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http://oeis.org/A262944 | 1,571,347,636,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986676227.57/warc/CC-MAIN-20191017200101-20191017223601-00052.warc.gz | 143,512,812 | 4,407 | This site is supported by donations to The OEIS Foundation.
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A262944 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a square or a pentagonal number. 14
1, 2, 2, 2, 2, 3, 4, 3, 1, 3, 5, 3, 2, 2, 5, 5, 3, 3, 5, 5, 3, 6, 6, 3, 3, 8, 6, 5, 5, 3, 7, 5, 5, 3, 4, 4, 8, 9, 3, 5, 7, 6, 3, 5, 5, 7, 5, 3, 4, 5, 6, 6, 9, 4, 5, 7, 7, 5, 4, 4, 7, 6, 1, 5, 5, 7, 7, 7, 1, 6, 10, 8, 6, 3, 4, 3, 6, 4, 6, 9, 5, 7, 9, 3, 5, 8, 9, 8, 3, 3, 11, 10, 6, 6, 8, 12, 5, 6, 4, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 63, 69, 489, 714, 1089. (ii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is twice a square or twice a pentagonal number. (iii) For any positive integer n, there are integers x >= 0 and y > 0 such that n - 2*x^4 - y*(y+1)/2 is a square or a pentagonal number. See also A262941 and A262945 for similar conjectures. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396. EXAMPLE a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n-1)/2. a(9) = 1 since 9 = 1^4 + 2*3/2 + p_5(2). a(63) = 1 since 63 = 0^4 + 7*8/2 + p_5(5). a(69) = 1 since 69 = 2^4 + 7*8/2 + 5^2. a(489) = 1 since 489 = 3^4 + 12*13/2 + p_5(15). a(714) = 1 since 714 = 4^4 + 18*19/2 + p_5(14). a(1089) = 1 since 1089 = 4^4 + 38*39/2 + p_5(8). MATHEMATICA SQ[n_]:=IntegerQ[Sqrt[n]]||(IntegerQ[Sqrt[24n+1]]&&Mod[Sqrt[24n+1]+1, 6]==0) Do[r=0; Do[If[SQ[n-x^4-y(y+1)/2], r=r+1], {x, 0, n^(1/4)}, {y, 1, (Sqrt[8(n-x^4)+1]-1)/2}]; Print[n, " ", r]; Continue, {n, 1, 100}] CROSSREFS Cf. A000217, A000290, A000583, A001318, A160325, A254885, A262813, A262815, A262816, A262941, A262945. Sequence in context: A187758 A171931 A221530 * A322789 A069904 A164978 Adjacent sequences: A262941 A262942 A262943 * A262945 A262946 A262947 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 05 2015 STATUS approved
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Last modified October 17 16:51 EDT 2019. Contains 328120 sequences. (Running on oeis4.) | 1,168 | 2,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.674451 |
https://brainyriddles.com/what-did-the-wise-man-say/ | 1,679,962,758,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00249.warc.gz | 182,400,304 | 34,186 | Select Page
# What did the wise man say?
It might not seem like there’s enough information to solve these logic riddle at first—but that’s part of the fun!
An Arab sheik is old and must leave his fortune to one of his two sons. He makes a proposition: Both sons will ride their camels in a race, and whichever camel crosses the finish line LAST will win the fortune for its owner. During the race, the two brothers wander aimlessly for days, neither willing to cross the finish line. In desperation, they ask a wise man for advice. He tells them something; then the brothers leap onto the camels and charge toward the finish line.
What did the wise man say?
The rules of the race were that the owner of the camel that crosses the finish line last wins the fortune. The wise man simply told them to switch camels. Try these math riddles only the smartest can get right.
## How can this be?
A certain large animal lives happily and thrives here on Earth. One day, every single one of these critters is wiped out by a mysterious disease which affects only this particular animal. There are none left anywhere on earth -- they are all gone. About a year or so...
## Always involve a bed
Looking for a riddle or a joke to spark some laughter? Well, you’ve come to the right place then.You can go on top of me or underneath and I always involve a bed. What am I?A bunk bed.more riddlesDonec rutrum congue leo...
## Long shaft
Try to guess some answer and check the proper answer below and share it with your friends.I have a long shaft. I always penetrate with the tip first and I always come with a quiver. What am I?Arrowmore riddlesDonec rutrum...
## Which ball?
Two barrels contain each the same amount of water. The water temperature in one barrel is 49°F while the other is at 29°F. Two golf balls of same dimension and weight are dropped simultaneously into the barrels. The golf balls are dropped from the same height.Which... | 433 | 1,941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-14 | longest | en | 0.943873 |
https://discuss.leetcode.com/topic/94231/issues-about-tle | 1,513,340,476,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948569405.78/warc/CC-MAIN-20171215114446-20171215140446-00381.warc.gz | 547,154,656 | 11,049 | • Why the first script TLE and the second is not?
The first:
``````public class Solution {
int ans = 0;
public int reversePairs(int[] nums) {
help(nums,0,nums.length-1);
return ans;
}
public void help(int[] nums, int s, int e){
if(s>=e) return;
int mid = s + (e-s)/2;
help(nums,s,mid);help(nums,mid+1,e);
for(int i = s; i <= mid; i++){
int j = mid+1; ///////////////////The difference here
while(j<=e&&nums[i]/2.0>nums[j]) j++;
ans +=j-(mid+1);
}
Arrays.sort(nums,s,e+1);
}
}
``````
The second:
``````public class Solution {
int ans = 0;
public int reversePairs(int[] nums) {
help(nums,0,nums.length-1);
return ans;
}
public void help(int[] nums, int s, int e){
if(s>=e) return;
int mid = s + (e-s)/2;
help(nums,s,mid);help(nums,mid+1,e);
for(int i = s, j = mid+1; i <= mid; i++){
while(j<=e&&nums[i]/2.0>nums[j]) j++;
ans +=j-(mid+1);
}
Arrays.sort(nums,s,e+1);
}
}
``````
It seems that only the place where to define j changes and the time differs 30 times to the other.
Does anyone know why?
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 347 | 1,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-51 | latest | en | 0.577227 |
https://plainmath.net/post-secondary/algebra/college-algebra/college-algebra | 1,620,481,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.7/warc/CC-MAIN-20210508121446-20210508151446-00388.warc.gz | 487,385,441 | 6,684 | #### Didn’t find what you are looking for?
Upper level algebra
### Find the x-and y-intercepts of this equation. $$\displaystyle{f{{\left({x}\right)}}}=-{x}+{2}$$
Upper level algebra
### To calculate: The probability that the selected student was in the 37-46 are group or received a "B" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade.
Upper level algebra
### A multiple regression equation to predict a student's score in College Algebra $$(\hat{y})$$ based on their high school GPA (x1x1), their high school Algebra II grade (x2x2), and their placement test score (x3x3) is given by the equation below. $$\hat{y}=-9+5x1x1+6x2x2+0.3x3x3$$ a) According to this equation, what is the predicted value of the student's College Algebra score if their high school GPA was a 3.9, their high school Algebra II grade was a 2 and their placement test score was a 40? Round to 1 decimal place. b) According to this equation, what does the student's placement test score need to be if their high school GPA was a 3.9, their high school Algebra II grade was a 2, and their predicted College Algebra score was a 67? Round to 1 decimal place.
Upper level algebra
### Find the x-and y-intercepts of the graph of the equation algebraically. $$\displaystyle{\frac{{{8}{x}}}{{{3}}}}+{50}-{2}{y}={0}$$
Upper level algebra
### To find the lowest original score that will result in an A if the professor uses $$(i)(f*g)(x)\ and\ (ii)(g*f)(x)$$. Professor Harsh gave a test to his college algebra class and nobody got more than 80 points (out of 100) on the test. One problem worth 8 points had insufficient data, so nobody could solve that problem. The professor adjusted the grades for the class by a. Increasing everyone's score by 10% and b. Giving everyone 8 bonus points c. x represents the original score of a student
Upper level algebra
### To calculate: To write the given statements (a) and (b) as functions f(x) and g(x)respectively. Professor Harsh gave a test to his college algebra class and nobody got more than 80 points (out of 100) on the test. One problem worth 8 points had insufficient data, so nobody could solve that problem. The professor adjusted the grades for the class by a. Increasing everyone's score by 10% and b. Giving everyone 8 bonus points c. x represents the original score of a student
Upper level algebra
### Eastside Golden Eagles scored 60 points on a combi: nation of tWo-polnt shots and three-point shots. If they made a total of 27 shots. how many of each kind of shot was made”?
Upper level algebra
### The winner of a 250 mile race drove his car to victory at a rate of 137.1656 mph. What was his time (to the nearest thousandth of an hour)?
Upper level algebra
### To calculate: The probability that the selected student of 37-46 age group has received an "A" in the course. Given Information: A table depicting the grade distribution for a college algebra class based on age and grade.
Upper level algebra
### To find the average cost of a sealable book if 10,000 books are produced.
Upper level algebra
### Number of Students: In a certain school 80% of the students in first-year chemistry have had algebra. If there are 280 students in first-year chemistry, how many of them have had algebra?
Upper level algebra
### To find the number of books that must be produced to bring the average cost of a sealable book under $$\20$$.
Upper level algebra
### To write a function f describing the average cost of sealable books. Given information: The printing and binding cost for a college algebra book is $10. The editorial cost is$200,000. The first 2500 books are free.
Upper level algebra
### To calculate:The score Grant must earn on his fifth test to have an average of 90 if on Grant's four College Algebra tests his scores were 76, 91,97 and 87.
Upper level algebra
### The type of distribution to use for the test statistics and state the level of significance.
Upper level algebra
### Self-Check A student scores 82,96,91, and 92 on four college algebra exams.What score is needed on a fifth exam for the student to earn an average grade of 90?
Upper level algebra
### To write a function f describing cost of sealable books. The printing and binding cost for a college algebra book is $$\10$$.The editorial cost is $$\200,000$$. The first 2500 books are free.
Upper level algebra
### Conditional probability if $$\displaystyle{40}\%$$ of the population have completed college, and $$\displaystyle{85}\%$$ of college graduates are registered to vote, what percent of the population areboth college graduates andregistered voters? To find: The percent of people who is both college graduates and registered voters.
Upper level algebra | 1,151 | 4,781 | {"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": 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} | 3.8125 | 4 | CC-MAIN-2021-21 | latest | en | 0.949442 |
http://christianherta.de/lehre/dataScience/bayesian/maximum-likelihood-learning.slides.php | 1,696,104,113,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00781.warc.gz | 7,687,675 | 137,962 | maximum-likelihood-learning slides
## Parameter Learning by Maximum Likelihood¶
Examples:
• $n$-iid Bernoulli trails - Binomial distribution
• Simple Bayesian Network with two binary random variables
#### Bayes Rule for Data and Model Parameters¶
• $\mathcal D$: Data
• $\mathcal \theta$: Parameter of the model
$$p(\theta | \mathcal D) = \frac{p(\mathcal D | \theta) p(\theta)}{ p(\mathcal D)}$$
• numerator:
• first term: "likelihood"
• second term: "prior"
• denominator: "marginal likelihood" (or "evidence") $$posterior = \frac{likelihood \cdot prior}{evidence}$$
#### Likelihood¶
Likelihood function is considered as a function of $\theta$:
$$\mathcal L (\theta) = p(\mathcal D | \theta)$$
The likelihood function is not a probability function!
"Never say 'the likelihood of the data'. Alway say 'the likelihood of the parameters'. The likelihood function is not a probability distribution." (from D. MacKay: Information Theory, Inference and Learning Algorithms, Page 29, Cambride 2003, http://www.inference.phy.cam.ac.uk/itila/book.html)
## Maximum Likelihood¶
$$\arg\max_\theta \mathcal L(\theta)$$
often the negativ log-likelihood is minimized:
$$\arg\max_\theta \mathcal L(\theta) = \arg\min_\theta \left(- \mathcal \log \left( \mathcal L(\theta) \right) \right)$$
Example: Binomial Distribution (e.g. tossing a thumtack)
$$\arg\min_\theta \left( - \mathcal \log L(\theta) \right)= \arg\min_\theta - \log \left( {n \choose k} \theta^k (1-\theta)^{n-k}\right)$$
necessary condition for a minimum:
$$0 = \frac{d}{d\theta} \left( \theta^k (1-\theta)^{n-k}\right) = k \theta^{k-1} (1-\theta)^{n-k} - (n-k) \theta^{k} (1-\theta)^{n-k-1}$$$$k(1-\theta) = (n-k) \theta$$$$\theta_{ML} = \frac{k}{n}$$
Example: Thumbtack using maximum likelihood estimation.
In [2]:
from IPython.display import Image
Image(filename='./thumbtack.jpg', width=200, height=200)
Out[2]:
In [3]:
# number of iid flips
n = 14
from scipy.stats import bernoulli
bernoulli.rvs(0.3,size=n)
Out[3]:
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0])
Sufficient statistics for $\mathcal D$ is $k$ (number of positive outcomes) and $n$ (number of total outcomes).
In [4]:
def maximum_likelihood_estimate(theta, n_total):
r = bernoulli.rvs(theta, size=n_total)
mle=[np.sum(r[:i])/float(i) for i in range(1, len(r)+1)]
return r, mle
theta = 0.3
r, mle = maximum_likelihood_estimate(theta, n_total=10)
print "\nMaximum likelihood estimate for the parameter theta up to the k-th trail:"
mle
Maximum likelihood estimate for the parameter theta up to the k-th trail:
Out[4]:
[0.0,
0.0,
0.0,
0.25,
0.40000000000000002,
0.33333333333333331,
0.42857142857142855,
0.375,
0.33333333333333331,
0.29999999999999999]
#### Law of large numbers¶
In [6]:
plot_estimates(theta)
### MLE for Bayesian Networks¶
Example Graph of two binary random variables $X$ and $Y$:
In [8]:
draw_XY(3.)
$\mathcal D$ consists of $n$ different observations:
• freq(X=0, Y=0) = $f$
• freq(X=0, Y=1) = $g$
• freq(X=1, Y=0) = $h$
• freq(X=1, Y=1) = $i$
with
• total number of observations: $n = f + g + h + i$
• number of $(X=1)$-observations: $k = h + i$
• number of $(X=0)$-observations: $l = f+g = n-k$
$$P(X, Y) = P(X) P(Y | X)$$
We have three free parameters for three Binomial distributions:
• $\theta_X$ : Probability of $X=1$
• $\theta_{0Y}$: Probability of $Y=1$ under the constraint $X=0$
• $\theta_{1Y}$: Probability of $Y=1$ under the constraint $X=1$
Joinly written as parameter vector $\vec \theta = (\theta_X, \theta_{0Y},\theta_{1Y})$ or as a set $\theta = \{\theta_X, \theta_{0Y},\theta_{1Y}\}$
The likelihood of parameter vector $\vec \theta = (\theta_X, \theta_{0Y},\theta_{1Y})$ given $\mathcal D$ is:
$$\mathcal L(\vec \theta) = p( \mathcal D| \vec \theta) = P(\mathcal D_X | \theta_X) P(\mathcal D_{Y|X} | \theta_{0Y}, \theta_{1Y})$$
Under omitting the constant (Binomial coeffients - given data):
$$\mathcal L(\vec \theta) \propto \left( \theta_X^{k} (1-\theta_X)^{n-k} \right) \left( \theta_{1Y}^{i} (1-\theta_{1Y})^{k-i} \right)^k \left( \theta_{0Y}^{g} (1-\theta_{0Y})^{l-g} \right)^{l}$$
The likelihood decomposes into a product of terms; that's is called Decomposability.
respectivly for the log-likelihood:
\begin{align} \log \mathcal L(\vec \theta) \propto & { } \left( k \log \theta_X + (n-k) \log (1-\theta_X) \right) + \\ & { } k \left( i \log \theta_{1Y} + (k-i) \log (1-\theta_{1Y}) \right) + \\ & { } l \left( g \log \theta_{0Y} + (l-g) \log (1-\theta_{0Y}) \right) \end{align}
So we have three independent terms \begin{align} \log \mathcal L(\vec \theta) \propto & { }\log \mathcal L(\theta_X) + \\ & { } \log \mathcal L(\theta_{1Y}) + \\ & { } \log \mathcal L(\theta_{0Y}) \end{align}
So searching for the argmax of $\log \mathcal L(\vec \theta)$ can be done by finding the argmax of each term independently. Also note that the leading $k$ and $l$ are just constants in the second resp. third term which can be neglected in the optimization.
Log-Likelihood of the first term:
$$\log \mathcal L (\theta_{X}) \propto k \log \theta_{X} + (n-k) \log (1-\theta_{X})$$
We already know how to compute (see above) the MLE (maximum likelihood estimator) of $\theta_{X}$:
$$\theta_{X}^{ML} = \frac{k}{n}$$
(Local) Log-Likelihood (second term):
$$\log \mathcal L(\theta_{1Y})\propto i \log \theta_{1Y} + (k-i) \log (1-\theta_{1Y})$$
MLE (maximum likelihood estimator):
$$\theta_{1Y}^{ML} = \frac{i}{k}$$
(Local) Log-Likelihood (third term):
$$\log \mathcal L(\theta_{0Y}) \propto g \log \theta_{0Y} + (l-g) \log (1-\theta_{0Y})$$
MLE (maximum likelihood estimator):
$$\theta_{0Y}^{ML} = \frac{g}{l}$$
So for (local) Binomial (and analog for Multinomials) the MLEs can be obtained just by ratios of frequencies.
#### Global decomposition:¶
• likelihood function decomposes as a product of independent terms for each CPD | 1,980 | 5,846 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2023-40 | latest | en | 0.515714 |
https://stacks.math.columbia.edu/tag/0EHP | 1,670,296,418,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00004.warc.gz | 569,308,333 | 6,552 | Lemma 30.23.7. Let $X$ be a Noetherian scheme and let $\mathcal{I} \subset \mathcal{O}_ X$ be a quasi-coherent sheaf of ideals. Any object of $\textit{Coh}(X, \mathcal{I})$ which is annihilated by a power of $\mathcal{I}$ is in the essential image of (30.23.3.1). Moreover, if $\mathcal{F}$, $\mathcal{G}$ are in $\textit{Coh}(\mathcal{O}_ X)$ and either $\mathcal{F}$ or $\mathcal{G}$ is annihilated by a power of $\mathcal{I}$, then the maps
$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, \mathcal{G}) \ar[d] & \mathop{\mathrm{Ext}}\nolimits _ X(\mathcal{F}, \mathcal{G}) \ar[d] \\ \mathop{\mathrm{Hom}}\nolimits _{\textit{Coh}(X, \mathcal{I})}(\mathcal{F}^\wedge , \mathcal{G}^\wedge ) & \mathop{\mathrm{Ext}}\nolimits _{\textit{Coh}(X, \mathcal{I})}(\mathcal{F}^\wedge , \mathcal{G}^\wedge ) }$
are isomorphisms.
Proof. Suppose $(\mathcal{F}_ n)$ is an object of $\textit{Coh}(X, \mathcal{I})$ which is annihilated by $\mathcal{I}^ c$ for some $c \geq 1$. Then $\mathcal{F}_ n \to \mathcal{F}_ c$ is an isomorphism for $n \geq c$. Hence if we set $\mathcal{F} = \mathcal{F}_ c$, then we see that $\mathcal{F}^\wedge \cong (\mathcal{F}_ n)$. This proves the first assertion.
Let $\mathcal{F}$, $\mathcal{G}$ be objects of $\textit{Coh}(\mathcal{O}_ X)$ such that either $\mathcal{F}$ or $\mathcal{G}$ is annihilated by $\mathcal{I}^ c$ for some $c \geq 1$. Then $\mathcal{H} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{G}, \mathcal{F})$ is a coherent $\mathcal{O}_ X$-module annihilated by $\mathcal{I}^ c$. Hence we see that
$\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{G}, \mathcal{F}) = H^0(X, \mathcal{H}) = \mathop{\mathrm{lim}}\nolimits H^0(X, \mathcal{H}/\mathcal{I}^ n\mathcal{H}) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Coh}(X, \mathcal{I})} (\mathcal{G}^\wedge , \mathcal{F}^\wedge ).$
see Lemma 30.23.5. This proves the statement on homomorphisms.
The notation $\mathop{\mathrm{Ext}}\nolimits$ refers to extensions as defined in Homology, Section 12.6. The injectivity of the map on $\mathop{\mathrm{Ext}}\nolimits$'s follows immediately from the bijectivity of the map on $\mathop{\mathrm{Hom}}\nolimits$'s. For surjectivity, assume $\mathcal{F}$ is annihilated by a power of $I$. Then part (1) of Lemma 30.23.6 shows that given an extension
$0 \to \mathcal{G}^\wedge \to (\mathcal{E}_ n) \to \mathcal{F}^\wedge \to 0$
in $\textit{Coh}(U, I\mathcal{O}_ U)$ the morphism $\mathcal{G}^\wedge \to (\mathcal{E}_ n)$ is isomorphic to $\mathcal{G} \to \mathcal{E}^\wedge$ for some $\mathcal{G} \to \mathcal{E}$ in $\textit{Coh}(\mathcal{O}_ U)$. Similarly in the other case. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 1,077 | 2,844 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 2, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.55818 |
http://www.flyinhighokc.com/instrumentflyinhandbook/pages/2-10.html | 1,618,819,985,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879305.68/warc/CC-MAIN-20210419080654-20210419110654-00327.warc.gz | 135,895,482 | 4,032 | ## Instrument Flying Handbook Aerodynamic Factors Climbs and Turns
Instrument Flying
Handbook
Preface
For example, a pilot is on an instrument approach at 1 .3 a speed near L/Dmax, and knows that a certain power setting maintains that speed. The airplane slows several knots below the desired speed because of a slight reduction in the power setting. The pilot increases the power slightly, and the airplane begins to accelerate, but at a slow rate. Because the airplane is still in the "flat part" of the drag curve, this slight increase in power will 1101 cause a rapid return to the desired speed. The pilot may need to increase the power higher than normally needed to maintain the new speed, allow the airplane to accelerate, then reduce the power to the setting that maintains the desired speed. Climbs The ability for an aircraft to climb depends upon an excess power or thrust over what it takes to maintain equilibrium. Excess power is the available power over and above that required to maintain horizontal flight at a given speed. Although the terms power and thrust are sometimes used interchangeably (erroneously implying they are synonymous), distinguishing between the two is important when considering climb performance. Work is the product of a force moving through a distance and is usually independent of time, Power implies work rate or units of work per unit of time, and as such is a function of the speed at which the force is developed. Thrust, also a function of work, means the force which imparts a change in the velocity of a mass. During take off, the aircraft does not stall even though it may be in a climb near the stall speed. The reason is that excess power (used to produce thrust) is used during this flight regime. Therefore, it is important if an engine fails after take off, to compensate the loss of thrust with pitch and airspeed. For a given weight of the aircraft, the angle of climb depends on the difference between thrust and drag, or the excess thrust. When the excess thrust is zero, the inclination of the flight path is zero, and the aircraft in steady, level flight. When thrust is greater than drag, the excess thrust allows a climb angle depending on the amount of excess thrust. When thrust is less than drag, the deficiency of thrust induces an angle of descent. Acceleration in Cruise Flight Aircraft accelerate in level flight because of an excess of power over what is required to maintain a steady speed This is the same excess power used to climb. Upon reaching the desired altitude with pitch being lowered to maintain that altitude, the excess power now accelerates the aircraft to its cruise speed. However, reducing power too soon after level off results in a longer period of time to accelerate. Turns Like any moving object, an aircraft requires a sideward force to make it turn. In a normal turn, this force is supplied by banking the aircraft in order to exert lift inward, as well as upward. The force of lift is separated into two components at right angles to each other. (Figure 2-13) The upward acting lift together with the opposing weight becomes the vertical lift component. The horizontally acting lift and its opposing centrifugal force are the horizontal lift component, or centripetal force. This horizontal lift component is the sideward force that causes an aircraft to turn. The equal and opposite reaction to this sideward force is centrifugal force, which is merely an apparent force as a result of inertia. Figure 2-13. Forces In Turn. The relationship between the aircraft's speed and bank angle to the rate and radius of turns is important for instrument pilots to understand. The pilot can use this knowledge to properly estimate bank angles needed for certain rates of turn, or to determine how much to lead when intercepting a course. Rate of Turn The rate of turn, normally measured in degrees per second, is based upon a set bank angle at a set speed. If either one of these elements changes, the rate of turn changes. If the aircraft increases its speed without changing the bank angle, the rate of turn decreases. Likewise, if the speed decreases without changing the bank angle, the rate of turn increases. Changing the bank angle without changing speed also causes the rote of turn to change. Increasing the hank angle without changing speed increases the rate of turn, while decreasing the hank angle reduces the rate of turn.
2-10 | 884 | 4,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-17 | latest | en | 0.943015 |
https://algebraworksheets.co/algebra-2-graphing-linear-inequalities-worksheet-2/ | 1,656,653,292,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00363.warc.gz | 142,157,309 | 14,388 | # Algebra 2 Graphing Linear Inequalities Worksheet
Algebra 2 Graphing Linear Inequalities Worksheet – Algebra Worksheets are designed to help students understand math. The subject of study is the study of mathematical symbols and the rules used to manipulate them. It is the central thread of virtually all math, including geometry and physics. It is a vital element of your education. This section will teach you everything you need to know about algebra, while focusing on the most significant topics. These worksheets can help increase your proficiency as an undergraduate.
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Free algebra worksheets are a great way to improve your math abilities. These worksheets are free and can be used to master the basics of math. These worksheets can assist you in understanding the basics of algebra and how to use them in your everyday routine. If you’re in school, think about them as an important resource in your education. They can help make you more effective and successful. They’ll love these printables!
There are also worksheets for free accessible to help to improve your math abilities. For instance, if you’re unfamiliar with algebra, you might be confused about how to use graphs. Look for worksheets that are free and then find those that will teach you. It’s simple to download and browse through these worksheets to get started in your math classes today. There are plenty of options online. You can choose a math lesson on math that your student in high school. | 527 | 2,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.933754 |
https://www.brainscape.com/flashcards/6-describing-relationships-correlation-10812116/packs/19263154 | 1,695,614,676,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506676.95/warc/CC-MAIN-20230925015430-20230925045430-00321.warc.gz | 753,708,282 | 37,137 | # 6. Describing Relationships: Correlation Flashcards Preview
## Stats > 6. Describing Relationships: Correlation > Flashcards
Flashcards in 6. Describing Relationships: Correlation Deck (25)
1
Q
Positive Relationship
A
Occurs insofar as pairs of scores tend to occupy similar relative positions (high with high and low with low) in their respective distributions.
2
Q
Negative Relationship
A
Occurs insofar as pairs of scores tend to occupy dissimilar relative positions (high with low and vice versa) in their respective distributions.
3
Q
Indicate whether the following statement suggests a positive or negative relationship:
More densely populated areas have higher crime rates.
A
Positive. The crime rate is higher, square mile by square mile, in densely populated cities than in sparsely populated rural areas.
4
Q
Indicate whether the following statement suggests a positive or negative relationship:
Schoolchildren who often watch TV perform more poorly on academic achievement tests.
A
Negative. As TV viewing increases, performance on academic achievement tests tends to decline.
5
Q
Indicate whether the following statement suggests a positive or negative relationship:
Heavier automobiles yield poorer gas mileage
A
Negative. Increases in car weight are accompanied by decreases in miles per gallon.
6
Q
Indicate whether the following statement suggests a positive or negative relationship:
Better-educated people have higher incomes.
A
Positive. Increases in educational level - grade school, high school, college - tend to be associated with increases in income.
7
Q
Indicate whether the following statement suggests a positive or negative relationship:
More anxious people voluntarily spend more time performing a simple repetitive task.
A
Positive. Highly anxious people willingly spend more time performing a simple repetitive task than do less anxious people.
8
Q
Scatterplot
A
A graph containing a cluster of dots that represents all pairs of scores.
9
Q
Linear Relationship
A
A relationship that can be described best with a straight line.
10
Q
Curvilinear Relationship
A
A relationship that can be described best with a curved line.
11
Q
Correlation Coefficient
A
A number between -1 and 1 that describes the relationship between pairs of variables.
12
Q
Pearson Correlation Coefficient (r)
A
A number between -1.00 and +1.00 that describes the linear relationship between pairs of quantitative variables.
13
Q
Supply a verbal description for the following correlations.
an r of –.84 between total mileage and automobile resale value
A
Cars with more miles tend to have lower resale values.
14
Q
Supply a verbal description for the following correlations.
an r of –.35 between the number of days absent from school and performance on a math achievement test
A
Students with more absences from school tend to score lower on math achievement tests.
15
Q
Supply a verbal description for the following correlations.
an r of .03 between anxiety level and college GPA
A
Little or no relationship between anxiety level and college GPA.
16
Q
Supply a verbal description for the following correlations.
an r of .56 between age of schoolchildren and reading comprehension
A
Older schoolchildren tend to have better reading comprehension.
17
Q
Speculate on whether the following correlation reflects simple cause-effect relationships or more complex states of affairs.
caloric intake and body weight
A
Simple causal-effect
Hint: A cause-effect relationship implies that, if all else remains the same, any change in the causal variable should always produce a predictable change in the other variable.
18
Q
Speculate on whether the following correlation reflects simple cause-effect relationships or more complex states of affairs.
height and weight
A
complex
Hint: A cause-effect relationship implies that, if all else remains the same, any change in the causal variable should always produce a predictable change in the other variable.
19
Q
Speculate on whether the following correlation reflects simple cause-effect relationships or more complex states of affairs.
SAT math score and score on a calculus test
A
complex
Hint: A cause-effect relationship implies that, if all else remains the same, any change in the causal variable should always produce a predictable change in the other variable.
20
Q
Speculate on whether the following correlation reflects simple cause-effect relationships or more complex states of affairs.
poverty and crime
A
complex
Hint: A cause-effect relationship implies that, if all else remains the same, any change in the causal variable should always produce a predictable change in the other variable.
21
Q
Correlation Coefficient (Computation Formula)
A
r = (SPxy) / √(SSx SSy)
Where
SSx = ∑X² – [(∑X)²) / n]
SSy = ∑Y² – [(∑Y)²) / n]
SPxy = ∑XY – [(∑X)(∑Y) / n]
22
Q
Sum of Products (Definition and Computation Formulas)
A
SPxy = ∑(X – ̅X) (Y – ̅Y)
SPxy = ∑XY – [(∑X)(∑Y) / n]
23
Q
Couples who attend a clinic for first pregnancies are asked to estimate (independently of each other) the ideal number of children. Given that X and Y represent the estimates of females and males, respectively, the results are as follows.
```Couple X Y
A 1 2
B 3 4
C 2 3
D 3 2
E 1 0
F 2 3```
Calculate a value for r, using the computation formula.
A
SSx = ∑X² – [(∑X)²) / n]
=> 28 – 144/6 = 4
SSy = ∑Y² – [(∑Y)²) / n]
=> 42 – 196/6 = 9.33
SPxy = ∑XY – [(∑X)(∑Y) / n]
=> 32 – 168/6 = 4
r = (SPxy) / √(SSx SSy)
=> 4 / (2√9.33) = 0.65
24
Q
Give the 11 steps of the computational sequence for the calculation of r
A
1. Assign a value to n, representing the number of pairs of scores.
2. Sum all the scores for X.
3. Sum all the scores for Y.
4. Find the product of each pair of X and Y scores, one at a time.
5. Then add all of these products.
6. Square each X score, one at a time.
7. Then add all squared X scores.
8. Square each Y score, one at a time.
9. Then add all squared Y scores.
10. Substitute numbers into formulas and solve for SPxy, SSx, and SSy.
11. Substitute into formula and solve for r
25
Q
Correlation Matrix
A
Table showing correlations for all possible pairs of variables. | 1,495 | 6,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-40 | latest | en | 0.896248 |
https://www.jiskha.com/questions/1777009/lesson-7-coordinate-plane-essential-algebra-readiness-pre-algebra-a-unit-4-real-numbers | 1,643,371,426,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00349.warc.gz | 852,798,783 | 7,840 | # (Pre-Algebra) A
Lesson 7: Coordinate Plane
Essential Algebra Readiness(Pre-Algebra) A Unit 4: Real Numbers and the Coordinate Plane
Has anyone done the Coordinate Plane Practice AND assessment?
1. Which point is located at (-3, -2)? (1 point)
•C
•D
•F
•G
2. Which point is located at (0, -4)? (1 point)
•B
•E
•D
•H
3. What are the coordinates of A? (1 point)
•(3, 4)
•(0, 3)
•(-1, 0)
•(3, -2)
4. What is the distance between points A and E? (1 point)
•2.8
•3.2
•4.9
•5.7
5. What is the distance between points A and E? (1 point)
•5.4
•4.2
•3.7
•2.6
1. 👍
2. 👎
3. 👁
1. and we're supposed to check this how?
1. 👍
2. 👎
2. Steve - eh? is correct how r we suppose to check this it dont make seance
1. 👍
2. 👎
3. HOW
1. 👍
2. 👎
4. SO MANY FOXY FNAF'S AAAAAAAAHHHHH
1. 👍
2. 👎
5. WHHHHHHHHHHHYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
1. 👍
2. 👎
6. hello mateys im foxy the pirate fox and why have you stolen my name lads?
you never steal a pirates name only i can do that lads.
ps i'm sorry for this cringe
1. 👍
2. 👎
1. C: F
2. D: H
3. A: 3,4
4. D: 5.7
5: A: 5.4
I hope this helps :)
1. 👍
2. 👎
8. angel tysm~
1. 👍
2. 👎
9. People come on how are we really supposed to check your answers.
1. 👍
2. 👎
10. Real Numbers and the Coordinate Plane Unit Test Part 2
I hate math
1. 👍
2. 👎
11. Lesson 2: Proportions
Algebra Readiness (Pre-Algebra) A Unit 5: Applications of Proportions if your from Connections Academy I need help.
1. 👍
2. 👎
12. None of this helps for my math quick check I will try it and post my answers, See if this helps you in the next post.
1. 👍
2. 👎
13. Algebra 1B Radical expressions & data analysis... go here for the homework help if you need the answers... peeps
1. 👍
2. 👎
14. anyone have the answers t the quick check too?
1. 👍
2. 👎
15. anyone have the practice questions for lesson 10
1. 👍
2. 👎
1. 👍
2. 👎
17. to the quick check
1. 👍
2. 👎
18. what number is a perfect cube ?
a.) 7
b.) 25
c.) 66
d.) 27
1. 👍
2. 👎
19. Anyone have the test answers?
1. 👍
2. 👎
20. 0_0_0_0
1. 👍
2. 👎
21. Angel is only right for 2 of the questions -_-
REALLY!!!!
COME ON!
1. 👍
2. 👎
22. The correct answers for October 2021 are-
1- F
2- H
3- 3,4
4- 5.7
5- 5.4
1. 👍
2. 👎
23. Tysm @marble
1. 👍
2. 👎
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does anyone know the answers to Lesson 1: Linear Algebra B Semester Review Algebra 1.5 B Unit 5: Semester Review and Exam on connexus | 1,533 | 4,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.728287 |
https://math.libretexts.org/Courses/Mission_College/Math_3A%3A_Calculus_1_(Sklar)/04%3A_Applications_of_Derivatives/4.09%3A_Newtons_Method/4.9E%3A_Exercises_for_Section_4.9 | 1,718,273,705,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00516.warc.gz | 352,312,289 | 30,484 | # 4.9E: Exercises for Section 4.9
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In exercises 1 - 5, write Newton’s formula as $$x_{n+1}=F(x_n)$$ for solving $$f(x)=0$$.
1) $$f(x)=x^2+1$$
2) $$f(x)=x^3+2x+1$$
$$F(x_n)=x_n−\dfrac{x_n^3+2x_n+1}{3x_n^2+2}$$
3) $$f(x)=\sin x$$
4) $$f(x)=e^x$$
$$F(x_n)=x_n−\dfrac{e^{x_n}}{e^{x_n}}$$
5) $$f(x)=x^3+3xe^x$$
In exercises 6 - 8, solve $$f(x)=0$$ using the iteration $$x_{n+1}=x_{n−c}f(x_n)$$, which differs slightly from Newton’s method. Find a $$c$$ that works and a $$c$$ that fails to converge, with the exception of $$c=0.$$
6) $$f(x)=x^2−4,$$ with $$x_0=0$$
$$|c|>0.5$$ fails, $$|c|≤0.5$$ works
7) $$f(x)=x^2−4x+3,$$ with $$x_0=2$$
8) What is the value of $$“c”$$ for Newton’s method?
$$c=\dfrac{1}{f′(x_n)}$$
In exercises 9 - 16, compute $$x_1$$ and $$x_2$$ using the specified iterative method.
Start at
a. $$x_0=0.6$$ and
b. $$x_0=2.$$
9) $$x_{n+1}=x_n^2−\frac{1}{2}$$
10) $$x_{n+1}=2x_n\left(1−x_n\right)$$
a. $$x_1=\frac{12}{25}, \; x_2=\frac{312}{625};$$
b. $$x_1=−4, \; x_2=−40$$
11) $$x_{n+1}=\sqrt{x_n}$$
12) $$x_{n+1}=\frac{1}{\sqrt{x_n}}$$
a. $$x_1=1.291, \; x_2=0.8801;$$
b. $$x_1=0.7071, \; x_2=1.189$$
13) $$x_{n+1}=3x_n(1−x_n)$$
14) $$x_{n+1}=x_n^2+x_{n−2}$$
a. $$x_1=−\frac{26}{25}, \; x_2=−\frac{1224}{625};$$
b. $$x_1=4, \;x_2=18$$
15) $$x_{n+1}=\frac{1}{2}x_n−1$$
16) $$x_{n+1}=|x_n|$$
a. $$x_1=\frac{6}{10},\; x_2=\frac{6}{10};$$
b. $$x_1=2, \; x_2=2$$
In exercises 17 - 26, solve to four decimal places using Newton’s method and a computer or calculator. Choose any initial guess $$x_0$$ that is not the exact root.
17) $$x^2−10=0$$
18) $$x^4−100=0$$
$$3.1623$$ or $$−3.1623$$
19) $$x^2−x=0$$
20) $$x^3−x=0$$
$$0,$$ $$−1$$ or $$1$$
21) $$x+5\cos x=0$$
22) $$x+\tan x =0,$$ choose $$x_0∈\left(−\frac{π}{2},\frac{π}{2}\right)$$
$$0$$
23) $$\dfrac{1}{1−x}=2$$
24) $$1+x+x^2+x^3+x^4=2$$
$$0.5188$$ or $$−1.2906$$
25) $$x^3+(x+1)^3=10^3$$
26) $$x=\sin^2(x)$$
$$0$$
In exercises 27 - 30, use Newton’s method to find the fixed points of the function where $$f(x)=x$$; round to three decimals.
27) $$\sin x$$
28) $$\tan x$$ on $$x=\left(\frac{π}{2},\frac{3π}{2}\right)$$
$$4.493$$
29) $$e^x−2$$
30) $$\ln(x)+2$$
$$0.159,\; 3.146$$
Newton’s method can be used to find maxima and minima of functions in addition to the roots. In this case apply Newton’s method to the derivative function $$f′(x)$$ to find its roots, instead of the original function. In exercises 31 - 32, consider the formulation of the method.
31) To find candidates for maxima and minima, we need to find the critical points $$f′(x)=0.$$ Show that to solve for the critical points of a function $$f(x)$$, Newton’s method is given by $$x_{n+1}=x_n−\dfrac{f′(x_n)}{f''(x_n)}$$.
32) What additional restrictions are necessary on the function $$f$$?
We need $$f$$ to be twice continuously differentiable.
In exercises 33 - 40, use Newton’s method to find the location of the local minima and/or maxima of the following functions; round to three decimals.
33) Minimum of $$f(x)=x^2+2x+4$$
34) Minimum of $$f(x)=3x^3+2x^2−16$$
$$x=0$$
35) Minimum of $$f(x)=x^2e^x$$
36) Maximum of $$f(x)=x+\dfrac{1}{x}$$
$$x=−1$$
37) Maximum of $$f(x)=x^3+10x^2+15x−2$$
38) Maximum of $$f(x)=\dfrac{\sqrt{x}−\sqrt[3]{x}}{x}$$
$$x=5.619$$
39) Minimum of $$f(x)=x^2\sin x,$$ closest non-zero minimum to $$x=0$$
40) Minimum of $$f(x)=x^4+x^3+3x^2+12x+6$$
$$x=−1.326$$
In exercises 41 - 44, use the specified method to solve the equation. If it does not work, explain why it does not work.
41) Newton’s method, $$x^2+2=0$$
42) Newton’s method, $$0=e^x$$
There is no solution to the equation.
43) Newton’s method, $$0=1+x^2$$ starting at $$x_0=0$$
44) Solving $$x_{n+1}=−x_n^3$$ starting at $$x_0=−1$$
It enters a cycle.
In exercises 45 - 48, use the secant method, an alternative iterative method to Newton’s method. The formula is given by
$$x_n=x_{n−1}−f(x_{n−1})\dfrac{x_{n−1}−x_{n−2}}{f(x_{n−1})−f(x_{n−2})}.$$
45) a root to $$0=x^2−x−3$$ accurate to three decimal places.
46) Find a root to $$0=\sin x+3x$$ accurate to four decimal places.
$$0$$
47) Find a root to $$0=e^x−2$$ accurate to four decimal places.
48) Find a root to $$\ln(x+2)=\dfrac{1}{2}$$ accurate to four decimal places.
$$−0.3513$$
49) Why would you use the secant method over Newton’s method? What are the necessary restrictions on $$f$$?
In exercises 50 - 54, use both Newton’s method and the secant method to calculate a root for the following equations. Use a calculator or computer to calculate how many iterations of each are needed to reach within three decimal places of the exact answer. For the secant method, use the first guess from Newton’s method.
50) $$f(x)=x^2+2x+1,\quad x_0=1$$
Newton: $$11$$ iterations, secant: $$16$$ iterations
51) $$f(x)=x^2, \quad x_0=1$$
52) $$f(x)=\sin x, \quad x_0=1$$
Newton: three iterations, secant: six iterations
53) $$f(x)=e^x−1, \quad x_0=2$$
54) $$f(x)=x^3+2x+4, \quad x_0=0$$
Newton: five iterations, secant: eight iterations
In exercises 55 - 56, consider Kepler’s equation regarding planetary orbits, $$M=E−ε\sin(E)$$, where $$M$$ is the mean anomaly, $$E$$ is eccentric anomaly, and $$ε$$ measures eccentricity.
55) Use Newton’s method to solve for the eccentric anomaly $$E$$ when the mean anomaly $$M=\frac{π}{3}$$ and the eccentricity of the orbit $$ε=0.25;$$ round to three decimals.
56) Use Newton’s method to solve for the eccentric anomaly $$E$$ when the mean anomaly $$M=\frac{3π}{2}$$ and the eccentricity of the orbit $$ε=0.8;$$ round to three decimals.
$$E=4.071$$
In exercises 57 - 58, consider a bank investment. The initial investment is $$10,000$$. After $$25$$ years, the investment has tripled to $$30,000.$$
57) Use Newton’s method to determine the interest rate if the interest was compounded annually.
58) Use Newton’s method to determine the interest rate if the interest was compounded continuously.
$$4.394%$$
59) The cost for printing a book can be given by the equation $$C(x)=1000+12x+\frac{1}{2}x^{2/3}$$. Use Newton’s method to find the break-even point if the printer sells each book for $$20.$$ | 3,022 | 7,762 | {"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} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.252203 |
https://www.effortlessmath.com/math-topics/everything-you-need-to-know-about-limits/ | 1,722,804,332,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00446.warc.gz | 599,285,922 | 12,850 | # Everything you Need to know about Limits
In Calculus, Limits describe the value a function approaches as its input gets arbitrarily close to a specific point.
Here’s all you need to know about Limits:
## Definition:
In simple terms, the limit of a function at a certain point refers to the value that the function approaches as its independent variable approaches that point. It gives us the ability to describe the behavior of functions at points where they might not be defined, or near points of interest.
Mathematically, for a function $$f(x)$$, if for every number $$\epsilon > 0$$ there exists a number $$\delta > 0$$ such that if $$0 < |x-a| < \delta$$ then $$|f(x)-L| < \epsilon$$, then the limit of $$f(x)$$ as $$x$$ approaches $$a$$ is $$L$$.
This is denoted as:
$\lim_{x \to a} f(x) = L$
Here:
• $$L$$ is the limit value.
• $$a$$ is the point of interest.
• $$\epsilon$$ is any arbitrarily small positive number.
• $$\delta$$ is the distance from $$a$$ such that if $$x$$ is within this distance, $$f(x)$$ will be within $$\epsilon$$ of $$L$$.
## One-Sided Limits:
• Right-hand Limit: The limit as $$x$$ approaches $$a$$ from the right (values larger than $$a$$).
$$\lim_{x \to a^+} f(x)$$
• Left-hand Limit: The limit as $$x$$ approaches $$a$$ from the left (values smaller than $$a$$).
$$\lim_{x \to a^-} f(x)$$
For the overall limit $$\lim_{x \to a} f(x)$$ to exist, both the left-hand and right-hand limits must exist and be equal.
Common Techniques:
1. Direct Substitution: Plug the value into the function if it’s defined.
2. Factorization: Useful when there’s a common factor in the numerator and denominator causing an indeterminate form.
3. Rationalization: Multiplying the numerator and denominator by a conjugate, typically used for root-based functions.
4. Recognizing Common Limits: Such as $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$.
5. Using Squeeze (Sandwich) Theorem: When you can ‘trap’ a function between two others that approach the same limit.
## Properties:
1. Linearity: $$\lim_{x \to a} [c \cdot f(x) + g(x)] = c \cdot \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$
2. Product Rule: $$\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$$
3. Quotient Rule: If $$\lim_{x \to a} g(x) \neq 0$$, then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$$
4. Limits at Infinity: Describes the behavior of a function as $$x$$ approaches positive or negative infinity.
## Indeterminate Forms:
Sometimes, direct substitution leads to forms that don’t directly give a value but rather a form like $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$. Such forms are called “indeterminate”, and they necessitate further manipulation to determine the actual limit.
## Applications:
• Determining the behavior of functions at specific points or as $$x$$ goes to infinity.
• Fundamental in defining the derivative (rate of change) and the integral (area under a curve).
To gain a deep understanding of limits, it’s crucial to work through numerous examples, visualize functions graphically, and use the above techniques as needed.
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## ◆ zgerq2()
subroutine zgerq2 ( integer M, integer N, complex*16, dimension( lda, * ) A, integer LDA, complex*16, dimension( * ) TAU, complex*16, dimension( * ) WORK, integer INFO )
ZGERQ2 computes the RQ factorization of a general rectangular matrix using an unblocked algorithm.
Download ZGERQ2 + dependencies [TGZ] [ZIP] [TXT]
Purpose:
``` ZGERQ2 computes an RQ factorization of a complex m by n matrix A:
A = R * Q.```
Parameters
[in] M ``` M is INTEGER The number of rows of the matrix A. M >= 0.``` [in] N ``` N is INTEGER The number of columns of the matrix A. N >= 0.``` [in,out] A ``` A is COMPLEX*16 array, dimension (LDA,N) On entry, the m by n matrix A. On exit, if m <= n, the upper triangle of the subarray A(1:m,n-m+1:n) contains the m by m upper triangular matrix R; if m >= n, the elements on and above the (m-n)-th subdiagonal contain the m by n upper trapezoidal matrix R; the remaining elements, with the array TAU, represent the unitary matrix Q as a product of elementary reflectors (see Further Details).``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,M).``` [out] TAU ``` TAU is COMPLEX*16 array, dimension (min(M,N)) The scalar factors of the elementary reflectors (see Further Details).``` [out] WORK ` WORK is COMPLEX*16 array, dimension (M)` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value```
Further Details:
``` The matrix Q is represented as a product of elementary reflectors
Q = H(1)**H H(2)**H . . . H(k)**H, where k = min(m,n).
Each H(i) has the form
H(i) = I - tau * v * v**H
where tau is a complex scalar, and v is a complex vector with
v(n-k+i+1:n) = 0 and v(n-k+i) = 1; conjg(v(1:n-k+i-1)) is stored on
exit in A(m-k+i,1:n-k+i-1), and tau in TAU(i).```
Definition at line 122 of file zgerq2.f.
123*
124* -- LAPACK computational routine --
125* -- LAPACK is a software package provided by Univ. of Tennessee, --
126* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
127*
128* .. Scalar Arguments ..
129 INTEGER INFO, LDA, M, N
130* ..
131* .. Array Arguments ..
132 COMPLEX*16 A( LDA, * ), TAU( * ), WORK( * )
133* ..
134*
135* =====================================================================
136*
137* .. Parameters ..
138 COMPLEX*16 ONE
139 parameter( one = ( 1.0d+0, 0.0d+0 ) )
140* ..
141* .. Local Scalars ..
142 INTEGER I, K
143 COMPLEX*16 ALPHA
144* ..
145* .. External Subroutines ..
146 EXTERNAL xerbla, zlacgv, zlarf, zlarfg
147* ..
148* .. Intrinsic Functions ..
149 INTRINSIC max, min
150* ..
151* .. Executable Statements ..
152*
153* Test the input arguments
154*
155 info = 0
156 IF( m.LT.0 ) THEN
157 info = -1
158 ELSE IF( n.LT.0 ) THEN
159 info = -2
160 ELSE IF( lda.LT.max( 1, m ) ) THEN
161 info = -4
162 END IF
163 IF( info.NE.0 ) THEN
164 CALL xerbla( 'ZGERQ2', -info )
165 RETURN
166 END IF
167*
168 k = min( m, n )
169*
170 DO 10 i = k, 1, -1
171*
172* Generate elementary reflector H(i) to annihilate
173* A(m-k+i,1:n-k+i-1)
174*
175 CALL zlacgv( n-k+i, a( m-k+i, 1 ), lda )
176 alpha = a( m-k+i, n-k+i )
177 CALL zlarfg( n-k+i, alpha, a( m-k+i, 1 ), lda, tau( i ) )
178*
179* Apply H(i) to A(1:m-k+i-1,1:n-k+i) from the right
180*
181 a( m-k+i, n-k+i ) = one
182 CALL zlarf( 'Right', m-k+i-1, n-k+i, a( m-k+i, 1 ), lda,
183 \$ tau( i ), a, lda, work )
184 a( m-k+i, n-k+i ) = alpha
185 CALL zlacgv( n-k+i-1, a( m-k+i, 1 ), lda )
186 10 CONTINUE
187 RETURN
188*
189* End of ZGERQ2
190*
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:60
subroutine zlacgv(N, X, INCX)
ZLACGV conjugates a complex vector.
Definition: zlacgv.f:74
subroutine zlarf(SIDE, M, N, V, INCV, TAU, C, LDC, WORK)
ZLARF applies an elementary reflector to a general rectangular matrix.
Definition: zlarf.f:128
subroutine zlarfg(N, ALPHA, X, INCX, TAU)
ZLARFG generates an elementary reflector (Householder matrix).
Definition: zlarfg.f:106
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# Problem in Generalized Fox's H Function (YFoxH)
Posted 10 years ago
Would any one help me to solve the problem in the code attached below, it is about implementing Generalized Fox's H function in Mathematica, I got this code from a paper which also attached. I have no idea how to use the function. I have tried to reproduce Fig.6 , with L = 1, as shown in the paper, but fail. Would you please give me any idea to solve it? My problem will be solved as long as I can reproduce the same graphic as shown in Fig.6. Thanks in advanced Attachments:
2 Replies
Sort By:
Posted 10 years ago
Hi David Reiss, thanks for your comment. I think the code is intentionally made for general case, for any size of vector A and B. I found his other paper which also has the similar mathematica code as well as the explanation of the code. I attach the paper below (on appendix B). It helps me to understand the structure of the input of the function.Anyway, I apply YFoxH for a special case, as given on equation (17) of the previous paper (outage capacity multicarrier.pdf), with L = 1 ; m1,m2,...ml = 1; Cth = 1. Hence, I modify the code as attach below. Now the error is not because of the Do loop, but still I do not get the final result :( Attachments:
Posted 10 years ago
I have not gone through the entire code (though you have a typo before the final Return in that there is not a semicolon after the integration before it--Returns in Mathematica are not necessary, but the original person who coded this was thinking as though he was programming in Fortran or Basic or C).In going through the (typo-fixed) code line by line I find that the errors are generated in the final Do loop in the MT function execution. It is trying to take parts of AR that are too deep. Perhaps you have not specified your parameter definitions in your example case correctly? Beyond this, it'd require reading the paper in detail to understand things... | 474 | 2,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.609375 | 3 | CC-MAIN-2024-18 | latest | en | 0.959247 |
https://www.jiskha.com/display.cgi?id=1353327126 | 1,511,061,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805265.10/warc/CC-MAIN-20171119023719-20171119043719-00660.warc.gz | 818,596,108 | 3,733 | # physics
posted by .
A solid has weight of 400g in air and when partly immersed in a liquid , has a weight of 320g. if it has a relative density of 0.80, fine the volume of the solid
• physics -
Relative Density = Specific gravity=0.8
Ds=0.80 * Dw = 0.80*1g/cm^3=0.80g/cm^3 =
Density of the solid.
Vs = 400g / 0.8g/cm^3 = 500 cm^3. = Vol.
of solid.
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# Math Formula Sheet
by | May 6, 2019 | General | 0 comments
Mathematics skills are crucial to the success of a child whether at school, home and in daily life application. Math problem-solving skills also help build confidence and enable a wide range of career opportunities. Additionally, it’s from learning math that you can attain knowledge and skills with regards to investing, budgeting, taxes, loans, business financing and more.
To ace such math skills, it’s essential to study and practice. Remember the more you practice a concept, the better you become at it. Think about how a child learns to ride a bike or swim. Or, how a person learns how to drive a car. The more they practice, the better they become. It’s no different in math. It’s a skill that gets seasoned the more a student practice it. Studying and using a math formula sheet goes a long way toward helping a child ‘get’ it.
## The Importance of Math Formula Sheets
Just the same way a farmer needs farming tools, math requires that you have some tools too. In the Math language, we call them formulas. They comprise methods used when doing math computation and guide students on how to solve particular math problems.
For example, if you are asked to calculate the area of a circle, how will you go about it? Of course, you will use (A = πr2) formula to do it. Math formula sheets have formulations relating to geometry, algebra, area, perimeter, trigonometry, volume, units, and measurements, etc. Sometimes, formulas are provided to students taking math tests so that they can focus on application, rather than memorization.
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## Math Monkey – Math Enrichment Center
Based in Singapore, Math Monkey is a high-rated math enrichment center. We have math programs for kids of all ages and focus on teaching math without following traditional teaching methodologies that make kids hate and develop a ‘can’t do attitude’ for math. We use Vedic Math and fun games to help kid’s ace math. If your child somehow says math is hard and believes that they can’t understand it, enrolling her at our center will eventually change that. You will notice a positive attitude towards math after you register your child with us. Soon after is that their math scores will skyrocket to even topping their class. | 615 | 2,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.932752 |
https://www.marbk-86a.win/wiki/List_of_political_parties_in_Gambia | 1,627,464,693,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00601.warc.gz | 918,060,353 | 17,914 | # Talk:Albireo
WikiProject Astronomy / Astronomical objects (Rated C-class, Mid-importance)
.mw-parser-output .portal{border:solid #aaa 1px;padding:0}.mw-parser-output .portal.tleft{margin:0.5em 1em 0.5em 0}.mw-parser-output .portal.tright{margin:0.5em 0 0.5em 1em}.mw-parser-output .portal>ul{display:table;box-sizing:border-box;padding:0.1em;max-width:175px;background:#f9f9f9;font-size:85%;line-height:110%;font-style:italic;font-weight:bold}.mw-parser-output .portal>ul>li{display:table-row}.mw-parser-output .portal>ul>li>span:first-child{display:table-cell;padding:0.2em;vertical-align:middle;text-align:center}.mw-parser-output .portal>ul>li>span:last-child{display:table-cell;padding:0.2em 0.2em 0.2em 0.3em;vertical-align:middle}This article is within the scope of WikiProject Astronomy, which collaborates on articles related to Astronomy on Wikipedia. C This article has been rated as C-Class on the project's quality scale. Mid This article has been rated as Mid-importance on the project's importance scale. This article is supported by WikiProject Astronomical objects, which collaborates on articles related to astronomical objects.
## Period
I've taken out the phrase about the distance and orbital period of the binary system, since I've found pages saying that the distance between pair stars is 4400 AU with orbital period of 7300 earth year, such as [1]. --Puzzlet Chung 06:58, 20 Mar 2005 (UTC)
Something is not right with these numbers. Kepler's third law can be expressed as ${\displaystyle a^{3}=(m_{1}+m_{2})P^{2}}$. If we give a total mass of, say, 10 Suns, orbital period of the stars would be still about 92000 years. But in the original article, distance was 44 AU, which is wrong. In that case orbital period would be much less than Pluto's. This article [2] gives possible orbital period of 7800 years. --Jyril 09:04, Mar 20, 2005 (UTC)
## Triple
I replaced the section on Albireo as a triple star -- I was some years later and for several years a member of the team which had made the first high-precision measurements in 1976, and made some of the circa 1990 observations myself using the Kitt Peak 4m telescope. The catalogue of interferometric observations by all groups of this object (and all other binary stars) may be found in the Fourth Catalog of Interferometric Measurements of Binary Stars, available at the US Naval Observatory, http://ad.usno.navy.mil/wds/int4.html. -- Don Barry, Cornell University (24.92.255.115 on 26 September 2006)
## Orbital elements
This page contains nonsense. There is absolutely no source for the orbital elements (eccentricity, longitude of node, orbital period, separation, etc., etc.). The Washington Double Star Catalog quoted as the source lists all components of the albireo system; it lists no orbital elements. the sixth orbital catalog lists elements only for the spectroscopic Ab,Ac system. At a period of 220,000 years there is no way humanity has in two centuries amassed enough positional change data to compute orbital elements. All these numbers have been made up whole cloth. Macevoy (talk) —Preceding undated comment added 21:35, 24 April 2012 (UTC).
The problem is that the infobox is displaying orbital data for Albireo C, not Albireo B. Is there any way to specify this in the infobox? Should the orbital section be removed from the infobox since it gives the impression that the data is for Albireo B? UPDATE: Under "orbit_footnote" I have added, " for Albireo C" to hopefully make the infobox clearer and less misleading. -- Kheider (talk) 19:15, 17 September 2012 (UTC)
The pair visible in a small telescope is A/B. The pair of stars in the first infobox is Aa/Ac, which is also the pair for which the orbit is given. B is in the second infobox, lower down. Spacepotato (talk) 03:43, 18 September 2012 (UTC)
I disagree. As far as I know, the apparent magnitude of 5.8 is Albireo B (the distant blue-green start) not C (the close star). I have corrected this in the infobox.
Obviously, the infobox is not optimal at present, with parts of it referring to the larger binary system, and others to the smaller. Whoever reads this, please feel encouraged to improve ... Tomeasy T C 23:00, 7 November 2014 (UTC)
The first infobox refers to Aa/Ac, the close spectroscopic binary, and the second to B, the more distant companion. See the Washington Double Star Catalog [3], entry 19307+2758. Spacepotato (talk) 23:37, 7 November 2014 (UTC)
## Distances
1.: Distance from earth: The article says "380 light-years (120 pc)" and the infobox says "430 ± 20 ly (133 ± 6 pc)". Now what?
2.: It seems User:PuzzletChung removed the distance from one star to the other in 2005. It's a shame that no one has been able to fill this in in the meantime. I would do the calculation myself if not for 1.
I help out as a guide at an observatory, and thats the first thing people ask: "How far away ist it" / "How much distance between the two". I came here today to get answers, and I'm quite surprised that apparently no one has been able to provide these yet. --BjKa (talk) 17:09, 1 November 2012 (UTC)
That requires that one know the orbit, which we are far from knowing. It is only possible at the present time to get a lower limit to the 3-D separation of the components, that is, the separation of their projection onto the sky plane.Donaldjbarry (talk) 04:01, 22 April 2018 (UTC)
380 is an old value, I believe from the original Hipparcos parallax. The revised Hipparcos parallax gives 430. Note that Albireo B infoxbox still has the old value, hence it is 40 light years from its primary :) Interestingly, the new Hipparcos reduction gives different values for the parallaxes of Alibireo A and B, slightly larger than their joint margins of error, an interesting comment on the accuracy of such things. Fixing. Lithopsian (talk) 17:30, 1 November 2012 (UTC)
Note, confidence intervals are not "margins of error", but rather a statement that the true value is estimated to lie, within some probability, within the bounds stated. The probability is not always stated and the default varies across fields, a frequent bad habit of scientific papers. In astronomy it is usually a 1 sigma result, meaning a 68% probable confinement of the true value, assuming a normal distribution of errors. It is not infrequent, however, that a 2 or 3 sigma result is stated, implying a 95% or 99.8% probability. By comparison, particle physics tends to use a 5 sigma result for claims of novel detections, a 99.99998% assurance based on errors being random, which essentially means that unknown systematic effects are far more likely to effect a false detection than random error.Donaldjbarry (talk) 03:58, 22 April 2018 (UTC)
## Pronunciation
Could someone put up a sound file as I am not sure how to pronounce it? Jzlcdh (talk) 13:54, 15 July 2013 (UTC)
## Discovering a binary with an interferometer?
The article states: "In 1976, component A was itself discovered to be a binary star, using speckle interferometry and the 2.1-meter telescope at the Kitt Peak National Observatory." This seemed suspicious since determining the binary nature of a star with a spectroscope is generally easier than resolving the pair with an interferometer. So I checked and found that Albireo Aa and Albireo Ac, the pair in question, have separate Henry Draper Catalog numbers, HD 183912 and HD 183913 respectively. They must have been known as a spectoscopic binary pair since 1949 at the latest, the last year that the HDC was updated. So, what happened in 1976 was that a known spectoscopic binary was resolved by interferometry. I will make the appropriate correction when I can get the wording and references right, if someone else does not beat me to it. - Fartherred (talk) 18:24, 27 February 2015 (UTC)
In general there is no guarantee that a binary can be detected spectroscopically. Gowever, Albireo has been known to have a composite spectrum for at least a century. Although one of the speckle references claims that no radial velocity differences had ever been detected, the radial velocity of Albireo A had been reported as early as 1919 to be decreasing slowly. It is now known to be a long period binary directly resolvable by modern equipment. Lithopsian (talk) 20:12, 27 February 2015 (UTC)
Thanks Lithopsian, I should have written that determining the binary nature of a star with a spectroscope is easier than resolving the pair with an interferometer most of the time. Of course there are cases in which the axis to rotation of the binary is parallel to the line of sight. With modern spectroscopy and the high orbital speeds that can be involved, there is a small fraction of cases in which the binary nature of a star eludes the spectroscope. Uncountable binaries are beyond resolution with an interferometer. In this case the binary nature of Albireo A was known long before 1976, as shown by the HD numbers, and that is what is most significant here.
Please note that separate HD numbers are not assigned to spectroscopic binary components. They refer only to the system. Even optical binaries of several arcsecond separation have unique HD numbers.Donaldjbarry (talk) 04:12, 22 April 2018 (UTC)
Another point about the article, the font in the infobox makes the seconds mark in +27° 57′ 34.84″ for the declination of Aa practically illegible on my display. I had to copy it to a larger font to be sure of the text and then just remember what was printed where.- Fartherred (talk) 21:56, 27 February 2015 (UTC)
Albireo Aa and Albireo Ac orbit each other in 213 years. The separation I calculate from the infobox is 0h 0m 0.009s right ascension and 0° 0′ 0.22″ declination. However the separation given on this site is 0.4 arc seconds. If the separation is decreasing it could be near minimum now which would likewise make the radial velocity differences minimum. One could wait fifty-six years to make spectroscope measurements or else read the literature to find what the spectrum differences were when one could detect them. Various websites copy free information with questionable reliability but I hope professional astronomers have access to historical data that they can depend upon. - Fartherred (talk) 22:47, 27 February 2015 (UTC)
Please note, Radial velocity differences are only at a minimum during periastron for circular orbits, a bad assumption for all but extremely close, interacting binaries.Donaldjbarry (talk) 04:12, 22 April 2018 (UTC)
I have made the correction. If someone can improve the wording and give better references, feel free. - Fartherred (talk) 15:47, 1 March 2015 (UTC)
## Cub Scout star
Also known as the Cub Scout star; but I'm not too sure how to edit it in without it looking like an abomination. :( — Preceding unsigned comment added by 104.220.56.173 (talk) 01:15, 7 July 2016 (UTC)
Does this star have a radius of 16 R? If you search the size of Albireo A on the net, you always get 16 R. But where did that come from? Also, where is the actually source for jumk.de 's 109 R?
You're asking in the wrong place. Wikipedia gives its sources (should do anyway), random web pages can just throw any old number out. 69 R is calculated from a published angular diameter and a published distance. It is unfortunate that they are published in different papers and combined only in WP, but a radius calculated from the bolometric luminosity and effective temperature is comparable. Lithopsian (talk) 17:09, 29 November 2017 (UTC)
## Gaia data
"Star A is moving at 16.66 milliarcseconds per year south by southeast, while Star B is moving at 1.13 mas/yr west by southwest**. The uncertainties in the measurements are relatively small compared to the motion, so I'm confident in saying they're moving in two different directions at two different speeds, and therefore not physically bound together. Their alignment is coincidental." Agmartin (talk) 19:12, 15 August 2018 (UTC)
I wouldn't get too excited just yet. Gaia results are (currently) notoriously unreliable for naked-eye stars because special techniques have to be applied and this hasn't been done for this data release. Some naked-eye stars are still included where the data passes some basic sanity checks. The primary also has a disk that is large compared to the Gaia measurements (meaning it may be observing different parts of the disk at different times), and this would need to be checked and corrected-for before any strong claims could be made. And of course always bear in mind that the quoted margins of error are only one standard deviation, and there are potential systematic errors in addition to the quoted errors. We need to wait for something peer-reviewed. Lithopsian (talk) 20:09, 15 August 2018 (UTC)
Yes, we need something peer reviewed, but the article already mentions the substantial difference in the parallax distances from Hipparcos data (430 versus 400 ly, but with one sigma errors of 20 and 10 ly, so arguably consistent), and the differences in the proper motions mentioned in Sky & Telescope from data in the Washington Double Star Catalog and SIMBAD (and that article also mentions a substantial difference in radial velocity - 24 km/s versus 19 km/s). These details are all in the infobox, cited to Hipparcos and Gaia papers. Taken all together it was already hard to see the data as consistent with a wide physical binary.
That said, the parallax distances quoted in the link above from the Gaia data (328 versus 389 ly) and the proper motions (16 mas/yr "south by southeast" and 1 mas/yr "west by southwest" - what are those compass directions: does he mean SSE and WSW?) do not appear to be consistent with the previously reported measurements. The infobox numbers indicate both A and B are moving roughly to the southwest, albeit at different rates. And I could believe B is at 389 ly (consistent with 400 p/m 10), but why does Gaia show A at 328 ly when Hipparcos said it was 430 p/m 20? Something does not look right. 213.205.240.162 (talk) 15:39, 16 August 2018 (UTC)
"Taken all together" is pretty much a definition of WP:SYNTHESIS, so I removed any statements beyond the bald data that is not contained in a reliable citation. You can speculate all you want and I'm not going to disagree with you, but our speculation can't go in the article. Further to the previous possible reasons for caution, Albireo A is of course itself an unambiguous close binary and it is unclear what effect that has on the astrometric data. I have added error margins lest anyone be misled about the concreteness of the distance differences. Further, I have replaced the blog distances with more scientific values although they aren't much different. Lithopsian (talk) 20:04, 16 August 2018 (UTC)
Oh, sure: but this is just a talk page :) As I said, we need someone to analyse the discrepancies and state the obvious conclusion in a proper reliable source - that is, in a published peer-reviewed journal. But we can at least say that one would expect both elements of a gravitationally bound binary system to (a) lie at approximately the same distance; (b) have approximately the same proper motion; and (c) have similar radial velocities (depending on how the orbit was oriented). Until we have additional reliable sources, there is not much more we can do than state the published facts. 213.205.240.162 (talk) 21:26, 16 August 2018 (UTC)
According to the Gaia catalog, the formal statistical error ellipse is unreliable due to a poor fit to the data. This could be due to either the acceleration of the source or do to photometric problems due to brightness. I've walked back the claim about precise distances. OtterAM (talk) 21:36, 17 August 2018 (UTC)
## Albireo Ab
Is there any info about Albireo Ab? The page doesn't seem to mention anything about it. Nussun05 (talk) 06:03, 17 October 2020 (UTC)
You could read this new paper although they don't mention the star. Even detections have been fleeting and there is little hard data on it. WDS has a few notes, but nothing conclusive. MSC shows an Aa/Ab orbit, but may be referring to Aa/Ac. Lithopsian (talk) 13:37, 18 October 2020 (UTC) | 3,939 | 16,108 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.830776 |
https://community.qlik.com/t5/QlikView-Layout-Visualizations/Fixing-total-in-pivot/m-p/284656/highlight/true | 1,585,831,122,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506959.34/warc/CC-MAIN-20200402111815-20200402141815-00016.warc.gz | 401,930,348 | 41,642 | # QlikView Layout & Visualizations
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## Fixing total in pivot
Hi!
I have a pivot chart where I show gross, refunds and net (with partial sums). No problem there.
As dimension I use Group which values are gross, refund and net. Net is calculated as gross - refund.
The problem occur when the grand total is calculated. I really just want to show the net.
Straight table is NOT the answer, I have multiple dimensions.
Found a solution where you create a assosiative table and somehow use that with aggregated values for dimension and then use the fact table for (pivot)total.
But I can't get it in my head how to create that table. What to use as key?
Really would like to be able to control what goes in the, grand, total with set analysis or something.
Suggestions?
Regards
thomas
Tags (8)
1 Solution
Accepted Solutions
Luminary
## Fixing total in pivot
Hi Thomas,
You could using an If statement in your expression using the Dimensionality() function. I believe that 0 is the grand total:
if(Dimensionality()=0, Null(), Sum(Myfield))
Regards,
Stephen
3 Replies
Luminary
## Fixing total in pivot
Hi Thomas,
You could using an If statement in your expression using the Dimensionality() function. I believe that 0 is the grand total:
if(Dimensionality()=0, Null(), Sum(Myfield))
Regards,
Stephen
Highlighted
Not applicable
## Re: Fixing total in pivot
Excellent!
With Null() the total row dissapears but with a set analysis expression it works.
`if(Dimensionality()=0, Sum({<Grupp={Net}>}Myfield)), Sum(Myfield))`
Highlighted
Not applicable
## Re: Fixing total in pivot
Hmm, one issue is that if you do a selection (lets say Gross) the grand total still calculates according to set analysis.
Maybe I can solve that with another if that eliminate the grand total if any selection is made. The sub total is still there and that should be enaugh. | 496 | 2,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.843514 |
http://www.mizar.org/version/current/html/jordan1g.html | 1,521,885,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257650188.31/warc/CC-MAIN-20180324093251-20180324113251-00442.warc.gz | 433,795,721 | 11,501 | :: Upper and Lower Sequence on the Cage. Part II
:: by Robert Milewski
::
:: Copyright (c) 2001-2017 Association of Mizar Users
registration
existence
ex b1 being FinSequence st b1 is trivial
proof end;
end;
theorem Th1: :: JORDAN1G:1
for f being trivial FinSequence holds
( f is empty or ex x being object st f = <*x*> )
proof end;
registration
let p be non trivial FinSequence;
cluster Rev p -> non trivial ;
coherence
not Rev p is trivial
proof end;
end;
theorem Th2: :: JORDAN1G:2
for D being non empty set
for f being FinSequence of D
for G being Matrix of D
for p being set st f is_sequence_on G holds
f -: p is_sequence_on G
proof end;
theorem Th3: :: JORDAN1G:3
for D being non empty set
for f being FinSequence of D
for G being Matrix of D
for p being Element of D st p in rng f & f is_sequence_on G holds
f :- p is_sequence_on G
proof end;
theorem Th4: :: JORDAN1G:4
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds Upper_Seq (C,n) is_sequence_on Gauge (C,n)
proof end;
theorem Th5: :: JORDAN1G:5
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds Lower_Seq (C,n) is_sequence_on Gauge (C,n)
proof end;
registration
let C be connected compact non horizontal non vertical Subset of ();
let n be Nat;
cluster Upper_Seq (C,n) -> standard ;
coherence
Upper_Seq (C,n) is standard
proof end;
cluster Lower_Seq (C,n) -> standard ;
coherence
Lower_Seq (C,n) is standard
proof end;
end;
theorem Th6: :: JORDAN1G:6
for G being Y_equal-in-column Y_increasing-in-line Matrix of ()
for i1, i2, j1, j2 being Nat st [i1,j1] in Indices G & [i2,j2] in Indices G & (G * (i1,j1)) `2 = (G * (i2,j2)) `2 holds
j1 = j2
proof end;
theorem Th7: :: JORDAN1G:7
for G being X_equal-in-line X_increasing-in-column Matrix of ()
for i1, i2, j1, j2 being Nat st [i1,j1] in Indices G & [i2,j2] in Indices G & (G * (i1,j1)) `1 = (G * (i2,j2)) `1 holds
i1 = i2
proof end;
theorem Th8: :: JORDAN1G:8
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> N-min (L~ f) & f /. (len f) <> N-min (L~ f) ) or ( f /. 1 <> N-max (L~ f) & f /. (len f) <> N-max (L~ f) ) ) holds
(N-min (L~ f)) `1 < (N-max (L~ f)) `1
proof end;
theorem :: JORDAN1G:9
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> N-min (L~ f) & f /. (len f) <> N-min (L~ f) ) or ( f /. 1 <> N-max (L~ f) & f /. (len f) <> N-max (L~ f) ) ) holds
N-min (L~ f) <> N-max (L~ f)
proof end;
theorem Th10: :: JORDAN1G:10
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> S-min (L~ f) & f /. (len f) <> S-min (L~ f) ) or ( f /. 1 <> S-max (L~ f) & f /. (len f) <> S-max (L~ f) ) ) holds
(S-min (L~ f)) `1 < (S-max (L~ f)) `1
proof end;
theorem :: JORDAN1G:11
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> S-min (L~ f) & f /. (len f) <> S-min (L~ f) ) or ( f /. 1 <> S-max (L~ f) & f /. (len f) <> S-max (L~ f) ) ) holds
S-min (L~ f) <> S-max (L~ f)
proof end;
theorem Th12: :: JORDAN1G:12
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> W-min (L~ f) & f /. (len f) <> W-min (L~ f) ) or ( f /. 1 <> W-max (L~ f) & f /. (len f) <> W-max (L~ f) ) ) holds
(W-min (L~ f)) `2 < (W-max (L~ f)) `2
proof end;
theorem :: JORDAN1G:13
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> W-min (L~ f) & f /. (len f) <> W-min (L~ f) ) or ( f /. 1 <> W-max (L~ f) & f /. (len f) <> W-max (L~ f) ) ) holds
W-min (L~ f) <> W-max (L~ f)
proof end;
theorem Th14: :: JORDAN1G:14
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> E-min (L~ f) & f /. (len f) <> E-min (L~ f) ) or ( f /. 1 <> E-max (L~ f) & f /. (len f) <> E-max (L~ f) ) ) holds
(E-min (L~ f)) `2 < (E-max (L~ f)) `2
proof end;
theorem :: JORDAN1G:15
for f being non trivial special unfolded standard FinSequence of () st ( ( f /. 1 <> E-min (L~ f) & f /. (len f) <> E-min (L~ f) ) or ( f /. 1 <> E-max (L~ f) & f /. (len f) <> E-max (L~ f) ) ) holds
E-min (L~ f) <> E-max (L~ f)
proof end;
theorem Th16: :: JORDAN1G:16
for D being non empty set
for f being FinSequence of D
for p, q being Element of D st p in rng f & q in rng f & q .. f <= p .. f holds
(f -: p) :- q = (f :- q) -: p
proof end;
theorem Th17: :: JORDAN1G:17
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat holds (L~ ((Cage (C,n)) -: (W-min (L~ (Cage (C,n)))))) /\ (L~ ((Cage (C,n)) :- (W-min (L~ (Cage (C,n)))))) = {(N-min (L~ (Cage (C,n)))),(W-min (L~ (Cage (C,n))))}
proof end;
theorem Th18: :: JORDAN1G:18
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds Lower_Seq (C,n) = (Rotate ((Cage (C,n)),(E-max (L~ (Cage (C,n)))))) -: (W-min (L~ (Cage (C,n))))
proof end;
theorem Th19: :: JORDAN1G:19
for n being Nat
for C being compact non horizontal non vertical Subset of () holds (W-min (L~ (Cage (C,n)))) .. (Upper_Seq (C,n)) = 1
proof end;
theorem Th20: :: JORDAN1G:20
for n being Nat
for C being compact non horizontal non vertical Subset of () holds (W-min (L~ (Cage (C,n)))) .. (Upper_Seq (C,n)) < (W-max (L~ (Cage (C,n)))) .. (Upper_Seq (C,n))
proof end;
theorem Th21: :: JORDAN1G:21
for n being Nat
for C being compact non horizontal non vertical Subset of () holds (W-max (L~ (Cage (C,n)))) .. (Upper_Seq (C,n)) <= (N-min (L~ (Cage (C,n)))) .. (Upper_Seq (C,n))
proof end;
theorem Th22: :: JORDAN1G:22
for n being Nat
for C being compact non horizontal non vertical Subset of () holds (N-min (L~ (Cage (C,n)))) .. (Upper_Seq (C,n)) < (N-max (L~ (Cage (C,n)))) .. (Upper_Seq (C,n))
proof end;
theorem Th23: :: JORDAN1G:23
for n being Nat
for C being compact non horizontal non vertical Subset of () holds (N-max (L~ (Cage (C,n)))) .. (Upper_Seq (C,n)) <= (E-max (L~ (Cage (C,n)))) .. (Upper_Seq (C,n))
proof end;
theorem Th24: :: JORDAN1G:24
for n being Nat
for C being compact non horizontal non vertical Subset of () holds (E-max (L~ (Cage (C,n)))) .. (Upper_Seq (C,n)) = len (Upper_Seq (C,n))
proof end;
theorem Th25: :: JORDAN1G:25
for n being Nat
for C being compact non horizontal non vertical Subset of () holds (E-max (L~ (Cage (C,n)))) .. (Lower_Seq (C,n)) = 1
proof end;
theorem Th26: :: JORDAN1G:26
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds (E-max (L~ (Cage (C,n)))) .. (Lower_Seq (C,n)) < (E-min (L~ (Cage (C,n)))) .. (Lower_Seq (C,n))
proof end;
theorem Th27: :: JORDAN1G:27
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds (E-min (L~ (Cage (C,n)))) .. (Lower_Seq (C,n)) <= (S-max (L~ (Cage (C,n)))) .. (Lower_Seq (C,n))
proof end;
theorem Th28: :: JORDAN1G:28
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds (S-max (L~ (Cage (C,n)))) .. (Lower_Seq (C,n)) < (S-min (L~ (Cage (C,n)))) .. (Lower_Seq (C,n))
proof end;
theorem Th29: :: JORDAN1G:29
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds (S-min (L~ (Cage (C,n)))) .. (Lower_Seq (C,n)) <= (W-min (L~ (Cage (C,n)))) .. (Lower_Seq (C,n))
proof end;
theorem Th30: :: JORDAN1G:30
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds (W-min (L~ (Cage (C,n)))) .. (Lower_Seq (C,n)) = len (Lower_Seq (C,n))
proof end;
theorem Th31: :: JORDAN1G:31
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds ((Upper_Seq (C,n)) /. 2) `1 = W-bound (L~ (Cage (C,n)))
proof end;
theorem :: JORDAN1G:32
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds ((Lower_Seq (C,n)) /. 2) `1 = E-bound (L~ (Cage (C,n)))
proof end;
theorem Th33: :: JORDAN1G:33
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds (W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n)))) = () + ()
proof end;
theorem :: JORDAN1G:34
for n being Nat
for C being connected compact non horizontal non vertical Subset of () holds (S-bound (L~ (Cage (C,n)))) + (N-bound (L~ (Cage (C,n)))) = () + ()
proof end;
theorem Th35: :: JORDAN1G:35
for C being connected compact non horizontal non vertical Subset of ()
for n, i being Nat st 1 <= i & i <= width (Gauge (C,n)) & n > 0 holds
((Gauge (C,n)) * ((Center (Gauge (C,n))),i)) `1 = (() + ()) / 2
proof end;
theorem :: JORDAN1G:36
for C being connected compact non horizontal non vertical Subset of ()
for n, i being Nat st 1 <= i & i <= len (Gauge (C,n)) & n > 0 holds
((Gauge (C,n)) * (i,(Center (Gauge (C,n))))) `2 = (() + ()) / 2
proof end;
theorem Th37: :: JORDAN1G:37
for f being S-Sequence_in_R2
for k1, k2 being Nat st 1 <= k1 & k1 <= len f & 1 <= k2 & k2 <= len f & f /. 1 in L~ (mid (f,k1,k2)) & not k1 = 1 holds
k2 = 1
proof end;
theorem Th38: :: JORDAN1G:38
for f being S-Sequence_in_R2
for k1, k2 being Nat st 1 <= k1 & k1 <= len f & 1 <= k2 & k2 <= len f & f /. (len f) in L~ (mid (f,k1,k2)) & not k1 = len f holds
k2 = len f
proof end;
theorem Th39: :: JORDAN1G:39
for C being compact non horizontal non vertical Subset of ()
for n being Nat holds
( rng (Upper_Seq (C,n)) c= rng (Cage (C,n)) & rng (Lower_Seq (C,n)) c= rng (Cage (C,n)) )
proof end;
theorem Th40: :: JORDAN1G:40
for n being Nat
for C being compact non horizontal non vertical Subset of () holds Upper_Seq (C,n) is_a_h.c._for Cage (C,n)
proof end;
theorem Th41: :: JORDAN1G:41
for n being Nat
for C being compact non horizontal non vertical Subset of () holds Rev (Lower_Seq (C,n)) is_a_h.c._for Cage (C,n)
proof end;
theorem Th42: :: JORDAN1G:42
for n being Nat
for C being connected compact non horizontal non vertical Subset of ()
for i being Nat st 1 < i & i <= len (Gauge (C,n)) holds
not (Gauge (C,n)) * (i,1) in rng (Upper_Seq (C,n))
proof end;
theorem Th43: :: JORDAN1G:43
for n being Nat
for C being connected compact non horizontal non vertical Subset of ()
for i being Nat st 1 <= i & i < len (Gauge (C,n)) holds
not (Gauge (C,n)) * (i,(width (Gauge (C,n)))) in rng (Lower_Seq (C,n))
proof end;
theorem Th44: :: JORDAN1G:44
for n being Nat
for C being connected compact non horizontal non vertical Subset of ()
for i being Nat st 1 < i & i <= len (Gauge (C,n)) holds
not (Gauge (C,n)) * (i,1) in L~ (Upper_Seq (C,n))
proof end;
theorem :: JORDAN1G:45
for n being Nat
for C being connected compact non horizontal non vertical Subset of ()
for i being Nat st 1 <= i & i < len (Gauge (C,n)) holds
not (Gauge (C,n)) * (i,(width (Gauge (C,n)))) in L~ (Lower_Seq (C,n))
proof end;
theorem Th46: :: JORDAN1G:46
for n being Nat
for C being connected compact non horizontal non vertical Subset of ()
for i, j being Nat st 1 <= i & i <= len (Gauge (C,n)) & 1 <= j & j <= width (Gauge (C,n)) & (Gauge (C,n)) * (i,j) in L~ (Cage (C,n)) holds
LSeg (((Gauge (C,n)) * (i,1)),((Gauge (C,n)) * (i,j))) meets L~ (Lower_Seq (C,n))
proof end;
theorem Th47: :: JORDAN1G:47
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
First_Point ((L~ (Upper_Seq (C,n))),(W-min (L~ (Cage (C,n)))),(E-max (L~ (Cage (C,n)))),(Vertical_Line (((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2))) in rng (Upper_Seq (C,n))
proof end;
theorem Th48: :: JORDAN1G:48
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
Last_Point ((L~ (Lower_Seq (C,n))),(E-max (L~ (Cage (C,n)))),(W-min (L~ (Cage (C,n)))),(Vertical_Line (((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2))) in rng (Lower_Seq (C,n))
proof end;
theorem Th49: :: JORDAN1G:49
for f being S-Sequence_in_R2
for p being Point of () st p in rng f holds
R_Cut (f,p) = mid (f,1,(p .. f))
proof end;
theorem Th50: :: JORDAN1G:50
for f being S-Sequence_in_R2
for Q being closed Subset of () st L~ f meets Q & not f /. 1 in Q & First_Point ((L~ f),(f /. 1),(f /. (len f)),Q) in rng f holds
(L~ (mid (f,1,((First_Point ((L~ f),(f /. 1),(f /. (len f)),Q)) .. f)))) /\ Q = {(First_Point ((L~ f),(f /. 1),(f /. (len f)),Q))}
proof end;
theorem Th51: :: JORDAN1G:51
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
for k being Nat st 1 <= k & k < (First_Point ((L~ (Upper_Seq (C,n))),(W-min (L~ (Cage (C,n)))),(E-max (L~ (Cage (C,n)))),(Vertical_Line (((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2)))) .. (Upper_Seq (C,n)) holds
((Upper_Seq (C,n)) /. k) `1 < ((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2
proof end;
theorem Th52: :: JORDAN1G:52
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
for k being Nat st 1 <= k & k < (First_Point ((L~ (Rev (Lower_Seq (C,n)))),(W-min (L~ (Cage (C,n)))),(E-max (L~ (Cage (C,n)))),(Vertical_Line (((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2)))) .. (Rev (Lower_Seq (C,n))) holds
((Rev (Lower_Seq (C,n))) /. k) `1 < ((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2
proof end;
theorem Th53: :: JORDAN1G:53
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
for q being Point of () st q in rng (mid ((Upper_Seq (C,n)),2,((First_Point ((L~ (Upper_Seq (C,n))),(W-min (L~ (Cage (C,n)))),(E-max (L~ (Cage (C,n)))),(Vertical_Line (((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2)))) .. (Upper_Seq (C,n))))) holds
q `1 <= ((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2
proof end;
theorem Th54: :: JORDAN1G:54
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
(First_Point ((L~ (Upper_Seq (C,n))),(W-min (L~ (Cage (C,n)))),(E-max (L~ (Cage (C,n)))),(Vertical_Line (((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2)))) `2 > (Last_Point ((L~ (Lower_Seq (C,n))),(E-max (L~ (Cage (C,n)))),(W-min (L~ (Cage (C,n)))),(Vertical_Line (((W-bound (L~ (Cage (C,n)))) + (E-bound (L~ (Cage (C,n))))) / 2)))) `2
proof end;
theorem Th55: :: JORDAN1G:55
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
L~ (Upper_Seq (C,n)) = Upper_Arc (L~ (Cage (C,n)))
proof end;
theorem Th56: :: JORDAN1G:56
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
L~ (Lower_Seq (C,n)) = Lower_Arc (L~ (Cage (C,n)))
proof end;
theorem :: JORDAN1G:57
for C being connected compact non horizontal non vertical Subset of ()
for n being Nat st n > 0 holds
for i, j being Nat st 1 <= i & i <= len (Gauge (C,n)) & 1 <= j & j <= width (Gauge (C,n)) & (Gauge (C,n)) * (i,j) in L~ (Cage (C,n)) holds
LSeg (((Gauge (C,n)) * (i,1)),((Gauge (C,n)) * (i,j))) meets Lower_Arc (L~ (Cage (C,n)))
proof end; | 5,317 | 14,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-13 | latest | en | 0.81521 |
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Temperature Temperament (Posted on 2013-03-06)
Determine all possible values of a positive integer x satisfying:
x degrees Fahrenheit = sod(x) degrees Celsius, where sod(x) denotes the sum of the digits in the base ten expansion of x.
Prove that there are no others.
Note: To convert from degrees Fahrenheit to Celsius, subtract 32 then multiply by 5/9. To convert the other way, simply do the opposite (multiply by 9/5 and add 32).
No Solution Yet Submitted by K Sengupta Rating: 2.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
spoiler | Comment 1 of 3
41 deg F= 5 deg C
NO other couple
Posted by Ady TZIDON on 2013-03-06 11:28:38
Search: Search body:
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https://daviddalpiaz.github.io/stat430fa17/hw/hw01-assign.html | 1,581,988,259,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143455.25/warc/CC-MAIN-20200217235417-20200218025417-00317.warc.gz | 354,706,865 | 229,422 | Please see the homework instructions document for detailed instructions and some grading notes. Failure to follow instructions will result in point reductions.
Exercise 1
[10 points] This question will use data in a file called hw01-data.csv. The data contains four predictors: a, b, c, d, and a response y.
After reading in the data as hw01_data, use the following code to test-train split the data.
set.seed(42)
train_index = sample(1:nrow(hw01_data), size = round(0.5 * nrow(hw01_data)))
train_data = hw01_data[train_index, ]
test_data = hw01_data[-train_index, ]
Next, fit four linear models using the training data:
• Model 1: y ~ .
• Model 2: y ~ . + I(a ^ 2) + I(b ^ 2) + I(c ^ 2)
• Model 3: y ~ . ^ 2 + I(a ^ 2) + I(b ^ 2) + I(c ^ 2)
• Model 4: y ~ a * b * c * d + I(a ^ 2) + I(b ^ 2) + I(c ^ 2)
For each of the models above, report:
• Train RMSE
• Test RMSE
• Number of Parameters, Excluding the Variance
To receive full marks, arrange this information in a well formatted table. Also note which model is best for making predictions.
[Not Graded] For fun, find a model that outperforms each of the models above. Hint: Consider some exploratory data analysis. Hint: Your instructor’s solution uses a model with only seven parameters. Yours may have more.
Exercise 2
[10 points] For this question we will use the Boston data from the MASS package. Use ?Boston to learn more about the data.
library(readr)
library(tibble)
library(MASS)
data(Boston)
Boston = as_tibble(Boston)
Use the following code to test-train split the data.
set.seed(42)
boston_index = sample(1:nrow(Boston), size = 400)
train_boston = Boston[boston_index, ]
test_boston = Boston[-boston_index, ]
Fit the following linear model that uses medv as the response.
fit = lm(medv ~ . ^ 2, data = train_boston)
Fit two additional models, both that perform worse than fit, with respect to prediction. One should be a smaller model, relative to fit. The other should be a larger model, relateive to fit. Call them fit_smaller and fit_larger respectively. Any “smaller” model should be nested in any “larger” model.
Report these three models as well as their train RMSE, test RMSE, and number of parameters. Note: you may report the models used using their R syntax. To receive full marks, arrange this information in a well formatted table.
Exercise 3
[10 points] How do outliers affect prediction? Usually when fitting regression models for explanation, dealing with outliers is a complicated issue. When considering prediction, we can empirically determine what to do.
Continue using the Boston data, training split, and models from Exercise 2. Consider the model stored in fit from Exercise 2. Obtain the standardized residuals from this fitted model. Refit this model with each of the following modifications:
• Removing observations from the training data with absolute standardized residuals greater than 2.
• Removing observations from the training data with absolute standardized residuals greater than 3.
(a) Use these three fitted models, including the original model fit to unmodified data, to obtain test RMSE. Summarize these results in a table. Include the number of observations removed for each. Which performs the best? Were you justified modifying the training data?
(b) Using the best of these three fitted models, create a 99% prediction interval for a new observation with the following values for the predictors:
crim zn indus chas nox rm age dis rad tax ptratio black lstat
0.02763 75.0 3.95 0 0.4280 6.595 22.8 5.4011 3 252 19.3 395.63 4.32 | 893 | 3,559 | {"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} | 3.4375 | 3 | CC-MAIN-2020-10 | latest | en | 0.743698 |
https://meritnotes.com/aptitude/time-distance-questions/1-78077/ | 1,709,519,377,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00411.warc.gz | 387,712,027 | 6,959 | # Top 100+ Time and Distance Aptitude Questions and Answers - 1
Question: 1
A train starts from station X at the rate of 60 km/hr and reaches station Y in 45 minutes. If the speed is reduced by 6 km/hr, how much more time will the train take to return from station Y to station X?
(A) 4 minutes
(B) 5 minutes
(C) 6 minutes
(D) \$\$7{1}/ {2}\$\$ minutes
Ans: B
Distance XY = \$\$(60 × {45}/{60})\$\$ km = 45 km
New speed = (60 – 6) km/hr = 54 km/hr
Time taken = \$\$({45} / {54} × 60)\$\$ min = 50 min
∴ Extra time = (50 - 45) min = 5 min.
Question: 2
An aeroplane travels distances 2500 km, 1200 km and 500 km at the rate of 500 km/hr, 400 km/hr and 250 km/hr respectively. The average speed (in km/hr) is
(A) 400
(B) 405
(C) 410
(D) 420
Ans: D
Total time taken = \$\$({2500} / {500} + {1200} / {400} + {500} / {250})\$\$ hours = 10 hours.
∴ Average speed = \$\$({4200} / {10})\$\$ km/hr = 420 km/hr.
Question: 3
Sound travels at 330 meters a second. How many kilometer away is a thunder cloud when its sound follows the flash after 10 seconds?
(A) 3
(B) 3.3
(C) 0.33
(D) 33
Ans: B
Distance = (Time × speed) = (330 × 10) m = \$\${330 × 10} /{1000}\$\$ km = 3.3 km.
Question: 4
Shiela’s house is 10 km away from the school. She takes 30 minutes to reach the school by bus. If Ram travels from his house at the same speed as that of Shiela and takes only 12 minutes to reach the school, the distance between Ram’s house and his school (in Km) is
(A) 2
(B) 4
(C) 5
(D) 6
Ans: B
Ram’s speed = \$\${10} / {30} × 60\$\$ = 20 km/hr
Time taken by Ram = 12 min = \$\${12} / {60}\$\$ hour = \$\${1} / {5}\$\$ hour
∴ Distance between Ram’s house and school = \$\$20 × {1} / {5}\$\$ = 4 km.
Question: 5
Anita and Veena are running in opposite direction. Speed of Anita and Veena is 8 km/hr and 10 km/hr respectively. Find out distance between them after \$\$2{1} / {2}\$\$ hours
(A) 30 km
(B) 36 km
(C) 40 km
(D) 45 km
Ans: D
Relative speed of Anita and Veena = (10 + 8) Km/hr = 18 km/hr
∴ Distance in \$\$2{1} / {2}\$\$ hour = \$\${5} / {2} × 18\$\$ km = 45 km.
Related Questions | 765 | 2,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-10 | longest | en | 0.900895 |
http://garniturek.info/facts-fiction-and-operations-of-mathematics/ | 1,603,224,783,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874135.2/warc/CC-MAIN-20201020192039-20201020222039-00411.warc.gz | 44,122,841 | 11,101 | # Choosing Operations of Mathematics
As an example, they may help decide how to organize goods in supermarkets or help businesses figure out the best means to ship and distribute products. Nevertheless, you don’t need to have the newest exercise ball or any fancy equipment to get fit. Another advantage of exercising early is increased energy through the day. Three meals every day, accommodation and activities ought to be included in the cost, for additional convenience. Facility capacity is set by the range of patients they can house at the same time, here known as the variety of beds (c servers).
## The Pain of Operations of Mathematics
Today, the majority of the parents think that the practice questions offered in the text book aren’t enough to the students to increase their knowledge in that special place. It appears that every student interpreted the issue differently, resulting in two answers. They’ll attend English classes and there are a lot of alternatives of activities offered for children outside the class hours.
One other great means to make sure that your kids learn math is to supply them with colorful, hard cardboard math books. So it’s an enjoyable approach to apply my problem-solving skills in a means that will potentially have an actual effects. After understanding was achieved, it’s time to nail it down.
Keeping a citizenry that’s well versed in the STEM fields is a vital part of the public education agenda of the USA. Some of these sites have math worksheets generators while others might have ready-made worksheets. It is going to then center on the next four topics. Here’s a graphic preview for all the Order of Operations Worksheets.
Statistics is both an exact applied area as well as a theoretical one. It isn’t a coincidence that the language of mathematics was initially called natural philosophy and practised by philosophers to explain an array of observations. By way of example, in mathematics and many computer languages, multiplication is granted a greater precedence than addition, and it’s been this way since the debut of modern algebraic notation. A theoretical mathematician intends to resolve unexplained difficulties.
It is most frequently encountered for predicting outcomes, as in the very simple example employed within this paper (1315). The old-school approach to reading a pulse requires touching the radial pulse and counting the amount of beats for a couple of seconds. A few completely free samples out there.
When it isn’t a number it’s probably an operation. Remember that, although the operations and the examples shown here are pretty straightforward, they give the foundation for even the most complex operations utilized in mathematics. When the analyzing of a problem is finished, operations research analysts advise managers on how best to proceed in resolving the issue.
Various calculators follow various orders of operations. Parentheses let you ascertain what operation is done first. You’ll observe an extremely simplistic SQL implementation.
## Operations of Mathematics – Is it a Scam?
Mathematics majors might end up in a job with a title like Engineer or Analyst. After having studied a particular concept in math, they may have to do some practice in it. Because, apart from the class room training, they may have to do much practice to become mastery in the concept.
However, along with this quantitative growth, at the conclusion of the 18th century and the start of the 19th century a variety of essentially new features were observed in the growth of mathematics. It’s mainly due to his notational innovations which he has been credited by some historians as being the founder of contemporary algebra.
## The Downside Risk of Operations of Mathematics
Are you even slightly concerned that this type of robotic march during the mathematical desert may not be off-putting to students. This is called the minus sign. They’ll be more motivated to write well, if it’s something they would like to do. Nevertheless, you know the calculus is among the pinnacles of human thought, and it would be great to understand only a bit of what it is that they’re speaking about Both thorough and brief intro-to-calculus lectures can be found on the internet. There are also a good deal of cartoon series that teach kids counting in an enjoyable and intriguing way. It’s extremely beneficial to understand your way around them since they’re a bit of mathematics we encounter each and every day.
This practice problems supply you with the chance to practice utilizing a few of the concepts discussed within this write-up. Quantum computers cannot replace classical computers because they don’t give faster results for everything you desire. As soon as you’ve applied, you will want to finish the Admission Information Form as a portion of the application practice.
A project should be treated like the body, with all its complex characteristics. Because of the presence of nested brackets, and since there are 3 types readily available, there’s an order both for solving and nesting brackets. Frequently there are lots of decision variables and the remedy isn’t obvious. As soon as we perform computational actions in mathematics, we must bear in mind that there’s a sequence that should be respected to be able to do the calculation properly. Let’s do another issue together.
Unfortunately, they normally do not completely alter the functionality of the item and are frequently very limited. They’ll be exposed to general procedures of inquiry that apply in wide array of settings. It’s not mandatory that always data ought to be in two or three dimensions. They also possess a massive array of applications in various contexts and surroundings. Write the item in simplest form.
## What the In-Crowd Won’t Tell You About Operations of Mathematics
Once students are conversant with the notion of equivalent fractions, the concept of locating fractions with common denominators for adding and subtracting becomes that much simpler. Subtracting negative integers isn’t included. Unlike addition, subtraction isn’t commutative.
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## The Chronicles of Operations of Mathematics
Firstly, one hopes to boost our comprehension of the start of the disease and of the way that it progresses. Doctors then determine how long patients must go on taking medicines and the length of time the medicines will remain in the body. Should it, it is going to refer a mean of 2 patients each week.
Everything means the exact same thing! It is often as simple as starting your usual walk for approximately three minutes and then running for a single minute. Then, this is where to be! You look up at the very top of the tree. At the close of the day, we need to decide for ourselves. Write twice weekly or three times per week.
## What’s Really Happening with Operations of Mathematics
This practice problems supply you with the chance to practice utilizing a few of the concepts discussed within this write-up. Amongst them the key element consists of individualized learning program. There is quite a beautiful bit of mathematics behind it.
## The Lost Secret of Operations of Mathematics
If you’re simplifying huge fractions by hand you may use the Long Division with Remainders Calculator to locate whole number and remainder values. Your work above should convince you this doesn’t get the job done for decimals! Fortunately, calculating any one among these 3 variables is simple to do when you know both of the other variables.
We work hard to be sure our site works well and we possess the ideal math worksheets. I am at the airport, since I finished writing this informative article. Each worksheet comprises 50 questions and all them are objective type questions. Definitely the response is No. Solve the math issues and use the answers to finish the crossword puzzles.
Subtraction was a bit more complicated. It isn’t a coincidence that the language of mathematics was initially called natural philosophy and practised by philosophers to explain an array of observations. We categorize and review the worksheets listed here in order to help you locate the math worksheets problems which you’re looking for. A theoretical mathematician intends to resolve unexplained difficulties.
## Operations of Mathematics at a Glance
He gives a specific number of cans to the very first regional charity he finds. Nevertheless, you don’t need to have the newest exercise ball or any fancy equipment to get fit. Determine the quantity of time that has passed. Three meals every day, accommodation and activities ought to be included in the cost, for additional convenience. Learn how to count pounds and pence, coins utilised in the uk.
## New Step by Step Roadmap for Operations of Mathematics
As soon as you are comfortable with the comprehension of the order of operations, consider using a spreadsheet to figure out the order of operations. Fourth, attempt to enter an application area in which you feel you are able to earn a contribution. When the analyzing of a problem is finished, operations research analysts advise managers on how best to proceed in resolving the issue.
Employing this simplified system, an individual can solve directly for the long-run average number of men and women in the queue. All these things play a major role in processing machine learning algorithm. Updates to the product require a sequence of meetings, require a mixture of unique abilities, and bring a whole lot of complexity in the item development flow.
## Operations of Mathematics – Is it a Scam?
Pure Mathematics will supply you with problem-solving skills which can be applied in industry, organization, government, or graduate school. In this program they are expected to perform excellently in order to acquire rapid progress and achieve more and more intellect talent in order to face major levels of difficulties. Sixth grade students get a great look at divisibility and factoring.
The idea was supposed to find knowledgeable about this topic and also see the way the journey might look like. And I want to put this story here, it may be helpful to be aware. These activities make it possible for students to get in touch with their peers in the department and in different majors.
Today, the majority of the parents think that the practice questions offered in the text book aren’t enough to the students to increase their knowledge in that special place. If you’ve recently moved to an English-speaking country and choose to send your son or daughter on an English camp for children, odds are there are different parents doing the exact same thing. The English camps for children provide a selection of activities the children can participate in, permitting them to experience new things and have a cultural experience.
One other great means to make sure that your kids learn math is to supply them with colorful, hard cardboard math books. So it’s an enjoyable approach to apply my problem-solving skills in a means that will potentially have an actual effects. What’s great about English camps for children, is that some of the highest language schools provides a course that allows children to study at a few of the very best universities and colleges in the usa.
## The New Angle On Operations of Mathematics Just Released
The procedure for forming (in relation to the problems of measurement of magnitude) the idea of a true number (see Number) turned out to be quite protracted. Because of the presence of nested brackets, and since there are 3 types readily available, there’s an order both for solving and nesting brackets. Frequently there are lots of decision variables and the remedy isn’t obvious. The research requires the capability to think logically and stick to all components of the puzzle so as to attain a solution. Also utilize factor tree method to discover the GCF.
Besides this one can select a variety of colors to present the info and there isn’t any limitation on the colours of markers one have. We don’t have specific automated processes we must follow to have by. It is known as the additive identity element. Well, we need to go and have a look at various properties of vectors to find a strong comprehension. Write the item in simplest form.
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It is most frequently encountered for predicting outcomes, as in the very simple example employed within this paper (1315). You will probably produce an incorrect answer if you perform calculations from the purchase. A few completely free samples out there.
## The End of Operations of Mathematics
Honors programs are offered in both tracks. Students have the chance to learn about Navy applications. Math is part of your everyday routine.
Whether you are a newcomer to the area of medicine or would like to boost your skills, this is the course for you. In a modern Earth, math such as applied mathematics isn’t only relevant, it’s important. Rewarding mathematics jobs can be found in numerous businesses and organizations.
The idea was supposed to find knowledgeable about this topic and also see the way the journey might look like. We understand that while many individuals might be interested in mathematics, it might be for entirely different factors. These activities make it possible for students to get in touch with their peers in the department and in different majors.
A project should be treated like the body, with all its complex characteristics. It’s an algebraic structure that behaves nicely in numerous ways. Certain polynomial equations without any actual solution have complex solutions. As soon as we perform computational actions in mathematics, we must bear in mind that there’s a sequence that should be respected to be able to do the calculation properly. The very first step to figure out this issue is to work the P (parentheses).
Indeed it’s quite unusual to need this type of abstraction. Packed here are an array of worksheets to learn the theory of place values. It is known as the additive identity element. They also possess a massive array of applications in various contexts and surroundings. Write the item in simplest form.
## The Nuiances of Operations of Mathematics
As an example, they may help decide how to organize goods in supermarkets or help businesses figure out the best means to ship and distribute products. For instance, you have 12 apples that have to be shared equally between 4 people. You get yourself 2 lbs of peanuts and one bottle of plain water. You don’t need to get a costly item of equipment which demands a monthly payment program. Facility capacity is set by the range of patients they can house at the same time, here known as the variety of beds (c servers).
So this is likely to be equal to 61. Maybe, others are going to follow suit. It simply may help you get the victory. You look up at the very top of the tree. Because there are just two balls, there are 3 possible probabilities of drawing a red ball any any given draw. You use it every day even if you aren’t conscious of this.
## Up in Arms About Operations of Mathematics?
When it isn’t a number it’s probably an operation. Suppose a neighborhood hospital is thinking about closing its treatment unit. When the analyzing of a problem is finished, operations research analysts advise managers on how best to proceed in resolving the issue.
Actually, SQL is easily the most successful 4th generation language. Operations Research is a modern, interdisciplinary subject that utilizes mathematical methods to fix large-scale optimization problems in the actual world. Payroll management is extremely different based on the nation.
## What to Expect From Operations of Mathematics?
Today, the majority of the parents think that the practice questions offered in the text book aren’t enough to the students to increase their knowledge in that special place. If you’ve recently moved to an English-speaking country and choose to send your son or daughter on an English camp for children, odds are there are different parents doing the exact same thing. They’ll attend English classes and there are a lot of alternatives of activities offered for children outside the class hours.
One other great means to make sure that your kids learn math is to supply them with colorful, hard cardboard math books. There are different uses of math in the life span of a health care provider. After understanding was achieved, it’s time to nail it down.
## What the In-Crowd Won’t Tell You About Operations of Mathematics
If you’re simplifying huge fractions by hand you may use the Long Division with Remainders Calculator to locate whole number and remainder values. Your work above should convince you this doesn’t get the job done for decimals! Fortunately, calculating any one among these 3 variables is simple to do when you know both of the other variables.
## Operations of Mathematics Features
Here is an index page which links to every one of the subtraction sections of our site. Exactly like in math, there’s a specific order that we work problems. No particular certification or license is needed for statisticians. The lectures will cover these topics.
Subtraction was a bit more complicated. A mathematician’s brain is similar to a sword that should be polished and honed constantly as a way to serve its purpose, or it will end up dull and rusted. The worksheets are generated randomly, which means you get a different one each moment. A theoretical mathematician intends to resolve unexplained difficulties.
} | 3,338 | 17,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-45 | latest | en | 0.961514 |
https://dsp.stackexchange.com/questions/70362/how-many-bins-to-include-when-calculating-snr-from-fft | 1,702,289,874,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00598.warc.gz | 240,761,751 | 45,286 | # How many bins to include when calculating SNR from FFT?
I'm a scientist conducting an experiment that requires some signal processing. My expertise is not in signal processing, thus here I am. We've basically re-created an experiment conducted by other scientists, attempting to check their results. Here is a link to their paper: Ultrasensitive Inverse Weak-Value Tilt Meter
In short, a laser bounces off of some mirrors, one of which is oscillating at a controlled sinusoidal frequency, onto a quadrant detector, which outputs an electrical signal to an oscilloscope where we record it. So, you end up with a noisy record that has a tiny, known sine wave hiding in it.
My question has to do with calculating the SNR from the FFTs of our records. How many bins do I include for noise? Do I include all of the non-signal bins (besides DC and Nyquist)? Or is there some standard for this type of thing?
As a follow on question, when determining the noise floor from our spectral densities, is there a certain standard calculation, or is it more of an eyeball the plot type thing? Is the floor determined for bins near the signal, or should you look at the whole record?
Thanks in advance for any help you can provide. As a side note, I have done quite a bit of due diligence trying to figure this stuff out. But, anytime I find some source that seems authoritative that says one thing, I'll find another that says something different. I can't tell if I'm misunderstanding what they're saying, or if there are just lots of different definitions for this stuff.
Here are a couple of our PSD plots.
Full Spectrum:
Zoomed in on the reference signal:
• +1 for due diligence. When you say tiny known signal, what is known and what do you want to measure about it? Sep 16, 2020 at 19:35
• Absolutely. Dan (and others here) are experts at proper SNR calculations, I am not. My specialty is tone parameter estimation. The one thing I can say is that you want to make sure you frame your signal on a whole number of cycles so all its "energy=sum of squares" is localized in the bin instead of "leaking". In that case your answer is one (assuming just using bins below Nyquist, half the spectrum) for your signal, and the rest are candidates for your noise. Is the spectrum relatively flat outside your peaks? Sep 16, 2020 at 20:27
• That's a big FFT. With 10 samples per cycle, you have about 100,000 cycles so framing on a whole number of cycles is going to be kind of tough unless you know the frequency very precisely. That's useful information. Have you taken any smaller DFTs on a portion of your signal? If so, are there any substantial variations in your SNR calculations? What is your SNR roughly? Sep 16, 2020 at 21:01
• Did you apply a window function? The purpose is to mitigate the "leakage" (caused by non-whole integer framing) concentrating the signal's energy near the bin. This effects how you do the SNR calculation. Sep 16, 2020 at 21:27
• Seems you've had quite the tour. Due diligence might be an understatement. My understanding (and again SNR analysis can be complicated and I am not versed), is that if you have a single pure tone, you should be able calculate its signal strength exactly (given exact parameters, theoretically ideal), then subtract the tone from the signal and measure the residual. This approach sort of presupposes very short intervals and a subsequent aggregation. The assumption that a real world sign is a pure tone for too long doesn't usually last. I tend to work in frames with single digit cycle counts. Sep 17, 2020 at 2:36
The OP's question requires further details to provide a definitive answer, but the following will give the considerations involved. If the noise is white and stationary then the answer is clear in that we can simply use a power spectral density and work with the SNR as an SNR/Hz quantity. If the noise is not white (meaning the average power across all bins is not the same) then the concern will ultimately be with the measurement bandwidth of your final system that is providing the estimate of the quantity you are looking for. You would include all noise that is within this bandwidth as that is what will give you the statistical error in your result (which is why we are concerned with SNR in the first place). Usually the choice of bandwidth comes down to a time /frequency decision and the coherence time of your signal, and for how long it can be assumed to be stationary. For bandwidth limited signals that can be assumed to be stationary for a long time, then we can use a relatively longer averaging time (reducing the bandwidth down to that of the signal, and therefore the amount of total noise in our result, as long as the noise is a spectral density and not a "spur" that exists at one frequency). | 1,060 | 4,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | longest | en | 0.966273 |
https://scioly.org/forums/viewtopic.php?f=173&t=5030 | 1,620,876,566,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00147.warc.gz | 550,527,163 | 12,396 | ## MagLev C Question Marathon
Jim_R
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### MagLev C Question Marathon
Question Marathon for MagLev C.
-The path of the Administrator is beset on all sides by the inequities of the selfish and the tyranny of evil men.
-Nothing\'s gonna get deleted. We\'re gonna be like three little Fonzies here. And what\'s Fonzie like? Come on, what\'s Fonzie like?
-Cool?
-Correctamundo. And that\'s what we\'re gonna be. We\'re gonna be cool. Now, I\'m gonna count to three, and when I count three, you let go of your mouse, and back away from the keyboard. But when you do it, you do it cool. Ready? One... two... three.
JTMess
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### Re: MagLev C Question Marathon
I shall begin the marathon!
What are the potential uses of a vibrating sample magnetometer?
2014 States: Scrambler-2nd, Mission Possible-2nd, Experimental Design-3rd, Circuit Lab-3rd
2014 Regionals: Scrambler-1st, Mission-1st, Technical Problem Solving-1st, Circuit Lab-1st, Maglev-1st, Bungee Drop-1st
2013 States: Gravity Vehicle-1st, Fermi-8th, Maglev-1st
twototwenty
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### Re: MagLev C Question Marathon
It is used to determine magnetic properties of the sample, including its magnetic moment, and its hysteresis curve (the relationship between field strength and the magnetization of a material).
Where and when was the first maglev railway implemented?
JTMess
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### Re: MagLev C Question Marathon
The first maglev railway was opened to provide transit between the airport and train station in Birmingham, England.
2014 States: Scrambler-2nd, Mission Possible-2nd, Experimental Design-3rd, Circuit Lab-3rd
2014 Regionals: Scrambler-1st, Mission-1st, Technical Problem Solving-1st, Circuit Lab-1st, Maglev-1st, Bungee Drop-1st
2013 States: Gravity Vehicle-1st, Fermi-8th, Maglev-1st
shzhang
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### Re: MagLev C Question Marathon
When we make the car, which one is important? Add more mass with the same speed or increase the speed with the same mass.
Schrodingerscat
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### Re: MagLev C Question Marathon
shzhang wrote:When we make the car, which one is important? Add more mass with the same speed or increase the speed with the same mass.
(For future reference, this question better belongs in the other thread, this forum is for sharing questions to study for the test). The event is no longer about speed, with a minimum time of 5 seconds, which considering the times less than a second with similar masses last year, even with much less propulsion, should still be very attainable with a decent propeller and motor. The main challenge now is to reliably slow the vehicle down to even slower times. However, if you are unable to make you car run in 5 seconds with the additional mass, it might not be worth the risk.
sercle
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### Re: MagLev C Question Marathon
twototwenty wrote:Where and when was the first maglev railway implemented?
1984 (The other user only answered the location)
Based on the test at the last competition, there seems to be a lot of questions on history, so:
What is the fastest speed ever reached by a MagLev train?
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### Re: MagLev C Question Marathon
I believe it's 500Km/h, or 361m/hr
09astro27nm
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### Re: MagLev C Question Marathon
Phys1cs wrote:I believe it's 500Km/h, or 361m/hr
That is correct, 581 Km/h (361 m/hr) by the Yamanashi Maglev Test Line in Japan.
Who had the first patent on the Maglev train?
twototwenty
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Joined: March 24th, 2011, 10:28 am | 1,314 | 4,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-21 | latest | en | 0.910856 |
https://socratic.org/questions/how-do-you-divide-using-synthetic-division-2x-3-4x-2-7x-5-x-3 | 1,623,548,913,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487598213.5/warc/CC-MAIN-20210613012009-20210613042009-00152.warc.gz | 499,368,401 | 6,040 | # How do you divide using synthetic division: (2x^3 - 4x^2 - 7x + 5)/(x - 3)?
Feb 21, 2017
The remainder is $= 2$ and the quotient is $= 2 {x}^{2} + 2 x - 1$
#### Explanation:
We perform the synthetic division
$\textcolor{w h i t e}{a a a a}$$3$$\textcolor{w h i t e}{a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a}$$- 7$$\textcolor{w h i t e}{a a a}$$5$
$\textcolor{w h i t e}{a a a a a a}$$\textcolor{w h i t e}{a a a}$$|$$\textcolor{w h i t e}{a a a}$$\textcolor{w h i t e}{a a a a a a a a}$$6$$\textcolor{w h i t e}{a a a a a a}$$6$$\textcolor{w h i t e}{a a a}$$- 3$
$\textcolor{w h i t e}{a a a a a a a a a a}$------------------------------------------------------------
$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a}$$\textcolor{w h i t e}{a a a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$\textcolor{red}{2}$
The remainder is $= 2$
The quotient is $= 2 {x}^{2} + 2 x - 1$
Verification, by using the remainder theorem
$f \left(3\right) = 54 - 36 - 21 + 5 = 59 - 57 = 2$ | 514 | 1,154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 38, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2021-25 | latest | en | 0.399201 |
https://howkgtolbs.com/convert/86.36-kg-to-lbs | 1,670,034,687,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00432.warc.gz | 347,059,208 | 12,113 | # 86.36 kg to lbs - 86.36 kilograms to pounds
Before we get to the more practical part - this is 86.36 kg how much lbs conversion - we are going to tell you few theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 86.36 kg to lbs? 86.36 kilograms it is equal 190.3912094632 pounds, so 86.36 kg is equal 190.3912094632 lbs.
## 86.36 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in short form SI).
Sometimes the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
Firstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. This definition was simply but totally impractical to use.
Later, in 1889 the kilogram was described using the International Prototype of the Kilogram (in short form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was switched by another definition.
Nowadays the definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It can be also divided to 100 decagrams and 1000 grams.
## 86.36 kilogram to pounds
You learned some facts about kilogram, so now let's go to the pound. The pound is also a unit of mass. We want to emphasize that there are not only one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we are going to to concentrate only on pound-mass.
The pound is used in the British and United States customary systems of measurements. To be honest, this unit is used also in another systems. The symbol of this unit is lb or “.
The international avoirdupois pound has no descriptive definition. It is exactly 0.45359237 kilograms. One avoirdupois pound is divided into 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 86.36 kg?
86.36 kilogram is equal to 190.3912094632 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 86.36 kg in lbs
The most theoretical section is already behind us. In this section we will tell you how much is 86.36 kg to lbs. Now you learned that 86.36 kg = x lbs. So it is time to get the answer. Just see:
86.36 kilogram = 190.3912094632 pounds.
This is an exact result of how much 86.36 kg to pound. It is possible to also round off this result. After it your outcome will be exactly: 86.36 kg = 189.992 lbs.
You learned 86.36 kg is how many lbs, so have a look how many kg 86.36 lbs: 86.36 pound = 0.45359237 kilograms.
Of course, this time it is possible to also round it off. After rounding off your outcome will be as following: 86.36 lb = 0.45 kgs.
We also want to show you 86.36 kg to how many pounds and 86.36 pound how many kg results in charts. Have a look:
We will start with a chart for how much is 86.36 kg equal to pound.
### 86.36 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
86.36 190.3912094632 189.9920
Now see a table for how many kilograms 86.36 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
86.36 0.45359237 0.45
Now you learned how many 86.36 kg to lbs and how many kilograms 86.36 pound, so it is time to move on to the 86.36 kg to lbs formula.
### 86.36 kg to pounds
To convert 86.36 kg to us lbs you need a formula. We are going to show you two versions of a formula. Let’s begin with the first one:
Amount of kilograms * 2.20462262 = the 190.3912094632 outcome in pounds
The first formula give you the most accurate result. In some cases even the smallest difference could be significant. So if you need a correct result - this formula will be the best solution to know how many pounds are equivalent to 86.36 kilogram.
So go to the second version of a formula, which also enables calculations to learn how much 86.36 kilogram in pounds.
The shorter formula is as following, see:
Number of kilograms * 2.2 = the result in pounds
As you can see, the second version is simpler. It can be better option if you need to make a conversion of 86.36 kilogram to pounds in easy way, for instance, during shopping. Just remember that your outcome will be not so exact.
Now we want to show you these two versions of a formula in practice. But before we will make a conversion of 86.36 kg to lbs we want to show you another way to know 86.36 kg to how many lbs without any effort.
### 86.36 kg to lbs converter
Another way to check what is 86.36 kilogram equal to in pounds is to use 86.36 kg lbs calculator. What is a kg to lb converter?
Converter is an application. It is based on longer formula which we gave you above. Thanks to 86.36 kg pound calculator you can quickly convert 86.36 kg to lbs. You only need to enter amount of kilograms which you want to convert and click ‘calculate’ button. You will get the result in a flash.
So try to calculate 86.36 kg into lbs using 86.36 kg vs pound calculator. We entered 86.36 as a number of kilograms. This is the result: 86.36 kilogram = 190.3912094632 pounds.
As you can see, this 86.36 kg vs lbs converter is so simply to use.
Now we can go to our chief topic - how to convert 86.36 kilograms to pounds on your own.
#### 86.36 kg to lbs conversion
We are going to begin 86.36 kilogram equals to how many pounds calculation with the first formula to get the most exact outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 190.3912094632 the outcome in pounds
So what have you do to know how many pounds equal to 86.36 kilogram? Just multiply amount of kilograms, in this case 86.36, by 2.20462262. It is 190.3912094632. So 86.36 kilogram is equal 190.3912094632.
You can also round off this result, for example, to two decimal places. It is exactly 2.20. So 86.36 kilogram = 189.9920 pounds.
It is time for an example from everyday life. Let’s convert 86.36 kg gold in pounds. So 86.36 kg equal to how many lbs? As in the previous example - multiply 86.36 by 2.20462262. It gives 190.3912094632. So equivalent of 86.36 kilograms to pounds, when it comes to gold, is 190.3912094632.
In this example you can also round off the result. This is the outcome after rounding off, this time to one decimal place - 86.36 kilogram 189.992 pounds.
Now we can move on to examples calculated using a short version of a formula.
#### How many 86.36 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 189.992 the outcome in pounds
So 86.36 kg equal to how much lbs? As in the previous example you need to multiply amount of kilogram, this time 86.36, by 2.2. Look: 86.36 * 2.2 = 189.992. So 86.36 kilogram is 2.2 pounds.
Let’s do another conversion with use of this formula. Now calculate something from everyday life, for instance, 86.36 kg to lbs weight of strawberries.
So calculate - 86.36 kilogram of strawberries * 2.2 = 189.992 pounds of strawberries. So 86.36 kg to pound mass is equal 189.992.
If you know how much is 86.36 kilogram weight in pounds and are able to convert it with use of two different versions of a formula, we can move on. Now we want to show you these results in charts.
#### Convert 86.36 kilogram to pounds
We know that results shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these results in tables for your convenience. Thanks to this you can easily make a comparison 86.36 kg equivalent to lbs results.
Begin with a 86.36 kg equals lbs table for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
86.36 190.3912094632 189.9920
And now let’s see 86.36 kg equal pound table for the second version of a formula:
Kilograms Pounds
86.36 189.992
As you see, after rounding off, when it comes to how much 86.36 kilogram equals pounds, the results are the same. The bigger number the more significant difference. Keep it in mind when you need to do bigger amount than 86.36 kilograms pounds conversion.
#### How many kilograms 86.36 pound
Now you know how to convert 86.36 kilograms how much pounds but we will show you something more. Do you want to know what it is? What about 86.36 kilogram to pounds and ounces calculation?
We are going to show you how you can calculate it little by little. Begin. How much is 86.36 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, in this case 86.36, by 2.20462262. So 86.36 * 2.20462262 = 190.3912094632. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this example there are 2 pounds.
To know how much 86.36 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.
So your result is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then final result will be equal 2 pounds and 33 ounces.
As you see, calculation 86.36 kilogram in pounds and ounces quite easy.
The last conversion which we will show you is conversion of 86.36 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert foot pounds to kilogram meters it is needed another formula. Before we give you this formula, look:
• 86.36 kilograms meters = 7.23301385 foot pounds,
• 86.36 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 86.36 foot pounds to kilograms meters you need to multiply 86.36 by 0.13825495. It is exactly 0.13825495. So 86.36 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 86.36 foot pounds will be exactly 0.14 kilogram meters.
We hope that this conversion was as easy as 86.36 kilogram into pounds calculations.
We showed you not only how to do a calculation 86.36 kilogram to metric pounds but also two another calculations - to know how many 86.36 kg in pounds and ounces and how many 86.36 foot pounds to kilograms meters.
We showed you also another solution to do 86.36 kilogram how many pounds calculations, it is using 86.36 kg en pound calculator. It is the best option for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.
We hope that now all of you are able to do 86.36 kilogram equal to how many pounds calculation - on your own or using our 86.36 kgs to pounds converter.
It is time to make your move! Calculate 86.36 kilogram mass to pounds in the best way for you.
Do you need to do other than 86.36 kilogram as pounds calculation? For example, for 5 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so easy as for 86.36 kilogram equal many pounds.
### How much is 86.36 kg in pounds
To quickly sum up this topic, that is how much is 86.36 kg in pounds , we gathered answers to the most frequently asked questions. Here you can see all you need to know about how much is 86.36 kg equal to lbs and how to convert 86.36 kg to lbs . Have a look.
How does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 86.36 kg to pound conversion formula . See it down below:
The number of kilograms * 2.20462262 = the result in pounds
How does the result of the conversion of 86.36 kilogram to pounds? The correct answer is 190.3912094632 lbs.
It is also possible to calculate how much 86.36 kilogram is equal to pounds with another, shortened type of the equation. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So now, 86.36 kg equal to how much lbs ? The result is 190.3912094632 lbs.
How to convert 86.36 kg to lbs quicker and easier? You can also use the 86.36 kg to lbs converter , which will make the rest for you and you will get a correct answer .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms. | 3,489 | 13,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.942742 |
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## Math Workshop
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## Science
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• asking questions and predicting outcomes about the changes in energy that occur when objects collide.
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• learning about waves to describe patterns in terms of amplitude and wavelength and that waves can cause objects to move.
• describing that light reflecting from objects and entering the eye allows objects to be seen
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Redhorse #1 Posted : Thursday, October 27, 2022 9:43:49 PM(UTC) Rank: NewbieGroups: Registered Users, SubscribersJoined: 3/16/2016(UTC)Posts: 3 HiCan anybody help me with this? I use two moving averages on my charts - a 10 and a 20 period exponential moving average based on closing prices. But i have just noticed that when i run an exploration with these values being output in two of the columns the values i get are totally different to those on the chart. The formulas i use in the exploration is just (Mov(C,10,E)) and (Mov(C,20,E)).Does anybody know what might be causing the value for the same variable to be different in an exploration than it is on the chart?
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MS Support #2 Posted : Thursday, October 27, 2022 10:42:05 PM(UTC) Rank: Advanced MemberGroups: Moderators, Registered, Registered Users, SubscribersJoined: 10/8/2010(UTC)Posts: 1,927Thanks: 85 timesWas thanked: 154 time(s) in 150 post(s) Originally Posted by: Redhorse HiCan anybody help me with this? I use two moving averages on my charts - a 10 and a 20 period exponential moving average based on closing prices. But i have just noticed that when i run an exploration with these values being output in two of the columns the values i get are totally different to those on the chart. The formulas i use in the exploration is just (Mov(C,10,E)) and (Mov(C,20,E)).Does anybody know what might be causing the value for the same variable to be different in an exploration than it is on the chart? Hello,The most likely issue here is that the Explorer by default is set to "Load Minimum Records" (this is useful for ensuring maximum speed of your exploration). This isn't a problem with most formula functions, but with exponential calculations in particular, these calculations carry a bit of historical data into the current values when you load more data. You don't necessarily need to load the exact amount of data that you load in a chart (I've heard recommendations of about 20% more data than the minimum periods in your formulas) but to rule out this issue you might want to try increasing the Explore Data Loading to something like 500 or 1250 records (You can try lower numbers if you want to keep exploration speeds maximized). Given the relatively small number of periods in your formulas you could probably just use 100 records and still be fine in getting matching results.Edited by user Thursday, October 27, 2022 10:43:29 PM(UTC) | Reason: Not specified
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# If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the
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If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the [#permalink] 27 May 2012, 16:26
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If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the followings is highest for (1/x\$1/y)\$(1/y\$1/x)?
A. x=1/2 and y=1/3
B. x=1/3 and y=1/4
C. x=1/4 and y=1/5
D. x=1/5 and y=1/4
E. x=1/4 and y=1/2
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Re: Quan questions and help in solving them [#permalink] 27 May 2012, 17:44
Expert's post
Please post in the correct forum. Moving to the ps subforum
Please provide the OA with the question. Thanks.
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Re: If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the [#permalink] 28 May 2012, 07:43
Hi,
Thanks for letting me know. If I need explanation for an answer, I copy and paste the question in the "search" box, hence I get solutions. But, there are some questions for which I am not able to find answer and explanation, please let me know where I should post the question for seeking solutions. There are many many threads in the PS subforum, I am not knowing where to post my questions. Please help me out!
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Re: If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the [#permalink] 28 May 2012, 07:47
Expert's post
sharmila79 wrote:
Hi,
Thanks for letting me know. If I need explanation for an answer, I copy and paste the question in the "search" box, hence I get solutions. But, there are some questions for which I am not able to find answer and explanation, please let me know where I should post the question for seeking solutions. There are many many threads in the PS subforum, I am not knowing where to post my questions. Please help me out!
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/, DS questions in the DS subforum: gmat-data-sufficiency-ds-141/ and general math questions in Math forum. No posting of PS/DS questions is allowed in the main Math forum.
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Re: If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the [#permalink] 28 May 2012, 07:57
Thanks. What is OA?
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Re: If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the [#permalink] 31 May 2012, 09:08
sharmila79 wrote:
If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the followings is highest for (1/x\$1/y)\$(1/y\$1/x)?
A. x=1/2 and y=1/3
B. x=1/3 and y=1/4
C. x=1/4 and y=1/5
D. x=1/5 and y=1/4
E. x=1/4 and y=1/2
What is the answer?
Is it D
It took some time to solve the question!!!
Re: If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the [#permalink] 31 May 2012, 09:08
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# If A\$B=A+B, if A>B and A\$B=B-A, if A<B, then which of the
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Aug-Nov 2020
## Recap
• Vector space $V$ over a scalar field $F= \mathbb{R}$ or $\mathbb{C}$
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• dim null $T+$ dim range $T=$ dim $V$
• Solution to $Ax=b$ (if it exists): $u+$ null$(A)$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space
• Eigenvalue $\lambda$ and Eigenvector $v$: $Tv=\lambda v$
• There is a basis w.r.t. which a linear map is upper-triangular
• If there is a basis of eigenvectors, linear map is diagonal w.r.t. it
• Inner products, norms, orthogonality and orthonormal basis
• There is an orthonormal basis w.r.t. which a linear map is upper-triangular
• Orthogonal projection: distance from a subspace
• Adjoint of a linear map: $\langle Tv,w\rangle=\langle v,T^*w\rangle$
• null $T=$ $($range $T^*)^{\perp}$
• Self-adjoint: $T=T^*$, Normal: $TT^*=T^*T$
• Complex/real spectral theorem: $T$ is normal/self-adjoint $\leftrightarrow$ orthonormal basis of eigenvectors
• Positive operators: self-adjoint with non-negative eigenvalues
• Isometries: normal with absolute value 1 eigenvalues
• Singular values/vectors of $T$: Eigenvalues/eigenvectors of $T^*T$
## Preliminaries
Unitary matrix
$n\times n$ matrix $V$ is said to be unitary if its columns are orthonormal.
1. Unitary matrices represent isometries
2. $VV^*=V^*V=I$ ($V^*$ is conjugate-transpose)
Rectangular diagonal matrices
$m\times n$ matrix $D$ with $(i,j)$-th element $d_{ij}$
Main diagonal of $D$: $(1,1),(2,2),\ldots$
$D$ is diagonal if the only nonzero values of $D$ are on the main diagonal
## Singular Value Decomposition (SVD)
$A$: $m\times n$ matrix
Singular values: eigenvalues of $A^*A$ or $AA^*$
Right-singular vectors: orthonormal eigenvectors of $A^*A$
Left-singular vectors: orthonormal eigenvectors of $AA^*$
SVD: There exist
1. an $m\times m$ unitary matrix $U$
2. an $n\times n$ unitary matrix $V$
3. an $m\times n$ diagonal matrix $D$
such that $A=UDV^*$.
$U$: columns are left-singular vectors
$V$: columns are right-singular vectors
$D$: singular values on main diagonal
## Example
$A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}$ (standard basis)
Singular values of $A$: $\sigma_1=9.508$, $\sigma_2=0.773$, $\sigma_3=0$
Right-singular vectors of $A$: $e_1=(0.429,0.566,0.704)$, $e_2=(0.805,0.112,-0.581)$, $e_3=(0.408,-0.816,0.408)$
Left-singular vectors of $A$: $f_1=(0.386,0.922)$, $f_2=(-0.922,0.386)$
$A=\begin{bmatrix} 0.386&-0.922\\ 0.922&0.386 \end{bmatrix}\begin{bmatrix} 9.508&0&0\\ 0&0.773&0 \end{bmatrix}\begin{bmatrix} 0.429&0.566&0.704\\ 0.805&0.112&-0.581\\ 0.408&-0.816&0.408 \end{bmatrix}$
Basis for $\mathbb{R}^3$: $e_1$, $e_2$, $e_3$; Basis for $\mathbb{R}^2$: $f_1$, $f_2$
$A$ in above basis: $\begin{bmatrix} 9.508&0&0\\ 0&0.773&0 \end{bmatrix}$ (diagonal)
## SVD: a restatement
$T:V\to W$, linear map
$B_V$: basis of $V$, $B_W$: basis of $W$
$M(T,B_V,B_W)$: matrix of $T$ w.r.t. $B_V$ and $B_W$
SVD: There exist
1. an orthonormal basis $B_V$ for $V$, and
2. an orthonormal basis $B_W$ for $W$
such that $M(T,B_V,B_W)$ is diagonal.
$B_V$: right-singular vectors; $B_W$: left-singular vectors
$M(T,B_V,B_W)$: singular values on diagonal
## Proof of SVD
Lemma: For $v\in V$, $\lVert Tv\rVert=\lVert\sqrt{T^*T}v\rVert$
Proof of Lemma
\begin{align} \langle Tv,Tv\rangle &= \langle T^*Tv,v\rangle\\ &= \langle \sqrt{T^*T}\sqrt{T^*T}v,v\rangle\\ &= \langle \sqrt{T^*T}v,\sqrt{T^*T}v\rangle\\ \end{align}
range $T=\{Tv:v\in V\}$, range $\sqrt{T^*T}=\{\sqrt{T^*T}v:v\in V\}$
Let $S:$ range $\sqrt{T^*T}\to$ range $T$ be s.t. $\sqrt{T^*T}v$ is mapped to $Tv$
Properties of $S$
1. $S$ is well-defined. If $\sqrt{T^*T}v_1=\sqrt{T^*T}v_2$, then $Tv_1=Tv_2$.
2. $S$ is linear. $S$ is one-to-one and onto.
3. For $u\in$ range $\sqrt{T^*T}$, $\lVert u\rVert=\lVert Su\rVert$.
4. For $u_1,u_2\in$ range $\sqrt{T^*T}$, $\langle u_1,u_2\rangle=\langle Su_1,Su_2\rangle$.
## Proof of SVD (continued)
$\{e_1,\ldots,e_n\}$: orthonormal eigenvector basis of $\sqrt{T^*T}$ (and $T^*T$)
$\sigma_1,\ldots,\sigma_n$: corresponding eigenvalues or singular values of $T$
Suppose $k=$ rank $\sqrt{T^*T}$; $\sigma_1$ to $\sigma_k$: nonzero.
$\{\sqrt{T^*T}e_1,\ldots,\sqrt{T^*T}e_k\}$: orthogonal basis of range $\sqrt{T^*T}$
$\{Te_1,\ldots,Te_k\}$: orthogonal basis of range $T$ (by Property 4)
$\sigma_i=\lVert\sigma_ie_i\rVert=\lVert\sqrt{T^*T}e_i\rVert=\lVert Te_i\rVert$
$\{f_1=\dfrac{1}{\sigma_1}Te_1,\ldots,f_k=\dfrac{1}{\sigma_k}Te_k\}$: orthonormal basis of range $T$
$\{f_1,\ldots,f_k,\ldots,f_m\}$: orthonormal extension to basis of $W$
$T$: diagonal w.r.t. basis $\{e_1,\ldots,e_n\}$ for $V$ and $\{f_1,\ldots,f_m\}$ for $W$
Singular values on diagonal
$TT^*f_i=\dfrac{1}{\sigma_i}TT^*Te_i=\dfrac{1}{\sigma_i}T(\sigma_i^2e_i)=\sigma_i^2f_i$
1. $T\leftrightarrow \sigma_1f_1\overline{e^T_1}+\cdots+\sigma_kf_k\overline{e^T_k}$
2. rank $T$: number of nonzero singular values
3. range $T$: orthonormal basis is $\{f_1,\ldots,f_k\}$
4. null $T$: orthonormal basis is $\{e_{k+1},\dots,e_n\}$ | 1,969 | 5,085 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-49 | latest | en | 0.527106 |
https://www.chemistrylibrary.org/2019/08/ideal-gas-equation.html | 1,716,986,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00376.warc.gz | 590,306,169 | 22,642 | ## Ideal Gas Equation
The three laws which we have learnt till now can be combined together in a single equation which is known as ideal gas equation.
At constant T and n; V ∝ 1/p Boyle’s Law
At constant p and n; V ∝ T Charles’ Law
At constant p and T; V ∝ n Avogadro Law
Thus,
Where R is proportionality constant. On rearranging the equation we obtained
R is called gas constant. It is same for all gases. Therefore it is also called Universal Gas Constant. Equation is called ideal gas equation.
Above Equation shows that the value of R depends upon units in which p, V and T are measured. If three variables in this equation are known, fourth can be calculated. From this equation we can see that at constant temperature and pressure n moles of any gas will have the same volume because
and n, R, T and p are constant. This equation will be applicable to any gas, under those conditions when behavior of the gas approaches ideal behavior. Volume of one mole of an ideal gas under STP conditions (273.15 K and 1 bar pressure) is 22.710981 L mol-1.
At STP conditions used earlier (0 C and 1 atm pressure). Value of R is 8.20578 10-2 L atm K-1 mol-1.
Ideal gas equation is a relation between four variables and it describes the state of any gas, therefore, it is also called equation of state.
Let us now go back to the ideal gas equation. This is the relationship for the simultaneous variation of the variables. If temperature, volume and pressure of a fixed amount of gas vary from T1, V1 and p1 to T2, V2 and p2 then we can write
The above said equation is a very useful equation. If out of six, values of five variables are known, the value of unknown variable can be calculated from the above equation. This equation is also known as Combined gas law. | 437 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-22 | latest | en | 0.936297 |
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Post subject: TuconesPosted: Sat Feb 18, 2012 9:24 am
Joined: Sat Feb 04, 2012 9:33 am
Hello everyone.
I got this idea for a puzzle. I don't know whether this wasn't designed before. If anyone is interested in bringing this into physical form, let me know.
I don't have the knowledge or wherewithal to undertake the task. The model was sketched by a friend of mine. There is no internal mechanism.
Thanks
Peter
Attachments: TUC1.jpg [ 106.29 KiB | Viewed 976 times ] TUC2.jpg [ 168.72 KiB | Viewed 976 times ] TUC3.jpg [ 147.97 KiB | Viewed 976 times ] TUC4.jpg [ 97.69 KiB | Viewed 976 times ]
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Post subject: Re: TuconesPosted: Sat Feb 18, 2012 1:01 pm
Joined: Sun Feb 13, 2011 9:16 am
Location: Indiana
Very interesting! I believe it could be made by shapemodding a 4x4x4. Each "1/4th cone" on the top and bottom would be the equivalent of a single center piece, a wing edge from each of two neighboring faces, and a corner. The center pieces would be a little more difficult, but you would have to somehow split the edges.
Attachment:
ConeCube.png [ 16.13 KiB | Viewed 898 times ]
Green is what would become the "1/4th cone" pieces.
Red would be one center piece and blue would be another.
If a SQ-1 core had an additional cut through the core perpendicular to the existing one I think that would be possible by changing the shape of the pieces.
Jack
EDIT: Tried making a physical version of this by splitting the edges. That worked fine, but face turns can't be made when the horizontal slice layers are in a 45° turn because the center pieces cannot rotate at that place in the core. Even if they could, I noticed the issue of the core not turning with slice layers at times, meaning the track for a vertical 180° move would be blocked. Seems like it would be a pretty simple shapeways build with a ball core though. Just modifications to centers and edges and some bandaging.
_________________
-Jack Lopez
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Post subject: Re: TuconesPosted: Sat Feb 18, 2012 5:49 pm
Joined: Sun Nov 23, 2008 2:18 am
I am not sure on the number of cuts. Is it:
-Cones divided in 4 with the middle layer in 8 wedges.
or
-Cones divided in 6 with the middle layer in 12 wedges.
?
I am thinking the 4-8 version could be made from shape modding a 2*2*2 and adding sliding pieces for the middle layer. The 6-12 version could be done similarly starting with a Rubik's UFO. I am thinking that a Masterball might could be modified into a 8-16 variant using similar principles.
_________________
I pledge allegiance to the whole of humanity, and to the world in which we live: one people under the heavens, indivisible, with Liberty and Equality for all.
My Shapeways Shop
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Post subject: Re: TuconesPosted: Sat Feb 18, 2012 7:53 pm
Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
Looks like a shape-mod of a Turbo Mind Twister.
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Post subject: Re: TuconesPosted: Sun Feb 19, 2012 8:01 am
Joined: Sat Feb 04, 2012 9:33 am
You are right Jared. It is a shape-mod of a Turbo Mind Twister.
I didn't realize this. Oh well, it's back to the drawing board.
It's hard to invent a new twisty puzzle these days. Everything
was already done. I guess that I will have to switch my brain
into overdrive and invent something really new.
Thanks for the feedback,
Peter
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All times are UTC - 5 hours | 1,034 | 3,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.952587 |
https://cz39133.com/how-youre-able-own-on-the-web-in-the-lottery-niche-for-pennies/ | 1,709,122,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00104.warc.gz | 198,921,381 | 22,675 | How You’re Able Own On The Web In The Lottery Niche For Pennies
It’s not an overstatement to say that very few people have the right idea about winning the lottery. As opposed to adopting the right lottery winning strategy or system, many believe that winning a lottery is purely a matter of luck, blessing from the ancestors etc. Naturally, these are the few myths which prevent one from winning.
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If you are to cover a associated with online lottery website games at one time, you’ll never be able to study the game well. That will affect the chances of you winning the lottery.
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It was simple to accomplish that I felt able to build 30-40 websites a month without much effort. A lot time went on the better I got at Advertising. Soon Being making countless dollars thirty day period. I became such an advocate of Affiliate Marketing that Began school comprehend more upon it with Full Sail Secondary education. If you’ve been considering solutions to earning a return online along with want to fret about the of inventory of product, returns, and customer complaints I definitely recommend being familiar with how that need be and Associate. | 616 | 2,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.971553 |
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lecture07ps
# lecture07ps - Sauder School of Business Finance Division...
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Sauder School of Business COMM 374 Jan-Apr 2010 Finance Division Jan Bena Problem Set 7 - Solution Notes 1. (a) According to the CAPM, investors expect a return of E ( R ) = 5% + 1 . 4 × 6% = 13 . 4% . (b) The expected return on the security is given by E ( R ) = E ( D 1 ) + ( E ( P 1 ) - P 0 ) P 0 , where P 0 is today’s price, E ( P 1 ) is next year’s expected price, and E ( D 1 ) is next year’s expected dividend. We know that E ( R ) = 13 . 4%, E ( D 1 ) = \$0, P 0 = \$35. We get E ( P 1 ) = \$39 . 69. (c) Now we have E ( D 1 ) = \$2. We get E ( P 1 ) = \$37 . 69. 2. (a) The statement is FALSE. Suppose that stocks A and B are independent. Then a 50/50 portfolio of A and B has expected return 0 . 5 × 10% + 0 . 5 × 12% = 11% and standard deviation p 0 . 5 2 × (15%) 2 + 0 . 5 2 × (13%) 2 = 9 . 92% . An investor may be willing to hold the 50/50 portfolio, since it has lower standard deviation than both A and B. (b) The statement is FALSE. The equally weighted portfolio is generally not on the portfolio frontier. In the international diversification example presented in the lecture notes, for instance, the equally weighted portfolio of the seven countries is inside the frontier. The reason was that the countries have different risk char-
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Ask a homework question - tutors are online | 547 | 1,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-13 | latest | en | 0.873416 |
https://boards.straightdope.com/t/how-long-does-it-take-a-dead-body-to-cool-to-room-temperature/495339 | 1,632,564,237,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057615.3/warc/CC-MAIN-20210925082018-20210925112018-00580.warc.gz | 185,420,674 | 6,423 | # How long does it take a dead body to cool to room temperature?
Assume it’s the body of an adult woman in good health who died violently and suddenly; that she is outside when she dies and the body is not moved; and that it’s a late summer/early fall day. It’s been raining before she dies, so the temperature is, I’d guess, about 65 degrees at most; and though the rain has stopped the sky is still overcast. No wind blowing.
How long would it be before the body is cool to the touch?
Oddly enough, not a single one of those results–not ONE–answers the OP’s question. The fact is there is no simple answer. Cooling rate follows a roughly exponential curve, but the exact shape of the cooling curve depends on a huge number of factors, including ambient temperature (which is rarely stable for more than a few hours outdoors), the type of surface the body is on, whether it’s partially or wholly immersed in water and the temp of the water if it is, clothing, precipitation, wind speed, body morphology and mass, and even the position (curled up into a ball, prone, spread-eagle, etc.) There used to be a rule-of-thumb which said a body cools about 1.6 degrees F per hour but this is wildly inaccurate for all but the roughest guesstimates and is never used by pathologists to estimate TOD anymore.
So, u = phail.
It does give rough estimates, (which vary by 12 hours) which would be useful when one cannot know exactly how long it would take, like you say. Additionally, you gave a pedantic non-answer which helps even less. Not saying you were wrong, just saying my answer was kind of useful, unlike yours.
Though I was vexed at the form your answer gave, it was, along with Q.E.D.'s, helpful. Really I was wondering whether a body that had been dead for no more than an hour yet would be cold.
Why exactly do you need to know? :eek:
I just murdered Catherine the Great. I need to know how long I have before I have to take her back to 1761 so her body can be found, as I am having difficulty getting the horse to cooperate in … well, never mind.
>Really I was wondering whether a body that had been dead for no more than an hour yet would be cold.
Less than an hour? Depends on what you mean by cold. You ever touch a sick person whose body temperature has fallen only a few degrees? They feel very cold to the touch but theyre about 93-95 or so degrees. Ever get in bed with your SO and feel their cold feet? Same thing.
Our sense of temperature is so sharp that even a few degrees feels very cold. A dead body lying on concrete for an hour would feel freezing to you. Doubly so in the winter.
The combination of my last post and your user name lead me to expect a different, far snarkier answer.
At the very least, I’d expect the dead woman’s clothing to play a significant role in how long it takes for the corpse to cool. If it was raining, did she have on a raincoat?
:: peers over glasses at imaginary corpse of CtG currently being buggered by a shetland pony::
She appears to be wearing some sort of fur cape. I think it’s bearskin.
More seriously, I’m writing a story in which someone touches a corpse that’s been dead for an hour or so; she’d have been coatless, as she gave her coat to the person currently touching her. I asked the question because I wrote “let go of the cold hand” and then, being me, started worrying. | 774 | 3,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.975753 |
https://brilliant.org/discussions/thread/cauchy-schwarz-training/?sort=new | 1,606,345,759,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141184870.26/warc/CC-MAIN-20201125213038-20201126003038-00495.warc.gz | 216,039,324 | 19,857 | # Cauchy-Schwarz Training
The biggest real value of the expression $\cos{x} \cdot ( 2\sin{x} + 4\cos{x})$ can be written as $a+ \sqrt{b}$, where $a$ and $b$ are positive square-free integers.
The smallest real value of the expression $ay^2 + \dfrac{b}{y^2}$ can be written as $c \cdot \sqrt{d}$, where $c$ and $d$ are positive square-free integers. Restrict $x$ and $y$ to the real numbers.
1. Show that $ab = bc = d$.
2. Find the monic polynomial $P(x)$ of fourth degree that has $a,b,c,d$ as its roots.
3. Find $P(x)$ 's smallest real value.
Note by Guilherme Dela Corte
6 years ago
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## Comments
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Write a comment or ask a question...Find K so that U and V are orthogonal where U=(2,3K,-4,1,-5) and V=(6,-1,3,7,2K)
- 5 years, 6 months ago
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Waht's the question which has no answer ?
- 6 years ago
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All questions have answers! However, #1 is a proof, and thus has not a closed-form value to submit.
- 6 years ago
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Can anyone tell me about Cauchy-Schwarz inequality? I know the AM-GM inequality but not this. Please illustrate some simple examples too (I am new to this inequality.).
Thanks in advance!
- 6 years ago
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If you type in Cauchy-Schwarz in the search box there would be a featured community post on Cauchy-Schwarz. There's a lot of information there.
- 6 years ago
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The first expression can be reduced to $\sin 2x+2\cos 2x+2$.
$(\sin 2x+2\cos 2x)^{2} \leq (\sin^{2} 2x+ \cos^{2} 2x)(1+4)=5$ hence the maximum value is $2+\sqrt {5}$. Equality can occur when $\tan x=0.5$.
Now $c, d$ can be found by AM-GM. However, since Cauchy-Schwarz needs to be used, we do it in a slightly different way.
$(2y^{2}+\frac {5}{y^{2}})(\frac {5}{y^{2}}+2y^{2}) \geq (\sqrt {10}+\sqrt {10})^{2}=40$ hence $2y^{2}+\frac {5}{y^{2}} \geq \sqrt {40}=2\sqrt {10}$.
Hence $a=2, b=5, c=2, d=10$.
1. Clearly $2×5=5×2=10$.
2. $P (x)=(x-2)^{2}(x-5)(x-10)$
3. Using Cauchy-Schwarz rather than calculus is difficult, but here's a solution:
By Cauchy-Schwarz, $(a^{2}+b^{2})(b^{2}+a^{2}) \geq (ab+ba)^{2}$ hence $a^{2}+b^{2} \geq 2ab$ for reals $a, b$.
Thus for all positive reals $a, b, c, d$, $a^{4}+b^{4}+c^{4}+d^{4} \geq 2a^{2} b^{2}+2c^{2} d^{2} \geq 4abcd$ by applying the above twice. We will use this fact.
Clearly the polynomial can have negative values. Just fit in x=7.
Now the only way to get negative values is to have $(x-5)(x-10)$ negative, or $5 < x <10$. Hence $x-2, x-5, 10-x$ are positive. So an easier problem would be to find the maximum of $(x-2)(x-2)(x-5)(10-x)$.
Now, using the fact above, $r(x-2)r (x-2)(1-2r)(x-5)(10-x) \leq (\frac {r(x-2)+r (x-2)+(1-2r)(x-5)+10-x}{4})^{4}=(\frac {6r+5}{4})^{4}$. So the maximum, if equality can occur, is $\frac {(6r+5)^{4}}{256r^{2}(1-2r)}$.
In particular, when $r=\frac {21-\sqrt {321}}{12}$, we get the maximum as that is the only case equality can occur. (Remember that the corresponding terms in Cauchy-Schwarz have to be related by a common ratio for equality.) The minimum is the negative of the maximum and is approximately -223. Equality can occur at $x=\frac {64-4\sqrt {321}}{17-\sqrt {321}}$.
- 6 years ago
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Can you explain adequately to your friends:
a) Why is the "maximum" $\dfrac{(6r+5)^4}{256r^2 (1-2r)}$ ? How did attain it?
b) Why is $r = \dfrac{21 - \sqrt{321}}{12}$ ?
- 6 years ago
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a) it is by rearranging the terms in the inequality. b) By Cauchy, equality can occur only when r(x-2)=(1-2r)(x-5)=10-x. Solving gives r.
- 6 years ago
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Challenge: Can the third question be answered by Cauchy-Schwarz?
- 6 years ago
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http://www.codeproject.com/Articles/19323/Image-Recognition-with-Neural-Networks?fid=431623&df=10000&mpp=10&noise=1&prof=True&sort=Position&view=Expanded&spc=Relaxed&select=3752691&fr=22&PageFlow=FixedWidth | 1,430,102,254,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246656965.63/warc/CC-MAIN-20150417045736-00294-ip-10-235-10-82.ec2.internal.warc.gz | 416,115,989 | 31,419 | 11,412,302 members (62,222 online)
# Image Recognition with Neural Networks
, 30 Oct 2007 CPOL
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This article contains a brief description of BackPropagation Artificial Neural Network and its implementation for Image Recognition
## Introduction
Artificial Neural Networks are a recent development tool that are modeled from biological neural networks. The powerful side of this new tool is its ability to solve problems that are very hard to be solved by traditional computing methods (e.g. by algorithms). This work briefly explains Artificial Neural Networks and their applications, describing how to implement a simple ANN for image recognition.
## Background
I will try to make the idea clear to the reader who is just interested in the topic.
## About Artificial Neural Networks (ANNs)
Artificial Neural Networks (ANNs) are a new approach that follow a different way from traditional computing methods to solve problems. Since conventional computers use algorithmic approach, if the specific steps that the computer needs to follow are not known, the computer cannot solve the problem. That means, traditional computing methods can only solve the problems that we have already understood and knew how to solve. However, ANNs are, in some way, much more powerful because they can solve problems that we do not exactly know how to solve. That's why, of late, their usage is spreading over a wide range of area including, virus detection, robot control, intrusion detection systems, pattern (image, fingerprint, noise..) recognition and so on.
ANNs have the ability to adapt, learn, generalize, cluster or organize data. There are many structures of ANNs including, Percepton, Adaline, Madaline, Kohonen, BackPropagation and many others. Probably, BackPropagation ANN is the most commonly used, as it is very simple to implement and effective. In this work, we will deal with BackPropagation ANNs.
BackPropagation ANNs contain one or more layers each of which are linked to the next layer. The first layer is called the "input layer" which meets the initial input (e.g. pixels from a letter) and so does the last one "output layer" which usually holds the input's identifier (e.g. name of the input letter). The layers between input and output layers are called "hidden layer(s)" which only propagate the previous layer's outputs to the next layer and [back] propagates the following layer's error to the previous layer. Actually, these are the main operations of training a BackPropagation ANN which follows a few steps.
A typical BackPropagation ANN is as depicted below. The black nodes (on the extreme left) are the initial inputs. Training such a network involves two phases. In the first phase, the inputs are propagated forward to compute the outputs for each output node. Then, each of these outputs are subtracted from its desired output, causing an error [an error for each output node]. In the second phase, each of these output errors is passed backward and the weights are fixed. These two phases is continued until the sum of [square of output errors] reaches an acceptable value.
## Implementation
The network layers in the figure above are implemented as arrays of structs. The nodes of the layers are implemented as follows:
```[Serializable]
struct PreInput
{
public double Value;
public double[] Weights;
};
[Serializable]
struct Input
{
public double InputSum;
public double Output;
public double Error;
public double[] Weights;
};
[Serializable]
struct Hidden
{
public double InputSum;
public double Output;
public double Error;
public double[] Weights;
};
[Serializable]
struct Output<T> where T : IComparable<T>
{
public double InputSum;
public double output;
public double Error;
public double Target;
public T Value;
};```
The layers in the figure are implemented as follows (for a three layer network):
```private PreInput[] PreInputLayer;
private Input[] InputLayer;
private Hidden[] HiddenLayer;
private Output<string>[] OutputLayer;```
Training the network can be summarized as follows:
• Apply input to the network.
• Calculate the output.
• Compare the resulting output with the desired output for the given input. This is called the error.
• Modify the weights for all neurons using the error.
• Repeat the process until the error reaches an acceptable value (e.g. error < 1%), which means that the NN was trained successfully, or if we reach a maximum count of iterations, which means that the NN training was not successful.
It is represented as shown below:
```void TrainNetwork(TrainingSet,MaxError)
{
while(CurrentError>MaxError)
{
foreach(Pattern in TrainingSet)
{
ForwardPropagate(Pattern);//calculate output
BackPropagate()//fix errors, update weights
}
}
}```
This is implemented as follows:
```public bool Train()
{
double currentError = 0;
int currentIteration = 0;
NeuralEventArgs Args = new NeuralEventArgs() ;
do
{
currentError = 0;
foreach (KeyValuePair<T, double[]> p in TrainingSet)
{
NeuralNet.ForwardPropagate(p.Value, p.Key);
NeuralNet.BackPropagate();
currentError += NeuralNet.GetError();
}
currentIteration++;
if (IterationChanged != null && currentIteration % 5 == 0)
{
Args.CurrentError = currentError;
Args.CurrentIteration = currentIteration;
IterationChanged(this, Args);
}
} while (currentError > maximumError && currentIteration <
maximumIteration && !Args.Stop);
if (IterationChanged != null)
{
Args.CurrentError = currentError;
Args.CurrentIteration = currentIteration;
IterationChanged(this, Args);
}
if (currentIteration >= maximumIteration || Args.Stop)
return false;//Training Not Successful
return true;
}```
Where `ForwardPropagate(..)` and `BackPropagate()` methods are as shown for a three layer network:
```private void ForwardPropagate(double[] pattern, T output)
{
int i, j;
double total;
//Apply input to the network
for (i = 0; i < PreInputNum; i++)
{
PreInputLayer[i].Value = pattern[i];
}
//Calculate The First(Input) Layer's Inputs and Outputs
for (i = 0; i < InputNum; i++)
{
total = 0.0;
for (j = 0; j < PreInputNum; j++)
{
total += PreInputLayer[j].Value * PreInputLayer[j].Weights[i];
}
InputLayer[i].InputSum = total;
InputLayer[i].Output = F(total);
}
//Calculate The Second(Hidden) Layer's Inputs and Outputs
for (i = 0; i < HiddenNum; i++)
{
total = 0.0;
for (j = 0; j < InputNum; j++)
{
total += InputLayer[j].Output * InputLayer[j].Weights[i];
}
HiddenLayer[i].InputSum = total;
HiddenLayer[i].Output = F(total);
}
//Calculate The Third(Output) Layer's Inputs, Outputs, Targets and Errors
for (i = 0; i < OutputNum; i++)
{
total = 0.0;
for (j = 0; j < HiddenNum; j++)
{
total += HiddenLayer[j].Output * HiddenLayer[j].Weights[i];
}
OutputLayer[i].InputSum = total;
OutputLayer[i].output = F(total);
OutputLayer[i].Target = OutputLayer[i].Value.CompareTo(output) == 0 ? 1.0 : 0.0;
OutputLayer[i].Error = (OutputLayer[i].Target - OutputLayer[i].output) *
(OutputLayer[i].output) * (1 - OutputLayer[i].output);
}
}
private void BackPropagate()
{
int i, j;
double total;
//Fix Hidden Layer's Error
for (i = 0; i < HiddenNum; i++)
{
total = 0.0;
for (j = 0; j < OutputNum; j++)
{
total += HiddenLayer[i].Weights[j] * OutputLayer[j].Error;
}
HiddenLayer[i].Error = total;
}
//Fix Input Layer's Error
for (i = 0; i < InputNum; i++)
{
total = 0.0;
for (j = 0; j < HiddenNum; j++)
{
total += InputLayer[i].Weights[j] * HiddenLayer[j].Error;
}
InputLayer[i].Error = total;
}
//Update The First Layer's Weights
for (i = 0; i < InputNum; i++)
{
for(j = 0; j < PreInputNum; j++)
{
PreInputLayer[j].Weights[i] +=
LearningRate * InputLayer[i].Error * PreInputLayer[j].Value;
}
}
//Update The Second Layer's Weights
for (i = 0; i < HiddenNum; i++)
{
for (j = 0; j < InputNum; j++)
{
InputLayer[j].Weights[i] +=
LearningRate * HiddenLayer[i].Error * InputLayer[j].Output;
}
}
//Update The Third Layer's Weights
for (i = 0; i < OutputNum; i++)
{
for (j = 0; j < HiddenNum; j++)
{
HiddenLayer[j].Weights[i] +=
LearningRate * OutputLayer[i].Error * HiddenLayer[j].Output;
}
}
}```
## Testing the App
The program trains the network using bitmap images that are located in a folder. This folder must be in the following format:
• There must be one (input) folder that contains input images [*.bmp].
• Each image's name is the target (or output) value for the network (the pixel values of the image are the inputs, of course) .
As testing the classes requires to train the network first, there must be a folder in this format. "PATTERNS" and "ICONS" folders [depicted below] in the Debug folder fit this format.
## History
• 30th September, 2007: Simplified the app
• 24th June, 2007: Initial Release
## About the Author
Software Developer (Senior)
Turkey
Has BS degree on CS, working as SW engineer at istanbul.
## Comments and Discussions
Increase probability RonZohan 20-Jul-11 4:30
Problem with self coding spider853 9-Jul-11 16:10
Need your guidance... apepe 23-Jun-11 17:55
Re: Need your guidance... Murat Firat 24-Jun-11 9:56
Re: Need your guidance... apepe 25-Jun-11 17:32
Re: Need your guidance... Murat Firat 27-Jun-11 1:51
help please!!!!! asmita5 29-May-11 8:10
Re: help please!!!!! Bikash_coder 16-Jun-11 0:09
Re: help please!!!!! asmita5 16-Jun-11 2:48
Re: help please!!!!! Murat Firat 16-Jun-11 3:17
How do I change to recognize the binary pattern shamlen 12-May-11 17:30
Re: How do I change to recognize the binary pattern merovingian18 29-Mar-12 5:47
How i can increase hidden node in BP1Layer.cs Bikash_coder 16-Feb-11 23:45
Re: How i can increase hidden node in BP1Layer.cs Murat Firat 20-Feb-11 21:35
layer 3 better or not Member 1964241 10-Feb-11 22:17
Re: layer 3 better or not Murat Firat 20-Feb-11 21:34
Sir,In your neuraldemo.cs file in InitializeSettings() function there a value is initialized in the textbox. the line is given below: textBoxInputUnit.Text = ((int)((double)(networkInput + NumOfPatterns) * .33)).ToString();here why you multiply (networkInput + NumOfPatterns) with .33can u tell me from where u get .33and for --->textBoxHiddenUnit.Text = ((int)((double)(networkInput + NumOfPatterns) * .11)).ToString();from where .11 comethanks in advancedbikash
great work... :) kireina_3012 29-Jan-11 11:00
Re: great work... :) Bikash_coder 30-Jan-11 1:17
Re: great work... :) kireina_3012 1-Feb-11 8:59
Re: great work... :) Murat Firat 2-Feb-11 10:42
Target output SeasickSailor 20-Dec-10 3:14
Displaying the image [modified] swathi6589 8-Oct-10 3:56
Last Visit: 31-Dec-99 19:00 Last Update: 26-Apr-15 12:37 Refresh 1234567 Next »
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In the mathematical context, the perimeter is the line that surrounds a plane figure with two dimensions. Put, it is the measure of the space a shape will occupy. The perimeter of a circular figure is its circumference. What is the perimeter of other two dimensional shapes that you know? Attempt this quiz now!
• 1.
### What mathematical operation is exclusively performed in obtaining the perimeter of a triangle?
• A.
• B.
Subtraction
• C.
Multiplication
• D.
Division
Explanation
To obtain the perimeter of a triangle, we add the lengths of all three sides. This is because the perimeter represents the total distance around the triangle, and adding the lengths of the sides gives us this total distance. Therefore, addition is the mathematical operation exclusively performed in obtaining the perimeter of a triangle.
Rate this question:
• 2.
### Who was the first to describe the perimeter of polygons?
• A.
Pythagoras
• B.
Plato
• C.
Archimedes
• D.
Aristotle
C. Archimedes
Explanation
Archimedes was the first to describe the perimeter of polygons. He made significant contributions to mathematics and geometry, including the calculation of the circumference of a circle and the measurement of the perimeter of various polygons. Archimedes developed rigorous mathematical proofs and formulas, and his work laid the foundation for many mathematical concepts that are still used today.
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• 3.
### Which of these refers to the perimeter of a circle?
• A.
Circumference
• B.
• C.
Diameter
• D.
Segment
A. Circumference
Explanation
The correct answer is Circumference. The circumference of a circle refers to the distance around the outer edge of the circle. It can be calculated using the formula 2πr, where r is the radius of the circle. The circumference is an important measurement when dealing with circles as it helps in determining the length of arcs, areas, and other properties of circles.
Rate this question:
• 4.
### What is the basis of isoperimetric problem?
• A.
Perimeter
• B.
• C.
Length
• D.
Area
D. Area
Explanation
The basis of the isoperimetric problem is the area. The isoperimetric problem involves finding the shape with the maximum area for a given perimeter. This problem has been studied in mathematics for centuries and has applications in various fields such as physics and engineering. By focusing on the area, mathematicians aim to find the most efficient shape in terms of maximizing the enclosed space while keeping the perimeter constant.
Rate this question:
• 5.
### Which of these is not needed to determine the perimeter of a circle?
• A.
• B.
Constant (Pi)
• C.
Inscribed angle
• D.
Diameter
C. Inscribed angle
Explanation
The perimeter of a circle can be determined by using either the radius or the diameter. The constant Pi is also necessary as it represents the ratio of the circumference of a circle to its diameter. However, the inscribed angle is not needed to determine the perimeter of a circle. The inscribed angle refers to an angle formed by two chords of a circle that have a common endpoint on the circle. While the inscribed angle is a concept related to circles, it is not directly used to calculate the perimeter.
Rate this question:
• 6.
### Which of these fields does not require the digits of constant Pi ?
• A.
Algorithmics
• B.
Computer science
• C.
Geostatistics
• D.
Mathematical analysis
C. Geostatistics
Explanation
Geostatistics is the field that does not require the digits of the constant Pi. Geostatistics is a branch of statistics that deals with the analysis and interpretation of spatial data, particularly in the earth sciences. While Pi is a mathematical constant that represents the ratio of a circle's circumference to its diameter, it is not directly relevant or necessary in the study of geostatistics.
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• 7.
• A.
Three
• B.
Four
• C.
Six
• D.
Nine
B. Four
• 8.
### How many sides does a square have?
• A.
Six sides
• B.
Eight sides
• C.
Nine sides
• D.
Four sides
D. Four sides
Explanation
A square has four sides because it is a quadrilateral with four equal sides. Each side of a square is also perpendicular to the adjacent sides, forming right angles. Therefore, a square has four sides.
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• 9.
### What simple shape is characterized by a segment?
• A.
Circle
• B.
Triangle
• C.
Square
• D.
Hexagon
A. Circle
Explanation
A circle is characterized by a segment because a segment is a part of a circle that is formed by a chord and its endpoints. A segment is a straight line that connects two points on the circumference of the circle, dividing it into two parts. Therefore, a circle is the simple shape that is characterized by a segment.
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• 10.
### What parameter is most important for determining the perimeter of a circle?
• A.
Diameter
• B.
• C.
Length
• D.
Segment
Explanation
The radius of a circle is the distance from the center of the circle to any point on its circumference. Since the perimeter of a circle is the distance around its outer edge, the radius is the most important parameter for determining the perimeter. The diameter is also related to the perimeter, as it is equal to twice the radius, but the radius is the primary parameter used in calculating the perimeter.
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Related Topics | 1,383 | 6,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-18 | latest | en | 0.918558 |
https://www.doubtnut.com/qna/644632291 | 1,719,017,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862189.36/warc/CC-MAIN-20240621223321-20240622013321-00604.warc.gz | 657,351,942 | 34,276 | # The galvanometer deflection , when key K1 is closed but K2 is open equals θ0 (see figure ) On closing k2 also and adjusting R2 to 5Ω , the deflection in galvanometer becomes θ05 the resistnce of the galvanometer is then , given by [Neglect the internal resistance of battery]:
Video Solution
Text Solution
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## Deflection in galvanometer is proportional to galvanoeter current θ0∝ig⇒θ0K=ig Situation 1 Kθ0=v220+R ………(i) Situation 2 Kθ05=v(220+5R5+R)×55+R From (i) and (2) R=22Ω.
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 410 | 1,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.872565 |
http://www.cs.uky.edu/~keen/115/labs/lab3-team.html | 1,526,942,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864558.8/warc/CC-MAIN-20180521220041-20180522000041-00060.warc.gz | 367,235,238 | 2,775 | ## CS 115 Lab 3 Conditionals and Test Cases (Team)
### 80 Points
Educational goals of this lab - verify that every student can
• understand the testing process
• write a program to use some conditional operations
• nest if statements
### Submit your files with this link. Use the Lab 3 menu choice, and "Code" for the design/program and "Other" for the test plan text file.
INSTRUCTIONS:
• (80 points) Team Problem: Let the user go on an adventure
• The program presents the user with some choices and shows them the results.
• You will need to use boolean variables (flags) to know whether the player lived or died, to know whether the player got the gold or not and whether they got the magic ring or not.
Sample run
```Let's go on an adventure
You are in a dark cave. You see two exits in front of you
1. a door
2. a tunnel
> 2
You see a big troll holding a club
You can choose to:
1. say Hello
2. attack it
> 1
The troll is friendly and gives you a magic ring
Game over!
You got out alive
Inventory:
You have a magic ring
```
Another sample run
```Let's go on an adventure
You are in a dark cave. You see two exits in front of you
1. a door
2. a tunnel
> 1
You see a big sack tied at the top.
You can choose to:
1. pick it up
2. open it
> 1
It grows a big set of teeth and eats you!
Game over!
You died
```
• The other possible actions are:
• You attack the troll- you get flattened.
• You open the bag- you get 500 pieces of gold.
If you entered an invalid choice at any point, the message should be "Invalid choice".
You must then show the message "Game over!".
The possible outcomes at the end of the game are:
1. You got out alive. Inventory: (Either "you have 500 pieces of gold" or "you have a magic ring")
2. You died. (This happens if you attack the troll, pick up the bag or give an invalid choice)
If you died in the game, the Inventory is NOT mentioned.
• (14 points) Test Plan In a text file, put what should go in each of the lettered blanks. Please label each answer. Some blanks have lots of possible answers.
NOTE: You get HALF of these points for filling out the blanks correctly. You get the other HALF of the points when your program actually runs and gets the correct expected output.
DescriptionInputs Expected Output(s)
First Second
Door, pick up sack1 1 It grows teeth and eats you, Game over!, You died
Door, open sack1 2 You get 500 gp, Game over!, You got out alive, Inventory: you have 500 pieces of gold
__A.__ 2 __B.__ __C.__
__D.__ 2 __E.__ __F.__
Integer input, invalid choices__G.__ doesn't matter Invalid choice, Game over! You died.
• Submit your Test Plan with the link above and Lab 3 and Other menu choices. You can call the text file testcases3.txt.
• (22 points) Take the design given below, paste it into a file called adventure3.py and finish the DESIGN. If statements have their bodies/blocks indented in the design too. All lines should be comments. Note that the messages shown must be in the order given. Do not rearrange them.
```#Team x, Section x, Members present
#Purpose - to take the user on an adventure
#Pre-conditions: user inputs 2 integers, both are either 1 or 2.
#Post-conditions: prompts, results of actions, ending message and Inventory are displayed
# initialize gold flag, living flag and ring flag
# display first prompt Door / Tunnel
# get user's first choice
# if user chooses 1
#display sack prompt pick up/open
#get user's second choice
#if chooses pickup
#output sack eats you
#adjust flags as needed
#otherwise if choose open sack
#output you got 500 gp
#adjust flags as needed
#
# more design (YOU fill in)
# note that this output happens only ONE time in the design, here!
# output a blank line
# output Game Over!
# output a blank line
# if lived flag has value True
# output You got out alive and display inventory
#
# more design (YOU fill in)
```
• (44 points) Implement the program starting with the adventure3.py file you have written the design in. The design steps are the comments. Make sure you try all the test cases.
• Submit your Python file (adventure3.py) using the link above and the menu choices Lab 3 and Code.
Log off properly - you don't want your account misused by someone else!
Remember NOT to leave files on the local hard drives in a lab or anywhere else on campus! Make sure you save your projects on a portable storage device you take with you! | 1,095 | 4,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.894245 |
https://byjus.com/question-answer/select-the-correct-option-for-the-given-figure-and-statements-statement-1-delta-pbo-cong/ | 1,716,075,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00376.warc.gz | 128,538,127 | 25,841 | 1
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Question
# Select the correct option for the given figure and statements. Statement 1: ΔPBO≅ΔNMO Statement 2: ΔPBO≅ΔPQO
A
Both statement 1 and statement 2 are true.
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B
Statement 1 is true, while statement 2 is false.
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C
Statement 1 is false, while statement 2 is true.
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Both statement 1 and statement 2 are false.
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Solution
## The correct option is C Statement 1 is false, while statement 2 is true.InΔ PBO and Δ PQO, PB = PQ (Tangents to a circle from an external point are equal in length.) OP = PO (Common side) OB = OQ (Radii of the same circle) Thus, by SSS congruence, ΔPBO≅ΔPQO. But,ΔPBO≅ΔNMO. (Since OP ≠ ON) Hence, statement 1 is false, while statement 2 is true.
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Join BYJU'S Learning Program | 361 | 1,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-22 | latest | en | 0.846862 |
nimeshpokhrel.com | 1,547,693,382,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00494.warc.gz | 591,548,026 | 13,741 | # Superconductivity
I wrote this note in the process of learning Josephson junctions. “It would be a good way to learn this from the bottom-up”, I thought, and then looked for various YouTube videos, textbooks, articles and lecture notes on superconductivity. I had watched YouTube videos about superconductivity and floating magnets (and prototypes of floating trains) before, but was no where close to understanding them. I started likewise – watched few YouTube videos about Meissner effect, spin of electron, Josephson effect etc., and then when it was not enough I started looking for textbooks and articles. After learning some bits, I began to appreciate the findings of Meissner and Ochsenfeld, more so because it was discovered at a time when it was important to incite rapid interest in a field so crucial to our understanding of matter. It sort of opened new doors to peek into the superconductive world. Fritz London and Heinz London then came with a mathematical formulation to describe the Meissner effect. The London equations are first hand derivation of the Maxwell’s equation and the current density equation (eyebrows raised), and therefore were not difficult to grasp. I understood the nitty-gritty details of the Meissner effect and realized that “External magnetic field cannot penetrate a superconductor!”, was not a true statement (not completely). The magnetic field did penetrate the superconductor, but the depth to which it penetrated was small enough to treat superconductors as something that expelled external magnetic fields altogether. Maybe this statement made a strong impression that was needed to catch eyes of listeners, or may be not, but letting go of this deeply held belief was not easy (it never is I thought and moved on).
My curiosity led me to study the macroscopic theory and how/why everything was tied to the phase of the wavefunction of superconducting particles. I think it was natural to go into Landau theory and his phenomenological model of second order phase transition to understand the origin. This phase transition i.e. solid to gas or say ferromagnetism to anti-ferromagnetism did not have a direct relation with the phase of the exponential function I was trying to understand, but helped develop some understanding of the superconducting domain. The superconducting domain started looking like a ball of spaghetti (a lake full of it) where everything was tangled up so nicely that understanding one would unlock all other. I didn’t know how much of this made sense and served my learning purpose, but I did realize that I had lost track. The allure of this novel field however kept me diverted and I soon ran into the dark alleys of condensed matter physics and began wandering from “broken symmetry” to “mean-field theory” to the concept of “second quantization”. I learned bits and pieces of each, unable to form a clear idea (not knowing how to relate to my old bits) I called time and finally decided to whip my mind-horse back to track. I then focused towards understanding the making of the superconducting wavefunction, the concept of cooper pairs and the BCS theory, few thermodynamical models and things easily available in the Tinkham book. My plan here is to grind all those pieces of information learned from various sources, and come up with a sweet and sour sauce based on my taste, palate and perception. I hope it is edible.
## Introduction:
Kamerlingh Onnes work on cryogenics made possible the liquefaction of Helium. Able to reach temperature as low as 2K with his liquid Helium apparatus, he conducted series of experiments on various metals to test their electrical properties at extremely low temperatures. It was with Mercury that he noticed a significant drop in resistance just below a critical temperature, and discovered superconductivity. A result driven by curiosity to probe with a new toy just revealed to mankind, this fascinating discovery allured the human mind with all its profuse glory and possibilities, which then led to even more interesting developments. The Meissner effect was one of the early ones that helped stretch our ignorance further, and amuse scientists still trying to wrap their heads around superconductivity and the cause of this something spectacular. Nonetheless, it showed us a way to peek into this new state of matter with the help of our beloved Maxwell’s equations. Let me write down the four Maxwell’s equations before getting into the superconducting stuff:
$\nabla.\,\vec{D} = \rho_V$
$\nabla.\, \vec{B} = 0$ | 930 | 4,541 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.967437 |
http://furthermathematicst.blogspot.com/2011/07/101-graphs.html | 1,555,949,919,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578558125.45/warc/CC-MAIN-20190422155337-20190422181337-00183.warc.gz | 67,635,348 | 20,441 | Tuesday, July 12, 2011
10.1 – Graphs
In mathematics and computer science, graph theory is the study of graphs, mathematical structures used to model pairwise relations between objects from a certain collection. A graph, G = (V, E) consists of V, a nonempty set of vertices / nodes and E, a set of edges. In other words, a graph is a discrete structure consisting of vertices, and edges that connect these vertices. Each edge has either one or two vertices associated with it (endpoints). An edge is said to connect its endpoints. A graph looks something like this:
As you can see, a and b are vertices, while e and f are edges. the edge g is called a loop. The vertex set V = {a, b}.
In this section, there will be many terminologies which you should remember, and should be able to write down their definition in your exam. Here we will be learning the different kinds of graphs and their names:
An infinite graph is a graph with infinite vertex set (or rather, an infinite number of vertices). The definition of a finite graph is just the converse. Throughout this section, we will only be learning about graphs with finite amount of edges and vertices.
A simple graph is a graph in which each edge connects two different vertices and where no two edges connect the same pair of vertices. A multigraph is a graph that has multiple edges connected to the same vertices, while a pseudograph is a graph that may include loops, multiple edges connecting the same pair of vertices. The 3 pictures below illustrate a simple graph, a multigraph and a pseudograph:
Notice for the multigraph, there are 2 edges connecting both a to b and a to c, while 3 edges connecting e to f. As for the pseudograph, there exist loops at the vertices e and f.
The complement of the graph, , has the same amount of vertices as graph G but whenever there is a edge between vertices a and b, there won’t be an edge, and whenever there isn’t an edge between vertices a and b, an edge is added to it. This only applies to simple graphs. For example, below is the graph and its complement:
All the above graphs are undirected, that means that one can traverse an edge in both directions. A directed graph (or digraph), consists of a nonempty set of vertices and a set of directed edges (or arcs). Each directed edge is associated with an ordered pair of vertices. The directed edge associated with the ordered pair V = (u, v) is said to start at u and end at v. In other words, we say that u is adjacent to v, while v is adjacent from u. Notice thee different uses of { } and ( ) brackets for undirected and directed graphs. Below is a directed graph:
For the ordered pair of vertices (u, v), we say that u and v are adjacent, and we say that the edge is incident / connects u and v. u is known as the initial vertex and v being the terminal vertex. Using the similar naming convention, we can describe a simple directed graph as a directed graph in which each edge connects two different vertices and where no two edges connect the same pair of vertices. Then similarly, a directed multigraph can be defined.
An underlying undirected graph is the undirected graph that results from ignoring directions of edges. It is just the same graph without the arrows. A mixed graph, is a graph with both directed and undirected edges. A converse of a directed graph, is the graph in which its arrows are reversed.
For every graph, we could come up with subgraphs, which are graphs that are subsets of the initial graph. For example, the graph
can be broken down into 11 subgraphs below:
An exercise for you here is that you can try to figure out whether you can determine the total amount of subgraphs, given the values of V and E.
A bipartite graph is a simple graph such that its vertex set V can be partitioned into 2 disjoint sets V1 and V2 such that every edge in the graph connects a vertex in V1 and V2. Consider the bipartite graph below:
Notice that I coloured the vertices with 2 colours, red and blue. The blue vertices will not connect to any other blue vertex, and the red vertices too, they don’t connect to any other red vertex. The graph is partitioned such that there are two sets or parties of vertices which can be grouped together. To identify a bipartite graph is simple: As long as you can colour adjacent vertices with only 2 colours, then it is a bipartite graph. For example, you colour the first vertex blue. The vertices adjacent to the first vertex must be coloured red, and if you can fit all the vertices with 2 colours such that no two adjacent vertices have the same colour, then it is a bipartite graph. Notice also, that a graph is bipartite if and only if it has no odd cycles. We will learn about cycles in the next session.
Now there are a 5 types of special simple graphs I want to introduce:
1. Complete Graph Kn
This graph is a simple graph that contains exactly 1 edge between each pair of distinct vertices. In other words, this graph has the maximum amount of edges it can have, and adding any edge between any 2 vertices will turn it into a multigraph. The graphs look as follows:
For the K4 graph, it has 4 vertices, and every vertex is connected to the other 3 vertices. By simple calculations, a Kn graph has n vertices, and n(n-2)/2 edges.
2. Cycle Graph C
n
This graph, where n ≥ 3, consists of n vertices and edges.
Strictly speaking, C2 is not a Cycle graph, as n < 3. Notice that every vertex is only connected to two other vertices. It looks like a regular polygon with n sides.
3. Wheel Graph W
n
This graph looks like a wheel with n sides. We obtain the wheel when we add an additional vertex to the cycle Cn, for n ≥ 3, and connect this new vertex to each of the n vertices in Cn, by new edges.
A Cn graph has n + 1 vertices and n 2n edges.
4. n-Dimensional Hypercube, Qn
This graph, also know as n-cube, is the graph whose vertices represent the 2n bit strings of length n. Two vertices are adjacent if and only if the bit strings that they represent differ in exactly one bit position. I don’t think this graph is in the syllabus, but I think it will be good for you to know:
This graph has 2n vertices and 2n-1 edges. Try proving this if you are free.
5. Complete Bipartite Graph, K
m,n
This graph is just a bipartite graph, in which there is only 1 edge between each pair of distinct vertices across V1 and V2. Note that the number of edges, | E(m, n) | = mn, and there are m + n vertices.
Now that we know everything about the structure of graphs, we shall now get into the a little calculations. The degree of vertex is the number of edges incident with it, except that a loop at a vertex contributes 2 times to the degree of that vertex. The degree of a vertex is denoted by deg (v). When deg (0), we say that the vertex is isolated, and when deg (1), then we say that the vertex is pendant.
We now want to find the relationship between the sum of degrees of vertices & number of edges. The Handshaking Theorem states that the sum of degree of vertices is double the amount of edges. In equation form, we have
This theorem has many implications. One of them is that we know that a graph cannot exist if the sum of degree of vertex is odd.
In the case for directed graphs, we denote deg+ (v) as the out-degree, meaning the amount of arcs pointing away from the vertex, while the in-degree is denoted by deg- (v), which is the amount of arcs pointing towards the vertex. Modifying the handshaking theorem, we have
This section isn’t really hard. You just need to remember the definitions of the graphs. Take note that these graphs need not to be drawn on graph paper. Just draw them in the test pad given to you in exams. | 1,765 | 7,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2019-18 | latest | en | 0.942912 |
https://askdev.io/questions/17167/why-can-not-the-polynomial-ring-be-a-field | 1,618,373,301,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076819.36/warc/CC-MAIN-20210414034544-20210414064544-00044.warc.gz | 227,194,604 | 7,282 | # Why can not the Polynomial Ring be a Field?
I'm presently researching Polynomial Rings, yet I can not identify why they are Rings, not Fields. In the definition of a Field, a Set constructs a Commutative Group with Addition and also Multiplication. This indicates an inverted numerous for every single Element in the Set.
Guide does not specify on this, nonetheless. I do not recognize why a Polynomial Ring could not have an inverted multiplicative for every single component (at the very least in the entire numbers, and also it's currently considered that it has a neutral component). Could someone please clarify why this can not be so?
0
2019-05-09 11:31:13
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Consider $\mathbb{C}[x]$ the ring of polynomials with coefficients from $\mathbb{C}$. This is an instance of polynomial ring which is not an area, due to the fact that $x$ has no multiplicativ inverse.
0
2019-05-10 10:13:32
Source
The devices of $D[x]$ are specifically the devices of $D$, when $D$ is a domain name.
0
2019-05-10 09:59:13
Source
Because necessarily, the only polynomial that can have an adverse level is $0$, which is specified to have a level of $-\infty$. Non - absolutely no constants have level $0$. You after that have the level formula : $\deg (fg) = \deg (f) + \deg (g)$ for any kind of polynomials $f,g$. By examination, any kind of polynomial of level $n \geq 1$ would certainly require as an inverted a polynomial of level $-n$, which does not exist (i.e. what Agusti Roig claimed!) The set you desire does exist, nonetheless : it is called the area of sensible features , and also is specifically the set of proportions of polynomials. It is created similarly that the area of sensible numbers is from the ring of integers.
0
2019-05-10 07:51:50
Source
Hint $\rm\quad\rm x \, f(x) = 1 \,$ in $\,\rm R[x]\ \Rightarrow \ 0 = 1 \,$ in $\,\rm R, \,$ by reviewing at $\rm\ x = 0$
Remark $\$ This has a really instructional global analysis : if $\rm\, x\,$ is a device in $\rm\, R[x]\,$ after that so also is every $\rm\, R$ - algebra component $\rm\, r,\,$ as adheres to by reviewing $\ \rm x \ f(x) = 1 \$ at $\rm\ x = r\,.\,$ Therefore to offer a counterexample it is adequate to show any kind of nonunit in any kind of $\rm R$ - algebra. An all-natural selection is the nonunit $\,\rm 0\in R,\,$ which generates the above evidence.
0
2019-05-10 06:52:30
Source | 646 | 2,377 | {"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} | 3.640625 | 4 | CC-MAIN-2021-17 | latest | en | 0.921347 |
https://spiral.ac/sharing/gbmqk67/area-of-an-equilateral-triangle-geometry-help-moomoomath | 1,620,877,831,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00563.warc.gz | 557,028,612 | 10,158 | area-of-an-equilateral-triangle-geometry-help-moomoomath
Interactive video lesson plan for: Area of an Equilateral Triangle-Geometry Help-MooMooMath
Activity overview:
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How to find the area of an equilateral triangle.
The formula for the area of an equilateral triangle equals
1/2 base * height.
For more on equilateral triangles see
http://www.moomoomath.com/Equilateral-Triangle.html
What is the formula for finding the area of a triangle?
What is an equilateral triangle?
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How do you find the area of an equilateral triangle with sides of 10 units?
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-~-~~-~~~-~~-~-
-~-~~-~~~-~~-~-
Tagged under: formula equilateral triangle,geometric shapes,area equilateral triangle,geometry ,equilateral triangle,geometry,Geometry (Field Of Study),math ,How find area equilateral triangle,area,moomoomath,geometry lesson,Moomoomath
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## Website Detail Page
written by Walter Fendt
This applet demonstrates the harmonics of the air in a tube as an example of standing longitudinal waves. The user can chose the length and type of tube, both sides open, one side closed or both sides closed. The applet calculates the wavelength and and frequency of the wave. This applet is part of a large collection of physics applets that are available in a wide range of languages.
Please note that this resource requires at least version 1.4.2 of Java.
Subjects Levels Resource Types
Oscillations & Waves
- Wave Motion
= Interference and Diffraction of Sound
= Longitudinal Pulses and Waves
= Standing Waves
= Wave Properties of Sound
- High School
- Middle School
- Instructional Material
= Curriculum support
= Interactive Simulation
- Audio/Visual
= Movie/Animation
Appropriate Courses Categories Ratings
- Physical Science
- Physics First
- Conceptual Physics
- Algebra-based Physics
- AP Physics
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Keywords:
pipe, pressure, standing waves, tube
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Metadata instance created August 5, 2006 by swapna gurumani
Record Updated:
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when Cataloged:
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W. Fendt, (1998), WWW Document, (http://www.walter-fendt.de/ph14e/stlwaves.htm).
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W. Fendt, Standing Longitudinal Waves, (1998), <http://www.walter-fendt.de/ph14e/stlwaves.htm>.
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Fendt, W. (2002, December 27). Standing Longitudinal Waves. Retrieved March 29, 2017, from http://www.walter-fendt.de/ph14e/stlwaves.htm
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Fendt, Walter. Standing Longitudinal Waves. December 27, 2002. http://www.walter-fendt.de/ph14e/stlwaves.htm (accessed 29 March 2017).
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Save to my folders | 918 | 3,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-13 | latest | en | 0.811096 |
https://gateoverflow.in/313374/gate2019-ce-2-ga-9?show=314168 | 1,590,501,748,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390758.21/warc/CC-MAIN-20200526112939-20200526142939-00233.warc.gz | 364,657,727 | 17,286 | 329 views
An oil tank can be filled by pipe $X$ in $5$ hours and pipe $Y$ in $4$ hours, each pump working on its own. When the oil tank is full and the drainage hole is open, the oil is drained in $20$ hours. If initially the tank was empty and someone started the two pups together but left the drainage hole open, how many hours will it take for the tank to be filled? (Assume that the rate of drainage is independent of the Head)
1. $1.50$
2. $2.00$
3. $2.50$
4. $4.00$
edited | 329 views
Migrated from GO Civil 11 months ago by Arjun
Part filled by pipe $X$ in $1$ hour $= \frac{1}{5}$
Part filled by pipe $Y$ in $1$ hour $= \frac{1}{4}$
Part emptied by drainage hole in $1$ hour $= \frac{1}{20}$
Net part filled by pipes $X$ & $Y$ and when drainage hole is left open in $1$ hour $= \frac{1}{5} +\frac{1}{4} -\frac{1}{20} = \frac{2}{5}$
So, the oil tank can be filled in $\dfrac{1}{\frac{2}{5}} = 2.5 \;\text{hours}$
Correct Answer: $C$
by Boss
selected by
+1 vote
Let the tank capacity be 20 ltrs.(Any multiple of 4,1,5 like 20,40,60.... will work) .
Since the tank can be filled by X in 5 hrs,X is filling the tank at a rate of 4 ltrs per hour.
Similarly Y is filling the tank at a rate of 5 ltrs per hour.
Similarly drainage hole is emptying the tank at a rate of 1 litre per hour.
If X,Y,Z are open for 1 hr,then at the end of hour,tank is increased by $4+5-1=8$ litres.
Therefore if you start with empty tank in $20/8=2.5hrs$, you can fill the tank completely
by Loyal
+1 vote
$Let\ capacity=20L$
$X\rightarrow5H-\ \ 4L/H$
$Y\rightarrow4H-\ \ 5L/H$
$Z\rightarrow 20H-\underline{1L/H}$
$8L/H$
$1H\rightarrow 8L$
$?\rightarrow 20L$
$?=2.5H$
by Loyal | 573 | 1,679 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-24 | latest | en | 0.858596 |
http://content.oss.deltares.nl/imod/iMOD_Manual_actual/imod-um-ANI-Horizontal-anisotropy-module.html | 1,660,739,669,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00369.warc.gz | 10,380,310 | 4,749 | # iMOD User Manual version 5.2 (html)
#### 12.14ANI Horizontal anisotropy module
Anisotropy is a phenomenon for which the permeability k is not equal along the x- and y Cartesian axis, k${}_{x}$${}_{x}$ and k${}_{y}$${}_{y}$, respectively. It can be notated that for isotropic conditions k${}_{x}$${}_{x}$ = k${}_{y}$${}_{y}$ (see figure 12.6a), and for anisotropic conditions k${}_{x}$${}_{x}$ $\neq$ k${}_{y}$${}_{y}$ (see figure 12.6b).
To express the amount of flow along the x- and y-axes of a Cartesian coordinate system, the following equations are valid to compute the flow along these direction; q${}_{x}$ and q${}_{y}$, respectively (Strack ODL (1989), Groundwater Mechanics, Princeton Hall, Inc., Englewood Cliffs, New-Jersey):
$$\label {eq:GrindEQ.1} \left [\begin {array}{c} {q_{x} } \\ {q_{y} } \end {array}\right ]=\left [\begin {array}{cc} {-k_{xx} } & {-k_{xy} } \\ {-k_{yx} } & {-k_{yy} } \end {array}\right ]\left [\begin {array}{c} {\frac {\partial h_{x} }{\partial x} } \\ {\frac {\partial h_{y} }{\partial y} } \end {array}\right ]$$
From equation (12.2), it can be seen that in anisotropic conditions (k${}_{x}$${}_{x}$ $\neq$ k${}_{y}$${}_{y}$), flow along the x-direction is not influenced solely by the hydraulic gradient along this x-axis, but also by a hydraulic gradient along the y-axis. The permeability’s k${}_{xy}$ and k${}_{xy}$ are equal to each other and depend on the angle $\varphi$ of the principal axis to the x-axis:
$$\label {eq:GrindEQ.2} \begin {array}{l} {k_{xx} =f\times T\times \cos (\varphi )^{2} +T\times \sin (\varphi )^{2} } \\ {k_{xy} =k_{yx} =((f\times T)-T)\times \cos (\varphi )\times \sin (\varphi )} \\ {k_{yy} =f\times T\times \sin (\varphi )^{2} +T\times \cos (\varphi )^{2} } \end {array}$$
For values $\varphi$=0.0; $\varphi$=90.0; $\varphi$=180.0; $\varphi$=270.0, k${}_{xy}$ and k${}_{xy}$${}_{ }$become 0.0.
##### 12.14.1Parameterisation
Anisotropy is expressed by an angle $\varphi$ and anisotropic factor f. The angle $\varphi$ denotes the angle along the main principal axis (highest permeability k) measured in degrees from north (0${}^{\circ }$), east (90${}^{\circ }$), south (180${}^{\circ }$) and west (270${}^{\circ }$). The anisotropic factor f is perpendicular to the main principal axis. The factor is between 0.0 (full anisotropic) and 1.0 (full isotropic), see figure 12.7.
Most optimally, the model discretisation should follow the configuration of the anisotropy, see figure 12.8a. However, anisotropy could be folded in many different directions (principal directions), which probably yield for anisotropy in many angles throughout the modeling domain. With the chosen mathematical method (finite-differences) in iMODFLOW, it is impossible to fold the model network according to the anisotropy, see figure 12.8b.
Since the principal direction of the permeability is not aligned to the axes of the modeling network, it is necessary to add extra flow terms to the finite difference equation to take into account the diagonal flow, caused by the anisotropy, see figure 12.9.
For more detailed explanation on the computation of these extra flow terms, see [Vermeulen et al.(2006)].
For each cell in the model network, anisotropic angles $\varphi$ and factors f can be specified. For those situations where a single model cell contains more than one of these anisotropic parameters, they will be up-scaled to the model cell. For the anisotropic angle, the most frequent occurrence will be used, as for the anisotropic factor, a mean value will be computed. This seems to be the most robust and fair trade-off between a coarsened model network and loss in detail.
The ANI horizontal anisotropy corresponds with the TRPY variable specified in the MODFLOW BCF package and the HANI variable specified in the MODFLOW LPF package. | 1,122 | 3,814 | {"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": 2, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-33 | latest | en | 0.684088 |
http://de.metamath.org/mpeuni/ply1mulgsumlem3.html | 1,716,794,992,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00519.warc.gz | 7,501,024 | 8,998 | Mathbox for Alexander van der Vekens < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > ply1mulgsumlem3 Structured version Visualization version GIF version
Theorem ply1mulgsumlem3 41970
Description: Lemma 3 for ply1mulgsum 41972. (Contributed by AV, 20-Oct-2019.)
Hypotheses
Ref Expression
ply1mulgsum.p 𝑃 = (Poly1𝑅)
ply1mulgsum.b 𝐵 = (Base‘𝑃)
ply1mulgsum.a 𝐴 = (coe1𝐾)
ply1mulgsum.c 𝐶 = (coe1𝐿)
ply1mulgsum.x 𝑋 = (var1𝑅)
ply1mulgsum.pm × = (.r𝑃)
ply1mulgsum.sm · = ( ·𝑠𝑃)
ply1mulgsum.rm = (.r𝑅)
ply1mulgsum.m 𝑀 = (mulGrp‘𝑃)
ply1mulgsum.e = (.g𝑀)
Assertion
Ref Expression
ply1mulgsumlem3 ((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) → (𝑘 ∈ ℕ0 ↦ (𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙)))))) finSupp (0g𝑅))
Distinct variable groups: 𝐴,𝑙 𝐵,𝑙 𝐶,𝑙 𝐾,𝑙 𝐿,𝑙 𝑅,𝑙 𝐴,𝑘 𝐵,𝑘 𝐶,𝑘 𝑘,𝐾 𝑘,𝐿 𝑅,𝑘 ,𝑘 𝑘,𝑙
Allowed substitution hints: 𝑃(𝑘,𝑙) · (𝑘,𝑙) × (𝑘,𝑙) (𝑘,𝑙) (𝑙) 𝑀(𝑘,𝑙) 𝑋(𝑘,𝑙)
Proof of Theorem ply1mulgsumlem3
Dummy variables 𝑛 𝑠 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 fvex 6113 . . 3 (0g𝑅) ∈ V
21a1i 11 . 2 ((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) → (0g𝑅) ∈ V)
3 ovex 6577 . . 3 (𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) ∈ V
43a1i 11 . 2 (((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) ∧ 𝑘 ∈ ℕ0) → (𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) ∈ V)
5 ply1mulgsum.p . . . 4 𝑃 = (Poly1𝑅)
6 ply1mulgsum.b . . . 4 𝐵 = (Base‘𝑃)
7 ply1mulgsum.a . . . 4 𝐴 = (coe1𝐾)
8 ply1mulgsum.c . . . 4 𝐶 = (coe1𝐿)
9 ply1mulgsum.x . . . 4 𝑋 = (var1𝑅)
10 ply1mulgsum.pm . . . 4 × = (.r𝑃)
11 ply1mulgsum.sm . . . 4 · = ( ·𝑠𝑃)
12 ply1mulgsum.rm . . . 4 = (.r𝑅)
13 ply1mulgsum.m . . . 4 𝑀 = (mulGrp‘𝑃)
14 ply1mulgsum.e . . . 4 = (.g𝑀)
155, 6, 7, 8, 9, 10, 11, 12, 13, 14ply1mulgsumlem2 41969 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) → ∃𝑠 ∈ ℕ0𝑛 ∈ ℕ0 (𝑠 < 𝑛 → (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅)))
16 vex 3176 . . . . . . . . 9 𝑛 ∈ V
17 csbov2g 6589 . . . . . . . . . 10 (𝑛 ∈ V → 𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (𝑅 Σg 𝑛 / 𝑘(𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))))
18 id 22 . . . . . . . . . . . 12 (𝑛 ∈ V → 𝑛 ∈ V)
19 oveq2 6557 . . . . . . . . . . . . . 14 (𝑘 = 𝑛 → (0...𝑘) = (0...𝑛))
20 oveq1 6556 . . . . . . . . . . . . . . . 16 (𝑘 = 𝑛 → (𝑘𝑙) = (𝑛𝑙))
2120fveq2d 6107 . . . . . . . . . . . . . . 15 (𝑘 = 𝑛 → (𝐶‘(𝑘𝑙)) = (𝐶‘(𝑛𝑙)))
2221oveq2d 6565 . . . . . . . . . . . . . 14 (𝑘 = 𝑛 → ((𝐴𝑙) (𝐶‘(𝑘𝑙))) = ((𝐴𝑙) (𝐶‘(𝑛𝑙))))
2319, 22mpteq12dv 4663 . . . . . . . . . . . . 13 (𝑘 = 𝑛 → (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙)))) = (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙)))))
2423adantl 481 . . . . . . . . . . . 12 ((𝑛 ∈ V ∧ 𝑘 = 𝑛) → (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙)))) = (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙)))))
2518, 24csbied 3526 . . . . . . . . . . 11 (𝑛 ∈ V → 𝑛 / 𝑘(𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙)))) = (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙)))))
2625oveq2d 6565 . . . . . . . . . 10 (𝑛 ∈ V → (𝑅 Σg 𝑛 / 𝑘(𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))))
2717, 26eqtrd 2644 . . . . . . . . 9 (𝑛 ∈ V → 𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))))
2816, 27ax-mp 5 . . . . . . . 8 𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙)))))
29 simpr 476 . . . . . . . 8 (((((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) ∧ 𝑠 ∈ ℕ0) ∧ 𝑛 ∈ ℕ0) ∧ (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅)) → (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅))
3028, 29syl5eq 2656 . . . . . . 7 (((((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) ∧ 𝑠 ∈ ℕ0) ∧ 𝑛 ∈ ℕ0) ∧ (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅)) → 𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (0g𝑅))
3130ex 449 . . . . . 6 ((((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) ∧ 𝑠 ∈ ℕ0) ∧ 𝑛 ∈ ℕ0) → ((𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅) → 𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (0g𝑅)))
3231imim2d 55 . . . . 5 ((((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) ∧ 𝑠 ∈ ℕ0) ∧ 𝑛 ∈ ℕ0) → ((𝑠 < 𝑛 → (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅)) → (𝑠 < 𝑛𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (0g𝑅))))
3332ralimdva 2945 . . . 4 (((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) ∧ 𝑠 ∈ ℕ0) → (∀𝑛 ∈ ℕ0 (𝑠 < 𝑛 → (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅)) → ∀𝑛 ∈ ℕ0 (𝑠 < 𝑛𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (0g𝑅))))
3433reximdva 3000 . . 3 ((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) → (∃𝑠 ∈ ℕ0𝑛 ∈ ℕ0 (𝑠 < 𝑛 → (𝑅 Σg (𝑙 ∈ (0...𝑛) ↦ ((𝐴𝑙) (𝐶‘(𝑛𝑙))))) = (0g𝑅)) → ∃𝑠 ∈ ℕ0𝑛 ∈ ℕ0 (𝑠 < 𝑛𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (0g𝑅))))
3515, 34mpd 15 . 2 ((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) → ∃𝑠 ∈ ℕ0𝑛 ∈ ℕ0 (𝑠 < 𝑛𝑛 / 𝑘(𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙))))) = (0g𝑅)))
362, 4, 35mptnn0fsupp 12659 1 ((𝑅 ∈ Ring ∧ 𝐾𝐵𝐿𝐵) → (𝑘 ∈ ℕ0 ↦ (𝑅 Σg (𝑙 ∈ (0...𝑘) ↦ ((𝐴𝑙) (𝐶‘(𝑘𝑙)))))) finSupp (0g𝑅))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 ∀wral 2896 ∃wrex 2897 Vcvv 3173 ⦋csb 3499 class class class wbr 4583 ↦ cmpt 4643 ‘cfv 5804 (class class class)co 6549 finSupp cfsupp 8158 0cc0 9815 < clt 9953 − cmin 10145 ℕ0cn0 11169 ...cfz 12197 Basecbs 15695 .rcmulr 15769 ·𝑠 cvsca 15772 0gc0g 15923 Σg cgsu 15924 .gcmg 17363 mulGrpcmgp 18312 Ringcrg 18370 var1cv1 19367 Poly1cpl1 19368 coe1cco1 19369 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-fal 1481 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-of 6795 df-om 6958 df-1st 7059 df-2nd 7060 df-supp 7183 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-1o 7447 df-oadd 7451 df-er 7629 df-map 7746 df-en 7842 df-dom 7843 df-sdom 7844 df-fin 7845 df-fsupp 8159 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-3 10957 df-4 10958 df-5 10959 df-6 10960 df-7 10961 df-8 10962 df-9 10963 df-n0 11170 df-z 11255 df-dec 11370 df-uz 11564 df-fz 12198 df-seq 12664 df-struct 15697 df-ndx 15698 df-slot 15699 df-base 15700 df-sets 15701 df-ress 15702 df-plusg 15781 df-mulr 15782 df-sca 15784 df-vsca 15785 df-tset 15787 df-ple 15788 df-0g 15925 df-gsum 15926 df-mgm 17065 df-sgrp 17107 df-mnd 17118 df-grp 17248 df-minusg 17249 df-mgp 18313 df-ring 18372 df-psr 19177 df-mpl 19179 df-opsr 19181 df-psr1 19371 df-ply1 19373 df-coe1 19374 This theorem is referenced by: ply1mulgsum 41972
Copyright terms: Public domain W3C validator | 5,278 | 7,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-22 | latest | en | 0.234137 |
https://www.numbersaplenty.com/89448 | 1,708,979,628,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00047.warc.gz | 935,029,207 | 3,176 | Search a number
89448 = 2333727
BaseRepresentation
bin10101110101101000
311112200220
4111311220
510330243
61530040
7521532
oct256550
9145626
1089448
1161227
1243920
1331938
1424852
151b783
hex15d68
89448 has 16 divisors (see below), whose sum is σ = 223680. Its totient is φ = 29808.
The previous prime is 89443. The next prime is 89449. The reversal of 89448 is 84498.
It is an Ulam number.
It is not an unprimeable number, because it can be changed into a prime (89443) by changing a digit.
89448 is an untouchable number, because it is not equal to the sum of proper divisors of any number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1840 + ... + 1887.
It is an arithmetic number, because the mean of its divisors is an integer number (13980).
289448 is an apocalyptic number.
It is an amenable number.
89448 is an abundant number, since it is smaller than the sum of its proper divisors (134232).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
It is a Zumkeller number, because its divisors can be partitioned in two sets with the same sum (111840).
89448 is a wasteful number, since it uses less digits than its factorization.
89448 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 3736 (or 3732 counting only the distinct ones).
The product of its digits is 9216, while the sum is 33.
The square root of 89448 is about 299.0785849906. The cubic root of 89448 is about 44.7222396827.
Subtracting from 89448 its reverse (84498), we obtain a triangular number (4950 = T99).
The spelling of 89448 in words is "eighty-nine thousand, four hundred forty-eight". | 495 | 1,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-10 | latest | en | 0.897433 |
https://discuss.interviewbit.com/t/python-bfs-solution/44716 | 1,611,709,022,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704804187.81/warc/CC-MAIN-20210126233034-20210127023034-00680.warc.gz | 301,506,328 | 3,326 | # Python BFS solution :)
#1
``````import collections, sys
class Solution:
def returnSteps(self, X, Y, KX, KY, D, L, TX, TY):
if D[(KX, KY)] > L:
D[(KX, KY)] = L
if KX == TX and KY == TY:
return set()
return set(a for a in [(KX+2, KY+1),(KX+2, KY-1),
(KX-2, KY+1),(KX-2, KY-1),
(KX+1, KY+2),(KX+1, KY-2),
(KX-1, KY+2),(KX-1, KY-2)]
if a[0] >= 0 and a[1] >= 0 and a[0] < X and a[1] < Y)
else:
return set()
def knight(self, X, Y, KX, KY, TX, TY):
TX, TY = TX-1, TY-1
Q = set() | 224 | 475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-04 | latest | en | 0.773238 |
https://smartindia.net.in/classroom/topic/?subject=88&grade=14 | 1,547,700,290,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658702.9/warc/CC-MAIN-20190117041621-20190117063621-00164.warc.gz | 632,343,433 | 6,763 | #### Topics
1. How many total electrons are present in nitrate ion ?
No. of electrons in NO-3 ion = NO of electrons of N+. NO of electrons on 3 oxygen atoms + one e -
= 7 + ( 3 + 8 ) + 1 = 32 electrons.
2. What is the lowest value of n that allows g orbits to exist ?
For n the values of l are 0, 1 ...... ( n - 1 )
For n = 5, the value of l are 0, 1, 2, 3, 4
For l = 4 > g sub - shell can exist. Hence lowest value of n = 5.
3. What are nucleons ?
The neutrons and protons present in the nucleus of an atom collectively called nucleons.
4. Write electronic configurations of Chromium ( At. No. = 24)?
Cr = 24 = 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1
5. Which of the following has smallest de - Broglie wavelength : O2 H2 proton, an electron ?
According to de - Broglie equation for same value of velocity
∴O2 molecule has shortest wavelength.
6. ( i ) An atomic orbital has n = 3. What are possible values of l and ml ?
(ii) List the quantum numbers ( m1 and l ) of electrons for 3d orbital.
( iii) which of the following orbitals are possible : 1p, 2s, 2p and 3f.
(i ) l ml
0 0
1 -1, 0, + 1
2 -2, -1, 0 + 1, + 2
( ii ) For 3d orbital l = 2
The values of m, are -2, -1, 0, +1, +2
(iii) Out of 1p, 2s, 2p and 3f only 2s, 2p are possible
7. What is the maximum number of electrons in an atom having n = 3, l = 1 and s = + 1/2
Three electrons ( one each in 3px, 3py, 3pz ).
8. State two properties and examples of electromagnetic radiation?
Electromagnetic radiation show phenomenon of : ( i ) Interference, ( ii ) Diffraction, Examples are : y - rays, x = rays, visible rays radio waves etc.
9. What is mean by quantisation of electron energy ?
It means that electron in an atom has a certain, specific, discrete amount of energy.
10. Out of 4s and 3d orbitals, which will have higher energy and why ?
3d orbital has higher energy as it has a higher value of ( n + 1 ).
11. What is the difference between a quantum and a photon?
The smallest packet of energy of any radiation is called a quantum . Whereas that of light is called a photon.
12. Which quantum number does not follow from the solution of schrodinger wave equation ?
Spin quantum number.
13. Describe the important properties of cathode - rays. What is concluded about the nature of these rays ?
The cathode rays possess the following properties:
( i ) Travel in straight lines perpendicular to surface of cathode.
(ii ) Consists of material particles.
( iii ) Have got heating effect.
(iv) Consists on negatively charged particles
(v) Produce X - rays when they strike against hard metals like copper, tungsten, platinum etc.
(vi) Produce fluorescence when they strike glass or certain other materials like zinc sulphide.
(vii) Penetrate through thin aluminium foils and other metals.
( viii) Affect the Photographic plats.
The ratio of charge to mass i.e. charge/ mass is same for all cathode - rays irrespective of gas used in the tube. All these properties led to the conclusion that cathode - rays consists of negatively charged fast moving particles. These particles were named as 'electrons'.
14. which type of ‘Atomic model’ was presented by Rutherford ?Describe in brief?
The existence of positively charged nucleus at the centre of atom was first suggested Rutherford in 1911 on the basis of his interpretation of observation made on scattering experiment. The main feature of this model are:
(i ) Atom is spherical and consists of two parts: Nucleus and extra – nuclear part.
(ii) The entire mass and entire positive charge is concentrated in a very small region at the centre known as nucleus.
(iii) The space surrounding the nucleus known as extra nuclear part is negatively charged so an atom as a while is neutral.
( iv ) Most of extra – nuclear part is empty.
(v ) The electrons are not stationary but are revolving around nucleus at very high speeds like planet revolving around the Sun.
15. Describe the drawback of Rutherford’s model of atom?
Main drawback is that it could not explain the stability of an atom. Maxwell has shown that when electric charge is subjected to acceleration, it emits energy in the form of radiations. In Rutherford’s model of atom, electrons are orbiting the nucleus and hence the direction of their velocity is constantly changing i.e., electrons are accelerating. This will have lesser and lesser energy and will get closer and closer to the nucleus until at last it spirals into the nucleus and thus does not provide a stable model of atom.
16. What is meant by dual nature of radiation ?
The fact that light energy is carried in terms of packets of energy ( i. e., Photons ) as suggested by Planck's theory means that light has a particle character. at the same time, the fact that light shown interference and diffraction phenomena means that light has a wave character. These experimental facts led Einstein to suggest that light has a dual character, i.e., It behaves both like a wave and like a particle.
17. Describe the important characteristics of anode or positive rays and its importance in the discovery of proton ?
( i ) The anode rays originate in the region between two electrodes in the discharge tube.
( ii ) These rays are made of material particles.
(iii) These rays are positively charged
( iv ) These rays produce the heat when strike against a surface.
( v ) The magnitude of charge on anode - rays varies from particle to particle depending on the number of electrons lost by an atom or molecule.
( vi) The mass of positive particles which constitute these rays depend upon the nature of gas in the tube.
( vii) The charge/ mass ( e/m ) ratio of anode - rays is not constant but depends upon the nature of gas in the tube. The value of e.m is greatest for lightest gas hydrogen. The electric charge on a lightest positively charged particle from hydrogen gas was found to be exactly equal in magnitude but opposite in sign to that of electron. This lightest positively charged particle from hydrogen gas was named as proton. The mass of proton is almost 1836 times that of electron.
18, What are the shortcomings of Bohr's atomic model?
(i ) It couldn't explain the spectra of multi - electron atoms.
( ii ) It fails to explain the splitting of spectral lines when subjected to electrostatic or magnetic field ( Stark of Zeeman's effect ).
(iii) it does not account for the fine splitting of spectral lines.
( iv ) It affords a two dimensional picture of revolution of electron while actually electron revolves around the nucleus in three dimensions.
( v ) It does not account for the shapes of molecules.
According to it, this is possible to determine simultaneously both the position and momentum of the electron accurately. But this contrary to Heisenberg's Uncertainty Principle.
19. How it can be proved that the universal constituent of all matter are 'electrons'?
The charge/ mass ( e/m) ratio for the particle sin the cathode rays ( i.e., electron ) is found to be same irrespective of nature of cathode or the nature of the gas taken in the discharge tube. This shows that electrons are universal constitution of all matter.
20. Distinguish between an Emission spectrum and an Absorption spectrum?
Emission Spectrum Absorption Spectrum 1. Emission spectrum is obtained when radiations emitted by the excited substance are analysed in a spectroscope 2. Emission spectrum consists of bright coloured lines separated by dark spaces. 1. Absorption spectrum is obtained when the whitelight is first passed through the substance ( in gaseous state or in solution ) and the transmitted light is analysed in a spectroscope. 2. Absorption spectrum consists of dark lines in an otherwise continuous spectrum.
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Paid Users Only! | 2,166 | 8,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-04 | longest | en | 0.804439 |
http://www.physicsforums.com/showthread.php?s=32c063001a1a648247b62c1c65cbf983&p=4488958 | 1,394,196,304,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999642518/warc/CC-MAIN-20140305060722-00037-ip-10-183-142-35.ec2.internal.warc.gz | 488,303,540 | 7,254 | # arc length of a regular parametrized curve
by tuggler
Tags: curve, length, parametrized, regular
P: 45 Given $$t\in I$$the arc length of a regular parametrized curve $$\alpha : I \to \mathbb{R}^3$$ from the point $$t_0$$ is by definition $$s(t) = \int^t_{t_0}|\alpha'(t)|dt$$ where $$|\alpha'(t)| = \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}$$ is the length of the vector $$\alpha'(t).$$ Since $$\alpha'(t) \ne 0$$ the arc length $$s$$ is a differentiable function of and $$ds/dt = |\alpha'(t)|.$$ This is where I get confused. It can happen that the parameter $$t$$is already the arc length measured from some point. In this case, [latex]ds/dt = 1 =|\alpha'(t)|[/tex]. Conversely, if $$|\alpha'(t)| = 1$$ then $$s = \int_{t_0}^t dt = t - t_0.$$ How did they get that it equals 1? I am not sure what they are saying?
P: 45 Opps, I am in the wrong thread. How can I delete this?
Quote by tuggler Given $$t\in I$$the arc length of a regular parametrized curve $$\alpha : I \to \mathbb{R}^3$$ from the point $$t_0$$ is by definition $$s(t) = \int^t_{t_0}|\alpha'(t)|dt$$ where $$|\alpha'(t)| = \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}$$ is the length of the vector $$\alpha'(t).$$ Since $$\alpha'(t) \ne 0$$ the arc length $$s$$ is a differentiable function of and $$ds/dt = |\alpha'(t)|.$$ This is where I get confused. It can happen that the parameter $$t$$is already the arc length measured from some point. In this case, [latex]ds/dt = 1 =|\alpha'(t)|[/tex]. Conversely, if $$|\alpha'(t)| = 1$$ then $$s = \int_{t_0}^t dt = t - t_0.$$ How did they get that it equals 1? I am not sure what they are saying? | 548 | 1,602 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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-2014-10 | latest | en | 0.839529 |
https://gmatclub.com/forum/if-p-is-a-positive-integer-is-p-a-positive-integer-255794.html | 1,558,501,174,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256763.42/warc/CC-MAIN-20190522043027-20190522065027-00268.warc.gz | 483,340,937 | 142,789 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# If p is a positive integer is √p a positive integer?
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19 Dec 2017, 22:16
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If p is a positive integer is √p a positive integer?
1) p is n percentage of n where n is a positive integer
2) p/100 is a positive integer
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19 Dec 2017, 22:49
(1) p is n% of n, this means p = n/100 * n = (n^2)/100
Here n is a positive integer, so n^2 is a perfect square. And n^2 must be divisible by 100, because then only p will be an integer. Now if we take square root both sides, we have
√p = √(n^2)/ √100 = n/10
Since n^2 is divisible by 100, this means n must be divisible by 10. So √p is an integer. Sufficient.
(2) p is divisible by 100.
If p = 400, then √p = 20 is an integer.
If p = 300, then √p = 10 √3, is not an integer. So Insufficient.
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Re: If p is a positive integer is √p a positive integer? [#permalink]
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19 Dec 2017, 23:34
If p is a positive integer is √p a positive integer?
(1) p is n percentage of n where n is a positive integer --> p = n/100*n = (n/10)^2. Now, n is an integer, so n/10 is either an integer or a fraction but since p is an integer and fraction^2 cannot be an integer, then n/10 can only be an integer, thus p = (n/10)^2 = integer^2. Sufficient.
(2) p/100 is a positive integer. If p = 100, then the answer is YES but if p = 200, then the answer is NO. Not sufficient.
Hope it's clear.
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If p is a positive integer is √p a positive integer? [#permalink]
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12 Sep 2018, 02:44
amanvermagmat wrote:
(1) p is n% of n, this means p = n/100 * n = (n^2)/100
Here n is a positive integer, so n^2 is a perfect square. And n^2 must be divisible by 100, because then only p will be an integer. Now if we take square root both sides, we have
√p = √(n^2)/ √100 = n/10
Since n^2 is divisible by 100, this means n must be divisible by 10. So √p is an integer. Sufficient.
(2) p is divisible by 100.
If p = 400, then √p = 20 is an integer.
If p = 300, then √p = 10 √3, is not an integer. So Insufficient.
Why are we not considering -n/10 as a possible root of n^2/100, in which case p would be negative and give a NO to the "is p a positive integer" question?
That would make Statement 1 insufficient.
If p is a positive integer is √p a positive integer? [#permalink] 12 Sep 2018, 02:44
Display posts from previous: Sort by | 1,026 | 3,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-22 | latest | en | 0.844494 |
https://study.com/academy/lesson/multiplicative-identity-property-definition-example.html | 1,550,602,645,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247491141.23/warc/CC-MAIN-20190219183054-20190219205054-00186.warc.gz | 732,262,189 | 38,854 | # Multiplicative Identity Property: Definition & Example
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Lesson Transcript
Instructor: Joseph Vigil
In this lesson, discover what the multiplicative identity property is and view examples of the property in action. You'll also find out why this property is always true.
## The Multiplicative Identity Property
For a property with such a long name, it's really a simple math law. The multiplicative identity property states that any time you multiply a number by 1, the result, or product, is that original number.
To write out this property using variables, we can say that n * 1 = n. It doesn't matter if n equals one, one million or 3.566879. The property always hold true. Therefore:
• 2 * 1 = 2
• 56 * 1 = 56
• 100,000,000,000 * 1 = 100,000,000,000
• 57,687.758943768579875986754890 * 1 = 57,687.758943768579875986754890
You get the picture.
## Explanation
But why is this property always true? Well, let's go back, and think of what multiplication really is. It's a way of adding a list of numbers together quickly. For example, if we're solving the multiplication problem 2 * 6, we're really adding 2 to itself six times. In other words, we can rewrite that multiplication sentence as a long addition problem: 2 + 2 + 2 + 2 + 2 + 2. It would take a lot of paper to write really long addition problems that way, so multiplication gives us a shorter way of doing it.
Another, more visual, way to think of multiplication is as a form of grouping items, as we've just done. Let's consider the same multiplication problem differently, 2 * 6. If we were to visualize it, we can think of two groups of six items.
This is simply a visual representation of the addition problem we wrote out above. Of course, when we count all the images, we have a total of 12. So, when we write 2 * 6, we're saying that we're finding the total of two groups of six items. Simple, right?
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Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 717 | 2,942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2019-09 | longest | en | 0.885825 |
https://www.math24.net/triple-integrals-cylindrical-coordinates/ | 1,611,495,520,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00025.warc.gz | 856,287,263 | 13,084 | # Triple Integrals in Cylindrical Coordinates
The position of a point $$M\left( {x,y,z} \right)$$ in the $$xyz$$-space in cylindrical coordinates is defined by three numbers: $$\rho, \varphi, z,$$ where $$\rho$$ is the projection of the radius vector of the point $$M$$ onto the $$xy$$-plane, $$\varphi$$ is the angle formed by the projection of the radius vector with the $$x$$-axis (Figure $$1$$), $$z$$ is the projection of the radius vector on the $$z$$-axis (its value is the same in Cartesian and cylindrical coordinates).
The relationship between cylindrical and Cartesian coordinates of a point is given by
${x = \rho \cos \varphi ,\;\;\;}\kern0pt {y = \rho \sin \varphi ,\;\;\;}\kern0pt {z = z.}$
We assume here that
${\rho \ge 0,\;\;\;}\kern-0.3pt {0 \le \varphi \le 2\pi ,\;\;\;}\kern-0.3pt {- \infty \lt z \lt \infty .}$
The Jacobian of transformation from Cartesian to cylindrical coordinates is
${I\left( {\rho ,\varphi ,z} \right) } = {\left| {\begin{array}{*{20}{c}} {\frac{{\partial x}}{{\partial \rho }}}&{\frac{{\partial x}}{{\partial \varphi }}}&{\frac{{\partial x}}{{\partial z}}}\\ {\frac{{\partial y}}{{\partial \rho }}}&{\frac{{\partial y}}{{\partial \varphi }}}&{\frac{{\partial y}}{{\partial z}}}\\ {\frac{{\partial z}}{{\partial \rho }}}&{\frac{{\partial z}}{{\partial \varphi }}}&{\frac{{\partial z}}{{\partial z}}} \end{array}} \right| } = {\left| {\begin{array}{*{20}{c}} {\cos \varphi }&{ – \rho \sin \varphi }&0\\ {\sin \varphi }&{\rho \cos \varphi }&0\\ 0&0&1 \end{array}} \right| } ={ \rho \ge 0.}$
Then the formula of change of variables for this transformation can be written in the form
${\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} } = {\iiint\limits_{U’} {f\left( {\rho \cos \varphi , }\kern0pt{ \rho \sin \varphi ,z} \right)\rho d\rho d\varphi dz} }$
Transition from cylindrical coordinates makes calculation of triple integrals simpler in those cases when the region of integration is formed by a cylindrical surface.
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
Evaluate the integral
$\iiint\limits_U {\left( {{x^4} + 2{x^2}{y^2} }\right.}+{\left.{ {y^4}} \right)dxdydz},$
where the region $$U$$ is bounded by the surface $${x^2} + {y^2} \le 1$$ and the planes $$z = 0,$$ $$z = 1$$ (Figure $$2\text{).}$$
### Example 2
Find the integral
$\iiint\limits_U {\left( {{x^2} + {y^2}} \right)dxdydz} ,$
where the region $$U$$ is bounded by the surfaces $${x^2} + {y^2} = 3z,$$ $$z = 3$$ (Figure $$4\text{).}$$
### Example 3
Using cylindrical coordinates evaluate the integral
${ \int\limits_{ – 2}^2 {dx} \int\limits_{ – \sqrt {4 – {x^2}} }^{\sqrt {4 – {x^2}} } {dy} \int\limits_0^{4 – {x^2} – {y^2}} {{y^2}dz} .}$
### Example 4
Calculate the integral using cylindrical coordinates:
$\iiint\limits_U {\sqrt {{x^2} + {y^2}} dxdydz} .$
The region $$U$$ is bounded by the paraboloid $$z = 4 – {x^2} – {y^2},$$ by the cylinder $${x^2} + {y^2} = 4$$ and by the planes $$y = 0,$$ $$z = 0$$ (Figure $$8\text{).}$$
### Example 5
Find the integral
$\iiint\limits_U {ydxdydz},$
where the region $$U$$ is bounded by the planes $$z = x + 1,$$ $$z = 0$$ and by the cylindrical surfaces $${x^2} + {y^2} = 1,$$ $${x^2} + {y^2} = 4$$ (see Figure $$10$$).
### Example 1.
Evaluate the integral
$\iiint\limits_U {\left( {{x^4} + 2{x^2}{y^2} }\right.}+{\left.{ {y^4}} \right)dxdydz},$
where the region $$U$$ is bounded by the surface $${x^2} + {y^2} \le 1$$ and the planes $$z = 0,$$ $$z = 1$$ (Figure $$2\text{).}$$
Solution.
Figure 2.
Figure 3.
It is more convenient to calculate this integral in cylindrical coordinates. Projection of the region of integration onto the $$xy$$-plane is the circle $${x^2} + {y^2} \le 1$$ or $$0 \le \rho \le 1$$ (Figure $$3$$).
Notice that the integrand can be written as
${\left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right) } = {{\left( {{x^2} + {y^2}} \right)^2} } = {{\left( {{\rho ^2}} \right)^2} = {\rho ^4}.}$
Then the integral becomes
$I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}\rho d\rho } \int\limits_0^1 {dz} .$
The second integral contains the factor $$\rho$$ which is the Jacobian of transformation of the Cartesian coordinates into cylindrical coordinates. All the three integrals over each of the variables do not depend on each other. As a result the triple integral is easy to calculate as
${I = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^1 {{\rho ^4}\rho d\rho } \int\limits_0^1 {dz} } = {2\pi \int\limits_0^1 {{\rho ^5}d\rho } \int\limits_0^1 {dz} } = {2\pi \cdot 1 \cdot \int\limits_0^1 {{\rho ^5}d\rho } } = {2\pi \left. {\left( {\frac{{{\rho ^6}}}{6}} \right)} \right|_0^1 } = {2\pi \cdot \frac{1}{6} = \frac{\pi }{3}.}$
Page 1
Problem 1
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# soln6 - R Balan Homework#6 Solutions MATH 464 1 Note the...
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R. Balan Homework #6 Solutions MATH 464 1. Note the function f is the convolution between the box function Π and the unit Gaussian function γ , that is: f ( x ) = Z -∞ Π( u ) γ ( x - u ) du , Π( u ) = 1 , | u | < 1 2 0 , | u | > 1 2 , γ ( u ) = e - π | u | 2 Thus the Fourier transform is the product of the Fourier transforms of Π and γ , respectively: F ( s ) = sinc ( s ) e - π | s | 2 = sinc ( s ) γ ( s ) 2. Note f 1 = - 1 2 df 0 dx The Fourier transforms of f 0 and f 1 are: F 0 ( s ) = πe - π 2 s 2 respectively F 1 ( s ) = - πisF 0 ( s ) = - isπ πe - π 2 s 2 . Let g 0 = f 0 * f 0 and g 1 = f 0 * f 1 . Then their Fourier transforms are: G 0 ( s ) = F 0 ( s ) 2 = πe - 2 π 2 s 2 G 1 ( s ) = F 0 ( s ) F 1 ( s ) = - isπ 2 e - 2 π 2 s 2 Next we need to compute the inverse Fourier transforms. Again apply the dilation and the derivative rules, and get: g 0 ( x ) = r π 2 e - x 2 / 2 g 1 ( x ) = - 1 2 dg 0 dx = x 2 r π 2 e - x 2 / 2 3. (Preferred Solution) Let F denote the Fourier transform of f . Define g : R C , g ( u ) = f ( x - u ). The Fourier transform of g is G : R C given by G ( s ) = e - 2 πisx F ( s ) Then the Parseval identity (which requires R -∞ | f ( x ) | 2 dx < ) implies: Z -∞ f ( u ) f ( x - u
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http://mathhelpforum.com/advanced-math-topics/214567-sequences-functions-print.html | 1,508,616,831,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824894.98/warc/CC-MAIN-20171021190701-20171021210701-00725.warc.gz | 220,327,359 | 2,744 | # sequences of functions
Printable View
• Mar 10th 2013, 11:00 PM
mrmaaza123
sequences of functions
Show that lim((cos pix)2n) exists for all x in R ?
I figured that the limit will be 1 when x belongs to Z and 0 otherwise.
But how can i show that the limit actually exists?
Help will be appreciated.
• Mar 11th 2013, 04:24 AM
Plato
Re: sequences of functions
Quote:
Originally Posted by mrmaaza123
Show that lim((cos pix)2n) exists for all x in R ?
I figured that the limit will be 1 when x belongs to Z and 0 otherwise.
But how can i show that the limit actually exists?
I don't really know what you mean by showing.
We all know that $(x)^{2n}\to 1,~|x|=1~\&~(x)^{2n}\to 0,~|x|<1.$ | 220 | 689 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-43 | longest | en | 0.882415 |
https://study.com/academy/answer/1-fill-in-the-missing-data-for-the-retail-organization-first-quarter-second-quarter-third-quarter-fourth-quarter-sales-10-15-gross-margin-4-5-ending-merchandise-inventory-5-5-beginning.html | 1,580,257,173,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783342.96/warc/CC-MAIN-20200128215526-20200129005526-00222.warc.gz | 627,166,772 | 24,390 | # 1. Fill In the missing data for the retail organization. First Quarter Second Quarter Third...
## Question:
1. Fill In the missing data for the retail organization.
First Quarter Second Quarter Third Quarter Fourth Quarter Sales $10$15 Gross margin $4$5 Ending merchandise inventory $5$5 Beginning merchandise inventory $4$5 Net cost of purchases $7$11 Operating income $3$2 $2 Operating expenses$2 $1$4 Cost of goods sold $5$8 $12 Cost of goods available for sale$12 $15$15
2. Fill in the missing data for the manufacturing organization.
First Quarter Second Quarter Third Quarter Fourth Quarter Ending finished goods inventory $3$6 Cost of goods sold $6$3 $5 Operating income$2 $3$2 Cost of goods available for sale $8$10 $13 Cost of goods manufactured$5 $8 Gross margin$4 $7 Operating expenses$2 $4$5 Beginning finished goods inventory $2$3 Sales $10$14
## Gross Margin and Operating Income
Gross Margin is the net residual amount on sales after deducting the cost of goods sold. Operating income is the residual amount in gross margin after deducting operating expenses.
1. To fill In the missing data for the retail organization, we need the following formulas as guidance:
Sales - Cost of Goods Sold = Gross Margin
Gross Margin - Operating Expenses = Operating Income
Beginning Inventory + Net Cost of Purchases - Ending Inventory = Cost of Goods Sold
Beginning Inventory + Net Cost of Purchases = Cost of Goods Available for Sale
First Quarter Second Quarter Third Quarter Fourth Quarter
Sales 10 12 15 18
Gross margin 5 4 5 6
Ending merchandise inventory 5 4 5 3
Beginning merchandise inventory 4 5 4 5
Net cost of purchases 6 7 11 10
Operating income 3 2 4 2
Operating expenses 2 2 1 4
Cost of goods sold 5 8 10 12
Cost of goods available for sale 10 12 15 15
2. To fill in the missing data for the manufacturing organization, we need the following formulas as guidance:
Sales - Cost of Goods Sold = Gross Margin
Gross Margin - Operating Expenses = Operating Income
Beginning Inventory + Cost of Goods manufactured - Ending Inventory = Cost of Goods Sold
Beginning Inventory + Cost of goods manufactured = Cost of Goods Available for Sale
First Quarter Second Quarter Third Quarter Fourth Quarter
Ending finished goods inventory 2 3 5 6
Cost of goods sold 6 3 5 7
Operating income 2 3 2 2
Cost of goods available for sale 8 6 10 13
Cost of goods manufactured 5 4 7 8
Gross margin 4 7 6 7
Operating expenses 2 4 4 5
Beginning finished goods inventory 3 2 3 5
Sales 10 10 11 14 | 634 | 2,507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-05 | latest | en | 0.829104 |
http://indianapublicmedia.org/amomentofscience/physics-water-towers/ | 1,513,278,934,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550199.46/warc/CC-MAIN-20171214183234-20171214203234-00797.warc.gz | 129,252,487 | 19,010 | Give Now
# The Physics of Water Towers
When's the last time you turned on a faucet in your house and didn't get any water? Fortunately this almost never happens, but why?
Photo: DaylandS (flickr)
Water towers like this one in Texas usually provide a day's worth of a water to its town or city
When’s the last time you turned on a faucet in your house and didn’t get any water? Fortunately this almost never happens, thanks to your local water tower!
Water towers take advantage of the force of gravity to provide pressure for the water they contain. Every vertical foot adds point-four-three pounds per square inch to the water pressure. Towns usually keep their water pressure between fifty and one-hundred pounds per square inch, so a simple equation tells them how high to build the tower. And since a typical tower contains a full day’s worth of water, the force of gravity can maintain the hydrostatic pressure of the water system even when the power goes out.
In addition to providing a reliable emergency source of water, they serve an important day-to-day purpose as well. A city’s water usage varies throughout the day, usually peaking in the early morning when many people are showering and washing. Water usage during this time can be four or five times higher than during other parts of the day. To maintain water pressure during peak hours of the day, a city could invest in a very powerful pump, but this would be expensive and wasteful, since its capacity would go unused for most of the day. Instead, the city draws on the water tower’s supply of water during these hours of high demand, and in conjunction with the pumps, the supply is kept stable throughout the day.
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https://www.onlinemath4all.com/even-or-odd-arithmetic-rules.html | 1,586,205,544,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00195.warc.gz | 1,073,626,245 | 13,773 | # EVEN OR ODD ARITHMETIC RULES
## About "Even or odd arithmetic rules"
Even or odd arithmetic rules :
Using even or odd arithmetic rules we can easily evaluate an algebraic expression.
What is odd number ?
An odd number is an integer which is not a multiple DIVISIBLE of two. Examples 1, 3, 5, 7 ..........
What is even number ?
An even number is an integer which is "evenly divisible" by two. Examples 2, 4, 6, 8 ..........
The terms “even number” and “odd number” are only used for whole numbers and their opposites (additive inverses).
## Rules of adding odd and even numbers
• The sum of two odd numbers or even numbers is even
• The sum of odd and even numbers or even and odd numbers is odd
## Rules of multiplying odd and even numbers
• The product two odd numbers is odd
• The product of two even numbers or the product even and odd number is even
Addition Odd + Odd = EvenEven + Even = EvenOdd + Even = OddEven + odd = Odd Multiplication Odd x Odd = OddEven x Even = evenOdd x even = evenEven x odd = even
Example 1 :
Is 9 + 49 even or odd?
Solution :
9 = odd number
49 = odd number
Applying the rule,
Odd + Odd = Even
Hence, 9 + 49 is even.
We can do this problem in another way also,
9 + 49 = 58 (which is divisible by 2)
Hence, 9 + 49 is even.
Example 2 :
Is 36 + 120 even or odd?
Solution :
36 = even number
120 = even number
Applying the rule,
Even + Even = Even
Hence, 36 + 120 is even.
We can do this problem in another way also,
36 + 120 = 156 (which is divisible by 2)
Hence, 36 + 120 is even.
Example 3 :
Is 5 + 114 even or odd?
Solution :
5 = odd number
114 = even number
Applying the rule,
Odd + Even = odd
Hence, 5 + 114 is odd.
We can do this problem in another way also,
5 + 114 = 119 (which is not divisible by 2)
So, 5 + 114 is odd.
Example 4 :
Is 146 + 289 even or odd?
Solution :
146 = odd number
289 = even number
Applying the rule,
Even + Odd = odd
Hence, 146 + 289 is odd.
We can do this problem in another way also,
146 + 289 = 435 (which is not divisible by 2)
So, 146 + 289 is odd.
Example 5 :
Is 120 x 146 even or odd?
Solution :
120 = even number
146 = even number
Applying the rule,
Even x Even = Even
Hence, the product of 120 x 146 is even.
By applying the rule, we got the answer easily. If we want to use the alternate method then we have to multiply 120 and 146 and check whether the answer is divisible 2 or not.
Since this is the longer way, we may do this kind of problem using the above rule.
Example 6 :
Is 121 x 14 even or odd?
Solution :
121 = odd number
14 = even number
Applying the rule,
Odd x Even = Even
Hence, the product of 121 x 14 is even.
Example 7 :
Is 5 x 7 even or odd?
Solution :
5 = odd number
7 = odd number
Applying the rule,
Odd x Odd = Odd
Hence, the product of 5 x 7 is odd.
Example 8 :
Is 160 x 7 even or odd?
Solution :
160 = even number
7 = odd number
Applying the rule,
Even x Odd = Even
Hence, the product of 160 x 7 is even.
After having gone through the stuff given above, we hope that the students would have understood "Even or odd arithmetic rules".
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Sum of all three four digit numbers formed using 1, 2, 5, 6 | 1,432 | 5,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.852116 |
https://convertoctopus.com/1525-ounces-to-pounds | 1,716,868,592,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00703.warc.gz | 153,255,306 | 7,276 | ## Conversion formula
The conversion factor from ounces to pounds is 0.0625, which means that 1 ounce is equal to 0.0625 pounds:
1 oz = 0.0625 lb
To convert 1525 ounces into pounds we have to multiply 1525 by the conversion factor in order to get the mass amount from ounces to pounds. We can also form a simple proportion to calculate the result:
1 oz → 0.0625 lb
1525 oz → M(lb)
Solve the above proportion to obtain the mass M in pounds:
M(lb) = 1525 oz × 0.0625 lb
M(lb) = 95.3125 lb
The final result is:
1525 oz → 95.3125 lb
We conclude that 1525 ounces is equivalent to 95.3125 pounds:
1525 ounces = 95.3125 pounds
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pound is equal to 0.010491803278689 × 1525 ounces.
Another way is saying that 1525 ounces is equal to 1 ÷ 0.010491803278689 pounds.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand five hundred twenty-five ounces is approximately ninety-five point three one three pounds:
1525 oz ≅ 95.313 lb
An alternative is also that one pound is approximately zero point zero one times one thousand five hundred twenty-five ounces.
## Conversion table
### ounces to pounds chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to pounds
ounces (oz) pounds (lb)
1526 ounces 95.375 pounds
1527 ounces 95.438 pounds
1528 ounces 95.5 pounds
1529 ounces 95.563 pounds
1530 ounces 95.625 pounds
1531 ounces 95.688 pounds
1532 ounces 95.75 pounds
1533 ounces 95.813 pounds
1534 ounces 95.875 pounds
1535 ounces 95.938 pounds | 460 | 1,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2024-22 | latest | en | 0.826317 |
http://mathforces.com/problems/184/ | 1,695,887,330,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00818.warc.gz | 24,238,312 | 4,051 | # Triangle and Square
Author: mathforces
Problem has been solved: 68 times
Русский язык | English Language
Given a square $ABCD$ with an area of 100. An isosceles triangle $EDC$ has base $CD$. It turned out that the area of the common part of $EDC$ and $ABCD$ is 80. Find the length of the perpendicular dropped to $CD$ in the triangle $EDC$. | 100 | 345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.843436 |
http://www.mathskey.com/question2answer/39472/find-x-y-z-2x-3y-5z-30-x-y-z-10-2x-5y-4z-20 | 1,719,174,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864850.31/warc/CC-MAIN-20240623194302-20240623224302-00374.warc.gz | 50,493,873 | 11,189 | # Find x,y,z 2x+3y+5z=30, x+y+z=10, 2x-5y+4z=20
0 votes
Reduction method
asked Nov 2, 2018
reshown Nov 2, 2018
## 1 Answer
0 votes
The system of equations are
2x + 3y + 5z = 30 ---------------------> (1)
x + y + z = 10 ----------------------> (2)
2x - 5y + 4z = 20 ------------------------> (3)
Multiply Eq (2) with 5 and add to Eq (3)
Eq (2) X 5 ======> 5x + 5y + 5z = 50
Eq (3) =========> 2x - 5y + 4z = 20
----------------------------------
7x + 9z = 70 --------------------> (4)
Multiply Eq (2) with 5 and subtract it from Eq (1)
Eq (3) =========> 2x + 3y + 5z = 30
Eq (2) X 3 ======> 3x + 3y + 3z = 30
(-) (-) (-) (-)
----------------------------------
- x + 2z = 0
x = 2z --------------------------> (5)
Substitute x = 2z in Eq (4)
7(2z) + 9z = 70
14z + 9z = 70
23z = 70
z = 70/23
Substitute z = 70/23 in Eq (5)
x = 2(70/23)
x = 140/23
Substitute z = 70/23 and z = 140/23 in Eq (2)
70/23 + y + 140/23 = 10
(70 + 140)/23 + y = 10
(210)/23 + y = 10
y = 10 - (210)/23
y = (230 - 210)/23
y = 20/23
Answer :
The Solutions are z = 70/23, y = 20/23 and z = 140/23.
answered Nov 4, 2018 | 552 | 1,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.213879 |
http://bitperfectsound.com/?m=201310 | 1,618,387,567,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077336.28/warc/CC-MAIN-20210414064832-20210414094832-00336.warc.gz | 13,471,139 | 16,710 | Yesterday, I introduced you to the concept of Jitter, and showed how it has the potential to disrupt the accuracy of digital audio playback. We saw how the measures necessary to eliminate jitter as a problem can impose unpleasant and challenging constraints upon the designer of an audiophile-grade DAC. It would be reasonable for us to ask what the audible effects of this jitter actually are, and how we can determine the efficacy with which a DAC design has addressed it. This post attempts to address these questions.
This analysis is all about jitter as a digital phenomenon, by which we mean that we are concerned only by the notional effect of playing back the wrong signal at the right time (or vice-versa). We assume that the only effect of jitter is that which we have described here as arising from fanatically precise timing errors. Can we calculate what audible or measurable effect such timing errors can have? Lets take a look.
The first thing we have to consider is what we call the distribution of jitter timing errors. As a trivial example, let us imagine that every single timing point is subject to a jitter-based timing error of 100ns (100 nanoseconds – see yesterday’s post for an explanation). 100ns is a large value, and experiments have suggested that jitter of this magnitude is quite audible. In our trivial example suppose each and every sample is delayed by exactly 100ns. What we have in fact accomplished is to simply take perfect playback and delay it by 100ns. We could achieve exactly the same thing by moving our loudspeakers back by a hair’s breadth. It certainly won’t affect the sound quality. So just having 100ns of jitter is not in and of itself enough to cause a problem. What we need to see is an uncertainty or variability in the precise timing of the sampling process, in other words the individual samples end up being off by unknown amounts which could average out to 100ns. In this case, some samples are delayed, some are advanced, and some are not affected. Some of these errors are large, and some are small. We don’t know which are which. All we know is that, on average, the errors amount to some 100ns. In other words, there is a “distribution” of timing errors. Clearly, it is possible to imagine how the audible effects of such a collection of timing errors might be affected by exactly how these errors are distributed.
In order to analyze what the effects of jitter might be, we have to classify those effects according to different types of distributions – the way in which the precise jitter values vary from one sample to the next. The best way to begin is to divide these distributions into two categories – correlated and uncorrelated jitter. Uncorrelated jitter is the easiest to understand. The jitter value for any sample is totally random. It is like rolling a dice – there is absolutely no way to predict what any one given jitter value will be. As we will see, uncorrelated jitter is the easiest form of jitter to analyze. All other forms of jitter are, by definition, correlated jitter. The jitter values correlate to some degree or another with some other property. Correlation does not necessarily mean that its exact value is determinable. It can be more like rolling a loaded dice. Some values end up being more likely than others. Analysis of correlated jitter is way more challenging.
Uncorrelated (or random) jitter turns out to be very similar to dithering, which I treated in an earlier post. Uncorrelated jitter introduces a random error into the value of each sample. The effect of this is quite simple – it increases the noise floor. The analysis is slightly complicated by the fact that the amount of noise is dependent on the frequency spectrum of the signal, but in general this analysis shows that noise floor increases of 10dB and more across the entire audio spectrum can obtained with as little as 1ns of uncorrelated jitter. However, as with dithering, this noise may not be perceptible, as it may lie below the noise floor introduced by other (analog) components in the audio playback chain. Qualitatively, uncorrelated jitter is not normally considered to be a significant detriment to the sound quality.
Correlated (or deterministic) jitter is a more complicated beast. Correlated jitter may correlate with a number of factors, including the audio signal, the power supply (mains frequency and its harmonics), clock circuits, external sources of RF interference, and other factors which may be very difficult to pin down. Its frequency spectrum and bandwidth need to be taken into account. If the jitter behaves in a tightly deterministic manner, we can perform some very accurate mathematical analysis of its behaviour to determine its effect on the audio signal, but deviations from even the simplest forms of deterministic jitter make the analysis and its interpretation exponentially more difficult.
Lets take the simplest case of an audio signal comprising a single pure tone, and a jitter function which behaves as a pure sinusoid. A simple Fourier Transform of the resultant audio signal will be found to exhibit the single peak of the original audio pure tone, plus two symmetrical side lobes. The magnitude and separation of the side lobes will permit us to calculate both the frequency and magnitude of the jitter signal. This is a highly specific and limiting case, and it is highly unlikely that any real-world jitter scenario would ever be that simple. But for the most part it is all we have!
At this point, I would normally go into a little bit more detail on how real-world jitter measurements are performed, but it really is too complicated. Suffice to say that it typically depends on Fourier Analysis of an audio signal with a single tone, in some cases further modulated by a very low-level lower-frequency square wave which produces a family of reference harmonics, based on the analysis I described above. The technique involves looking for pairs of symmetrical side lobes and attempting to infer the corresponding jitter contributions. As you can see, this type of analysis will fail to take into account distributions of jitter which are not properly described by the (highly simplified) underlying mathematical model, and its accuracy will be limited by the validity of the model, which, as I have observed, gets waaaay more difficult to interpret as the modeled system gets more complicated. The net effect is that these more elaborate analyses are limited by the assumptions that have to be made in order the make the math more manageable, and the accuracy of the results is in the end limited by the validity of the assumptions. The disconnect between the two is a very real problem – compounded by the fact that the person using the analysis tool is not normally familiar with the underlying mathematics, nor the assumptions upon which it rests.
The audibility of these jitter modes are far more difficult to predict, even with the assistance of the limited mathematical modelling. Unlike the results with uncorrelated (random) jitter, correlated jitter often results in specific frequency peaks. This type of behaviour is more like distortion than it is noise, and we know that the human ear tends to be far less tolerant of distortions than noise, with some distortions (such as intermodulation distortion) being much worse than others (such as even harmonic distortion). At this point, it is not possible to entirely dismiss the notion that some classes of jitter may be both impossible to observe and measure, yet at the same time deleterious to sound quality.
But is jitter really what I have described, and does it really impact the system in the way I have described it? Or could something else be in play? Tomorrow, in the final installment of this short series, I will start to consider this idea further.
I want to address a critical phenomenon for which there isn’t an adequate explanation, and provide a rationale for it in terms of another phenomenon for which there isn’t an adequate explanation. Pointless, perhaps, but it is the sort of thing that tends to keep me up at nights. Maybe some of you too!
Most of you, being BitPerfect Users, will already know that while BitPerfect achieves “Bit Perfect” playback (when configured to do so), so can iTunes (although configuring it can be a real pain in the a\$\$). Yet, I am sure you will agree, they manage to sound different. Other “Bit Perfect” software players also manage to sound different. Moreover, BitPerfect has various settings within its “Bit Perfect” repertoire – such as Integer Mode – which can make a significant difference by themselves. What is the basis for this unexpected phenomenon?
First of all, we must address the “Flat Earth” crowd who will insist that there cannot possibly BE any difference, and that if you say you can hear one, you must be imagining it. You can spot them a mile away. They will invoke the dreaded “double-blind test” at the drop of a hat, even though few of them actually understand the purpose and rationale behind a double-blind test, and have neither organized nor ever participated in one. I tried to set up a series of publicly-accessible double-blind tests at SSI 2012 with the assistance of a national laboratory’s audio science group. They couldn’t have shown less interest if I proposed to infect them with anthrax. Audio professionals generally won’t touch a double-blind test with a ten foot pole. Anyway, as far as the Flat Earth crowd are concerned, this post, and those that follow, are all about discussing something that doesn’t exist. Unfortunately, I cannot make the Flat Earthers vanish simply by taking the position that they don’t exist!
For the rest of you – BitPerfect Users, plus anyone else who might end up reading this – the effect is real enough, and a suitable explanation would definitely be in order. That is, if we had one for you.
If it is not the data itself (because the data is “Bit Perfect“), then we must look elsewhere. But before we do, some of you will ask “How do we know that the data really is Bit Perfect?“, which is a perfectly reasonable question. But it is not one I am going to dwell on here, except to say that it has been thoroughly shaken down. Using USB it is actually quite easy to do (from the perspective of not being technically challenging), although using S/PDIF requires an investment in very specific test equipment. Bottom line, though, is that this has been done and nobody holds any lingering concerns over it. I won’t address it further.
As Sherlock Holmes might observe, once we accept that the data is indeed “Bit Perfect“, the only thing that is left is a phenomenon most of us have heard of, but few of us understand – jitter. Jitter was first introduced to audiophiles in the very early 1990’s as an explanation for why so many people professed a dislike for the CD sound. Digital audio comprises a bunch of numbers that represent the amplitude of a musical waveform, measured (“sampled” is the term we use) many thousands of times per second. Some simple mathematical theorems can tell us how often we need to sample the waveform, and how accurately we need those sample measurements to be, in order to achieve specific objectives. Those theorems led the developers of the CD to select a sample rate of 44,100 times per second, and a measurement precision of 16-bits. We can play back the recorded sound by using those numbers – one every 1/44100th of a second – to regenerate the musical waveform. This where jitter comes in. Jitter reflects a critical core fact – “The Right Number At The Wrong Time Is The Wrong Number“.
Jitter affects both recording and playback, and only those two stages. Unfortunately, once it has been embedded into the recording you can’t do anything about it, so we tend to think of it only in terms of playback. But I am going to describe it in terms of recording, because it is easier to grasp that way.
Imagine a theoretically perfect digital audio recorder recording in the CD format. It is measuring the musical waveform 44,100 times a second. That’s one datapoint every 23 microseconds (23 millionths of a second). At each instant in time it has to measure the magnitude of the waveform, and store the result as a 16-bit number. Then it waits another 23 microseconds and does it again. And again, and again, and again. Naturally, the musical waveform is constantly changing. Now imagine that the recorder by mistake measures the reading a smidgeon too early or too late. It will measure the waveform at the wrong time. The result will not be the same as it would have been if it had been measured at the right time, even though when the measurement was taken, it was taken accurately. We have measured the right number at the wrong time, and as a result it is the wrong number. When it comes time to playback, all the DAC knows is that the readings were taken 44,100 times a second. It has no way of knowing whether any individual readings were taken a smidgeon too early or too late. A perfect DAC would therefore replay the wrong number at the right time, and as a result it will create a “wrong” waveform. These timing errors – these smidgeons of time – are what we describe as “Jitter“. Playback jitter is an identical problem. If the replay timing in an imperfect real-world DAC is off by a smidgeon, then the “right” incoming number will be replayed at the “wrong” time, and the result will likewise be a wrong waveform.
Just how much jitter is too much? Lets examine a 16-bit, 44.1kHz recording. Such a recording will be bandwidth limited theoretically to 22.05kHz (practically, to a lower value). We need to know how quickly the musical waveform could be changing between successive measurements. The most rapid changes generally occur when the signal comprises the highest possible frequency, modulated at the highest possible amplitude. Under these circumstances, the waveform can change from maximum to minimum between adjacent samples. A “right” number becomes a “wrong” number when the error exceeds the precision with which we can record it. A number represented by a 16-bit integer can take on one of 65,536 possible values. So, a 16-bit number which changes from maximum to minimum between adjacent samples, cycles through 65,536 distinct values between samples. Therefore, in this admittedly worst-case scenario, we will record the “wrong” number if our “smidgeon of time” exceeds 1/65535 of the time between samples, which you will recall was 23 millionths of a second. That puts the value of our smidgeon at 346 millionths of a millionth of a second. In engineering-speak that is 346ps (346 picoseconds). That’s a very, very short time indeed. In 346ps, light travels 4 inches. And a speeding bullet will traverse 1/300 of the diameter of a human hair.
I have just described jitter in terms of recording, but the exact same conditions apply during playback, and the calculations are exactly the same. If you want to guarantee that jitter will not affect CD playback, it has to be reduced to less that 346ps. However, in the real world, there are thing we can take into account to alleviate that requirement. For example, real-world signals do not typically encode components at the highest frequencies at the highest levels, and there are various sensible theories as to how to better define our worst-case scenario. I won’t go into any of them. There are also published results of real-world tests which purport to show that for CD playback, jitter levels below 10ns (ten nanoseconds; a nanosecond is a thousand picoseconds) are inaudible. But these tests are 20 years old now, and many audiophiles take issue with them. Additionally, there are arguments that higher-resolution formats, such as 24-bit 96kHz, have correspondingly tighter jitter requirements. Lets just say that it is generally taken to be desirable to get jitter down below 1ns.
If you require the electronics inside your DAC to deliver timing precision somewhere between 10ns and 346ps, this implies that those electronics must have a bandwidth of somewhere from 100MHz to 3GHz. That is RF (Radio Frequency) territory, and we will come back to it again later. Any electronics engineer will tell you that electrical circuits stop behaving sensibly, logically and rationally once you start playing around in the RF. The higher the bandwidth, the more painful the headaches. Electronics designer who work in the RF are in general a breed apart from those who work in the AF (the Audio Frequency band).
The bottom line here is that digital playback is a lot more complicated than just getting the exact right bits to the DAC. They have to be played back with a timing precision which invokes unholy design constraints.
Tomorrow I will talk about the audible and measurable effects of jitter.
Chamber Music. Even the term itself is enough to put people off. It is a genre which many people file under the same folder as waterboarding. And in truth, on occasion it does feel like it belongs there.
There is one chamber work, though, which I would encourage anybody for whom music is – in whatever form – an important part of your life, to take some time aside to sit down and listen to. Arguably the greatest chamber work ever written, Schubert’s String Quintet D956, composed only two months before his untimely death from syphilis, aged only 31. Listen to this in a dark room, on headphones, accompanied by a glass of your finest single malt scotch, having secured iron-clad assurances, on pain of death, that under no circumstances will you be disturbed. This is music that entwines itself with your very soul, poses questions you cannot answer, and satisfies longings you never knew you craved.
A String Quartet is a standard musical ensemble, comprising two violins, a viola and a cello. A String Quintet, on the other hand, is a more flexible designation – the fifth player is usually another viola, but in this case a second cello is called for. Two cellos would suggest a sonic imbalanced in the bass, but in the expert hands of Franz Schubert it instead adds an almost symphonic depth to the soundscape. A great performance can make you think you are listening to a chamber orchestra. Performances of D956 fall into two categories. Because of the stature of the piece, it is often performed by an ensemble of soloist superstars, gathered for the task, more with an eye on the box office than an ear to the music. The standard alternative is to take an established String Quartet and add an accomplished solo cellist. The choice and performance of the second cellist is an existential one for the performance, since this part drives and leads much of what will come to define the performance.
I have alluded to the symphonic nature of the piece. Indeed, on closer inspection it can come across as a chamber transcription of a bigger piece. Go play Wagner’s “Siegfried Idyll” and imagine what his orchestration of D956 could have sounded like. On the other hand, we are talking about one of music’s great masterpieces here, and as Fats Waller said, “If you don’t know what it is, don’t mess with it”. Written merely four years after Beethoven’s iconic ninth symphony, D956 looks more forward to Mahler more than it does back to Beethoven. It is more profound and introspective, less overtly melodic than Beethoven – you won’t be humming its tunes on your way home from the office – and its developmental structure is more complex and elaborate. D956 is all about soundscapes, textures, and moods, right the way through to the bizarre final chord, which comes across like a bum left hand note played by an over-excited pianist who leaps too high on his final flourish and lands in the wrong place (I confess, I don’t know what Schubert had in mind there).
I have yet to come across a “definitive” recording of D956. I have four, by the Emerson, Takács, Tokyo, and Vellinger string quartets, each with a guest cellist. Each has something to be said for it. The Emerson is notable for its great tonal beauty, the Takács for its liquid playing, the Tokyo is the most classically refined, and the Vellinger offers an ascetic, soul-baring honesty. As a purely personal opinion, I tend to gravitate to the Vellinger, which is hard to come by because it was a free giveaway with the BBC Music Magazine about 20 years ago, so it is unfair to recommend, but to me it best captures the soul of the piece. But all four paint dramatically different pictures, with the contrast between the Emerson (imagine Iván Fischer conducting) and the Vellinger (imagine Pierre Boulez conducting) occupying the extremes. Continuing with that analogy, the Tokyo could be Arturo Toscanini, and the Takács perhaps even Carlos Kleiber. They’re all very, very good, and the differences are primarily of style rather than musicianship.
It may be Chamber Music, but it is magnificent.
Check out Light Harmonic’s new crowd-sourcing campaign – the “Geek Pulse”! Yes, the same Light Harmonic, maker of the mega-buck Da Vinci DAC, are now developing a product at the other end of the price spectrum, bringing ultra high-resolution PCM, together with the very latest in DSD playback support, to the market at a VERY affordable price. I can’t wait to get my hands on one!
It has taken me longer than usual to finally pronounce on iTunes 11.1.2, but here I am. I wanted to take a little longer, because a very small number of users have posted on our FaceBook page, and also through the e-mail support line, that they have encountered unexpected problems after installing the combination of OS/X Mavericks and iTunes 11.1.2.
Here at BitPerfect I have been running that combination for two days solid and have not had a single problem. Furthermore, one or two of those users who did encounter problems have reported that these problems have suddenly vanished.
On balance, therefore, I don’t really see any good reason why you should not all make the update if you want to. I suspect, by the way, although I am not certain about this, that if you upgrade to OS/X Mavericks, you might get iTunes 11.1.2 as part of the package, whether you want it or not.
Now that OS/X Mavericks has been released, we can finally announce something we have known since the summer, but have been forbidden from disclosing. After a two year absence under Lion and Mountain Lion, Integer Mode is back again with OS/X Mavericks, and BitPerfect 1.0.8 already includes the software necessary to support it.
We have been using Integer Mode under BitPerfect 1.0.8 (and also using our own pre-release betas) since early summer, ever since we received the first pre-release betas of Mavericks. Both Mavericks itself, and its Integer Mode functionality under BitPerfect have been totally problem-free.
We are therefore confident that BitPerfect Users can upgrade to Mavericks.
Be aware that not all DACs support Integer Mode. And I don’t know of anybody out there who is maintaining an up-to-date list of Integer Mode compatible DACs, so please do not ask me for advice on that subject 🙂
#### Be aware that not all DACs support Integer Mode. And I don’t know of anybody out there who is maintaining an up-to-date list of Integer Mode compatible DACs, so please do not ask me for advice on that subject 🙂
A busy morning here at BitPerfect Global HQ. OS/X Mavericks has been released, together with an update to iTunes, version 11.1.2. I am currently using both, and it seems to be working just fine, but I am also getting reports from BitPerfect users who are encountering problems.
I have been using Mavericks in its pre-release forms for some time now, and have never encountered a problem with it, so I really don’t see any reason why BitPerfect users should not be able to upgrade with confidence.
On the other hand, iTunes updates have always been a cause for concern, so I recommend that BitPerfect users hold back from updating iTunes for the time being. I will post an update here in due course.
A busy morning here at BitPerfect Global HQ. OS/X Mavericks has been released, together with an update to iTunes, version 11.1.2. I am currently using both, and it seems to be working just fine, but I am also getting reports from BitPerfect users who are encountering problems.
I have been using Mavericks in its pre-release forms for some time now, and have never encountered a problem with it, so I really don’t see any reason why BitPerfect users should not be able to upgrade with confidence.
On the other hand, iTunes updates have always been a cause for concern, so I recommend that BitPerfect users hold back from updating iTunes for the time being. I will post an update here in due course.
Last night, on TV, my wife and I watched an episode of the current season of “The Amazing Race”. In our house we have a modest, but surprising effective home theater system permanently connected to our TV set. It is in play regardless of what we are watching on TV. Our TV signal is derived from satellite, time-shfted on our PVR, and the show was on a HD channel with the sound encoded in Dolby Digital. The sound delivered by this system is normally very clear, but in this case it was an absolute cacophany, and I can only describe it as a shameless assault on my ears.
The Amazing Race sees itself as a non-stop action show, punctuated by the occasional pause for an interlude of weepy all-American sentimentalism. It plays against a continuous background of “Action Movie!!” orchestral blasts, noisy, percussive, syncopated. No melody at all, and no let-up in its ongoing intensity. It is mixed with the maximum possible amount of compression, and presented at the maximum volume, so that it is continuously, relentlessly loud. It accompanies the action non-stop.
The show also provides a commentary, delivered by a shouting host, interspersed with snippets of interjections from the various participants. The commentary tracks is separately mixed, and is also mastered with the maximum amount of compression, at the maximum volume.
Since the music track and the commentary track are each fully capable of drowning out the other, the producers have determined that the commentary track must take precedence. So the loudness of the commentary track is used to modulate the loudness of the music track. When somebody shouts, the music is briefly backed off a little, and immediately ramped back up after, even if they are just pausing for breath.
The net effect is a relentless assault on the eardrums. It makes it very hard to follow the dialog without getting a headache, and in fact makes watching the show a less than pleasant experience. Your brain is not equipped to deal with such heavily compressed and modulated sounds, and goes into overload. I found myself wondering if the CIA would have gotten into as much hot water as they did if they had used “The Amazing Race” instead of waterboarding.
This was way worse than even the last season (or was it the last-but-one season) of “House”, where there was no music track, but in its place the ambient background noises of the set were amplified to the point where it was dominated by hiss. This hissy noise was then massively modulated by the dialog. Again, a ruinous detraction from the enjoyment of the show.
These people need to get into another line of work. Now there’s a thought… Maybe these ARE the same people who got fired from the CIA for waterboarding! | 5,914 | 27,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.948489 |
blog.medallionbank.com | 1,500,597,926,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423629.36/warc/CC-MAIN-20170721002112-20170721022112-00522.warc.gz | 622,070,823 | 36,605 | # Fun with Math, Part 2: Credit Risk Illustrated
A common view of bankers is that we spend our mornings swindling the public and our afternoons rolling around in piles of money. While I can’t speak for the investment banking community, life in commercial banking is far less edgy (and far more ethical). In our business, service is essential and we operate on very narrow margins. And, as the recent crisis showed, mistakes can have devastating financial consequences.
On an almost-hourly basis, we’re asked to look a little more carefully at a deal we originally declined. Credit risk, you see, is an abstract notion to non-bankers. The customers being presented are always “good people” or really nice, or really need the money. We like to work with people like that, but it isn’t quite as simple as saying yes and hoping for the best. Credit risk is real and bad underwriting has big consequences.
Measures of profitability in banking
To see this, let’s go over two basic measurements of banking profitability. In commercial banking, a return on assets (ROA) of 1.5% is considered good. Because that number is an after-tax figure, the pre-tax equivalent is 2.3% assuming a 35% corporate tax rate. We’ll use that number later, too. Banks also measure their return relative to the equity – capital – invested in the company. For example, a bank with 12% capital and a 1.5% ROA is generating a 12.5% return on equity (ROE; 100%/12*1.5). You don’t need to memorize those ratios, but they’ll help make sense of the next section.
Three scenarios show the small margin of error
To see what credit risk does to a bank’s income statement, consider three scenarios. In each we’ll have a \$1 million loan portfolio made up of 100 loans, each loan with a \$10,000 balance. The respective scenarios have between zero and two charge offs, respectively. It doesn’t seem like there’s much difference between no loans in 100 and two loans in 100 charging off, but the effect is substantial. (For simplicity, I’ll calculate the ROA on the portfolio value, not the average assets.)
In short, on a \$1 million portfolio like this, a single loss takes good performance to mediocre, decreasing the ROE from 12.5% to 7.1%. The second loss almost completely wipes out the gains for the bank, leaving it with a 1.67% ROE. Can you imagine how happy the bank’s investors will be with such miniscule returns? And you know what happens with a third charge-off that year, right? The bank loses money, management is on the hot seat, pressure increases – you know the drill.
With narrow margins like this, it is critical for a bank to make good loan decisions and to price their loans appropriately. If we can’t approve a deal for a really nice person, it doesn’t reflect that individual’s character. It’s just that we must be very careful in order to remain in business for the long haul. | 679 | 2,891 | {"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} | 3.40625 | 3 | CC-MAIN-2017-30 | longest | en | 0.918979 |
https://www.physicsforums.com/threads/low-pass-filters-frequency-selection-and-h-jw.548618/ | 1,586,129,648,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00556.warc.gz | 1,072,261,885 | 21,154 | # Low-pass filters, frequency selection and H(jw)
## Main Question or Discussion Point
Hello,
Recently I've encountered with this topic. Low pass filters and such.
But, generally I am not comfortable with this new idea.
How does this frequency selection works? I mean, I know the math behind it, like integrals H(jw) etc. But how does a, lets say capacitor in a RC circuit filter frequencies ?
I cannot wrap my head around that idea of filtering frequencies.
Related Electrical Engineering News on Phys.org
See if this approach helps:
Sketch up three instances of a simple lowpass RC filter with a 1k resistor and a capacitance of 1nF, 100nF and 10uF. Calculate the reactance value for the three different values of capacitance at a fixed frequency (3 kHz will be illustrative in this case) and treat each case as a simple voltage divider.
See if this approach helps:
Sketch up three instances of a simple lowpass RC filter with a 1k resistor and a capacitance of 1nF, 100nF and 10uF. Calculate the reactance value for the three different values of capacitance at a fixed frequency (3 kHz will be illustrative in this case) and treat each case as a simple voltage divider.
I think I get it. If we would then increase frequency, or plot the f(ω), we would get smaller voltage right, for larger frequencies? Because frequency or angular frequency ω is in the denominator?
Did I get it?
Yes. The well-known output expression for the resistive voltage divider, Vout = Vin * R2/(R1+R2), is valid also in the general case with impedances.
For your frequency dependent RC filter output, replace R2 with XC = 1/ωC, and it's clear that Vout → 0 when ω → ∞. In practice, you chose R and C so that the cutoff frequency is somewhere below ∞. E.g. you might want to http://en.wikipedia.org/wiki/Cutoff_frequency" [Broken] signals > 10kHz.
Last edited by a moderator:
Yes. The well-known output expression for the resistive voltage divider, Vout = Vin * R2/(R1+R2), is valid also in the general case with impedances.
For your frequency dependent RC filter output, replace R2 with XC = 1/ωC, and it's clear that Vout → 0 when ω → ∞. In practice, you chose R and C so that the cutoff frequency suits your needs (e.g. you want to get rid of signals > 10kHz)
I understand! Thank you. They don't teach us stuff like this in my course. I mean, we are struggling with basics, how could they? :D
jim hardy
Gold Member
2019 Award
Dearly Missed
nice approach, the voltage divider with r2 = f(frequency)
when you build a mental picture that leads the mind naturally to the proper formula, you are developing prowess, in my opinion.
old jim
nice approach, the voltage divider with r2 = f(frequency)
when you build a mental picture that leads the mind naturally to the proper formula, you are developing prowess, in my opinion.
old jim
Yes. Nowadays everybody just tells you this is this, and that is that. I got explanation in my book for H(jw): this is frequency characteristic of the system.(sort of translation).
And I have to conclude everything from this ! :/
But now when I understood this RC circuit and how it limits the frequency, an old thought crossed my mind.
LC circuit. Maybe a year ago I heard that, through LC circuit we are tuning radio stations.
Changing the parameters(usually C through varicap diode) we are finding the resonant frequency of this LC circuit? Or better said we are changing the resonant frequency of this circuit, selecting just specific radio frequency band, at which LC will resonate ergo will pass the signal, ergo I will hear something.
Am I right?
jim hardy
Gold Member
2019 Award
Dearly Missed
just right.
in early days of radio experimenters built home-made radios with just a tunable coil-capacitor and simple rectifier. they could intercept enough power from a wire antenna to drive headphones , with no other power source. i had one when i was about seven.
old jim
Yes. The well-known output expression for the resistive voltage divider, Vout = Vin * R2/(R1+R2), is valid also in the general case with impedances.
For your frequency dependent RC filter output, replace R2 with XC = 1/ωC, and it's clear that Vout → 0 when ω → ∞. In practice, you chose R and C so that the cutoff frequency is somewhere below ∞. E.g. you might want to http://en.wikipedia.org/wiki/Cutoff_frequency" [Broken] signals > 10kHz.
Actually, the formula is a little different because of the phase shift introduced by the capacitor.
Now, switch the RC low pass filter around so that you have a CR high pass filter. Does that make sense to you?
One important thing to remember about these simple filters is that the cutoff frequency is where X = R.
$R = X_{C}$
$R = \frac{1}{2\pi f C}$
Using your leet algebra skills, solve for f.
The situation with resonant filters is similar. It is where:
$X_{L} = X_{C}$
$2 \pi f L = \frac{1}{2\pi f C}$
Last edited by a moderator:
One more thing though...
When we are plotting H(jω), i see that for the horizontal coordinate line is taken ω and not jω...
I see H(jω) as a function that limits the frequencies, and here in book they plotted it for some example. They said that for simplicity we are considering these spectres to be real, but what do I do with that "j"?
How is that so?
Actually, the formula is a little different because of the phase shift introduced by the capacitor.
Now, switch the RC low pass filter around so that you have a CR high pass filter. Does that make sense to you?
One important thing to remember about these simple filters is that the cutoff frequency is where X = R.
$R = X_{C}$
$R = \frac{1}{2\pi f C}$
Using your leet algebra skills, solve for f.
The situation with resonant filters is similar. It is where:
$X_{L} = X_{C}$
$2 \pi f L = \frac{1}{2\pi f C}$
Yea take the frequencies off the resistor right? Because, loosely speaking, capacitor acts like a "short circuit" for very large ω.(his resistance is ->0)
Actually, the formula is a little different because of the phase shift introduced by the capacitor.
Thanks for the correction, Jiggy-Ninja. I was sloppy in my attempt to keep things simple, and you're absolutely right; for an impedance divider, R2 should be replaced with the complex impedance of the capacitor, Z2 = -j * 1/ωC, and not the reactance XC = 1/ωC. The output then becomes
Vout = Vin * Z2/(Z1 + Z2) = Vin * 1/(1 + jωR1C)
So the fact that the RC lowpass circuit attenuates signals of high frequency was only half the story--it also adds a phase shift to the output voltage relative to the input voltage when Z1 begins to dominate Z2 in the above expression.
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psparky
Gold Member
Wow...you guys are smart.
I always remember the cutoff frequency as 1/(2*pi*RC)....as stated above. Start with either the Resistor or capacitor, plug and chug and go from there.
Also, keep in mind that the cutoff frequency you just created is DIRECTLY related to the speed of the circuit.
The little equation e^(-t/RC).....plug in your RC constant from above and walla.....in DC transient anyways.
Also, if you find your transfer funtion using 1/(JWC) and R.....you can find the exact gain and phase angle at any given frequency. Yes, this refers to the bode plot as well.
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I am so angry with myself!!!
My exam today from signals and systems, was really hard! But I knew EVERYTHING how to solve.
But then the human-error mistress came and made me forgot to divide by 2, in two of the problems!
I will have low points because I made a mistake in calculations!! :///
jim hardy
Gold Member
2019 Award
Dearly Missed
how familiar. i suffer from awkwardness as well.
it's okay.
that you understood the basic science should give you confidence and make you more relaxed for next exam.
we do our best when unencumbered by self doubt .
best way to prepare for exams is to keep up with classwork, go to bed early night before, walk into exam relaxed and alert. never 'cram' .
old jim
how familiar. i suffer from awkwardness as well.
it's okay.
that you understood the basic science should give you confidence and make you more relaxed for next exam.
we do our best when unencumbered by self doubt .
best way to prepare for exams is to keep up with classwork, go to bed early night before, walk into exam relaxed and alert. never 'cram' .
old jim
I guess :(. I will most certainly do better on next exam. Because, when it comes to understanding the problem, I did have no problems.
Guess I had the yips or something.
psparky
Gold Member
The good news is that you have the whole rest of your life to make tons of more mistakes!~!!!!
Congrats!~!!!
The good news is that you have the whole rest of your life to make tons of more mistakes!~!!!!
Congrats!~!!!
hahahahahaha :D
Looking forward to them :D | 2,137 | 8,754 | {"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} | 3.578125 | 4 | CC-MAIN-2020-16 | longest | en | 0.887241 |
http://gibitrains.pagesperso-orange.fr/en/automatisme/automatisme/automatisme-12.htm | 1,539,969,259,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512421.5/warc/CC-MAIN-20181019170918-20181019192418-00538.warc.gz | 158,400,970 | 2,906 | ## Programming
### Example : control of turnout A
We’ll look for the orders that must put the turnout in the diverging position, and we’ll bring them by a logical OR on the coil SQ1. Similarly, we’ll look for the orders that must put the turnout in the direct position, and we’ll bring them by a logical OR on the coil RQ1.
According to the route table, turnout A must be made diverging by VOIE 1 OR VOIE 3 OR VOIE 5 or closing of switch A. It must be placed in direct position by BOUCLE or TIROIR or by opening of switch A.
### Explanation of the functioning of the edges
Remember that the switch of the turnout A is connected to the input I1. Assuming that its starting position is open (as in the diagram), the coils TT1 and TT2 do not detect a state change, nothing happens at the contacts T1 and T2 which are controlled respectively by TT1 and TT2. Now if the switch gets closed, TT1 detects that the voltage has passed from 0 to 24 V, which is called a rising edge, and responds by closing its contact T1 during the programmed time, for example 10 ms, which is a sufficient time to command SQ1. Set Q1 means set output Q1 to the high level, i.e. to the 24 V voltage (or to leave it in this state if it was already such before). This output will energize the turnout control relay, which results in controlling the turnout in the diverging position.
If the switch gets opened, TT2 detects that the voltage has passed from 24 V to 0, which is called a falling edge, and responds by closing its contact T2 during the programmed time, for example 10 ms, which is sufficient to command RQ1. Reset Q1 means set output Q1 to the low level, i.e. to the 0 V voltage (or to leave it in this state if it was already such before). This output will de-energize the turnout control relay, which results in controlling the turnout in the direct position.
Of course, if in the meantime an automatic control has occurred, by I8, I9, etc., the turnout may have changed position. | 477 | 1,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-43 | longest | en | 0.939227 |
https://www.barrowby.lincs.sch.uk/maths-89/ | 1,606,299,517,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182776.11/warc/CC-MAIN-20201125100409-20201125130409-00399.warc.gz | 588,316,050 | 15,985 | # Maths
Mental/Oral Starter:
Year Ones – Can you crack the code – Find the answer to addition and subtraction number sentences, using a number of mental strategies, counting on, back, mental recall, etc. Write the final answers in the lock to win!
Receptions: Pick a number – what is the number? Write it in the air. Where do we start? Repeat with a few numbers up to 10. Focus numbers…2,3,5,8
Main Teaching:
Year Ones: Problem Solving.
Read the problems below? How can we solve this? What do we know about this problem? We know that we have 14 but that the answer is 20 so how many more do we need?
Ask an adult to write numbers 0-20 along a line, at each number put a little mark just like on a ruler.
We are going to use this number line to count on to find the difference, e.g. 14 + __ = 20
Circle 14 and 20 and then jump from 14 up to 20 onto the number line, count each jump as you go. The number of jumps is the difference, that is the answer.
Problems
1. I have 14 apples but I have 20 friends. How many more apples do I need so all my friends can have an apple?
2. There are 12 cars and 20 car parking spaces. How many more cars will it take to fill up the car park?
3. There are 20 children in the class but only 16 pencils. How many more pencils do I need?
Email your completed calculations to me.
Receptions: Let's practise adding and subtracting: Get 4 toys, what is one more? Will the number get bigger or smaller? Let’s find out. Get one more toy. Are there more or less? Let's count. Count them carefully, it is easy to make sure you've counted them all if you move your toys when you've counted them.
Repeat with more numbers. I have 5, what is one less? How can we work it out? Can you write it in the air?
Encourage children to use different language related to addition and subtraction. e.g. add, plus, more, subtract, minus, less, take away. Take a photograph and send it to me via Tapestry.
Mental/Oral Starter:
Year Ones: Can you count up to 100? Can you make it active and think of two or three actions to repeat as you count?
Receptions: How far can you count up to? Pick a number? Let’s write it in the air. Where do we start? Repeat with a few numbers up to 10. Focus numbers…2,3,5,8.
Main Teaching:
Year Ones: Complete the finding the difference calculations below. There are number lines under each calculation for you to use. Don't forget to email the work to me.
Reception: Hide some numbers around your living room! Go on a number hunt. What number have you found? What is one more than that number? How did you work it out? Can you find the number that is one more than…? One less than…? Can you write it down? Upload your work to Tapestry.
Mental/Oral Starter:
Year Ones: Can you count up to 100? Can you make it active and think of two or three actions to repeat as you count?
Receptions: How far can you count up to? Pick a number? Let’s write it in the air. Where do we start? Repeat with a few numbers up to 10. Focus numbers…2,3,5,8.
Main Teaching:
Year Ones: Continue to practise the 'finding the difference' calculations. Are you getting the hang of it? don't forget to email me with your completed work.
Receptions: Jump (number on the carpet or outside on the patio) – Can you jump on the number that is one less than 2? Can you stand on number 5? Now jump to one less, what is the answer? Repeat!
Mental/Oral Starter:
Year Ones: Can you count up to 100? Can you make it active and think of two or three actions to repeat as you count?
Receptions: How far can you count up to? Pick a number? Let’s write it in the air. Where do we start?
Repeat with a few numbers up to 10. Focus numbers…2,3,5,8.
Main Teaching:
Year Ones: Use building blocks/cubes. Let’s find the difference in a different way – we have been using a number line. Let’s use the cubes. Roll two dice on the board. Make two towers representing these numbers. What is the difference between these numbers? Stand them next to each other. Look at the cubes. How many more cubes are there on the taller tower? That is the difference. How can you make the towers the same size? Either take away the difference from the tallest tower or add the difference to the shortest tower. Have another go, roll the dice again.
Receptions: Number your Duplo bricks/blocks – Can you put the bricks in order according to the numbers? Can you make a tower that is one less than…? Can you make a tower that is one more than?
Mental/Oral Starter:
All: Ask your adult to say a number. Can you tell them a number that is one more and/or one less? Jump that many times! Repeat with different starting numbers and different actions.
Main Teaching:
Year Ones: Pick 2 numbers (one from a ‘teen number’ bag and one from a ‘under 10’ bag). Build a duplo tower for each number. Can you see the difference between the two towers? What is the difference? Use a number line to check your answer. Circle the lowest number, circle the highest number. Jump from the low to the high and count the ‘jumps’. Were you correct? Write down the calculation. Email me with your completed calculations.
Reception: Number formation. Can you write over the top of numbers an adult has written for you? Can you find the number that is one less than 2? How many number twos can you see? Can you spot it anywhere else outside? Can you find 2 objects? What is one more?
A great way to complete this is writing numbers on to slabs outside with chalk and using water and a paint brush to write over the top of the chalk. It's great fun and it even washes the chalk away!
Top | 1,442 | 5,613 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2020-50 | latest | en | 0.920718 |
https://ece4uplp.com/2022/11/ | 1,686,258,794,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00158.warc.gz | 252,249,629 | 21,190 | ## Normal incidence on a perfect conductor
Normal incidence on a perfect conductor
whenever an EM Wave travelling in one medium impinges second medium the wave gets partially transmitted and partially reflected depending up on the type of the second medium.
Assume the first case in Normal incidence that is Normal incidence on a Perfect conductor.
i.e an EM wave propagating in free space strikes suddenly a conducting Boundary which means the other medium is a conductor.
The figure shows a plane Wave which is incident normally upon a boundary between free space and a perfect conductor.
assume the wave is propagating in positive z-axis and the boundary is z=0 plane.
The transmitted wave since the electric field intensity inside a perfect conductor is zero.
The incident and reflected waves are in the medium 1 that is free space.
The energy transmitted is zero so the energy absorbed by the conductor is zero and entire wave is reflected to the same medium
now incident wave is
in free space for medium 1
( a wave propagating in positive z-direction) and the reflected wave is ( a wave propagating in positive z-direction).
.
by using tangential components .
The resultant wave is .
.
the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations
.
This is the wave equation which represents standing wave , which is the contribution of incident and reflected waves. as this wave is stationary it does not progress.
it has maximum amplitude at odd multiples of and minimum amplitude at multiples of .
Similarly The resultant Magnetic field is
The resultant wave is .
.
the above equation is in phasor notation , converting the above equation into time-harmonic (or) sinusoidal variations
.
this wave is a stationary wave it has minimum amplitude at odd multiples of and maximum amplitude at multiples of .
## Nature of Magnetic materials
In order to find out the various types of materials in magnetic fields and their behaviour we use the knowledge of the action of magnetic field on a current loop with a simple model of an atom
Magnetic materials are classified on the basis of presence of magnetic dipole moments in the materials.
a charged particle with angular momentum always contributes to the permanent contributions to the angular moment of an atom
1. orbital magnetic dipole moment.
2. electron spin moment.
3. Nuclear spin magnetic moment.
Orbital Magnetic dipole Moment:-
The simple atomic model is one which assumes that there is a central positive nucleus surrounded by electrons in various circular orbits.
an electron in an orbit is analogous to a small current loop and as such experiences a torque in an external magnetic field, the torque tending to align the magnetic field produced by the orbiting electron with the external magnetic field.
Thus the resulting magnetic field at any point in the material would be greater than it would be at that point when the other moments were not considered.
so there are Quantum numbers which describes the orbital state of notion of electron in an atom there are n,l and ml
n-principal Quantum number, which determines the energy of an electron.
l-Orbital Quantum number which determines the angular momentum of orbit.
ml-magnetic Quantum number which determines the component of magnetic moment along the direction of an electric field.
electron spin Magnetic Moment:-
The angular momentum of an electron is called spin of the electron. as electron is a charged particle the spin of the electron produces magnetic dipole moment because electron is spinning about it’s own axis and thus generates a magnetic dipole moment.
is the value of electron spin when we consider an atom those electrons which are in shells which are not completely filled with contribute to a magnetic moment for the atom.
Nuclear spin Magnetic Moment:-
a third contribution of the moment of an atom is caused by nuclear spin this provides a negligible effect on the overall magnetic properties of material
That is the mass of the nucleus is much larger than an electron thus the dipole moments due to nuclear spin are very small.
so the total magnetic dipole moment of an atom is nothing but the summation of all the above mentioned .
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https://arc2climate.org/and-pdf/842-tensor-calculus-and-riemannian-geometry-pdf-50-516.php | 1,643,206,487,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00226.warc.gz | 162,632,022 | 7,148 | # Tensor Calculus And Riemannian Geometry Pdf
By Logan H.
In and pdf
02.12.2020 at 20:00
File Name: tensor calculus and riemannian geometry .zip
Size: 21793Kb
Published: 02.12.2020
Coupled with these.
## differential geometry - Two Definitions of the Weyl Tensor
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A Some Basic Rules of Tensor Calculus The tensor calculus is a powerful tool for the description of the fundamentals in con-tinuum mechanics and the derivation of the governing equations for applied prob-lems. New opportunities for me to make tons of typos and for everyone to point the. From OeisWiki. In mathematics, a tensor is an arbitrarily complex geometric object that maps in a multi-. Synge, , Dover Publications edition, in English. Observe that. What is a Tensor?
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To browse Academia. Skip to main content. By using our site, you agree to our collection of information through the use of cookies.
Thump, thump, thump with the fat heel of his hand, and his lips now all big and loose with pleasure at the lie he had told. He was the one who bought the Midnight Oil tape. Some of my people back in Alexandria know how to make weapons from it, like my sword, and to make other more useful tools as well. But these people, these Yslanders must know far more than we do.
Фонтейн насчитал уже шесть гудков. Бринкерхофф и Мидж смотрели, как он нервно шагает по комнате, волоча за собой телефонный провод. Директор АНБ напоминал тигра на привязи.
Мужчины оглянулись. В дверях стояла Росио Ева Гранада. Это было впечатляющее зрелище. Длинные ниспадающие рыжие волосы, идеальная иберийская кожа, темно-карие глаза, высокий ровный лоб.
Хейл поклялся, что никогда больше не переступит порога тюрьмы, и сдержал слово, предпочтя смерть. - Дэвид… - всхлипывала. - Дэвид.
Соши лихорадочно прогоняла текст на мониторе в обратном направлений и наконец нашла то, что искала. - Да. Здесь говорится о другом изотопе урана. Мидж изумленно всплеснула руками.
ГЛАВА 70 Дэвид Беккер почувствовал, что у него подкашиваются ноги. Он смотрел на девушку, понимая, что его поиски подошли к концу. Она вымыла голову и переоделась - быть может, считая, что так легче будет продать кольцо, - но в Нью-Йорк не улетела.
Narkis S.
03.12.2020 at 12:15 - Reply
Tensor Calculus and Riemannian Geometry. The attentive reader probably noticed that the concept of a Riemann metric on an open subset of R n which we. | 1,117 | 4,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-05 | latest | en | 0.906016 |
https://serverlogic3.com/which-data-structure-you-use-for-bfs-and-dfs/ | 1,713,406,296,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00463.warc.gz | 466,867,565 | 15,274 | # Which Data Structure You Use for BFS and DFS?
//
Angela Bailey
Data structures play a crucial role in various algorithms and operations. When it comes to algorithms like Breadth-First Search (BFS) and Depth-First Search (DFS), choosing the right data structure can significantly impact the efficiency and performance of these algorithms. In this article, we will explore the data structures commonly used for BFS and DFS.
BFS is an algorithm used to traverse or search a graph or tree data structure. It starts at the root node and explores all the neighbor nodes at the present depth before moving on to nodes at the next depth level.
The most commonly used data structure for implementing BFS is a queue. A queue follows the First-In-First-Out (FIFO) principle, which perfectly aligns with how BFS explores nodes level by level. Each time we visit a node during BFS, we enqueue its neighboring nodes into the queue.
To implement a queue in your code, you can use an array or a linked list. However, it’s important to note that using an array might require resizing if we encounter a large graph or tree with an unknown number of nodes.
## Depth-First Search (DFS)
DFS is another traversal algorithm used for searching or traversing graphs and trees. Unlike BFS, DFS explores as far as possible before backtracking.
When it comes to implementing DFS, two common data structures come into play: a stack and a recursive call stack.
A stack, which follows the Last-In-First-Out (LIFO) principle, is used to keep track of visited nodes during DFS. As we traverse through each node in DFS, we push its neighboring unvisited nodes onto the stack. This ensures that we visit the deepest nodes first before backtracking.
The other option is to use the recursive call stack provided by programming languages. When we encounter a node during DFS, we make a recursive call to explore its neighboring unvisited nodes. The recursive call stack implicitly acts as a stack data structure, maintaining the order of visited nodes.
### Summary
In summary, for BFS, we utilize a queue, which follows the FIFO principle. On the other hand, for DFS, we can choose between a stack or use the recursive call stack. Each of these data structures is suited to their respective algorithms and helps in efficient traversal or search.
Remember that choosing the right data structure is crucial in optimizing your algorithms and achieving better performance. Understanding how different data structures align with specific algorithms like BFS and DFS will undoubtedly enhance your programming skills.
Now that you know which data structures are commonly used for BFS and DFS, you can confidently implement these algorithms in your code! | 533 | 2,725 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | latest | en | 0.900499 |
https://www.math.hmc.edu/funfacts/ffiles/20002.8.shtml | 1,563,510,403,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525974.74/warc/CC-MAIN-20190719032721-20190719054721-00287.warc.gz | 774,994,023 | 7,175 | hosted by the Harvey Mudd College Math Department created, authored and ©1999-2010 by Francis Su
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From the Fun Fact files, here is a Fun Fact at the Medium level:
# Vickrey Auction
Suppose you are hosting a silent auction to sell your antique car. The rules are: (1) prospective buyers bid for your car by placing their bids in sealed envelopes, (2) then after collecting all bids, you sell the car to the highest bidder for the price that he bid.
This may seem like a reasonable way to conduct a silent auction, but there is a risk involved for you, the seller. If your car is highly valuable, but all buyers think they are the only ones who recognize that it is, they may actually make bids that are LESS than they think the car is actually worth. As a result, your bids will be lower, and you suffer.
Are there rules for a silent auction that will induce people to bid what they think an object is truly worth?
The answer is yes, and is given by a Vickrey auction. The rules specify that each player make bids, and the car goes to the highest bidder, but at the second-highest bid!
Can you figure out why this induces people to bid truthfully?
Presentation Suggestions:
This may be fun for students to ponder, or try as a group experiment, before discussing the answer. Additionally, you could ask students to think about what behavior other kinds of bidding rules would induce.
The Math Behind the Fact:
Here is the reason it works--- we'll show that, when holding all other bids fixed (and unknown to a given bidder), that bidder's optimal strategy is to bid what she thinks the car is worth.
Suppose the bidder's name is Alice. Let V be the amount that Alice thinks the car is actually worth, and B the bid that she actually makes. Let M be the maximum of all other bids. If M is more than V, then Alice should set her bid B less than or equal to V, so that she does not get the car for more than she thinks it is worth. If M is less than V, then Alice should set B=V, because if she bids any less, she will not get the car any cheaper, and she may lose the car.
The study of mathematical models for decision making is called game theory.
Su, Francis E., et al. "Vickrey Auction." Math Fun Facts. <http://www.math.hmc.edu/funfacts>.
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Keywords: game theory
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Fun Fact suggested by: Francis Su
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For more fun, tour the Mathematics Department at Harvey Mudd College! | 692 | 2,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-30 | latest | en | 0.960548 |
http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=P4849&l=en | 1,532,346,937,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596336.96/warc/CC-MAIN-20180723110342-20180723130342-00113.warc.gz | 491,004,379 | 5,308 | Mathematical and Physical Journal
for High Schools
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# Problem P. 4849. (May 2016)
P. 4849. The total mass of an easily moveable cart and a hemisphere shaped container of radius $\displaystyle R=0.3$ m attached to the cart is $\displaystyle M=2$ kg. Initially a point-like object of mass $\displaystyle m=1$ kg is held at the rim of the hemisphere (such that it touched the rim), and then it is released from rest.
$\displaystyle a)$ What are the velocities of the small object and the cart when the object descended a height of $\displaystyle h=R/2$? (Give both directions and speeds.)
$\displaystyle b)$ What is the radius of the curvature of the path of the motion of the object with respect to the ground, when it reaches its lowest position? (Friction is negligible everywhere.)
(5 pont)
Deadline expired on June 10, 2016.
### Statistics:
40 students sent a solution. 5 points: Asztalos Bogdán, Balogh Menyhért, Bartók Imre, Büki Máté, Csorba Benjámin, Csuha Boglárka, Di Giovanni András, Fehér 169 Szilveszter, Fekete Balázs Attila, Forrai Botond, Ghada Alshalan, Iván Balázs, Jakus Balázs István, Kluèka Vivien, Kovács Péter Tamás, Krasznai Anna, Marozsák Tóbiás , Molnár Mátyás, Olosz Adél, Páhoki Tamás, Pázmán Előd, Póta Balázs, Sal Kristóf, Szentivánszki Soma , Tófalusi Ádám. 4 points: Bekes Nándor, Berke Martin, Kasza Bence, Makovsky Mihály, Nagy 555 Botond, Németh 777 Róbert. 3 points: 5 students. 2 points: 3 students. 0 point: 1 student.
Problems in Physics of KöMaL, May 2016 | 519 | 1,569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-30 | latest | en | 0.622136 |
https://chouprojects.com/decimal-excel/ | 1,718,389,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00663.warc.gz | 154,732,350 | 27,004 | Published on
Written by Jacky Chou
# Decimal: Excel Formulae Explained
## Key Takeaway:
• Excel allows users to easily work with decimal numbers in a variety of ways, from adjusting the number of decimal places in formatting to performing arithmetic calculations with decimal numbers.
• To format numbers to specific decimal places, users can utilize the number formatting feature in Excel. Rounding decimal numbers to specific places is also possible within the number formatting options.
• Performing arithmetic calculations with decimal numbers is straightforward in Excel. Users can simply use the addition, subtraction, multiplication, and division operators to work with decimal numbers.
• The DECIMAL function in Excel allows users to convert text values that represent decimal numbers into actual decimal numbers for use in calculations and data analysis.
• Common errors with decimal numbers in Excel include issues with the decimal point and comma, which can cause miscalculations and errors in data analysis. To avoid errors with decimal calculations, it is important to double-check the formatting and precision of decimal numbers and to use caution when working with large datasets.
Struggling with decimal calculations in Excel? You’re not alone. Learn how to use the DECIMAL function to perform this tedious task with ease. From beginners to advanced users, this article offers insights into this powerful Excel formulae.
## Overview of Decimal in Excel
Excel is a powerful tool that offers remarkable precision in calculations, including the use of decimal points. Decimal in Excel refers to the usage of numbers with a decimal point and is vital in financial and scientific calculations. Decimal formatting in Excel determines the number of decimal places displayed and is adjustable according to the user’s preference. Additionally, for better data accuracy, it is advisable to round off the results to a specific decimal point as needed. Understanding the significance of decimals and their formatting options in Excel can lead to better accuracy and efficiency in spreadsheet calculations.
A unique feature of decimal in Excel is that the decimal separator can vary depending on regional settings. In some countries, a “comma” functions as a decimal separator, while in others, a “period” is used. This variation can cause compatibility issues when using Excel spreadsheets across different regions. To avoid such issues, use the Decimal data type, which automatically adapts to the user’s regional settings.
To maintain data accuracy, it is critical to avoid adding decimal points manually, as they may be subject to human error. Instead, include formulas that automatically calculate the results with the correct decimal points. One way to ensure data precision is to use “IF” and “ROUND” functions that enable automatic rounding off to the desired decimal places. Additionally, you can use the “CEILING” function to round up to the nearest desired decimal point.
## Decimal Places in Number Formatting
Need to format or round numbers in Excel? The ‘Decimal Places in Number Formatting‘ section in ‘DECIMAL: Excel Formulae Explained‘ has got you covered! It divides into two helpful sub-sections:
1. ‘Formatting Numbers to Specific Decimal Places’
2. ‘Rounding Numbers to Specific Decimal Places’
Check them out for some Excel formula enhancing tips!
### Formatting Numbers to Specific Decimal Places
When working with numbers in Excel, it is often necessary to format them to a specific number of decimal places. Whether it’s for financial reports or scientific data, precision is crucial. Here’s how you can achieve this:
1. Select the cell(s) containing the number(s) you want to format.
2. Right-click on the selected cells and choose ‘Format Cells’.
3. In the ‘Format Cells’ dialog box, select ‘Number’ in the Category list.
4. In the Decimal Places field, enter the number of decimal places you want to display.
5. Click ‘OK’ to apply your changes and close the dialog box.
6. Your numbers should now be formatted with the specified number of decimal places.
It’s important to remember that rounding errors can occur when working with decimals in Excel. If accuracy is critical, consider using a formula instead of manually formatting cells.
In addition to formatting numbers with fixed decimal places, Excel also provides options for scientific notation and rounding up/down based on a certain threshold value.
Don’t miss out on accurate calculations and professional-looking reports – make sure you’re familiar with formatting numbers in Excel!
Round to the nearest decimal place, or risk being rounded up by the accounting department’s disapproving glares.
### Rounding Numbers to Specific Decimal Places
When working with numbers, it is often necessary to round them to a specific number of decimal places. This process is called Decimal Places in Number Formatting. This technique is useful for situations where you want to simplify or clarify your data.
Rounding Numbers to Specific Decimal Places can be easily achieved using a four-step guide:
1. Select the cell(s) containing the number you want to round.
2. Click on the ‘Home’ tab on your Excel spreadsheet and then click on ‘Number Format’.
3. From the drop-down menu that appears, choose the number format that best suits your rounding requirements.
4. Finally, hit enter or press OK.
It’s important to note that when rounding numbers, you may encounter some discrepancies. For example, if you round up from 7.5 to 8 using one decimal place, it will display as 7.5 still because Excel defaults down rather than up in situations such as these.
The concept of Decimal Places in Number Formatting has been around for centuries. However, it only became popular with the widespread adoption of computers and electronic devices where precise calculations were needed.
To conclude, understanding how to Round Numbers to Specific Decimal Places is an essential skill for anyone who works extensively with numerical values. By following a simple four-step guide such as outlined here, you can quickly and accurately perform this important function in Excel while avoiding common pitfalls along the way.
I may not be good at math, but I can still appreciate the beauty of arithmetic calculations with decimal numbers.
## Arithmetic Calculations with Decimal Numbers
Do you need to calculate with decimals? Check out the ‘Addition and Subtraction of Decimal Numbers‘ and ‘Multiplication and Division of Decimal Numbers‘ sub-sections! Here, you’ll learn how to use these functions for precise and effective calculations.
### Addition and Subtraction of Decimal Numbers
When working with decimal numbers, it is crucial to understand how to add and subtract them accurately. Here’s a guide on performing this operation professionally:
1. Step 1: Ensure that all numbers have the same decimal point placement before adding or subtracting them.
2. Step 2: Add or subtract as you would with whole numbers, but keep the decimal in line throughout your calculations.
3. Step 3: Check your final answer to ensure that the decimal point is in its correct position.
It’s important to note that when adding or subtracting decimals, it’s good practice to carry extra digits beyond the intended precision to avoid rounding errors later.
To prevent confusion while carrying out arithmetic operations involving decimal points, it may be helpful to use zeros as placeholders when necessary.
By following these steps, you can confidently perform addition and subtraction with decimal numbers in Excel without making costly mistakes.
Don’t let fear of making an error discourage you from mastering these arithmetic calculations. With enough practice and attention to detail, anyone can become proficient at working with decimal numbers!
Why did the decimal number break up with the whole number? It just couldn’t handle the division.
### Multiplication and Division of Decimal Numbers
When multiplying or dividing decimal numbers, precise and accurate calculations are essential. Understanding different techniques can simplify these operations with more complex decimals and large numbers.
1. To multiply decimal numbers, follow the steps below:
1. Multiply the significant digits of each value as if they are integer values.
2. Count the total number of digits to the right of the decimal point in both values.
3. Add this count from step 2 to get the number of digits to place after the decimal point in your answer.
2. To divide decimal numbers, follow the steps below:
1. Move the decimal point in both values so that there are no decimals in the divisor. Count how many times you move it, and perform an equal movement on both values.
2. Divide as if you were working with whole numbers.
3. Count how many digits were behind the decimal originally for each value. Add those two counts together, then place that amount of decimals back into your answer from left to right.
3. A helpful tip when performing multiplication and division operations is to round answers only after all calculations have been completed. Round to achieve a specific level of accuracy required.
In addition, when working with large decimal numbers or long computations, grouping digits by every three can help maintain clarity throughout long calculations. Furthermore, taking breaks between challenging problems allows for better focus and accuracy during re-engagement with computations.
Pro Tip: Remember that rounding too soon may result in inaccurate answers. To avoid mistakes in critical computation series, set up specific protocols for rounded results at certain intervals or thresholds based on precision requirements.
You don’t need to be a math genius to use the DECIMAL function, but it certainly helps if you can count to ten without using your fingers.
## Using the DECIMAL Function
Gain a solid understanding of the powerful DECIMAL function in Excel! To help you, this guide explains the syntax and gives examples. Learn quickly by using the sub-sections:
• Syntax of DECIMAL
• Examples of Using it in Excel
### Syntax of the DECIMAL Function
The DECIMAL function in Excel is a mathematical formula that converts a text string to an actual decimal number. It uses two arguments- the text/string that needs to be converted and the number of digits to round off after decimal.
To apply this function, start with typing “=DECIMAL” in the cell where you want the answer. Then, put the first argument inside double quotes, followed by a comma, and then add the second argument (number of digits) without quotes. Press “Enter” to get the result.
It is important to note that the DECIMAL function returns an error if it detects letters or symbols in place of numbers within double quotes.
Incorporating this formula into your data analysis can optimize your workflow and significantly reduce manual errors.
Don’t miss out on the benefits of using Excel functions like DECIMAL for effortless data analysis and presentation. Embrace technology to improve efficiency and accuracy in your work today!
DECIMAL function: the perfect tool for Excel users who want to add precision to their numbers, unlike those who think 2+2=5.
### Examples of Using the DECIMAL Function in Excel
The DECIMAL Function in Excel can be used in various ways. Here’s how to utilize this function effectively:
1. Identify the cell where you want to display the result.
2. Type `=DECIMAL(number, radix)` and replace ‘number’ with the cell reference or value you want to convert to decimal and ‘radix’ with the base from which you are converting.
3. Press Enter, and the result will appear in decimal format.
4. You can also drag the formula down to multiple cells if needed.
5. To change the displayed decimal places, right-click on the cell, select Format Cells, choose Number tab, enter the desired number of decimal places under Decimal Places, click OK.
Using DECIMAL Function provides precise decimal conversion for non-decimal based systems like binary, octal or hexadecimal.
A little-known fact about using DECIMAL Function is that it was first introduced in Excel 2013 as a part of Office 365 subscription update. Before that version, one had to use other functions like HEX2DEC and OCT2DEC for similar conversions.
Decimal may be precise, but Excel users still manage to find creative ways to screw it up.
## Common Errors with Decimal in Excel
Avoiding mistakes in decimal calculations in Excel can be tricky. To help, let’s explore the sections “Issues with Decimal Point and Comma” and “Avoiding Errors with Decimal Calculations in Excel” of the guide “DECIMAL: Excel Formulae Explained.” These sections will provide tips to keep errors away.
### Issues with Decimal Point and Comma
When working with decimal numbers in Excel, there are various issues that can arise. One common problem is the misuse or misunderstanding of the decimal point and comma. This may lead to errors in calculations and misinterpretations of data.
To illustrate this issue clearly, a table can be created to show how different uses of decimal point and comma impact calculations. For instance, consider the following table:
ValueDecimal Point UsedComma Used
10.2710.27ERROR
5,874ERROR5874
The above table highlights the difference between using the decimal point versus a comma in two sets of values. As evident from the table, using a comma where a decimal point should be used leads to an error in calculation or interpretation.
While issues with Decimal Point and Comma are quite common, another potential problem one might encounter when dealing with decimals in Excel is rounding errors. Inaccurate rounding can lead to differences between expected and actual results.
Interestingly, as per historical accounts, Microsoft introduced support for multiple languages in Office products to help international users overcome their challenges with variations like decimal separators across different cultures. This made it easier for people who prefer to use commas instead of periods as decimals or vice versa while using MS-Office applications like Excel.
### Avoiding Errors with Decimal Calculations in Excel
When working with decimals in Excel, appropriate attention must be taken to prevent computational errors. Precision is key in making correct calculations and avoiding mistakes. Here’s a helpful guide on how to avoid errors when dealing with decimal points in Excel.
1. When inputting numbers, utilize the Decimal format under the “Number Format” category.
2. If utilizing mathematical formulae that require a high degree of accuracy, use ROUND functions to ensure precision to the desired decimal place.
3. Double-check calculations involving division by 0. As this operation is undefined mathematically, it can lead to big discrepancies in your data set.
4. Avoid using floating point arithmetic as much as possible, as it may result in small inaccuracies.
5. Use parentheses to group certain parts of your calculation which require strict adherence and follow the PEMDAS order of operations (Parentheses, Exponents, Multiplication and Division — left-to-right — Addition and Subtraction — left-to-right).
It’s important to remember that even small errors can have profound effects on the final outcome you wish to achieve. By following these guidelines for precision when handling decimals in Excel, you will be able to minimize inaccuracies and ultimately increase efficiency.
A common mistake made during data entry involves inputting too many or too few zeros before or after a decimal point. This leads to wrong numbers appearing on your spreadsheets despite appearing alright at first glance. Misinterpreting or forgetting these positions might also result from other factors such as human error or copying conditions from one cell to another without double-checking prior calculations.
Excel has been instrumental in data analysis since its inception and only becomes more valuable with each upgrade; however, it has not always been perfect especially when handling decimal values, leading to occasional inaccuracies that may have gone unnoticed until long after essential decisions have been made based on wrong computations.
## Five Facts About DECIMAL: Excel Formulae Explained:
• ✅ DECIMAL is an Excel formula that rounds a number to a specified number of decimal places. (Source: ExcelJet)
• ✅ DECIMAL can be used in combination with other functions, such as SUM and AVERAGE. (Source: Ablebits)
• ✅ DECIMAL allows for greater control and precision when working with numerical data in Excel. (Source: Microsoft Support)
• ✅ The syntax for DECIMAL is =DECIMAL(number, decimal_places). (Source: Excel Easy)
• ✅ DECIMAL is a versatile tool for financial analysis, budgeting, and data visualization in Excel. (Source: Investopedia)
## FAQs about Decimal: Excel Formulae Explained
### What is ‘DECIMAL: Excel Formulae Explained’?
‘DECIMAL: Excel Formulae Explained’ is a topic that aims to help users understand how to work with decimal numbers in Excel. It focuses on providing formulae to make calculations more efficient and less time-consuming.
### What is the difference between ROUND and ROUNDUP?
The ROUND and ROUNDUP formulae both round off decimal numbers, but their functionality is slightly different. ROUND rounds the number to the nearest multiple based on the decimal point, whereas ROUNDUP rounds the number to the nearest multiple greater than the number.
### How do I convert a decimal to a percentage?
To convert a decimal to a percentage, you can multiply the decimal by 100 or use the formula ‘=VALUE*100%’. For example, if you have a decimal value of 0.75, multiplying it by 100 or using the formula would give you a percentage value of 75%.
### How do I find the average of decimal numbers?
To find the average of decimal numbers, you can use the formula ‘=AVERAGE(range)’, where ‘range’ is the range of cells you want to find the average of. For example, if you want to find the average of the numbers in cells A1 to A5, the formula would be ‘=AVERAGE(A1:A5)’.
### How do I add decimals in Excel?
To add decimals in Excel, you can simply use the ‘+’ operator. For example, if you want to add the numbers in cells A1 and A2, the formula would be ‘=A1+A2’.
### How do I format decimal numbers in Excel?
To format decimal numbers in Excel, you can use the ‘Number Format’ option. Click on the cell(s) you want to format, then go to the ‘Home’ tab and select the ‘Number Format’ drop-down menu. From there, you can select the type of decimal format you want to use, such as ‘Number’, ‘Currency’, or ‘Accounting’.
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Key Takeaway: The MATCH function in Excel is used to ... | 3,722 | 18,956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.853224 |
https://dev.to/zeeter/talking-clock-challenge-17p9 | 1,721,031,556,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00724.warc.gz | 179,665,007 | 22,961 | D Smith
Posted on
# Introduction
Challenge can be found here.
# Description
Create a program that can convert time strings into text.
## Input Description
Input is an hour (0-23) followed by a colon followed by a minute (0-59)
``````00:00
01:30
12:05
14:01
20:29
21:00
``````
## Output Description
Program should output time in words, using the 12 hour format. (12:00 am)
``````It's twelve am
It's one thirty am
It's twelve oh five pm
It's two oh one pm
It's eight twenty nine pm
It's nine pm
``````
## Optional Challenge
Make the program actually talk using sound clips, like the ones found here.
# My Solution
So I have a solution for this that doesn't involve testing. However I have been trying to get into the habit of
following Test Driven Development. Also I will be using a literate programming technique to write the program
inside this article. The other thing I want to address is that I almost definately overengineered this too,
I get a little zealous when I'm writting.
## Functional Test
First things first, the following code block is the functional test for this challenge.
``````def talkingclock(time):
hour,minute = map(int, time.split(":"))
if hour >= 12:
period = "pm"
else:
period = "am"
numnames = (
'one','two','three',
'four','five','six',
'seven','eight','nine',
'ten','eleven','twelve')
if hour > 12:
hour -= 12
else:
pass
hour_string = numnames[hour-1]
minute_string = ""
onetable = ("one","two","three",
"four", "five", "six",
"seven","eight","nine")
teentable = ("ten","eleven","tweleve","thirteen",
"fourteen", "fifteen","sixteen",
"seventeen","eighteen","nineteen")
tentable = ("","ten","twenty","thirty","forty","fifty",)
if minute in range(1,10):
minute_string = "oh " + onetable[minute-1]
elif minute in range(10,20):
minute_string = teentable[minute-10]
elif minute in range(20,60):
ones = minute % 10
tens = minute - ones
if ones > 0:
minute_string = tentable[tens//10] + " " + onetable[ones-1]
else:
minute_string = tentable[tens//10]
output = "It's " + hour_string + ' '
if minute_string == '':
output += period
else:
output += minute_string + ' ' + period
return output
testtable = [i for i in zip(si.split(),so.split('\n\t'))]
results = [["Input","Expected Output","Actual Output", "Passed"]]
for test in testtable:
si,so = test
to = talkingclock(si)
results.append([si,so,to,so==to])
return results
``````
This test works by gathering the sample input and sample output as described in the challenge section, it splits them into the proper lines
and stores them as a list comprehension. I would have used tuples here, as the data never changes. However python creates generator objects
if you replace square brackets with parenthesis when defining a listcomp. At that point I would be just as well using the stand alone zip fuction.
Anyway the input string is fed into the talkingclock function and the output is compaired against the test output.
This creates the above results, and if any of them are false, we can look at which line it occures on and test that input separately.
While this is not the most elegant solution for testing, it is a rather practical test as we get a nice table out of it.
## Parsing the input
The input that the talking clock is given is in the format of (HH:MM) in 24 hour time.
To do this, we take the string and split it at the colon using the string's split method.
This produces a of the strings for hour and minute. These are then turned into ints and assigned to
the variables hour and minute respectively.
``````hour,minute = map(int, time.split(":"))
``````
To test our parser works, we need to look for three things.
1. It returns an hour value
2. It returns a minute value
3. Both hour and minute are ints
def testparser(time):
hour,minute = map(int, time.split(":"))
return hour,minute
th,tm = testparser(test)
isint = all(map(lambda x: type(x) == int, (th,tm)))
hashour = th != None
hasmin = tm != None
results=(
("Returns hour: ", hashour),
("Returns minutes: ", hasmin),
("All are ints: ", isint))
return results
Assuming when you run this test they all passed, we know that the parser is good to go.
## Converting intigers to words
Now that we have the input parsed into the correct form, we need to process it to get out the correct strings.
### Getting Hour String
Getting the hour string has a few requirements.
1. The hour string will always be between 1 and 12.
2. The hour string will be the english name for numbers (e.g) one
numnames = (
'one','two','three',
'four','five','six',
'seven','eight','nine',
'ten','eleven','twelve')
if hour > 12:
hour -= 12
else:
pass
hour_string = numnames[hour-1]
def testhourstring(hour):
numnames = (
'one','two','three',
'four','five','six',
'seven','eight','nine',
'ten','eleven','twelve')
``````if hour > 12:
hour -= 12
else:
pass
hour_string = numnames[hour-1]
return hour_string
``````
nname = (
'one','two','three',
'four','five','six',
'seven','eight','nine',
'ten','eleven','twelve')
zeroistwelve = (testhourstring(0) == 'tweleve')
oneisoneect = all([testhourstring(i) == nname[(i-1)] for i in range(1,13)])
thirteenisoneect = all([testhourstring(i) == nname[(i-1)] for i in range(1,13)])
result = (
("Test", "Passed"),
("0 yields Tweleve: ", zeroistwelve),
("1-13 yield correct name: ", oneisoneect),
("13-23 yield correct name: ", thirteenisoneect)
)
return result
### Getting Minute String
The minute string is a bit more complex than the hour string.
Effectively I have broken it up into a few parts
1. a ones table which runs from 1 to 9.
2. a teens table which runs from 10 to 19
3. a tens table which is 10 to 50
For numbers that are 1-9 the string is made by fetching the number of off the ones table and appending it to the "oh" phrase.
Because english is awful, I had to make the teen table, because the numbers that are 11-19 all have to be special and not conform to other numeric standards.
Finally for numbers that are between 20-59 the number is made by fetching the tens space, and if it has a number in the ones space, apending that to it.
The tens table also includes 10 to make some of my math easier, at the cost of making the tuple an element longer, which I think is a valid
``````minute_string = ""
onetable = ("one","two","three",
"four", "five", "six",
"seven","eight","nine")
teentable = ("ten","eleven","tweleve","thirteen",
"fourteen", "fifteen","sixteen",
"seventeen","eighteen","nineteen")
tentable = ("","ten","twenty","thirty","forty","fifty",)
if minute in range(1,10):
minute_string = "oh " + onetable[minute-1]
elif minute in range(10,20):
minute_string = teentable[minute-10]
elif minute in range(20,60):
ones = minute % 10
tens = minute - ones
if ones > 0:
minute_string = tentable[tens//10] + " " + onetable[ones-1]
else:
minute_string = tentable[tens//10]
def testminstring(minute):
minute_string = ""
onetable = ("one","two","three",
"four", "five", "six",
"seven","eight","nine")
teentable = ("ten","eleven","tweleve","thirteen",
"fourteen", "fifteen","sixteen",
"seventeen","eighteen","nineteen")
tentable = ("","ten","twenty","thirty","forty","fifty",)
if minute in range(1,10):
minute_string = "oh " + onetable[minute-1]
elif minute in range(10,20):
minute_string = teentable[minute-10]
elif minute in range(20,60):
ones = minute % 10
tens = minute - ones
if ones > 0:
minute_string = tentable[tens//10] + " " + onetable[ones-1]
else:
minute_string = tentable[tens//10]
return minute_string
singledigit = testminstring(5) == "oh five"
teens = testminstring(15) == "fifteen"
justtens = testminstring(20) == "twenty"
tensandones = testminstring(35) == "thirty five"
doesnone = testminstring(0) == ""
results = (("test","passed"),
("Does single digits: ", singledigit),
("Does teens: ", teens),
("Does just tens: ", justtens),
("Does tens: ", tensandones),
("Does zero: ", doesnone)
)
return results
``````
This test used some hard coded values in it, unlike the hours test. However I needed to test for some specific conditions, and I figured
a less robust test would work here better. Like with my hour tests, you could likely have it test all possible valid inputs with a list comp.
But in this case, I'm not sure that it would be worth the effort.
### Finding the period
To find the period, (am or pm) is a pretty simple task.
This is done by taking the hour variable, and checking if it is greater than or equal to 12.
If so the period is pm, otherwise it is am.
What is important is that this piece must be run before getting the hour string, because the hour variable is change from that operation.
This could be mitgated by storing the original value, however I think that would just be overcomplicating the issue.
``````if hour >= 12:
period = "pm"
else:
period = "am"
def testgetperiod(hour):
if hour >= 12:
period = "pm"
else:
period = "am"
return period
amworks = all([testgetperiod(t) == 'am' for t in range(0,12)])
pmworks = all([testgetperiod(t) == 'pm' for t in range(12,23)])
results = (("test","passed"),
("Am works: ",amworks),
("Pm works: ",pmworks),
)
return results
``````
With a part this simple, tests really aren't needed. But I want to stay true to the bleepings of the testing goat.
## Putting it all together
Now that we have all the dependant parts, we just need to string them together in the right order.
The main thing I look for here is if the minute string is empty or not.
If it is, there should only be one space between the hour and the period.
In one of my last attempts at this problem, I got tripped up at this stage, hense my caution.
``````def talkingclock(time):
hour,minute = map(int, time.split(":"))
if hour >= 12:
period = "pm"
else:
period = "am"
numnames = (
'one','two','three',
'four','five','six',
'seven','eight','nine',
'ten','eleven','twelve')
if hour > 12:
hour -= 12
else:
pass
hour_string = numnames[hour-1]
minute_string = ""
onetable = ("one","two","three",
"four", "five", "six",
"seven","eight","nine")
teentable = ("ten","eleven","tweleve","thirteen",
"fourteen", "fifteen","sixteen",
"seventeen","eighteen","nineteen")
tentable = ("","ten","twenty","thirty","forty","fifty",)
if minute in range(1,10):
minute_string = "oh " + onetable[minute-1]
elif minute in range(10,20):
minute_string = teentable[minute-10]
elif minute in range(20,60):
ones = minute % 10
tens = minute - ones
if ones > 0:
minute_string = tentable[tens//10] + " " + onetable[ones-1]
else:
minute_string = tentable[tens//10]
output = "It's " + hour_string + ' '
if minute_string == '':
output += period
else:
output += minute_string + ' ' + period
return output
``````
Now with this version of the program, the output string is not printed to the
stdout. I left that out for a few reasons. If I decide to do the optional challenge
in the future, printing directly to the stdout won't be super helpful.
Second, I wanted to keep the tests as simple as possible. If you notice,
the above section never import code, this includes unittest.
Unittest is a super helpful framework for more complicated tests, but I haven't
figured out how to use it to make pretty tables. I may look into using nose in the future, but I haven't quite got there yet. | 3,094 | 11,159 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.842977 |
http://www.yougonumbers.com/roman-numeral/cxxvi | 1,503,343,699,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109525.95/warc/CC-MAIN-20170821191703-20170821211703-00708.warc.gz | 732,957,551 | 6,335 | ### Read and Write Roman Numeral CXXVI
Roman numeral symbols are written and read from left to right, from highest to lowest values. If CXXVI is text, it is the number 126.
### Miscellaneous & Conversions For 126
Written English One Hundred Twenty-Six Roman Numeral 126 CXXVI RGB Colour (# 07e or #00007e)
Decimal / Number 126 Binary 1111110 Ternary 11200 Hexadecimal (Hex) 0x7e Quaternary 1332 Quinary 1001 Senary (Heximal) 330 Octal 176 Duodecimal (Dozenal) a6
### Mathematics
Prime Factors: 2 x 3 x 3 x 7 = 126 Factorial: 126! 2.37217324288E+211 Harshad Number: 126 True Triangular Number Of 126 8001 Square Root: √126 11.2249721603 Cube Root (Root 3): 3√126 5.01329793496 Tesseract Root (Root 4): 4√126 3.35036895883 Quintal / nth Root. (Root 5) 5√126 2.63071686526 ¾ Root 37.6077982897 Fibonnaci # 126 9.6151855463E+25 Recipricol 1/ 126 0.00793650793651 Quarter 1/4 31.50 or 31.50/126 Half 1/2 63 or 63/126 Third 1/3 42 or 42/126
### Roman Numeral Chart. Read & Write 126
Reading and writing roman numerals isn't common knowledge. However, this roman numeral chart should help you read and write numbers in roman numerals from 1 to 1,000,000. Using an overline symbol is times 1,000. ( Example: X means to add 3 zeros to your number. So X which is 5, but written as X with the overline is 5,000.
• When a symbol appears after a larger symbol it is added: Example: VI = V + I = 5 + 1 = 6
• When a symbol appears before a larger symbol it is subtracted: Example: IX = X - I = 10 - 1 = 9
• Do not use the same symbol more than 3 times. But, IIII is sometimes used for 4
1 I 10 x 500 D 1,001 MI 2 II 20 XX 501 DI 1,002 MII 3 III 30 XXX 530 DXXX 1,030 MXXX 4 IV 40 XL 550 DL 1,550 MDL 5 V 50 L 600 DC 1,555 MDLV 6 VI 60 LX 650 DCL 5,000 V 7 VII 70 LXX 700 DCC 10,000 X 8 VIII 80 LXXX 800 DCCC 50,000 L 9 IX 90 XC 890 DCCCXC 100,000 C 10 X 95 XCV 900 CM 500,000 D 11 XI 99 XCIX 950 CML 1,000,000 M 12 XII... 100 C 1000 M
39 XXXIX 37 XXXVII
2% Fibonacci
38% Numbers
20% Binary | 735 | 1,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.657568 |
https://www.studymode.com/essays/Baron-Coburg-65140891.html | 1,709,457,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00683.warc.gz | 995,397,090 | 15,853 | Baron Coburg.
Good Essays
468 Words
Grammar
Plagiarism
Writing
Score
Baron Coburg.
Conclusions
I am concluding that Frederick is the better farmer/businessman. When comparing the financials of the two farmers, Frederick's consistently demonstrated higher production per acre of land allotted him. Ivan ended up with higher net income and more assets at the end of the year, but when calculated into common terms was actually less productive overall.
Finding of Facts.
Since the two farmers were not given equal plots, in order to make them comparable a common factor must be used in their analyses. In order to put both Ivan and Frederick in common terms, acres of land were used as the common denominator.
Ivan: At the end of the year Ivan's net income was 214 bushels of wheat. After determining net income of 214 bushels, I divided that by the 20 acres that he was allotted to determine that he produced 10.7 bushels per acre. Ivan's operating expenses per acre and total expenses per acre were \$0.35 and \$1.45 respectively. Ivan's balance sheet depicted a 318% positive change from the beginning of the year to the end; however, if that is demonstrated in terms of percent change per acre for comparability Ivan has an increase of 15.9% per acre. His balance sheet also shows his assets and liabilities to be 259 bushels and total 12.95 stated as bushels per acre. On top of this, Ivan still owes Feyador 3 bushels for his plow.
Frederick: At the end of the year Frederick's net income totaled 119 bushels--significantly less than Ivan's net income because of the land allotment differences. Frederick produced a net income of 11.9 bushels per acre, a favorable difference of 11% over Ivan's 10.7. His operating expenses per acre and total expenses per acre were \$0.50 and \$1.60, which is higher than Ivan's largely due to the fixed costs of the plow and ox. Frederick's gross margin per acre is 12.4 bushels compared to Ivan's 11.05 bushels, a difference of 12.2%. Frederick'sbalance sheet total of 143 bushels is also less when looking at the raw data. However, when the
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Satisfactory Essays | 2,184 | 8,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.960591 |
https://buy-essay.com/the-value-of-pi-in-matlab-pi-can-be-computed-with-the-sequence-pi-lim_n-rightarrow-infinity-2pi-1-pi_n-wher/ | 1,674,874,791,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00245.warc.gz | 163,857,135 | 13,212 | # The value of pi (in matlab pi) can be computed with the sequence: pi = lim_n rightarrow infinity 2^pi + 1 pi_n wher
IN MATLAB
The value of pi (in matlab pi) can be computed with the sequence: pi = lim_n rightarrow infinity 2^pi + 1 pi_n where pi_n is defined using the iteration pi_n = squareroot (1/2 pi_n – 1)^2 + [1 – squareroot (1/2 pi_n – 1)^2]^2 pi_0 = squareroot 2 (J. Munkhammar, pers.comm., April 27, 2000: http: //mathworld.wolfram.com/PiFormulas.html). Compute what term of the sequence produce an error
Functon file
========================================================================
function [picalc,err,k]=iPi(N)
pin=sqrt(2);
k=0;
for i =1:N
k=k+1;
pin=sqrt((.5*pin)^2+(1-(sqrt(1-(.5*pin)^2)))^2);
picalc=2^(i+1)*pin;
err=abs(pi-picalc);
end
end
Main file
===============================================================
clear;
clc;
fprintf(‘iteration computed pi computed errorn’);
N=1;
while(N>0)
[pi_calc,err,i]=iPi(N);
fprintf(‘%9i%15.7f%14.7fn’,i,pi_calc,err);
N=N+1;
if(err<10^-6)
break;
end
end
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At this stage, our editor will go through your essay and make sure your writer did meet all the instructions. | 731 | 2,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.81809 |
https://stats.stackexchange.com/questions/381453/do-interaction-terms-in-a-linear-regression-model-increase-its-predictive-power | 1,713,541,634,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817438.43/warc/CC-MAIN-20240419141145-20240419171145-00775.warc.gz | 482,691,354 | 40,814 | # Do interaction terms in a linear regression model increase its predictive power?
Interaction terms are sometimes added to linear regression models when the effect of one variable depends on the value of another variable. But will the inclusion of such interaction terms increase the model's predictive power? Or is the only effect to yield a model that can be better interpreted?
Or put another way, if I only care about the model's performance, and otherwise treat it as a black box, do I need to think about interaction terms?
• Yes, adding interactions to a statistical model can increase its predictive power, if those interactions are actually present in the data. (Don't have time to write a better answer right now.) Dec 11, 2018 at 13:55
• yes interaction terms and any other added nonlinear transformation of the inputs can increase model's predictive power Dec 11, 2018 at 13:55
• Generally, higher power terms and interactions are harder to interpret (except perhaps 2 way interactions which are easy to interpret). For instance, taking a kernel implicitly maps your data into higher dimension by taking various kind of high order interactions raised to some power (depending on which kernel) but becomes a black box. To answer your question, it can lead to higher performance but could also lead to overfitting more easily, so be sure to have some regularisation.
– Tom
Dec 11, 2018 at 13:57
• How to define "predictive power"? Dec 11, 2018 at 15:34
For linear regression, we have the hypothesis space $$\mathcal{F}_1$$, the set of all functions whose output is a linear combination of the input variables. Including interaction terms gives the hypothesis space $$\mathcal{F}_2$$, the set of all functions whose output is a linear combination of the input variables and their interaction terms. Note that $$\mathcal{F_1}$$ is a subset of $$\mathcal{F}_2$$. That is, every function in $$\mathcal{F}_1$$ is also in $$\mathcal{F}_2$$ (because we can always set the coefficients for the interaction terms to zero), but $$\mathcal{F}_2$$ contains functions that are not in $$\mathcal{F}_1$$. This means that including interaction terms gives us the possibility of fitting a wider variety of functions. In particular, $$\mathcal{F}_2$$ contains functions that are nonlinear with respect to the original input variables.
When fit to a particular dataset, a model that includes interaction terms must fit at least as well as a model that does not. This follows from the fact that, if the model with lowest error on the training data is in $$\mathcal{F_1}$$, it's also in $$\mathcal{F}_2$$, as above. But, whether this yields an increase in predictive power depends on the problem. Including interaction terms can increase predictive power in some settings, but decrease it in others. | 630 | 2,793 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-18 | latest | en | 0.909099 |
https://www.jiskha.com/questions/1077646/Posted-by-rfvv-on-Friday-July-11-2014-at-3-01am-1-Exchange-points-with-another | 1,534,611,178,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213691.59/warc/CC-MAIN-20180818154147-20180818174147-00420.warc.gz | 928,924,450 | 5,025 | English
Posted by rfvv on Friday, July 11, 2014 at 3:01am.
1. Exchange points with another team.
2. Swap points with another team.
(Are both the same?)
3. Take away 300 points from one team.
4. Bring 300 points from another team.
5. Take away 30 points from each team.
(Are they the same in meaning?)
(Do we have to use 'one' or 'another' if there are three teams and they are playing a game related to getting points by answering questions on slides? Do you have some more suitable expressions we can use in such a game?)
English - Ms. Sue, Friday, July 11, 2014 at 12:11pm
1 and 2 are both correct.
3 is correct. 4 is wrong. 5 is different; it means that you're taking points from all of the teams.
====================
Thank you for your help. Are the following grammatical and the same, and can we use them in a game? Can we use 'another' instead of 'one'?
6. Take away 300 points from one team.
7. Remove 300 points from one team.
8. Get rid of 300 points from one team.
9. Delete 300 points from one team.
10. Take out 300 points from one team.
1. 6 - 10 are all grammatical. Yes, we can use "another" instead of "one."
posted by Ms. Sue
2. Thank you,and they are he same in meaning, aren't they?
posted by rfvv
3. Yes, they all mean about the same.
posted by Ms. Sue
Similar Questions
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Posted by rfvv on Friday, July 11, 2014 at 3:01am. 1. Exchange points with another team. 2. Swap points with another team. (Are both the same?) 3. Take away 300 points from one team. 4. Bring 300 points from another team. 5. Take
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More Similar Questions | 1,079 | 3,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-34 | latest | en | 0.954991 |
https://www.iodraw.com/en/blog/220450644 | 1,653,511,270,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00581.warc.gz | 937,616,809 | 4,817 | <> front
I think so. I think most of the roast people have roast on the Internet
If you don't know what happened in this session , And want to know what happened to arouse the group's anger
You can go to the Blue Bridge Cup , Found on 2022 year 4 month 9 A few days before the th , Or search “ carpooling ”
You can see how many people can be embarrassed to cheat
<> in
Seriously, let's talk about my summary
I don't think it's done very well , Vaguely
oh dear I still don't want to explain the problem Tell me some knowledge I think you may have （ Because that's what I did ）
Prefix and （ A naked question ）, But in fact, the last question is to query the sum of intervals, which can also be understood as the prefix and sum in the forest （ I hope it's right ）
I think the tree climber may be an expectation dp,dp It's a weakness , No direct practice g Yes , It seems that he gave a special judgment and tried to cheat some points
There is a dichotomy , I think the idea is also very simple , But when you finish debugging , feel dev There is a problem with this version of
It's him lower_bound and upper_bound I always have unexpected results （ Or I forgot how to use this thing ）
Then I handwritten two points, it felt easy to make mistakes , Just put it first , First, later , I came back and found there was no time , I didn't hand in this question
There is also a simulation problem , It's a circle , Feel save the order , It should be ok if you run it again
The last piece I ran , When searching for connected blocks, you can finish the prefix and , The edge is bidirectional, one positive and one negative , Because there is no contradictory solution , So no matter which way you go , The absolute value of the distance between two points is equal
You can find the representative yuan when looking for the Unicom block , Deal with this
（ I don't know, right , Anyway, I wrote this in a daze at that time , The samples were handed in directly ）
What? You asked me why I didn't fill in the blanks ?
How cheap it is to fill in the blanks , I got an answer and handed it in directly / open the mouth and show the teeth
Anyway, I have time to fight all the violence , But I don't know whether violence is right or not
I didn't seem to wake up , Random answer / awkward
<> after
Anyway , This year's Blue Bridge Cup is over , I won't fight again , I feel like I've been cut into leeks
“ It's hard for me to say , On the one hand, the topic is really difficult , On the other hand, I don't want to face the fact that I am more backward than before …”
Technology
Daily Recommendation | 613 | 2,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | longest | en | 0.965298 |
http://web2.0calc.com/questions/how-on-earth | 1,513,449,715,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588420.68/warc/CC-MAIN-20171216181940-20171216203940-00209.warc.gz | 301,617,687 | 6,181 | +0
# How on earth ?
+4
146
10
+108
can someone help me learn how figure out how to do questions like my last 2 posts? It's these questions: -19y = -15 – 16y
ItsTayha May 10, 2017
Sort:
#1
+280
+4
Sure!
-19y = -15 – 16y
First, add 16y to the left side and the right side.
-19y + 16y = -15 - 16y + 16y
This cancels out - 16y from the left, making it easier to solve.
-3y = -15y
Change both signs to a positive (Only if both sides are negative or already positive)
3y = 15
Divide by 3 on both sides.
3y/3 = 15/3
Solve.
y = 5
When you divide 3y by 3, it cancels out the three, because 3/3 is 1, and 1y = y.
So your answer is y = 5
Jelly May 10, 2017
#2
+108
+1
What if it has 4 numbers like this : 10m – 2 = 9m – 14
ItsTayha May 10, 2017
#3
+280
+1
I will show you!
Jelly May 10, 2017
#4
+280
+2
10m – 2 = 9m – 14
Subtract 9m from both sides.
10m - 9m – 2 = 9m - 9m – 14
Solve.
m - 2 = (-14)
Add 2 to both sides. (Do the opposite of the sign it shows. Negative = Positive. Times = Divide)
m - 2 + 2 = (-14) + 2
Solve.
m = (-12)
Always make sure m positive.
Jelly May 10, 2017
#5
+108
+1
Wow! I have keep reading that over till I know how to do it!
ItsTayha May 10, 2017
#6
+280
+2
I am sure you will get it!
Just focus, and keep trying.
Jelly May 10, 2017
#7
+108
+1
Yeah I might post more of those up, they are in a bio I have for math class
ItsTayha May 10, 2017
#8
+280
0
I will try to solve all the ones you do put up.
Just for the sake of doing math during school hours xD
Jelly May 10, 2017
#9
+108
+1
My math teacher doesn't really help us out
ItsTayha May 10, 2017
#10
+280
0
That isn't a very good teacher =/
Jelly May 10, 2017
### 4 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 750 | 1,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-51 | latest | en | 0.866829 |
https://metanumbers.com/215272 | 1,638,352,554,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359976.94/warc/CC-MAIN-20211201083001-20211201113001-00254.warc.gz | 465,799,040 | 7,467 | # 215272 (number)
215,272 (two hundred fifteen thousand two hundred seventy-two) is an even six-digits composite number following 215271 and preceding 215273. In scientific notation, it is written as 2.15272 × 105. The sum of its digits is 19. It has a total of 5 prime factors and 16 positive divisors. There are 105,840 positive integers (up to 215272) that are relatively prime to 215272.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 19
• Digital Root 1
## Name
Short name 215 thousand 272 two hundred fifteen thousand two hundred seventy-two
## Notation
Scientific notation 2.15272 × 105 215.272 × 103
## Prime Factorization of 215272
Prime Factorization 23 × 71 × 379
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 53818 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 215,272 is 23 × 71 × 379. Since it has a total of 5 prime factors, 215,272 is a composite number.
## Divisors of 215272
16 divisors
Even divisors 12 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 410400 Sum of all the positive divisors of n s(n) 195128 Sum of the proper positive divisors of n A(n) 25650 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 463.974 Returns the nth root of the product of n divisors H(n) 8.39267 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 215,272 can be divided by 16 positive divisors (out of which 12 are even, and 4 are odd). The sum of these divisors (counting 215,272) is 410,400, the average is 25,650.
## Other Arithmetic Functions (n = 215272)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 105840 Total number of positive integers not greater than n that are coprime to n λ(n) 3780 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 19179 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 105,840 positive integers (less than 215,272) that are coprime with 215,272. And there are approximately 19,179 prime numbers less than or equal to 215,272.
## Divisibility of 215272
m n mod m 2 3 4 5 6 7 8 9 0 1 0 2 4 1 0 1
The number 215,272 is divisible by 2, 4 and 8.
## Classification of 215272
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Pentagonal
## Base conversion (215272)
Base System Value
2 Binary 110100100011101000
3 Ternary 101221022001
4 Quaternary 310203220
5 Quinary 23342042
6 Senary 4340344
8 Octal 644350
10 Decimal 215272
12 Duodecimal a46b4
20 Vigesimal 16i3c
36 Base36 4m3s
## Basic calculations (n = 215272)
### Multiplication
n×y
n×2 430544 645816 861088 1076360
### Division
n÷y
n÷2 107636 71757.3 53818 43054.4
### Exponentiation
ny
n2 46342033984 9976142339803648 2147584113774210912256 462314727340401931503173632
### Nth Root
y√n
2√n 463.974 59.9325 21.5401 11.6573
## 215272 as geometric shapes
### Circle
Diameter 430544 1.35259e+06 1.45588e+11
### Sphere
Volume 4.1788e+16 5.82351e+11 1.35259e+06
### Square
Length = n
Perimeter 861088 4.6342e+10 304441
### Cube
Length = n
Surface area 2.78052e+11 9.97614e+15 372862
### Equilateral Triangle
Length = n
Perimeter 645816 2.00667e+10 186431
### Triangular Pyramid
Length = n
Surface area 8.02668e+10 1.1757e+15 175769
## Cryptographic Hash Functions
md5 35bf7cd4d8eac0594629f390254f18da 4b86503ecb9180dcb75e8b9c50b40f1eee4818e8 3881bd0659c42e127b92a0563bdb80790f54baf11026e261a3abc1f7977f8ed0 73e682fa3965c11a0ae91d44e026e0371762acce3c78d2ef88b7496c44c0df40bff189d95611cc9c64ec823d97a556a48c78cf790a30ffddbea43eed8665aab8 6af67f5b0584b1fc77bcf27e59ded13858e3386b | 1,462 | 4,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.806361 |
https://mathybeagle.com/2014/01/27/pattern-power/?replytocom=90 | 1,621,271,830,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991258.68/warc/CC-MAIN-20210517150020-20210517180020-00081.warc.gz | 401,390,162 | 21,755 | Pattern Power
If you have little kids and you’ve been privy to an episode of Team Umizoomi, then perhaps the title of this post evoked a little jingle in your head. You’re welcome; I’m here all day.
My daughter, although she doesn’t choose Umizoomi over Mickey Mouse as often as I’d like, picked up on patterns relatively quickly after watching this show a couple of times. She’s 3 years old, and she finds patterns all over the place. Mostly color and shape patterns, but a string of alternating letters can usually get her attention as well. These observations of hers made me realize that pattern seeking is something that is innate and our built-in desire for order seeks it out.
High school students search patterns out as well. For example, I put the numbers 4, 4, 5, 5, 5, 6, 4 so that the custodian knew how many desks should be in each row after it was swept. It drove students absolutely CRAZY trying to figure out what these numbers meant. I almost didn’t want to tell them what it really was as I knew they’d be disappointed that it lacked any real mathematical structure.
I’m not as familiar with the elementary and middle school math standards as perhaps I should be, but I’m confident that patterns are almost completely absent from most high school curriculum. Why are most high school math classes completely devoid of something that is so natural for us?
Dan Meyer tossed out some quotes from David Pimm’s Speaking Mathematically for us to ponder. This one in particular sheds light on this absence of pattern working in high school mathematics:
Premature symbolization is a common feature of mathematics in schools, and has as much to do with questions of status as with those of need or advantage. (pg. 128)
In other words, we jump to an abstract version of mathematical ideas and see patterns as lacking the “sophistication” that higher-level math is known for. To be completely honest, this mathematical snobbery is one of the reasons I discounted Visual Patterns at first. Maybe it was Fawn Nguyen’s charisma that drew me back there, but those patterns have allowed for some pretty powerful interactions in my classroom. I’ve used them in every class I teach, from remedial mathematics up to college algebra because they are so easy to differentiate.
I think high school kids can gain a more conceptual understanding of algebraic functions with the use of patterns. For example, this Nrich task asks students to maximize the area of a pen with a given perimeter. The students were able to use their pattern-seeking skills to generalize the area of the pen much more easily than if they had jumped right from the problem context to the abstract formula.
I also notice that the great high school math textbooks include patterns as a foundation for their algebra curriculum. For example, Discovering Advanced Algebra begins with recursively defined sequences. IMP also starts with a unit titled Patterns. I think these programs highlight what a lot of traditional math curriculums too quickly dismiss: patterns need to be not only elementary noticings of young math learners but also valued as an integral part of a rich high school classroom.
1. True we should not discount the innateness of patterns. Humans are naturally drawn to patterns because of the complexity of our natural environment. The human brain is incapable of absorbing and responding to all of the information it takes in. To compensate, we begin to not only see but also to hear, feel, taste and smell patterns as filters for that information. It is a way to make sense of the world and survive in it without going into information overload. So we begin to see patterns in letters (and then associated sounds); letters that are transformed into words; clouds that predict the weather, seasonal changes, movement of the stars, animal movement, shapes that tell us one thing is a tree while other shapes are mountains. We are dependent upon patterns.
We have become so inured to patterns that we do not often realize that the importance and significance they have in our lives is¬- in the simplest explanation—to keep us alive.
2. You have some mighty math powers, Ms Schmidt. I hope you’re ready to umi-shake. | 901 | 4,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.964609 |
https://www.physicsoverflow.org/32677/evaluation-of-the-spin-sum-for-a-massive-spin-1-particle?show=32678 | 1,713,609,260,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817576.41/warc/CC-MAIN-20240420091126-20240420121126-00576.warc.gz | 843,120,697 | 22,941 | # Evaluation of the spin-sum for a massive spin-1 particle?
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Assuming that a massive spin-1 particle has momentum only in the z-direction, the polarization vectors are given by
$$\varepsilon_{\mu}(J_z = +1) = (0,-\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$
$$\varepsilon_{\mu}(J_z = 0) = (\frac{p}{m},0,0, \frac{E}{m})$$
$$\varepsilon_{\mu}(J_z = -1) = (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$
The so-called spin-sum is the claimed to be
$$\sum\limits_{J_z = -1,0,+1} \varepsilon_{\mu}\varepsilon_{\nu}^* = g_{\mu\nu} + \frac{p_{\mu}p_{\nu}}{m^2}$$
I absolutely dont understand how this spin-sum is evaluated.
What does $\varepsilon_{\mu}\varepsilon_{\nu}^*$ even exactly mean? Is it a scalar product between two of the three above polarization vectors, or is it a "tensor-product" between the components of a single polarization vector which results in a 4x4 matrix and one has finally to sum all such matrices for the three possible values of $J_z$?
I would really appreciate it if somebody can explain to me what this spin-sum exactly means and how it is evaluated step-by-step.869
asked Jul 29, 2015
It is a tensor-product, as you could have seen from the right hand side. Just evaluate the three matrices explicitly and sum them to get the result. It is pure linear algebra.
To add to Arnold, here might be a even straighter way for OP to see this is a elementary linear algebra calculation: Think of $\varepsilon_\mu(J_z)$ as a 4 by 3 matrix $\varepsilon$, rows labeled by $\mu$ and columns labeled by $J_z$, then the spin sum is nothing but $\varepsilon \varepsilon^\dagger$, which is a 4 by 4 matrix.
## Your answer
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https://nl.mathworks.com/matlabcentral/cody/problems/2018-side-of-a-rhombus/solutions/435740 | 1,606,286,577,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00179.warc.gz | 429,119,046 | 16,898 | Cody
# Problem 2018. Side of a rhombus
Solution 435740
Submitted on 27 Apr 2014 by Brad
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; y_correct = sqrt(5)/2; tolerance = 1e-12; assert(abs(rhombus_side(x)-y_correct)<tolerance)
2 Pass
%% x = 3; y_correct = 5/2; tolerance = 1e-12; assert(abs(rhombus_side(x)-y_correct)<tolerance)
3 Pass
%% x = 2; y_correct = sqrt(13)/2; tolerance = 1e-12; assert(abs(rhombus_side(x)-y_correct)<tolerance)
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 218 | 704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-50 | latest | en | 0.70917 |
https://science-memo.blogspot.com/2014/05/is-ergodicity-reasonable-hypothesis.html | 1,680,110,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00364.warc.gz | 579,837,407 | 14,497 | ## Tuesday, 13 May 2014
### Is ergodicity a reasonable hypothesis? Understanding Boltzmann's ergodic hypothesis
Ergodic vs. non-ergodic trajectories (Wikipedia)
Many undergraduate Physics students barely study Ergodic Hypothesis in detail. It is usually manifested as ensemble averages being equal to time averages. While the concept of the statistical ensemble maybe accessible to students, when it comes to ergodic theory and theorems, where higher level mathematical jargon kicks in, it maybe confusing for the novice reader or even practicing Physicists and educator what does ergodicity really mean. For example recent pre-print titled "Is ergodicity a reasonable hypothesis?" defines the ergodicity as follows:
...In the physics literature "ergodicity" is taken to mean that a system, including a macroscopic one, visits all microscopic states in a relatively short time...[link]
Visiting all microscopic states is not a pre-condition for ergodicity from statistical physics stand point. This form of the theory is the manifestation of strong ergodic hypothesis because of the Birkhoff theorem and may not reflect the physical meaning of ergodicity. However, the originator of ergodic hypothesis, Boltzmann, had a different thing in mind in explaining how a system approaches to thermodynamic equilibrium. One of the best explanations are given in the book of J. R. Dorfman, titled An introduction to Chaos and Nonequilibrium Statistical Mechanics [link], in section 1.3, Dorfman explains what Boltzmann had in mind:
...Boltzmann then made the hypothesis that a mechanical system's trajectory in phase-space will spend equal times in regions of equal phase-space measure. If this is true, then any dynamical system will spend most of its time in phase-space region where the values of the interesting macroscopic properties are extremely close to the equilibrium values...[link]
Saying this, Boltzmann did not suggest that a system should visit ALL microscopic states. His argument only suggests that only states which are close the equilibrium has more likelihood to be visited.
Postscript (June 2022)
The sufficiency of Sparse Visits: Physical states are rarely fine-grained
A requirement for attaining ergodicity is visiting all possible states or regions due to the ergodic theorems of Birkhoff and von Neumann. This requirement is not correct for Physics. The key concepts here are coarse-graining and the sufficiency of sparse visits. Most of the physical systems have equally likely states.
The generated dynamics would rarely need to visit all accessible states or regions. Physical systems are rarely fine-grained and have a degree of sparseness, reducing their astronomically large number of states to a handful. In summary, visiting all physical states or regions in time averages is not strictly needed for the physics definition of ergodicity.
A collection of regions or multiple states with a higher probability will need to be covered to achieve thermodynamic equilibrium. A concept of “sufficiency of sparse visits”. This approach makes physical experiments possible over a finite time consistent with thermodynamics. | 639 | 3,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-14 | latest | en | 0.899862 |
https://www.instructables.com/Circuit-Board-1/ | 1,701,919,017,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100632.0/warc/CC-MAIN-20231207022257-20231207052257-00041.warc.gz | 922,915,153 | 30,207 | ## Introduction: Complete Circuit, Circuit Board Project
This Instructable covers standards in Science for grades 3,4, and 7. I use this activity with my fourth graders and have used it in the fifth and sixth grade several years back.
Use your circuit tester (direction to build here) to help build a circuit board. We will build a simple circuit board to “spark” students interest and discuss key concepts.
Below are 3rd and 4th grade standards covered by this Instructable!
Heat, electrical energy, light, sound and magnetic energy are forms of energy. (3) … Electrical circuit or solar panel models can be used to demonstrate different forms of energy and the source of the energy… (3) Energy can be transformed from one form to another or can be transferred from one location to another. (4) Electric circuits require a complete loop of conducting materials through which an electrical energy can be transferred. (4) Electrical energy in circuits can be transformed to other forms of energy, including light, heat, sound and motion. (4) Electrical conductors are materials through which electricity can flow easily. Electricity introduced to one part of the object spreads to other parts of the object. (4) Electrical insulators are materials through which electricity cannot flow easily. Electricity introduced to one part of the object does not spread to other parts of the object. (4) In order for electricity to flow through a circuit, there must be a complete loop through which the electricity can pass. When an electrical device (e.g., lamp, buzzer, motor) is not part of a complete loop, the device will not work. Electric circuits must be introduced in the laboratory by testing different combinations of electrical components. When an electrical device is a part of a complete loop, the electrical energy can be changed into light, sound, heat or magnetic energy. Electrical devices in a working circuit often get warmer. (4)
Electrical energy transfers when an electrical source is connected in a complete electrical circuit to an electrical device. (7) An electric circuit exists when an energy source (e.g., battery, generator, solar cell) is connected to an electrical device (e.g., light bulb, motor) in a closed circuit. The energy source transfers energy to charges in the circuit. Charges flow through the circuit. Electric potential is a measure of the potential electrical energy of each charge. (7)
## Step 1: Materials Needed
For this activity each student or group of students will need a single hole punch, making tape, piece of paper (card stock works best), and strips of foil. Circuit Tester from up coming Instructable or Voltmeter.
Attached is a template similar to ours.
## Step 2: Punch
Punch holes along one side of the paper and then the other. Try to have the holes across from each other on the other side of the paper. You will end up with two columns on opposite sides of the paper.
## Step 3: Run Foil
Run one piece of thin foil from one hole to another on opposite side. Use a piece of tape to hold the ends of the foil in place.
Make sure the foil is visible through the hole on the other side of the paper.
## Step 4: Cover
Cover the one piece completely with masking tape.
The foil will act as a conductor, allowing the electrons to flow through.
The masking tape acts as an insulator, preventing the electrons to flow through.
It is important to completely cover the foil with tape on this side. If even the slightest part of foil touches the other foil pieces, we will be putting on later, there will be a short. The connection will not work they way it is intended.
According to Google, "A short circuit is a problem in an electrical circuit where two or more wires that are not supposed to come in contact with each other touch. A short circuit can result in a very high current flowing through the circuit. ... A "short circuit" also happens when there is a bypass of electrical current."
## Step 5: Continue
Continue adding foil to the paper and completely covering with masking tape.
## Step 6: Complete! ;-)
Now that the foil (conductor) and the masking tape (insulator) is all done, check connections. You could use a voltmeter or a circuit tester like ours. Touch one hole on the left to one hole at a time on the right. The circuit will complete and you will have a complete circuit.
We have our students make up their own review questions and answers to match the complete circuits. After all the circuit boards are done we pass them from student to student to check for shorts and to review.
## Step 7: Video Step by Step
Here is a step by step video with suggestions for teachers.
Participated in the
Classroom Science Contest | 986 | 4,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-50 | latest | en | 0.92634 |
https://il.tradingview.com/script/vnpB6Jmq-Keltner-Center-Of-Gravity-Channel-KeltCOG/ | 1,721,577,959,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517747.98/warc/CC-MAIN-20240721152016-20240721182016-00773.warc.gz | 252,407,375 | 70,118 | # Keltner Center Of Gravity Channel ( KeltCOG )
מעודכן
I have the ambition to create a ‘landscape’ which enables the user to see the ‘mood’ of the market about the price of an instrument, simply by looking where the candles go. Prices are a simple phenomenon , they go up or down or stay the same. This is represented quite well for the short term by a candle. I recommend to study candle patterns. Prices not only fluctuate but also trend up, down or go sideways. The user should analyze this by determining the COG (Center Of Gravity) and the ‘normal’ current range by using the historical data in a lookback period.
As a COG the center line of a Donchian Channel is often used. I.m.o. a COG should be a zone, in this channel I use the gray zone of my Donchian Fibonacci Channel, The ‘normal’ range is a multiple of Average True Range, as used in a Keltner Channel. Combining the two can give a cumbersome result, as one can see in my Keltner Fibonacci Channel. In this KeltCOG channel I solved this by not using all Fibonacci levels and by making the Keltner lines strictly parallel to the nearest COG line. To do this, I use the fact that the COG lines have horizontal stretches, there I make the Keltner lines horizontal too. Only where the COG lines change value, the Keltner lines are recalculated. This way the channel gets a very regular shape with three clear zones.
Interpretation of a chart by using the KeltCOG channel.
Overbought: If the candles go higher then the blue zone, the market is hyper enthusiast, creating an overbought situation. This is often followed by a reversion to the COG.
Uptrend: If the candles form in the blue zone, the market is enthusiast and willing to pay more.
Hopeful: If the candles form in or near the upper uncolored zone, the market is hopeful and is thinking about paying more. Sometimes prices go a little up.
Content: If the candles form in the gray zone, which represents COG, the market is happy with the current prices, so these move sideways
Disappointed: If the candles form in or near the lower uncolored zone, the market is disappointed and contemplates paying less, sometimes prices go a little down.
Downtrend: If the candles form in red zone, the market doesn’t like the instrument at all, rejects the current price and is only prepared to pay less.
Oversold: If the candles form below the red zone, the market overdoes its disgust, creating an oversold situation, often followed by a reversion to the COG.
הערות שחרור:
I added a function that automatically adjusts the width of the blue and red zones when the user changes the look back periods.
For my own use I set the look back to 14 for weekly charts, to 21 for dailies, to 28 for 3hourlies and to 35 to hourly charts.
For finer intraday charts I guess the user will like even longer look back periods, but that is to his, hers or whatevers discretion.
הערות שחרור:
After comparing mij Keltner Fibonacci Channel with prvious versions of this KeltCOG Channel I came to the conclusion that all fibonacci lines are relevant because these give support and resistance. So I made it possible to use these as innerlow and innerhigh. In the inputs this can be switched off, then 0.5*ATR will be used to distance it from the nearest COG-line. Also I now think about the uncolored zones as 'accumulation zone' below the COG and 'distribution zone' above the COG, giving regard to the method of Wyckoff.
הערות שחרור:
I gave the uncolored zones the colors green and yellow and also names, i.e.: "optimistic accumulation and distribution area", which is made green and "pessimistic accumulation and distribution area" which I gave a yellow color. The gray zone is made more transparant. You can change these in the inputs if you don't like them.
הערות שחרור:
The possibility to add extra Keltner lines is added. These lines have a distance of 2 ATR from the border of the channel. These can be used to provide values for setting a take profit target. I also show extra lines in intraday timeframes to function as 'pivot points' in attempts to predict future quotes. Default setting is off , you can switch showing individual extra lines on in the inputs. Enjoy.
הערות שחרור:
1. The script chooses a lookback period that matches the timeframe. This feature can be switched off in the inputs, allowing the user to enter the lookback manually
Remarks on the connection between lookback and timeframe: The lookback should i.m.o. be equal to the ‘look forward’. Examples: If you choose a minute chart, you probably are a day trader and you wish to estimate where the quotes might go in the next hour. Then you should choose a lookback near 60, because the cannel will then provide the range and the COG generated by the previous hour and show the relative place of the current price in this range. If you choose an hourly chart, your horizon is probably a few days, so 35 is a proper choise, If you choose a monthly chart, you might want to estimate possibilities for the coming half year, so 7 is a proper value for this argument. The scrips chooses multiples of seven due to some superstition of mine. The script values reveal my personal subjective opinion, so the user can choose other arguments.
2. The script chooses a width that matches the lookback period. This feature can be switched off in the inputs, allowing the user to enter the width manually.
Remarks on the desirable width of the channel: The ‘landscape’ of the channel is meant to encompass most candles with some of those touching the border, then one can reasonably claim that the channel shows the ‘natural range’, providing an argument to call and ‘overbought’ or ‘oversold’ if candles move outside. Interdependency with lookback is caused by the fact that both the COG and the natural range are relatively wider when the lookback is further. However because this is not a linear connection, it took me a few hours to create a non linear formula that seems to works fine.
3. A feedback label is created to show the values that are used for lookback and width and whether these originate from the script or the user. This label can be switched off. The standard arguments label provided by Trandingview only show input defaults or entries while the compiler blocks attempts to change these with the script, so some other way to inform the user on the actual values for these parameters was needed.
The purpose of these additions is that you can throw the KeltCOG in whatever timeframe without worries about the proper settings. e.g. you can toggle through different timeframes trusting that this channel remains relevant.
הערות שחרור:
I added a label “KeltCOG width ## %” in the right upper side. It shows the percent of the width of the channel from the value of its middle line. This line is sort of an average price of the instrument in the lookback period.
You can switch it off in the inputs, you can also just show the number.
It can be taken as a sort of measurement of the volatility, the wider the cannel, the more volatile the price.
I need this feature to limit risk in my system of momentum investing. In this system I use the value of the outerlow ( the purple lowest line of the channel) as stoploss level. I have eight positions in different liquid stocks of equal size. I open a position when the price moves in the blue uptrend area. To limit the risk per trade to 2 % of the total invested finances, the maximum risk of one out of eight equal positions should not be higher 16 %. If the channel is wider than 16 %, I might prefer to use the higher red innerlow, or if the channel is too wide, not take the trade.
Also changes were made to the minimum lookback period for the script setting to 14, the formula for calculating width by the script, The green area is renamed to “prudent buyers area”, the yellow area to “prudent sellers area”.
הערות שחרור:
I added two features to the channel,
The first is the option to show the last 20 of a 40 period Hull Moving Average.
Strategy: I am a fan of this line because it is very smooth, doesn’t give many false signals and everything about it has a meaning: its direction (up, sideways or down) and the position of the bars towards this line (under=down, over=up). For me the last 20 will do.
The second feature is the option to reduce the channel to a step line with colored patches. The advantage of this option is that the indicator takes no extra space in the panel, giving a better view of the candles. Disadvantage is that you don’t have values in the price scale.
I use this option in the layout with 4 panels showing 4 different timeframes of an instrument.
In the inputs you find three switches for this option under the Extra Lines heading:
“Show just nearby Fibonacci level and color patches” Switching this on will reduce the channel to a step line and the zones become color patches in which the candle moves, zones in which no candle moves are not shown as patches.
“Toggle supportive / resistive level” If you toggle this, the step line will either have a preference to run below the candles and take the color lime, this is supportive level, or to run above the candles and take the color maroon, this is resistive level. Of course the blue patches are higher than any Fibonacci line and the red patches lower, so when the candle is in such a patch both levels are equal.
“Show both channel and nearby fib level with patches” This is added to reassure the user that one form is really a variation of the other.
Creating this option required a lot of changes in the script code. I added two sets of the six lines of the channel, the values for both sets are calculated and put in variables named F1 through F6 in the top section of the script.
In the middle section I give these values to permanent lines when the proper condition is met, these retain the names Outerhigh through Outerlow and are plotted exactly as in the original script.
In the final section I give these values to patch lines when more complicated conditions are met. I named these conditions CL1 through CL6 to apply to L1 through L6 when these values are given to variables with a somewhat truncated name Outhi through Outlow. These are plotted with the display.none condition, as the plot only serves for the fill command that creates the patches. If the variable has no value, nothing is plotted and nothing is filled, this how the patches are created only when a candle in them. To ensure that there is at least on clickable item in the indicator I created the nearby fib line by checking the before mentioned CL1 throug CL6 conditions again.
Enjoy, Eykpunter.
הערות שחרור:
Example chart showing both complete channel and Nearby Fib line with patches and HMA40
הערות שחרור:
The label which states the height of the channel as percent of its middle, has been moved to a more practical place in the lower right corner. Also the layout is changed into a smaller size and a silver background.
In the inputs the text can still be brought back to just the number.
הערות שחרור:
Different modes: Channelmode and Supertrendmode
After discovering that the “Supertrend” indicator combines beautifully with the KeltCOG with de setting “show just nearby fibonacci level and color patches” activated, I decided to integrate Supertrend into this option and rename it in “Supertrendmode”.
Integration is easy because the algorithms for KeltCOG and Supertrend use the same settings (i.e. ‘multiplier’ and ‘lookback’) in the same way. I copied lines from the code by Kivanc Ozbilgic (www.tradingview.com/script/r6dAP7yi/) made it work with the KeltCOG settings.
The option "toggle supportive / resistive level" is abolished because now this nearby fifonacci level will automatically turn to supportive and change color to lime if Supertrend indicates uptrend and in downtrend to resistive colored maroon. Because the highlight color for Supertrend is between the “Support/Resistance level” and the Supertrendline, the patches of the channel colors in which the bars run, keep their colors unchanged. The colors or the highlighters are yellowish for the uptrend and reddish for the downtrend.
The inputs are regrouped, now the first option is "toggle channel mode (off) supertrend mode (on)"
The option “use Fibonacci lines as inner high and low” is abolished, the channel now always gives Fibonacci zones.
The use of this mode is the same as for supertrend, but you get the additional information of the relevant Fibonacci level and channel zone. The outside channel situation, which occurs when the market is over enthusiastic or fearful are clearly visible by the absence of patch color.
enjoy
הערות שחרור:
Added checkbox in inputs to provide an option to show just outer zones combined with supertrend | 2,876 | 12,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.916241 |
https://us.metamath.org/ileuni/ixpeq2dva.html | 1,708,494,306,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00732.warc.gz | 611,991,034 | 3,755 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > ixpeq2dva GIF version
Theorem ixpeq2dva 6573
Description: Equality theorem for infinite Cartesian product. (Contributed by Mario Carneiro, 11-Jun-2016.)
Hypothesis
Ref Expression
ixpeq2dva.1 ((𝜑𝑥𝐴) → 𝐵 = 𝐶)
Assertion
Ref Expression
ixpeq2dva (𝜑X𝑥𝐴 𝐵 = X𝑥𝐴 𝐶)
Distinct variable group: 𝜑,𝑥
Allowed substitution hints: 𝐴(𝑥) 𝐵(𝑥) 𝐶(𝑥)
Proof of Theorem ixpeq2dva
StepHypRef Expression
1 ixpeq2dva.1 . . 3 ((𝜑𝑥𝐴) → 𝐵 = 𝐶)
21ralrimiva 2480 . 2 (𝜑 → ∀𝑥𝐴 𝐵 = 𝐶)
3 ixpeq2 6572 . 2 (∀𝑥𝐴 𝐵 = 𝐶X𝑥𝐴 𝐵 = X𝑥𝐴 𝐶)
42, 3syl 14 1 (𝜑X𝑥𝐴 𝐵 = X𝑥𝐴 𝐶)
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 103 = wceq 1314 ∈ wcel 1463 ∀wral 2391 Xcixp 6558 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 681 ax-5 1406 ax-7 1407 ax-gen 1408 ax-ie1 1452 ax-ie2 1453 ax-8 1465 ax-10 1466 ax-11 1467 ax-i12 1468 ax-bndl 1469 ax-4 1470 ax-17 1489 ax-i9 1493 ax-ial 1497 ax-i5r 1498 ax-ext 2097 This theorem depends on definitions: df-bi 116 df-tru 1317 df-nf 1420 df-sb 1719 df-clab 2102 df-cleq 2108 df-clel 2111 df-nfc 2245 df-ral 2396 df-in 3045 df-ss 3052 df-ixp 6559 This theorem is referenced by: ixpeq2dv 6574
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https://domyessayforme.org/bus-475-week-4-knowledge-check/ | 1,638,901,694,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363405.77/warc/CC-MAIN-20211207170825-20211207200825-00370.warc.gz | 289,936,395 | 11,503 | # Bus 475 week 4 knowledge check
BUS 475 Week 4 Knowledge Check
University of Phoenix
Strategy Implementation
20 Questions
1. A fast-food restaurant asks customers to evaluate the drive-thru service as good, average, or poor. What level of data measurement is this classification?
2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was \$250 with a standard deviation of \$25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November
3. A shipment of 20 DVDs has arrived at a video rental store. Based on past experience, the manager knows that 10% of all new DVDs sent to the store have a visible defect. The manager tells you to begin inspecting the new DVDs one at a time at random until you find the first DVD that has a defect. If 10% of the DVDs have a visible defect in the new shipment, what is the probability that the first DVD that has a defect is the 3rd one that you inspect? (Round your answer to 3 decimal places.)
4. An insurance agent has selected a sample of drivers that she insures whose ages are in the range from 16–42 years old. For each driver, she records the age of the driver and the dollar amount of claims that the driver filed in the previous 12 months. A scatterplot showing the dollar amount of claims as the response variable and the age as the predictor shows a linear regression line with: y^=3710 – 55.4x. If the age of a driver increases by 1 year, by how much and in what direction would the dollar amount of claims be predicted to change for the driver?
5. A researcher claims that the proportion of employees who play video games in the workplace is higher than it was 10 years ago. You might be willing to reject the null hypothesis of no change with a = 0.10 or larger. The p-value for this test is 0.15. In this case the researcher should
6. The general plan of major actions through which a firm intends to achieve is long-term objectives is called its
7. How valuable a low-cost leader’s cost advantage is depends on
8. Which of the following represent marketing capabilities at the growth stage of industry evolution?
9. __________ is an organization structured around the idea of sharing knowledge, seeking knowledge, and creating opportunities to create new knowledge……………………. | 539 | 2,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-49 | latest | en | 0.948178 |
https://ist.ac.at/en/news/decades-old-erdos-conjecture-cracked/ | 1,716,850,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00101.warc.gz | 276,773,774 | 22,616 | March 14, 2022
## Youngest ISTA professor achieves mathematical breakthrough in combinatorial design theory
Some of the most famous problems in mathematics remain unsolved for centuries. For Erdős’ conjecture, it took fifty years for a solution to be found. Professor Matthew Kwan from the Institute of Science and Technology Austria (ISTA) and mathematicians from Harvard and MIT present a proof that shows the existence of so-called high-girth Steiner triple systems.
When Matthew Kwan heard about the Erdős conjecture during his studies, he did not expect to be part of proving this infamous mathematical theorem. In a new paper, Kwan and colleagues prove the existence of Steiner triple systems with arbitrarily high girth. For the sophisticated proof, the new ISTA faculty member Kwan collaborated with colleague Michael Simkin from Harvard University and two prodigy graduate students Ashwin Sah and Mehtaab Sawhney from the Massachusetts Institute of Technology (MIT).
Steiner triple systems are fundamental types of combinatorial designs, which have their roots in the design of scientific experiments. For example, it is important to know which grain types can flourish in various soils with different properties. How to probe all these combinations efficiently? You can in fact reduce the number of experiments with clever combinatorics. Nowadays, researchers in combinatorial design theory study more abstract settings than agriculture. Their results are relevant to computer programming and coding theory. Yet, even fundamental problems have remained unsolved. This included Erdős’ conjecture – until now.
## The meaning of Erdős’ conjecture
The conjecture is concerned with so-called Steiner triple systems. Assume seven people want to form triples. Each person can be part of multiple triples, yet each pair of persons can only be part of exactly one triple. What does that mean? One individual A can be part of three triples then, forming the first triple with B and C, the second one with E and F, and the third one with D and G. At the same time, everyone else searches for partners. B for instance would not be allowed to join any triples that contain A or C because they already met them in the initial ABC triple. However, B can form two other triples still. In fact, with seven people – or points, as they are called in designs –, exactly seven triples are possible, making it a Steiner triple system.
The mathematicians showed that Steiner triple systems with the desirable property of high girth indeed exist. High girth is a statistical condition. When many triples span over a small number of points the girth is low. “Many triples across few points, such dense spots inevitably appear with algebraic designs,” explains Kwan. “Erdős wondered whether you could avoid them. Where triples have no such configurations, the system is said to possess high girth. To prove their existence, you must avoid algebra and bring in probabilistic methods. That´s what we managed to do.”
## Interdisciplinarity inside mathematics
In design theory, perhaps oldest field of combinatorics, there has been limited progress on fundamental questions until eight years ago. Related to Kwan’s background of probabilistic combinatorics, several probabilistic results revolutionized the field since game-changing advances in 2014. The sophisticated new proof comprises a wide array of techniques at the frontier of extremal and probabilistic combinatorics. “Additionally, we used two novel methods: Retrospective analysis, which allows us to keep track of the randomness in previous steps, and sparsification, which deals with all the obstacles that relate to that,” summarizes Kwan the technical aspects of the result.
More
Retrospective analysis
Retrospective analysis circumvents a unique property of the Erdős conjecture that you cannot merge disjoint systems conserving certain properties. Instead of remembering all information of previous steps, the novel method remembers only the randomness. “This is a vital component for the proof. Our analysis would otherwise be completely intractable,” says Kwan.
Sparsification
Sparsification on the other hand addresses the limits of iterative absorption, a method, which repeatedly executes random processes to build more and more triples in a design, while conserving the Steiner property. “You accumulate constraints through iterative absorption. By sparsifying the setting before the random processes, you get away with fewer constraints, and that allows you to reach a conclusion.”
With his group at ISTA, Kwan seeks to get a better fundamental understanding of designs, especially from a viewpoint of probability. “My philosophy of mathematics: I try to work on different things. It may be tempting to really focus on just one subfield, but when you work on various problems throughout your life, you discover techniques that help you succeed in other areas.”
Publication:
Kwan E, Sah A, Sawhney M and Simkin M. 2022. High-girth Steiner triple systems. Preprint. arXiv:2201.04554v3
Erica Klarreich (2015): A Design Dilemma Solved, Minus Designs. Quanta Magazine. – A popular account of Steiner triple systems and recent breakthroughs in design theory.
Yufei, Zhao (2022): Kwan-Sah-Sawhney-Simkin: High-girth Steiner triple systems.Detailed blog entry.
Share | 1,092 | 5,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.936939 |
https://ar5iv.labs.arxiv.org/html/0807.2052 | 1,717,083,531,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00526.warc.gz | 80,921,986 | 115,505 | ###### Abstract
In certain classes of subharmonic functions $u$ on $\mathbb{C}$ distinguished in terms of lower bounds for the Riesz measure of $u$, a sharp estimate is obtained for the rate of approximation by functions of the form $\log|f(z)|$, where $f$ is an entire function. The results complement and generalize those recently obtained by Yu. Lyubarskii and Eu. Malinnikova.
\setkeys
Gindraft=false
APPROXIMATION OF SUBHARMONIC FUNCTIONS
I. Chyzhykov
Faculty of Mechanics and Mathematics,
Lviv National University, Lviv, Ukraine
ichyzh@lviv.farlep.net
## 1 Introduction
We use the standard notions of the subharmonic function theory [1]. Let us introduce some notation. Let $D_{z}(t)=\{\zeta\in{\mathbb{C}}:|\zeta-z|, $z\in{\mathbb{C}}$, $t>0$. For a subharmonic function $u$ in ${\mathbb{C}}$ we write $B(r,u)=\max\{u(z):|z|=r\}$, $r>0$, and define the order $\rho[u]$ by the equality $\rho[u]=\limsup\limits_{r\to+\infty}\log B(r,u)/\log r$. Let also $\mu_{u}$ denote the Riesz measure associated with the subharmonic function $u$, $m$ be the plane Lebesgue measure. The symbol $C$ with indices stands for some positive constants. If $u=\log|f|$, where $f$ is an entire function with the zeros $\{a_{k}\}$, then $\mu_{u}=\sum\limits_{k}n_{k}\delta(z-a_{k})$, where $n_{k}$ is the multiplicity of the zero $a_{k}$, $\delta(z-a_{k})$ is the Dirac function concentrated at the pointΒ $a_{k}$. However, the class of functions subharmonic in ${\mathbb{C}}$ is broader than those of the form $\log|f|$, where $f$ is an entire function. Since it is often easier to construct a subharmonic function rather than an entire one with desired asymptotic properties, a natural problem arises of approximation of subharmonic functions by the logarithms of the moduli of entire functions. Apparently, V.Β S.Β Azarin [2] was the first to investigate this problem in the general form in the class of subharmonic in the plane functions and of finite order of growth. The results cited below have numerous applications in the function theory and the potential theory (see, e.g., [3β6]).
In 1985 R.Β S.Β Yulmukhametov [7] obtained the following remarkable result. For each function $u$ subharmonic in ${\mathbb{C}}$ and of order $\rho\in(0,+\infty)$, and $\alpha>\rho$ there exist an entire function $f$ and a set $E_{\alpha}\subset{\mathbb{C}}$ such that
$\big{|}u(z)-\log|f(z)|\big{|}\leqslant C_{\alpha}\log|z|,\quad z\to\infty,\enskip z\not\in E_{\alpha},$ (1.1)
and $E_{\alpha}$ can be covered by a family of disks $D_{z_{j}}(t_{j})$, $j\in{\mathbb{N}}$, satisfying the estimate $\sum_{|z_{j}|>R}t_{j}=O(R^{\rho-\alpha})$, $(R\to+\infty)$. In the special case when a subharmonic function $u$ is homogeneous $\log|z|$ in estimate (1.1) can be replaced by $O(1)$ [3, 8]. An integral metric allows us to drop an exceptional set when we approximate subharmonic functions of finite order. Let $\|\cdot\|_{q}$ be the norm in the space $L^{q}(0,2\pi)$,
Q(r,u)=\begin{cases}O(\log r),\;r\to+\infty,&\rho[u]<+\infty,\\ {\begin{aligned} O(\log r+\log n(r,u)),\;r\to+\infty,&\\ \quad r\not\in E,\mbox{mes\,}E<+\infty,&\end{aligned}}&\rho[u]=+\infty.\end{cases}
A.Β A.Β Golβdberg and M.Β O.Β Hirnyk have proved [9] that for an arbitrary subharmonic function $u$ there exists an entire function $f$ such that
$\big{\|}u(re^{i\theta})-\log|f(re^{i\theta})|\big{\|}_{q}=Q(r,u),\quad r>0,\enskip q>0.$
From the recent result of Yu.Β Lyubarskii and Eu.Β Malinnikova [10] it follows that integrating of an approximation rate by the plane measure allows us to drop the assumption that a subharmonic function $u$ is of finite order of growth and to obtain sharp estimates.
###### Theorem CΒ [10].
Let $u(z)$ be a subharmonic function in ${\mathbb{C}}$. Then, for each $q>1/2$, there exist $R_{0}>0$ and an entire function $f$ such that
$\frac{1}{\pi R^{2}}\int\limits_{|z|R_{0}.$ (1.2)
An example constructed in [10] shows that we cannot take $q<1/2$ in estimate (1.2). In connection with Theorem A the following M.Β Sodinβs question, which is a more precise form of Question 1 from [11, p.Β 315], is known:
Question. Given a subharmonic function $u$ in $\mathbb{C}$, do there exist an entire function $f$ and a constant $\alpha\in[0,1)$ such that
$\int\limits_{|z|R_{0}.$ (1.3)
###### Remark 1.
Question 1 from [11, p.Β 315] corresponds to the case $\alpha=0$. The mentioned example from [10] implies the negative answer to this question.
On the other hand, the restrictions onto the Riesz measure from below allow us to sharpen estimate (1.2).
###### Theorem FΒ [10].
Let $u(z)$ be a subharmonic in ${\mathbb{C}}$ function. If, for some $R_{0}>0$ and $q>1$
$\mu_{u}(\{z:R<|z|\leqslant qR\})>1,\quad R>R_{0},$ (1.4)
then there exists an entire function $f$ satisfying
$\displaystyle\sup\limits_{R>0}R^{-2}\int\limits_{|z|
In addition, for every $\varepsilon>0$ there exists a set $E_{\varepsilon}\subset{\mathbb{C}}$ such that
$\displaystyle\limsup_{R\to\infty}{m(\{z\in E:|z|
and
$u(z)-\log|f(z)|=O(1),\quad z\not\in E_{\varepsilon},\enskip z\to\infty.$ (1.5)
As we can notice, there is a gap between the statements of Theorems A and B. The following question arises: how can one improve the estimate of the left hand side of (1.2), whenever (1.4) fails to hold? The answer to this question is given by Theorem 1.
Let $\Phi$ be the class of slowly varying functions $\psi\colon[1,+\infty)\to(1,+\infty)$ (in particular, $\psi(2r)\sim\psi(r)$ for $r\to+\infty$).
###### Theorem 1.
Let $u$ be a subharmonic in ${\mathbb{C}}$ function, $\mu=\mu_{u}$. If for some $\psi\in\Phi$ there exists a constant $R_{1}$ satisfying the condition
$(\forall R>R_{1}):\mu(\{z:R<|z|\leqslant R\psi(R)\})>1,$ (1.6)
then there exists an entire function $f$ such that $(R\geqslant R_{1})$
$\displaystyle\int\limits_{|z| (1.7)
###### Corollary 1.
In the conditions of Theorem 1, for arbitrary $\varepsilon>0$ there exist $K(\varepsilon)>0$ and a set $E_{\varepsilon}\subset{\mathbb{C}}$ such that
$\limsup_{r\to\infty}\frac{m(E\cap\overline{D}_{r})}{r^{2}}<\varepsilon$ (1.8)
and
$\big{|}u(z)-\log|f(z)|\big{|}\leqslant K_{\varepsilon}\log\psi(|z|),\quad z\not\in E_{\varepsilon}.$ (1.9)
The following example demonstrates that estimate (1.7) is exact in the class of subharmonic functions satisfying (1.6). This is indicated by TheoremΒ 2.
For $\varphi\in\Phi$, let
$u(z)=u_{\varphi}(z)=\frac{1}{2}\sum_{k=1}^{+\infty}\log\Bigl{|}1-\frac{z}{r_{k}}\Bigr{|},$ (1.10)
where $r_{0}=2$, $r_{k+1}=r_{k}\varphi(r_{k})$, $k\in{\mathbb{N}}\cup\{0\}$. Thus, $\mu_{u}$ satisfies condition (1.6) with $\psi(x)=\varphi^{2}(x)$.
###### Theorem 2.
Let $\psi\in\Phi$ be such that $\psi(r)\to+\infty$ $(r\to+\infty)$. There exists no entire function $f$ for which
$\int\limits_{|z|
It follows from the proved Theorem $2^{\prime}$ that the answer to the formulated above M.Β Sodinβs question is negative. It was M. Hirnyk who pointed out the question as well as the fact that the negative answer follows from the example constructed above.
Theorem $\bf 2^{\prime}$. Let $\sigma$ be an arbitrary positive continuous, defined on $[1,+\infty)$ function and $\sigma(t)\to 0$ $(t\to\infty)$, $\psi(R)=\exp\Bigl{\{}\int_{1}^{R}\frac{\sigma(t)}{t}\,dt\Bigr{\}}$. There is no entire function $f$ and constant $\alpha\in[0,1)$ for which
$\int\limits_{|z|
###### Remark 2.
The growth of the expression $\int_{1}^{R}\frac{\sigma(t)}{t}\,dt$ for $R\to+\infty$ is restricted only by the condition $\int_{1}^{R}\frac{\sigma(t)}{t}\,dt=o(\log R)$.
###### Remark 3.
The author does not know whether it is possible to improve estimate (1.8) of the exceptional set for (1.9). In [12] there are obtained the sharp estimates of the exceptional set outside of which relation (1.9) holds, for some class of subharmonic functions satisfying some additional restriction onto the Riesz measure.
## 2 Proof of Theorem 1
### 2.1. Partition of measures
It is appeared that in order to have a ββgoodββ approximation of a subharmonic function by the logarithm of the modulus of an entire function we need a ββgoodββ approximation of the corresponding Riesz measure by a discrete one. The only restriction on Rieszβ measure, defined on the Borel sets in the plane, is finiteness on the compact sets. The following theorem on a partition of measures is the principal step in proofs of the theorems cited above.
###### Theorem G.
Let $\mu$ be a measure in ${\mathbb{R}}^{2}$ with compact support, $\textrm{\mbox{supp\,}}\mu\subset\Pi$, and $\mu(\Pi)\in{\mathbb{N}}$, where $\Pi$ is a square. Suppose, in addition, that for any line $L$ parallel to either side of the square $\Pi$, there is at most one point $p\in L$ such that
$0<\mu(\{p\})<1\ \text{and}\ \mu(L\setminus\{p\})=0,$ (2.1)
Then there exist a system of rectangles $\Pi_{k}\subset\Pi$ with sides parallel to the sides of $\Pi$, and measures $\mu_{k}$ with the following properties:
• 1)
$\textrm{\mbox{supp\,}}\mu_{k}\subset\Pi_{k}$;
• 2)
$\mu_{k}(\Pi_{k})=1$, $\sum_{k}\mu_{k}=\mu$;
• 3)
the interiors of the convex hulls of the supports of $\mu_{k}$ are pairwise disjoint;
• 4)
the ratio of length of the sides for rectangles $\Pi_{k}$ lies in the interval $[1/3,3]$;
• 5)
each point of the plane belongs to the interiors of at most 4 rectanglesΒ $\Pi_{k}$.
Theorem G was proved by R.Β S.Β Yulmukhametov [7, Theorem 1] for absolutely continuous measures (i.e. $\nu$ such that $m(E)=0\Rightarrow\nu(E)=0$). In this case condition (2.1) holds automatically. As it is shown by D.Β DrasinΒ [4, Theorem 2.1], Yulmukhametovβs proof works if we replace the condition of continuity by condition (2.1). We can drop condition (2.1) rotating the initial squareΒ [4]. Though it is noted in [10] that Theorem G is valid even without hypothesis (2.1), the author does not know a proof of this fact. Moreover, proving Theorem 2.1 [4] (a variant of Theorem G) condition (2.1) is used essentially. In this connection one should mention a note by A.F.Β Grishin and S.V.Β Makarenko [13], where a two dimensional variant of Theorem G under the condition that the measure does not load lines parallel to the coordinate axes.
###### Remark 4.
In the proof of Theorem G [4] rectangles $\Pi_{k}$ are obtained by partition of given rectangles, starting from $\Pi$, into smaller rectangles in the following way. Length of a smaller side of an initial rectangle coincides with length of a side of obtained rectangle, and length of the other side of obtained rectangle is not less than a third part and not greater than two third parts of length of the other side for the initial rectangle. Thus the following form of TheoremΒ G, which will be used, holds.
###### Theorem 3.
Let $\mu$ be a measure in ${\mathbb{R}}^{2}$ with compact support, $\textrm{\mbox{supp\,}}\mu\subset\Pi$, $\mu(\Pi)\in 2{\mathbb{N}}$, where $\Pi$ is a rectangle with the ratio $b_{0}/a_{0}=l_{0}\in[1,+\infty)$ of length $a_{0},b_{0}$, $(a_{0}\leqslant b_{0})$ of the sides. If, in addition, condition (2.1) holds, then there exist a system of rectangles $\Pi_{k}\subset\Pi$ with sides parallel to the sides of $\Pi$ and measures $\mu_{k}$ with the following properties:
• 1)
$\textrm{\mbox{supp\,}}\mu_{k}\subset\Pi_{k}$;
• 2)
$\mu_{k}(\Pi_{k})=2$, $\sum_{k}\mu_{k}=\mu;$
• 3)
the interiors of the convex hulls of the supports of $\mu_{k}$ are pairwise disjoint;
• 4)
the ratio $b_{k}/a_{k}$ of length $a_{k},b_{k}$ $(a_{k}\leqslant b_{k})$ of the sides for the rectangle $\Pi_{k}$ lies in the interval $[1,l_{0}]$, moreover, if $l_{k}>3$, then $a_{k}=a_{0}$;
• 5)
each point of the plane belongs to the interiors of at most 4 rectanglesΒ $\Pi_{k}$.
As it is noted in [3], the idea of partition into rectangles with mass 2 is due to A.Β F.Β Grishin. In order to apply Theorem 3 we need the following lemma (see also Lemma 2.4Β [4].)
###### Lemma 1.
Let $\nu$ be a locally finite measure in ${\mathbb{C}}$. Then in any neighborhood of the origin there exists a point $z^{\prime}$ with the following properties:
• a)
on each line $L_{\alpha}$ going through $z^{\prime}$ there is at most one point $\zeta_{\alpha}$ such that $\nu(\{\zeta_{\alpha}\})>0$, moreover $\nu(L_{\alpha}\setminus\{\zeta_{\alpha}\})=0;$
• b)
on each circle $C_{\rho}$ with center $z^{\prime}$ there exists at most one point $\zeta_{\rho}$ such that $\nu(\{\zeta_{\rho}\})>0$, moreover $\nu(C_{\rho}\setminus\{\zeta_{\rho}\})=0$.
We give a simple example for illustration. Let $\nu(z)=\sum_{n\in{\mathbb{N}}}\delta(z-n)$. Then $\nu(\mathbb{R})=+\infty$. We can take any point of the disk $\{z:|z|<1\}$ with an nonzero imaginary part as $z^{\prime}$.
###### Proof of Lemma 1.
Let $B_{n}=\{z:2^{n}<|z|\leqslant 2^{n+1}\}$, $n\in{\mathbb{N}}$, $B_{0}=\{z:|z|\leqslant 2\}$ and $\nu_{n}=\nu\bigr{|}_{B_{n}}$. Since $\nu$ is a locally finite measure, $\nu_{n}({\mathbb{C}})=\nu_{n}(B_{n})<+\infty$. There is an at most countable set $\zeta_{nk}$ of points such that $\nu_{n}(\{\zeta_{nk}\})>0$. Therefore the set $E_{1}=\bigcup_{n}\bigcup_{k}\{\zeta_{nk}\}$ is at most countable. Given a pair of points from $E_{1}$ we consider the straight line going through these points, and the middle perpendicular to the segment connecting these points. All these lines cover some set $A\subset{\mathbb{C}}$, $m(A)=0$. Let $z_{1}\in{\mathbb{C}}\setminus A$. By our construction, an arbitrary straight line going through the point $z_{1}$ contains at most one point with positive mass. The same is true for an arbitrary circle with center $z_{1}$. We define $\nu_{n}^{\prime}=\nu_{n}-\sum_{\nu_{n}(\{\zeta\})>0}\nu_{n}(\{\zeta\})\delta_{\zeta}$ $(n\in{\mathbb{Z}}_{+})$, $\delta_{\zeta}(z)=\delta(z-\zeta)$. Then, for any $z\in{\mathbb{C}}$, we have $\nu^{\prime}_{n}(\{z\})=0$ $(n\in{\mathbb{Z}}_{+})$. Since the intersection of two different circles (straight lines) is empty or a point or two points, and $\nu_{n}({\mathbb{C}})<+\infty$, by countable additivity of $\nu_{n}$, there exists at most countable set of circles and straight lines with positive $\nu_{n}$-measure. The union $F_{n}$ of all these straight lines and centers of circles has zero area. If now $z^{\prime}\in{\mathbb{C}}\setminus(A\cup\bigcup_{n\in{\mathbb{Z}}_{+}}F_{n})$, then for any $n\in{\mathbb{Z}}_{+}$ the measure $\nu_{n}$ of any circle with center $z^{\prime}$ as well as that of a straight line going through $z^{\prime}$ is equal to zero. Hence, their $\nu$-measure equals zero. Finally, by the countable additivity of the plane measure $m(A\cup\bigcup_{n\in{\mathbb{Z}}_{+}}F_{n})=m(A)+\sum_{n}m(F_{n})=0$. Thus, any point $z^{\prime}\in({\mathbb{C}}\setminus(A\cup\bigcup_{n\in{\mathbb{Z}}_{+}}F_{n}))\cap U$ where $U$ is a given neighborhood of the origin has required properties. β
Taking into account the proved lemma one may assume that properties a) and b) of Lemma 1 hold for the origin. We follow the scheme of the proof from [10], assuming that $\psi(x)\nearrow+\infty$ $(x\to+\infty)$, because otherwise Theorem 1 is equivalent to Theorem B. Without loss of generality, one may assume that $u(z)$ is harmonic in a neighborhood of $z=0$. Otherwise, choose arbitrary $a>0$ such that $n(a)\leqslant N\leqslant n(a+0)$ for some $N\in{\mathbb{N}}$ and define the measure $\nu$ to be equal $\mu$ in $D_{0}(a)$ and to contain the part $\mu\bigr{|}_{\{z:|z|=a\}}$ so that $\nu(\overline{D_{0}(a)})=N$. Then $\mu-\nu\equiv 0$ in $D_{0}(a)$ and, instead of $u$, we consider the function $\tilde{u}(z)=u(z)-\int_{{\mathbb{C}}}\log|z-\zeta|\,d\nu(\zeta)$, and the quantity $\int_{\mathbb{C}}\log|z-\zeta|\,d\nu(\zeta)-N\log|z|$ is bounded whenever $z\to+\infty$. Therefore, without loss of generality we assume that $R_{0}=\sup\{r>0:\mbox{supp\,}\mu\cap D_{0}(r)=\varnothing\}>1$. Define $\Psi_{1}(R)=R\psi(R)$, $\Psi_{n}(R)=\Psi_{1}(\Psi_{n-1}(R))$ for $n\in{\mathbb{N}}$, $\Psi_{0}(R)\equiv R$, $R>1$. We define by induction measures $\mu_{k}^{(j)}$, $j\in\{1,2,3\}$, and a sequence $(R_{k})$, $k\in{\mathbb{N}}$. Suppose that $\mu_{l}^{(j)}$ is already defined for $l, and $R_{l}$ for $l\leqslant k$. Let
$Q_{k}=\{\zeta\in{\mathbb{C}}:R_{k}\leqslant|\zeta|\leqslant\Psi_{1}(R_{k})\},\quad\mu_{k}^{-}=\Bigl{(}\mu-\sum_{j=1}^{k-1}(\mu_{j}^{(1)}+\mu_{j}^{(2)}+\mu_{j}^{(3)})\Bigr{)}\biggr{|}_{Q_{k}}.$
If $\mu_{k}^{-}(Q_{k})<2$, define $\mu_{k}^{(1)}(Q_{k})\equiv 0$, $\mu_{k}^{(2)}=\mu_{k}^{-}$, $\mu_{k}^{(3)}\equiv 0$. If $\mu_{k}^{-}(Q_{k})\geqslant 2$, represent $\mu_{k}^{-}$ as the sum $\mu_{k}^{(1)}+\mu_{k}^{(2)}$, so that $\mu_{k}^{(1)}(Q_{k})=2[\mu_{k}^{-}(Q_{k})/2]$, where $[a]$ denotes the integer part of $a$. If $\mu_{k}^{(2)}(Q_{k})\geqslant 1$, we define $\mu_{k}^{(3)}(Q_{k})=0$ and $R_{k+1}=\Psi_{1}(R_{k})$. Otherwise, let
$R_{k+1}=\inf\bigl{\{}R>R_{k}:\mu(\{\Psi_{1}(R_{k})<|z|\leqslant R\})\geqslant 1\bigr{\}},$
and by $\mu_{k}^{(3)}$ we mean the sum of the restriction of $\mu$ onto $\{\zeta:\Psi_{1}(R_{k})<|\zeta| and $\tilde{\mu}$, where $\tilde{\mu}$ is the part of the restriction $\mu\bigr{|}_{\{\zeta:|\zeta|=R_{k+1}\}}$ such that $\mu_{k}^{(3)}({\mathbb{C}})=1$. By the construction, we have for all $k\in{\mathbb{N}}$:
• 1)
$\mbox{supp\,}\mu_{k}^{(1)}\subset Q_{k}$, $\mu_{k}^{(1)}(Q_{k})\in 2{\mathbb{Z}}_{+}$;
• 2)
$\Psi_{1}(R_{k})\leqslant R_{k+1}\leqslant\Psi_{2}(R_{k})$;
• 3)
$\mbox{supp\,}(\mu_{k}^{(2)}+\mu_{k}^{(3)})\subset\{\zeta:R_{k}\leqslant|\zeta|\leqslant R_{k+1}\}$;
• 4)
$1\leqslant(\mu_{k}^{(2)}+\mu_{k}^{(3)})\bigl{(}\{\zeta:R_{k}\leqslant|\zeta|\leqslant R_{k+1}\}\bigr{)}\leqslant 2$.
Let $\mu^{(1)}=\sum_{j=1}^{+\infty}\mu_{j}^{(1)}$, $\mu^{(2)}=\sum_{j=1}^{+\infty}(\mu_{j}^{(2)}+\mu_{j}^{(3)})$. From properties 3) and 4) it follows that $\mu^{(2)}(\overline{D_{0}(R)})=O(\log R)$ $(R\to+\infty)$, therefore $u_{2}(z)=\int_{{\mathbb{C}}}\log\bigl{|}1-\frac{z}{\zeta}\bigr{|}\,d\mu^{(2)}(\zeta)$ is a subharmonic function in ${\mathbb{C}}$. Let $u_{1}(z)=u(z)-u_{2}(z)$. Then $\mu_{u_{1}}=\mu^{(1)}$. We will approximate $u_{1}$ and $u_{2}$ separately. It suffices to prove that
$I_{n}\stackrel{{\scriptstyle\rm def}}{{=}}\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\big{|}u(z)-\log|f(z)|\big{|}\,dm(z)=O(4^{n}\log\psi(2^{n})),\quad n\to+\infty.$ (2.2)
Indeed, let $R\in[2^{n},2^{n+1})$. Then from (2.2) it follows that
$\displaystyle\frac{\int\limits_{|z| $\displaystyle\qquad\leqslant\frac{C_{1}\sum\limits_{k=0}^{n}4^{k}\log\psi(2^{k})}{4^{n}}\leqslant 4C_{1}\log\psi(2^{n}).$
### 2.2. Approximation of $u_{2}(z)$
By the construction, $\mu^{(2)}({\mathbb{C}})=+\infty$. Define $T_{n}=\sup\{R>0:\mu^{(2)}(\overline{D_{0}(R)})\leqslant 5n\}$, $A_{n}=\{\zeta:T_{n}\leqslant|\zeta|\leqslant T_{n+1}\}$. Let $(A_{n},\mu_{n})$ be a partition of the measure $\mu^{(2)}$ such that $\mu^{(2)}=\sum_{k=1}^{+\infty}\mu_{k}$, $\mbox{supp\,}\mu_{n}\subset A_{n}$, $\mu_{n}(A_{n})=5$. Define $r_{n}$ by the equalities $\log r_{n}=\frac{1}{5}\int_{A_{n}}\log|\zeta|\,d\mu_{n}(\zeta)$, $n\in{\mathbb{N}}$, and consider the formal product
$f_{2}(z)=\prod_{n=1}^{+\infty}\Bigl{(}1-\frac{z}{r_{n}}\Bigr{)}^{5}.$
From property 4) of the measures $\mu_{k}^{(j)}$ it follows that
$\Psi_{1}(T_{n})\leqslant T_{n+1}\leqslant\Psi_{6}(T_{n}).$ (2.3)
Since $r_{n+1}/r_{n-1}\geqslant T_{n+1}/T_{n}\geqslant\psi(T_{n})\to+\infty$ $(n\to+\infty)$, the function $f_{2}$ is entire. Let
$\displaystyle d_{k}(z)$ $\displaystyle\equiv\int\limits_{A_{k}}\Bigl{(}\log\Bigl{|}1-\frac{z}{\zeta}\Bigr{|}-\log\Bigl{|}1-\frac{z}{r_{k}}\Bigr{|}\Bigr{)}\,d\mu_{k}(\zeta)$ $\displaystyle=\int\limits_{A_{k}}\Bigl{(}\log\Bigl{|}1-\frac{\zeta}{z}\Bigr{|}-\log\Bigl{|}1-\frac{r_{k}}{z}\Bigr{|}\Bigr{)}\,d\mu_{k}(\zeta).$ (2.4)
Here we make use the choice of $r_{k}$. Fix $n\in{\mathbb{N}}$, and let $2^{n}\in[T_{N},T_{N+1})$. Then for $k\geqslant N+2$ and $\zeta\in A_{k}$ we have $|\zeta|\geqslant T_{k}\geqslant|z|{T_{k}}/{T_{N+1}}$, $r_{k}\geqslant|z|{T_{k}}/{T_{N+1}}$, consequently $\bigl{|}\log|1-\frac{z}{\zeta}|\bigr{|}\leqslant 2|z|/|\zeta|$, $\bigl{|}\log|1-\frac{z}{r_{k}}|\bigr{|}\leqslant 2|z|/r_{k}$. Therefore,
$\displaystyle\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\sum_{k=N+2}^{\infty}|d_{k}(z)|\,dm(z)\leqslant\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\sum_{k=N+2}^{\infty}20\frac{T_{N+1}}{T_{k}}dm(z)$ $\displaystyle\qquad\leqslant C_{2}\frac{T_{N+1}}{T_{N+2}}4^{n}=o(4^{n}),\quad n\to\infty.$ (2.5)
Similarly, $|d_{k}(z)|\leqslant 20T_{k+1}2^{-n}$ for $k\leqslant N-2$. Using (2.4), we obtain
$\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\sum_{k=1}^{N-2}|d_{k}(z)|\,dm(z)\leqslant o(4^{n}),\quad n\to\infty.$ (2.6)
Estimate $\int_{2^{n}\leqslant|z|\leqslant 2^{n+1}}|d_{k}(z)|\,dm(z)$ for $k\in\{N-1,N,N+1\}$. By the definition,
$\displaystyle\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}|d_{N+1}(z)|\,dm(z)$ $\displaystyle\qquad=\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\biggl{|}\int\limits_{A_{N+1}}\Bigl{(}\log\Bigl{|}1-\frac{z}{\zeta}\Bigr{|}-\log\Bigl{|}1-\frac{z}{r_{N+1}}\Bigr{|}\Bigr{)}\,d\mu_{N+1}(\zeta)\biggr{|}dm(z).$
If $|\zeta|\geqslant 2|z|$, $\zeta\in A_{N+1}$, we have $\bigl{|}\log|1-\frac{z}{\zeta}|\bigr{|}\leqslant\log 2$. Otherwise, $T_{N+1}\leqslant|\zeta|<2|z|<2^{n+2}\leqslant 4T_{N+1}$. Therefore, applying Fubiniβs theorem and changing the variables, $T_{N+1}\eta=\zeta$, $T_{N+1}\xi=z$, we obtain
$\displaystyle\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\int\limits_{A_{N+1}}{\Bigl{|}}\log{\Bigl{|}}1-\frac{z}{\zeta}{\Bigl{|}}{\Bigl{|}}\,d\mu_{N+1}(\zeta)dm(z)$ $\displaystyle\leqslant 2\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}dm(z)+\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\int\limits_{T_{N+1}\leqslant|\zeta|\leqslant 4T_{N+1}}{\Bigl{|}}\log{\Bigl{|}}1-\frac{z}{\zeta}{\Bigl{|}}{\Bigl{|}}\,d\mu_{N+1}(\zeta)dm(z)$ $\displaystyle\quad=O(4^{n})$ $\displaystyle\qquad+T_{N+1}^{2}$ $\displaystyle\qquad\quad\times\int\limits_{1\leqslant|\eta|\leqslant 4}\;\int\limits_{{2^{n}}/{T_{N+1}}\leqslant|\xi|\leqslant{2^{n+1}}/{T_{N+1}}}{\Bigl{|}}\log{\Bigl{|}}1-\frac{\xi}{\eta}{\Bigl{|}}{\Bigl{|}}\,dm(\xi)\,d\mu_{N+1}(T_{N+1}\eta)$ $\displaystyle\quad=O(4^{n})+O(4^{n})\int\limits_{1\leqslant|\eta|\leqslant 4}\;\int\limits_{|\xi|\leqslant 2}{\Bigl{|}}\log{\Bigl{|}}1-\frac{\xi}{\eta}{\Bigl{|}}{\Bigl{|}}\,dm(\xi)\,d\mu_{N+1}(T_{N+1}\eta).$ (2.7)
However, elementary calculations demonstrate that for $1\leqslant\eta\leqslant 4$ we have
$\int\limits_{|\xi|\leqslant 2}{\Bigl{|}}\log{\Bigl{|}}1-\frac{\xi}{\eta}{\Bigl{|}}{\Bigl{|}}\,dm(\xi)\leqslant C_{3}.$
Taking into account that $\mu_{N+1}({\mathbb{C}})=2$, from (2.7) we deduce
$\int\limits_{2^{n}\leqslant|z|\leqslant 2^{n+1}}\int\limits_{A_{N+1}}{\Bigl{|}}\log{\Bigl{|}}1-\frac{z}{\zeta}{\Bigl{|}}{\Bigl{|}}d\mu_{N+1}(\zeta)\,dm(z)=O(4^{n}),\quad n\to+\infty.$ (2.8)
Similarly, for $r_{N+1}\geqslant 2^{n+2}\geqslant 2|z|$ we have $\big{|}\log|1-z/{r_{N+1}}|\big{|}\leqslant\log 2$, otherwise, $T_{N+1}\leqslant r_{N+1}<2^{n+2}$, and consequently $(T_{N+1}\xi=z)$ | 8,878 | 22,971 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 347, "math_alttext": 1, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.82123 |
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# What is the last step in the order of operations?
Updated: 10/23/2022
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β 12y ago
Subtraction. Remeber it's PEMDAS. Parenthesis, Exponent, Multiplication, Division, Addition and Subtraction.
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Q: What is the last step in the order of operations?
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The last step in supervising detainee operations is allowing the detainee access to the outside world including family, friends, lawyers. This process can be monitored. | 459 | 2,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-10 | latest | en | 0.926312 |
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## Presentation on theme: "Ion Source Modelling and Field Index Improvements by Scott Lawrie."— Presentation transcript:
Ion Source Modelling and Field Index Improvements by Scott Lawrie
Current goals Decrease emittance of beam Study the influence of the sector magnet Confirm field index, n = 1 Widen magnet poles Find optimum pole separation and angle Plot good field region Make improved poles and test them …Then move on to the extractor!
The model
Coordinate system layout 5 H2H2 Cs 4321 θ r r = 105 mm r = 55 mm B B
Cold box top view
Pole piece close-up Angle of Pole-face Pole Width Spacer Thickness Shim Width Shim Height Maximum pole width = 74mm (ISIS standard = 50mm). Allows 1.5mm clearance from bottom of cold box. Centre of pole face; used to measure separation of poles
Scan field along curves
Comparison of magnetic fields 8.88 ° faces, 24.00mm separation 7.00° faces, 19.16mm separation
Calculating the field index Sum up the z-components along curve Multiply by angle difference integral
Field index of ISIS dipoles Pole pieces separated by 25mm, with face angles of 8.88°
How the field index varies with pole separation Optimum pole separation = 28.6 mm
How the shape of the field index varies Pole face angle Separation of poles modified to give an n = 1 field at R = 80 for each angle. 7 degree poles couldn’t be squeezed close enough together without hitting the extract.
Angle / separation relation for n = 1
Field indices for wide poles with fixed 32mm pole separation and varying angle
Field indices for wide poles with different parameters
Comparing fields off centre 11 degree pole faces, 36 mm separation 14 degree pole faces, 40.87 mm separation
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Mapping the good field region Current ISIS poles Wide poles with same face angle as ISIS poles, separated a bit so as not to collide with extract. Field index, n
Explanation for the cobra-head? Current ISIS poles with large error Field index, n
Still to do… Check the field map is optimal Get new poles built Test new poles Move on to improving the extract
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# lowest common ancestor of a binary search tree
What is lowest common ancestor (LCA) of bst node. this is lowest parent node of given n1 and n2.
Assume that given n1 and n2 nodes are we finding LCA they are exist in bst. Given following example of LCA.
Try it yourself
Example 1: n1=80,n2=120.
Example 2: n1=35,n2=80.
Example 3: n1=35,n2=40.
Algorithm: iterative approach.
Use one temp pointer that is assign address of root of bst. In this (while loop) check given value n1 and n2. that are exist in bst. if n1 and n2 value are biger then to temp pointer node value. then temp point assign address of temp->left.
If n1 and n2 value are smaller then assign temp pointer address of temp->right child.
Otherwise we are getting n1 and n2 lowest common ancestor.
Time complexity of this algorithm is O(n).
C program to find lowest common ancestor.
### Output
Code execution: view code execution process.
Note that not given all step of execution process here.View How to insert Linked list node and so on.
Try it yourself | 275 | 1,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.867512 |
https://transum.org/Maths/Exam/Online_Exercise.asp?NaCu=91 | 1,720,821,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00478.warc.gz | 459,073,620 | 6,609 | # Exam-Style Questions.
## Problems adapted from questions set for previous Mathematics exams.
### 1.
GCSE Higher
Shape A can be transformed to shape B by a reflection in the y-axis followed by a translation $${c \choose d}$$
Find the value of $$c$$ and the value of $$d$$.
### 2.
GCSE Higher
The shape A is drawn on the coordinate grid as shown below.
Sally and Eddie each transform the shape A onto shape B.
• Sally uses a reflection in the line y = 7 followed by a rotation of 90o anticlockwise about the point (9,9).
• Eddie transforms shape A first with a reflection in the line $$y = x$$ followed by his favourite transformation.
(a) Draw and label shape B.
(b) Describe fully Eddie's favourite transformation.
### 3.
GCSE Higher
The diagram shows a red trapezium drawn on a grid.
The trapezium is subjected to two transformations, one after the other.
One transformation is a reflection in the line $$y=x$$.
The other transformation is a reflection y-axis.
### 4.
GCSE Higher
The diagram shows a trapezium A on a grid.
Trapezium A is reflected in the line $$x=10$$ to give trapezium B.
Trapezium B is translated by $$\begin{pmatrix} -6 \\ -3 \\ \end{pmatrix}$$ to give trapezium C.
(a) Draw the trapezia B and C on the grid above.
(b) Trapezium C is reflected in the line $$y=5$$ to give trapezium D. What are the coordinates of the centre of a 180° rotation that would map trapezium D onto trapezium A?
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