% chance of underbid competitors if the bid price is \$180K. The company estimates that the cost of placing a bid is \$7.5K and the cost of developing the equipment, given that it wins the contract, is \$150K. So, for example, if the company bids \$170K and wins the bid, its profit will be \$20K less the cost of placing the bid.
Show your calculations by drawing a decision tree. EMV=Probability x Impacts
Order Now And Get A 20% Discount!
Pages (550 words)
Approximate price: -
Advantages of using our writing services
Custom Writing From Scratch
All our custom papers are written by qualified writers according to your instructions, thus evading any case of plagiarism. Our team consists of native writers from the USA, Canada, and the Uk, making it convenient for us to find the best to handle your order.
Unlimited Free Revisions
If you feel your paper didn't meet all your requirements, we won't stop till it's perfect. You're entitled to request a free revision within 7 days after we submit your paper.
Quality Writing In Any Format
If you have issues with citing sources and referencing, you need not worry. Our writers are highly knowledgeable in referencing, including APA/MLA/Havard/Chicago/Turabian and all other formatting styles.
All our custom papers are delivered on time, even the most urgent. If we need more time to perfect your paper, we may contact you via email or phone regarding the deadline extension.
Originality & Security
At Master Homework, your security and privacy is our greatest concern. For this reason, we never share your personal information with third parties. We use several writing tools to ensure your paper is original and free from plagiarism.
Our agents are online 24 hours a day, 7 days a week, and are always ready to serve you. Feel free to contact us through email or talk to our live agents whenever you need assistance with your order.
Try it now!
## Calculate the price of your order
We'll send you the first draft for approval by at
Total price:
\$0.00
How it works?
Fill in the order form and provide all details of your assignment.
Proceed with the payment
Choose the payment system that suits you most.
Our Services
We work nonstop to see the best client experience.
## Flexible Pricing
We offer pocket-friendly prices that coincide with the preferred client's deadline. | 929 | 4,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-45 | latest | en | 0.918826 |
satunutakuwy.histoires-etranges.com | 1,579,963,549,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00108.warc.gz | 629,628,942 | 2,716 | # Write a program to test various operations on a binary search tree
If the given key is equal to the key at the root, we return the number of keys t in the left subtree; if the given key is less than the key at the root, we return the rank of the key in the left subtree; and if the given key is larger than the key at the root, we return t plus one to count the key at the root plus the rank of the key in the right subtree.
Why not use the predecessor? Add to BST. In-order traversal: In this traversal mode, one starts visiting with the left child, followed by its parent and then the right child.
First, we need to find the node that will replace the deleted node. If the left link of the root is null, the smallest key in a BST is the key at the root; if the left link is not null, the smallest key in the BST is the smallest key in the subtree rooted at the node referenced by the left link.
The possible substitute would be either 1 i.
## Binary search tree c++
Finding the maximum key is similar, moving to the right instead of to the left. The three most commonly used traversing methods are: In-order Pre-order Post-order 1. Here since 11 is the root we will not visit it but will move to its left child that is 2. If a given key key is less than the key at the root of a BST, then the floor of key the largest key in the BST less than or equal to key must be in the left subtree. The cases to consider for the element to be deleted are as follows: Case 1: The element to be deleted is a leaf node. We'll use the same tree that we used before and we'll show the traversal order for each case. Reverse a BST. For delete the minimum, we go left until finding a node that that has a null left link and then replace the link to that node by its right link. That again 2 is a root and it also has a left child which is 1 so first we visit 1, and move to the right child of 2. The right child here is 9, but is a root and also has a left child thus we traverse to the left child 3. In worst case, the time it takes to search an element is 0 n. The diagram above depicts the situation described as under case 1. We keep going to either right subtree or left subtree depending on the value and when we reach a point left or right subtree is null, we put the new node there. After this, we move on to the right sub-tree.
From the example quoted above, we delete the element
Rated 5/10 based on 100 review | 570 | 2,414 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-05 | latest | en | 0.927308 |
https://www.gamedev.net/forums/topic/454914-solved-thanks-so-much-srand-and-rand--questions/ | 1,531,794,597,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589537.21/warc/CC-MAIN-20180717012034-20180717032034-00104.warc.gz | 883,246,182 | 25,945 | Public Group
# (solved, thanks so much!) srand() and rand() + questions.
This topic is 4026 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Could someone explain to me how this works? I've read 3 or 4 tutorials but I simply cannot understand how to use random numbers. And random numbers are a critical part of an idea i have. I mostly don't understand "seeding". Also, (sorry if I cannot explain this clearly) what If I have many sets of "dice roll from ___ to ___?" Oh one last thing. Why is it a bad idea to use system pause and rest function? [Edited by - yurian on July 8, 2007 9:56:36 PM]
##### Share on other sites
From what I have worked out with seeding and random numbers:
Imagine your computer has a set of random numbers for example:
23 65 71 276 94 1 83 2293 82 73
when you dont seed a random number and start calling the rand function it returns the values in order starting from those imaginary numbers, so if you called rand 3 times you would get:
23, 65 and 71
when you seed a random number, you get a new position in those random numbers. for examle after seeding the start position might be 94. Then if you call rand another 3 times you would get:
94, 1 and 83
This may be incorrect but from what I have read thats what i can gather.
Also by seeding using the system time you can be sure to get random numbers each time your program is run.
##### Share on other sites
Call
srand(time(0));
and then call
rand() % 12 + 1;
Replace 12 with any number you want. Does that help?
##### Share on other sites
... what? I don't understand.
##### Share on other sites
Going with the above, I know that if you replace 12 with 100 you will get a random number between 1-100. If you replace with 1000 you get a random number between 1-1000. The percent sign is the mudulos operator. Look it up to see how to implement it better.
srand(time(0)) just starts the random number generator. Rand gets the number which could be really large. You then have to put the number in the range you want by using modulos.
##### Share on other sites
cNoob has the right idea, just using rand() will get you the same "random" numbers each and every time you run your application. Using srand() you start at another place in the sequence. Although, if you supply the same seed for srand() every time you will get the the same result as with rand() only, with different numbers.
The idea to seed srand() with time() is that you will get a new seed every time you start the application as the time will never be the same (or that's the idea, anyway).
So for your application, call srand() once before you use any rand() and then you can forget about srand() for the rest of the application, normally.
##### Share on other sites
Quote:
Original post by csscplayerGoing with the above, I know that if you replace 12 with 100 you will get a random number between 1-100. If you replace with 1000 you get a random number between 1-1000. The percent sign is the mudulos operator. Look it up to see how to implement it better.srand(time(0)) just starts the random number generator. Rand gets the number which could be really large. You then have to put the number in the range you want by using modulos.
Thanks so much! But whats the time for?
##### Share on other sites
Quote:
Original post by PzccNoob has the right idea, just using rand() will get you the same "random" numbers each and every time you run your application. Using srand() you start at another place in the sequence. Although, if you supply the same seed for srand() every time you will get the the same result as with rand() only, with different numbers.The idea to seed srand() with time() is that you will get a new seed every time you start the application as the time will never be the same (or that's the idea, anyway).So for your application, call srand() once before you use any rand() and then you can forget about srand() for the rest of the application, normally.
Oh I see thank you very much! =)
1. 1
Rutin
24
2. 2
3. 3
JoeJ
18
4. 4
5. 5
• 38
• 23
• 13
• 13
• 17
• ### Forum Statistics
• Total Topics
631712
• Total Posts
3001849
× | 1,047 | 4,189 | {"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} | 3.125 | 3 | CC-MAIN-2018-30 | latest | en | 0.909897 |
http://jmlilly.net/doc/trajchunk.html | 1,539,903,048,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512015.74/warc/CC-MAIN-20181018214747-20181019000247-00386.warc.gz | 202,133,867 | 2,271 | # TRAJCHUNK is the jOceans module of jLab.
``` TRAJCHUNK Converts Lagrangian trajectories into chunks based on the Coriolis period.
TRAJCHUNK is used to split float or drifter data into chuncks such that
the length of each chunk is a fixed multiple of the average Coriolis
frequency. This is useful in spectral analysis.
[NUMO,LATO]=TRAJCHUNK(NUM,LAT,P), where NUM and LAT are date number and
latitude for Lagangian float or drifter data, re-organizes these into
chunks such that the average Coriolis frequency f_C in each chunk is at
least P times the Rayleigh frequency f_R for that chunk.
Recall that the Rayleigh frequency is f_R=2*pi/(DT*N), in units of
radians per unit time, where DT is the sample interval and N is the
number of samples.
The input fields NUM and LAT may either be numerical arrays, or
cell arrays of numerical arrays, e.g. NUM{1}=NUM1, NUM{2}=NUM2, etc.
The output variables NUM and LAT are now cell arrays of numerical
arrays, with each cell is truncated such that its length is just long
enough such that f_C > P * f_R.
Trajectories that are not long enough to satisfy this criterion are
discarded, as are short residual segments at the end of trajectories.
Each input trajectory is thus split into zero, one, or more than one
cells in the output variables.
TRAJCHUNK(...,P,LMIN) additionally specifies a mininum number of points
LMIN for each chunk.
__________________________________________________________________
Multiple input arguments
[NUMO,LATO,Y1,Y2,...,YM]=TRAJCHUNK(NUM,LAT,X1,X2,...XM,N) chunks the M
input arrays X1, X2,... XM in the same manner, and returns these as Y1,
Y2,... YM. The input variables may either all be numerical arrays of
all the same size, or cell arrays of numerical arrays.
In the case of cell array input, some of the XM may be numerical arrays
of the same length as the cells NUM and LAT. The corresponding output
variable will then also be a numerical array. An example of such a
field is the identification number used in FLOATS.MAT and DRIFTERS.MAT.
TRAJCHUNK with no output arguments overwrites the original named output
variables.
__________________________________________________________________
Keeping short data segments
By default, any data is cells shorter than the specified length are
discarded, as are data segments at the end of the trajectories.
TRAJCHUNK(...,'keep') keeps these instead. Short cells are returned in
their own chunks, and leftover segments are appended to the end of the
preceding chunk.
This preserves the number of data points, while favoring a requested
length, if possible, from each trajectory.
__________________________________________________________________
Overlap
TRAJCHUNK(...,'overlap',PCT) outputs chunks with a percentage PCT
overlap. For example, TRAJCHUNK(...,'overlap',50) outputs chunks that
overlap by 50%. The default behavior gives chunks with no overlap.
__________________________________________________________________
'trajchunk --t' runs a test.
Usage: [num,lat]=trajchunk(num,lat,P);
[num,lat,lon,cv]=trajchunk(num,lat,lon,cv,P);
[num,lat,lon,cv]=trajchunk(num,lat,lon,cv,P,lmin);
[num,lat,lon,cv]=trajchunk(num,lat,lon,cv,P,'overlap',50);
trajchunk(num,lat,lon,cv,P);
__________________________________________________________________
This is part of JLAB --- type 'help jlab' for more information
(C) 2014--2015 J.M. Lilly --- type 'help jlab_license' for details
``` | 807 | 3,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-43 | longest | en | 0.821432 |
http://iflorinsky.impb.ru/e.htm | 1,582,537,599,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145910.53/warc/CC-MAIN-20200224071540-20200224101540-00433.warc.gz | 71,697,169 | 5,990 | en ru
###### Difference curvature
Difference curvature (E) is a half-difference of horizontal and vertical curvatures*. The unit of measurement is m-1.
Once elevations are given by , where x and y are plane Cartesian co-ordinates, difference curvature is a function of the partial derivatives of z:
,
where kh and kv are horizontal and vertical curvatures, correspondingly, , , , , .
Two mechanisms of flow accumulation, convergence and deceleration, are controlled by horizontal and vertical curvatures, correspondingly. So, difference curvature shows to what extent flow deceleration is higher than flow convergence at the given point of the topographic surface.
Like other local morphometric variables, difference curvature can be derived from a digital elevation model (DEM) by a universal spectral analytical method as well as finite-difference methods (e.g., method 1, method 2, and method 3).
Example**. A model of difference curvature was derived from a DEM of Mount Ararat by the universal spectral analytical method. The model includes 779,401 points (the matrix 1081 x 721); the grid spacing is 1". To deal with the large dynamic range of this variable, its values were logarithmically transformed. The vertical exaggeration of the 3D model is 2x. The data processing and modelling were carried out using the software Matlab R2008b.
#### ** Florinsky, I.V., 2016. An illustrated introduction to geomorphometry. Almamac Space and Time, 11 (1): 20 p. (in Russian, with English abstract). Article at the journal website
For details and other examples, see:
DIGITAL TERRAIN ANALYSIS IN SOIL SCIENCE AND GEOLOGY 2nd revised edition I.V. Florinsky Elsevier / Academic Press, 2016 Amsterdam, 486 p. ISBN 978-0-12-804632-6 | 404 | 1,750 | {"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-2020-10 | longest | en | 0.8464 |
https://pfeiffertheface.com/how-do-you-convert-f-to-c-mentally/ | 1,679,405,320,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00268.warc.gz | 530,711,755 | 10,077 | Pfeiffertheface.com
Discover the world with our lifehacks
# How do you convert F to C mentally?
## How do you convert F to C mentally?
How to approximate the conversion of Fahrenheit temperatures to Celsius with mental math.
1. Subtract 32o to adjust for the offset in the Fahrenheit scale.
2. Divide the Celsius temperature by 2 (multiply by 0.5).
3. Take 1/10 of this number (0.5 * 1/10 = 0.05) and add it from the number above.
How do you determine C from F?
Key Points
1. Celsius and Fahrenheit are two important temperature scales that are commonly misspelled as Celcius and Farenheit.
2. The formula to find a Celsius temperature from Fahrenheit is: °F = (°C × 9/5) + 32.
3. The formula to find a Fahrenheit temperature from Celsius is: °F = (°C × 9/5) + 32.
### How do you calculate temperature?
The temperature change depends on the amount of heat released or absorbed. Temperature difference = amount of heat absorbed or released/ mass of the body* specific heat of the body. ΔT = Q/m*c.
What temperature is 75 Fahrenheit in Celsius?
75 degrees Fahrenheit equals 23.89 degrees Celsius!
## What’s the difference in temperature between Fahrenheit and Celsius?
In the Celsius scale there are 100 degrees between the freezing point and the boiling point of water compared to 180 degrees in the Fahrenheit scale. This means that 1 °C = 1.8 °F (check the section about temperature differences below).
What is difference between Celsius and Fahrenheit?
You see that Celsius has 100 degrees between the freezing and boiling point, whereas Fahrenheit has 180 degrees between these two points. One degree Celsius is 1.8 times larger than one degree Fahrenheit.
### Is 23 degrees cold or hot?
23 degrees Celsius is a Goldilocks temperature for humans: not too hot, not too cold but just right.
What temperature is 77 Fahrenheit in Celsius?
77 degrees Fahrenheit equals 25 degrees Celsius!
## Which is colder 32 F or 0 C?
When the temperature in Celcius is 0°C, the temperature in Fahrenheit will be 32°F. And when the temperature in Fahrenheit is 0°F, the temperature in Celcius will be -17.8°C. The lesser the temperature, the colder it becomes. Here, 0°F is less than 0°C. | 524 | 2,194 | {"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-2023-14 | longest | en | 0.799387 |
http://www.mcs.anl.gov/~fathom/moab-docs/html/contents.html | 1,505,840,564,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685912.14/warc/CC-MAIN-20170919164651-20170919184651-00594.warc.gz | 506,700,444 | 31,195 | moab
1.Introduction
2.MOAB Data Model
2.1. MOAB Interface
2.2. Mesh Entities
2.3. Entity Sets
2.4. Tags
3.MOAB API Design Philosophy and Summary
4.Related Mesh Services
4.1. Visualization
4.2. Parallel Decomposition
4.3. Skinner
4.4. Tree Decompositions
4.6. File Readers/Writers Packaged With MOAB
5.Parallel Mesh Representation and Query
5.1. Nomenclature & Representation
5.2. Parallel Mesh Initialization
5.3. Parallel Mesh Query Functions
5.4. Parallel Mesh Communication
6.Building MOAB-Based Applications
7.iMesh (ITAPS Mesh Interface) Implementation in MOAB
8.Structured Mesh Representation
9.Spectral Element Meshes
9.1. Representations
9.3. MOAB Representation
10.Performance and Using MOAB Efficiently from Applications
11.Conclusions and Future Plans
12.References
## 1.Introduction
In scientific computing, systems of partial differential equations (PDEs) are solved on computers. One of the most widely used methods to solve PDEs numerically is to solve over discrete neighborhoods or “elements” of the domain. Popular discretization methods include Finite Difference (FD), Finite Element (FE), and Finite Volume (FV). These methods require the decomposition of the domain into a discretized representation, which is referred to as a “mesh”. The mesh is one of the fundamental types of data linking the various tools in the analysis process (mesh generation, analysis, visualization, etc.). Thus, the representation of mesh data and operations on those data play a very important role in PDE-based simulations.
MOAB is a component for representing and evaluating mesh data. MOAB can store structured and unstructured mesh, consisting of elements in the finite element “zoo”, along with polygons and polyhedra. The functional interface to MOAB is simple, consisting of only four fundamental data types. This data is quite powerful, allowing the representation of most types of metadata commonly found on the mesh. Internally MOAB uses array-based storage for fine-grained data, which in many cases provides more efficient access, especially for large portions of mesh and associated data. MOAB is optimized for efficiency in space and time, based on access to mesh in chunks rather than through individual entities, while also versatile enough to support individual entity access.
The MOAB data model consists of the following four fundamental types: mesh interface instance, mesh entities (vertex, edge, tri, etc.), sets, and tags. Entities are addressed through handles rather than pointers, to allow the underlying representation of an entity to change without changing the handle to that entity. Sets are arbitrary groupings of mesh entities and other sets. Sets also support parent/child relationships as a relation distinct from sets containing other sets. The directed graph provided by set parent/child relationships is useful for embedding graphs whose nodes include collections of mesh entities; this approach has been used to represent a wide variety of application-specific data, including geometric model topology, processor partitions, and various types of search trees. Tags are named data which can be assigned to the mesh as a whole, individual entities, or sets. Tags are a mechanism for attaching data to individual entities, and sets are a mechanism for describing relations between entities; the combination of these two mechanisms is a powerful yet simple interface for representing metadata or application-specific data.
Various mesh-related tools are provided with MOAB or can be used directly with MOAB. These tools can be used for mesh format translation (mbconvert), mesh skinning (Skinner class), solution transfer between meshes (MBCoupler tool), ray tracing and other geometric searches (OrientedBoxTreeTool, AdaptiveKDTree), visualization (vtkMOABReader tool), and relation between mesh and geometric models (the separately-packed Lasso tool). These tools are described later in this document.
MOAB is written in the C++ programming language, with applications interacting with MOAB mostly through its moab::Interface class. All of the MOAB functions and classes are isolated in and accessed through the moab namespace1. The remainder of this report gives class and function names without the “moab::” namespace qualification; unless otherwise noted, the namespace qualifier should be added to all class and function names referenced here. MOAB also implements the iMesh interface, which is specified in C but can be called directly from other languages. Almost all of the functionality in MOAB can be accessed through the iMesh interface. MOAB is developed and supported on the Linux and MacOS operating systems, as well as various HPC operating systems. MOAB can be used on parallel computing systems as well, including both clusters and high-end parallel systems like IBM BG/P and Cray systems. MOAB is released under a standard LGPL open source software license.
MOAB is used in several ways in various applications. MOAB serves as the underlying mesh data representation in several scientific computing applications [1]. MOAB can also be used as a mesh format translator, using readers and writers included in MOAB. MOAB has also been used as a bridge to couple results in multi-physics analysis and to link these applications with other mesh services [2].
The remainder of this report is organized as follows. Section 2, “Getting Started”, provides a few simple examples of using MOAB to perform simple tasks on a mesh. Section 3 discusses the MOAB data model in more detail, including some aspects of the implementation. Section 4 summarizes the MOAB function API. Section 5 describes some of the tools included with MOAB, and the implementation of mesh readers/writers for MOAB. Section 6 describes how to build MOAB-based applications. Section 7 contains a brief description of MOAB’s relation to the iMesh mesh interface. Sections 8 and 9 discuss MOAB's representations of structured and spectral element meshes, respectively. Section 10 gives helpful hints for accessing MOAB in an efficient manner from applications. Section 11 gives a conclusion and future plans for MOAB development. Section 12 gives references cited in this report.
Several other sources of information about MOAB may also be of interest to readers. Meta-data conventions define how sets and /or tags are used together to represent various commonly-used simulation constructs; conventions used by MOAB are described in Ref [4], which is also included in the MOAB source distribution. This document is maintained separately from this document, since it is expected to change over time. The MOAB project maintains a wiki [5], which links to most MOAB-related information. MOAB also uses several mailing lists [6],[7] for MOAB-related discussions. Potential users are encouraged to interact with the MOAB team using these mailing lists.
1 Non-namespaced names are also provided for backward compatibility, with the “MB” prefix added to the class or variable name.
## 2.MOAB Data Model
The MOAB data model describes the basic types used in MOAB and the language used to communicate that data to applications. This chapter describes that data model, along with some of the reasons for some of the design choices in MOAB.
<tab>
### 2.1. MOAB Interface
</tab> MOAB is written in C++. The primary interface with applications is through member functions of the abstract base class Interface. The MOAB library is created by instantiating Core, which implements the Interface API. Multiple instances of MOAB can exist concurrently in the same application; mesh entities are not shared between these instancesi2. MOAB is most easily viewed as a database of mesh objects accessed through the instance. No other assumptions explicitly made about the nature of the mesh stored there; for example, there is no fundamental requirement that elements fill space or do not overlap each other geometrically.
2 One exception to this statement is when the parallel interface to MOAB is used; in this case, entity sharing between instances is handled explicitly using message passing. This is described in more detail in Section 5 of this document.
### 2.2. Mesh Entities
MOAB represents the following topological mesh entities: vertex, edge, triangle, quadrilateral, polygon, tetrahedron, pyramid, prism, knife, hexahedron, polyhedron. MOAB uses the EntityType enumeration to refer to these entity types (see Table 1). This enumeration has several special characteristics, chosen intentionally: the types begin with vertex, entity types are grouped by topological dimension, with lower-dimensional entities appearing before higher dimensions; the enumeration includes an entity type for sets (described in the next section); and MBMAXTYPE is included at the end of this enumeration, and can be used to terminate loops over type. In addition to these defined values, the an increment operator (++) is defined such that variables of type EntityType can be used as iterators in loops. MOAB refers to entities using “handles”. Handles are implemented as long integer data types, with the four highest-order bits used to store the entity type (mesh vertex, edge, tri, etc.) and the remaining bits storing the entity id. This scheme is convenient for applications because:
• Handles sort lexicographically by type and dimension; this can be useful for grouping and iterating over entities by type.
• The type of an entity is indicated by the handle itself, without needing to call a function.
• Entities allocated in sequence will typically have contiguous handles; this characteristic can be used to efficiently store and operate on large lists of handles.
This handle implementation is exposed to applications intentionally, because of optimizations that it enables, and is unlikely to change in future versions.
### Table 1: Values defined for the EntityType enumerated type.
MBVERTEX = 0 MBPRISM MBEDGE MBKNIFE MBTRI MBHEX MBQUAD MBPOLYHEDRON MBPOLYGON MBENTITYSET MBTET MBMAXTYPE MBPYRAMID
MOAB defines a special class for storing lists of entity handles, named Range. This class stores handles as a series of (start_handle, end_handle) subrange tuples. If a list of handles has large contiguous ranges, it can be represented in almost constant size using Range. Since entities are typically created in groups, e.g. during mesh generation or file import, a high degree of contiguity in handle space is typical. Range provides an interface similar to C++ STL containers like std::vector, containing iterator data types and functions for initializing and iterating over entity handles stored in the range. Range also provides functions for efficient Boolean operations like subtraction and intersection. Most API functions in MOAB come in both range-based and vector-based variants. By definition, a list of entities stored in an Range is always sorted, and can contain a given entity handle only once. Range cannot store the handle 0 (zero).
Typical usage of an Range object would look like:
```using namespace moab;
int my_function(Range &from_range) {
int num_in_range = from_range.size();
Range to_range;
Range::iterator rit;
for (rit = from_range.begin(); rit != from_range.end(); rit++) {
EntityHandle this_ent = *rit;
to_range.insert(this_ent);
}
}
```
Here, the range is iterated similar to how std::vector is iterated.
The term adjacencies is used to refer to those entities topologically connected to a given entity, e.g. the faces bounded by a given edge or the vertices bounding a given region. The same term is used for both higher-dimensional (or bounded) and lower-dimensional (or bounding) adjacent entities. MOAB provides functions for querying adjacent entities by target dimension, using the same functions for higher- and lower-dimension adjacencies. By default, MOAB stores the minimum data necessary to recover adjacencies between entities. When a mesh is initially loaded into MOAB, only entity-vertex (i.e. “downward”) adjacencies are stored, in the form of entity connectivity. When “upward” adjacencies are requested for the first time, e.g. from vertices to regions, MOAB stores all vertex-entity adjacencies explicitly, for all entities in the mesh. Non-vertex entity to entity adjacencies are never stored, unless explicitly requested by the application.
In its most fundamental form, a mesh need only be represented by its vertices and the entities of maximal topological dimension. For example, a hexahedral mesh can be represented as the connectivity of the hex elements and the vertices forming the hexes. Edges and faces in a 3D mesh need not be explicitly represented. We refer to such entities as “AEntities”, where ‘A’ refers to “Auxiliary”, “Ancillary”, and a number of other words mostly beginning with ‘A’. Individual AEntities are created only when requested by applications, either using mesh modification functions or by requesting adjacencies with a special “create if missing” flag passed as “true”. This reduces the overall memory usage when representing large meshes. Note entities must be explicitly represented before they can be assigned tag values or added to entity sets (described in following Sections).
### 2.3. Entity Sets
Entity sets are used to store arbitrary collections of entities and other sets. Sets are used for a variety of things in mesh-based applications, from the set of entities discretizing a given geometric model entity to the entities partitioned to a specific processor in a parallel finite element application. MOAB entity sets can also store parent/child relations with other entity sets, with these relations distinct from contains relations. Parent/child relations are useful for building directed graphs with graph nodes representing collections of mesh entities; this construct can be used, for example, to represent an interface of mesh faces shared by two distinct collections of mesh regions. MOAB also defines one special set, the “root set” or the interface itself; all entities are part of this set by definition. Defining a root set allows the use of a single set of MOAB API functions to query entities in the overall mesh as well as its subsets.
MOAB entity sets can be one of two distinct types: list-type entity sets preserve the order in which entities are added to the set, and can store a given entity handle multiple times in the same set; set-type sets are always ordered by handle, regardless of the order of addition to the set, and can store a given entity handle only once. This characteristic is assigned when the set is created, and cannot be changed during the set’s lifetime.
MOAB provides the option to track or not track entities in a set. When entities (and sets) are deleted by other operations in MOAB, they will also be removed from containing sets for which tracking has been enabled. This behavior is assigned when the set is created, and cannot be changed during the set’s lifetime. The cost of turning tracking on for a given set is sizeof(EntityHandle) for each entity added to the set; MOAB stores containing sets in the same list which stores adjacencies to other entities.
Using an entity set looks like the following:
```using namespace moab;
// load a file using MOAB, putting the loaded mesh into a file set
EntityHandle file_set;
ErrorCode rval = moab->create_meshset(MESHSET_SET, file_set);
Range set_ents;
// get all the 3D entities in the set
rval = moab->get_entities_by_dimension(file_set, 3, set_ents);
```
Entity sets are often used in conjunction with tags (described in the next section), and provide a powerful mechanism to store a variety of meta-data with meshes.
### 2.4. Tags
Applications of a mesh database often need to attach data to mesh entities. The types of attached data are often not known at compile time, and can vary across individual entities and entity types. MOAB refers to this attached data as a “tag”. Tags can be thought of loosely as a variable, which can be given a distinct value for individual entities, entity sets, or for the interface itself. A tag is referenced using a handle, similarly to how entities are referenced in MOAB. Each MOAB tag has the following characteristics, which can be queried through the MOAB interface:
• Name
• Size (in bytes)
• Storage type
• Data type (integer, double, opaque, entity handle)
• Handle
The storage type determines how tag values are stored on entities.
• Dense: Dense tag values are stored in arrays which match arrays of contiguous entity handles. Dense tags are more efficient in both storage and memory if large numbers of entities are assigned the same tag. Storage for a given dense tag is not allocated until a tag value is set on an entity; memory for a given dense tag is allocated for all entities in a given sequence at the same time.
• Sparse: Sparse tags are stored as a list of (entity handle, tag value) tuples, one list per sparse tag, sorted by entity handle.
• Bit: Bit tags are stored similarly to dense tags, but with special handling to allow allocation in bit-size amounts per entity.
MOAB also supports variable-length tags, which can have a different length for each entity they are assigned to. Variable length tags are stored similarly to sparse tags.
The data type of a tag can either be one understood at compile time (integer, double, entity handle), in which case the tag value can be saved and restored properly to/from files and between computers of different architecture (MOAB provides a native HDF5-based save/restore format for this purpose; see Section 4.6). The opaque data type is used for character strings, or for allocating “raw memory” for use by applications (e.g. for storage application-defined structures or other abstract data types). These tags are saved and restored as raw memory, with no special handling for endian or precision differences.
An application would use the following code to attach a double-precision tag to vertices in a mesh, e.g. to assign a temperature field to those vertices:
```using namespace moab;
// load a file using MOAB and get the vertices
Range verts;
rval = moab->get_entities_by_dimension(0, 0, verts);
// create a tag called “TEMPERATURE”
Tag temperature;
double def_val = -1.0d-300, new_val = 273.0;
rval = moab->tag_create(“TEMPERATURE”, sizeof(double), MB_TAG_DENSE,
MB_TYPE_DOUBLE, temperature, &def_val);
// assign a value to vertices
for (Range::iterator vit = verts.begin();
vit != verts.end(); vit++)
rval = moab->tag_set_data(temperature, &(*rit), 1, &new_val);
```
The semantic meaning of a tag is determined by applications using it. However, to promote interoperability between applications, there are a number of tag names reserved by MOAB which are intended to be used by convention. Mesh readers and writers in MOAB use these tag conventions, and applications can use them as well to access the same data. Ref. [4] maintains an up-to-date list of conventions for meta-data usage in MOAB.
## 3.MOAB API Design Philosophy and Summary
This section describes the design philosophy behind MOAB, and summarizes the functions, data types and enumerated variables in the MOAB API. A complete description of the MOAB API is available in online documentation in the MOAB distribution [8].
MOAB is designed to operate efficiently on collections of entities. Entities are often created or referenced in groups (e.g. the mesh faces discretizing a given geometric face, the 3D elements read from a file), with those groups having some form of temporal or spatial locality. The interface provides special mechanisms for reading data directly into the native storage used in MOAB, and for writing large collections of entities directly from that storage, to avoid data copies. MOAB applications structured to take advantage of that locality will typically operate more efficiently.
MOAB has been designed to maximize the flexibility of mesh data which can be represented. There is no explicit constraint on the geometric structure of meshes represented in MOAB, or on the connectivity between elements. In particular, MOAB allows the representation of multiple entities with the same exact connectivity; however, in these cases, explicit adjacencies must be used to distinguish adjacencies with AEntities bounding such entities.
The number of vertices used to represent a given topological entity can vary, depending on analysis needs; this is often the case in FEA. For example, applications often use “quadratic” or 10-vertex tetrahedral, with vertices at edge midpoints as well as corners. MOAB does not distinguish these variants by entity type, referring to all variants as “tetrahedra”. The number of vertices for a given entity is used to distinguish the variants, with canonical numbering conventions used to determine placement of the vertices [9]. This is similar to how such variations are represented in the Exodus [10] and Patran [11] file formats. In practice, we find that this simplifies coding in applications, since in many cases the handling of entities depends only on the number of corner vertices in the element. Some MOAB API functions provide a flag which determines whether corner or all vertices are requested.
The MOAB API is designed to balance complexity and ease of use. This balance is evident in the following general design characteristics:
• Entity lists: Lists of entities are passed to and from MOAB in a variety of forms. Lists output from MOAB are passed as either STL vector or Range data types. Either of these constructs may be more efficient in both time and memory, depending on the semantics of the data being requested. Input lists are passed as either Range’s, or as a pointer to EntityHandle and a size. The latter allows the same function to be used when passing individual entities, without requiring construction of an otherwise unneeded STL vector.
• Entity sets: Most query functions accept an entity set as input. Applications can pass zero to indicate a request for the whole interface. Note that this convention applies only to query functions; attempts to add or subtract entities to/from the interface using set-based modification functions, or to add parents or children to the interface set, will fail. Allowing specification of the interface set in this manner avoids the need for a separate set of API functions to query the database as a whole.
• Implicit Booleans in output lists: A number of query functions in MOAB allow specification of a Boolean operation (Interface::INTERSECT or Interface::UNION). This operation is applied to the results of the query, often eliminating the need for code the application would need to otherwise implement. For example, to find the set of vertices shared by a collection of quadrilaterals, the application would pass that list of quadrilaterals to a request for vertex adjacencies, with Interface::INTERSECT passed for the Boolean flag. The list of vertices returned would be the same as if the application called that function for each individual entity, and computed the intersection of the results over all the quadrilaterals. Applications may also input non-empty lists to store the results, in which case the intersection is also performed with entities already in the list. In many cases, this allows optimizations in both time and memory inside the MOAB implementation.
Since these objectives are at odds with each other, tradeoffs had to be made between them. Some specific issues that came up are:
• Using ranges: Where possible, entities can be referenced using either ranges (which allow efficient storage of long lists) or STL vectors (which allow list order to be preserved), in both input and output arguments.
• Entities in sets: Accessing the entities in a set is done using the same functions which access entities in the entire mesh. The whole mesh is referenced by specifying a set handle of zero3.
• Entity vectors on input: Functions which could normally take a single entity as input are specified to take a vector of handles instead. Single entities are specified by taking the address of that entity handle and specifying a list length of one. This minimizes the number of functions, while preserving the ability to input single entities.4
Table 2 lists basic data types and enumerated variables defined and used by MOAB. Values of the ErrorCode enumeration are returned from most MOAB functions, and can be compared to those listed in Appendix [ref-appendix].
MOAB uses several pre-defined tag names to define data commonly found in various mesh-based analyses. Ref. [4] describes these meta-data conventions in more detail. These conventions will be added to as new conventions emerge for using sets and tags in MOAB applications.
### Table 2: Basic data types and enums defined in MOAB.
Enum / Type Description
ErrorCode Specific error codes returned from MOAB
EntityHandle Type used to represent entity handles
Tag Type used to represent tag handles
TagType Type used to represent tag storage type
DataType Type used to represent tag data type
Table 3 lists the various groups of functions that comprise the MOAB API. This is listed here strictly as a reference to the various types of functionality supported by MOAB; for a more detailed description of the scope and syntax of the MOAB API, see the online documentation [8].
### Table 3: Groups of functions in MOAB API. See Ref. [8] for more details.
Function group Examples Description
Constructor, destructor, interface Interface, ~Core, query_interface Construct/destroy interface; get pointer to read/write interface
Entity query get_entities_by_dimension, get_entities_by_handle Get entities by dimension, type, etc.
Vertex coordinates get_coords, set_coords Get/set vertex coordinates
Connectivity get_connectivity, set_connectivity Get/set connectivity of non-vertex entities
Tags tag_get_data, tag_create Create, read, write tag data
Handles type_from_handle, id_from_handle Go between handles and types/ids
Geometric dimension get_dimension, set_dimension Get/set geometric dimension of mesh
Mesh modification create_vertex, delete_entity Create or delete mesh entities
Information list_entities, get_last_error Get or print certain information
High-order nodes high_order_node Get information on high-order nodes
Canonical numbering side_number Get canonical numbering information
3In iMesh, the whole mesh is specified by a special entity set handle, referred to as the “root set”.
4Note that STL vectors of entity handles can be input in this manner by using &vector[0] and vector.size() for the 1d vector address and size, respectively.
## 4.Related Mesh Services
A number of mesh-based services are often used in conjunction with a mesh library. For example, parallel applications often need to visualize the mesh and associated data. Other services, like spatial interpolation or finding the faces on the “skin” of a 3D mesh, can be implemented more efficiently using knowledge of specific data structures in MOAB. Several of these services provided with MOAB are described in this chapter.
### 4.1. Visualization
Visualization is one of the most common needs associated with meshes. The primary tool used to visualize MOAB meshes is VisIt [12]. Users can specify that VisIt read mesh directly out of the MOAB instance, by specifying the ITAPS-MOABC mesh format and a file readable by MOAB (see xxx).
There are some initial capabilities in VisIt for limited viewing and manipulation of tag data and some types of entity sets. Tag data is visualized using the same mechanisms used to view other field data in VisIt, e.g. using a pseudocolor plot; sets are viewed using VisIt’s SIL window, accessed by selecting the SIL icon in the data selection window. xxx shows a vertex-based radiation temperature field computed by the Cooper rad-hydro code [1] for a subset of geometric volumes in a mesh.
Reorganization of VisIt’s set handling is also underway, to increase versatility and flexibility of this important mechanism.
### 4.2. Parallel Decomposition
To support parallel simulation, applications often need to partition a mesh into parts, designed to balance the load and minimize communication between sets. MOAB includes the MBZoltan tool for this purpose, constructed on the well-known Zoltan partitioning library [13]. After computing the partition using Zoltan, MBZoltan stores the partition as either tags on individual entities in the partition, or as tagged sets, one set per part. Since a partition often exhibits locality similar to how the entities were created, storing it as sets (based on Range’s) is often more memory-efficient than an entity tag-based representation. Xxx shows a partition computed with MBZoltan (and visualized in VisIt).
### 4.3. Skinner
An operation commonly applied to mesh is to compute the outermost “skin” bounding a contiguous block of elements. This skin consists of elements of one fewer topological dimension, arranged in one or more topological balls on the boundary of the elements. The Skinner tool computes the skin of a mesh in a memory-efficient manner. Skinner uses knowledge about whether vertex-entity adjacencies and AEntities exist to minimize memory requirements and searching time required during the skinning process. This skin can be provided as a single collection of entities, or as sets of entities distinguished by forward and reverse orientation with respect to higher-dimensional entities in the set being skinned.
The following code fragment shows how Skinner can be used to compute the skin of a range of hex elements:
```using namespace moab;
Range hexes, faces;
ErrorCode rval = moab->get_entities_by_dimension(0, 3, hexes);
Skinner myskinner(moab);
bool verts_too = false;
ErrorCode rval = myskinner.find_skin(hexes, verts_too, faces);
```
Skinner can also skin a mesh based on geometric topology groupings imported with the mesh. The geometric topology groupings contain information about the mesh “owned” by each of the entities in the geometric model, e.g. the model vertices, edges, etc. Links between the mesh sets corresponding to those entities can be inferred directly from the mesh. Skinning a mesh this way will typically be much faster than doing so on the actual mesh elements, because there is no need to create and destroy interior faces on the mesh.
### 4.4. Tree Decompositions
MOAB provides several mechanisms for spatial decomposition and searching in a mesh:
• AdaptiveKDTree: Adaptive KD tree, a space-filling decomposition with axis-aligned splitting planes, enabling fast searching.
• BSPTree: Binary Space Partition tree, with non-axis-aligned partitions, for fast spatial searches with slightly better memory efficiency than KD trees.
• OrientedBoxTreeTool: Oriented Bounding Box tree hierarchy, useful for fast ray-tracing on collections of mesh facets.
These trees have various space and time searching efficiencies. All are implemented based on entity sets and parent/child relations between those sets, allowing storage of a tree decomposition using MOAB’s native file storage mechanism (see Section 4.6.1). MOAB’s entity set implementation is specialized for memory efficiency when representing binary trees. Tree decompositions in MOAB have been used to implement fast ray tracing to support radiation transport [14], solution coupling between meshes [2], and embedded boundary mesh generation [15]. MOAB also includes the DAGMC tool, supporting Monte Carlo radiation transport.
The following code fragment shows very basic use of AdaptiveKDTree. A range of entities is put in the tree; the leaf containing a given point is found, and the entities in that leaf are returned.
```using namespace moab;
// create the adaptive kd tree from a range of tets
EntityHandle tree_root
ErrorCode rval = myTree.build_tree(tets, tree_root);
// get the overall bounding box corners
double boxmax[3], boxmin;
rval = myTree.get_tree_box(tree_root, boxmax, boxmin);
// get the tree leaf containing point xyz, and the tets in that leaf
rval = myTree.leaf_containing_point(tree_root, xyz, treeiter);
Range leaf_tets;
rval = moab->get_entities_by_dimension(treeiter.handle(), 3,
leaf_tets, false);
```
More detailed examples of using the various tree decompositions in MOAB can be found in [ref-treeexamples].
Mesh readers and writers communicate mesh into/out of MOAB from/to disk files. Reading a mesh often involves importing large sets of data, for example coordinates of all the nodes in the mesh. Normally, this process would involve reading data from the file into a temporary data buffer, then copying data from there into its destination in MOAB. To avoid the expense of copying data, MOAB has implemented a reader/writer interface that provides direct access to blocks of memory used to represent mesh.
The reader interface, declared in ReadUtilIface, is used to request blocks of memory for storing coordinate positions and element connectivity. The pointers returned from these functions point to the actual memory used to represent those data in MOAB. Once data is written to that memory, no further copying is done. This not only saves time, but it also eliminates the need to allocate a large memory buffer for intermediate storage of these data.
MOAB allocates memory for nodes and elements (and their corresponding dense tags) in chunks, to avoid frequent allocation/de-allocation of small chunks of memory. The chunk size used depends on from where the mesh is being created, and can strongly affect the performance (and memory layout) of MOAB. Since dense tags are allocated at the chunk size, this can also affect overall memory usage in cases where the mesh size is small but the number of dense tags or dense tag size is large. When creating vertices and elements through the normal MOAB API, default chunk sizes defined in the SequenceManager class are used. However, most of the file readers in MOAB allocate the exact amount of space necessary to represent the mesh being read. There are also a few exceptions to this:
• When compiled in parallel, this space is increased by a factor of 1.5, to allow subsequent creation of ghost vertices/elements in the same chunk as the original mesh.
• The .cub file reader, which creates nodes and elements for individual geometric entities in separate calls, allocates using the default vertex/element sequence sizes, which are defined in the SequenceManager class in MOAB.
Applications calling the reader interface functions directly can specify the allocation chunk size as an optional parameter.
The reader interface consists of the following functions:
• get_node_coords: Given the number of vertices requested, the number of geometric dimensions, and a requested start id, allocates a block of vertex handles and returns pointers to coordinate arrays in memory, along with the actual start handle for that block of vertices.
• get_element_connect: Given the number of elements requested, the number of vertices per element, the element type and the requested start id, allocates the block of elements, and returns a pointer to the connectivity array for those elements and the actual start handle for that block. The number of vertices per element is necessary because those elements may include higher-order nodes, and MOAB stores these as part of the normal connectivity array.
• update_adjacencies: This function takes the start handle for a block of elements and the connectivity of those elements, and updates adjacencies for those elements. Which adjacencies are updated depends on the options set in AEntityFactory.
The following code fragment illustrates the use of ReadUtilIface to read a mesh directly into MOAB’s native representation. This code assumes that connectivity is specified in terms of vertex indices, with vertex indices starting from 1.
```// get the read iface from moab
// allocate a block of vertex handles and read xyz’s into them
std::vector<double*> arrays;
EntityHandle startv, *starth;
rval = iface->get_node_coords(3, num_nodes, 0, startv, arrays);
for (int i = 0; i < num_nodes; i++)
infile >> arrays[0][i] >> arrays[1][i] >> arrays[2][i];
// allocate block of hex handles and read connectivity into them
rval = iface->get_element_connect(num_hexes, 8, MBHEX, 0, starth);
for (int i = 0; i < 8*num_hexes; i++)
infile >> starth[i];
// change connectivity indices to vertex handles
for (int i = 0; i < 8*num_hexes; i++)
starth[i] += startv-1;
```
The writer interface, declared in WriteUtilIface, provides functions that support writing vertex coordinates and element connectivity to storage locations input by the application. Assembling these data is a common task for writing mesh, and can be non-trivial when exporting only subsets of a mesh. The writer interface declares the following functions:
• get_node_coords: Given already-allocated memory and the number of vertices and dimensions, and a range of vertices, this function writes vertex coordinates to that memory. If a tag is input, that tag is also written with integer vertex ids, starting with 1, corresponding to the order the vertices appear in that sequence (these ids are used to write the connectivity array in the form of vertex indices).
• get_element_connect: Given a range of elements and the tag holding vertex ids, and a pointer to memory, the connectivity of the specified elements are written to that memory, in terms of the indices referenced by the specified tag. Again, the number of vertices per element is input, to allow the direct output of higher-order vertices.
• gather_nodes_from_elements: Given a range of elements, this function returns the range of vertices used by those elements. If a bit-type tag is input, vertices returned are also marked with 0x1 using that tag. If no tag is input, the implementation of this function uses its own bit tag for marking, to avoid using an n2 algorithm for gathering vertices.
• reorder: Given a permutation vector, this function reorders the connectivity for entities with specified type and number of vertices per entity to match that permutation. This function is needed for writing connectivity into numbering systems other than that used internally in MOAB.
The following code fragment shows how to use WriteUtilIface to write the vertex coordinates and connectivity indices for a subset of entities.
```using namespace moab;
// get the write iface from moab
WriteUtilIface *iface;
ErrorCode rval = moab->get_interface("WriteUtilIface", &iface);
// get all hexes the model, and choose the first 10 of those
Range tmp_hexes, hexes, verts;
rval = moab->get_entities_by_type(0, MBHEX, tmp_hexes);
for (int i = 0; i < 10; i++) hexes.insert(tmp_hexes[i]);
rval = iface->gather_nodes_from_elements(hexes, 0, verts);
// assign vertex ids
iface->assign_ids(verts, 0, 1);
// allocate space for coordinates & write them
std::vector<double*> arrays(3);
for (int i = 0; i < 3; i++) arrays[i] = new double[verts.size()];
iface->get_node_coords(3, verts.size(), verts, 0, 1, arrays);
// put connect’y in array, in the form of indices into vertex array
std::vector<int> conn(8*hexes.size());
iface->get_element_connect(hexes.size(), 8, 0, hexes, 0, 1, &conn[0]);
```
### 4.6. File Readers/Writers Packaged With MOAB
MOAB has been designed to efficiently represent data and metadata commonly found in finite element mesh files. Readers and writers are included with MOAB which import/export specific types of metadata in terms of MOAB sets and tags, as described earlier in this document. The number of readers and writers in MOAB will probably grow over time, and so they are not enumerated here. See the src/io/README file in the MOAB source distribution for a current list of supported formats.
Because of its generic support for readers and writers, described in the previous section, MOAB is also a good environment for constructing new mesh readers and writers. The ReadTemplate and WriteTemplate classes in src/io are useful starting points for constructing new file readers/writers; applications are encouraged to submit their own readers/writers for inclusion in MOAB’s contrib/io directory in the MOAB source.
The usefulness of a file reader/writer is determined not only by its ability to read and write nodes and elements, but also in its ability to store the various types of meta-data included with the typical mesh. MOAB readers and writers are distinguished by their ability to preserve meta-data in meshes that they read and write. For example, MOAB’s CUB reader imports not only the mesh saved from CUBIT, but also the grouping of mesh entities into sets which reflect the geometric topology of the model used to generate the mesh. See [4] for a more detailed description of meta-data conventions used in MOAB’s file readers and writers, and the individual file reader/writer header files in src/io for details about the specific readers and writers.
Three specific file readers in MOAB bear further discussion: MOAB’s native HDF5-based file reader/writer; the CUB reader, used to import mesh and meta-data represented in CUBIT; and the CGM reader, which imports geometric models. These are described next.
A mesh database must be able to save and restore the data stored in its data model, at least to the extent to which it understands the semantics of that data. MOAB defines an HDF5-based file format that can store data embedded in MOAB. By convention, these files are given an “.h5m” file extension. When reading or writing large amounts of data, it is recommended to use this file format, as it is the most complete and also the most efficient of the file readers/writers in MOAB.
CUBIT is a toolkit for generating tetrahedral and hexahedral finite element meshes from solid model geometry [16]. This tool saves and restores data in a custom “.cub” file, which stores both mesh and geometry (and data relating the two). The CUB reader in MOAB can import and interpret much of the meta-data information saved in .cub files. Ref. [4] describes the conventions used to store this meta-data in the MOAB data model. The information read from .cub files, and stored in the MOAB data model, includes:
• Geometric model entities and topology
• Model entity names and ids
• Groups, element blocks, nodesets, and sidesets, including model entities stored in them
• Mesh scheme and interval size information assigned to model entities
Note that although information about model entities is recovered, MOAB by default does not depend on a solid modeling engine; this information is stored in the form of entity sets and parent/child relations between them. See Ref. [4] for more information.
The Common Geometry Module (CGM) [17] is a library for representing solid model and other types of solid geometry data. The CUBIT mesh generation toolkit uses CGM for its geometric modeling support, and CGM can restore geometric models in the exact state in which they were represented in CUBIT. MOAB contains a CGM reader, which can be enabled with a configure option. Using this reader, MOAB can read geometric models, and represent their model topology using entity sets linked by parent/child relations. The mesh in these models comes directly from the modeling engine faceting routines; these are the same facets used to visualize solid models in other graphics engines. When used in conjunction with the VisIt visualization tool (see Section ), this provides a solution for visualizing geometric models. Xxx shows a model imported using MOAB’s CGM reader and visualized with VisIt.
## 5.Parallel Mesh Representation and Query
A parallel mesh representation must strike a careful balance between providing an interface to mesh similar to that of a serial mesh, while also allowing the discovery of parallel aspects of the mesh and performance of parallel mesh-based operations efficiently. MOAB supports a spatial domain-decomposed view of a parallel mesh, where each subdomain is assigned to a processor, lower-dimensional entities on interfaces between subdomains are shared between processors, and ghost entities can be exchanged with neighboring processors. Locally, each subdomain, along with any locally-represented ghost entities, are accessed through a local MOAB instance. Parallel aspects of the mesh, e.g. whether entities are shared, on an interface, or ghost entities, are embedded in the same data model (entities, sets, tags, interface) used in the rest of MOAB. MOAB provides a suite of parallel functions for initializing and communicating with a parallel mesh, along with functions to query the parallel aspects of the mesh.
### 5.1. Nomenclature & Representation
Before discussing how to access parallel aspects of a mesh, several terms need to be defined:
Shared entity: An entity shared by one or several other processors.
Multi-shared entity: An entity shared by more than two processors.
Owning Processor: Each shared entity is owned by exactly one processor. This processor has the right to set tag values on the entity and have those values propagated to any sharing processors.
Part: The collection of entities assigned to a given processor. When reading mesh in parallel, the entities in a Part, along with any lower-dimensional entities adjacent to those, will be read onto the assigned processor.
Partition: A collection of Parts which take part in parallel collective communication, usually associated with an MPI communicator.
Interface: A collection of mesh entities bounding entities in multiple parts. Interface entities are owned by a single processor, but are represented on all parts/processors they bound.
Ghost entity: A shared, non-interface, non-owned entity.
Parallel status: A characteristic of entities and sets represented in parallel. The parallel status, or “pstatus”, is represented by a bit field stored in an unsigned character, with bit values as described in Table 4.
### Table 4: Bits representing various parallel characteristics of a mesh. Also listed are enumerated values that can be used in bitmask expressions; these enumerated variables are declared in MBParallelConventions.h.
Bit Name Represents
0x1 PSTATUS_NOT_OWNED Not owned by the local processor
0x2 PSTATUS_SHARED Shared by exactly two processorstd>
0x4 PSTATUS_MULTISHARED Shared by three or more processors
0x8 PSTATUS_INTERFACE Part of lower-dimensional interface shared by multiple processors
0x10 PSTATUS_GHOST Non-owned, non-interface entities represented locally
Parallel functionality is described in the following sections. First, methods to load a mesh into a parallel representation are described; next, functions for accessing parallel aspects of a mesh are described; functions for communicating mesh and tag data are described.
### 5.2. Parallel Mesh Initialization
Parallel mesh is initialized in MOAB in several steps:
1. Establish a local mesh on each processor, either by reading the mesh into that representation from disk, or by creating mesh locally through the normal MOAB interface.
2. Find vertices, then other entities, shared between processors, based on a globally-consistent vertex numbering stored on the GLOBAL_ID tag.
3. Exchange ghost or halo elements within a certain depth of processor interfaces with neighboring processors.
These steps can be executed by a single call to MOAB’s load_file function, using the procedure described in Section 5.2.1. Or, they can be executed in smaller increments calling specific functions in MOAB’s ParallelComm class, as described in Section 5.2.2. Closely related to the latter method is the handling of communicators, described in more detail in Section.
In the file reading approach, a mesh must contain some definition of the partition (the assignment of mesh, usually regions, to processors). Partitions can be derived from other set structures already on the mesh, or can be computed explicitly for that purpose by tools like mbzoltan (see Section 4.2). For example, geometric volumes used to generate the mesh, and region-based material type assignments, are both acceptable partitions (see Ref. [4] for information about this and other meta-data often accompanying mesh). In addition to defining the groupings of regions into parts, the assignment of specific parts to processors can be done implicitly or using additional data stored with the partition.
MOAB implements several specific methods for loading mesh into a parallel representation:
• READ_PART: each processor reads only the mesh used by its part(s).
• READ_DELETE: each processor reads the entire mesh, then deletes the mesh not used by its part(s).
• BCAST_DELETE: the root processor reads and broadcasts the mesh; each processor then deletes the mesh not used by its part(s).
The READ_DELETE and BCAST_DELETE methods are supported for all file types MOAB is able to read, while READ_PART is only implemented for MOAB’s native HDF5-based file format.
Various other options control the selection of part sets or other details of the parallel read process. For example, the application can specify the tags, and optionally tag values, which identify parts, and whether those parts should be distributed according to tag value or using round-robin assignment.
The options used to specify loading method, the data used to identify parts, and other parameters controlling the parallel read process, are shown in Table 5.
### Table 5: Options passed to MOAB’s load_file function identifying the partition and other parameters controlling the parallel read of mesh data. Options and values should appear in option string as “option=val”, with a delimiter (usually “;”) between options.
Option Value Description
PARTITION <tag_name> Sets with the specified tag name should be used as part sets
PARTITION_VAL <val1, val2-val3, ...> Integer values to be combined with tag name, with ranges input using val1, val2-val3. Not meaningful unless PARTITION option is also given.
PARTITION_DISTRIBUTE (none) If present, or values are not input using PARTITION_VAL, sets with tag indicated in PARTITION option are partitioned across processors in round-robin fashion.
PARALLEL_RESOLVE_SHARED_ENTS <pd.sd> Resolve entities shared between processors, where partition is made up of pd- dimensional entities, and entities of dimension sd and lower should be resolved.
PARALLEL_GHOSTS <gd.bd.nl[.ad]> Exchange ghost elements at shared inter-processor interfaces. Ghost elements of dimension gd will be exchanged. Ghost elements are chosen going through bd-dimensional interface entities. Number of layers of ghost elements is specified in nl. If ad is present, lower-dimensional entities bounding exchanged ghost entities will also be exchanged; allowed values for ad are 1 (exchange bounding edges), 2 (faces), or 3 (edges and faces). PARALLEL_COMM <id> Use the ParallelComm with index <id>. Index for a ParallelComm object can be checked with ParallelComm::get_id(), and a ParallelComm with a given index can be retrieved using ParallelComm::get_pcomm(id).
CPUTIME
(none)
Print cpu time required for each step of parallel read & initialization.
MPI_IO_RANK
<r>
If read method requires reading mesh onto a single processor, processor with rank r is used to do that read.
Several example option strings controlling parallel reading and initialization are:
“PARALLEL=READ_DELETE; PARTITION=MATERIAL_SET; PARTITION_VAL=100, 200, 600-700”: The whole mesh is read by every processor; this processor keeps mesh in sets assigned the tag whose name is “MATERIAL_SET” and whose value is any one of 100, 200, and 600-700 inclusive.
“PARALLEL=READ_PART; PARTITION=PARALLEL_PARTITION, PARTITION_VAL=2”: Each processor reads only its mesh; this processor, whose rank is 2, is responsible for elements in a set with the PARALLEL_PARTITION tag whose value is 2. This would by typical input for a mesh which had already been partitioned with e.g. Zoltan or Parmetis.
“PARALLEL=BCAST_DELETE; PARTITION=GEOM_DIMENSION, PARTITION_VAL=3, PARTITION_DISTRIBUTE”: The root processor reads the file and broadcasts the whole mesh to all processors. If a list is constructed with entity sets whose GEOM_DIMENSION tag is 3, i.e. sets corresponding to geometric volumes in the original geometric model, this processor is responsible for all elements with index R+iP, i >= 0 (i.e. a round-robin distribution).
### 5.2.2. Parallel Mesh Initialization Using Functions
After creating the local mesh on each processor, an application can call the following functions in ParallelComm to establish information on shared mesh entities. See the [ref-directparmesh example] in the MOAB source tree for a complete example of how this is done from an application.
• ParallelComm::resolve_shared_entities (collective): Resolves shared entities between processors, based on GLOBAL_ID tag values of vertices. Various forms are available, based on entities to be evaluated and maximum dimension for which entity sharing should be found.
• ParallelComm::exchange_ghost_cells (collective): Exchange ghost entities with processors sharing an interface with this processor, based on specified ghost dimension (dimension of ghost entities exchanged), bridge dimension, number of layers, and type of adjacencies to ghost entities. An entity is sent as a ghost if it is within that number of layers, across entities of the bridge dimension, with entities owned by the receiving processor, or if it is a lower-dimensional entity adjacent to a ghost entity and that option is requested.
### 5.2.3. Communicator Handling
The ParallelComm constructor takes arguments for an MPI communicator and a MOAB instance. The ParallelComm instance stores the MPI communicator, and registers itself with the MOAB instance. Applications can specify the ParallelComm index to be used for a given file operation, thereby specifying the MPI communicator for that parallel operation. For example:
```using namespace moab;
// pass a communicator to the constructor, getting back the index
MPI_Comm my_mpicomm;
int pcomm_index;
ParallelComm my_pcomm(moab, my_mpicomm, &pcomm_index);
// write the pcomm index into a string option
pcomm_index);
// specify that option in a parallel read operation
```
In the above code fragment, the ParallelComm instance with index pcomm_index will be used in the parallel file read, so that the operation executes over the specified MPI communicator. If no ParallelComm instance is specified for a parallel file operation, a default instance will be defined, using MPI_COMM_WORLD.
Applications needing to retrieve a ParallelComm instance created previously and stored with the MOAB instance, e.g. by a different code component, can do so using a static function on ParallelComm:
```ParallelComm *my_pcomm = ParallelComm::get_pcomm(moab, pcomm_index);
```
ParallelComm also provides the ParallelComm::get_all_pcomm function, for retrieving all ParallelComm instances stored with a MOAB instance. For syntax and usage of this function, see the MOAB online documentation for ParallelComm.hpp [8].
### 5.3. Parallel Mesh Query Functions
Various functions are commonly used in parallel mesh-based applications. Functions marked as being collective must be called collectively for all processors that are members of the communicator associated with the ParallelComm instance used for the call.
ParallelComm::get_pstatus: Get the parallel status for the entity.
ParallelComm::get_pstatus_entities: Get all entities whose pstatus satisfies (pstatus & val).
ParallelComm::get_owner: Get the rank of the owning processor for the specified entity.
ParallelComm::get_owner_handle: Get the rank of the owning processor for the specified entity, and the entity's handle on the owning processor.
ParallelComm::get_sharing_data: Get the sharing processor(s) and handle(s) for an entity or entities. Various overloaded versions are available, some with an optional “operation” argument, where Interface::INTERSECT or Interface::UNION can be specified. This is similar to the operation arguments to Interface::get_adjacencies.
ParallelComm::get_shared_entities: Get entities shared with the given processor, or with all processors. This function has optional arguments for specifying dimension, whether interface entities are requested, and whether to return just owned entities.
ParallelComm::get_interface_procs: Return all processors with whom this processor shares an interface.
ParallelComm::get_comm_procs: Return all processors with whom this processor communicates.
### 5.4. Parallel Mesh Communication
Once a parallel mesh has been initialized, applications can call the ParallelComm::exchange_tags function for exchanging tag values between processors. This function causes the owning processor to send the specified tag values for all shared, owned entities to other processors sharing those entities. Asynchronous communication is used to hide latency, and only point-to-point communication is used in these calls.
## 6.Building MOAB-Based Applications
There are two primary mechanisms supported by MOAB for building applications, one based on MOAB-defined make variables, and the other based on the use of libtool and autoconf. Both assume the use of a “make”-based build system.
The easiest way to incorporate MOAB into an application’s build process is to include the “moab.make” file into the application’s Makefile, adding the make variables MOAB_INCLUDES and MOAB_LIBS_LINK to application’s compile and link commands, respectively. MOAB_INCLUDES contains compiler options specifying the location of MOAB include files, and any preprocessor definitions required by MOAB. MOAB_LIBS_LINK contains both the options telling where libraries can be found, and the link options which incorporate those libraries into the application. Any libraries depended on by the particular configuration of MOAB are included in that definition, e.g. the HDF5 library. Using this method to incorporate MOAB is the most straightforward; for example, the following Makefile is used to build one of the example problems packaged with the MOAB source:
```include \${MOAB_LIB_DIR}/moab.make
GetEntities : GetEntities.o
.cpp.o :
\${CXX} \${MOAB_INCLUDES} -c \$<
```
Here, the MOAB_LIB_DIR environment variable or make argument definition specifies where the MOAB library is installed; this is also the location of the moab.make file. Once that file has been included, MOAB_INCLUDES and MOAB_LIBS_LINK can be used, as shown.
Other make variables are defined in the moab.make file which simplify building applications:
• MOAB_LIBDIR, MOAB_INCLUDEDIR: the directories into which MOAB libraries and include files will be installed, respectively. Note that some include files are put in a subdirectory named “moab” below that directory, to reflect namespace naming conventions used in MOAB.
• MOAB_CXXFLAGS, MOAB_CFLAGS, MOAB_LDFLAGS: Options passed to the C++ and C compilers and the linker, respectively.
• MOAB_CXX, MOAB_CC, MOAB_FC: C++, C, and Fortran compilers specified to MOAB at configure time, respectively.
The second method for incorporating MOAB into an application’s build system is to use autoconf and libtool. MOAB is configured using these tools, and generates the “.la” files that hold information on library dependencies that can be used in application build systems also based on autoconf and libtool. Further information on this subject is beyond the scope of this User’s Guide; see the “.la” files as installed by MOAB, and contact the MOAB developer’s mailing list [6] for more details.
## 7.iMesh (ITAPS Mesh Interface) Implementation in MOAB
iMesh is a common API to mesh data developed as part of the Interoperable Tools for Advanced Petascale Simulations (ITAPS) project [18]. Applications using the iMesh interface can operate on any implementation of that interface, including MOAB. MOAB-based applications can take advantage of other services implemented on top of iMesh, including the MESQUITE mesh improvement toolkit [19] and the GRUMMP mesh improvement library [20].
MOAB’s native interface is accessed through the Interface abstract C++ base class. Wrappers are not provided in other languages; rather, applications wanting to access MOAB from those languages should do so through iMesh. In most cases, the data models and functionality available through MOAB and iMesh are identical. However, there are a few differences, subtle and not-so-subtle, between the two:
SPARSE tags used by default: MOAB’s iMesh implementation creates SPARSE tags by default, because of semantic requirements of other tag-related functions in iMesh. To create DENSE tags through iMesh, use the iMesh_createTagWithOptions extension function (see below).
Higher-order elements: ITAPS currently handles higher-order elements (e.g. a 10-node tetrahedron) usi[21]5. As described in [sec-entities], MOAB supports higher-order entities by allowing various numbers of vertices to define topological entities like quadrilateral or tetrahedron. Applications can specify flags to the connectivity and adjacency functions specifying whether corner or all vertices are requested.
Self-adjacencies: In MOAB’s native interface, entities are always self-adjacent6; that is, adjacencies of equal dimension requested from an entity will always include that entity, while from iMesh will not include that entity.
Option strings: The iMesh specification requires that options in the options string passed to various functions (e.g. iMesh_load) be prepended with the implementation name required to parse them, and delimited with spaces. Thus, a MOAB-targeted option would appear as “moab:PARALLEL=READ_PART moab:PARTITION=MATERIAL_SET”.
To provide complete MOAB support from other languages through iMesh, a collection of iMesh extension functions are also available. A general description of these extensions appears below; for a complete description, see the online documentation for iMesh-extensions.h [8].
• Recursive get_entities functions: There are many cases where sets include other sets (see [4] for more information). MOAB provides iMesh_getEntitiesRec, and other recursive-supporting functions, to get all non-set entities of a given type or topology accessible from input set(s). Similar functions are available for number of entities of a given type/topology.
• Get entities by tag, and optionally tag value: It is common to search for entities with a given tag, and possibly tag value(s); functions like iMesh_getEntitiesByTag are provided for this purpose.
• Options to createTag: To provide more control over the tag type, the iMesh_createTagWithOptions is provided. The storage type is controlled with the “
• MBCNType: Canonical numbering evaluations are commonly needed by applications, e.g. to apply boundary conditions locally. The MBCN package provides these evaluations in terms of entity types defined in MOAB [9]; the getMBCNType is required to translate between iMesh_Topology and MBCN type.
• Iterator step: Step an iterator a specified number of entities; allows advancement of an iterator without needing to allocate memory to hold the entity handles stepped over.
• Direct access to tag storage: The Interface::tag_iterate function allows an application get a pointer to the memory used to store a given tag. For dense tags on contiguous ranges of entities, this provides more efficient access to tags. The iMesh functionn iMesh_tagIterate provides access to this functionlity. See examples/TagIterateC.c and examples/TagIterateF.F for examples of how to use this from C and Fortran, respectively.
As required by the iMesh specification, MOAB generates the “iMesh-Defs.inc” file and installs it with the iMesh and MOAB libraries. This file defines make variables which can be used to build iMesh-based applications. The method used here is quite similar to that used for MOAB itself (see Section 6). In particular, the IMESH_INCLUDES and IMESH_LIBS make variables can be used with application compile and link commands, respectively, with other make variables similar to those provided in moab.make also available.
Note that using the iMesh interface from Fortran-based applications requires a compiler that supports Cray pointers, along with the pass-by-value (VAL) extension. Almost all compilers support those extensions; however, if using the gcc series of compilers, you must use gfortran 4.3 or later.
5There are currently no implementations of this interface.
6iMesh and MOAB both define adjacencies using the topological concept of closure. Since the closure of an entity includes the entity itself, the d-dimensional entities on the closure of a given entity should include the entity itself.
## 8.Structured Mesh Representation
A structured mesh is defined as a D-dimensional mesh whose interior vertices have 2D connected edges. Structured mesh can be stored without connectivity, if certain information is kept about the parametric space of each structured block of mesh. MOAB can represent structured mesh with implicit connectivity, saving approximately 57% of the storage cost compared to an unstructured representation7. Since connectivity must be computed on the fly, these queries execute a bit slower than those for unstructured mesh. More information on the theory behind MOAB's structured mesh representation can be found in “MOAB-SD: Integrated structured and unstructured mesh representation”[17].
Currently, MOAB's structured mesh representation can only be used by creating structured mesh at runtime; that is, structured mesh is saved/restored in an unstructured format in MOAB's HDF5-based native save format. For more details on how to use MOAB's structured mesh representation, see the scdseq_test.cpp source file in the test/ directory.
7 This assumes vertex coordinates are represented explicitly, and that there are approximately the same number of vertices and hexahedra in a structured hex mesh.
## 9.Spectral Element Meshes
The Spectral Element Method (SEM) is a high-order method, using a polynomial Legendre interpolation basis with Gauss-Lobatto quadrature points, in contrast to the Lagrange basis used in (linear) finite elements [20]. SEM obtains exponential convergence with decreasing mesh characteristic sizes, and codes implementing this method typically have high floating-point intensity, making the method highly efficient on modern CPUs. Most Nth-order SEM codes require tensor product cuboid (quad/hex) meshes, with each d-dimensional element containing (N+1)d degrees of freedom (DOFs). There are various methods for representing SEM meshes and solution fields on them; this document discusses these methods and the tradeoffs between them. The mesh parts of this discussion are given in terms of the iMesh mesh interface and its implementation by the MOAB mesh library [21].
The figure above shows a two-dimensional 3rd-order SEM mesh consisting of four quadrilaterals. For this mesh, each quadrilateral has (N+1)^2=16 DOFs, with corner and edge degrees of freedom shared between neighboring quadrilaterals.
### 9.1. Representations
There are various representations of this mesh in a mesh database like MOAB, depending on how DOFs are related to mesh entities and tags on those entities. We mention several possible representations:
1. Corner vertices, element-based DOFs: Each quadrilateral is defined by four vertices, ordered in CCW order typical of FE meshes. DOFs are stored as tags on quadrilaterals, with size (N+1)^2 values, ordered lexicographically (i.e. as a 2D array tag(i,j) with i varying faster than j.) In the figure above, the connectivity for face 1 would be (1, 4, 16, 13), and DOFs would be ordered (1..16). Note that in this representation, tag values for DOFs shared by neighboring elements must be set multiple times, since there are as many copies of these DOFs as elements sharing them.
2. High-order FE-like elements: Each DOF is represented by a mesh vertex. Quadrilaterals each have (N+1)^2 vertices, ordered as they would be for high-order finite elements (corner vertices first, then mid-edge and mid-face elements; see [22]). Mid -face, -edge, and -region vertices for a given edge/face/region would be ordered lexicographically, according to positive direction in a corresponding reference element. In the figure above, the connectivity array for face 1 would be (1, 4, 16, 13, 2, 3, 8, 12, 14, 15, 5, 9, 6, 7, 10, 11). DOF values are stored as tags on vertices. Since DOFs are uniquely associated with vertices and vertices are shared by neighboring elements, tag values only need to be set once. Full vertex-quadrilateral adjacencies are available, for all vertices.
3. Linear FE-like elements, one vertex per DOF, array with DOF vertices: Each quadrilateral is defined by four (corner) vertices, with additional vertices representing mid-edge and mid-face DOFs. An additional “DOF array” tag is assigned to each quadrilateral, storing the array of vertices representing the (N+1)^2 DOFs for the quadrilateral, ordered lexicographically. For the figure above, the connectivity array for face 1 would be (1, 4, 16, 13), and the DOF array would be (1..16), assuming that vertex handles are integers as shown in the figure. DOF values are stored as tags on vertices, and lexicographically-ordered arrays of DOFs can be retrieved using the DOF array tag as input to the tag_get_data function in MOAB. Adjacency functions would only be meaningful for corner vertices, but tag values would only need to be set once per DOF.
4. High-order FE-like elements, array with DOF vertices: This is a combination of options 2 and 3. The advantage would be full vertex-quad adjacency support and direct availability of lexicographically-ordered vertex arrays, at the expense of more memory.
5. Convert to linear mesh: Since a spectral element is a cuboid with higher-order vertices, it can always be converted to N^2 linear cuboids using the high-order vertices as corners of the finer quads/hexes. This is how readers in ParaView and VisIt typically import spectral meshes (CAM-SE also exports connectivity in this form).
As a convenience for applications, functions could also be provided for important tasks, like assembling the vertex handles for an entity in lexographic order (useful for option 2 above), and getting an array of tag values in lexicographic order (for option 3 above).
There are various competing tradeoffs in the various representation types. These include:
• Adjacencies: being able to retrieve the element(s) using a given (corner or higher-order) vertex.
• Connectivity list: being able to retrieve the connectivity of a given element, consisting of all (corner + higher-order) vertices in the element, usually in lexicographical order. This is closely linked with being able to access the connectivity list as a const*, i.e. using the list straight from memory without needing to copy it.
• Memory vs. time: There is a memory vs. execution time tradeoff between duplicating interface vertex solution/tag variables in neighboring elements (more memory but more time-efficient and allows direct access to tag storage by applications) versus using vertex-based tags (less memory but requires assembly of variables into lexicographically-ordered arrays, and prevents direct access from applications).
The lower-memory option (storing variables on vertices and assembling into lexicographically-ordered arrays for application use) usually ends up costing more in memory anyway, since applications must allocate their own storage for these arrays. On the other hand, certain applications will always choose to do that, instead of sharing storage with MOAB for these variables. In the case where applications do share memory with MOAB, other tools would need to interpret the lexicographically-ordered field arrays specially, instead of simply treating the vertex tags as a point-based field.
### 9.3. MOAB Representation
In choosing the right MOAB representation for spectral meshes, we are trying to balance a) minimal memory usage, b) access to properly-ordered and -aligned tag storage, and c) maximal compatibility with tools likely to use MOAB. The solution we propose is to use a representation most like option 2) above, with a few optional behaviors based on application requirements.
In brief, we propose to represent elements using the linear, FE-ordered connectivity list (containing only corner vertices from the spectral element), with field variables written to either vertices, lexicographically-ordered arrays on elements, or both, and with a lexicographically-ordered array (stored on tag SPECTRAL_VERTICES) of all (corner+higher-order) vertices stored on elements. In the either/or case, the choice will be evident from the tag size and the entities on which the tag is set. In the both case, the tag name will have a “-LEX” suffix for the element tags, and the size of the element tag will be (N+1)^2 times that of the vertex-based tag. Finally, the file set containing the spectral elements (or the root set, if no file set was input to the read) will contain a “SPECTRAL_ORDER” tag whose value is N. These conventions are described in the “Metadata Information” document distributed with the MOAB source code.
## 10.Performance and Using MOAB Efficiently from Applications
MOAB is designed to operate efficiently on groups of entities and for large meshes. Applications will be most efficient when they operate on entities in groups, especially groups which are close in their order of creation. The MOAB API is structured to encourage operations on groups of entities. Conversely, MOAB will not perform as well as other libraries if there are frequent deletion and creation of entities. For those types of applications, a mesh library using a C++ object-based representation is more appropriate. In this section, performance of MOAB when executing a variety of tasks is described, and compared to that of other representations. Of course, these metrics are based on the particular models and environments where they are run, and may or may not be representative of other application types.
One useful measure of MOAB performance is in the representation and query of a large mesh. MOAB includes a performance test, located in the test/perf directory, in which a single rectangular region of hexahedral elements is created then queried; the following steps are performed:
• Create the vertices and hexes in the mesh
• For each vertex, get the set of connected hexahedra
• For each hex, get the connected vertices, their coordinates, average them, and assign them as a tag on the hexes
This test can be run on your system to determine the runtime and memory performance for these queries in MOAB.
## 11.Conclusions and Future Plans
MOAB, a Mesh-Oriented datABase, provides a simple but powerful data abstraction to structured and unstructured mesh, and makes that abstraction available through a function API. MOAB provides the mesh representation for the VERDE mesh verification tool, which demonstrates some of the powerful mesh metadata representation capabilities in MOAB. MOAB includes modules that import mesh in the ExodusII, CUBIT .cub and Vtk file formats, as well as the capability to write mesh to ExodusII, all without licensing restrictions normally found in ExodusII-based applications. MOAB also has the capability to represent and query structured mesh in a way that optimizes storage space using the parametric space of a structured mesh; see Ref. [17] for details.
Initial results have demonstrated that the data abstraction provided by MOAB is powerful enough to represent many different kinds of mesh data found in real applications, including geometric topology groupings and relations, boundary condition groupings, and inter-processor interface representation. Our future plans are to further explore how these abstractions can be used in the design through analysis process.
## 12.References
[2] T.J. Tautges and A. Caceres, “Scalable parallel solution coupling for multiphysics reactor simulation,” Journal of Physics: Conference Series, vol. 180, 2009.
[3] T.J. Tautges, MOAB Meta-Data Information, 2010.
[4] T.J. Tautges, “MOAB - ITAPS – Trac.”, http://trac.mcs.anl.gov/projects/ITAPS/wiki/MOAB
[5] “MOAB Developers Email List.”, moab-dev@mcs.anl.gov.
[6] “MOAB Users Email List.”, moab@mcs.anl.gov.
[7] “MOAB online documentation.”, http://gnep.mcs.anl.gov:8010/moab-docs/
[8] T.J. Tautges, “Canonical numbering systems for finite-element codes,” Communications in Numerical Methods in Engineering, vol. Online, Mar. 2009.
[9] L.A. Schoof and V.R. Yarberry, EXODUS II: A Finite Element Data Model, Albuquerque, NM: Sandia National Laboratories, 1994.
[10] M. PATRAN, “PATRAN User’s Manual,” 2005.
[11] VisIt User's Guide.
[12] K. Devine, E. Boman, R. Heaphy, B. Hendrickson, and C. Vaughan, “Zoltan Data Management Services for Parallel Dynamic Applications,” Computing in Science and Engineering, vol. 4, 2002, pp. 90–97.
[13] T.J. Tautges, P.P.H. Wilson, J. Kraftcheck, B.F. Smith, and D.L. Henderson, “Acceleration Techniques for Direct Use of CAD-Based Geometries in Monte Carlo Radiation Transport,” International Conference on Mathematics, Computational Methods & Reactor Physics (M&C 2009), Saratoga Springs, NY: American Nuclear Society, 2009.
[14] H. Kim and T. Tautges, “EBMesh: An Embedded Boundary Meshing Tool,” in preparation.
[15] G.D. Sjaardema, T.J. Tautges, T.J. Wilson, S.J. Owen, T.D. Blacker, W.J. Bohnhoff, T.L. Edwards, J.R. Hipp, R.R. Lober, and S.A. Mitchell, CUBIT mesh generation environment Volume 1: Users manual, Sandia National Laboratories, May 1994, 1994.
[16] T.J. Tautges, “CGM: A geometry interface for mesh generation, analysis and other applications,” Engineering with Computers, vol. 17, 2001, pp. 299-314.
[17] T. J. Tautges, MOAB-SD: Integrated structured and unstructured mesh representation, Engineering with Computers, vol. 20, no. 3, pp. 286-293, 2004.
[18] “Interoperable Technologies for Advanced Petascale Simulations (ITAPS),” Interoperable Technologies for Advanced Petascale Simulations (ITAPS).
[19] P. Knupp, “Mesh quality improvement for SciDAC applications,” Journal of Physics: Conference Series, vol. 46, 2006, pp. 458-462.
[20] M. O. Deville, P. F. Fischer, and E. H. Mund, High-order methods for incompressible fluid flow. Cambridge, UK; New York: Cambridge University Press, 2002.
[21] T. J. Tautges, “MOAB Wiki.” [Online]. Available: http://trac.mcs.anl.gov/projects/ITAPS/wiki/MOAB. [Accessed: 30-Oct-2012].
[22] T. J. Tautges, “Canonical numbering systems for finite-element codes,” International Journal for Numerical Methods in Biomedical Engineering, vol. 26, no. 12, pp. 1559–1572, 2010. | 16,591 | 78,652 | {"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-2017-39 | latest | en | 0.828719 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.