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https://www.coursehero.com/file/228191/Chapter-22/ | 1,516,505,020,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889917.49/warc/CC-MAIN-20180121021136-20180121041136-00228.warc.gz | 871,773,653 | 62,793 | # Chapter 22 - Chapter 22 1 Picture the Problem A proton...
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Chapter 22 1. Picture the Problem: A proton moves in a magnetic field that is directed at right angles to its velocity. Strategy: Combine Newton's Second Law with the magnetic force (equation 22-1) to find the acceleration of the particle. Solution: Set the magnetic force equal to the mass multiplied by the acceleration and solve for a : Insight: If the magnetic field were parallel to the velocity, the angle θ = 0° and the force and acceleration would be zero. 8. Picture the Problem: An electron moves in a region in which a magnetic field exists. Strategy: The ion experiences no magnetic force when it is moving in the direction, so we conclude that the magnetic field also points along either the direction. We can use either the Left-Hand Rule for negative charges, or use the Right-Hand Rule and remember to reverse the direction of the force because the charge on the electron is negative. Try pointing your left thumb downwards ( direction) and the fingers of your left hand in the direction to find that the magnetic field must point in the direction. Then use equation 22-1 and the force experienced by the electron when it moves in the direction to find the magnitude of Solution: Use equation 22-1 to find Insight: If instead the electron were to travel in the direction, it would experience a force 9. Picture the Problem: Two charged particles travel in a magnetic field along the same direction and experience the same magnetic force, but they travel at different speeds. Strategy: Use equation 22-1 to calculate the ratio of the speeds of the particles. Solution: 1. (a) Since the magnetic force is directly proportional to both the charge and the speed of the particles, and since the particles experience the same force, particle 2 must have a greater speed because particle 1 has the greater charge. 2. (b) Use equation 22-1 to find the ratio: Insight: Suppose both charges were allowed to travel in circles as described in section 22-3. Assuming both charges have the same mass, we find that
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Ask a homework question - tutors are online | 557 | 2,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-05 | latest | en | 0.9185 |
www.b-eac.com | 1,580,277,826,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00219.warc.gz | 187,476,736 | 6,220 | # (zhuan) Variational Autoencoder: Intuition and Implementation
• 时间:
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Our objective here is to model the data, hence we want to find P(X)P(X). Using the law of probability, we could find it in relation with zz as follows:
Let’s define some notions:
In this post, we will look at the intuition of VAE model and its implementation in Keras.
Now if only we know P(X,z)P(X,z), or equivalently, P(X|z)P(X|z) and P(z)P(z)…
that is, we marginalize out zz from the joint probability distribution P(X,z)P(X,z).
Alright, now let’s say we want to infer P(z|X)P(z|X) using Q(z|X)Q(z|X). The KL divergence then formulated as follows:
But the problem is, we have to infer that distribution P(z|X)P(z|X), as we don’t know it yet. In VAE, as it name suggests, we infer P(z|X)P(z|X) using a method called Variational Inference (VI). VI is one of the popular choice of method in bayesian inference, the other one being MCMC method. The main idea of VI is to pose the inference by approach it as an optimization problem. How? By modeling the true distribution P(z|X)P(z|X) using simpler distribution that is easy to evaluate, e.g. Gaussian, and minimize the difference between those two distribution using KL divergence metric, which tells us how difference it is PP and QQ.
Suppose we want to generate a data. Good way to do it is first to decide what kind of data we want to generate, then actually generate the data. For example, say, we want to generate an animal. First, we imagine the animal: it must have four legs, and it must be able to swim. Having those criteria, we could then actually generate the animal by sampling from the animal kingdom. Lo and behold, we get Platypus!
Alright, that fable is great and all, but how do we model that? Well, let’s talk about probability distribution.
From the story above, our imagination is analogous to latent variable. It is often useful to decide the latent variable first in generative models, as latent variable could describe our data. Without latent variable, it is as if we just generate data blindly. And this is the difference between GAN and VAE: VAE uses latent variable, hence it’s an expressive model.
Recall the notations above, there are two things that we haven’t use, namely P(X)P(X), P(X|z)P(X|z), and P(z)P(z). But, with Bayes’ rule, we could make it appear in the equation:
There are two generative models facing neck to neck in the data generation business right now: Generative Adversarial Nets (GAN) and Variational Autoencoder (VAE). These two models have different take on how the models are trained. GAN is rooted in game theory, its objective is to find the Nash Equilibrium between discriminator net and generator net. On the other hand, VAE is rooted in bayesian inference, i.e. it wants to model the underlying probability distribution of data so that it could sample new data from that distribution.
The idea of VAE is to infer P(z)P(z) using P(z|X)P(z|X). This is make a lot of sense if we think about it: we want to make our latent variable likely under our data. Talking in term of our fable example, we want to limit our imagination only on animal kingdom domain, so we shouldn’t imagine about things like root, leaf, tyre, glass, GPU, refrigerator, doormat, … as it’s unlikely that those things have anything to do with things that come from the animal kingdom. Right?
P(X)=P(X|z)P(z)dzP(X)=∫P(X|z)P(z)dz
01-29
01-29
01-29
01-29
01-29
01-28
01-27
01-27 | 882 | 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} | 3.140625 | 3 | CC-MAIN-2020-05 | longest | en | 0.897889 |
https://www.mathworks.com/matlabcentral/cody/problems/1224-flexible-anonymous-function/solutions/195872 | 1,495,977,758,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609817.29/warc/CC-MAIN-20170528120617-20170528140617-00381.warc.gz | 1,148,188,091 | 11,686 | Cody
# Problem 1224. Flexible Anonymous Function
Solution 195872
Submitted on 24 Jan 2013 by Ken Deeley
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
%% myf=@(x) det(x); yourf = flexf(myf); x={[1 2;3 4],7,[1 2 3; 4 5 6; 2 3 1],3,[2 -1 ; 1 -1]}; [a,~,b,c]=yourf(x{:}); y_correct = {-2 9 3}; assert(isequal({a b c},y_correct));
``` ```
2 Pass
%% myf=@(x) x.^2; yourf = flexf(myf); x={[1 2;3 4],7,[1 2 3; 4 5 6; 2 3 1],3,[2 -1 ; 1 -1]}; [a,~,b,c]=yourf(x{:}); y_correct = cellfun(myf,x,'uni',0); assert(isequal({a b c},y_correct([1 3 4])));
``` ``` | 282 | 675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.56138 |
https://buy-essay.com/answered-essay-wood-corporation-owns-70-percent-of-carter-companys-voting-shares-on-january-1-20x3-carter-sold-bonds-with/ | 1,670,035,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710918.58/warc/CC-MAIN-20221203011523-20221203041523-00291.warc.gz | 193,358,765 | 13,222 | # Answered Essay: Wood Corporation owns 70 percent of Carter Company’s voting shares. On January 1, 20X3, Carter sold bonds with
Wood Corporation owns 70 percent of Carter Company’s voting shares. On January 1, 20X3, Carter sold bonds with a par value of \$742,500 at 98. Wood purchased \$495,000 par value of the bonds; the remainder was sold to nonaffiliates. The bonds mature in five years and pay an annual interest rate of 8 percent. Interest is paid semiannually on January 1 and July 1.
a. What amount of interest expense should be reported in the 20X4 consolidated income statement?
b.1-b.3. Prepare the journal entries Wood recorded during 20X4 with regard to its investment in Carter bonds. (If no entry is required for a transaction/event, select “No journal entry required” in the first account field. Round your market rate of interest to 3 decimals. For example, .0547523 should be rounded to 5.475%)
b.1 January 01: Record the interest received on the bonds.
b.2 July 01: Record the interest received on the bonds.
b.3 December 31: Record the interest receivable on the bonds.
c1-c2: Prepare all worksheet consolidation entries needed to remove the effects of the intercorporate bond ownership in preparing consolidated financial statements for 20X4.
c1. Record the entry to eliminate the effects of the intercompany ownership in the bonds.
c2. Record the entry to eliminate the intercompany interest receivables/payables.
a. What amount of interest expense should be reported in the 20X4 consolidated income statement?
Par value of bond \$ 742,500 annual interest rate of 8 percent \$ 59,400 Payable to Wood Corporations \$ 39,600 Payable to nonaffiliates. \$ 19,800 interest expense to be reported in the 20X4 consolidated income statement \$ 19,800
b.1-b.3. Prepare the journal entries Wood recorded during 20X4 with regard to its investment in Carter bonds.
Date General Journal in woods corporations Debit Credit January 1, 20X4 Cash A/c \$ 19,800 To Intesrest on Bonds \$ 19,800 ( Record the interest received on the bonds. ) July 1, 20X4 Cash A/c \$ 19,800 To Intesrest on Bonds \$ 19,800 ( Record the interest received on the bonds. ) December 31, 20X4 Cash A/c \$ 59,400 To Intesrest on Bonds \$ 59,400 ( Record the interest receivable on the bonds )
c1-c2: Prepare all worksheet consolidation entries needed to remove the effects of the intercorporate bond ownership in preparing consolidated financial statements for 20X4.
c1. Record the entry to eliminate the effects of the intercompany ownership in the bonds.
No entry required
c2. Record the entry to eliminate the intercompany interest receivables/payables.
January 1, 20X4 Carter Company \$ 39,600 To Wood Corporations \$ 39,600 ( Record the interest received on the bonds. )
Pages (550 words)
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At this stage, our editor will go through your essay and make sure your writer did meet all the instructions. | 1,077 | 4,653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.899364 |
http://physics.stackexchange.com/questions/tagged/material-science+elasticity | 1,406,270,889,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997893881.91/warc/CC-MAIN-20140722025813-00134-ip-10-33-131-23.ec2.internal.warc.gz | 293,667,018 | 15,010 | # Tagged Questions
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http://experiment-ufa.ru/how_to_write_85_in_roman_numerals | 1,524,331,240,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945272.41/warc/CC-MAIN-20180421164646-20180421184646-00085.warc.gz | 107,203,568 | 6,022 | # How to write 85 in roman numerals
## Write 85 in roman numerals.
If it's not what You are looking for, type in into the box below your integer to convert it into roman numerals.
## 85 in roman numerals:
When you convert 85 into roman numerals you get
LXXXV
In summary 85 in roman numerals is LXXXV
Now when you know how to write 85 in roman numerals you can use it whenever you want.
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There are lots of number games for children. These games are pretty easy to play but not so easy to
make. We will discuss about an interesting game here. Each player will be given N positive integer.
(S)He can make a big integer by appending those integers after one another. Such as if there are
4 integers as 123, 124, 56, 90 then the following integers can be made — 1231245690, 1241235690,
5612312490, 9012312456, 9056124123, etc. In fact 24 such integers can be made. But one thing is sure
that 9056124123 is the largest possible integer which can be made.
You may think that it’s very easy to find out the answer but will it be easy for a child who has just
got the idea of number?
Input
Each input starts with a positive integer N (≤ 50). In next lines there are N positive integers. Input
is terminated by N = 0, which should not be processed.
Output
For each input set, you have to print the largest possible integer which can be made by appending all
the N integers.
## Code Examples
### #1 Code Example with C Programming
Code - C Programming
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t,m,n,q;
int a,b;
char grid[101][101];
cin >> t;
while(t--){
cin >> m >> n >> q;
for(int i=0;i < m;i++) scanf("%s",grid[i]);
printf("%d %d %d\n",m,n,q);
while(q--){
cin >> a >> b;
bool valid = true;
int len = 3,res=1;
char cur = grid[a][b];
while(valid){
int rowStart = a-len/2;
int rowEnd = a+len/2;
int colStart = b-len/2;
int colEnd = b+len/2;
if(rowStart < 0 || colStart < 0> valid = false;
if(rowEnd >= m || colEnd >= n) valid = false;
for(int i=rowStart; i<=rowEnd && valid; i++){
if(grid[i][colStart] != cur) valid = false;
if(grid[i][colEnd] != cur) valid = false;
}
for(int i=colStart; i < =colEnd && valid; i++){
if(grid[rowStart][i] != cur) valid = false;
if(grid[rowEnd][i] != cur) valid = false;
}
if(valid> res = len;
len += 2;
}
cout << res << endl;
}
}
}
Copy The Code &
Input
cmd
4
123 124 56 90
5
123 124 56 90 9
5
9 9 9 9 9
0
Output
cmd
9056124123
99056124123
99999 | 623 | 2,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.663071 |
https://gmatclub.com/forum/if-a-b-and-c-are-positive-integers-such-that-a-is-divisible-by-b-an-209958.html | 1,550,677,390,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247495147.61/warc/CC-MAIN-20190220150139-20190220172139-00219.warc.gz | 562,967,030 | 150,015 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# If a, b, and c are positive integers such that a is divisible by b, an
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If a, b, and c are positive integers such that a is divisible by b, and c is divisible by a, which of the following is NOT necessarily an integer?
A. (a + c)/b
B. (c - a)/b
C. ca/b
D. (c + b)/a
E. cb/a
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Re: If a, b, and c are positive integers such that a is divisible by b, an [#permalink]
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13 Dec 2015, 09:14
[quote="Bunuel"]If a, b, and c are positive integers such that a is divisible by b, and c is divisible by a, which of the following is NOT necessarily an integer?
A. (a + c)/b
B. (c - a)/b
C. ca/b
D. (c + b)/a
E. cb/a
Ans:-D
a=bx.........(1)
c=aY.........(2)
a. (A+C)/B = (bx+ay)/b = (bx+ bxy)/b =x+xy=integer
b. (c-a)/b = (ay-a)/b = a(y-1)/b = bx(y-1)/b = integer.
similarly
All options were integer except D.
d.(c+b)/a
ay+b)/a .....not integer......
Ans D
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Re: If a, b, and c are positive integers such that a is divisible by b, an [#permalink]
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13 Dec 2015, 09:28
as per the question.
let a be 15
b be 5
and c be 45
plugging in the values in answer choices only option D turns out not to be an integer.
verify the result by plugging in a=9,b=3 and c=27 or a=14, b =7 and c=35
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Re: If a, b, and c are positive integers such that a is divisible by b, an [#permalink]
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Re: If a, b, and c are positive integers such that a is divisible by b, an [#permalink] 04 Feb 2019, 05:33
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https://se.mathworks.com/matlabcentral/answers/84508-cut-vector-in-set-range?s_tid=prof_contriblnk | 1,679,956,539,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00459.warc.gz | 583,192,916 | 25,626 | # Cut vector in set range
55 views (last 30 days)
Jonasz on 11 Aug 2013
I have about 10 thousand vectors and I want to cut them all to special range set by my min and max . eg. have only vectors from 600 - 1000 values How to implement it in the best efficient way.
Image Analyst on 11 Aug 2013
Try this for each of the vectors:
minValue = 600; % Or whatever you want.
maxValue = 1000; % Or whatever you want.
indexesInRange = data > minValue & data < maxValue;
subVector = data(indexesInRange);
That will, for a vector called data, extract all the values between minValue (which you can define to be 600) and maxValue (which you can define to be 1000) into a new vector called subVector. Do this 10 thousand times for each of the vectors (most likely done in a loop).
Azzi Abdelmalek on 11 Aug 2013
How your vectors are stored? in a cell array? If yes
Example
v=arrayfun(@(x) randi(2000,1,100),1:10,'un',0) % your cell array
for k=1:numel(v)
out{k}=v{k}(v{k}>600 & v{k}<1000)
end | 283 | 982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.815155 |
https://jp.mathworks.com/matlabcentral/answers/525758-how-can-we-store-many-matrices-z-from-for-loop-in-a-single-matrix-d-say-in-my-problem | 1,624,284,406,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00571.warc.gz | 319,506,569 | 23,594 | # How can we store many matrices(z) from for loop in a single matrix D(say) in my problem.
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MOHD UWAIS 2020 年 5 月 15 日
x=[1 2 3];
y=1./x;
for l=1:3;
I=eye(3,3);
z=y(l)*I;
end
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### 回答 (1 件)
Prasad Reddy 2020 年 5 月 17 日
% If you want your matrices z1,z2,z3 to be stored in side by side ie D=[z1 z2 z3] the following code helps
clc
clear all
x=[1 2 3];
y=1./x;
D=[];
for l=1:3;
I=eye(3,3);
z=y(l)*I;
D=[D,z]
end
% If you want your matrices z1,z2,z3 to be stored in a column manner ie D=[z1
% z2
% z3]
%the following code helps
clc
clear all
x=[1 2 3];
y=1./x;
D=[];
for l=1:3;
I=eye(3,3);
z=y(l)*I;
D=[D;z]
end
% Please give a up thumb if this answer works for you. Thank you.
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posted by .
use factoring to solve the equation
4x²-23x=6
I think the answer is x=-1/4,x=6
Just want to make sure that's right
• Algebra -
4 x^2 -23 x - 6 = 0
(4 x + 1)(x - 6) = 0
x = -1/4 or x = 6
agree
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More Similar Questions | 726 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-47 | latest | en | 0.931055 |
http://roperescuetraining.com/en/physics_angles.asp | 1,524,455,626,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945724.44/warc/CC-MAIN-20180423031429-20180423051429-00194.warc.gz | 273,421,078 | 5,733 | # Rope Angles
The angle of ropes that are connected to anchors can have a tremendous affect on forces.
## 0° Angle
This system has a 100-pound load suspended from two anchors. The two ropes are basically parallel which creates an angle of approximately 0°. In this case, each anchor is holding half of the load (i.e., 50 pounds).
(You can mentally replace the word "pounds" with "kilograms" throughout this page.)
## 90° Angle
This system has the same 100-pound load suspended from two anchors, but this time the angle between the two ropes is 90°. In this case, each anchor is holding approximately 70 pounds. Yes, the 100‑pound load is actually generating 140 pounds of force (i.e., 2 x 70 pounds)!
## 150° Angle
As the angle between the two ropes increases, the force on the anchors increases—dramatically. This next system has the same 100-pound load, but this time the angle between the ropes is 150 degrees. The force on each anchor (and on each rope) is now more than 190 pounds—that is almost double the 100-pound load.
## Angles and Forces
Forces by Rope Angle
50 lb.
45° 54 lb.
90° 71 lb.
120° 100 lb.
140° 146 lb.
150° 193 lb.
160° 288 lb.
170° 574 lb.
This table shows the force-multiplying effects of rope angles. The left-hand column is the angle between the ropes and the right hand column is the force that would be applied to each rope and anchor. (Again, you can mentally replace the term "pounds" in this table with "kilograms.")
You can see that when the angle between the ropes is 120 degrees, each rope will receive forces equivalent to the entire load (in this case, 100 pounds). The 120-degree angle is sometimes referred to as the "critical angle" to remind rescuers that exceeding 120 degrees will result in more than 100% of the load being applied to each rope.
Rescuers do not need to memorize this table. Knowing that 50% of the load is on each strand when the ropes are side-by-side (i.e., 0 degrees) and that 100% of the load is on each strand when the ropes are 120 degrees is adequate. As a general rule of thumb, most rescuers ensure that the angle between the ropes never exceeds 90 degrees. The choice of 90 degrees is selected because it is easy for people to visualize and it provides an additional safety margin from the "critical angle" of 120 degrees.
This chart shows how forces increase dramatically as the angles between the ropes increase.
## Handy Protractor
With your thumb spread widely, the angle between your thumb and index finger is approximately 90 degrees (generally considered the maximum angle). When all of your fingers are spread widely, the angle between your thumb and little finger is approximately 120 degrees (the "critical angle").
## Calculator
You can use this calculator to calculate forces that are generated by any load at any angle.
Load: kilograms pounds Angle between ropes: Exerts xxx xxx force on each rope
## Nerdy Details
So how can the 100-pound load in this 150-degree system generate more than 190 pounds on two anchors? That is a total of ~380 pounds being generated by a single 100-pound load!
In fact, each anchor is still receiving only 50 pounds of vertical force from the 100-pound load. However, the load is also pulling the two anchors horizontally toward each other as shown in this illustration.
The 50 pounds of vertical force, run through a little trigonometry, generates a net diagonal force of 193 pounds.
The following formulas show the math to calculate the net force on each strand of rope. Each strand is holding 50 pounds of vertical force at a 75° angle (one-half of the 150 degrees) which generates ~193 pounds of tension on each strand of rope.
These next formulas show how you can calculate only the horizontal force on each of the anchors if you know the vertical force (50 pounds in our example) and the net force on each strand of rope (193.2 pounds in our example). These formulas (and the above illustration) show that there are 186.6 pounds of horizontal force on each of the anchors. The 186.6 pounds of horizontal force plus the 50 pounds of vertical force results in 193 pounds of total force on each of the anchors.
You do not need to understand the trigonometry that is used to calculate these forces (and vRigger can do this math for you), but every rescuer must know that wide-rope-angles can generate massive forces that can easily exceed the load's weight (e.g., in the above example, the 100-pound load is generating approximately 380 pounds of force).
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## JKBOSE Class 12th Statistics Notes
JKBOSE Class 12th Statistics Notes PDF Download. If you are the students of Jammu and Kashmir and are looking for important questions and Notes of Statistics Subject then you are at right place. Get JKBOSE important Study Materials Notes of all the subjects for Class 12th in this site but in this article we will provide you Statistics Notes for Class 12th. So keep visiting and get the free and best notes.
## JKBOSE Class 12th Statistics Notes Unitwise
Unit- 1 Probability – I
Introduction and Objectives
Probability: Probability is a fundamental concept in mathematics and statistics that quantifies the likelihood of an event occurring. It provides a framework for reasoning and making predictions in uncertain situations.
Probability is defined as a number between 0 and 1 where 0 represents impossibility and 1 represents certainty. The probability of an event A denoted as P(A) measures the relative likelihood of A occurring. It is determined by considering all possible outcomes and assigning numerical values to them based on their likelihood.
Axion of Peobability: The axioms of probability are a set of fundamental principles that govern the behavior of probabilities. There are three axioms:
Non-Negativity: The probability of any event is a non-negative number i.e. P(A) ≥ 0 for any event A.
Normalization: The probability of the entire sample space denoted as S is equal to 1 i.e. P(S) = 1.
Additivity: For any collection of mutually exclusive events (events that cannot occur simultaneously) the probability of their union is equal to the sum of their individual probabilities. If A and B are mutually exclusive events then P(A ∪ B) = P(A) + P(B).
Concept of Conditional Probability
Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B) where A and B are two events. The conditional probability of A given B is calculated by dividing the probability of the intersection of A and B by the probability of event B assuming that P(B) > 0. It is defined as P(A|B) = P(A ∩ B) / P(B).
Conditional probability allows us to update our knowledge or beliefs about an event based on new information or evidence. It plays a crucial role in various fields including statistics machine learning and decision theory.
Unit- 2 Probability – II
Introduction and Objectives
Random variable: A random variable in probability refers to a numerical quantity whose value is determined by the outcome of a random event or experiment. It is a way of quantifying uncertainty and capturing the variability in a particular situation. Random variables are typically denoted by capital letters such as X or Y and can represent a wide range of quantities such as the number of heads obtained when flipping a coin or the temperature measured in a specific location.
Discrete variable: A discrete variable is a type of random variable that can only take on a countable number of distinct values. In other words its values are typically integers or whole numbers. For example the number of students in a classroom the outcomes of rolling a die or the number of cars passing through a toll booth in a given time period are all examples of discrete variables. The probability distribution of a discrete random variable can be represented by a probability mass function (PMF) which assigns probabilities to each possible value of the variable.
Continuous Random Variable: A continuous random variable can take on any value within a certain interval or range. It is not restricted to specific individual values. Examples of continuous random variables include measurements like height weight time or temperature. The probability distribution of a continuous random variable is described by a probability density function (PDF) which specifies the relative likelihood of the variable taking on different values. Unlike the PMF of a discrete variable the PDF does not assign probabilities to specific values but instead provides the likelihood of the variable falling within certain intervals.
Unit- 3 Regression Analysis
Introduction and Objectives
Concept of Regression: Regression is a statistical concept used to analyze the relationship between variables. It aims to predict the value of a dependent variable based on one or more independent variables. In other words it helps us understand how changes in independent variables affect the dependent variable. Regression is widely used in various fields such as economics social sciences finance and machine learning.
In regression analysis we often use a technique called linear regression. It involves fitting a line (regression line) to a scatterplot of data points to approximate the relationship between variables. The regression line represents the best fit to the data minimizing the distance between the line and the actual data points.
Regression lines: The regression line is defined by an equation of the form y = a + bx where y is the dependent variable x is the independent variable a is the y-intercept and b is the slope of the line. The y-intercept (a) represents the predicted value of the dependent variable when the independent variable is zero while the slope (b) represents the change in the dependent variable for a unit change in the independent variable.
Regression Coefficients: Regression coefficients refer to the values of a and b in the regression equation. The coefficient a is also known as the intercept coefficient while the coefficient b is the slope coefficient. These coefficients provide important information about the relationship between variables. A positive slope coefficient indicates a positive relationship between the variables while a negative slope coefficient indicates a negative relationship. The coefficients can be used to make predictions by plugging in values for the independent variable into the regression equation.
Unit- 4 Theory of Attributes
Introduction and Objectives
Manifolds Classifications: Manifolds are fundamental objects in mathematics that play a crucial role in various fields including differential geometry topology and physics. A manifold can be loosely described as a space that locally looks like Euclidean space. One of the key aspects of studying manifolds is their classification.
Manifolds can be classified in different ways based on various criteria. One common classification is based on the dimension of the manifold. For example a 1-dimensional manifold is a curve a 2-dimensional manifold is a surface and a 3-dimensional manifold is our familiar three-dimensional space. Beyond three dimensions manifolds are more difficult to visualize but can still be mathematically defined.
Another classification is based on the smoothness of the manifold. A smooth manifold is one where there are no abrupt changes or singularities and the transition between local Euclidean spaces is smooth. These smooth manifolds are further classified into different categories such as compact manifolds (those that are closed and bounded) and non-compact manifolds (those that are either open or unbounded).
There are also specific types of manifolds that are studied extensively such as Riemannian manifolds which have a metric tensor that defines a notion of distance and angle. These play a crucial role in differential geometry and the formulation of physical theories like general relativity.
Ultimate Class Frequency: The term "ultimate class frequency" is not a standard term or concept in the context of manifold classification. It does not have a defined meaning and thus it is not possible to provide a precise explanation for it.
Unit- 5 Index Numbers
Introduction and Objectives
Index Number: Index numbers are statistical tools that measure changes in a variable over time. They provide a way to compare different observations or sets of data relative to a base period or reference point. Index numbers are commonly used in economics finance and other fields to track changes in prices quantities or other measurable factors.
Characteristics of Index numbers: The characteristics of index numbers include:
Base period: Index numbers require a base period which serves as a reference point for comparison. All subsequent observations are measured relative to this base period.
• Relative measurement: Index numbers provide a relative measurement by expressing the change in a variable as a percentage or ratio compared to the base period. This allows for meaningful comparisons between different periods or groups.
• Weighting: Depending on the application index numbers may incorporate weighting to reflect the importance of different components within a dataset. Weighting assigns greater significance to certain variables or groups resulting in a more accurate representation of the overall change.
• Uniqueness: Index numbers are unique to the variable they measure and the purpose they serve. Different variables may require different formulas and methodologies to construct appropriate index numbers.
Uses of Index numbers: Some common uses of index numbers include:
• Inflation measurement: Consumer price indices (CPI) are widely used to measure changes in the average price level of a basket of goods and services over time. These indices help track inflation rates and are essential for economic policy-making.
• Stock Market Analysis: Stock indices such as the S&P 500 or Dow Jones Industrial Average provide a snapshot of the overall performance of a selected group of stocks. Investors and analysts use these indices to assess market trends and make investment decisions.
• Economic Indicators: Index numbers are used to track changes in various economic indicators like industrial production employment levels and business activity. These indicators provide insights into the overall health and performance of an economy.
• Cost-of-living Adjustments: Index numbers are used to calculate cost-of-living adjustments (COLAs) for wages pensions and social security benefits. They ensure that income levels keep pace with changes in the cost of living maintaining the purchasing power of individuals over time.
Unit- 6 Vital Statistics
Introduction and Objectives
Vital Statistics: Vital statistics refer to numerical data and information related to events that are vital or essential to individuals and populations. They primarily focus on three main aspects: births deaths and marriages. These statistics provide valuable insights into population dynamics health and demographic characteristics which are crucial for planning and policy-making.
Nature of Vital Statistics: The nature of vital statistics is primarily quantitative as they involve numerical measurements and analysis. They are collected through the registration of vital events by governmental authorities such as birth and death certificates marriage licenses and related documents. These records contain important details such as names dates locations and other demographic information.
Uses of Vital Statistics: The uses of vital statistics are multifaceted. First and foremost they are essential for demographic analysis and studying population trends. They provide information on birth rates death rates and marriage rates allowing researchers and policymakers to understand population growth fertility patterns mortality rates and changes in marital status.
Vital statistics also play a crucial role in public health. They provide data on causes of death which helps in the identification and monitoring of diseases identifying risk factors and developing appropriate public health interventions. They also aid in assessing the effectiveness of healthcare systems and interventions.
Furthermore vital statistics are vital for administrative purposes. They serve as legal documents for individuals allowing them to establish their identity citizenship and other legal rights. They also facilitate the functioning of government agencies and enable the allocation of resources and services based on population needs.
Unit- 7 Sampling Theory
Introduction and Objectives
Meaning of Sampling: Sampling is a statistical technique used to gather data from a subset of a larger group known as a population in order to make inferences about the whole population. It is impractical and time-consuming to collect data from an entire population so sampling allows researchers to study a representative sample and draw conclusions about the population as a whole.
Objectives of Sampling: The primary objective of sampling is to obtain accurate and reliable information about a population while minimizing costs and resources. By selecting a sample that is representative of the population researchers aim to generalize the findings from the sample to the larger population. This is possible when the sample is chosen using randomization techniques and when the sample size is sufficiently large to minimize sampling errors.
Concept of Statistical Population: The concept of a statistical population refers to the entire group of individuals objects or events that researchers are interested in studying. It represents the larger target group from which a sample is drawn. The population can be finite such as the number of students in a school or infinite such as all the possible outcomes when rolling a dice.
It is crucial to define the population accurately to ensure the findings from the sample can be applied to the intended population. The characteristics of the population such as size variability and homogeneity influence the sampling process. Researchers use various sampling methods including random sampling stratified sampling cluster sampling and convenience sampling depending on the research objectives and constraints.
Sampling involves selecting a subset of a larger population to gather data and make inferences about the whole population. The objective is to obtain reliable information while minimizing costs and the statistical population refers to the entire group of interest from which the sample is drawn.
Unit- 8 Time Series and Computers
Introduction and Objectives
Time series analysis is a statistical technique that deals with data collected over time at regular intervals. It involves studying the pattern behavior and trends exhibited by the data to make predictions and forecasts about future values. Time series analysis is widely used in various fields such as finance economics weather forecasting stock market analysis sales forecasting and many others.
The importance of time series analysis lies in its ability to uncover valuable insights and patterns hidden within the temporal data. By analyzing historical data time series models can capture seasonality trends and cyclic patterns allowing us to understand the underlying dynamics of a system and make informed decisions. Here are some key reasons why time series analysis is significant:
• Forecasting: Time series analysis helps in predicting future values based on historical patterns and trends. By understanding the past behavior of a time series we can develop accurate forecasts for future time points. This is particularly valuable in areas such as sales forecasting demand planning and resource allocation.
• Pattern recognition: Time series analysis allows us to identify recurring patterns and regularities in the data. These patterns could include seasonal effects cyclical variations or long-term trends. Recognizing these patterns helps in understanding the behavior of the system and can provide insights for decision-making.
• Anomaly detection: Time series analysis can identify outliers and anomalies in the data. By comparing the observed values with predicted values we can identify unexpected variations or events that deviate from the regular pattern. Anomaly detection is crucial in various domains such as fraud detection network monitoring and quality control.
• Descriptive analysis: Time series analysis helps in describing the characteristics of a time series. It involves analyzing statistical properties such as mean variance autocorrelation and stationarity which provide insights into the underlying dynamics of the data. Descriptive analysis helps in understanding the data structure and can guide the selection of appropriate modeling techniques
• Decision-making and planning: Time series analysis provides a foundation for data-driven decision-making. By analyzing historical data identifying trends and making forecasts businesses can make informed decisions about resource allocation inventory management budgeting and investment strategies. Time series analysis also assists in policy planning and formulation by providing insights into economic indicators and trends.
The time series analysis plays a vital role in understanding the dynamics of temporal data making accurate predictions detecting anomalies and supporting informed decision-making. Its applications span across various industries and domains making it an essential tool for analyzing and leveraging time-dependent information.
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## JKBOSE Class 12th Statistics Important Textual Questions
#### What is probability and how is it defined?
Probability is a fundamental concept in mathematics and statistics that quantifies the likelihood of an event occurring. It is defined as a number between 0 and 1 where 0 represents impossibility and 1 represents certainty. The probability of an event A denoted as P(A) measures the relative likelihood of A occurring.
#### What are the axioms of probability?
The axioms of probability are a set of fundamental principles that govern the behavior of probabilities. There are three axioms: non-negativity (probability of any event is a non-negative number) normalization (probability of the entire sample space is equal to 1) and additivity (probability of the union of mutually exclusive events is equal to the sum of their individual probabilities).
#### What is conditional probability and how is it calculated?
Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. It is denoted as P(A|B) where A and B are two events. The conditional probability of A given B is calculated by dividing the probability of the intersection of A and B by the probability of event B assuming that P(B) > 0. It is defined as P(A|B) = P(A ∩ B) / P(B).
#### What is regression analysis and how does it work?
Regression analysis is a statistical concept used to analyze the relationship between variables. It aims to predict the value of a dependent variable based on one or more independent variables. In linear regression a line (regression line) is fitted to a scatterplot of data points to approximate the relationship between variables. The regression line represents the best fit to the data minimizing the distance between the line and the actual data points.
#### What are index numbers and how are they used?
Index numbers are statistical tools that measure changes in a variable over time. They provide a way to compare different observations or sets of data relative to a base period or reference point. Index numbers are commonly used in economics finance and other fields to track changes in prices quantities or other measurable factors. They help in measuring inflation analyzing stock market performance tracking economic indicators and calculating cost-of-living adjustments.
JKBOSE CLASS 10TH NOTES
JKBOSE CLASS 9TH NOTES | 3,675 | 20,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-50 | latest | en | 0.940854 |
https://www.physicsforums.com/threads/local-gauge-invariance.885152/ | 1,679,625,938,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00388.warc.gz | 1,032,589,056 | 17,846 | # Local Gauge invariance
• I
Silviu
Hello! Can someone explain to me what exactly a local gauge invariance is? I am reading my first particle physics book and it seems that putting this local gauge invariance to different lagrangians you obtain most of the standard model. The math makes sense to me, I just don't see what is the physical meaning of this local gauge invariance and why would you come up with it? Like, it seems such an easy thing to do, just make the phase of wave function depends on position, but doing this you obtain everything up to Higgs field, and I am not sure I understand, physically, why. Thank you
## Answers and Replies
ohad
Gauge invariance is about local symmetry of a field.
If you have any field and there is some global symmetry which is preserved by the lagrangian, than requiring this symmetry to be local (i.e. to be position dependent) usually breaks the symmetry, for that you need to introduce the gauge field to make the symmetry possible.
the best example is electrodynamics, you can see that by requiring the local symmetry of the fermion (i.e electron) field you must have the em field which is the gauge field - the photon.
So you got 'for free' the photon and the charge mechanism of the fermions just by requiring the local symmetry.
Gold Member
well some fun way to try and imagine things...
With a global gauge symmetry: in each point of spacetime you have the same arc (for the phase degrees) that you are "rotating" your field... Then you can immediately see that you can perform a parallel transport on such a setup without anything (as the flat spacetime).
With a local gauge symmetry, each point gets a different phase that rotates the field... In order to perform a parallel transport, you introduce the gauge fields, as when in GR you made the metric spacetime dependent you introduced the Christoffel Symbols.
That's why the gauge fields can also be seen as connections, for they allow you to define a parallel transport (remember how the partial derivatives change to covariant derivatives).
Why local? because global would be photonless...
my2cts
Those are all indirect reasons for a fundamental requirement for a field theory.
Global symmetry enforces a corresponding conservation law.
There is no conservation law that requires gauge symmetry.
Last edited:
Science Advisor
The local gauge principle is just a way of guessing the form of the interaction, also called "minimal coupling". In general relativity, the principle of equivalence" is a minimal coupling type of guess.
https://arxiv.org/abs/1305.0017
Science Advisor
Gold Member
2022 Award
One reason for "local gauge invariance" is the realization of the proper orthochronous Poincare group in terms of massless vector fields. If you want to have a realization with only discrete intrinsic ("spin like") quantum numbers, which is in accordance with observation, you necessarily have to represent the null rotations of the corresponding "little group" trivially, and this leads to the necessity to represent the massless vector fields on a quotient space, i.e., to envoke local (Abelian) gauge invariance. That's how the gauge invariance of electrodyamics comes into view from a group theoretical approach to relativistic QT.
The extension of this insight to non-Abelian local gauge symmetry was an ingenious discovery of Yang and Mills. It's just a natural generalization of the local gauge principle to non-abelian gauge groups. Sometimes in the history of science the mathematically beautiful turns out to be of great utility in physics. That's for sure the case for non-Abelian gauge theories underlying the Standard Model of elementary particles which is more successful than wanted ;-)).
dextercioby
Lapidus
Hi Silviu,
gauge symmetry is arguably the central tenet of particle physics and qft. But in many older textbooks its physical meaning is only badly explained. Check out Schwartz QFT textbook chapter 8 for a deep yet readable explanation.
One of the best layment yet precise explanation of it can be found in Randall "Warped Passages" in the "Symmetry and Forces" chapter. | 879 | 4,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-14 | latest | en | 0.935325 |
https://www.wired.com/2012/08/optimal-lawn-mowing-patterns/ | 1,686,194,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654031.92/warc/CC-MAIN-20230608003500-20230608033500-00327.warc.gz | 1,165,947,009 | 166,339 | Optimal Lawn Mowing Patterns
How do you mow a perfect lawn in the least amount of time? Dot Physics blogger Rhett Allain uses video analysis and physics to find the optimal pattern.
What do you do when you mow the lawn? I like to listen to podcasts and think of new blog ideas. This one has been in my head for a long time. What is the best way to mow the lawn? Should I just go back and forth, or should I make a box-shaped loop? These questions must be answered.
Video Data
I need some data. How fast does a lawn mower move? How long does a turn take? How fast do you go when pulling the lawn mower backwards? Without these answers, really the only thing I could do to optimize the lawn mowing would be to make sure I don't go over the same spot twice.
Video time. Yes, I made a video. If you really, really, really want to see it - fine. Why would you even watch that? Oh, I guess I should also say something about the lawn mower. This is one of those "self propelled" models. You still have to push - so it doesn't really propel itself, but it does make a difference. Oh, if you look at the video, the cones are 1 meter apart.
Here is my first plot. This shows the horizontal position of the lawn mower. The first thing to look at is the acceleration.
This shows an acceleration of about 0.6 m/s2 for about 1 second. After that, the velocity seems fairly constant.
This shows an approximate speed of about 1.67 m/s. The acceleration during the stopping part of the motion is quite a bit larger than the starting acceleration. It seems to be somewhere around 4 m/s2.
In the video, I did several things. In the first case, I went a few meters and then turned the mower around. In the next case, I went a few meters and then just pulled the mower backwards without turning it around. So, are the forward and backward speeds significantly different? Since I went back and forth several times, I can make a histogram showing the distributions of speeds for both forward motion and backwards motion.
I know it isn't as much data as I would like, but I can only mow the same spot so many times before I get bored. Also, there are more "forwards" than "backwards" because whenever I was turning around I would go forward twice. Anyway, the speeds seem to be clearly different with an average forward speed of 1.607 m/s and a backwards average of 1.255 m/s.
How long does it take to turn around? Looking at the data (which I won't plot since that gets boring), I get an average turn around time of 2.213 seconds.
What about a right angle turn? How long does this take? I get an average time of 1.326 seconds. Ok - I think that is all the data I need.
Finally, what about the time to stop and go backwards if I am just pulling the mower back and not turning it around? That would be about 0.893 seconds.
Modeling a Lawn Mow
Let me now start with a simple square lawn that is 30 meters by 30 meters. I think I need one more piece of information, the cutting width. My particular mower has a 22 inch blade. This gives and approximate cutting width of about 0.52 meters (I cut off some of the width for a little bit of overlap).
Here is my first cutting strategy, back and forth.
How long does this lawn take to mow? For this case, it isn't too difficult to figure out. If the width of the mowed path is s, then the mower would need to make L/s cuts to finish the lawn. If the number of paths doesn't fit perfectly, round up. The total mowing time would then be:
The number of turn around times is 1 less than the number of rows. If I have 8 rows, I will have to turn around 7 times. So, back to my 30m square yard (with no trees or anything in the middle). Using my values from the video, this lawn would take 20.14 minutes to mow.
What about the other common mowing pattern - the spiraling square? (I just made that name up)
First, how many of these square patterns will be needed? This is similar to the back and forth case, but there will be half as many "squares" as rows in the previous case (I think). That would put N at L/(2s). For each square, there will be 4 right angle turns. The thing that changes is the length of each side for the successive squares. If the first square is L x L, the next square will be (L - 2s) x (L - 2s).
But how would you calculate this? One way is a with a super-simple python program. Like this:
In the loop, the program calculates the time for mowing in a square including 4 right angle turns. It then decrease the size of the square and repeats the time calculation. Using this method, I get a mowing time of 21.14 minutes. Just slightly longer than the back and forth method. Why does it take longer? Although a right angle turn is quicker than a full turn around, there are more of them. And yes, I know sometimes this square pattern is better for mowing since you can push all the grass clippings into the center (if you are into collecting grass clippings).
What about a non-square yard? I think (but not completely sure) that for non-square yards, the spiral-square method will still be at a disadvantage. Think about two square yards that are placed side by side. For both methods, you would still have the same number of turns as the square case. Perhaps the only way the square-spiral method could win would be if it was a non-square yard and you use the back and forth method in the direction of the small side of the yard.
Suppose you get to a small part of your yard. Maybe it is that small section on the other side of your driveway. What is the best strategy to mow this part? Should you do a short run, turn around and then do it again? Or is it quicker to do one side and then pull the lawn mower backwards? The answer probably depends on the length of the yard area to cut.
Let me first consider the case for turning around to make the cut. If I consider one row to include one turn around, then the total time for this one row would be:
But really, to compare to the pull back method I need to look at two rows. This time would just be twice the value for one row. Now, what about the pull back case. Here is the time for two rows:
Perhaps my notation is a bit confusing. Here, I am using Δt~pull back~ to represent the time to go from pushing the mower forward to pulling it backwards. The Δt~turn around~ is the time to turn the mower around. The velocity with the f subscript is the moving forward velocity and the b subscript is the pulling back velocity.
Using the velocities and turn around times from above, here is a plot of the time for the two rows as a function of the length of the row.
So here you go. If the length is less than about 5 meters, you would save time by just pulling the lawn mower back. If the row is longer than 5 meters, your best bet is to turn around. Perhaps I should change that to "my best bet is to turn around" since you will have different speeds and turn around times with your own lawn mower. Probably.
Of course there are more things to explore (always true). What if it is an irregularly shaped lawn? Are there any other lawn mowing strategies that would be more efficient? | 1,653 | 7,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.972199 |
https://hsm.stackexchange.com/questions/tagged/complex-analysis | 1,721,760,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00809.warc.gz | 253,049,576 | 49,735 | Questions tagged [complex-analysis]
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How did Roger Cotes come up with logarithm form of Euler formula?
I have been trying to get my head around how Roger Cotes first discovered Euler Formula. I knew how Euler did it, but I wanted a new perspective, especially from someone who discovered it earlier. ...
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361 views
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• 33
935 views
Who was the first person to notice logarithms of negatives numbers and for what reason?
Who was the first person to notice logarithms of negatives numbers and why? When did they first arise naturally? I thought I saw somewhere that it had to do with integration but I can't find it ...
242 views
How much ground was prepared for Riemann so that he could conjecture Riemann hypothesis?
Although I do not doubt in Riemann˙s originality, I would like to know how much complex analysis was developed up to the day when Riemann conjectured what is today called Riemann hypothesis and how ...
529 views
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117 views
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1k views
What was the motivation for Cauchy's Integral Theorem?
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• 345
278 views
Why is Riemann's dissertation (from 1851) considered a turning point in the history of the theory of conformal mappings?
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572 views
Mathematics development can sometimes **exceed** the practical needs, right?
I read below paragraph from the book "A Friendly Introduction to Number Theory": The use of "$i$" to denote the square root of negative $1$ dates back to the days when people viewed such numbers ...
• 255
295 views
What is the modern interpretation of Gauss's "Summatorische Function"?
In Buhler's biography of Gauss (Gauss: A Biographical Study), at the chapter on modular forms and hypergeometric series, he mentions a function that Gauss called "Summatorische Function", which he ...
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gauss' opinion on de moivre's theorem
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87 views
Variants in graphical presentation of real and complex numbers
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841 views
Did Gauss know the residues theorem in complex analysis in 1811?
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227 views
Were complex number first considered of limited usefulness?
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483 views
Who came up with the link between the spectrum of an operator and the poles of a meromorphic function?
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815 views
What is the history of using $i$/$\iota$ as the imaginary unit?
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• 153
584 views
History of complex analysis
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700 views
Introduction of $\imath$ and $\jmath$ notations for the imaginary unit
The imaginary unit is generally denoted $i$ or $\imath$. I have learned that the term imaginary ("imaginaires") was coined by R. Descartes in 1637, and the "i" notation was introduced by L. Euler (cf. ...
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Origin of Klein's $j$-invariant
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• 161
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Euler's first proof of $e^{ix}=\cos(x)+i\sin(x)$
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Analytic and holomorphic functions, definitions and foundations?
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432 views
First papers on holomorphic functions
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• 1,242
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Did Dedekind show any evidence of pictorial geometric sense?
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• 2,441
654 views
Who named the fugacity, who coined the variable name and did it already relate to complex analysis?
In Riemanns monumental paper, he expresses a prime counting function as an inverse Mellin transform of the log of the function he analytically continued into the complex plane \Pi(x) = \frac{1}{2\...
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What was Euler's motivation for introducing $i$ for $\sqrt{-1}$?
[Mauro Allegranza has answered the question of who introduced the notation $i$ (Euler, followed later by Gauss), so I have changed the title. I have also edited the question in other ways to make it ... | 1,938 | 8,412 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-30 | latest | en | 0.937772 |
https://justaaa.com/physics/250183-an-automobile-moves-on-a-level-horizontal-road-in | 1,685,398,923,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00182.warc.gz | 381,364,231 | 10,236 | Question
An automobile moves on a level horizontal road in a circular path of radius 30.5m. The...
An automobile moves on a level horizontal road in a circular path of radius 30.5m. The coefficients of friction between the tires and the road are Ps = 0.50 and Pk = 0.1, respectively. The maximum speed with which this car can round this curve is?
The car is traveling in a circle, assumed at a constant speed, and thus requires an acceleration, inward. That acceleration is centripetal acceleration:
a = v^2/r
What causes this acceleration? Traction...also known as static friction.
What must exist for there to be any type of dry friction? A normal force.
The normal force will do what ever is necessary to prevent the car from sinking in to the asphalt. On an unbanked, flat roadway, the normal force is only responsible for opposing the force of gravity. It is thus:
N = m*g
Relating to maximum deliverable force of static friction:
F = mu_s*m*g
And Newton's 2nd law:
Fnet = F = m*a
Substitute centripetal acceleration and parameters of maximum friction:
mu_s*m*g = m*v^2/r
solve for v (notice m's cancel and result is independent of mass):
v = sqrt(mu_s*r*g)
Data:
mu_s:=0.5; r:=30.5 m; g:=9.8 N/kg;
Result:
v = 12.22 meters/second | 320 | 1,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-23 | latest | en | 0.891966 |
https://programsandcourses.anu.edu.au/2014/course/MATH1113 | 1,723,750,941,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641311225.98/warc/CC-MAIN-20240815173031-20240815203031-00807.warc.gz | 364,169,145 | 12,010 | • Offered by Mathematical Sciences Institute
• ANU College ANU Joint Colleges of Science
• Course subject Mathematics
• Course convener
• Dr Bai-Ling Wang
• Mode of delivery In Person
• Offered in Second Semester 2014
Mathematical Foundations for Actuarial Studies (MATH1113)
This course provides a study of the fundamental concepts of calculus and linear algebra. The use and understanding of proof and abstract ideas, will allow students to develop analytical skills which will form a foundation for further study in the quantitative areas of actuarial studies.
Topics to be covered include: Calculus - limits, continuity, differentiation, inverse functions, transcendental functions, extrema, concavity and inflections, applications of derivatives, Taylor Polynomials, integration, differential equations, functions of several variables, partial derivatives, optimality, gradient and the second derivative test in two variables, double integrals; Linear Algebra - complex numbers, solving linear equations, matrix equations, linear independence, linear transformations, matrix operations, matrix inverses, subspaces, dimension and rank, determinants, Cramer's rule, volumes, eigenvalues, eigenvectors.
## Learning Outcomes
Upon successful completion, students will have the knowledge and skills to:
On satisfying the requirements of this course, students will have the knowledge and skills to:
1. Explain the fundamental concepts of calculus and linear algebra and their role in modern mathematics and applied contexts. (LO1)
2. Demonstrate accurate and efficient use of calculus and linear algebra techniques. (LO2)
3. Demonstrate capacity for mathematical reasoning through analyzing, proving and explaining concepts from calculus and linear algebra. (LO3)
4. Apply problem-solving using calculus and linear algebra techniques applied to situations in statistics, physics, engineering and other mathematical contexts. (LO4)
## Other Information
Secondary School Prerequisite: A satisfactory result in ACT Specialist Mathematics Major-Minor or NSW HSC Mathematics Extension 1 or equivalent. Students with a good pass in ACT Specialist Mathematics Major or NSW HSC Mathematics or equivalent will be considered. For students with a level of maths equivalent to ACT Mathematical Methods, MATH1003 is required to be completed before enrolling.
## Indicative Assessment
Assessment will be based on:
• Nine assignments (20% in total: LO 1-4)
• Mid-semester test (30%; LO1-4)
• Final examination (50%; LO 1-4)
The ANU uses Turnitin to enhance student citation and referencing techniques, and to assess assignment submissions as a component of the University's approach to managing Academic Integrity. While the use of Turnitin is not mandatory, the ANU highly recommends Turnitin is used by both teaching staff and students. For additional information regarding Turnitin please visit the ANU Online website.
48 lectures and 10 hours of laboratory and tutorial sessions
## Requisite and Incompatibility
You are not able to enrol in this course if you have previously completed MATH1013 or MATH1115.
## Prescribed Texts
Linear Algebra (3rd edition or 3rd edition Update, or 4th edition) by David Lay
Calculus: A Complete Course, eighth edition by Robert A. Adams
## Fees
Tuition fees are for the academic year indicated at the top of the page.
If you are a domestic graduate coursework or international student you will be required to pay tuition fees. Students continuing in their current program of study will have their tuition fees indexed annually from the year in which you commenced your program. Further information for domestic and international students about tuition and other fees can be found at Fees.
Student Contribution Band:
2
Unit value:
6 units
If you are an undergraduate student and have been offered a Commonwealth supported place, your fees are set by the Australian Government for each course. At ANU 1 EFTSL is 48 units (normally 8 x 6-unit courses). You can find your student contribution amount for each course at Fees. Where there is a unit range displayed for this course, not all unit options below may be available.
Units EFTSL
6.00 0.12500
## Course fees
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Year Fee Description
1994-2003 \$1650
2014 \$2946
2013 \$2946
2012 \$2946
2011 \$2946
2010 \$2916
2009 \$2916
2008 \$2916
2007 \$2520
2006 \$2520
2005 \$2298
2004 \$1926
International fee paying students
Year Fee
1994-2003 \$3390
2014 \$3762
2013 \$3756
2012 \$3756
2011 \$3756
2010 \$3750
2009 \$3618
2008 \$3618
2007 \$3618
2006 \$3618
2005 \$3450
2004 \$3450
Note: Please note that fee information is for current year only.
## Offerings, Dates and Class Summary Links
ANU utilises MyTimetable to enable students to view the timetable for their enrolled courses, browse, then self-allocate to small teaching activities / tutorials so they can better plan their time. Find out more on the Timetable webpage.
The list of offerings for future years is indicative only.
Class summaries, if available, can be accessed by clicking on the View link for the relevant class number.
### Second Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
8158 21 Jul 2014 01 Aug 2014 31 Aug 2014 30 Oct 2014 In Person N/A | 1,185 | 5,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.840705 |
https://www.physicsforums.com/threads/what-is-a-point-dipole.595449/ | 1,537,543,624,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157216.50/warc/CC-MAIN-20180921151328-20180921171728-00473.warc.gz | 836,895,366 | 12,782 | # What is a point dipole?
1. Apr 11, 2012
### ShayanJ
Mathematically,its easy to say that when the distance between the points tends to zero,for the dipole moment to stay finite,the charge should tend to infinity.
My question is,how should I interpret it?
What is a point dipole physically?
Could you give examples?
thanks
2. Apr 12, 2012
### Staff: Mentor
You can interpret it as a dipole which is so small that its size does not matter. This implies that all relevant distances in your problem are much larger than the size of the dipole, and higher moments (quadrupole, ...) do not matter.
3. Apr 12, 2012
### ashishsinghal
When we say a point dipole we mean that the distance at which are calculating electric field is much greater than the distance between the charges .
eg 1cm is very small before 10km
But it does not mean that the distance itself is infinitely small. Hence charge would also be finite. | 227 | 921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2018-39 | latest | en | 0.95441 |
https://www.kaysonseducation.co.in/questions/p-span-sty_3374 | 1,701,947,719,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00429.warc.gz | 922,499,179 | 11,906 | : Kaysons Education
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## Question
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Correct option is
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From the given equation we get
#### SIMILAR QUESTIONS
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https://www.oxfordscholarship.com/view/10.1093/acprof:oso/9780198568209.001.0001/acprof-9780198568209-chapter-4 | 1,563,399,153,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525402.30/warc/CC-MAIN-20190717201828-20190717223828-00029.warc.gz | 797,883,197 | 27,687 | Users without a subscription are not able to see the full content.
## Jorge L. Ramírez Alfonsín
Print publication date: 2005
Print ISBN-13: 9780198568209
Published to Oxford Scholarship Online: September 2007
DOI: 10.1093/acprof:oso/9780198568209.001.0001
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# Sylvester denumerant
Chapter:
(p.71) 4 Sylvester denumerant
Source:
The Diophantine Frobenius Problem
Publisher:
Oxford University Press
DOI:10.1093/acprof:oso/9780198568209.003.0004
In 1857, while investigating the partition number function, J. J. Sylvester defined the function d(m; a1, . . . , an), called the denumerant, as the number of nonnegative integer representations of m by a1, . . . , an. This chapter is devoted to the study of the denumerant and related functions. After discussing briefly some basic properties of the partition function and its relation with denumerants, the general behaviour of d(m; a1, . . . , an) and its connection to g(a1, . . . , an) are analyzed. Two interesting methods for computing denumerants — one based on a decomposition of the rational fraction into partial fractions and another due to E. T. Bell — are described. An exact value of d(m; p, q) — first found by T. Popoviciu in 1953 — is proved, and the known results when n = 2 and n = 3 are summarized. The calculation of g(a1, . . . , an) by using Hilbert series via free resolutions, and the use of this approach to show an explicit formula for g(a1, a2, a3), are shown. The connection among denumerants, FP, and Ehrhart polynomial as well as two variants of d(m; a1, . . . , an) are discussed.
Oxford Scholarship Online requires a subscription or purchase to access the full text of books within the service. Public users can however freely search the site and view the abstracts and keywords for each book and chapter. | 571 | 2,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.839907 |
https://unlearningmath.com/020-inches-to-mm/ | 1,721,842,739,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00726.warc.gz | 521,078,764 | 10,570 | # 020 inches to mm (Inches to Millimeters)
By / Under Inches To Millimeter / Published on
Should be an SEO optimised description including the original keyword just once
## Here is how to easily convert 020 inches to mm
020 inches to mm is 508 millimeters.
When dealing with measurements in different units, converting from inches to millimeters can be incredibly important for precision, especially in fields like engineering, manufacturing, and everyday projects. Understanding these conversions helps in ensuring accurate results and avoiding costly errors.
To convert 020 inches to mm, you use the conversion factor where 1 inch equals 25.4 millimeters. Therefore, 020 inches translates directly to 508 millimeters. This conversion is critical whether you're working on simple DIY projects or complex industrial designs.
### Importance of Accurate Conversion
In the world of measurements, precision matters a lot. A miscalculation, even by a fraction, can lead to incorrect fitting in manufacturing or design flaws. For example, in the automotive industry, accurate measurements are crucial for parts to fit perfectly, ensuring efficiency and safety.
Moreover, in manual crafting or DIY projects, understanding these conversions ensures that your builds and adjustments will align correctly with specified measurements, fostering both safety and usability.
### Real Life Example
Let's picture an analogy: Imagine you are a carpenter building a custom piece of furniture. You decide to convert 020 inches of a wooden plank into millimeters, ensuring it fits precisely into your design. An incorrect conversion could mean that shelves won't align, or joints may not fit snugly, causing instability in the structure.
### Did you know?
A recent survey in the engineering sector revealed that over 90% of professionals rely on precise measurement conversions daily. Similarly, educational studies have shown that students who grasp unit conversions early on perform better in scientific subjects.
### FAQs
#### What is 020 inches in mm?
020 inches is equivalent to 508 millimeters.
#### How do you convert 020 inches to millimeters?
To convert 020 inches to millimeters, multiply 020 by 25.4 (since 1 inch = 25.4 mm). This gives you 508 mm.
#### Why is understanding inches to mm conversion essential?
Accurate conversions are crucial in many fields, from engineering to everyday projects, ensuring precision and avoiding costly errors. | 486 | 2,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-30 | latest | en | 0.875993 |
https://briefencounters.ca/57939/parallel-and-perpendicular-lines-worksheet-answer-key/ | 1,719,214,052,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00462.warc.gz | 129,635,622 | 20,850 | # Parallel and Perpendicular Lines Worksheet Answer Key
A few days ago, I was looking for a quick and easy to understand Parallel and Perpendicular Lines Worksheet Answer Key. It was difficult to find one anywhere online, and even harder to understand what the key would be. After searching for a while, I found one in a small print size which contained a very cryptic key. I knew exactly what it was, but still could not make much sense of it. Any help would be much appreciated.
Slope Intercept form A Line Unique Parallel and Perpendicular from parallel and perpendicular lines worksheet answer key , source:datform.co
One of the problems with using the Keyword function is that you will not always get the exact value or result you are searching for. For example, if you use the keyword “Subtotal” and then use the Keyword “Parallel Line” you will get a different number than if you use the Keyword and “Parallel Line”. Using keywords is great because it will often bring up similar results, however, using them randomly will not give you the exact result you are looking for. If you have a worksheet that does not have a lot of formulas inside it you should use the keyword that contains the answer or formula you are trying to find.
When I was working on a class project for a class assignment, I came across many different problems that were solved using the Keyword and Parallel Lines functions. One problem was when I was trying to calculate the value of the slope of an unknown line. I could not figure out how to calculate this without using the Keyword and Parallel Lines functions. I then decided to search for the answer using the Yahoo! Answers site and sure enough, there were many people who had already asked this exact same question.
Parallel and Perpendicular Lines from parallel and perpendicular lines worksheet answer key , source:saylordotorg.github.io
My next step was to try to find the answer by using the Google search engine. Sure enough, there were several pages that had the same question and solution. Then I remembered that in college we used to do projects on handwritten research papers and would often draw or cut charts that represented our answer to the problem. The paper then would be turned in and the professor would make suggestions or corrections to what we had drawn using the chart tools. Using charts in this manner is a great way to save time and get the job done.
Peripatic and Parallel Lines are both great tools to help with working the problem in an effective and efficient manner. Peripatic uses the actual vertical movement of the line to show how the line is going to move. Parallel Lines works in a different way. Rather than showing where the line is actually going you can see the end point of the line. Using the Parallel Line worksheet you can quickly see the end point of any line you want.
Mr Lee MrLeeGeometry from parallel and perpendicular lines worksheet answer key , source:twitter.com
The final key to working with the Peripatic and Parallel Lines worksheet is the function. This worksheet function allows you to turn a horizontal or vertical bar to the different types of lines you can use in your charts. For example you can create a line chart or a bar chart with the use of this function. You can also use the function to do other things such as creating scatter charts. These scatter charts allow you to multiple columns in the worksheet and create a chart of varying sizes. You can also have the function do a scatter graph on a column.
Understanding how to use the Peripatic and Parallel Lines worksheet will take some time and practice. These are some of the harder worksheets to learn because they will require some advanced skills of how to use Microsoft Excel. Once you learn these skills, learning how to use the Peripatic and Parallel Lines worksheet is easy and fun. You should start to see an increase in productivity with your data analysis projects once you get comfortable with these functions.
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Parallel and Perpendicular Lines Worksheet Fresh Parallel and from parallel and perpendicular lines worksheet answer key , source:chrisonomicon.com | 964 | 5,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.967114 |
http://gmatclub.com/forum/ps-crystal-ball-77004.html?kudos=1 | 1,436,083,697,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097354.86/warc/CC-MAIN-20150627031817-00035-ip-10-179-60-89.ec2.internal.warc.gz | 116,207,686 | 47,389 | Find all School-related info fast with the new School-Specific MBA Forum
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PS - Crystal Ball
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PS - Crystal Ball [#permalink] 25 Mar 2009, 12:12
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A crystal ball is contained in a hexagon gift box. If the radius of the ball is 2inches, what is the minimum surface area of the box, in square inches? [Assume that thickness of the box could be negligible]
A 48*root3
B 40*root3
C 36*root3
D 24*root3
E 20*root3
I got 16*root3...
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Re: PS - Crystal Ball [#permalink] 25 Mar 2009, 22:44
A? I got 48root3. what is OA?
circle is inscribed in a hexagon, draw a line from center of circle to one side of hexagon, its length is 2. so u get a 30-60-90 triangle. base of this triange is 2/root3. So the one side of hexagon is 2*(2/root3) = 4/root3.
now, to calculate area of the lid we divide into 6 equilateral triangles. We need to find the area of the one equilateral triangle and multiply by 6.
Area of our 30-60-90 triangle is 2/root3. so area of each equilateral triangle is 4/root3. Multiplying by 6 we get 8root3. Two lids = 16root3.
Each side is 4/root3 and height is 4. so total area of 6 sides is 4/root3*(4)(6) = 32root3.
hence, total area is lids + sides = 48root3.
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Re: PS - Crystal Ball [#permalink] 26 Mar 2009, 09:41
Perfect....OA is A, I missed the second part. (I assumed thickness = width of box...shame!!)
thank you for nice explanation.
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Re: PS - Crystal Ball [#permalink] 26 Mar 2009, 11:58
Thanks for the explanation Economist.
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Re: PS - Crystal Ball [#permalink] 26 Mar 2009, 11:58
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Discussion in 'Physics & Math' started by arfa brane, Sep 18, 2016.
1. ### arfa branecall me arfValued Senior Member
Messages:
5,637
Suppose you'd like to map the rotation group of a square to a square matrix.
First of all a 2 x 2 matrix should suffice to describe the four vertices, i.e. something like:
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$
or if you want to represent a 4-cycle: (clockwise labeling):
$\begin{pmatrix} a & b \\ d & c \end{pmatrix}$
The permutation matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, acting on the left of any 2 x 2 matrix will swap the rows. Acting on the right it will swap the columns.
$\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ d & c \end{pmatrix} = \begin{pmatrix} d & c \\ a & b \end{pmatrix}$
$\begin{pmatrix} a & b \\ d & c \end{pmatrix} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix}b & a \\ c & d \end{pmatrix}$
So with the matrix transpose, a rotation or cyclic permutation of the elements of a 2 x 2 matrix can be defined:
$\begin{pmatrix} a & b \\ d & c \end{pmatrix}^{\intercal}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix}d & a \\ c & b \end{pmatrix} = \biggl {(} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ d & c \end{pmatrix} \biggr {)}^{\intercal}$
But you can also map the four vertices {a,b,c,d} to a column vector:
$\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}$
Then rotations are represented by 4 x 4 matrices, a subset of the 24 possible permutations of four symmetric objects. Can the above representation of rotating a square be extended to the whole of $S_4$, which is the symmetry group of a solid cube ? A cubic matrix will be needed, and a way to define rotations of the entire matrix (or equivalently, rotations of an abstract system of coordinates, given by a set of indices).
$\begin{pmatrix} e & f \\ h & g \end{pmatrix}$
With connections between the rows/columns of this matrix to the first one. Index-wise, this is just a pair of 2 x 2 matrices connected, or 'sliced' as $M_{1ij} + M_{2ij}$. Obviously the indices will allow three distinct slices of the 2 x 2 x 2 matrix.
Last edited: Sep 18, 2016
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5. ### arfa branecall me arfValued Senior Member
Messages:
5,637
Not sure what you mean by "encapsulated" so much, but to be a tensor, a multidimensional array has to have certain transformation properties. You need, I think, to start with vector spaces and these have to have co- and contravariant properties.
A bare 2 x 2 x 2 cubic array of "numbers" or just symbols like a, b, c . . . with the transformation I've defined so far, might not qualify as a tensor. I need to define a direction vector somewhere along with what happens to it when it gets transported around under the rotation I've defined.
7. ### rpennerFully WiredRegistered Senior Member
Messages:
4,833
What do you mean?
The cyclic permutation group of 4 vertices? The rotational group of a square with distinct vertices in a 2-D embedding space? The rotational group of a square with indistinguishable vertices in a 2-D embedding space? The rotational group of a square with distinct vertices in a 2-D embedding space? The rotational group of an oriented square with indistinguishable vertices in a 3-D embedding space? The rotational group of an unoriented square with indistinguishable vertices in a 3-D embedding space?
8. ### arfa branecall me arfValued Senior Member
Messages:
5,637
Hmm. The rotation group or subgroup of symmetries of a square, or any polygon in a Euclidean domain, is closed under composition, any composition of rotations is a rotation.
This isn't true for reflections in the plane because compositions of reflections are rotations, and transposing a 2 x 2 matrix is analogous to reflecting two vertices of a square about a diagonal line.
In a matrix, all the elements are distinguished with indices. The transpose operation with another 'reflection' through a horizontal or vertical line of symmetry gives a 'rotation'.
So if say,
$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$
Then if A is any 2 x 2 matrix, we have:
$(A^{\intercal}\sigma_x)^2 = \sigma_x A \sigma_x$
and $\sigma_x = \sigma_x^{-1}$
Last edited: Sep 21, 2016
9. ### arfa branecall me arfValued Senior Member
Messages:
5,637
$(A^{\intercal}\sigma_x)^2 = \sigma_x A \sigma_x$
and $\sigma_x = \sigma_x^{-1}$
Screwed up the notation. It should be:
$(A^{\intercal}\sigma_x)^{\intercal}\sigma_x = \sigma_x A \sigma_x$
Still thinking about how to "repeat" either expression so you get the identity permutation on both sides.
10. ### arfa branecall me arfValued Senior Member
Messages:
5,637
Well, there are the equalities since, from the above:
$(A^{\intercal} \sigma_x)^{\intercal} = \sigma_x A$, so
$(A^{\intercal} \sigma_x) = (\sigma_x A) ^{\intercal}$,
by a symmetry, we have:
$(\sigma_x A^{\intercal}) = (A \sigma_x )^{\intercal}$
But $\sigma_x A \sigma_x$ "twice", is
$\sigma_x (\sigma_x A \sigma_x ) \sigma_x = \sigma_x^2 A \sigma_x^2 = A$
11. ### arfa branecall me arfValued Senior Member
Messages:
5,637
Just a thought about what $\sigma_x A \sigma_x$ means.
Mathematically the rows are exchanged (transposed) and the columns are exchanged. It doesn't matter which is "done first".
However, it should be clear that
$\sigma_x (A \sigma_x) = ( \sigma_x A) \sigma_x$
Which although the parentheses are redundant, shows there is a rule of composition.
12. ### arfa branecall me arfValued Senior Member
Messages:
5,637
With the matrix transpose, we have a symmetric kind of operation:
$(\sigma_x A \sigma_x)^{\intercal} = \sigma_x A^{\intercal} \sigma_x$
But when $\sigma_x$ appears just once, either on the left or the right of A, the matrix transpose also transposes the position of $\sigma_x$:
$(\sigma_x A)^{\intercal} = A^{\intercal}\sigma_x,\; (A\sigma_x)^{\intercal} = \sigma_x A^{\intercal}$ | 1,723 | 5,937 | {"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} | 4.15625 | 4 | CC-MAIN-2018-30 | latest | en | 0.642045 |
https://www.mathworks.com/matlabcentral/cody/problems/3-find-the-sum-of-all-the-numbers-of-the-input-vector/solutions/965255 | 1,604,133,967,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107916776.80/warc/CC-MAIN-20201031062721-20201031092721-00572.warc.gz | 774,010,980 | 18,731 | Cody
# Problem 3. Find the sum of all the numbers of the input vector
Solution 965255
Submitted on 11 Sep 2016 by Aaron
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 = 1; assert(isequal(vecsum(x),y_correct))
2 Pass
x = [1 2 3 5]; y_correct = 11; assert(isequal(vecsum(x),y_correct))
3 Pass
x = [1 2 3 5]; y_correct = 11; assert(isequal(vecsum(x),y_correct))
4 Pass
x = 1:100; y_correct = 5050; assert(isequal(vecsum(x),y_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 213 | 709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.671182 |
https://zbmath.org/?q=an:07228770 | 1,611,250,549,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00154.warc.gz | 1,071,195,788 | 11,253 | # zbMATH — the first resource for mathematics
The boundary element method applied to the solution of the diffusion-wave problem. (English) Zbl 07228770
Summary: A Boundary Element Method formulation is developed for the solution of the two-dimensional diffusion-wave problem, which is governed by a partial differential equation presenting a time fractional derivative of order $$\alpha$$, with $$1.0<\alpha<2.0$$. In the proposed formulation, the fractional derivative is transferred to the Laplacian through the Riemann-Liouville integro-differential operator; then, the basic integral equation of the method is obtained through the Weighted Residual Method, with the fundamental solution of the Laplace equation as the weighting function. In the final expression, the presence of additional terms containing the history contribution of the boundary variables constitutes the main difference between the proposed formulation and the standard one. The proposed formulation, however, works well for $$1.5\leq\alpha<2.0$$, producing results with good agreement with the analytical solutions and with the Finite Difference ones.
##### MSC:
65 Numerical analysis 35 Partial differential equations
FODE
Full Text:
##### References:
[1] Miller, KS; Ross, B., An introduction to the fractional calculus and fractional differential equations (1993), Wiley-Interscience [2] Ortigueira, MD, Fractional calculus for scientist and engineers, 84 (2011), Springer, Lectures Notes in Electrical Engineering [3] Mainardi, F.; Paradisi, P., A model of diffusive waves in viscoelasticity, (Proceedings of the 36th conference on decision & control (1997), San Diego, California, USA), 4961-4966 [4] Mainardi, F.; Paradisi, P., Fractional diffusive waves, J Comput Acoust, 9, 1417-1436 (2001) · Zbl 1360.76272 [5] El-Saka H, Ahmed E. A fractional order network model for ZIKA. bioRxiv 2016 doi: 10.1101/039917. [6] Dalir, M.; Bashour, M., Applications of fractional calculus, Appl Math Sci, 4, 1021-1032 (2010) · Zbl 1195.26011 [7] Sun, H.; Zhang, Y.; Baleanu, D.; Chen, W.; Chen, Y., A new collection of real world applications of fractional calculus in science and engineering, Commun Nonlinear Sci Numer Simul, 64, 213-231 (2018) [8] Carrer, JAM; Seaid, M.; Trevelyan, J.; Solheid, BS, The boundary element method applied to the solution of the anomalous diffusion, Eng Anal Bound Elem, 109, 129-142 (2019) · Zbl 07127496 [9] Brebbia, CA; Telles, JCF; Wrobel, LC, Boundary element techniques: theory and application in engineering (1984), Springer Verlag [10] Murillo, JQ; Yuste, SB, An explicit difference method for solving fractional diffusion and diffusion-wave equations in the Caputo form, J Comput Nonlinear Dyn, 6, Article 021014 pp. (2011), 6 pages doi:101115/1.4002687 [11] Yuste, SB; Acedo, L., An explicit finite difference method and a new von Neumann-type stability analysis for fractional diffusion equations, SIAM J Numer Anal, 42, 1862-1874 (2005) · Zbl 1119.65379 [12] Murio, DA, Implicit finite difference approximation for time fractional diffusion equations, Comput Math Appl, 56, 1138-1145 (2008) · Zbl 1155.65372 [13] Meerschaert, MM; Tadjeran, C., Finite difference approximations for fractional advection-dispertion flow equations, J Comput Appl Math, 172, 65-77 (2004) · Zbl 1126.76346 [14] Li, W.; Li, C., Second order explicit difference schemes for the space fractional advection diffusion equation, Appl Math Comput, 257, 446-457 (2015) · Zbl 1339.65129 [15] Deng, WH, Finite element method for the space and time fractional Fokker-Planck equations, SIAM J Numer Anal, 47, 204-226 (2008) [16] Roop, JP, Computational aspects of FEM Approximation of fractional advection dispersion equations on bounded domains in ℜ^2, J Comput Appl Math, 193, 243-268 (2006) · Zbl 1092.65122 [17] Huang, Q.; Huang, G.; Zhan, H., A finite element solution for the fractional advection-dispersion equations, Adv Water Res, 31, 1578-1589 (2008) [18] Katsikadelis, JT, The BEM for numerical solution of partial fractional differential equations, Comput Math Appl, 62, 891-901 (2011) · Zbl 1228.74103 [19] Dehghan, M.; Safarpoor, M., The dual reciprocity boundary elements method for the linear and nonlinear two-dimensional time-fractional partial differential equations, Math Methods Appl Sci, 39, 3979-3995 (2016) · Zbl 1347.65182 [20] Carrer, JAM; Mansur, WJ; Vanzuit, RJ, Scalar Wave equation by the boundary element method: a D-BEM approach with non-homogeneous initial conditions, Comput Mech, 44, 31-44 (2009) · Zbl 1162.74483 [21] Houbolt, JC, A recurrence matrix solution for the dynamic response of elastic aircraft, J Aeronaut Sci, 17, 540-550 (1950) [22] Ray, SS, Exact solutions for time-fractional diffusion-wave equations by decomposition method, Phys Scr, 75, 53-61 (2007) · Zbl 1197.35147 [23] Mansur WJ. A Time-Stepping Technique to Solve Wave Propagation Problems Using the Boundary Element Method, Ph.D. Thesis, University of Southampton, 1983. [24] Carrer, JAM; Costa, VL, Boundary element method formulation for the solution of the Scalar Wave equation in one-dimensional problems, J Braz Soc Mech Sci Eng, 37, 959-971 (2015)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 1,475 | 5,515 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-04 | latest | en | 0.811564 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-linear-algebra-7th-edition/chapter-1-systems-of-linear-equations-1-3-applications-of-systems-of-linear-equations-1-3-exercises-page-32/1 | 1,575,878,919,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518337.65/warc/CC-MAIN-20191209065626-20191209093626-00547.warc.gz | 728,570,358 | 12,184 | # Chapter 1 - Systems of Linear Equations - 1.3 Applications of Systems of Linear Equations - 1.3 Exercises - Page 32: 1
$$p(x)=29-18 x+3 x^{2}.$$
#### Work Step by Step
Suppose that $$p(x)=a_{0}+a_{1} x+a_{2} x^{2}.$$ Now, we have $$\begin{array}{l}{p(2)=a_{0}+a_{1}(2)+a_{2}(2)^{2}=a_{0}+2a_{1}+4a_{2}=5} \\ {p(3)=a_{0}+a_{1}(3)+a_{2}(3)^{2}=a_{0}+3 a_{1}+9 a_{2}=2} \\ {p(4)=a_{0}+a_{1}(4)+a_{2}(4)^{2}=a_{0}+4 a_{1}+16 a_{2}=5} \end{array}$$ The above system gas the solution $$a_{0}=29, \quad a_{1}=-18, \quad a_{2}=3.$$ Hence, $$p(x)=29-18 x+3 x^{2}.$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 314 | 722 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-51 | latest | en | 0.650165 |
https://www.justcrackinterview.com/interviews/pimlico-plumbers/ | 1,726,399,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651622.79/warc/CC-MAIN-20240915084859-20240915114859-00476.warc.gz | 787,199,201 | 18,197 | # Pimlico plumbers Interview Questions, Process, and Tips
Ques:- What is the full form of ATM?
Ques:- A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s
Distance = Speed*Time
=>800+x=653∗60=>800+x=20∗65=1300=>x=1300−800=500
Ques:- What does “customer care service” mean to you?
Ques:- Who is your Favorite Musician
Ques:- What is your dream company?
Ques:- Verbal, logical reasoning, numerical ability.
Ques:- A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group?
600
Ques:- Exploring capability to manage multiple projects and provide recommendations
Ques:- A sales person multiplied by a number and get the answer 3. Instead of that number divided by 3. What is the answer she actually has to get.
1/3
Ques:- What have you learned from mistakes on the job?
Ques:- Out of 80 coins, one is counterfeit. What is the minimum number of weighings needed to find out the counterfeit coin?
79
Ques:- What might make you leave this job?
Ques:- 80% pass in english, 70%pass in maths , 10%fail in both , 144 pass in both . How many all appeared to the test?
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
Ques:- What goals have you achieved in your present career?And How did you plan to have achieved these goals?
Ques:- 24. 13 kigs and 6 libs can produce 510 tors in 10 hrs, 8 kigs and 14 libs can produce 484 tors in 12 hrs.Find the rate of production of tors for kigs and libs. Express the answer in tors/hr.
13 Kigs & 6 Libs –> 510tors in 10hrs
” –> 51 tors/hr
14 Libs –> 484 tors in 12hrs
” –> 121/3 tors/hr
1 Lib –> 121/42 tors/hr
==> 6 Libs –> 121/7 tors/hr
Now,
13 kigs and 6 Libs –> 51 tors/hr
Only 13 Kigs –> 51 – (121/7) tors/hr
= 236/91 tors/hr
Ques:- The distance between Delhi and Mathura is 110 kms. A starts from Delhi with a speed of 20 kmph at 7 a.m. for Mathura and B starts from Mathura with a speed of 25 kmph at 8 p.m. from Delhi. When will they meet?
Distance covered by B to meet A=Total distance – Distance covered by A hrs
[using : distance = speed x time]
Putting value of from equation (1),
hrs
Therefore, time at which both A and B will meet is = 7 a.m. + 3 hrs =10 am
Ques:- A train 280 m long is moving at 60 km/hr. Time taken by the train to cross a tunnel 220 m long is
A. 15 sec
B. 20 sec
C. 25 sec
D. 30 sec
30 Sec
Ques:- Five persons – P, Q, R, S and T are being compared in weight and height. The second heaviest person. S is the shortest. P is the 2nd tallest and shorter than T, The heaviest person is the third tallest person. There is only one person shorter than Q, who is lighter than T and P respectively.Who is the heaviest person?
R
T,P,R,Q,S
Ques:- What is black money?
Ques:- In a mixture of 60 liters, the ratio of milk and water is 2:1. What amount of water must be added to make the ratio of milk and water as 1:2?
60 liter
Ques:- Of the 50 researchers in a workgroup 40% will be assigned to Team A and the remaining 60% to Team B. However, 70% of the researchers prefer Team A and 30% prefer Team B. What is the least possible number of researchers who will not be assigned to the team they prefer?
A. 15
B. 20
C. 30
D. 35
15
Ques:- What is Team Player.
Ques:- The greatest number that will divided 187, 233 and 279 leaving the same remainder in each case is
A. 16
B. 26
C. 36
D. 46
The greatest number that will divide 187, 233 and 279 leaving the same remainder in each case.
To find : The number ?
Solution :
First we find the difference between these numbers.
The required numbers are
233-187=46
279-233=46
279-187=92
Now, We find the HCF of 46 and 92.
Therefore, The required largest number is 46.
Ques:- Have you ever tried to delay any decision-making? What were the consequences of this on both your company and customers?
Ques:- The first three terms of a proportion are 3, 9 and 12. The fourth term is?
24
Ques:- Find the C.I. on a sum of Rs.1600 for 9 months at 20% per annum, interest being compounded quarterly?
Ques:- If a person walks at 4/5th of his usual spee he reaches 40min late. If he walks athis usual speed how much time does he travels.
let x=speed
t=time taken when speed is x so…
xt=4/5x(t+40)
t=160 minutes
2 hr 40 minutes
Ques:- A man sells a horse for Rs.800 and loses something, if he had sold it for Rs.980, his gain would have been double the former loss. Find the cost price of the horse? | 1,465 | 4,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.898248 |
http://www.r-bloggers.com/an-easier-to-use-iv-regression-command-in-r/ | 1,469,653,798,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827079.61/warc/CC-MAIN-20160723071027-00036-ip-10-185-27-174.ec2.internal.warc.gz | 672,435,460 | 15,620 | # An easier to use IV regression command in R
March 10, 2011
By
(This article was first published on Coffee and Econometrics in the Morning, and kindly contributed to R-bloggers)
Update: I have added some functionality to my ivregress() command. Check out my newer post here.
After I posted my last video tutorial on how to use my IV regression function, I received a comment asking why I didn’t write the command a different way to make the syntax easier to read.
The answer is that I didn’t know how to write an easier to use function a year ago (when I wrote the ivreg() function). After some digging, I figured out how to work with “formula objects” in R and the result is an easier to use IV regression function (called ivregress()).
How to “install” ivregress()
Here’s the code you need to run to define ivregress() and its companion summary command sum.iv(). This will provide instrumental variables regression estimates if you have one endogenous regressor with one or more instruments.
You only need to run the above code once to define the function object ivregress() for all future uses (as long as you don’t write over it or clear your workspace).
How to use ivregress()
To create an “IV object” simply use the ivregress command as follows (the command relies on the car library):
library(car)
myivobject = ivregress(Y~X1+X2+…+Xk, Xj ~ Z1+Z2+…+Zl, dataframe)
where X1, X2,…Xk are second stage regressors, Xj is an endogenous regressor for which you would like to instrument, and Z1, Z2,…,Zl are instruments.
Then use sum.iv() from the previous posts to produce summary output. Sometime soon, I will post a video tutorial showing how to use ivregress() to perform IV regression.
To leave a comment for the author, please follow the link and comment on their blog: Coffee and Econometrics in the Morning.
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Click here to close (This popup will not appear again) | 612 | 2,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.853596 |
https://cracku.in/blog/data-sufficiency-questions-cat/ | 1,656,299,637,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00247.warc.gz | 247,488,197 | 36,773 | # Data Sufficiency Questions for CAT
0
2318
Data Sufficiency Questions for CAT:
Instructions:
Directions for the following four questions:
Each question is followed by two statements, A and B. Answer each question using the following instructions:
Question 1:
In a particular school, sixty students were athletes. Ten among them were also among the top academic performers. How many top academic performers were in the school?
A. Sixty per cent of the top academic performers were not athletes.
B. All the top academic performers were not necessarily athletes.
a) The question can be answered by using the statement A alone but not by using the statement B alone.
b) The question can be answered by using the statement B alone but not by using the statement A alone.
c) The question can be answered by using either of the statements alone
d) The question can be answered by using both the statements together but not by either of the statements alone.
e)The question cannot be answered on the basis of the two statements.
Question 2:
Five students Atul, Bala, Chetan, Dev and Ernesto were the only ones who participated in a quiz contest. They were ranked based on their scores in the contest. Dev got a higher rank as compared to Ernesto, while Bala got a higher rank as compared to Chetan. Chetan’s rank was lower than the median. Who among the five got the highest rank?
A. Atul was the last rank holder.
B. Bala was not among the top two rank holders.
a) The question can be answered by using the statement A alone but not by using the statement B alone.
b) The question can be answered by using the statement B alone but not by using the statement A alone.
c) The question can be answered by using either of the statements alone.
d) The question can be answered by using both the statements together but not by either of the statements alone.
e)The question cannot be answered on the basis of the two statements.
Question 3:
Thirty per cent of the employees of a call centre are males. Ten per cent of the female employees have an engineering background. What is the percentage of male employees with engineering background?
A. Twenty five per cent of the employees have engineering background.
B. Number of male employees having an engineering background is 20% more than the number of female employees having an engineering background
a) The question can be answered by using the statement A alone but not by using the statement B alone.
b) The question can be answered by using the statement B alone but not by using the statement A alone.
c) The question can be answered by using either of the statements alone.
d) The question can be answered by using both the statements together but not by either of the statements alone.
e)The question cannot be answered on the basis of the two statements.
Question 4:
In a football match, at the half-time, Mahindra and Mahindra Club was trailing by three goals. Did it win the match?
A. In the second-half Mahindra and Mahindra Club scored four goals.
B. The opponent scored four goals in the match.
a) The question can be answered by using the statement A alone but not by using the statement B alone.
b) The question can be answered by using the statement B alone but not by using the statement A alone.
c) The question can be answered by using either of the statements alone.
d) The question can be answered by using both the statements together but not by either of the statements alone.
e)The question cannot be answered on the basis of the two statements.
Logical Reasoning solved CAT questions
Data Interpretation solved CAT questions
Answers and Explanations for Data Sufficiency Questions for CAT:
Solutions:
No. of athletes which are top academic performers: 10
According to statement a) Sixty per cent of the top academic performers were not athletes so 40% of top academic performers are athletes which we know is 10. So total no. of top academic performers are 25.
Hence statement a is alone to answer the question.
while we cant deduce anything from statement b .
From the information given in the question, D has a better rank than E and B has a better rank than C. Also, C = 4 or 5.
From statement A, A = 5 => C = 4. But we do not know if D or B is number 1.
From statement B, B = 3 or 4. Again, we cannot determine the number 1 rank precisely.
By using both the statements together, we know C = 4, A = 5, B = 3 and D has a better rank than E.
So, D = 1 and E = 2. Therefore, the question can be answered by using both the statements together.
Hence, option D.
Let the total number of employees be 100k.
Number of males = 30k. Number of females = 70k.
Number of female engineers = 7k.
From statement A, number of male engineers = 25k – 7k = 18k.
From statement B, number of male engineers = (6/5)*7k.
Therefore, percentage of male engineers can be calculated.
So, the question can be answered by either statement alone.
From statement A, we know that the club scored four goals in the second half but we do not know the number of goals scored by the opposition. So, we cannot answer the question.
From statement B, we know that the opposition scored 4 goals in the match but we do not know the number of goals scored by Mahindra and Mahindra club. So, we cannot answer the question.
Using both the statements, we know that the opposition scored 4 goals in the match and Mahindra and Mahindra scored 4 goals in the second half. But, we do not know if the score was 3-0 or 4-1 at the end of first half. If it was 4-1, then Mahindra and Mahindra would win the match after the second half. But if the score was 3-0 at the end of first half, the match would end in a draw after the second half.
Therefore, the question cannot be answered even by using both the statements together.
Hope this practice test on Data Sufficiency Questions for CAT will be very much useful to you. Attend for free CAT Online preparation classes to crack CAT 2017/2018 | 1,353 | 5,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | longest | en | 0.977027 |
https://pastecode.io/s/2d7d82hw | 1,720,885,521,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00679.warc.gz | 369,015,600 | 59,060 | # Untitled
unknown
matlab
2 years ago
3.0 kB
2
Indexable
Never
```%% comp3_24
close all
clear all
N=500;
[P,u] = markov(N, 7/8, 7/8);
var_ee = 1;
var_ww = 4;
b=20;
rng (0);
N = 10000;
ee = var_ee*randn(N, 1 ) ;
ww = var_ww*rand(N, 1);
A0 = 1;
A1 = linspace( A0,A0, N);
y = zeros(N,1);
Re = [ 1e-6 0 ; 0 1e-6];
Rw = 0.1;
for k=2:N % Implement filter by hand to allow a1 to change.
y(k) = ee(k) + ww(k) + b*u(k) + A0*y(k-1);
end
% Plot realisation.
figure
plot(y)
title( 'Realisation of an hittepåsakmojäng-process' )
ylabel('Amplitude')
xlabel('Time')
xlim([1 N])
figure
subplot(2,1,1)
plot(y(end-100-2:end-2))
subplot(2,1,2)
plot(u(end-100-2:end-2))
%%
%y = tar2;
% Define the state space equations .
A = [1 0 : 0 1];
Re = [ 0.004 0 ; 0 0 ] ; % State covariance matrix
Rw = 1.25; % Observation variance
% Set some initialvalues
Rxx_1 = 10*eye ( 2 ) ; % Initialstate variance
xtt_1 = [ 0 0 ]'; % Initialstatevalues %xtt_1 --> x_{t|t-1}
%xt = zeros(2,N); % x_{t|t} initial state
% Vectors to store values in
Xsave = zeros( 2 ,N) ; % Stored states
ehat = zeros( 1 ,N) ; % Prediction residual
yt1 = zeros( 1 ,N) ; % One s t e p p r e d i c t i o n
yt2 = zeros( 1 ,N) ; % Two s t e p p r e d i c t i o n
% The f i l t e r use data up to t ime t-1 to predictvalue at t,
% then update using the prediction error. Why do we start
% from t=3? Why stop at N-2?
for t=3:N-2
Ct = [ -y(t-1) -y(t-2) ] ; %C_{t|t-1}
yhat(t) = Ct*xtt_1; %y_{t|t-1} = Ct*x_{t|t-1}
ehat(t) = y(t)-yhat(t); % e_t = y_t - y_{t|t-1}
xtt_1 = A.*Xsave(:,t-1) + b.*u(:,t-1); % x_{t|t-1} = A x_{t-1|t-1}
% Update
Ryy = Ct*Rxx_1*Ct' + Rw; % R^{yy}_{t|t-1} = Ct*R_{t|t-1}^{x,x} + Rw
Kt = Rxx_1*Ct'/Ryy; % K_t = R^{x,x}_{t|t-1} C^T inv( R_{t|t-1}^{y,y} )
xt_t = xtt_1 + Kt*(ehat(t)); % x_{t|t} = x_{t|t-1} + K_t ( y_t - Cx_{t|t-1} )
Rxx = Rxx_1 - Kt*Ryy*Kt'; % R{xx}_{t|t} = R^{x,x}_{t|t-1} - K_t R_{t|t-1}^{y,y} K_t^T
% Predict the nextstate
%xt_t1 = A*xt_t; % x_{t+1|t} HÄR KAN VARA EN B*U_t term om man har input???
Rxx_1 = A*Rxx*A' + Re; % R^{x,x}_{t+1|t} = A R^{x,x}_{t|t} A^T + Re
% Form 2 stepprediction . Ignore this part at first .
% Ct1 = [ -y(t) -y(t-1) ] ;
% yt1(t+1) = Ct1*xt_t;
% Ct2 = [ -yt1(t+1) -y(t) ] ;
% yt2 (t+2) = Ct2*A*xt_t
% % Store the statevector
Xsave(:, t) = xt_t;
end
%% Examine the estimated parameters.
Q0 = [thx(:,1) thx(:,2)]; % These are the true parameters we seek.
figure
plot( Xsave' )
hold on
plot(thx(:,1), 'Color','red','LineStyle',':')
hold on
plot(thx(:,2), 'Color','red','LineStyle',':')
title(sprintf('Estimated parameters, with Re = %7.6f and Rw = %4.3f', Re, Rw))
xlabel('Time')
ylim([-2 2])
%% Plot
figure
plot(Xsave(1,:))
hold on
plot(Xsave(2,:))
legend('a1', 'a2', 'Location','SW')
figure
plot(ehat)
title('residual')
sumsqr(ehat)``` | 1,268 | 3,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.356127 |
http://www.rasteredge.com/gallery/c36/233/ | 1,619,001,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039536858.83/warc/CC-MAIN-20210421100029-20210421130029-00381.warc.gz | 163,547,619 | 36,991 |
# c# pdf free : Add hyperlink pdf document control software platform web page winforms .net web browser TRENCH_REAL_ANALYSIS18-part235
Section3.5
173
(Exercise3.5.1).Ifa<x<b,theoscillationoff atxisdefinedby
w
f
.x/D lim
h!0C
W
f
.xh;xCh/:
ThecorrespondingdefinitionsforxDaandxDbare
w
f
.a/D lim
h!0C
W
f
.a;aCh/ and
w
f
.b/D lim
h!0C
W
f
.bh;b/:
Forafixedxin.a;b/,W
f
.xh;xCh/isanonnegativeandnondecreasingfunction
ofhfor0 < < h < min.xa;bx/; ; therefore,w
f
.x/existsandisnonnegative, by
Theorem2.1.9.Similarargumentsapplytow
f
.a/andw
f
.b/.
Theorem3.5.2
Letf bedefinedonŒa;b:Thenf iscontinuousatx
0
inŒa;bif
andonlyifw
f
.x
0
/ D D 0:.Continuityataorbmeanscontinuityfromtherightorleft,
respectively./
Proof
Supposethata<x
0
<b.First,supposethatw
f
.x
0
/D0and>0.Then
W
f
Œx
0
h;x
0
Ch<
forsomeh>0,so
jf.x/f.x
0
/j< if x
0
hx;x
0
x
0
Ch:
Lettingx
0
Dx
0
,weconcludethat
jf.x/f.x
0
/j< if jxx
0
j<h:
Therefore,f iscontinuousatx
0
.
Conversely,iff iscontinuousatx
0
and>0,thereisaı>0suchthat
jf.x/f.x
0
/j<
2
and jf.x
0
/f.x
0
/j<
2
ifx
0
ıx,xx
0
Cı.Fromthetriangleinequality,
jf.x/f.x
0
/jjf.x/f.x
0
/jCjf.x
0
/f.x
0
/j<;
so
W
f
Œx
0
h;x
0
Ch if h<ıI
therefore,w
f
.x
0
/D0.Similarargumentsapplyifx
0
Daorx
0
Db.
Lemma3.5.3
If w
f
.x/ < for a x b; ; thenthere is aı > 0suchthat
W
f
Œa
1
;b
1
;providedthatŒa
1
;b
1
Œa;bandb
1
a
1
<ı:
Proof
WeusetheHeine–Boreltheorem(Theorem1.3.7). Ifw
f
.x/ < ,thereisan
h
x
>0suchthat
jf.x
0
/f.x
00
/j<
(3.5.1)
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174 Chapter3
IntegralCalculusofFunctionsofOneVariable
if
x2h
x
<x
0
;x
00
<xC2h
x
and x
0
;x
00
2Œa;b:
(3.5.2)
IfI
x
D.xh
x
;xCh
x
/,thenthecollection
HD
˚
I
x
ˇ
ˇ
axb
isanopencoveringofŒa;b,sotheHeine–Boreltheoremimpliesthattherearefinitely
manypointsx
1
,x
2
,...,x
n
inŒa;bsuchthatI
x
1
,I
x
2
,...,I
x
n
coverŒa;b.Let
hD min
1in
h
x
i
andsupposethatŒa
1
;b
1
Œa;bandb
1
a
1
< h. Ifx
0
andx
00
areinŒa
1
;b
1
,then
x
0
2I
x
r
forsomer.1rn/,so
jx
0
x
r
j<h
x
r
:
Therefore,
jx
00
x
r
jjx
00
x
0
jCjx
0
x
r
j<b
1
a
1
Ch
x
r
< hCh
x
r
2h
x
r
:
Thus,anytwopointsxandx00 inŒa
1
;b
1
satisfy(3.5.2)withxDx
r
,sotheyalsosatisfy
(3.5.1).Therefore,isanupperboundfortheset
˚
jf.x
0
/f.x
00
/j
ˇ
ˇ
x
0
;x
00
2Œa
1
;b
1
;
whichhasthesupremumW
f
Œa
1
;b
1
.Hence,W
f
Œa
1
;b
1
.
Inthefollowing,L.I/isthelengthoftheintervalI.
Lemma3.5.4
Letf beboundedonŒa;banddefine
E
D
˚
x2Œa;b
ˇ
ˇ
w
f
.x/
:
ThenE
isclosed;andf isintegrableonŒa;bifandonlyifforeverypairofpositive
numbersandı;E
canbecoveredbyfinitelymanyopenintervalsI
1
;I
2
;...;I
p
such
that
Xp
jD1
L.I
j
/<ı:
(3.5.3)
Proof
WefirstshowthatE
isclosed.Supposethatx
0
isalimitpointofE
.Ifh>0,
thereisan
xfromE
in.x
0
h;x
0
Ch/. SinceŒ
xh
1
;
xCh
1
Œx
0
h;x
0
Chfor
sufficientlysmallh
1
andW
f
Œ
xh
1
;
xCh
1
,itfollowsthatW
f
Œx
0
h;x
0
Ch
forallh>0.Thisimpliesthatx
0
2E
,soE
isclosed(Corollary1.3.6).
Nowwewillshowthatthestatedconditioninnecessaryforintegrability.Supposethat
theconditionisnotsatisfied;thatis,thereisa>0andaı>0suchthat
p
X
jD1
L.I
j
/ı
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Section3.5
175
foreveryfinitesetfI
1
;I
2
;:::;I
p
gofopenintervalscoveringE
.IfP Dfx
0
;x
1
;:::;x
n
g
isapartitionofŒa;b,then
S.P/s.P/D
X
j2A
.M
j
m
j
/.x
j
x
j1
/C
X
j2B
.M
j
m
j
/.x
j
x
j1
/; (3.5.4)
where
˚
j
ˇ
ˇ
Œx
j1
;x
j
\E
¤;
and BD
˚
j
ˇ
ˇ
Œx
j1
;x
j
\E
D;
:
Since
S
j2A
.x
j1
;x
j
/containsallpointsofE
exceptanyofx
0
,x
1
,...,x
n
thatmay
beinE
,andeachofthesefinitelymanypossibleexceptionscanbecoveredbyanopen
impliesthat
X
j2A
.x
j
x
j1
/ı:
Moreover,ifj 2A,then
M
j
m
j
;
so(3.5.4)impliesthat
S.P/s.P/
X
j2A
.x
j
x
j1
/ı:
SincethisholdsforeverypartitionofŒa;b,fisnotintegrableonŒa;b,byTheorem3.2.7.
Thisprovesthatthestatedconditionisnecessaryforintegrability.
Forsufficiency,letandıbepositivenumbersandletI
1
,I
2
,...,I
p
beopenintervals
thatcoverE
andsatisfy(3.5.3).Let
e
I
j
DŒa;b\
I
j
:
(
I
j
DclosureofI.)Aftercombininganyof
e
I
1
,
e
I
2
,...,
e
I
p
thatoverlap,weobtainaset
ofpairwisedisjointclosedsubintervals
C
j
DŒ˛
j
j
; 1j j q.p/;
ofŒa;bsuchthat
a˛
1
1
2
2
<˛
q1
q1
q
q
b;
(3.5.5)
q
X
iD1
i
˛
i
/<ı
(3.5.6)
and
w
f
.x/<; ˇ
j
x˛
jC1
; 1j j q1:
Also,w
f
.x/<forax˛
1
ifa<˛
1
andforˇ
q
xbifˇ
q
<b.
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176 Chapter3
IntegralCalculusofFunctionsofOneVariable
LetP
0
bethepartitionofŒa;bwiththepartitionpointsindicatedin(3.5.5),andrefine
P
0
bypartitioningeachsubintervalŒˇ
j
jC1
(aswellasŒa;˛
1
ifa < < ˛
1
andŒˇ
q
;b
ifˇ
q
<b)intosubintervalsonwhichtheoscillationoff isnotgreaterthan. . Thisis
possiblebyLemma3.5.3. Inthisway, , afterrenamingtheentirecollectionofpartition
points,weobtainapartitionPDfx
0
;x
1
;:::;x
n
gofŒa;bforwhichS.P/s.P/canbe
writtenasin(3.5.4),with
X
j2A
.x
j
x
j1
/D
q
X
iD1
i
˛
i
/<ı
(see(3.5.6))and
M
j
m
j
; j j 2B:
Forthispartition,
X
j2A
.M
j
m
j
/.x
j
x
j1
/2K
X
j2A
.x
j
x
j1
/<2Kı;
whereKisanupperboundforjfjonŒa;band
X
j2B
.M
j
m
j
/.x
j
x
j1
/.ba/:
Wehavenowshownthatifandıarearbitrarypositivenumbers,thereisapartitionP of
Œa;bsuchthat
S.P/s.P/<2KıC.ba/:
(3.5.7)
If>0,let
ıD
4K
and D
2.ba/
:
Then(3.5.7)yields
S.P/s.P/<;
andTheorem3.2.7impliesthatf isintegrableonŒa;b.
WeneedthenextdefinitiontostateLebesgue’sintegrabilitycondition.
Definition3.5.5
AsubsetS ofthereallineisofLebesguemeasurezeroifforevery
>0thereisafiniteorinfinitesequenceofopenintervalsI
1
,I
2
,...suchthat
S
[
j
I
j
(3.5.8)
and
Xn
jD1
L.I
j
/<; n1:
(3.5.9)
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Section3.5
177
NotethatanysubsetofasetofLebesguemeasurezeroisalsoofLebesguemeasurezero.
(Why?)
Example3.5.1
TheemptysetisofLebesguemeasurezero,sinceitiscontainedin
anyopeninterval.
Example3.5.2
AnyfinitesetS D D fx
1
;x
2
;:::;x
n
g isofLebesguemeasure zero,
sincewecanchooseopenintervalsI
1
,I
2
,...,I
n
suchthatx
j
2 I
j
andL.I
j
/< =n,
1j n.
Example3.5.3
Aninfinitesetisdenumerableifitsmemberscanbelistedinase-
quence(thatis,inaone-to-onecorrespondencewiththepositiveintegers);thus,
SDfx
1
;x
2
;:::;x
n
;:::g:
(3.5.10)
Aninfinitesetthatdoesnothavethispropertyisnondenumerable. Anydenumerableset
(3.5.10)isofLebesguemeasurezero,sinceif>0,itispossibletochooseopenintervals
I
1
,I
2
,...,sothatx
j
2I
j
andL.I
j
/<2j,j 1.Then(3.5.9)holdsbecause
1
2
C
1
22
C
1
23
CC
1
2n
D1
1
2n
<1:
(3.5.11)
TherearealsonondenumerablesetsofLebesguemeasurezero,butitisbeyondthescope
ofthisbooktodiscussexamples.
Thenexttheoremisthemainresultofthissection.
Theorem3.5.6
Aboundedfunctionf isintegrableonafiniteintervalŒa;bifand
onlyifthesetSofdiscontinuitiesoff inŒa;bisofLebesguemeasurezero:
Proof
FromTheorem3.5.2,
S D
˚
x2Œa;b
ˇ
ˇ
w
f
.x/>0
:
Sincew
f
.x/>0ifandonlyifw
f
.x/1=iforsomepositiveintegeri,wecanwrite
SD
[1
iD1
S
i
;
(3.5.12)
where
S
i
D
˚
x2Œa;b
ˇ
ˇ
w
f
.x/1=i
:
Nowsupposethatf isintegrableonŒa;band>0.FromLemma3.5.4,eachS
i
can
becoveredbyafinitenumberofopenintervalsI
i1
,I
i2
,...,I
in
oftotallengthlessthan
=2i.Wesimplyrenumbertheseintervalsconsecutively;thus,
I
1
;I
2
;DI
11
;:::;I
1n
1
;I
21
;:::;I
2n
2
;:::;I
i1
;:::;I
in
i
;::::
Now(3.5.8)and(3.5.9)holdbecauseof(3.5.11)and(3.5.12),andwehaveshownthatthe
statedconditionisnecessaryforintegrability.
VB.NET Create PDF from Word Library to convert docx, doc to PDF in
Change Word hyperlink to PDF hyperlink and bookmark. Add necessary references doc As DOCXDocument = New DOCXDocument(inputFilePath) ' Convert it to PDF document.
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178 Chapter3
IntegralCalculusofFunctionsofOneVariable
Forsufficiency, supposethatthestatedconditionholdsand > 0. ThenS can n be
coveredbyopenintervalsI
1
;I
2
;:::thatsatisfy(3.5.9).If>0,thentheset
E
D
˚
x2Œa;b
ˇ
ˇ
w
f
.x/
ofLemma3.5.4iscontainedinS(Theorem3.5.2),andthereforeE
iscoveredbyI
1
;I
2
;:::.
SinceE
isclosed(Lemma3.5.4)andbounded,theHeine–BoreltheoremimpliesthatE
iscoveredbyafinitenumberofintervalsfromI
1
;I
2
;:::. Thesumofthelengthsofthe
latterislessthan,soLemma3.5.4impliesthatf isintegrableonŒa;b.
3.5Exercises
1.
InconnectionwithDefinition3.5.1,showthat
sup
x;x
0
2Œa;b
jf.x/f.x
0
/jD sup
axb
f.x/ inf
axb
f.x/:
2.
UseTheorem3.5.6toshowthatiff isintegrableonŒa;b,thensoisjfjand,if
f.x/>0.axb/,sois1=f.
3.
Prove: TheunionoftwosetsofLebesguemeasurezeroisofLebesguemeasure
zero.
4.
UseTheorem3.5.6andExercise3.5.3toshowthatiff andgareintegrableon
Œa;b,thensoarefCgandfg.
5.
Supposef isintegrableonŒa;b,˛D D inf
axb
f.x/,andˇ D sup
axb
f.x/.
LetgbecontinuousonŒ˛;ˇ.ShowthatthecompositionhDgıf isintegrable
onŒa;b.
6.
Letf beintegrableonŒa;b,let˛Dinf
axb
f.x/andˇDsup
axb
f.x/,and
supposethatGiscontinuousonŒ˛;ˇ.Foreachn1,let
aC
.j 1/.ba/
n
u
jn
;v
jn
aC
j.ba/
n
; 1j j n:
Showthat
lim
n!1
1
n
Xn
jD1
jG.f.u
jn
//G.f.v
jn
//jD0:
7.
Leth.x/ D 0forallx inŒa;bexceptforx inasetofLebesguemeasurezero.
Showthatif
R
b
a
h.x/dxexists,itequalszero.H
INT
:Anysubsetofasetofmeasure
zeroisalsoofmeasurezero:
8.
Supposethatf andgareintegrableonŒa;bandf.x/Dg.x/exceptforxinaset
ofLebesguemeasurezero.Showthat
Z
b
a
f.x/dxD
Z
b
a
g.x/dx:
CHAPTER4
InfiniteSequences andSeries
INTHISCHAPTERweconsiderinfinitesequencesandseriesofconstantsandfunctions
ofarealvariable.
SECTION4.1introducesinfinitesequencesofrealnumbers. Theconceptofalimitofa
sequenceisdefined,asistheconceptofdivergenceofasequenceto˙1. Wediscuss
boundedsequencesandmonotonicsequences. Thelimitinferiorandlimitsuperiorofa
sequencearedefined. WeprovetheCauchyconvergencecriterionforsequencesofreal
numbers.
SECTION4.2definesasubsequenceofaninfinitesequence. Weshowthatifasequence
convergestoalimitordivergesto˙1,thensodoallsubsequencesofthesequence.Limit
pointsandboundednessofasetofrealnumbersarediscussedintermsofsequencesof
membersoftheset.Continuityandboundednessofafunctionarediscussedintermsofthe
valuesofthefunctionatsequencesofpointsinitsdomain.
SECTION4.3introducesconceptsofconvergenceanddivergenceto˙1forinfiniteseries
ofconstants. WeproveCauchy’sconvergencecriterionforaseriesofconstants. . Incon-
nectionwithseriesofpositiveterms,weconsiderthecomparisontest,theintegraltest,the
ratiotest,andRaabe’stest. Forgeneralseries,weconsiderabsoluteandconditionalcon-
vergence,Dirichlet’stest,rearrangementofterms,andmultiplicationofoneinfiniteseries
byanother.
SECTION4.4dealswithpointwiseanduniformconvergenceofsequencesandseriesof
functions. Cauchy’suniformconvergencecriteriaforsequencesandseriesareproved,as
isDirichlet’stestforuniformconvergenceofaseries. Wegivesufficientconditionsfor
thelimitofasequenceoffunctionsorthesum ofaninfiniteseriesoffunctionstobe
continuous,integrable,ordifferentiable.
SECTION4.5considerspowerseries. Itisshownthatapowerseriesthatconvergeson
anopenintervaldefinesaninfinitelydifferentiablefunctiononthatinterval. Wedefine
theTaylorseriesofaninfinitelydifferentiablefunction,andgivesufficientconditionsfor
theTaylorseriestoconvergetothefunctiononsomeinterval.Arithmeticoperationswith
powerseriesarediscussed.
178
Section4.1
SequencesofRealNumbers
179
4.1SEQUENCESOFREALNUMBERS
Aninfinitesequence(morebriefly,asequence)ofrealnumbersisareal-valuedfunction
definedonasetofintegers
˚
n
ˇ
ˇ
nk
.Wecallthevaluesofthefunctionthetermsofthe
sequence.Wedenoteasequencebylistingitstermsinorder;thus,
fs
n
g
1
k
Dfs
k
;s
kC1
;:::g:
(4.1.1)
Forexample,
1
n2C1
1
0
D
1;
1
2
;
1
5
;:::;
1
n2C1
;:::
;
f.1/
n
g
1
0
Df1;1;1;:::;.1/
n
;:::g;
and
1
n2
1
3
D
1;
1
2
;
1
3
;:::;
1
n2
;:::
:
Therealnumbers
n
isthenthtermofthesequence. Usuallyweareinterestedonlyinthe
termsofasequenceandtheorderinwhichtheyappear,butnotintheparticularvalueofk
in(4.1.1).Therefore,weregardthesequences
1
n2
1
3
and
1
n
1
1
asidentical.
Wewillusuallywritefs
n
gratherthanfs
n
g
1
k
. Intheabsenceofanyindicationtothe
contrary,wetakekD0unlesss
n
isgivenbyarulethatisinvalidforsomenonnegative
integer,inwhichcasekisunderstoodtobethesmallestpositiveintegersuchthats
n
is
definedforallnk.Forexample,if
s
n
D
1
.n1/.n5/
;
thenkD6.
n
gconcernthebehaviorofs
n
forlargen.
LimitofaSequence
Definition4.1.1
Asequencefs
n
gconvergestoalimitsifforevery>0thereisan
integerNsuchthat
js
n
sj< if nN:
(4.1.2)
Inthiscasewesaythatfs
n
gisconvergentandwrite
lim
n!1
s
n
Ds:
Asequencethatdoesnotconvergediverges,orisdivergent | 6,890 | 13,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2021-17 | longest | en | 0.351338 |
https://www.scoop.it/topic/emdrive/p/4025764205/2014/08/05/no-nasa-has-not-verified-an-impossible-space-drive | 1,556,129,157,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578655155.88/warc/CC-MAIN-20190424174425-20190424200425-00497.warc.gz | 827,221,915 | 33,406 | EmDrive (Propelantless microwave resonant reactor by Roger Shawyer)
46.4K views | +0 today
Scooped by Alain Coetmeur onto EmDrive (Propelantless microwave resonant reactor by Roger Shawyer)
No, NASA has not verified an impossible space drive!
We are being told that NASA has successfully tested a device which can push along a space vehicle without consuming any propellant. If true, this would be an astonishing discovery, possibly allowing easy access to the worlds of the Solar System. But are these reports true?
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EmDrive (Propelantless microwave resonant reactor by Roger Shawyer)
Just a tech-watch on a possible revolution with EmDrive (Propelantless microwave resonant reactor by Roger Shawyer). Not sure it works. Not sure it cannot work. Just watching. by AlainCo of http://www.lenr-forum.com , @alain_co https://twitter.com/alain_co
Curated by Alain Coetmeur
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EmDrive - FAQ
Alain Coetmeur's insight:
FAQ
Theory
1.
Q. Is the thrust produced by the EmDrive a reactionless force?
A. No, the thrust is the result of the reaction between the end plates of the waveguide and the Electromagnetic wave propagated within it.
2.
Q. How can a net force be produced by a closed waveguide?
A. At the propagation velocities (greater than one tenth the speed of light) the effects of special relativity must be considered. Different reference planes have to be used for the EM wave and the waveguide itself. The thruster is therefore an open system and a net force can be produced.
3.
Q. Why does the net force not get balanced out by the axial component of the sidewall force?
A. The net force is not balanced out by the axial component of the sidewall force because there is a highly non linear relationship between waveguide diameter and group velocity. (e.g. at cut off diameter, the group velocity is zero, the guide wavelength is infinity, but the diameter is clearly not zero.) The design of the cavity is such that the ratio of end wall forces is maximised, whilst the axial component of the sidewall force is reduced to a negligible value.
4.
Q. Does the theory of the EmDrive contravene the accepted laws of physics or electromagnetic theory?
A. The EmDrive does not violate any known law of physics. The basic laws that are applied in the theory of the EmDrive operation are as follows:
Newton’s laws are applied in the derivation of the basic static thrust equation (Equation 11 in the theory paper) and have also been demonstrated to apply to the EmDrive experimentally.
The law of conservation of momentum is the basis of Newtons laws and therefore applies to the EmDrive. It is satisfied both theoretically and experimentally.
The law of conservation of energy is the basis of the dynamic thrust equation which applies to the EmDrive under acceleration,(see Equation 16 in the theory paper).
The principles of electromagnetic theory are used to derive the basic design equations.
5.
Q. Why does the EmDrive not contravene the conservation of momentum when it operates in free space?
A. The EmDrive cannot violate the conservation of momentum. The electromagnetic wave momentum is built up in the resonating cavity, and is transferred to the end walls upon reflection. The momentum gained by the EmDrive plus the momentum lost by the electromagnetic wave equals zero. The direction and acceleration that is measured, when the EmDrive is tested on a dynamic test rig, comply with Newtons laws and confirm that the law of conservation of momentum is satisfied.
6.
Q. Is the EmDrive a form of perpetual motion machine?
A. The EmDrive obeys the law of conservation of energy and is therefore not a perpetual motion machine. Energy must be expended to accelerate the EmDrive (see Equation 16 of the theory paper). Once the EmDrive is switched off, Newton’s laws ensure that motion is constant unless it is acted upon by another force.
7.
Q. Why does the thrust decrease as the spacecraft velocity along the thrust vector increases?
A. As the spacecraft accelerates along the thrust vector, energy is lost by the engine and gained as additional kinetic energy by the spacecraft. This energy can be defined as the thrust multiplied by the distance through which the thrust acts. For a given acceleration period, the higher the mean velocity, the longer the distance travelled, hence the higher the energy lost by the engine.
This loss of stored energy from the resonant cavity leads to a reduction in Q and hence a reduction of thrust.
Test procedures
8.
Q. Has buoyancy been allowed for?
A. Buoyancy has been allowed for in the initial experiments and then eliminated by hermetically sealing the thruster.
9.
Q. Are there any convection currents which might affect the results?
A. Convection currents did not affect the results, as measurements were taken with the thrust vector up, down and horizontal. Test runs were also carried out using a thermal simulation heater to quantify the effects of change of coolant temperature.
10.
Q. Has stiffness in cables and pipes been allowed for?
A. The only connections to the balance were high flex electrical links
11.
Q. Has friction in any pivots been allowed for?
A. Static thrust measurements were carried out using 3 different techniques – a counterbalance rig with a knife edge pivot, a direct weighing method using a 16kg balance (0.1 gm resolution), and with the thruster suspended from a spring balance with the weight partly offloaded on to an electronic balance.
12.
Q. Have electromagnetic effects been taken into account? These include interactions between current-carrying conductors and between such conductors carrying RF currents and nearby metallic structures in which currents might be induced.
A. Stray electromagnetic effects were eliminated by using different test rigs, by testing two thrusters with very different mounting structures, and by changing the orientation by 90 degrees to eliminate the Earth’s magnetic field.
13.
Q. Is there any ionization within the air, which might cause electrostatic charging and resulting forces?
A. Electrostatic charges were eliminated by the comprehensive earthing required for safety reasons, and to provide the return path for the magnetron anode current.
14.
Q. Could RF pick-up measurement circuits have produced erroneous results?
A. EMC tests were carried out on the instrumentation to eliminate the effects of RF pick up.
15.
Q. Could acceleration be caused by spurious torques generated by the air bearing?
A. Dynamic tests are preceded by an acceleration calibration test, using standard weights to determine the air bearing friction.
16.
Q. Could acceleration be caused by anomalous thermal or electromagnetic effects?
A. Acceleration and deceleration tests have been carried out in both clockwise and anti-clockwise directions Acceleration from rest only starts when the magnetron output frequency matches the resonant frequency of the engine, following an initial warm-up period.
Applications
17.
Q. Can the technology be qualified for space applications?
A. Yes, all the basic microwave, power supply, thermal and control technologies are similar to flight equipment currently used on high power communication satellites.
18.
Q. How can the EmDrive produce enough thrust for terrestrial applications?
A. The second generation engines will be capable of producing a specific thrust of 30kN/kW. Thus for 1 kilowatt (typical of the power in a microwave oven) a static thrust of 3 tonnes can be obtained, which is enough to support a large car. This is clearly adequate for terrestrial transport applications.
The static thrust/power ratio is calculated assuming a superconducting EmDrive with a Q of 5 x 109. This Q value is routinely achieved in superconducting cavities.
Note however, because the EmDrive obeys the law of conservation of energy, this thrust/power ratio rapidly decreases if the EmDrive is used to accelerate the vehicle along the thrust vector. (See Equation 16 of the theory paper). Whilst the EmDrive can provide lift to counter gravity, (and is therefore not losing kinetic energy), auxiliary propulsion is required to provide the kinetic energy to accelerate the vehicle.
Alain Coetmeur's comment, August 24, 2013 7:38 AM
It seems hard to believe, but if they did as they say, it seems experimentally hard to criticize.
For theoretical reasons, they seems to have well studied (and have been criticized), and incredulity is probably based on misunderstanding of relativistic effects.
If EmDrive is erroneous, it is probably both a strange experimental artifact that survives many checking, and tricky detail in the computations, that were not noticed by critics.
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Whisper From the First Stars Sets Off Loud Dark Matter Debate | Quanta Magazine
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Physics from the edge: A paper on QI & cold fusion
I've just published a paper on cold fusion, in Progress in Physics which is a nice open access journal that has the laudable goal o
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EmDrive Presentation by Roger Shawyer Part 2 of 3 opt - YouTube
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Hawking effect and Unruh effect from the uncertainty principle - IOPscience
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Alain Coetmeur's insight:
in spanish
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Leaked NASA papers prove ‘Impossible Fuel free Engine’, aka EmDrive WORKS
According to the leaked NASA documents, the 'impossible' fuel-free engine that could take mankind to Mars in just 10 weeks WORKS. According to researchers, the revolutionary engine creates thrust by 'bouncing' microwaves in a closed chamber and is powered by solar energy. The possibility of a space engine that does not require fuel –it actually transforms electricity into impulse by moving microwaves inside an enclosed chamber - seems to be closer than ever thanks to leaked NASA documents. Not long ago, several independent laboratories concluded that somehow, the ‘impossible’ fuel-free engine called EmDrive that could transport astronauts to Mars in just 10 weeks works, even though no one known how or why it works. The revolutionary engine creates thrust by bouncing microwaves in a closed chamber and is powered by solar energy. The engine that has been dubbed as impossible since it defies everything we thought we knew about physics has created quite a buzz in the last couple of months.
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(14) SSI APW 2017: 04. Dr. Martin Tajmar - YouTube
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Expanding cosmos hints at new physics
A discrepancy in measurements of the Universe's expansion rate has now become "pretty serious".
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How quantised inertia gets rid of dark matter -- Sott.net
It is well-known that galaxies rotate so fast that centrifugal forces should have long ago beaten the gravitational self-attraction of all the visible mass we see, and torn them apart, but galaxies sit there apparently not being torn apart
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How Quantised Inertia Gets Rid Of Dark Matter
It is well-known that galaxies rotate so fast that centrifugal forces should have long ago beaten the gravitational self-attraction of all the visible mass
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World EM-Drive Team Crowd Fund
To raise awareness and funding in creating a citizen based space platform for launching and controlling drones for deep outer space exploration.
Alain Coetmeur's insight:
Make your due diligence first, as usual!
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Emdrive, the motor impossible. Original article by Alessandro Brizzi. - Cognitio
Emdrive, the motor impossible. Original article by Alessandro Brizzi. NASA has released the results of a test that would confirm th
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How quantised inertia gets rid of dark matter | Mike McCulloch | TEDxPlymouthUniversity - YouTube
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Trajectoires des sondes spatiales : de nouvelle anomalies - Science & Vie
Après les sondes Pioneer qui ralentissaient, c'est au tour de Juno, Galileo, Near, Cassini et Rosetta d'accélérer de manière étrange... les scientifiques ont-ils trouvé l'explication ?
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Emdrive Presentation By Roger Shawyer Part 3 of 3 opt - YouTube
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EmDrive Presentation by Roger Shawyer Part 1 of 3. The future of spacetravel. - YouTube
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(14) EmDrive and a new interpretation of photon energy (1) - YouTube
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EmDrive : le propulseur de l’espace qui défie les lois de la physique fonctionne dans le vide
Il faut croire que votre Guru perd la tête, mais il était sûr d’avoir évoqué dans un précèdent article le propulseur qui défient les lois de la physique, q
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The SpaceDrive Project – Developing Revolutionary Propulsion at TU Dresden
Propellantless propulsion is believed to be the best option for interstellar travel. However, photon rockets or solar sails have thrusts so low that maybe only nano-scaled spacecraft may reach the next star within our lifetime using very high-power laser beams. Since 2012, a dedicated breakthrough propulsion physics group was founded at the Institute of Aerospace Engineering at TU Dresden to investigate different concepts based on non-classical/revolutionary propulsion ideas that claim to be at least an order of magnitude more efficient in producing thrust compared to photon rockets. Most of these schemes rely on modifying the inertial mass, which in turn could lead to a new propellantless propulsion method. Our intention is to develop an excellent research infrastructure to test new ideas and measure thrusts and/or artefacts with high confidence to determine if a concept works and if it does how to scale it up. At present, we are focusing on two possible revolutionary concepts: The EMDrive and the Mach-Effect Thruster. The first concept uses microwaves in a truncated cone-shaped cavity that is claimed to produce thrust. Although it is not clear on which theoretical basis this can work, several experimental tests have been reported in the literature, which warrants a closer examination. We are building several models of different sizes to understand scaling laws and the interaction with the test environment. The second concept is theoretically much better understood and is believed to generate mass fluctuations in a piezo-crystal stack that creates non-zero time-averaged thrusts. Apart from theoretical models, we are testing and building several such thrusters in novel setups to further investigate their thrust capability. In addition, we are performing side-experiments to investigate other experimental areas that may be promising for revolutionary propulsion. To improve our testing capabilities, several cutting-edge thrust balances are under development to compare thrust measurements in different measurement setups to gain confidence and to identify experimental artefacts
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How QI gets rid of dark matter
Many people have asked me for a simple, graphical explanation of how quantised inertia (QI) gets rid of the awful dark matter, so here i
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Impossible Physics: Meet NASA’s Design for a Warp Drive Ship
efore we jump into this, you should know that a number of scientists are currently researching the feasibility of warp drive (and EMdrive and a number of other modes of faster than light travel); however,
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Assessing the EM-Drive claims
Follow project: Assessing the EM-Drive claims by Jose Rodal on ResearchGate, the professional network for scientists.
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Crowdfunding: World Em-Drive Team - EmDrive Forum
I just found this website proposing a crowdfunding of EmDrive research
emdriveteam.com/
It looks enthusiastic and human sized, with a californian address, and a yet to come address in Las Vegas...
As Usual before puting money, even love money, it is…
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"Dark Matter is a Mythical Beast" --A New Theory of Gravity and the Dark Side of the Cosmos (A 'Galaxy' 2017 Most Viewed)
The classical theory of gravity is in dire need of new approaches, since it doesn't combine well with quantum physics. Both theories, crown jewels of 20th century physics, cannot be true at the same time. A Dutch theoretical physicis
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Microwave Energy Injection into a Conical Frustum - NSF1701 Ohase 1 Test Report - 2015 Dave Distler
Microwave Energy Injection into a Conical Frustum: The NSF-1701 Phase I Test Report by Dave Distler rfmwguy@gmail.com October 4th, 2015 2:01 PM EST Chagrin Fal…
No comment yet. | 3,734 | 16,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-18 | latest | en | 0.906109 |
https://socratic.org/questions/an-object-is-thrown-vertically-from-a-height-of-7-m-at-3-m-s-how-long-will-it-ta | 1,576,052,506,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540530452.95/warc/CC-MAIN-20191211074417-20191211102417-00552.warc.gz | 542,322,060 | 6,843 | # An object is thrown vertically from a height of 7 m at 3 m/s. How long will it take for the object to hit the ground?
Jun 20, 2017
$t = 1.54$ $\text{s}$
#### Explanation:
We're asked to find the time $t$ when the object hits the ground, given that it was thrown vertically upward with a speed of $3 \text{m"/"s}$, at a height of $7$ $\text{m}$.
As soon as the object is let go, it is in a state of free-fall, and the kinematics equations can predict its motion, such as its maximum height reached, time of flight, etc.
To find this time, we can use the equation
$y = {y}_{0} + {v}_{0 y} t + \frac{1}{2} {a}_{y} {t}^{2}$
For this equation,
• The height $y$ is ground level, which I'll call $0$ $\text{m}$
• The initial ${y}_{0}$ is $7$ $\text{m}$
• The initial $y$-velocity ${v}_{0 y}$ is $3 \text{m"/"s}$
• The time $t$ is when it is at height $y = 0$, which is what we're trying to find
• The $y$-acceleration ${a}_{y}$ is equal to $- g$, which is -9.8"m"/("s"^2)
Plugging in our known values, we have
$0$ $\text{m} = 7$ $\text{m}$ + (3"m"/"s")t + 1/2(-9.8"m"/("s"^2))t^2
$\left(- 4.9 \text{m"/("s"^2))t^2 + (3"m"/"s}\right) t + 7$ $\text{m}$ $= 0$
Using the quadratic formula:
t = (-3+-sqrt((3)^2 - 4(-4.9)(7)))/(2(-4.9)) = color(red)(1.54 color(red)("s" (positive solution)
Thus, the object will strike the ground after color(red)(1.54 sfcolor(red)("seconds". | 503 | 1,384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 33, "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} | 4.8125 | 5 | CC-MAIN-2019-51 | longest | en | 0.878547 |
https://mathoverflow.net/questions/215923/is-mathbbr3-setminus-mathbbq3-simply-connected/215930 | 1,723,158,565,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640741453.47/warc/CC-MAIN-20240808222119-20240809012119-00293.warc.gz | 303,738,965 | 35,276 | # Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?
Similarly is the complement of any countable set in $\mathbb R^3$ simply connected?
Reading around I found plenty of articles discussing the path connectedness $\mathbb R^2 \setminus \mathbb Q^2$ and even an approach using cofiltered limits to approach that problem, but I am not read enough in that literature to see if this could be applied here.
• There's a direct argument for this using the transversality-extension theorem. Commented Aug 30, 2015 at 6:22
• While I harbor some doubts that one could do better than Martin's answer below, why not explain this alternative argument? It could be educational, even for professionals. Commented Aug 30, 2015 at 16:58
• @MikeMiller Why not write up the argument carefully as an answer before the question is closed? The comment as written is a little too telegraphic for me to follow easily, and I don't know which thing of Hirsch you mean. A careful, detailed, self-contained answer would be peachy. Commented Aug 30, 2015 at 21:14
• Is it obvious that transversality arguments work for weird non-differentiable curves? In a manifold you know any curve is homotopic to a differentiable one, but showing that fact in $\mathbb{R}^3 \setminus \mathbb{Q}^3$ doesn't seem any easier than the original question. Commented Aug 30, 2015 at 22:02
• Adams mentions in "Lectures on Lie Groups" that there is a good theory of transversality and homotopy formulated using Hausdorff dimension, but he does not give a reference and I have never seen one. Assuming that that is correct, it would immediately answer the question asked here. Commented Aug 31, 2015 at 10:19
Yes, the complement of any countable set in $$\mathbb{R}^3$$ is simply connected, by the Baire category theorem.
Say your set is $$X = \{x_1, x_2, ... \}$$, and let $$y$$ be any point in $$\mathbb{R}^3 \setminus X.$$
Let $$f:S^1 \rightarrow \mathbb{R}^3 \setminus X$$, and consider the space of homotopies $$h:S^1 \times [0,1] \rightarrow \mathbb{R}^3$$, where $$h(x, 0) = f(x)$$ and $$h(x, 1) = y$$. With the natural topology the space of homotopies is a Baire space, and for each $$n$$, the set of homotopies that avoid the point $$x_n$$ is open and dense. So the set of homotopies that miss all of $$X$$ is nonempty.
• Does this argument generalize beyond countable to show also that $\mathbb{R}^3\setminus X$ is simply connected whenever $X$ has size less than $\text{cov}(\cal M)$? (The cardinal characteristic $\text{cov}(\cal M)$ is the covering number of the meager ideal, or in other words, the smallest number of meager sets that union to the whole space; it is more interesting when the continuum hypothesis fails.) Commented Aug 29, 2015 at 20:05
• I think that some essential details are lacking. Where does this argument use that the ambient space is $\Bbb R^3$? Because the statement would be false for $\Bbb R$. Commented Aug 29, 2015 at 21:22
• @Victor The argument takes for granted that the ambient space stays simply connected (even connected) when you remove just a finite set of points--fine in $\mathbb{R}^3$ but not $\mathbb{R}$ or $\mathbb{R}^2$. Commented Aug 29, 2015 at 21:47
• @JoelDavidHamkins I think so. It might be interesting to understand this number more fully. One should be able to replace the space of homotopies with a finite-dimensional space and understand exactly what sort of spaces are being removed to get a better lower bound. Commented Aug 30, 2015 at 0:28
• Great solution! I believe this also immediately generalizes to show $\pi_{n-2}(\mathbb{R}^n \setminus \mathbb{Q}^n) = 0$. Commented Aug 30, 2015 at 6:10
$$\newcommand\R{\mathbb{R}} \newcommand\cH{\mathcal{H}} \newcommand\bbD{\mathbb{D}}$$ I'm marking this community wiki since it's not so much an answer but an attempt to make it clear to the casual reader that despite Nick Mendler's answer it really is obvious that for a finite set $$S \subset \R^3$$, a continuous curve $$\gamma\colon S^1 \rightarrow \R^3 \setminus S$$, and a point $$y \in \R^3 \setminus S$$ the set of homotopies from $$\gamma$$ to the constant map $$y$$ that avoid $$S$$ is open and dense in the set of all homotopies. Here "obvious" means "immediate from standard tools/arguments that have little to do with $$\R^3$$". All I will use about $$\R^3$$ is that it is a smooth 1-connected manifold of dimension at least $$3$$.
The most convenient way to think of such homotopies is as the set $$\cH$$ of all continuous maps $$f\colon \bbD^2 \rightarrow \R^3$$ with $$f|_{S^1} = \gamma$$ and $$f(0) = y$$. Let $$\cH_S$$ be the subspace of $$\cH$$ consisting of such $$f$$ whose image is disjoint from $$S$$. Give $$\cH$$ the compact-open topology. We want to prove that $$\cH_S$$ is open and dense in $$\cH$$. That it is open is genuinely obvious, so the key thing to prove is that it is dense.
Since $$\mathbb{R}^3$$ is $$1$$-connected, the set $$\cH$$ is nonempty. Consider some $$f_0 \in \cH$$. Our goal is to perturb $$f_0$$ slightly (I'm not going to bother keeping track of $$\epsilon$$'s) to move it into $$\cH_S$$. Choose three nested proper open annuli $$A_1 \subset A_2 \subset A_3$$ in $$\bbD^2$$ with $$\overline{A}_1 \subset A_2$$ and $$\overline{A}_2 \subset A_3$$ such that $$f_0(\bbD^2 \setminus A_1)$$ is disjoint from $$S$$. Here "proper open annuli" means that $$0 \notin A_1$$ and $$S^1 \cap \overline{A}_3 = \emptyset$$. That we can do this is clear: just make $$A_1$$ nearly all of $$\bbD^2$$.
The first standard fact we appeal to is that we can perturb $$f_0$$ to some $$f_1 \in \cH$$ with the following properties:
1. $$f_1$$ is smooth on $$\overline{A}_2$$; and
2. $$f_1 = f_0$$ on $$\bbD^2 \setminus A_3$$, which implies that $$f_1 \in \cH$$.
By making this perturbation small enough, we can assume that it is still the case that $$f_1(\bbD^2 \setminus A_1)$$ is disjoint from $$S$$.
The second standard fact we appeal to is that we can perturb $$f_1$$ to some $$f_2 \in \cH$$ with the following properties:
1. $$f_2$$ is smooth on $$\overline{A}_2$$ and is transverse to $$S$$ on $$A_1$$; and
2. $$f_2 = f_1$$ on $$\bbD^2 \setminus A_2$$, which implies that $$f_2 \in \cH$$.
Again, by making this perturbation small enough, we can assume that it is still the case that $$f_2(\bbD^2 \setminus A_1)$$ is disjoint from $$S$$. Since on the $$2$$-manifold $$A_1$$ the map $$f$$ is transverse to the $$0$$-submanifold $$S$$ of the $$3$$-manifold $$\R^3$$, we actually have that $$f_2(A_1)$$ is disjoint from $$S$$. We conclude that $$f \in \cH_S$$, as desired.
• Yes, such details are best to leave as an exercise to a reader, they are at a homework level in an algebraic topology class. Commented Jun 3 at 16:57
• @MoisheKohan: I agree. Commented Jun 3 at 16:59
• After all, if $h^s(t)$ is a free homotopy in $\mathbb R^3$ between loops $h^0(t)$ and $h^1(t)$ in $\mathbb R^3\setminus \{P\}$, and its image has no interior (e.g. it is Lipschitz) then a small translate $h^s(t)+v$ avoids $P$, and $h^0$ and $h^1$ are homotopic in $\mathbb R^3\setminus \{P\}$ resp. to $h^0+v$ and $h^1+v$ , therefore $h^0$ and $h^1$ are homotopic in $\mathbb R^3\setminus \{P\}$ too. Commented Jun 4 at 19:09
It's a good idea to use the Baire Category Theorem, but there is still work to be done to verify that the sets are open and dense.
Here's a summary of the proof for fast readers (for more clarification, I provide all the gory details afterward):
The set of homotopies avoiding $$x_n$$ is clearly open. Let $$h:\mathbb{D}\rightarrow \mathbb{R}^d$$ be a homotopy from $$\gamma$$ to $$y$$ and suppose $$x_n\not\in \gamma(S^1).$$ Approximate $$h$$ by a piecewise-linear (PWL) map $$h^*,$$ and let $$g^t$$ be the straight-line homotopy from $$h$$ to $$h^*.$$ Let $$\rho:\mathbb{D}\rightarrow[0,1]$$ be a continuous map with $$\rho|_{\partial\mathbb{D}}=0$$ and $$\rho|_U=1$$ where $$U\subset \textrm{int}(\mathbb{D})$$ is an open neighborhood of $$h^{-1}(x_n).$$ Defining $$H = g^{\rho(-)}(-),$$ we then have that $$H$$ is close to $$h,$$ $$H|_{\partial \mathbb{D}}=\gamma,$$ and $$H|_U$$ is PWL. Moreover, we can take $$U$$ sufficiently large so that $$H(\mathbb{D}\setminus U)$$ is sufficiently close to $$\gamma(S^1),$$ so that $$x_n\not\in H(\mathbb{D}\setminus U)$$. Because $$H|_U$$ is PWL, $$H(U)$$ has no interior, so it follows that $$x_n$$ is not in the interior of $$H(\mathbb{D}).$$ Hence we can find a point $$y\not\in H(\mathbb{D})$$ that is close to $$x_n.$$ In particular $$B_\varepsilon(y) \cap H(\mathbb{D})=\emptyset.$$ Then, apply a perturbation $$f_y:\mathbb{R}^d\rightarrow \mathbb{R}^d,$$ that enlarges the neighborhood $$B_\varepsilon(y)$$ so that $$x_n\in f_y(B_\varepsilon(y)).$$ We can certainly choose $$f_y$$ to be small and equal to identity on $$\gamma(S^1).$$ Hence the perturbed homotopy $$f_y\circ H$$ is close to $$h,$$ avoids $$x_n,$$ and $$(f_y\circ H)|_{\partial \mathbb{D}} = \gamma$$ as desired.
Formal proof:
Let $$\{x_1,x_2,\dotsc\}\subset \mathbb{R}^d$$ be given, and let $$\gamma:S^1\rightarrow \mathbb{R}^d\setminus \{x_1,x_2,\dotsc\}$$ be a loop.
Denote $$\mathbb{D} = \{z\in\mathbb{C}\mid \lvert z\rvert\leq 1\}$$, and identify $$\partial \mathbb{D} = S^1$$, and let $$C(\mathbb{D},\mathbb{R}^d)$$ be the space of continuous functions from $$\mathbb{D}$$ to $$\mathbb{R}^d$$ equipped with the uniform norm. Denote the subset $$\mathcal{H} = \{h\in C(\mathbb{D},\mathbb{R}^d)\mid h\rvert_{\partial \mathbb{D}} = \gamma\}.$$ Thus, $$\mathcal{H}$$ can be viewed as the space of all homotopies of $$\gamma$$ to a point. The uniform limit theorem ensures that $$\mathcal{H}$$ is complete. In particular, $$\mathcal{H}$$ is a Baire-space.
For each $$x_n$$, define $$S_n = \{h\in \mathcal{H}\mid x_n\notin h(\mathbb{D})\}.$$
If $$h_m$$ is a sequence in $$\mathcal{H}\setminus S_n$$, then there is a sequence $$z_m\in \mathbb{D}$$ for which $$h_m(z_m)=x_n$$. And, if $$h_m\rightarrow h\in \mathcal{H}$$, then—because convergence is uniform—we have that for every $$\varepsilon$$ there is an $$m$$ large enough so that $$d(h(z_m),x_n) = d(h(z_m),h_{m}(z_{m}))<\varepsilon.$$ So for any convergent subsequence $$z_{m_j}\rightarrow z$$ we have that $$h(z) = h(\lim_{j\rightarrow\infty}z_{m_j})= \lim_{j\rightarrow\infty}h(z_{m_j}) = x_n.$$ In particular, $$h\notin S_n$$. Thus $$H\setminus S_n$$ is closed, and so $$S_n$$ is open.
To see that $$S_n$$ is dense in $$\mathcal{H}$$, let $$h\in \mathcal{H}\setminus S_n$$ and $$\varepsilon>0$$ be given, and suppose that $$\varepsilon$$ is sufficiently small so that the open ball $$B_{3\varepsilon}(x_n)$$ is disjoint from $$\gamma(S^1)$$ (this can be done because $$\gamma(S^1)$$ is closed and disjoint from every $$x_n$$).
For every $$m\in\mathbb{N}$$, let $$T_m$$ be a triangulation of $$\mathbb{D}$$ so that every triangle $$\tau\in T_m$$ has diameter $$\lvert\tau\rvert<2^{-m}$$. Let $$h_m$$ be the piecewise linear (PWL) approximation of $$h$$; that is, $$h_m$$ is linear on every triangle $$\tau\in T_m$$, and $$h_m$$ agrees with $$h$$ on every vertex of every triangle.
Then, because $$h$$ is uniformly continuous, and the vertices of the triangles in $$T_m$$ are $$2^{-m}$$ dense in $$\mathbb{D}$$, we can certainly take an $$m\in \mathbb{N}$$ sufficiently large so that
$$\lVert h_m - h\rVert_{\infty}\leq\varepsilon.$$
If $$h_m\in S_n$$, we are done. So suppose otherwise that $$x_n\in h_m(\mathbb{D})$$.
As $$h$$ is uniformly continuous, we can take $$0 sufficiently large so that $$h(\mathbb{D}\setminus B_R(0))\subset B_\varepsilon(\gamma(S^1)).$$ Also, let $$R$$ be large enough so that $$h^{-1}(x_n)\subset B_R(0).$$
Let $$g^t$$ be the straight-line homotopy from $$h$$ to $$h_m,$$ and let $$\rho:\mathbb{D}\rightarrow[0,1]$$ be a continuous map with $$\rho|_{\partial\mathbb{D}}=0$$ and $$\rho|_{B_R(0)}=1.$$ Defining $$H = g^{\rho(-)}(-),$$ we then have that
\begin{aligned}||H-h_m||_\infty &= ||(1-\rho)h + \rho h_m - h_m||_\infty \\ &\leq ||1-\rho||_\infty\cdot||h-h_m||_\infty \\&\leq \varepsilon. \end{aligned}
Also, for every $$z\in \mathbb{D}\setminus B_R(0),$$ we have \begin{aligned}\inf_{s\in[0,1]}|H(z)-\gamma(s)| &= |H(z) - h_m(z)+h_m(z)-h(z)+h(z)-\gamma(s)| \\ &\leq \inf_{s\in[0,1]}|h(z)-\gamma(s)| + 2\varepsilon \\ &< 3\varepsilon,\end{aligned} and because $$B_{3\varepsilon}(x_n)\cap \gamma(S^1)=\emptyset,$$ this shows that $$x_n\not \in H(\mathbb{D}\setminus B_R(0)).$$
Now, consider that $$H(B_R(0)) = \bigcup_{\tau\in T_m}h_m(\tau\cap B_R(0))$$, and each $$h_m(\tau\cap B_R(0))$$ has no interior in $$\mathbb{R}^d$$ (this is where we need $$d\geq 3$$). Also, $$T_m$$ contains finitely many triangles, and so $$H(B_R(0))$$ is a finite union of sets with no interior, and hence has no interior.
Because $$x_n\not\in H(\mathbb{D}\setminus B_R(0)),$$ and $$H(B_R(0))$$ has no interior, we can find a $$y\not\in H(\mathbb{D})$$ for which $$\lvert x_n-y\rvert<\varepsilon$$.
Next, define
$$f_{y}:\mathbb{R}^d\setminus\{y\}\rightarrow\mathbb{R}^d: f_{y}(v) = (\lvert v-y\rvert+\varepsilon^*)\frac{(v-y)}{\lvert v-y\rvert} + y,$$ where $$\varepsilon^* = \begin{cases}\varepsilon &: \lvert v-y\rvert<\varepsilon\\ 2\varepsilon - \lvert v-y\rvert &: \varepsilon\leq \lvert v-y\rvert<2\varepsilon \\ 0&:2\varepsilon\leq \lvert v-y\rvert.\end{cases}$$
Consider that $$f_y$$ is continuous on $$\mathbb{R}^d\setminus\{y\}$$. And since $$y\notin H(\mathbb{D})$$, it follows that $$f_y\circ H$$ is continuous. Also $$f_y$$ is the identity on $$\mathbb{R}^d\setminus B_{2\varepsilon}(x_n)$$; so the fact that $$B_{2\varepsilon}(x_n)$$ is disjoint from $$\gamma(S^1)$$ implies that $$(f_y\circ H)\rvert_{\partial \mathbb{D}} = f_y\circ \gamma=\gamma$$, and so we have that $$f_y\circ H \in \mathcal{H}$$.
Also note that $$\lvert f_{y}(v)-v\rvert \leq \varepsilon$$ for every $$v\in \mathbb{R}^d\setminus\{y\}$$, and so $$\lVert(f_y\circ H) - H\rVert_\infty \leq \varepsilon.$$ Thus the triangle inequality yields $$\lVert(f_y\circ H) - h\rVert_{\infty} \leq 3\varepsilon.$$
Finally, it is clear that the open ball $$B_\varepsilon(y)$$ is disjoint from the image of $$f_y$$, and so $$B_\varepsilon(y)$$ must be disjoint from $$(f_y\circ H)(\mathbb{D})$$. But $$x_n\in B_\varepsilon(y)$$, so we have that $$x_n\notin (f_y\circ H)(\mathbb{D})$$, which shows that $$(f_y\circ H)\in S_n$$.
Thus it has been shown that each $$S_n$$ is open and dense in $$\mathcal{H}.$$ And because $$\mathcal{H}$$ is a Baire-space, the BCT implies that $$\bigcap_{n\in\mathbb{N}} S_n\neq \emptyset$$. Any $$h\in \bigcap_{n\in\mathbb{N}} S_n$$ is a homotopy in $$\mathbb{R}^n\setminus \{x_1,x_2,\dotsc\}$$ from $$\gamma$$ to a point.
NOTES: as to @JoelDavidHamkins's comment, this argument clearly generalizes for $$\{x_1,x_2,\dotsc\}$$ replaced by any set of cardinality less than $$\operatorname{cov}(\mathcal{M})$$.
This argument does not explicitly use the fact that $$\mathbb{R}^d$$ minus a finite set is simply connected, but it does use the fact that, the linear image of a $$(d-1)$$-dimensional polygon has no interior in $$\mathbb{R}^d.$$
• Is $f_y \circ h_m \in H$? It seems to me that it does not agree with $h$ on the boundary. Commented Jun 5 at 10:24
• The problem occurs already for $h_m$, unless $\gamma$ is already piecewise linear. Commented Jun 5 at 18:37
• Your right, it had to be fixed. I fixed it by blending $h_m$ with $h.$ Commented Jun 5 at 21:27
• A flag has been raised with respect to the large number of edits. The trouble is that each edit bumps the post to the front page, at the expense of other posts also vying for attention, and this is annoying to the community. Please try to condense as many edits as you can into a single revision. Commented Jun 9 at 22:59
• Got it, thanks for letting me know Commented Jun 11 at 23:48
Since the conversation seems to focus on the density issue in Martin M. W.'s proof: yes, density is really as obvious as the openness. If $$h^s(t)$$ is a free homotopy in $$\mathbb R^3$$ between loops $$h^0(t)$$ and $$h^1(t)$$ in $$\mathbb R^3\setminus \{P\}$$ and its image has no interior (e.g. $$h$$ it is Lipschitz) then a small translate $$h^s(t)+v$$ avoids $$P$$, and $$h^0+v$$ and $$h^1+v$$ are homotopic in $$\mathbb R^3 \setminus\{P\}$$ resp. to $$h^0$$ and $$h^1$$, therefore $$h^0$$ and $$h^1$$ are homotopic in $$\mathbb R^3\setminus \{P\}$$ too.
• I agree that if you assume any kind of regularity on your homotopies it becomes utterly trivial, but for density you need to handle homotopies that are arbitrary. This means you need some initial step regularizing your homotopy. And if you don’t assume some regularity on your loops (their images might have nonempty interior), then you are quickly led to something like what I did. Commented Jun 5 at 12:51
• Yes, but regularization of h is clear by Weierstrass, isn't it? since $h$ is inside an open set $H$ of a Banach space there is a smooth $\tilde h\in B(h,\epsilon)\subset H$, so one can go from $h^0$ to $\tilde h^0$ by an affine homotopy etc. Commented Jun 5 at 14:59
• Weierstrass is one way, I suppose. There are many ways to do this, and the disadvantage of Weierstrass is that it's more tied to open subsets of Euclidean space rather than general smooth manifolds. Commented Jun 5 at 15:05
• I think this method by translation is certainly the simplest, and therefore best, way of doing the perturbation. But I still think using piecewise linear approximations is the most elementary way of ensuring no interior. Commented Jun 5 at 18:25
• over here @TarasBanakh makes a brilliant argument without using Baire Category theorem. Also, he shows that $R^d \setminus C$ is simply connected as long as $C$ has cardinality less than the continuum. If the continuum hypothesis fails, this is a stronger result than can be gotten with the Baire Category theorem Commented Jun 5 at 22:23 | 5,918 | 17,789 | {"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": 252, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.915647 |
http://xehugariwykice.killarney10mile.com/write-a-quadratic-equation-with-imaginary-numbers-in-math-78548byk9856.html | 1,544,861,861,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826800.31/warc/CC-MAIN-20181215061532-20181215083532-00047.warc.gz | 521,898,019 | 4,987 | # Write a quadratic equation with imaginary numbers in math
Simplifying the expression under the radical gives: Algebra scares me very much as it does at times seem like hieroglyphics to me, so I am really going to try and OWN this material.
After what the students have done today, this should be fairly straightforward practice. You cannot simplify your fractions any further, because the numerator and denominator do not have any common factors other than 1; therefore, this is your final solution: These lessons are awesome!
The variables a, b, and c in the quadratic formula correspond to the coefficients in the quadratic equation. Given a quadratic equation then the roots of the equation can be found by completing the square as below: Problem 4 asks students to apply this new notion of factoring to the problem of the sum of two squares MP8.
I found it a great value considering to learn in college setting costs significantly more. When we combine two AC currents they may not match properly, and it can be very hard to figure out the new current. In this lesson, we chose to expose you to this situation, but not to provide details on imaginary and complex solutions.
So, absolutely I would recommend this site. It is part of a subject called "Signal Processing". In cases such as this, when solving quadratic equations with non-real solutions, you learned that you can use the imaginary unit i to write the solutions of the quadratic equation as complex numbers.
Quadratic Formula If the expression under the square root is negative, then the quadratic equation will have zero real solutions. I got an 89 on my test the next day, and I have to credit your website for that.
Homework 3 minutes Regardless of whether or not everyone has completed problems 3 and 4, I call a halt to all work a couple of minutes before the period ends.
Next, simplify your numerator, starting with the expression underneath the square root. Very clear, concise explanations and the practice problems have the option to see the problem worked out with or without his guidance. And that is also how the name " Real Numbers " came about real is not imaginary.
This website was so helpful — I learned so much here as I was able to pause, write the problem down and unpause again, repeating until it made sense.
I was able to pass college algebra with an A. I especially like the live teacher showing you how to begin the problem. But using complex numbers makes it a lot easier to do the calculations. Yep, Complex Numbers are used to calculate them!
Notice that there is a negative number under the square root symbol. My answers for homework are now correct. I am depending on your help as the semester rolls along.
Now, substituting this into the Quadratic Formula: Therefore, your expression is: This is a general formula which can be used to solve for the roots of any quadratic equation. I have an A in College Algebra, after being a poor math student in high school.
I will be taking college algebra this coming semester, so I wanted to get a heads-up on what I will be facing. Problem 3 asks students to carefully examine the structure of a factored quadratic and recognize that once the roots of any quadratic are known - complex or real - that quadratic can be written in a factored form.
Someone here at the College told me about your program and I love it. For example, if asked to find the roots of the given quadratic equation looking at only and 1 is greater than zero, we can conclude that the quadratic equation has real roots, which is proved by finding the roots of the equation using the quadratic formula.
In the next example, try using the quadratic formula to solve the quadratic equation: To factor a sum of two squares, they might set the sum of two squares equal to zero, find the solutions, and then write the factored form. Today you reviewed imaginary numbers, recalling that the square root of a negative number is non-real because any real number squared will not be negative.
From this derivation, we can generalize a few equalities based on the formula. I wish I had known about your site when I started in July — it would have made my life easier.Python Program to Solve Quadratic Equation This program computes roots of a quadratic equation when coefficients a, b and c are known.
To understand this example, you should have the knowledge of following Python programming topics. Simplifying Square Roots That Contain Whole Numbers Solving Quadratic Equations by Completing the Square Quadratic Equations with Imaginary Solutions.
Example. Step 1: Write the quadratic equation in standard form. Subtract 3x from both sides of the equation. 2x 2 - 3x + 7 = 0: Step 2: Identify the values of a, b, and c. Quadratic Formula and Functions. We've run out of actual numbers to throw at you, so now we're just going to make some numbers up?
Not really. Imaginary numbers, despite the name, are totally l Completing the Square. Sample Problem Solve the quadratic equation x2 + 2x + 5 = 0. Hmm, now this is tricky.
This equation is—most. How to Use the Quadratic Formula to Solve a Quadratic Equation By Contributor; Updated April 24, More advanced algebra classes will require you to solve all kinds of different equations.
I am not a ‘math person’, but I have been able to keep a high ‘A’ average in College Algebra with the help of your service.” Imaginary Numbers (Adding and Subtracting Square Roots of Negatives) Write Quadratic Equation with Roots (6 + root 10)/2 and (6.
In elementary algebra, the quadratic formula is the solution of the quadratic equation. There are other ways to solve the quadratic equation instead of using the quadratic formula, such as factoring, completing the square, or graphing.
Write a quadratic equation with imaginary numbers in math
Rated 4/5 based on 77 review | 1,217 | 5,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.945732 |
https://analgorithmaday.blogspot.com/2011/05/construct-binary-tree-from-pre-order.html | 1,714,000,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00415.warc.gz | 78,442,248 | 17,355 | ## Wednesday 25 May 2011
### Construct a binary tree from pre-order and in-order traversals given
Question
I came to know about this question from ihas1337code.com. The problem is to construct a normal binary tree from the given pre-order and in-order arrays.
For example,
say, you are given,
preorder = {7,10,4,3,1,2,8,11} inorder = {4,10,3,1,7,11,8,2}
if you need to get a tree like this,
_______7______
/ \
__10__ ___2
/ \ /
4 3 _8
\ /
1 11
How would you do that ? Just guess before going into above link
Approach
I came up with the approach. But feed-up with the recursion logic and broke my head with vectors & stacks. This is where the power of recursion is fully used. There are many problems like this which you must be used to get ready for interviews.The approach for this problem is like this,
• We know that the first node of the pre-order is definitely the root of the tree
• Subsequent nodes listed in the pre-order arrays are all roots of sub trees, coming with format left-right-root
• Also we know that, in-order traversal has left first, root and then right
• So, simply take a element in the pre-order array and search in the in-order array to get the indices. After you get the index, the left is the left sub tree, right is the right sub tree
For example, Consider the above input set.
The first iteration will be like,
_______7______
/ \
4 10 3 1 11 8 2
So, detect 7 as the root.. search in the in-order array and find the index.. what you got is that, {4 10 3 1} are the nodes of the left sub tree and {11, 8, 2} are nodes of the right sub tree.
After you detect this, you need recursively do the same to in left and right lists to get the full tree constructed.
That is, the next root after seven is nothing but 10.. search for 10 in the left first.. having 10 as the root, 4 becomes the left and {3,1} becomes the right tree.
But how to do this approach ?
Approach – for implementation
The implementation of this approach is little bit tricky. Firstly, we should be very clear about what all we need replication for each tree creation. The following data is needed in each iteration,
• The root node!! its different for each sub-tree that we create
• the pre-order array movement and the in-order array movements.. we need to split the in-order arrays with root node found from pre-order array as the middle element.
• root->left and root->right should get the nodes which are processed in the future!! :(
After doing this exercise, i understood that, i need to restudy recursion again in a more clearer way :)
Recursive call of a function does the following things
• the arguments are put into a stack.. You can clearly see the arguments part of the call stack frame
• the local variables in the function are all replicated and if you return local variable value in the recursion.. it gets shared with the previous recursive function call in the stack.. :)
The local variables & return value based recursion is also known as tail recursion. We will see about this in detail separately.
We have to use tail recursion to successfully implement this code.
Code
` `
`struct BinaryTree `
`{`
` int data;`
` BinaryTree* left;`
` BinaryTree* right;`
`};`
` `
`BinaryTree* newTreeNode(int data)`
`{`
` BinaryTree* newNode = new BinaryTree;`
` newNode->data = data;`
` newNode->left = NULL;`
` newNode->right = NULL;`
` `
` return newNode;`
`}`
` `
`int getIndex(int* arr, int val, int size)`
`{`
` for(int i=0; i<size;i++) {`
` if(arr[i] == val) `
` return i;`
` }`
` return -1;`
`}`
` `
` `
` `
`BinaryTree* create_bintree(int* parray, int* iarray, int size)`
`{`
` if(size == 0) return NULL;`
` `
` int rootVal = parray[0];`
` BinaryTree* root = newTreeNode(rootVal);`
` int newIdx = getIndex(iarray, rootVal, size);`
` root->left = create_bintree(parray+1, iarray, newIdx);`
` root->right = create_bintree(parray+newIdx+1, iarray+newIdx+1, size-newIdx-1);`
` return root; `
`}`
`void main()`
`{`
` int preorder[] = {7,10,4,3,1,2,8,11};`
` int inorder[] = {4,10,3,1,7,11,8,2};`
` `
` BinaryTree* tree = create_bintree(preorder, inorder, 8);`
` `
`}`
• We have taken some code logic from ihas1337code. But we don’t use mapIndex and offsets, as it complicates things and also adds a restriction to the element set
• Note that, elements should be unique!!.. Otherwise, this approach will never workout.. :)
• NEVER EVER TRY WITH vectors, stacks without recursion. It will screw up heavily!!
• every node gets a new tree Node with newTreeNode function inside create_bintree
• we do the tail recursion, parray+1 to get the root->left. and we will end this recursion only after we have also seen the right nodes in the tree. :)
• That is, when we reach 0 in root->left.. we check for root->right as well!!.. End condition is newIdx is 0 (i.e, size = 0)
• parray+1 and parray+newIdx+1 is simple logic
• You get the newIdx from inorder array.. which is the middle element and the root.
• 0 to newIdx-1 will be left and newIdx+1 to end will be the right
• iarray and iarray+newIdx+1 works uniformly with parray so that.. root->left gets only left side nodes and root->right gets only the right side nodes.
• size-newIdx-1 in this, –1 is just because the array index starts at 0. size-newIdx is the actual size of right half. that is, end – newIdx+1 is the size of the right side in all cases
• There are lots of invariant in this code!!.. and its not verified with all kinds of Input. Handle with care!! :)
Unknown said...
can we make a tree if we are only given preorder expression.
Unknown said...
Only pre order wont suffice to construct an unordered tree.
thakursaab said...
THANks | 1,528 | 5,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-18 | latest | en | 0.889765 |
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Helper II
## Time calculation
Hi,
Another query on calculating dates/times, but I haven't seen anything similar to what I'm trying to do, so I'd appreciate any help!
When we place an order with a carrier, we speicify the delivery date/time that we want. We have cut off times agreed with our carriers for when we will place the order for them to then deliver by the requested date, and I need to see how often we are failing to hit that cut off.
The data is set by times in one column and day number in another; today is Day 0, tomorrow is Day 1, next day is Day 2 etc
Order Cut Off Day Delivery Time Point A to Point B 10:00 1 16:00 Point A to Point C 14:00 2 09:00
In this example, if we want an item delivering to Point B tomorrow, we would need to book the order by 10:00 today. Point C is further away - we need to book before 14:00 Day 0 to ensure delivery at 09:00 Day 2.
Essentially, Point B requires 30 hours lead time, point C needs 43 hours. How do I calculate that from this date in a new column?
Many thanks!!
2 ACCEPTED SOLUTIONS
Resolver IV
Thanks for detailed explaination
Use below formula in Calculated Column
Spoiler
```Duration (Hours) =
Var StartDate = TODAY()+'Table'[Order Cut Off]
Var EndDate = TODAY()+'Table'[Delivery Time] + 'Table'[Day]
RETURN
(EndDate - StartDate) * 24```
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Proud to be Datanaut!
Super User
If Delivery Time and Order Cut Off are formatted as Time and Day as whole number, you can simply do
`NewColumn = 24*(Table1[Delivery Time] + Table1[Day] - Table1[Order Cut Off] )`
3 REPLIES 3
Resolver IV
Thanks for detailed explaination
Use below formula in Calculated Column
Spoiler
```Duration (Hours) =
Var StartDate = TODAY()+'Table'[Order Cut Off]
Var EndDate = TODAY()+'Table'[Delivery Time] + 'Table'[Day]
RETURN
(EndDate - StartDate) * 24```
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Proud to be Datanaut!
Helper II
Thank you!
Super User
If Delivery Time and Order Cut Off are formatted as Time and Day as whole number, you can simply do
`NewColumn = 24*(Table1[Delivery Time] + Table1[Day] - Table1[Order Cut Off] )`
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Before learning about the dummy variable trap, let’s first understand what actually dummy variable is.
Dummy Variable in Regression Models:
In statistics, especially in regression models, we deal with various kinds of data. The data may be quantitative (numerical) or qualitative (categorical). The numerical data can be easily handled in regression models but we can’t use categorical data directly, it needs to be transformed in some way.
For transforming categorical attributes to numerical attributes, we can use the label encoding procedure (label encoding assigns a unique integer to each category of data). But this procedure is not alone that suitable, hence, One hot encoding is used in regression models following label encoding. This enables us to create new attributes according to the number of classes present in the categorical attribute i.e if there are n number of categories in categorical attribute, n new attributes will be created. These attributes created are called Dummy Variables. Hence, dummy variables are “proxy” variables for categorical data in regression models.
These dummy variables will be created with one-hot encoding and each attribute will have a value of either 0 or 1, representing the presence or absence of that attribute.
Dummy Variable Trap:
The Dummy variable trap is a scenario where there are attributes that are highly correlated (Multicollinear) and one variable predicts the value of others. When we use one-hot encoding for handling the categorical data, then one dummy variable (attribute) can be predicted with the help of other dummy variables. Hence, one dummy variable is highly correlated with other dummy variables. Using all dummy variables for regression models leads to a dummy variable trap. So, the regression models should be designed to exclude one dummy variable.
For Example –
Let’s consider the case of gender having two values male (0 or 1) and female (1 or 0). Including both the dummy variable can cause redundancy because if a person is not male in such case that person is a female, hence, we don’t need to use both the variables in regression models. This will protect us from the dummy variable trap.
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Write down your answers and use the answer key below to check if they are right.Three-digit multiplication word problems Solve these well-researched word problems that involve three-digit multiplication.Word problems take math understanding to the next level. These worksheets contain simple multiplication word problems.
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Math Word Problems (by Type)These word problems are sorted by type: addition, subtraction, multiplication, division, fractions and more.
Mixed Skills: Word Problems These worksheets, sorted by grade level, cover a mix of skills from the curriculum.
Multiplication doesn’t have to be all drills and timed tests!
In this engaging third-grade worksheet, kids will put their multiplication prowess to the test with one-digit multiplication word problems.
Answer the word problems based on three fascinating themes - Winter Season, Ice rink and Library.
Multiplication word problems: Three-digit times two-digit Read the word problems featured in these worksheets to find the product of three-digit and two-digit numbers.
Another set of worksheets hone children's multiplication skill by multiplying large numbers. Single-digit multiplication word problems The printable PDF worksheets presented here involve single-digit multiplication word problems.
Each worksheet carries five word problems based on day-to-day scenarios.
Multiply a three or four-digit number by a single-digit multiplier to find the correct product.
Multi-digit word problems: Multiplying large numbers Sharpen your skills by solving these engaging multi-digit word problems.
• ###### Multiplication Word Problem Worksheets 3rd Grade
Each sheet involves solving a range of written multiplication problems. There are 3 levels of difficulty for each worksheet below A,B and C. Worksheet A is the.…
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Simple multiplication word problems in ready to print PDFs! These are perfect for third grade or fourth grade applied math. Worksheet 3 · Easy Multiplication.…
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Multiplication word problems for grade 3 students. Each worksheet has a number of word problems and an answer sheet. All worksheets are pdf documents and.…
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This is a complete lesson for third grade with teaching & word problems with the aim of teaching children some basics about multiplication word problems. The basic idea. 7 days × 3 meals a day = _____ meals in a normal week. This Friday.…
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Printable third grade word problem worksheets including addition, subtraction, multiplication, division and fraction word problems. Answer sheets can also be.…
• ###### Problem Solving Grade 3 Mathematics - hcpss
Instead problem solving is a skill that favors every mathematics lesson. Problem solving. Deck B Grades 3-4 Student Copy. Multiplication/Division Problems.… | 708 | 3,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-17 | latest | en | 0.864531 |
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posted by .
The amount of money in an account with continuously compounded interest is given by the formula A = Pert, where P is the principal, r is the annual interest rate, and t is the time in years. Calculate to the hundredth of a year how long it takes for an amount of money to double if interest is compounded continuously at 6.5%.
the formula is A=Pe^rt, I assume yours was a typo
Consider any Principal, eg P=100
so you want
200 = 100e^(.065t)
2 = e^.065t
.065t=ln 2
I got t=10.66 using my calculator
so you can use anything you want for the principal?
yes
we could have simply used the principal P
Since we want it to "double" the amount would then be 2P
2P = P e^(.065t), divide by P and we get
2 = e^.065t like before
oh okay, thanks!
• math -
The amount of money in an account with continuously compounded interest is given by the formula A=Pe^rt , where P is the principal, r is the annual interest rate, and t is the time in years. Calculate to the nearest hundredth of a year how long it takes for an amount of money to double if interest is compounded continuously at 6.2%. Round to the nearest tenth.
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www.rathleens.ie | 1,659,966,469,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570827.41/warc/CC-MAIN-20220808122331-20220808152331-00654.warc.gz | 838,034,831 | 206,509 | # Primary Maths & Science
## Step 2 - Technology
We have many other older experiments to share with you from our Science programme in 2017-2018 still on our blog - please visit and see the work of our super scientists and mathematicans. - click here
## Panpipes
Strand: Energy and Forces
Unit: Sound
Maths: measures – length
We made panpipes from plastic drinking straws.
### What you need
• 8 plastic drinking straws
• sellotape
• scissors
• a ruler
• pen or pencil
What we did:
We started by cutting the first straw at 20cm. Then the next straw we cut 2cm shorter, at 18cm. We cut each straw 2cm shorter than the last until we had cut them all.
We lined up the straws longest to shortest.
We used sellotape to stick them together in the pan pipes formation.
We blew across the top of the straws to get a sound. The longest straws were the deepest sound and the shortest straws had the highest sound
measuring the straw
measuring the straw
measuring
measuring
measuring
measuring
cutting the straw
cutting the straw
panpipes
panpipes
panpipes
panpipes
Sound is a type of energy made by vibrations. When any object vibrates, it causes movement in the air particles.
You will need: clingfilm, cowl, dry rice, baking tray, wooden spoon
1. Stretch the clingfilm tightly over the bowl
2. Sprinkle a small amount of rice on the clingfilm.
3. Hold the baking tray above the rice and hit the tray with the wooden spoon. The sound makes the air vibrate and when the sound waves reach the clingfilm they make the it vibrate. This makes the rice dance.
## Energy and Forces
Sound
How do sound waves travel?
## String Telephones
Energy and Forces: Sound
What you’ll need:
• 2 paper cups
• A sharp pencil and blutac to pierce holes
• String
Instructions:
1. Cut a piece of string 1 metre in length.
2. Pierce a small hole in the bottom of each cup.
3. Thread the string through each cup and tie knots at each end to stop it pulling through the cup.
4. Move into position with you and a friend holding the cups at a distance. Make sure the string is tight.
5. One person talks into the cup while the other puts the cup to their ear and listens.
What happened:
Speaking into the cup creates sound waves which are converted into vibrations at the bottom of the cup. The vibrations travel along the string and are converted back into sound waves at the other end so your friend can hear what you said. Sound travels through the air but it travels even better through solids such as your cup and string, allowing you to hear sounds that might be too far away when traveling through the air.
As with any science experiment it did not work for a few of the pupils but we discovered that the pupils did not have the string tight
### Maths, History and Art
We used three arcs to create a Celtic knot design for St. Patrick’s Day. Many Celtic designs feature interlocking shapes. The Celtic knot is a very simple design that uses weaving to connect the three arcs. We were learning about the Book of Kells in History and the circle in Maths.
The children drew the arcs by drawing concentric circles with a compass and then cutting them out, them weaving them together to form Celtic knots.
### Strand Unit: Materials
Strand unit: properties and characteristics of materials
Materials needed: 1 boiled egg, bowl, clingfilm, white vinegar, red food colouring.
Method:
1. Place the boiled egg in a bowl.
2. Cover the boiled egg with white vinegar.
3. Add a few drops of food colouring to the vinegar.
4. Cover the bowl with clingfilm.
5. Leave the egg in the bowl of vinegar for 3 days. After remove the egg.
6. Rinse the shell of the egg gently under a tap, The shell will be removed.
Result:
The vinegar (an acid) reacts with the calcium carbonate producing a salt and a gas called carbon dioxide (these are the bubbles you see). The vinegar will keep reacting with the calcium carbonate until it is all gone, leaving the egg contained in just the cell membrane.
A delicate, but bouncy egg.
Inniu Máirt na hInide agus mar sin rinneamar pancóga inniu. Bhí siad go hálainn. Bhí siad milis agus blasta. D’itheamar
iad le siúcra, líomóid, Nutella agus síoróip. Bhi sé éasca go leor na pancóga a dhéanamh agus níor thóg sé mórán ama. Míle buíochas do Ber chun cabhraigh linn.
## pancakes (2)
### Energy and Forces
A force is a push or a pull on an object. A push moves an object away from another object; a pull moves it towards another object. Forces can make objects move, get faster, slow down, stop move in a different direction or change their shape. When an object moves along a surface or another object, a force known as friction occurs. Friction makes it more difficult for the object to move.
We observed and investigated the movement of toy cars along rough and smooth surfaces.
## Vitruvian Man
The Vitruvian Man is a drawing by the artist Leonardo Da Vinci based on the work of Vitruvius, a Roman architect.
Leonardo Da Vinci was born near Florence in Italy in 1452. He became a world famous artist. He died in 1519.
He was not just an artist, he was a scientist, a sculptor, a mathematician and an inventor!
Our objective was to recognise what the Vitruvian Man is and its significance to science and geometry.
The Vitruvian theory contains a total of 10 ratios between different parts of the body. As you can see we tested many of the ratios.
Seakeeper's Project - The Basking Shark
Seakeeper's Project - The Basking Shark
Seakeeper's Project - The Basking Shark
Report Writing
Seakeeper's Project - The Basking Shark
Report Writing
Report Writing
Our school is participating in The SeaKeepers Project which is a new initiative of the Green-Schools Global Citizenship Marine Environment theme. This is part of our work towards our 8th Green Flag. This project focuses on the ecology of six native Irish marine species such as The Basking Shark. As we are situated near the Atlantic ocean we are finding these lessons informative and fun. We each completed a report and some artwork on the Basking Shark which is displayed in the classroom. This month we are learning about the Grey Seal which is very common in Irish waters.
## STEM
### The Marshmallow Challenge
Objective: To construct a tower as high as possible using only spaghetti and masking tape. A marshmallow must be placed on top of the tower. The tallest tower standing unaided wins.
Time:20 minutes
Materials: 90 cm Masking tape, 25 sticks of spaghetti and a marshmallow.
This task involved teamwork, creativity, innovation and problem solving. Group 3 built the tallest tower but did not manage to hold the marshmallow on top.
All of the pupils were engaged in the task and worked very well in their groups. | 1,603 | 6,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-33 | longest | en | 0.88775 |
https://moviecultists.com/when-inputs-of-nand-gate-are-connected-together | 1,675,630,292,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500288.69/warc/CC-MAIN-20230205193202-20230205223202-00497.warc.gz | 413,760,013 | 13,585 | When inputs of nand gate are connected together?
NOT Gate: You may simply connect the two inputs in the NAND gate together to create a NOT Gate from the NAND Gate. Since the two inputs of the NAND gate are connected, only two input combinations can be used. The NAND Gate will emit a LOW if any input is HIGH. The NAND gate would be output HIGH if all inputs are LOW.
What happens if the two inputs of NAND gate are connected together?
Because the two inputs of the NAND gate are tied together, only two input combinations are possible: both HIGH or both LOW. If both inputs are HIGH, the NAND gate will output a LOW. If both inputs are LOW, the NAND gate will output HIGH. Thus, the circuit behaves exactly as a NOT gate would.
When connecting the two inputs in a NAND gate or in a NOR gate What is the equivalent gate?
A NOR gate is equivalent to an inverted-input AND gate. An OR gate is equivalent to an inverted-input NAND gate. Two NOT gates in series are same as a buffer because they cancel each other as A'' = A.
What is the output of a NAND gate when both its inputs are 1?
The NAND (Not – AND) gate has an output that is normally at logic level “1” and only goes “LOW” to logic level “0” when ALL of its inputs are at logic level “1”.
How many two input NAND gate are required to perform the action of a two input OR gate and its draw?
The answer is 3 NAND gates.
When all the inputs of a NAND gate are connected together, the resulting circuit is :-
40 related questions found
What is NAND gate in physics?
The NAND gate is a combination of an AND gate and NOT gate. They are connected in cascade form. It is also called Negated And gate. The NAND gate provides the false or low output only when their outputs is high or true.
Which gate will give an output of 1 when both its input is 1 or both inputs is 0 but gives an output of 0 when any of its input is different from the other?
An XOR gate is normally two inputs logic gate where the output is only logical 1 when only one input is logical 1. When both inputs are equal, either are 1 or both are 0, the output will be logical 0. This is the reason an XOR gate is also called an anti-coincidence gate or inequality detector.
Which logic gate gives output 1 only if both the inputs are 1?
Combinational circuits are built of five basic logic gates: AND gate - output is 1 if BOTH inputs are 1. OR gate - output is 1 if AT LEAST one input is 1. XOR gate - output is 1 if ONLY one input is 1.
What is the output of an OR gate if the inputs are 1 and 0?
If the input is 1, then the output is 0. If the input is 0, then the output is 1.
Why NAND and NOR gates are universal gate?
Therefore, an AND gate is made by inverting the inputs of a NOR gate. ... The same goes for the NOR gate too. ∴ NAND and NOR gates are called universal gates because they can be combined to produce any of the other gates like OR, AND, and NOT gates.
Why are NAND and NOR gates more popular?
NAND and NOR gates are more popular as these are less expensive and easier to design. Also other functions NOT AND OR can easily be implemented using NAND/NOR gates. Thus NAND NOR gates are also referred to as Universal Gates.
Can both OR and AND gates can have only two inputs?
Both OR and AND gates can have only two inputs. Explanation: Any number of inputs are possible.
How many two input AND gates and two input OR gates are required to realize y BD CE AB?
How many two input AND gates and two input OR gates are required to realize Y = BD + CE + AB? Explanation: There are three product terms. So, three AND gates of two inputs are required.
What are the three fundamental logic gates?
All digital systems can be constructed by only three basic logic gates. These basic gates are called the AND gate, the OR gate, and the NOT gate. Some textbooks also include the NAND gate, the NOR gate and the EOR gate as the members of the family of basic logic gates.
Which of the following logic gates has only two inputs?
Answer: The XOR gate is true when the inputs are opposite of each other, but false when they are equal. The XNOR gate follows the same conventions as above, and acts like an XOR gate whose output is then fed into a NOT gate. Therefore, an XNOR gate is true only when the two inputs are the same.
Under what conditions output of two inputs and gate is one?
2-input Ex-OR Gate
If these two inputs, A and B are both at logic level “1” or both at logic level “0” the output is a “0” making the gate an “odd but not the even gate”. In other words, the output is “1” when there are an odd number of 1's in the inputs.
When anyone of inputs is at 1 the value of output of OR gate will be?
When either input is a logic 1, the output will be 0. The only combination of inputs that results in an output of 1 is A = 0 and B = 0. The two-input NOR gate can be implemented in CMOS as shown in Fig.
Which gate produces a 1 only if all its inputs are 1 and a 0 otherwise?
A full adder takes the carry-in value into account. Produces a 1 only if all its inputs are 1 and a 0 otherwise. Produces a 0 only of its inputs are the same and a 1 otherwise. How many transistors does it take for a NOR gate?
What is the status of the inputs A and B of a two input NAND gate if its output?
For a 2-input NAND gate, the output Q is NOT true if BOTH input A and input B are true, giving the Boolean Expression of: ( Q = not(A AND B) ).
When the inputs to a 3 input AND gate are 001 the output is 1?
LOW. When the inputs to a 3-input OR gate are 001, the output is 1. An application requires a 3-input AND gate; however, all three of the inputs actually produce a LOW when the inputs are ON.
How are NAND and NOR gates implemented?
NAND and NOR implementation
1. If bubbles are introduced at AND gates output and OR gates inputs (the same for NOR gates), the above circuit becomes as shown in figure.
2. Now replace OR gate with input bubble with the NAND gate.
How are logic gates implemented?
Logic gates are primarily implemented using diodes or transistors acting as electronic switches, but can also be constructed using vacuum tubes, electromagnetic relays (relay logic), fluidic logic, pneumatic logic, optics, molecules, or even mechanical elements.
Which of the following gates are added to the inputs of the OR gate to convert it to the NAND gate?
The NOT gates are to be added to the inputs of OR gate, it will give the NAND gate output. Because, bubbled OR gate will acts as a NAND gate. | 1,523 | 6,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-06 | latest | en | 0.930009 |
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Showing posts with label divide. Show all posts
Showing posts with label divide. Show all posts
## Wednesday, May 1, 2013
### Elementary Algebra Exam #4
Click on the 10 question exam covering topics in chapter 7 (Rational Expressions and Equations). Give yourself one hour to try all of the problems and then come back and check your answers.
Simplify (Assume all denominators are nonzero.)
Perform the operations and state the restrictions.
Solve.
10. The sum of the reciprocals of two consecutive odd integers is 4/3. Set up an algebraic equation and use it to find the two integers.
---
## Sunday, April 21, 2013
### Elementary Algebra Exam #3
Click on the 10 question exam covering topics in chapters 5 and 6. Give yourself one hour to try all of the problems and then come back and check your answers.
10. The length of a rectangle is 4 centimeters less than twice its width. The area is 96 square centimeters. Find the length and width. (Set up an algebraic equation then solve it)
---
## Monday, November 19, 2012
### Multiplying and Dividing Radical Expressions
As long as the indices are the same, we can multiply the radicands together using the following property.
Since multiplication is commutative, you can multiply the coefficients and the radicands together and then simplify.
Multiply.
Instructional Video: Multiplying Radicals
Take care to be sure that the indices are the same before multiplying. We will assume that all variables are positive.
Simplify.
Divide radicals using the following property.
Divide. (Assume all variables are positive.)
Rationalizing the Denominator
A simplified radical expression cannot have a radical in the denominator. When the denominator has a radical in it, we must multiply the entire expression by some form of 1 to eliminate it. The basic steps follow.
Rationalize the denominator:
Multiply numerator and denominator by the 5th root of of factors that will result in 5th powers of each factor in the radicand of the denominator.
Notice that all the factors in the radicand of the denominator have powers that match the index. This was the desired result, now simplify.
Rationalize the denominator.
This technique does not work when dividing by a binomial containing a radical. A new technique is introduced to deal with this situation.
Rationalize the denominator:
Multiply numerator and denominator by the conjugate of the denominator.
And then simplify. The goal is to eliminate all radicals from the denominator.
Instructional Video: Dividing Radicals
Rationalize the denominator.
Video Examples on YouTube:
J. Redden on G+ | 583 | 2,622 | {"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} | 4.59375 | 5 | CC-MAIN-2018-30 | latest | en | 0.909571 |
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Είσοδος
Βιβλία Βιβλία
RULE. — Multiply all the numerators together for a new numerator, and all the denominators for a new denominator ; then reduce the new fraction to its lowest terms.
Practical arithmetic - Σελίδα 91
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...all whole and mixed numbers to their equira* lent fractions. 2. Multiply all the numerators together for a new numerator, and all the denominators for a new denominator j Mid they will form the fraction required. EXAMPLE8. 1 . Reduce J of f of | of .^ to a simple fraction,... | 1,218 | 4,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.526432 |
https://www.nyfamily-digital.com/b-intercept-formula-15-precautions-you-must-take-before-attending-b-intercept-formula-311733 | 1,582,857,181,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146940.95/warc/CC-MAIN-20200228012313-20200228042313-00241.warc.gz | 829,769,323 | 5,918 | # B Intercept Formula 15 Precautions You Must Take Before Attending B Intercept Formula
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Printing Job Order Form How Printing Job Order Form Is Going To Change Your Business Strategies 15 Form Less Than 15 The Modern Rules Of 15 Form Less Than 15 Typical Meeting Agenda Template 10 Things To Expect When Attending Typical Meeting Agenda Template Cover Letter Template To Whom It May Concern Seven Things You Most Likely Didn’t Know About Cover Letter Template To Whom It May Concern Generic Purchase Order Form The Miracle Of Generic Purchase Order Form Gtu Syllabus 13rd Sem How To Have A Fantastic Gtu Syllabus 113rd Sem With Minimal Spending Vertex Form Homework 17 Clarifications On Vertex Form Homework Membership Form Declaration Here’s Why You Should Attend Membership Form Declaration Standard Form Definition How To Get People To Like Standard Form Definition | 704 | 3,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-10 | latest | en | 0.820993 |
https://fr.maplesoft.com/support/help/addons/view.aspx?path=type%2Frational | 1,716,243,865,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00430.warc.gz | 224,869,498 | 21,341 | extended_rational - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
# Online Help
###### All Products Maple MapleSim
type/rational
check for an object of type rational
Calling Sequence type(x, rational)
Parameters
x - any expression
Description
• The type(x, rational) function returns true if x is an integer or fraction. Otherwise, it returns false.
Subtypes
•
Supertypes
Examples
> $\mathrm{type}\left(1,\mathrm{rational}\right)$
${\mathrm{true}}$ (1)
> $\mathrm{type}\left(\frac{1}{2},\mathrm{rational}\right)$
${\mathrm{true}}$ (2)
> $\mathrm{type}\left(0.5,\mathrm{rational}\right)$
${\mathrm{false}}$ (3)
> $\mathrm{type}\left(\mathrm{Name},\mathrm{rational}\right)$
${\mathrm{false}}$ (4)
> $\mathrm{type}\left(\frac{a}{b},\mathrm{rational}\right)$
${\mathrm{false}}$ (5)
> $\mathrm{type}\left(\mathrm{∞},\mathrm{extended_rational}\right)$
${\mathrm{true}}$ (6)
See Also | 296 | 977 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.243295 |
https://www.coursehero.com/file/14068280/Quiz11-16solutions/ | 1,545,131,940,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829140.81/warc/CC-MAIN-20181218102019-20181218124019-00593.warc.gz | 836,794,624 | 272,270 | Quiz11_16solutions
# Quiz11_16solutions - Physics 221 Quiz 11 Spring 2016 Name...
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Physics 221 Quiz 11 Spring 2016 Name & Section Quiz 11A 1. The periods of two pendula, one on Earth and the other on Mars, are the same. If the acceleration due to gravity on Earth and Mars are g E = 9.80 m/s 2 and g M =3.73 m/s 2 , respectively, and the length of the pendulum on Earth is 1.0 m, what is the length of the pendulum on Mars? Solution: T E =2 π ! ! ! ! - the period on Earth T M =2 π ! ! ! ! – the period on Mars T E = T M = 4s= ! ! ! ! = ! ! ! ! = > ࠵? ! = ! ! × ! ! ! ! = 0 . 38 m 2. A 200 g block is attached to a horizontal spring and executes simple harmonic motion with period of 0.25 s. The total energy of the system is 2.0 J. Find: a) the force constant of the spring b) the amplitude of the motion Solution: a) m = 200 g, T = 0.25s, ࠵? = ! ! ! = ! ! ! . !" = 25 . 1 rad/s ࠵? = ࠵? ࠵? ! = 0 . 2 00 ࠵?࠵? × 25 . 1 !"# ! ! = 126 N/m b) ࠵? = ! ! ! ! = > ࠵? = ! ! ! = ! × ! . ! !"# = 0 . 18 m
This preview has intentionally blurred sections. Sign up to view the full version. | 422 | 1,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-51 | latest | en | 0.749922 |
jericaoblak.com | 1,713,195,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00528.warc.gz | 20,024,728 | 1,931 | # The Augmentation Matrix
#### The Exponential and Recursive Structure of the Harmonic Series
Since the harmonic series consists of frequencies ascending through the integral multiples of the fundamental, there is a clear relationship between the fundamental and its upper partials. If f1 indicates the frequency of the fundamental, then the frequencies of its overtones equal 2f1, 3f1, 4f1, 5f1, etc. If any of these frequencies are substituted by n, therefore 2f1 = n, 3f1 = n, 4f1 = n, or 5f1 = n, etc., it follows that each order of n, 2n, 3n, 4n, 5n, etc. creates a transposition of a given series within the series itself. It means that each harmonic of the harmonic series is a fundamental of a new harmonic series found in the given series through the multiples of n (where n indicates the frequency value of the harmonic or the position of the harmonic in the given series). It also follows that the order of the octaves in the harmonic series is determined by powers of 2 and if n again indicates the frequency value of the harmonic or its position in the series, then its octave repetitions equal 2n, 22n, 23n, 24n, 25n, etc. | 285 | 1,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-18 | latest | en | 0.911552 |
https://stage.geogebra.org/m/peJKazrs | 1,679,409,187,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00713.warc.gz | 624,353,957 | 10,203 | # Example 1
A school tracks the total number of students enrolled each year. The school uses the change in the total number of students to estimate how many students have been enrolled in the school each year since 2000. If is the number of years after 2000, the total number of students, , can be estimated using the function . How is the total number of students changing each year? Is the total number of students growing or decaying?
1. Identify the yearly rate of change in the function.
2. Describe how the rate of change relates to the change of the dependent quantity.
3. Determine whether the dependent quantity is growing or decaying.
This applet is provided by Walch Education as supplemental material for their mathematics programs. Visit www.walch.com for more information. | 163 | 787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-14 | latest | en | 0.922501 |
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## Need help with creating a key index
Hello,
So basically I have a FactSheet containing production data and I need to calculate the total shift time per production line.
Each productionline follows a D (Day) shift and a A (Night) shift. Since the staff occupation is highly flexible shift times are different for each production line.
Following the sample data below on 9 oct 2018 for production line 22 consider the day shift to start at 11:29 and end on 13:30.
Hence I used to create a KeyIndex:
```DateShiftLijnIndex =
CONCATENATE(FactProductie[ProductieDatum];FactProductie[Shift])&FactProductie[Lijn]```
For now all is fine, however when the night shift works over midnight the production date will have 2 values (the startday and the next day). In order to solve this I have to calculate the min date for each DateShiftLijnIndex.
Below I created the measure that I believe will solve the issue, but I need somebody to complete it
```MinDate =
VAR MinDate = MIN(FactProductie[ProductieDatum])
RETURN
CALCULATE(MinDate;
FILTER(VALUES(FactProductie[DateShiftLijnIndex]```
1 ACCEPTED SOLUTION
Accepted Solutions
Member
## Re: Need help with creating a key index
Hi again,
1. You are missing the filter function around the table argument in the MINX function
2. You should not compair the date column to the variable, but the Index column to the variable.
This is how it should be:
VAR index = IndexColumn
RETURN
MINX(
FILTER(
Table,
IndexColumn = index
);
DateColumn
)
Regards,
Kristjan
4 REPLIES 4
Member
## Re: Need help with creating a key index
I am not sure if I understand this well enough, but why not use the start date in your key? But on the other hand you can get the min start date for your key with
```MinDate =
VAR key = FactProductie[DateShiftLijnIndex]
RETURN
MINX(
FILTER(
FactProductie;
FactProductie[DateShiftLijnIndex] = key
);
FactProductie[ProductieDatum] // or the startdate
)```
Regards,
Kristjan
Frequent Visitor
## Re: Need help with creating a key index
Hi @Kristjan76,
Thanks for rapid answer. Agree with your logic, only issue left is that DAX is unable to compare values of date with string. The error considers the use of VALUE or FORMAT. Any clue on how to apply this to the formula?
Member
## Re: Need help with creating a key index
Hi again,
1. You are missing the filter function around the table argument in the MINX function
2. You should not compair the date column to the variable, but the Index column to the variable.
This is how it should be:
VAR index = IndexColumn
RETURN
MINX(
FILTER(
Table,
IndexColumn = index
);
DateColumn
)
Regards,
Kristjan
Frequent Visitor
## Re: Need help with creating a key index
Fantastic it works thanks so much | 690 | 2,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-18 | latest | en | 0.640187 |
http://tailieu.vn/doc/character-animation-with-direct3d-p20-333405.html | 1,518,938,124,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811794.67/warc/CC-MAIN-20180218062032-20180218082032-00035.warc.gz | 338,792,491 | 25,481 | # Character Animation with Direct3D- P20
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## Character Animation with Direct3D- P20
Mô tả tài liệu
Character Animation with Direct3D- P20:This book is primarily aimed at teaching indie and hobby game developers how to create character animation with Direct3D. Also, the seasoned professional game developer may find some interesting things in this book. You will need a solid understanding of the C++ programming language as well as general object-oriented programming skills.
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## Nội dung Text: Character Animation with Direct3D- P20
1. 366 Character Animation with Direct3D //Get position from the four control hairs float3 ch1 = GetHairPos(0, IN.hairIndices[0], IN.hairIndices[1], IN.position.z); float3 ch2 = GetHairPos(1, IN.hairIndices[0], IN.hairIndices[1], IN.position.z); float3 ch3 = GetHairPos(2, IN.hairIndices[0], IN.hairIndices[1], IN.position.z); float3 ch4 = GetHairPos(3, IN.hairIndices[0], IN.hairIndices[1], IN.position.z); //Blend linearly in 2D float3 px1 = ch2 * IN.position.x + ch1 * (1.0f - IN.position.x); float3 px2 = ch3 * IN.position.x + ch4 * (1.0f - IN.position.x); float3 pos = px2 * IN.position.y + px1 * (1.0f - IN.position.y); //Transform to world coordinates float4 posWorld = mul(float4(pos.xyz, 1), matW); OUT.position = mul(posWorld, matVP); //Copy texture coordinates OUT.tex0 = IN.tex0; return OUT; } Here, the exact same operations are done as described earlier in the GetBlended- Point() function of the HairPatch class. The blended position of each of the four control hairs is obtained from the ControlHairTable using the GetHairPos() helper function. Note that the indices passed to the helper function have been pre-computed and stored in the vertex data. Next, the points returned from the control hairs are blended depending on the 2D position of the hair vertex. The resulting point will be in object space, so to get it to the correct position on the screen it is multiplied with the world and the view-projection matrix. As said before, the beauty of going the long way about this is that you now can update the control hairs on the CPU and the mesh in the hair patch will just follow suit “automagically.” Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
2. Chapter 15 Hair Animation 367 EXAMPLE 15.1 In Example 15.1 a single hair patch is created and rendered. The four control hairs are animated with a simple noise function, and the mesh is updated and animated completely on the GPU. C REATING A H AIRCUT So how would you go about creating these 10 to 20 control hairs needed to create a decent-looking haircut for a character? Well, the most obvious way is to enter them manually as was done with the four control hairs in Example 15.1. Although it doesn’t take many minutes to realize that to model a set of 3D lines with a text editor is probably not the way to go. The best way is to use 3D modeling software (Max, Maya, or whatever other flavor you prefer). Figure 15.7 shows an image of a “haircut” created in 3D Studio Max as a set of lines. ease purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
3. 368 Character Animation with Direct3D FIGURE 15.7 Control hairs used to create a haircut. I was lucky enough to know Sami Vanhatalo (Senior Technical Artist at Remedy Entertainment) who was kind enough to write an exporter for 3D Studio Max for me. With this exporter it becomes very easy to dump a set of lines from 3D Studio Max to either a text or a binary file. I won’t cover the exporter here in this book since it is written in Max Script and is out of the scope of this book. However, you’ll find both the text and binary version of the exporter on the accompanying CD-ROM, along with detailed instructions on how to use them. For the next example I’ll use the data that has been outputted from the binary exporter. The data is in Table 15.1. Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
4. Chapter 15 Hair Animation 369 TABLE 15.1 THE BINARY HAIR FORMAT Type Description long File version number long Number of lines long Number of points for Line 1 float X value Line 1, Point 1 float Y value Line 1, Point 1 float Z value Line 1, Point 1 float X value Line 1, Point 2 float Y value Line 1, Point 2 float Z value Line 1, Point 2 … float X value Line N, Point 1 float Y value Line N, Point 1 float Z value Line N, Point 1 The file structure is very simple and doesn’t contain any superfluous information. Feel free to modify the exporter according to your own needs. In any case, to read a set of lines from a binary file with this format, I’ve created the LoadHair() function to the Hair class. The following code segment is an excerpt from this function showing how to read one of these haircut data files: ifstream in(hairFile, ios::binary); if(!in.good()) return; //Version number long version; in.read((char*)&version, sizeof(long)); //Number splines long numSplines = 0; in.read((char*)&numSplines, sizeof(long)); ease purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
5. 370 Character Animation with Direct3D //Read splines for(int i=0; i
6. Chapter 15 Hair Animation 371 FIGURE 15.8 Physical setup of the hair animation. I’ve created a very simple bounding sphere class to be used for the character head representation. The BoundingSphere class is defined as follows: class BoundingSphere { public: BoundingSphere(D3DXVECTOR3 pos, float radius); void Render(); bool ResolveCollision(D3DXVECTOR3 &hairPos, float hairRadius); private: static ID3DXMesh* sm_pSphereMesh; D3DXVECTOR3 m_position; float m_radius; }; Not so surprisingly, this class contains a position and a radius used to define the bounding sphere. The static sm_pSphereMesh and the Render() function is just for debugging and visualization purposes. The most important function in this class is the ResolveCollision() function. This function tests whether a point collides with the bounding sphere, and, if so, it moves the point away from the sphere until the point no longer touches the sphere: ease purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
7. 372 Character Animation with Direct3D bool BoundingSphere::ResolveCollision(D3DXVECTOR3 &hairPos, float hairRadius) { //Difference between control hair point and sphere center D3DXVECTOR3 diff = hairPos - m_position; //Distance between points float dist = D3DXVec3Length(&diff); if(dist < (m_radius + hairRadius)) { //Collision has occurred; move hair away from bounding sphere D3DXVec3Normalize(&diff, &diff); hairPos = m_position + diff * (m_radius + hairRadius); return true; } return false; } Once you have this function up and running and a few spheres placed to roughly represent the character head, you can start simulating in some fashion. To get a hair animation to look good, there’s a whole lot of black magic needed. For instance, you can’t use gravity in the same way you would for other physical simulations. If you used only gravity for simulating the hairs, the result would end up looking like a drenched cat with hair hanging straight down. In reality, haircuts tend to roughly stay in their original shape. To quickly emulate this, I’ve stored the original points of the control hair points as the haircut is created. Then, at run time, I have a small spring force between the current control hair’s position and its original position. This will keep the haircut from deforming completely. To show a quick hair simulation in action, I’ve added the UpdateSimulation() function to the ControlHair class: void ControlHair::UpdateSimulation( float deltaTime, vector &headSpheres) { const float SPRING_STRENGTH = 10.0f; const D3DXVECTOR3 WIND(-0.2f, 0.0f, 0.0f); for(int i=1; i
8. Chapter 15 Hair Animation 373 D3DXVECTOR3 diff = m_originalPoints[i] - m_points[i]; float length = D3DXVec3Length(&diff); D3DXVec3Normalize(&diff, &diff); //Update velocity of hair control point (random wind) float random = rand()%1000 / 1000.0f; D3DXVECTOR3 springForce = diff * length * SPRING_STRENGTH; D3DXVECTOR3 windForce = WIND * prc * random; m_velocities[i] += (springForce + windForce) * deltaTime; //Update position m_points[i] += m_velocities[i] * deltaTime; //Resolve head collisions for(int h=0; h
9. 374 Character Animation with Direct3D class Hair { public: Hair(); ~Hair(); void LoadHair(const char* hairFile); void CreatePatch(int index1, int index2, int index3, int index4); void Update(float deltaTime); void Render(D3DXVECTOR3 &camPos); int GetNumVertices(); int GetNumFaces(); public: vector m_controlHairs; vector m_hairPatches; IDirect3DTexture9* m_pHairTexture; vector m_headSpheres; }; One important thing to mention is that the Render() function in this class sorts all the hair patches in Z depth from the camera before rendering them to the screen. The Update() function takes care of calling the UpdateSimulation() function in the ControlHair objects, providing the m_headSpheres vector for the physical simulation. If you want to extend this example, it would probably be in the Hair class that you would, for example, keep a pointer to the head bone of the character. Using this bone pointer you would then update the position of all the control hairs as the head moves around. Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
10. Chapter 15 Hair Animation 375 EXAMPLE 15.2 In this example a complete haircut is created and simulated in real time. You can control the angle of the camera by clicking and dragging the mouse. By pressing space, you will see the physical representation of the head along with the control hairs as they are animated. ease purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
11. 376 Character Animation with Direct3D Figure 15.9 shows a series of frames taken from the animated hair. FIGURE 15.9 A haircut animated with the system covered in this chapter. C ONCLUSIONS After reading this chapter, you should have a good understanding of the difficulties you’re faced with when trying to implement hair animation. To increase the quality of the hair animation presented in this chapter, there are two obvious improve- ments. First, create a proper physical simulation instead of the “random wind” force I’ve used here. Second, attach the hair to the character head bone and let it inherit motion from the head of the character. When doing this you’ll see some wonderful secondary motion as the character moves and then suddenly stops, etc. Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark.
12. Chapter 15 Hair Animation 377 Another thing I haven’t covered in this chapter is the rendering of hair. There are many advanced models of how light scatters on a hair surface. A really great next stop for you is the thesis titled “Real-Time Hair Simulation and Visualization for Games” by Henrik Halen [Halen07]. You can find this thesis online using the following URL: http://graphics.cs.lth.se/theses/projects/hair/ In the next and final chapter of this book, I’ll put most of the concepts covered throughout this book into a single character class. C HAPTER 15 E XERCISES Create a Hair class supporting multiple levels of details. Scale the number of segments used in hair strips and the number of hair strips used in total. Try to create a longer haircut. (This may require a more detailed physical representation of the character.) Attach the haircut to a moving character head. Implement a triangle hair patch relying on three control hairs instead of four. F URTHER R EADING [Halen07] Halen, Henrik, “Real-Time Hair Simulation and Visualization for Games.” Available online at: http://graphics.cs.lth.se/theses/projects/hair/ report.pdf, 2007. [Jung07] Jung, Yvonne, “Real Time Rendering and Animation of Virtual Characters.” Available online at http://www.ijvr.org/issues/issue4-2007/6.pdf, 2007. [Nguyen05] Nguyen, Hubert, “Hair Animation and Rendering in the Nalu Demo.” Available online at http://http.developer.nvidia.com/GPUGems2/gpugems2_ chapter23.html, 2005. [Ward05] Ward, Kelly, “Modeling Hair Using Levels-of-Detail.” Available online at: http://www.cs.unc.edu/Research/ProjectSummaries/hair05.pdf, 2005. ease purchase PDF Split-Merge on www.verypdf.com to remove this watermark. | 2,967 | 12,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-09 | latest | en | 0.763237 |
http://www.nelson.com/nelson/school/elementary/mathK8/math8/quizzes/math8quizzes/m8ch9l7.htm | 1,521,845,149,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649095.35/warc/CC-MAIN-20180323220107-20180324000107-00205.warc.gz | 451,233,684 | 8,219 | Name: Try It Out -- Chapter 9, Lesson 7
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
1.
Select the division expression that the picture represents.
a. b.
2.
Select the division expression that the picture represents.
a. b.
3.
Calculate the quotient using equivalent fractions.
a. b. 1 c. 24
4.
Calculate the quotient using equivalent fractions.
a. b. 5 c. 4
5.
Bill is cooking his famous ribs. He cooks them slowly over low heat for 3 hours. He checks them every 20 minutes or h. How many times will Bill check the ribs?
a. times b. 10 times c. 10 times
6.
Calculate.
a. b. c. 3
7.
Calculate.
a. 2 c. b. 7 d. 8
8.
Karen takes 19 minutes to type 9 pages. How many pages does she type per minute?
a. 2 pages/minute b. 1 page/minute c. page/minute
9.
Nathan needs to measure 3 cups. How many times must he fill a cup measuring cup?
a. times b. 13 times c. 12 times
10.
Joanne has a roll of ribbon which is 6 m long. She needs to cut pieces that are m long. If she cuts the whole roll how many pieces will she have?
a. 8 c. 7 b. d. | 327 | 1,137 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.80399 |
http://www.instructables.com/id/Charlieplexing-LEDs--The-theory/step6/Some-Practical-matters/ | 1,419,158,850,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802770860.97/warc/CC-MAIN-20141217075250-00096-ip-10-231-17-201.ec2.internal.warc.gz | 589,169,013 | 33,645 | # Charlieplexing LEDs- The theory
## Step 6: Some Practical matters
The magic of charlieplexing relies on the fact the individual voltage presented across multiple LEDs in series will always be less than that across one single LED when the single LED is in parrallel with the series combination. If the voltage is less, then the current is less, and hopefully the current in the series combination will be so low that the LED will not light.
This isn't always the case however.
Lets say you had two red LEDs with a typical forward voltage of 1.9V in your matrix and a blue LED with a forward voltage of 3.5V (say LED1=red, LED3=red, LED5=blue in our 6 LED example). If you lit up the blue LED, you would end up with 3.5/2 = 1.75V for each of the red LEDs. This may be very close to the dim operating area of the LED. You might find the red LEDs will glow dimly when the blue is illuminated.
It is a good idea therefore to make sure the forward voltage of any different coloured LEDs in your matrix are roughly the same at the operating current, or else use the same coloured LEDs in a matrix.
In my Microdot/Minidot projects I didnt have to worry about this, I used high efficiency blue/green SMD LEDs which fortunately have much the same forward voltage as the reds/yellows. However if I implemented the same thing with 5mm LEDs the result would have more problematical. In this case I would have implemented a blue/green charlieplex matrix and a red/yellow matix seperately. I'd have needed to use more pins....but there you go.
Another issue is to look at your current draw from the micro and how bright you want the LED. If you have a big matrix, and are rapidally scanning it, then each LED is on for only a brief time. This it will appear relatively dim compared to a static display. You can cheat by increasing the current through the LED by reducing the current limiting resistors, but only to a point. If you draw too much current from the micro for too long you'll damage the output pins.
If you have a slowly moving matrix, say a status or cyclon display, you could keep the current down to a safe level but still have a bright LED display because each LED is on for a longer time, possibly static (in the case of a status indicator).
Some advantages of charlieplexing:
- uses only a few pins on a microcontroller to control many LEDs
- reduces component count as you don't need lots of driver chips/resistors etc
- your micro firmware will need to handle setting both voltage state and input/output state of the pins
- need to be careful with mixing different colours
- PCB layout is difficult, because the LED matrix is more complex.
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ebeccarayray3 years ago
Thank you so much for posting this! I am working on my electrical engineering senior design project which is going to be a bicycle with a lighting system along with other features. I am looking into Charlieplexing the taillight which is made up of 3 premade LED matrices.
My main question is, what is the actual time that each LED is lit up? I understand that the human eye takes in a new image 25 time per second, so I'm wondering how long should each LED stay lit for.
I'm also concerned that with the LED matrix I'm using, I won't actually be able to wire all of them up in the way that you've shown. This is the part I'm using: http://www.futurlec.com/LED/LEDM88RGCA.shtml | 780 | 3,387 | {"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} | 3.1875 | 3 | CC-MAIN-2014-52 | longest | en | 0.9471 |
https://touringplans.com/blog/disney-dining-plan-will-it-end-up-costing-you-more/ | 1,718,728,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00539.warc.gz | 506,576,407 | 60,157 | Disney Dining Plan: Will It End Up Costing You More?
The Disney Dining Plan is coming back in 2024, and many vacationers are vocally excited about the return of this once-valuable program. But with the new price increases and coverage decreases, what are the chances that you’ll actually save money with the Dining Plan? Are you more likely to come out ahead, or end up paying Disney more for the same food and drink? Yesterday we looked at some high-level math, but today we’re diving into the details! We’ve pulled every single menu item from every possible dining location at Walt Disney World to tell you just how often you can expect to break even with your Dining Plan credits … and how often Disney will come out ahead.
How Much is a Dining Plan Credit Worth?
In order to figure out the various menu combinations and have something to compare them to, we need to know not just what you’re paying for day of the dining plan per person – we need to know exactly how much each dining credit is worth. And to do that, we’re going to make two quick assumptions up-front:
1. Snacks are worth an average of \$6
2. The refillable mug is treated as a free throw-in
Both of these assumptions could be a little shaky – maybe you like expensive snacks, or maybe you just go healthy with a piece of fruit every time. And technically a refillable mug is worth \$21.99. But you could use it 40 times a day (and probably be miserable), or you could use it not at all. And that throws off our math. So we’re calling it a nice freebie.
With those two assumptions, we can use some middle school algebra to figure out the value of an adult Quick Service dining credit and an adult Table Service credit.
You can grade your own answers, because here is the answer key: a Quick Service credit is worth \$25.50 (but we’ll call it \$25 to be nice), and that means a Table Service credit is worth \$63.
Those values tell us exactly how much our menu combinations need to cost in order to break even or save us money when using a dining plan credit!
What’s Included in a Credit?
A Quick Service credit only includes a drink and an entrée. So those will be easy math – we can just pick beverages and combine them with entrees.
A Table Service credit is a little more tricky – because with those you get a drink and an entrée and a dessert. Or sometimes just a buffet or prix fixe meal.
That all means if I’m going to figure out how many meals will save you money on the Disney Dining plan, I need to know some average beverage and dessert costs! That’s what we’ll calculate next.
Average Beverage and Dessert Costs
Quick Service Beverage Cost
After going through every beverage available at every quick service location on the Disney Dining Plan, here are the average costs, by beverage type:
• Cocktail = \$13.00 (need a \$12 entrée to break even)
• Wine = \$11.00 (need a \$14 entrée to break even)
• Beer = \$9.50 (need a \$15.50 entrée to break even)
• Specialty Beverage = \$4.50 (need a \$20.50 entrée to break even)
• Soda = \$4.50 (need a \$20.50 entrée to break even)
• Water = Free! (need a \$25 entrée to break even)
Table Service Dessert Cost
When averaging the cost of every possible ala carte dessert at Table Service locations on the Disney Dining Plan, the average cost is \$10.50. So for Table Service credits that aren’t prix fixe or buffets, we’ll take this into consideration.
Table Service Beverage Cost
After going through every beverage available at every quick service location on the Disney Dining Plan, here are the average costs, by beverage type:
• Cocktail = \$15.50 (need a \$37 entrée or \$47.50 prix fixe to break even)
• Wine = \$14.50 (need a \$38 entrée or \$48.50 prix fixe to break even)
• Beer = \$9.50 (need a \$44 entrée or \$54.50 prix fixe to break even)
• Specialty Beverage = \$7.00 (need a \$45.50 entrée or \$56 prix fixe to break even)
• Soda = \$4.50 (need a \$48 entrée or \$58.50 prix fixe to break even)
• Water = Free! (need a \$52.50 entrée or \$63 prix fixe to break even)
How Often Will You Save Money on Quick Service with the Disney Dining Plan?
Now that we have our “win conditions”, we can see how often we’ll win the Dining Plan game vs how often Disney will come out ahead!
Quick Service Breakfast
• If you manage to find a quick-service breakfast cocktail with your meal, 51 out of 258 potential breakfast entrees will let you break even. That’s a 20% chance of saving money.
• If you’re fancy and prefer wine with your breakfast, only three quick service breakfast entrees will be expensive enough to save you money by using a dining credit. That’s a 1% chance of saving money.
• Let’s say you’re someone who enjoys breakfast beer. Who am I to judge? In that case, two entrees are expensive enough for you. That’s a 1% chance of saving money.
• If you’d like a breakfast specialty beverage (like a coffee or juice or milk) or soda, 1 lonely breakfast entrée can earn you money on the dining plan. That’s rounded to a 0% chance of saving money.
• And if you’d just like some tap water with your breakfast, there are no quick service breakfast entrees worth \$25 that will allow you to break even. Once again, a 0% chance of saving money.
Quick Service Lunch or Dinner
• If you enjoy a quick-service cocktail with your lunch or dinner, 612 out of 1311 potential lunch or dinner entrees will let you break even by using a dining plan credit. That’s not too bad! It’s a 47% chance of saving money.
• But things start to go downhill if you order wine instead, when only 242 quick service lunch or dinner entrees will be expensive enough to save you money by using a dining credit. That’s an 18% chance of saving money.
• Maybe you think you’d rather drink a beer. In that scenario, 122 entrees are expensive enough for you. That’s a 9% chance of saving money.
• If you’d like a specialty beverage (like a coffee or juice or milk) or soda with your lunch or dinner, only 7 lunch and dinner quick service entrees across all of Walt Disney World can earn you money on the dining plan. That’s a 1% chance of saving money.
• And if you’d just like some tap water with your meal, to hydrate in the middle of a hot park day, there are 5 quick service lunch or dinner entrees worth the \$25 that will allow you to break even. That rounds back down to a 0% chance of saving money.
How Often Will You Save Money on Table Service with the Disney Dining Plan?
Things get a little trickier with table service credits. First, we’re going to just go ahead and say upfront that no two-credit meals are ever “worth it”. You’re not going to spend \$126 on a single drink/entrée/dessert combo, even at the fanciest restaurants on the dining plan.
Next, in each of these scenarios, trust that I’m counting prix fixe options, buffets, and entrée/dessert combos in the numerator and the denominator, and I’ll just call them meal choices.
Table Service Breakfast
• If you enjoy a lovely table-service breakfast cocktail with your meal, 3 out of 310 potential breakfast meal options will let you break even. That’s a 1% chance of saving money. Out of the gate with 1% success. Don’t do table service breakfast on the dining plan please.
• If you’re fancy and prefer wine with your breakfast, those same three table service breakfast meal options will still be expensive enough to save you money by using a dining credit. That’s a 1% chance of saving money.
• If you “downgrade” to breakfast beer, two meal options are expensive enough for you. That’s a 1% chance of saving money.
• If you’d like a breakfast specialty beverage (like a coffee or juice or milk) or soda, 1 lonely table service meal options can earn you money on the dining plan. That’s rounded to a 0% chance of saving money.
• And if you’d just like some tap water with your breakfast, there are no table service meal options worth \$63 that will allow you to break even. Once again, a 0% chance of saving money.
Table Service Lunch or Dinner
So Table Service breakfast is a no-go on the Disney Dining Plan. But maybe
• If you order a table-service cocktail with your lunch or dinner, 124 out of 1596 potential lunch or dinner meal options will let you break even by using a dining plan credit. Not nearly as good as our quick-service numbers It’s a 8% chance of saving money.
• And things get worse from there – if you order wine with your lunch or dinner, then only 109 table service lunch or dinner meal options will be expensive enough to save you money by using a dining credit. That’s a 7% chance of saving money.
• If you’d rather have a beer with your lunch or dinner, 41 meal options are expensive enough for you. That’s a 3% chance of saving money.
• If you’d like a specialty beverage (like a coffee or juice or milk) or soda with your lunch or dinner, only 23 or 17 (respectively) lunch and dinner table service meal options can earn you money on the dining plan. That’s a 1% chance of saving money.
• And, finally, if you’d just like some tap water with your relaxing table service lunch or dinner, there are a whopping 9 meal options worth the \$63 that will allow you to break even. That rounds to a 1% chance of saving money.
Disney Dining Plan Summary
Let’s put all of those numbers in one easy spot to refer to as we draw our conclusions.
1. None of these numbers is ever above 50%. You’ve probably heard the gambling term “the house always wins” – and this is the Disney Dining Plan equivalent. The Mouse House always wins. If you just order randomly off of every menu in Walt Disney World, you will statistically lose money by paying for the Dining Plan instead of paying out of pocket. Can you come out ahead in Vegas? Sure, without too much alcohol, and with a lot of skill. Can you come out ahead in Disney? Sure, with a lot of alcohol, and with a lot of skill.
2. The population that really really can almost never come out ahead are non-drinkers and anyone between the ages of 10 and 20. If your kids are over the age of 9, and you opt to purchase the Dining Plan for your room, they’re only going to be working from the bottom half of this adult credit chart. Where, at best, they’ve got 1% of menu options to choose from if you don’t want to lose money on their plan. If you limit them to only choosing things that save money on the Dining Plan, they have 1 quick-service breakfast and 1 table-service breakfast to choose from for the entire trip. Or 7 quick-service lunch/dinner options and 23 table-service lunch/dinner options. That’s across all parks, resorts, and Disney Springs.
3. If you really really want to use the Dining Plan because you like pre-paying for your food or having that all-inclusive “feel” – the Quick Service Dining Plan is your best option, and you should only be using those credits for lunch and dinner. Plus you should be ordering cocktails at every meal. Then, across property, you’ll be able to select from about half of the available dining options and still make money on your plan.
4. Keep in mind that with the dining plan, you still have to pay gratuity out of pocket at every meal. And you’ll be ordering those expensive drinks and meals, which means higher gratuity than you might otherwise pay without the Dining Plan.
5. One of the advertised benefits of the Dining Plan is getting to eat whatever you want whenever you want and not worrying about the cost. But if you put in the work to select those few dining locations where the odds are in your favor – especially at table service locations – you’ll be competing for reservations with everyone else that thinks the same way you do. That 6 am reservation opening window is going to be very important for you.
Have you priced out the Disney Dining Plan for any of your past or upcoming vacations? Is the convenience worth the potential additional cost? Or do you look to maximize your money while at Disney? Let us know in the comments!
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Becky Gandillon
Becky Gandillon was trained in biomedical engineering, but is now a full-time data and analytics nerd. She loves problem solving and travelling. She and her husband, Jeff, live in St. Louis with their two daughters and they have Disney family movie night every Saturday. You can follow her on LinkedIn: https://www.linkedin.com/in/becky-gandillon/ or instagram @raisingminniemes
8 thoughts on “Disney Dining Plan: Will It End Up Costing You More?”
• Thanks for such a great analysis. The dining plan seems like a tremendous amount of planning with very little chance for savings. Probably a better move to focus on “hacking the system” to purchase discounted Disney gift cards and using those to pay for dining. Too bad the Deluxe Dining Plan is not back. It was easier to extract value from that plan.
Thanks for the great analysis!
Isabel
• Fantastic analysis. I realize you were trying to be positive about how you could save money. But the flipside is you will lose money if you don’t go to those very limited choices. Basically if you want easy but pay more then get the dining plan. I liked your suggestion you put elsewhere about putting that money into gift cards. Then it is prepaid. You can order appetizers and what you want. And pay tip and buy yourself a gift at the end with the leftover.
• I was wondering if this math takes into account any price increases in menus over the coming months. How often does Disney raise menu prices? Would it be worth it to buy a QS dining plan now for a trip planned in summer 2024?
• Unfortunately, I’m not a future-seer regarding price increases. Menu prices do historically increase sometime between October and the end of each calendar year. In October 2022, we saw hundreds of price increases, but almost all of the snack and quick service increases were something like 30 cents per item. At one-credit table service locations, I believe the _biggest_ entree price increase we saw was \$2. We saw bigger price increases at what are now two-credit dining locations.
Even if we see similar price increases again this year (I’m sure prices will increase, but inflation has cooled down, so hopefully they won’t increase by as much), the percentages that are presented in this article wont change significantly.
• I hope everyone that reads this thinking, “Well, I’m just going to eat at restaurants that have the most expensive food and cocktails to maximize the DDP!” reads number 5 on the summary. Even if they do manage to get all the TS restaurants they want, both plans use QS restaurants, and everyone trying to go to those will find long lines and/or chaotic return times for mobile ordering.
• When my daughters were 6 & 4 back in 2008, we did the dining plan. At that point, it included 3 sit down meals.
We did it because the girls were more interested in characters and they couldn’t ride a lot of the big kid rides.
It was a ton of food, but it worked because we did a ton of character meals, and had some nice dinners as well.
But they were kids, so the math worked.
This is the only way I would ever do it. Loading up on character meals with kids. I don’t think it makes sense any other way. | 3,429 | 15,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.941016 |
https://community.deeplearning.ai/t/doubt-sequence-models-w2-lecture/283642 | 1,708,555,613,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00563.warc.gz | 189,075,212 | 6,156 | Doubt - Sequence Models W2 lecture
In Sequence models course, “Embedding Matrix” video at time 1:10 Andrew creates 1 hot vector O6257 of size (10k, 1).
In Matrix E, O6257 was orange word and it’s size is (300, 1) so how it went from (300, 1) to (10k, 1)?
Thanks in advance Mitesh
Hello @Mitesh!
We appreciate your post, but it is in the incorrect category. To ensure that your question receives a prompt and accurate response, I suggest that you edit your post by clicking the pencil icon located next to the title and move it to the relevant course category (as shown in attached image). Our mentors will be happy to assist you. In this case, it seems to be in DLS 5 category.
Best,
Saif.
Oh Sorry and Thanks.
I believe you’re referring to this slide from the Week 2 lecture.
The operator I’ve pointed to with the arrow is a dot product.
I know that is a dot product.
But I am confused about the term O6257 in E (Dot Product) O6257.
If you see in slide, on the right side Andrew has drawn vector O6257 with size (10000, 1) But in Matrix E, if you see Orange 6257 is of length 300 only.
My thinking is that if Andrew has drawn O6257 from Matrix E then it should be of size (300,1). | 317 | 1,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.926668 |
https://brainly.com/question/271732 | 1,485,263,997,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284411.66/warc/CC-MAIN-20170116095124-00462-ip-10-171-10-70.ec2.internal.warc.gz | 806,236,681 | 8,431 | • 047
• Ambitious
2015-01-26T03:15:51-05:00
2x = y - 4
or y = 2x + 4
for x=0, y = 2×0 + 4 = 4
for x=1, y = 2×1 + 4 = 6
for x=3, y = 2×3 + 4 = 10
for x=5, y = 2×5 + 4 = 14
for x=-1, y = 2×(-1) + 4 = 2
for x=-3, y = 2×(-3) + 4 = -2
for x=-5, y = 2×(-5) + 4 = -6
Plot the points and join with a line. See the attachment. | 189 | 318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-04 | latest | en | 0.670195 |
https://sciencedocbox.com/Physics/86040082-Minimum-spanning-trees.html | 1,628,102,518,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154897.82/warc/CC-MAIN-20210804174229-20210804204229-00160.warc.gz | 471,931,506 | 28,851 | # Minimum Spanning Trees
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Transcription
1 Minimum Spnning Trs
2 Minimum Spnning Trs Problm A town hs st of houss nd st of rods A rod conncts nd only houss A rod conncting houss u nd v hs rpir cost w(u, v) Gol: Rpir nough (nd no mor) rods such tht: 1. Evryon stys connctd: cn rch vry hous from ll othr houss, nd. Totl rpir cost is minimum 11 b c d i 6 1 h g f 1
3 Minimum Spnning Trs A connctd, undirctd grph: Vrtics = houss, Edgs = rods A wight w(u, v) on ch dg (u, v) E Find T E such tht: 1. T conncts ll vrtics. w(t) = Σ (u,v) T w(u, v) is minimizd b c d 11 i 1 h 1 g f 3
4 Minimum Spnning Trs T forms tr = spnning tr A spnning tr whos wight is minimum ovr ll spnning trs is clld minimum spnning tr, or MST. b c d 11 i 1 g 1 g f
5 Proprtis of Minimum Spnning Trs Minimum spnning trs r not uniqu Cn rplc (b, c) with (, h) to obtin diffrnt spnning tr with th sm cost MST hv no cycls W cn tk out n dg of cycl, nd still hv th vrtics connctd whil rducing th cost b c d 11 i 1 h 1 g f # of dgs in MST: V - 1 5
6 Growing MST Minimum-spnning-tr problm: find MST for connctd, undirctd grph, with wight function ssocitd with its dgs A gnric solution: Build st A of dgs (initilly mpty) Incrmntlly dd dgs to A such tht thy would blong to MST An dg (u, v) is sf for A if nd only if A {(u, v)} is lso subst of som MST 11 b c d i 6 1 h g f 1 W will dd only sf dgs 6
7 Gnric MST lgorithm 1. A. whil A is not spnning tr 3. do find n dg (u, v) tht is sf for A. A A {(u, v)} 5. rturn A How do w find sf dgs? b c d 11 i 1 h 1 g f
8 Finding Sf Edgs Lt s look t dg (h, g) Is it sf for A initilly? Ltr on: S b c d 11 i 1 h 1 g f V - S Lt S V b ny st of vrtics tht includs h but not g (so tht g is in V - S) In ny MST, thr hs to b on dg (t lst) tht conncts S with V - S Why not choos th dg with minimum wight (h,g)?
9 Dfinitions A cut (S, V - S) is prtition of vrtics into disjoint sts S nd V - S An dg crosss th cut b c d S 11 i 1 S V- S V- S h g f 1 (S, V - S) if on ndpoint is in S nd th othr in V S A cut rspcts st A of dgs no dg in A crosss th cut An dg is light dg crossing cut its wight is minimum ovr ll dgs crossing th cut For givn cut, thr cn b > 1 light dg crossing it
10 Thorm Lt A b subst of som MST, (S, V - S) b cut tht rspcts A, nd (u, v) b light dg crossing (S, V - S). Thn (u, v) is sf for A. Proof: Lt T b MST tht includs A Edgs in A r shdd Assum T dos not includ th dg (u, v) Id: construct nothr MST T tht includs A {(u, v)} u v S V - S
11 Thorm - Proof T contins uniqu pth p btwn u nd v (u, v) forms cycl with dgs on p S (u, v) crosss th cut pth p x must cross th cut (S, V - S) t lst onc: lt (x, y) b tht dg Lt s rmov (x, y) brks T into two componnts. u v p V - S y Adding (u, v) rconncts th componnts T = T - {(x, y)} {(u, v)} 11
12 Thorm Proof (cont.) T = T - {(x, y)} {(u, v)} Hv to show tht T is MST: S (u, v) is light dg w(u, v) w(x, y) w(t ) = w(t) - w(x, y) + w(u, v) w(t) Sinc T is spnning tr u v p x V - S y w(t) w(t ) T must b n MST s wll 1
13 Thorm Proof (cont.) Nd to show tht (u, v) is sf for A: i.., (u, v) cn b prt of MST A T nd (x, y) A A T S x A {(u, v)} T Sinc T is n MST u p y (u, v) is sf for A v V - S 13
14 Discussion In GENERIC-MST: A is forst contining connctd componnts Initilly, ch componnt is singl vrtx Any sf dg mrgs two of ths componnts into on Ech componnt is tr Sinc n MST hs xctly V - 1 dgs - ftr itrting V - 1 tims, w hv only on componnt 1
15 Th Algorithm of Kruskl Strt with ch vrtx bing its own componnt Rptdly mrg two componnts into on by choosing th light dg tht conncts thm 11 b c d i 6 1 h g f 1 W would dd dg (c, f) Scn th st of dgs in monotoniclly incrsing ordr by wight Uss disjoint-st dt structur to dtrmin whthr n dg conncts vrtics in diffrnt componnts 15
16 Oprtions on Disjoint Dt Sts MAKE-SET(u) crts nw st whos only mmbr is u FIND-SET(u) rturns rprsnttiv lmnt from th st tht contins u My b ny of th lmnts of th st tht hs prticulr proprty E.g.: S u = {r, s, t, u}, th proprty is tht th lmnt b th first on lphbticlly FIND-SET(u) = r FIND-SET(s) = r FIND-SET hs to rturn th sm vlu for givn st 16
17 Oprtions on Disjoint Dt Sts UNION(u, v) units th dynmic sts tht contin u nd v, sy S u nd S v E.g.: S u = {r, s, t, u}, S v = {v, x, y} UNION (u, v) = {r, s, t, u, v, x, y} 1
18 KRUSKAL(V, E, w) 1. A. for ch vrtx v V 3. do MAKE-SET(v). sort E into non-dcrsing ordr by wight w 5. for ch (u, v) tkn from th sortd list 6. do if FIND-SET(u) FIND-SET(v). thn A A {(u, v)}. UNION(u, v). rturn A Running tim: O(E lgv) dpndnt on th implmnttion of th disjoint-st dt structur 1
19 Exmpl 11 1: (h, g) b c d : (c, i), (g, f) : (, b), (c, f) 6: (i, g) : (c, d), (i, h) i 6 1 h g f 1 : (, h), (b, c) : (d, ) : (, f) 11: (b, h) 1: (d, f) {}, {b}, {c}, {d}, {}, {f}, {g}, {h}, {i} 1. Add (h, g). Add (c, i) 3. Add (g, f). Add (, b) 5. Add (c, f) 6. Ignor (i, g). Add (c, d). Ignor (i, h). Add (, h). Ignor (b, c) 11. Add (d, ) 1. Ignor (, f) 13. Ignor (b, h) 1. Ignor (d, f) {g, h}, {}, {b}, {c}, {d}, {}, {f}, {i} {g, h}, {c, i}, {}, {b}, {d}, {}, {f} {g, h, f}, {c, i}, {}, {b}, {d}, {} {g, h, f}, {c, i}, {, b}, {d}, {} {g, h, f, c, i}, {, b}, {d}, {} {g, h, f, c, i}, {, b}, {d}, {} {g, h, f, c, i, d}, {, b}, {} {g, h, f, c, i, d}, {, b}, {} {g, h, f, c, i, d,, b}, {} {g, h, f, c, i, d,, b}, {} {g, h, f, c, i, d,, b, } {g, h, f, c, i, d,, b, } {g, h, f, c, i, d,, b, } {g, h, f, c, i, d,, b, } 1
20 Th lgorithm of Prim Th dgs in st A lwys form singl tr Strts from n rbitrry root : V A = {} At ch stp: Find light dg crossing cut (V A, V - V A ) Add this dg to A Rpt until th tr spns ll vrtics Grdy strtgy b c d 11 i 1 h 1 g f At ch stp th dg ddd contributs th minimum mount possibl to th wight of th tr 0
21 How to Find Light Edgs Quickly? Us priority quu Q: Contins ll vrtics not yt includd in th tr (V V A ) V = {}, Q = {b, c, d,, f, g, h, i} With ch vrtx w ssocit ky: b c d 11 i 1 h g f 1 ky[v] = minimum wight of ny dg (u, v) conncting v to vrtx in th tr Ky of v is if v is not djcnt to ny vrtics in V A Aftr dding nw nod to V A w updt th wights of ll th nods djcnt to it W ddd nod ky[b] =, ky[h] = 1
22 PRIM(V, E, w, r) 1. Q b c d. for ch u V 0 3. do ky[u] 11 i 1. π[u] NIL 5. INSERT(Q, u) h g f 1 6. DECREASE-KEY(Q, r, 0) 0. whil Q Q = {, b, c, d,, f, g, h, i}. do u EXTRACT-MIN(Q) V A =. for ch v Adj[u] Extrct-MIN(Q). do if v Q nd w(u, v) < ky[v] 11. thn π[v] u 1. DECREASE-KEY(Q, v, w(u, v))
23 Exmpl b c d 11 i 1 h g f 1 0 Q = {, b, c, d,, f, g, h, i} V A = Extrct-MIN(Q) b c d 11 i 1 h g f 1 ky [b] = π [b] = ky [h] = π [h] = Q = {b, c, d,, f, g, h, i} V A = {} Extrct-MIN(Q) b 3
24 Exmpl b c d 11 i 1 h g f 1 b c d 11 i 1 h g f 1 ky [c] = π [c] = b ky [h] = π [h] = - unchngd Q = {c, d,, f, g, h, i} V A = {, b} Extrct-MIN(Q) c ky [d] = π [d] = c ky [f] = π [f] = c ky [i] = π [i] = c Q = {d,, f, g, h, i} V A = {, b, c} Extrct-MIN(Q) i
25 Exmpl b c d 11 i 1 h g f 1 b c d 11 i 1 h 1 g f ky [h] = π [h] = i ky [g] = 6 π [g] = i Q = {d,, f, g, h} V A = {, b, c, i} Extrct-MIN(Q) f ky [g] = π [g] = f ky [d] = π [d] = c unchngd ky [] = π [] = f Q = {d,, g, h} V A = {, b, c, i, f} Extrct-MIN(Q) g 5
26 Exmpl b c d 11 i 1 h 1 g f 1 b c d 11 i 1 h 1 g f 1 ky [h] = 1 π [h] = g 1 Q = {d,, h} V A = {, b, c, i, f, g} Extrct-MIN(Q) h Q = {d, } V A = {, b, c, i, f, g, h} Extrct-MIN(Q) d 6
27 Exmpl b c d 11 i 1 h 1 g f 1 ky [] = π [] = f Q = {} V A = {, b, c, i, f, g, h, d} Extrct-MIN(Q) Q = V A = {, b, c, i, f, g, h, d, }
28 PRIM(V, E, w, r) 1. Q. for ch u V 3. do ky[u]. π[u] NIL 5. INSERT(Q, u) 6. DECREASE-KEY(Q, r, 0) ky[r] 0. whil Q. do u EXTRACT-MIN(Q). for ch v Adj[u]. do if v Q nd w(u, v) < ky[v] 11. thn π[v] u Totl tim: O(VlgV + ElgV) = O(ElgV) O(V) if Q is implmntd s min-hp Excutd V tims Tks O(lgV) Excutd O(E) tims Constnt 1. DECREASE-KEY(Q, v, w(u, v)) Min-hp oprtions: O(VlgV) Tks O(lgV) O(ElgV)
### Minimum Spanning Trees
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Wintr 2016 COMP-250: Introduction to Computr Scinc Lctur 23, April 5, 2016 Commnt out input siz 2) Writ ny lgorithm tht runs in tim Θ(n 2 log 2 n) in wors cs. Explin why this is its running tim. I don | 13,176 | 35,947 | {"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} | 3.234375 | 3 | CC-MAIN-2021-31 | latest | en | 0.772841 |
https://www.crapspit.org/how-dealers-keep-track-of-everyones-bets/ | 1,723,277,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00476.warc.gz | 579,778,779 | 42,049 | # How Dealers Keep Track of Everyone’s Bets
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When the table gets hot and filled with 8 players squeezed in at each end, the point boxes on the layout are typically full of chips that appear at first glance to be scattered everywhere at the craps table. It looks like chaos to the untrained eye. How in the world do the dealers keep track of everyone’s bets? It’s actually a very well organized and simple process based on each player’s table position (i.e., the physical location where each player places his chips in the chip rack). In most cases, the player stands directly behind his chips. However, you might find a rare occasion, for example, where a drunk is at an empty table with his chips in the chip rack next to the stickman but he has staggered to the end of the table to throw the dice. In this unusual case, the physical location of his chip stack doesn’t correspond to the physical location of where he stands. So to keep it simple, let’s just say the process that the dealers use to track players’ chips is based on where each player stands at the table.
This concept of positioning players’ chips in and around the point boxes based on where the players stand at the craps table applies to Come bets, Don’t Come bets, Buy bets, and Place bets. Let’s look specifically at the Place bet. The illustration below shows the eight player positions at each end of the table. Position #1 is beside the stickman, and position #8 is at the end nearest the dealer. Note in the illustration that each point box as two rectangles for Place bets: (1) one rectangle at the top of the point box, and (2) one rectangle at the bottom of the point box. In the illustration below, we use the 4 point box to identify the location within each Place rectangle that corresponds to the player positions around the table. Looking at the 4 point box, we can see that players in positions #1 through #4 have their Place bets positioned in the Place rectangle that’s at the bottom of the point box, and the players in positions #5 through #8 have their Place bets positioned in the Place rectangle that’s at the top of the point box. Those relative positions as shown in the 4 point box are the same for all the other point boxes (i.e., the 5, 6, 8, 9, and 10). For example, the player standing in table position #1 (i.e., next to the stickman) always has his Place bets, regardless of which number he’s Place betting, positioned in the bottom Place rectangle at the far-left side of the rectangle. Let’s take a closer look at the illustration below for clarification. Look at the point boxes for the 5, 6, 8, 9, and 10, and notice the chips in the Place rectangles at the top and bottom of the point boxes. Let’s see if you’re paying attention…if the chips for the player in table position #1 are always positioned in the bottom Place rectangle at the far-left side, how many Place bets does player #1 have in the illustration below, and on what numbers are those Place bets? Very good! We see in the illustration below that the 9 and 10 have chips at the far-left sides of the bottom Place rectangles. Therefore, in this example, we can quickly see at a glance that the player in table position #1 has two Place bets, and they are on the 9 and 10.
Looking at the illustration below, we see that the point is 4 because of the white “ON” puck, and we see the following Place bets.
• Place bets for player #1 are on the 9 and 10. We know this because only the 9 and 10 have chips at the far-left sides of the bottom Place rectangles.
• Place bets for player #2 are on the 6 and 8. The 6 and 8 are the only numbers that have chips in the #2 position of the bottom Place rectangles.
• Place bets for player #3 are on the 5 and 8. The 5 and 8 are the only numbers that have chips in the #3 position of the bottom Place rectangles.
• Place bets for player #4 are on the 5 and 10. The 5 and 10 are the only numbers that have chips in the #4 position (i.e., the far-right sides) of the bottom Place rectangles.
• Place bets for player #5 are on the 5, 6, and 8. The 5, 6, and 8 are the only numbers that have chips in the #5 position (i.e., the far-right sides) of the top Place rectangle.
• There are no Place bets for player #6 because no chips are positioned in the #6 position of in the top Place rectangle for any of the numbers.
• The Place bet for player #7 is on the 9. The 9 is the only number that has chips in the #7 position of the top Place rectangle.
What about player #8? Does she have any Place bets? As shown in the 4 point box in the illustration, if player #8 had any Place bets, they’d be positioned at the far-left side of the top Place rectangle. We see in the illustration above that there are no chips for any number positioned at the far-left sides of the top Place rectangles. Therefore, in the illustration, player #8 does not have any Place bets. Or does she? That was a trick question to see if you’ve been reading and studying our other articles. Yes, player #8 does, indeed, have a Place bet, and it’s on the point number, which is 4 (we know the point is 4 because of the “ON” puck, as shown in the illustration). Player #8’s Place bet on the point is positioned on the back line of the Pass Line directly in front of where she stands. Also notice how the \$5 chip straddles the back line. That is, the chip is not in the apron and it is not inside the Pass Line. By perfectly straddling the back line, the dealers know her chip is a Place bet on the point number instead of a Pass Line bet. So, if you want to Place bet the point, you can make the bet yourself without the dealer’s help by positioning your chip(s) so it straddles the back line. Note that this can be done only when Placing the point. Place bets on any non-point number must be positioned by the dealer in the Place rectangles adjacent to the point boxes.
The way dealers pay off winning bets is just as organized and simple. The dealers follow a strict sequence for paying off bets and players. They never deviate from the sequence. The process is strictly adhered to because it helps the boxman and pit boss track payouts to ensure the right people are paid the right amounts. Pass Line bets are paid first, then all Come bets are paid, and then all Place bets are paid. Place bets are paid in sequence according to the players’ table positions, starting with the player in position #1 (next to the stickman) and ending with the player in position #8 (nearest the dealer).
Bets for the center section of the layout (e.g., the Hardways, etc.) are positioned in their little rectangles similarly to Place bets in that the chip positions inside the rectangle identify which bets belong to whom.
The next time you walk up to a crowded table and see a sea of chips scattered across the layout, be assured that the dealers (and the experienced players at the table) know exactly who owns every single chip on the table. It looks like chaos, but it is, in fact, the exact opposite. Check out some reliable online casinos such as Sunpalace, Casino Max, or slotsplus. | 1,674 | 7,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.934695 |
https://forum.allaboutcircuits.com/threads/learning-project-pwm.6266/ | 1,709,173,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474775.80/warc/CC-MAIN-20240229003536-20240229033536-00811.warc.gz | 251,555,371 | 27,554 | # Learning - project pwm
#### MrWoo
Joined Jun 18, 2007
11
Hi.
I like to learn all sorts of things, and the latest is some electric theory. I have decided to learn using a rather common project, a fan controller for cpu 12v dc fans.
At this point, I have poured over sites and schematics and datasheets looking at examples, as well as scads of tutorials and the like to learn a bit. I feel I have an OK understanding of linear models as well as pulsed analog ones. I am using multisim 10 for now until I graduate to a "non-frying of components" level.
For the basic linear voltage circuit, I am going to pick a bit of a challenge. Designing circuits to handle a 12v source off a psu has been done. There is plenty of current available and only the control circuit needs to handle the heat generated by the load on the fan. So, I choose to control 3 fans with a motherboard 3pin fan header. Research is rather sketchy on the actual max load the motherboard headers can handle. I have seen an average of 6 watts per all headers, as well as 3watts per header, as well as some documentation stating that a single header is capable of up to 6watts.
So, not knowing what the actual upper threshold of current is for a motherboard fan header, erring on side of caution. In a linear voltage output from a mobo header, I have been playing with a very simple circuit that uses a pair of transistors, specifically a 2n3904 and TIP32a. Simply, 12V source is applied to collector of TIP32a. 2N3904 collector to TIP32a base. 2N3904 base to V from fan header. Both transistors emitters go to the fan V in. This effectively opening the TIP32a linearly, with the slight voltage drop, while isolating the acutual load from the fan header.
This is fairly straight forward, as one can calculate the load on both sides. However, there is also the trend to include PWM circuits into the fan headers. Research has shown both low and high side circuits. The low side is generally what is seen for the DIY PWM circuit. Understanding that a transistor or FET is being induced to allow the voltage from the fan to ground, at a rate, is not too hard to grasp. The reverse where the voltage is actually modulated is also not too terribly hard to grasp.
But here is where most of the grasping really starts.
For instance, on my motherboard, the Intel 975xbx2, I have the Andigilog aSC7621 chip onboard. This particular chip has 4 fan outputs (PWM). All 4 are capable of utilizing the newer 4pin configuration, where the 12v is fed continuous, and the fan itself uses the 4th pin, the SIGNAL, to modulate the duty cycle. The problem here is that while in the datasheets, the example shcematic tells of simply using an external MOSFET to convert to 3pin configuration, it does not state implicitly that it is a high or low side circuit. I am assuming low side, but it is hard to tell, because there is no schematic. The PWM signal, as stated in the datasheet is up to 5.5v. To make matters worse, or better, the frequency of the PWM is adjustable. Using SpeedFan, one can have a fan header @ 10Hz, up to 30kHz. While I find that 10Hz can run my 120mm 12v .30A fan down at 5% duty cycle, there is a definiate rotor click, as well as LED fan "blink". Tuning the frequency to 23kHz fixes the rotor click and the LED "blink", but forces the fan to stall at 5%. 10% duty and it starts and maintains.
Now, heh, do a voltage reading. On the 4pin header, reading the 12v is solid 12v, no matter the duty. Reading the SIGNAL, @10Hz, with neg lead on the case, and there are from 32 to 0 mV. Read the same SIGNAL @23kHz, and it is 0 to 180 mV. Now also read the 3pin voltage. Here it is a bit different, @10Hz, jumping around like crazy, @23kHz, solid 12v. Now put pos lead on pos 3pin, neg lead on ground of 3pin, and you will see @10Hz jumping around, @23kHz linear voltage looking, 12 - 0v.
Assuming that Intel has already used a MOSFET in converting the PWM Signal to a 3pin configuration, how does one tell if it a high or low side PWM signal? Couple that to this little issue, that the 4pin header is (assuming data sheet is correct) only @5.5v max, while the 3pin is @12v.
Ah, more learning. I have some circuits now that use an OP AMP to drive a 12v fan using a 5v feed. There is a MOSFET in it, a BUZ71L. I have tried other MOSFET's as well as transistors and found lot's of working designs.
But we have not only a voltage difference, but also a possible method of PWM implementation. Assume we deal with the 12v 3pin header. It is pulsing the voltage or the ground path. The modulation is already there. A quick changing MOSFET should mimic this. I have that working in sim now. NOTE: Please bear in mind that my project goal is to LEARN, in this case, how to isolate the actual load to 3 fans and protect the origin from overload, while maintaining the source "speed". With that in mind, ahem, the MOSFET chosen should handle the load of 3 fans, arbitrarily let's say a 3A model. So, the PWM signal is put to a MOSFET, high or low side, and it replicates the gate signal, pulse on @ % duty.
Next, we have the issue where we have a total of 5.5v, and the assumption is that on a 4pin header, the signal is NOT pulsing, but is a signal for a PWM fan. Now, I don't know, nor know how to detect if this 5.5v signal is linear or pulsed. Either way, learning both cannot be bad.
So let's say the 5.5v is pulsed. Again, probably low side. So, we use a MOSFET to control a 12v signal, and it should work. An intersting note is that many schemtics show a 'kick start' capacitor, as well as a pull-up resistor and some diodes. Lot's of different designs out there. In protecting the motherboard, how does one design such a circuit for protection?
And then we can guess that maybe the 5.5v is actually a linear voltage, designed to trigger the circuitry in a PWM fan. Now one must build a PWM circuit, and use the 5.5v through most likely an OP Amp, and output the PWM 12v according to the 5.5v input.
Whew. Hopefully that is enough detail to describe what I am dealing with. The actual reason for this post is to, obviously, glean some knowledge from those who know more than I.
Specifically, I wish to develop the following circuits, with each circuit capable of withstanding a 1.5A load, while protecting the signal (motherboard fan header):
1. 5.5v linear input to 12v PWM output
2. 5.5v PWM input to 12v PWM output
3. 12v PWM input to 12v PWM output (simply keep same, but isolate fan header)
4. 12v linear to 12v PWM output
5. 12v linear to 12v linear (simply keep same, but isolate fan header)
I am open to learning anything ATM. PIC I would like to learn, but first wish to understand these concepts. I welcome any advice or criticism that helps to learn. I welcome any reference to schematics or components that may work better.
Again, I am just wishing to learn something, and this is nothing new under the sun, but does present itself as a great project because I have access to many different computers with different chips, as well as many different fans. It would seem a logical choice.
Thank you to whomever may wish to divulge some info.
Respectfully,
MrWoo
#### MrWoo
Joined Jun 18, 2007
11
Hmm. I found some more data on the 4pin headers. It would seem that there is what is labeled as "low frequency" and "high frequency" PWM. The lower frequency, say <20kHz, within the "audible" level, you can hear commutation noise from the fan. The >20kHz would seem to eliminate that.
Evidentily the newer chips which support high frequency PWM need not supply an external FET, in the fact that being designed for a 4pin fan, the FET is built into the fan. So, that COULD mean that my mobo's 4pin header is applying a contstant voltage, expecting there to be a 4pin fan with the FET built-in. That being the case, one would have to design a PWM circuit to drive the 3 proposed fans off of that 5.5v signal. One schematic I looked at shows the PWM signal OUT, going to the PWN signal IN pin on the 4 wire fan, with a 3.3v source fed through a 2k Ohm resistor BEFORE it reaches the fan. I am left to assume, because I don't know, that the PWM output is less than 3.3v, and that the FET inside the 4 wire fan requires say 2v minimum on gate, so that 3.3v ensures FET starts to operate?
But, on the 3pin headers, there must already be an external FET in place on the mobo somewhere that is using the PWM SIGNAL out for it's related channel(s). What ramification does that have on me building a circuit for it? I would presume none, as it is already pulsing through a FET somewhere. All I would need to do is have my FET take it and open it's gates to run the 3 fans whilst isolating the source signal. Hmm.
Which leads us to another need. How not only to design a circuit that uses (I think) a 5.5v signal and convert that to a 12v PWM, but also how does one go about creating low or high frequency PWM models? I think the higher frequency, from my playing with it on my board, would be the better choice. But, what do you say?
Thank you again,
MrWoo
#### MrWoo
Joined Jun 18, 2007
11
Yet again, this time an example based on a TI MSP430F417 microcontroller. The schematics relate to the PWM output of the MCU to a 3wire or 4wire fan. While this is not really like the aSC7621, it does provide an example wiring diagram.
Here is the pdf file that has the schematics.
Thank you,
MrWoo
#### Attachments
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#### beenthere
Joined Apr 20, 2004
15,819
I have to give a shudder when thinking of adding pulse-generating circuitry inside a computer. I would not do it. Motherboards have enough problems with their own signals.
This is a technician's point of view, by the way. It may be attractive to an engineer and/or hobbyist to regulate the cooling fan speed to keep some temperature maintained inside the computer case. I just want to extract the heat, so running the fans at full speed seems to me to be the way to do it. As much cooling as possible is just enough.
That said, Microchip has quite a bit of stuff dealing with speed regulation of brushless DC motors - http://www.microchip.com/stellent/idcplg?IdcService=SS_GET_PAGE&nodeId=1523.
Joined Jan 10, 2006
614
An option I have used in the past is to switch between the 12 volt and 5 volt supplys depending on temperature. Just make sure the fan initialises at 12 volts, as some wont reliably start with a 5 volt input.
#### MrWoo
Joined Jun 18, 2007
11
I am sincerely thankful for those thoughts. The link is a good one for lot's of info I had not found yet on thier site.
Let me state that I don't really want a fan controller. I am just using it as a good project to learn some new stuff with. I could have chosen some robotics or light dimming or whatever. I thought though the whole isolating the motherboard, learning how to read datasheets for components, and the fact that I have access easily to different versions (different mobo's with different chips) would be a good place to dive in.
I have been looking into the interference that PWM circuits put into place. I did not realize that could happen.
While I do wish to build the things I speak of, it is not that important if I make it and it is at best a homemade solution that poses problems with say the cpu or hdd or something. What is more important is that I learn by hands on work, and this is something that could turn out to be of benefit to me or my friends (my guinea pigs lol) in the way of fan control. Or it could be that I build it, more importantly understand it, and progress to other projects that actually do something. Such as inline probes for pH & EC readings in water, with a calorimetric flowmeter, outputting 4-20mA and then with a DAQ writing code that converts it to real values and logs it into a database that then has a front end to graph the results. lol, which is what I have. But, I want to do more automation. More controlling. More computer interface.
Thank you again for your time,
MrWoo | 3,039 | 11,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.937135 |
https://www.quesba.com/questions/7-sorter-enterprises-a-major-real-estate-developer-recently-accepted-a-16-0-2716 | 1,653,136,733,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00524.warc.gz | 1,094,434,382 | 41,243 | # 7. Sorter Enterprises, a major real estate developer, recently accepted a \$16,000,000, five-year, 4%
7. Sorter Enterprises, a major real estate developer, recently accepted a \$16,000,000, five-year, 4% note receivable in exchange for products sold. Interest is paid annually. The current market rate of interest is 7%. The note is not publicly traded. 3 (Click the icon to view the Future Value of \$1 2 table.) (Click the icon to view the Future Value of an 4 Ordinary Annuity table.) (Click the icon to view the Future Value of an 6 Annuity Due table.) (Click the icon to view the Present Value of \$1 table.) (Click the icon to view the Present Value of an Ordinary Annuity table.) (Click the icon to view the Present Value of an Annuity Due table.) 5 " value Prepare the journal entry to record the sale. Ignore cost of goods sold. (Record debits first, then credits. Exclude explanations from any journal entries. Use the present value and future value tables, the formula method, a financial calculator, or a spreadsheet for your calculations. If using present and future value tables or the formula method, use factor amounts rounded to five decimal places, X.XXXXX. Round your final answers to the nearest whole dollar) Current Year Account (1) Notes Receivable (2) Discount on Notes Receivable (3) Sales 1: Future Value of \$1 Future Value of \$1 Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% 1 1.01000 1.020001.03000 1.04000 1.05000 1.06000 1.07000 1.08000 1.09000 1.10000 1.11000 1.12000 1.13000 1.14000 1.15000 2 1.020101.040401.06090 1.08160 1.10250 1.12360 1.14490 1.16640 1.18810 1.21000 1.23210 1.25440 1.27690 1.299601 32250 1.030301.061211.09273 1.12486 1.15763 1.19102 1.22504 1.25971 1.29503 1.33100 1.36763 1.40493 1.44290 1.48154 1.52088 4 1.040601.08243 1.12551 1.16986 1.21551 1.26248 1.31080 1.36049 1.41158 1.46410 1.51807 1.57352 1.63047 1.68896 1.74901 1.05101|1.10408|1.15927 1.21665 1.27628| 1.33823 1.40255 1.46933 1.53862 1.61051 1.68506 1.76234 1.84244 1.92541 2.01136 1.061521.126161.19405 1.26532 1.34010 1.41852 1.50073 1.58687 1.67710 1.77156 1.87041 1.97382 2.08195 2.19497 2.31306 1.072141.148691.22987 1.31593 1.40710 1.50363 1.60578 1.71382 1.82804 1.94872 2.07616 2.21068 2.35261 2.50227 2.66002 8 1.08286 1.171661.26677 1.36857 1.47746 1.59385 1.71819 1.85093 1.992561 2.14359 2.30454 2.47596 2.65844 2.85259 3.05902 11.09369|1.195091.30477 1.42331 1.55133 1.68948 1.83846| 1.99900 2.17189 2.35795 2.55804 2.77308 3.00404 3.25195 3.51788 1.104621.218991.34392 1.48024 1.62889 1.79085 1.96715 2.15892 2.36736 2.59374 2.83942 3.10585 3.39457 3.70722 4.04556 11 1.11567|1.24337|1.38423 1.53945 1.71034 1.89830 2.10485 2.33164 2.58043 2.85312 3.15176 3.47855 3.83586 4.22623 4.65239 11.12683/1.268241.42576 1.60103 1.79586| 2.01220 2.25219 2.51817 2.81266 3.13843 3.49845 3.89598 4.33452 4.81790 5.35025 13 1.138091.293611.46853 1.66507 1.88565 2.13293 2.40985 2.71962 3.06580 3.45227 3.88328 4.36349 4.89801 5.49241 6.15279 14 11.14947 1.31948 1.51259 1.73168 1.97993 2.26090 2.57853 2.93719 3.34173 3.79750 4.310441 4.88711 5.53475 6.26135) 7.07571 (1.16097|1,34587/1.55797 1.80094 2.07893 2.39656 2.75903 3.17217 3.64248 4.17725 4.78459 5.47357 6.25427 7.13794 8.13706 1.172581.372791.60471 1.87298 2.18287 2.54035 2.95216 3.42594 3.97031 4.59497 5.31089 6.13039 7.06733 8.13725 9.35762 17 1.184301.400241.65285 1.94790 2.29202 2.69277 3.15882 3.70002 4.32763 5.05447) 5.89509 6.86604 7.98608 9.27646 10.76126 18 1.196151.42825 1.70243 2.02582 2.40662 2.85434 3.37993 3.99602 4.71712 5.55992 6.54355 7.68997 9.02427 10.57517 12.37545 19 1.208111.45681 1.75351 2.10685 2.52695 3.02560 3.61653 4.31570 5.141666.11591 7.26334 8.61276) 10.19742 12.05569 14.23177 1.22019 1.48595/1.80611 2.19112 2.65330 3.20714 3.86968 4.66096 5.60441 6.72750 8.06231 9.64629 11.52309 13.74349 16.36654 1.28243 1.640612.09378 2.66584 3.38635 4.29187 5.42743 6.84848 8.62308 10.83471 13.58546 17.00006 21.23054 26.46192 32.91895 30 1.347851.81136/2.42726 3.24340 4.32194 5.74349 7.61226 10.06266 13.26768 17.44940 22.89230 29.95992 39.11590 50.95016 66.21177 1.41660 1.99989 2.81386 3.94609 5.51602 7.6860910.67658 14.78534 20.41397 28.10244 38.57485 52.79962 72.06851 98.10018 133.17552 1.48886 2.20804 3.26204 4.80102 7.03999 10.28572 14.97446 21.72452 31.40942 45.25926 65.00087 93.05097 132.78155 188.88351 267.86355 1.564812.437853.78160 5.84118 8.98501 13.76461|21.00245 31.92045 48.32729 72.89048 109.53024163.98760 244.64140 363.67907 538.76927 50 1.644632.691594.38391 7.1066811.46740 18.4201529.45703 46.90161 74.35752 117.39085184.56483 289.00219 450.73593 700.23299 1,083.65744 60 1.81670 3.281035.89160 10.51963 18.67919 32.98769|57.94643101.25706 176.03129304.48164|524.05724897.59693 1,530.053472,595.918664,383.99875 Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% 2: Present Value of \$1 We were unable to transcribe this imageFuture Value of an ordinary Annuity Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 1.000000 1.000000 1.00000 100000 100000 1.000000 1.000000 1.000000 1.000000 100000 1.000001 1.00000 1.000000 2.01000 2.02000 2.03000 2.04000 2.05000 2.06000 2.070002.08000 2.09000 2.10000) 2.11000 2.12000 2.130001 3 3.03010 3.06040 3.09090 3.12160 3.15250 3.18360 3.21490 3.24640 3.27810 3.31000 3.34210 3.37440 3.40690 4 4.06040 4.12161 4.18363 4.24646 4.310134.37462 4.43994 4.50611 4.57313 4.64100 4.70973 4.77933 4.84980 5 5.10101 5.20404 5.30914 5.41632 5.52563 5.63709 5.75074 5.86660 5.98471 6.10510 6.22780 6.35285 6.48027 6.15202 6.30812 6.46841 6.63298 6.80191 6.97532 7.15329 7.33593 7.52333 7.71561 7.91286 8.11519 8.32271 7.21354 7.43428 7.66246 7.89829 8.14201 8.39384 8.65402 8.92280 9.20043 9.48717 9.78327 10.08901 10.40466 8 8.28567 8.58297 8.89234 9.21423 9.54911 9.89747 10.25980 10.63663 11.02847 11.43589 11.85943 12.29969 12.75726) 9 9 .36853 9.75463 10.15911 10.58280 11.02656 11.49132) 11.97799 12.48756 13.02104 13.579481 14.16397| 14.77566| 15.415711 10.46221 10.94972 11.46388 12.00611 12.57789 13.18079 13.81645 14.48656| 15.19293 15.93742 16.72201 17.54874 18.41975 11 11.56683 12.16872 12.80780 13.48635 14.20679 14.97164 15.78360 16.64549 17.56029 18.53117 19.56143 20.65458 21.81432 12.68250 13.41209 14.19203 15.02581 15.91713 16.86994 17.88845 18.97713| 20.14072 21.38428 22.71319 24.13313 25.65018 13 13.80933 14.68033 15.61779 16.62684 17.71298 18.88214 20.14064 21.49530 22.95338 24.52271 26.21164 28.02911 29.98470 14 14.94742 15.97394 17.08632 18.29191 19.59863 21.01507 22.55049 24.21492 26.01919 27.97498 30.09492 32.39260 34.88271 15 16.09690 17.29342 18.59891 20.02359 21.57856 23.27597 25.12902 27.15211 29.36092 31.77248 34.40536 37.27971 40.41746) 17.25786) 18.63929 20.15688 21.82453 23.65749 25.67253 27.88805 30.32428 33.00340 35.94973 39.18995 42.75328 46.67173 17 18.43044 20.01207) 21.76159 23.69751 25.84037 28.21288 30.84022 33.75023 36.97370 40.54470 44.50084 48.88367 53.73906 119.61475 21.41231 23.41444 25.64541 28.13238] 30.90565 33.99903) 37.45024 41.30134 45.59917 50.39594 55.74971 61.72514 19 20.81090 22.84056 25.11687 27.67123 30.53900 33.75999 37.37896) 41.44626 46.01846 51.15909 56.93949 63.43968 70.74941 20 2.01900 24.29737 26.87037 29.77808 33.06595 36.78559 40.99549 45.76196 51.16012 57.27500 64.20283 72.05244 80.94683 25 28.24320 32.03030 36.45926 41.64591 47.72710 54.86451 63.24904 73.10594 84.70090 98.34706 114.41331 133.33387 155.61956 1: 30 34.78489 40.56808 47.57542 56.08494 66.43885 79.05819 94.46079 113.28321 136.30754 164.49402 199.02088 241.33268 293.19922 3 41.66028 49.99448 60.46208) 73.65222 90.32031 111.43478/138.23688 172.31680 215.71075 271.024371 341.58955 431.66350 546.68082 6! 18.88637 60.40198 75.40126 95.02552120.79977 154.76197199.63511 259.05652 337.88245 442.59256 581.82607 767.09142 1,013.70424 1,34 45 56.48107) 71.89271 92.71986|121.02939159.70016212.74351285.74931 386.50562 525.85873 718.90484 986.638561,358.23003 1.874.16463) 2,5 50 \$4.46318 84.57940112.79687152.66708209.34800290.33590|406.52893 573.77016815.083561,163.908531,668.77115/2,400.01825 3,459.50712 4,99 60 61.66967 114.05154 163.05344237.99069 353.58372533.12818|813.52038 1,253.21330|1,944.792133,034.816404,755.065847,471.64111 11,761.94979 18,5 Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 4: Present Value of an Ordinary Annuity Present Value of an ordinary Annuity Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% 0.99010 0.98039 0.97087 0.96154 0.95238] 0.94340 0.93458 0.92593 0.91743 0.90909 0.90090 0.89286 0.88496 0.87719 0.86957) 0.1 1.97040 1.94156 1.91347 1.88609 1.85941 1.83339 1.80802 1.78326 1.75911 1.73554 1.71252 1.69005) 1.66810 1.64666 1.62571 1.6 2.94099 2.88388 2.82861 2.77509 2.72325 2.67301 2.62432 2.57710 2.53129 2.48685 2.44371 2.40183 2.36115 2.32163 2.28323 2.: 3.90197 3.80773 3.71710 3.62990 3.54595 3.46511 3.38721 3.31213 3.23972 3.16987 3.10245 3.03735 2.97447 2.91371 2.85498 2. 4.85343| 4.71346 4.57971 4.45182 4.32948 4.21236 4.10020 3.99271 3.88965 3.79079 3.69590 3.60478 3.51723 3.43308 3.35216 3. 5.79548 5.60143 5.41719 5.24214 5.07569 4.91732 4.76654 4.62288 4.48592 4.35526 4.23054 4.11141 3.99755 3.88867 3.78448 3. 6.72819 6.47199 6.23028 6.00205 5.78637 5.58238 5.38929 5.20637 5.03295 4.86842 4.71220 4.56376 4.42261 4.288304.16042 4.1 7.65168 7.32548 7.01969 6.73274 6.46321 6.20979 5.97130 5.74664 5.53482 5.33493 5.14612 4.96764 4.79877 4.63886 4.48732 4.5 8.56602 8.16224 7.78611 7.43533 7.10782 6.80169 6.51523 6.24689 5.99525 5.75902 5.53705 5.32825 5.13166 4.94637 4.77158 4. 9.47130 8.98259 8.53020 8.11090 7.721737.36009 7.02358 6.71008 6.417666.14457 5.88923 5.65022 5.42624 5.216125.01877 4. 11 10.36763 9.78685 9.25262 8.76048 8.30641 7.88687 7.49867 7.13896 6.80519 6.49506 6.20652 5.93770 5.68694 5.45273 5.23371 5.6 12 11.25508 10.57534 9.95400 9.38507 8.863258.38384 7.94269 7.53608 7.16073 6.81369 6.49236 6.19437 5.91765 5.66029 5.42062 5. 12.13374 11.3483710.63496 9.98565 9.39357 8.85268 8.35765 7.90378 7.48690 7.10336 6.74987 6.42355 6.121815.84236 5.58315 5. 14 13.00370 12.10625/11.2960710.56312] 9.89864 9.294988.74547| 8.24424 7.78615 7.36669 6.98187 6.62817 6.30249 6.00207 5.72448 5. 13.86505 12.84926 11.9379411.11839 10.37966 9.71225 9.10791 8.55948 8.06069 7.60608 7.19087 6.810866.46238 6.14217 5.84737 5.! 14.71787 13.57771 12.56110 11.65230 10.83777 10.10590 9.44665 8.85137 8.31256 7.82371 7.37916 6.97399 6.60388 6.265065.95423 5. 17 15.5622514.29187|13.1661212.16567 11.27407 10.47726 9.76322 9.12164 8.54363 8.02155 7.54879 7.11963 6.72909 6.37286 6.04716 5: 18 16.39827 14.9920313.7535112.65930) 11.68959 10.82760 10.05909 9.37189 8.75563 8.20141 7.701627.24967 6.83991 6.46742 6.12797 5. 19 17.22601 15.6784614.32380 13.13394 12.08532 11.15812 10.33560 9.60360 8.95011 8.36492 7.839297.36578 6.93797 6.55037 6.19823 5. 18.04555 16.35143 14.8774713.59033 12.46221 11.46992 10.59401 9.81815 9.12855 8.51356 7.96333 7.46944 7.02475 6.62313 6.25933 5. 25 22.0231619.52346 17.4131515.62208 14.09394 12.7833611.65358 10.67478 9.82258 9.07704 8.42174 7.84314 7.32998 6.87293 6.46415 6.6 30 25.80771 22.3964619.60044 17.29203 15.37245 13.76483 12.40904 11.25778 10.27365 9.42691 8.69379 8.05518 7.49565 7.00266 6.56598 6. 35 29.40858|24.99862 21.48722 18.66461 16.37419 14.49825 12.94767 11.6545710.56682 9.64416) 8.85524 8.17550 7.58557) 7.07005) 6.616616. 32.8346927.35548|23.11477 19.79277 17.15909 15.04630 13.33171 11.92461 10.75736 9.77905 8.95105 8.24378 7.63438 7.10504 6.64178 6. 36.09451 29.49016|24.51871 20.72004 17.77407 15.45583 13.60552 12.10840 10.88120 9.86281 9.00791 8.28252 7.66086 7.12322 6.65429 6.4 50 39.19612 31.4236125.72976/21.48218 18.2559315.76186 13.80075 12.23348 10.961689.91481 9.041658.30450 7.67524 7.13266l 6.660511 6. 60 44.95504 34.7608927.67556|22.6234918.92929 16.16143 14.03918 12.37655 11.04799 9.96716 9.07356 8.32405 7.68728 7.14011 6.66515 6.1 Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% 5: Future Value of an Annuity Due Future Value of an Annuity Due Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 1.01000 1.02000 1.03000 1.04000 1.05000 1.06000 1.07000 1.08000 1.09000 1.10000 1.11000 1.12000 1.13000 2 2.03010 2.06040 2.09090 2.12160 2.15250 2.18360 2.21490 2.24640 2.27810 2.31000 2.34210 2.37440 2.40690 3 3.06040 3.12161 3.18363 3.24646 3.31013 3.37462 3.43994 3.50611 3.57313 3.64100 3.70973 3.77933 3.84980 4 4.10101 4.20404 4.30914 4.41632) 4.52563 4.63709 4.750741 4.86660 4.984711 5.105101 5.22780 5.35285 5.48027 5 5.15202 5.30812 5.46841 5.63298 5.80191 5.97532 6.15329 6.33593 6.52333 6.71561 6.91286 7.11519 7.32271 6.21354 6.43428 6.662466.89829 7.14201 7.39384 7.65402 7.92280 8.20043 8.48717 8.78327 9.08901 9.40466 7.28567 7.58297 7.89234 8.21423 8.54911 8.89747 9.25980| 9.63663 10.02847 10.43589 10.85943 11.29969 11.75726 8.36853 8.75463 9.15911 9.58280 10.02656 10.49132 10.97799| 11.48756 12.02104 12.57948 13.16397 13.77566 14.41571 9.46221 9.94972 10.46388 11.00611 11.57789 12.18079 12.81645 13.48656 14.19293 14.93742 15.722011 16.54874 17.41975 10.56683 11.16872 11.80780 12.48635 13.20679 13.97164 14.78360 15.64549 16.56029 17.53117 18.56143 19.65458 20.81432 11 111.68250 12.41209 13.19203 14.02581 14.91713 15.86994 16.88845 17.97713 19.14072 20.38428 21.71319 23.13313 24.65018 12 12.80933 13.68033 14.61779 15.62684 16.71298 17.88214 19.14064 20.49530 21.95338 23.522711 25.21164 27.02911 28.98470 13 13.94742 14.97394 16.08632 17.29191 18.59863 20.01507) 21.55049 23.21492 25.01919 26.97498 29.09492 31.392601 33.88271 : 14 15.09690 16.29342 17.59891 19.02359 20.57856 22.27597 24.12902 26.15211 28.36092 30.77248 33.40536 36.27971 39.41746 16.25786 17.63929 19.15688 20.82453 22.65749 24.67253 26.88805 29.32428 32.00340 34.94973 38.18995 41.75328 45.67173 16 17.43044 19.01207 20.76159 22.69751 24.84037 27.21288) 29.84022 32.75023 35.97370 39.54470 43.50084 47.88367 52.73906 18.61475 20.41231 22.41444 24.64541 27.13238 29.90565 32.99903 36.45024 40.30134 44.59917 49.39594 54.74971 60.725141 18 19.81090 21.84056] 24.116870 26.67123 29.53900 32.75999 36.378961 40.44626 45.01846] 50.15909 55.93949 62.43968 69.749411 21.01900 23.29737 25.87037 28.77808 32.06595 35.78559 39.99549 44.76196 50.16012 56.27500 63.20283 71.05244 79.94683 22.23919 24.78332 27.67649 30.96920 34.71925 38.99273 43.86518| 49.42292 55.76453 63.00250 71.26514 80.69874 91.46992 11 28.52563 32.67091 37.55304 43.31174 50.1134558.15638 67.67647 78.95442 92.32398 108.18177 126.99877 149.33393 175.85010 21 30 35.13274 41.37944 49.00268 58.32834 69.76079 83.80168101.07304 122.34587 148.57522| 180.94342 220.91317 270.29261 331.31511 41 42.07688 50.99437 62.27594 76.59831 94.83632 118.12087147.91346 186.10215 235.12472 298.12681 379.16441 483.46312 617.74933 7! 40 49.37524 61.61002 77.6633098.82654 126.83976164.04768213.60957 279.78104 368.29187 486.85181 645.82693 859.14239 1,145.48579 1,5 45 17.04589 73.33056 95.5014625.87057167.68516225.50812305.75176 417.42607 573.18602 790.795321,095.16880/1,521.21764 2,117.80603 2,9 55.10781 86.27099116.18077158.77377219.81540307.75606434.98595 619.67177 888.441081,280.29938 1,852.335982,688.02044 3,909.24304 5,6! 60 \$2.48637 116.33257167.94504247.51031 371.26290565.11587 870.46681 1,353.470362,119.823423,338.298035,278.123088,368.2380513,291.0032721,1; Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 6: Present Value of an Annuity Due 11 Present Value of an Annuity Due Periods) 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% | 1.00000 1.00000) 1.00000 1.00000 1.00000) 1.00000) 1.00000 1.00000) 1.00000 1.00000 1.00000 1.00000) 1.00000 1.00000 1.00000 1. | 1.99010/ 1.98039) 1.97087) 1.96154) 1.95238) 1.94340) 1.93458) 1.92593| 1.91743) 1.90909) 1.90090) 1.89286| 1.88496| 1.87719| 1.86957) 1. 2.97040| 2.94156) 2.91347) 2.88609) 2.85941) 2.83339| 2.80802 2.78326) 2.75911) 2.73554) 2.71252 2.69005) 2.66810| 2.64666] 2.62571| 2. | 3,94099| 3,88388| 3.82861) 3.77509) 3,72325) 3.67301) 3.62432 3.57710| 3.53129) 3,48685) 3.44371) 3,40183) 3.36115| 3.32163) 3.28323| 3. 4.90197 4.80773 4.71710 4.62990) 4.54595 4.46511) 4.38721| 4.31213| 4.23972| 4.16987) 4.10245) 4.03735) 3.97447) 3.91371) 3.85498 5.85343| 5.71346) 5.57971) 5.45182) 5.32948) 5.21236| 5.10020] 4.99271) 4.88965) 4.79079] 4.69590) 4.60478| 4.51723 4.43308) 4.35216) 4. 6.79548| 6.60143) 6.41719) 6.24214) 6.07569) 5.91732| 5.76654 5.62288| 5.48592| 5.35526| 5.23054 5.11141) 4.99755) 4.88867) 4.78448) 4. 7.72819 7.47199) 7.23028 7.00205) 6.78637 6.58238| 6.389291 6.20637 6.03295) 5.86842| 5.71220) 5.56376 5.42261) 5.28830) 5.16042 5. | 8.65168) 8.32548) 8.01969) 7.73274) 7,46321) 7.20979) 6.97130) 6.74664) 6.53482| 6.33493) 6.14612| 5,96764) 5.79877) 5.63886| 5.48732 5. 9.56602) 9.16224) 8.78611) 8.43533| 8.10782 7.80169) 7.51523| 7.24689) 6.99525) 6.75902| 6.53705 6.32825) 6.13166) 5.94637) 5.77158| 10.47130) 9.98259) 9.53020| 9.11090| 8.72173 8.36009] 8.02358| 7.71008| 7.41766| 7.14457) 6.88923 6.65022| 6.42624 6.21612| 6.01877] 5, |11.36763| 10.78685|10.25262 9.76048| 9.30641) 8.88687) 8,49867) 8.13896] 7,80519] 7,49506) 7.20652 6.93770) 6.68694) 6.45273| 6.23371) 13 [12.25508| 11.57534| 10.95400| 10.38507) 9.86325] 9.38384) 8.94269) 8.53608| 8,16073| 7.81369) 7.49236| 7.19437) 6.91765) 6.66029) 6.42062 6. 14 |13,13374| 12.34837] 11.63496| 10.98565|10.39357) 9.85268) 9.35765) 8.90378| 8,48690 8.103361 7.749871 7.42355) 7.12181) 6.84236) 6.58315) 6. 14.00370| 13.10625|12.29607) 11.56312|10.89864) 10.29498) 9.74547) 9.24424/ 8.78615] 8.36669) 7.98187) 7.62817| 7.30249) 7.00207) 6.72448| 14.86505| 13.84926 12.93794| 12.11839|11.37966| 10.71225|10.10791) 9.55948| 9.06069/ 8.60608| 8.19087) 7.81086| 7.46238| 7.14217) 6.84737) 6. 15.71787) 14.57771|13.56110| 12.65230| 11.83777|11.10590|10.44665/ 9.85137) 9.31256| 8.82371) 8.37916| 7.97399) 7.60388| 7.26506| 6.95423) |16.562251 15.29187) 14.16612|13.16567|12.27407|11.47726|10.76322 10.12164) 9.54363) 9.02155) 8.54879| 8.11963| 7.72909| 7.37286) 7.04716| 6. |17.39827| 15.99203|14.75351| 13.65930 12.68959) 11.82760|11.05909) 10.37189/ 9.75563/ 9.20141) 8.70162 8.24967) 7.83991) 7.46742| 7.12797) 18.22601) 16.67846|15.32380| 14.13394| 13.08532| 12.15812|11.33560| 10.60360) 9.95011| 9.36492| 8.83929] 8.36578| 7.93797| 7.55037] 7.19823| 6, 22.24339| 19.91393|17.93554) 16.24696| 14.79864) 13.55036| 12.46933| 11.52876|10.70661) 9.98474) 9.34814) 8.78432| 8.28288) 7.83514) 7.43377] 7, 26.06579| 22.84438|20.18845) 17.98371|16.14107) 14.59072|13.27767) 12.15841|11.19828| 10.36961) 9.65011) 9.02181) 8.47009| 7.98304) 7.55088) 7. 29.70267| 25.49859|22.13184) 19.41120|17.19290| 15.36814|13.85401) 12.58693|11.51784| 10.60857) 9.82932 9.15656) 8.57170) 8.05985) 7.60910| 7. 33.16303| 27,90259|23.80822| 20.58448| 18.01704) 15.94907|14.26493| 12.87858|11.72552|10.75696) 9.93567) 9.23303) 8.62684) 8.09975| 7.63805) 7. |36.45545| 30.07996|25.25427| 21.54884|18.66277|16.38318|14,55791| 13.07707|11.86051| 10.84909) 9.99878| 9,27642| 8.65678| 8.12047) 7.65244 7. (39.58808| 32.05208/26.50166| 22.34147| 19.16872| 16.70757) 14.76680| 13.21216|11.94823| 10.90630|10.03624 9.30104) 8.67302| 8.13123| 7.65959| 7. 45.40459| 35.45610|28.50583| 23.52843) 19.87575| 17.13111|15.02192| 13.36668|12.04231) 10.96387|10.07165] 9.32294) 8.68663) 8.13972| 7.66492| 7. Periods 1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 15% - - - - - - - - - - - ம் ம்
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# Reorganizing the corporate structure of Nitescheleds will be a cause
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Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink]
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31 Dec 2018, 11:58
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Project SC Butler: Day 55 Sentence Correction (SC2)
Reorganizing the corporate structure of Nitescheleds will be a cause for concern for the employees; just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers, so restructuring the corporation will cause employees to feel anxious.
A) just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers,
B) like suddenly hiring a large number of new workers does lead to a sense of instability and uncertainty among company employees,
C) the company employees will have a sense of instability an uncertainty following the sudden hiring of a large number of new workers,
D) as suddenly hiring a large number of new workers which will lead to a sense of instability and uncertainty among company employees,
E) as suddenly hiring a large number of new workers will lead to a sense of instability and uncertainty among company employees,
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Re: Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink]
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31 Dec 2018, 15:15
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Quote:
Reorganizing the corporate structure of Nitescheleds will be a cause for concern for the employees; just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers, so restructuring the corporation will cause employees to feel anxious.
The sentence presents a few facts. It says that reorganizing the corporate structure of Nitescheleds cause concern in employees.
Then it goes on to support with an example:
Just as hiring a large number of new workers will lead to a sense of instability and uncertainty among company employees,
restructuring the corporation will cause employees to feel anxious.
Quote:
B) like suddenly hiring a large number of new workers does lead to a sense of instability and uncertainty among company employees,
We can not use like to denote examples, A and B are out.
Idiom tested, Just as X, so Y where X and Y have to be parallel.
Quote:
C) the company employees will have a sense of instability an uncertainty following the sudden hiring of a large number of new workers,
Parallelism issue, X = noun (employees) , Y = restructuring (verb-ing denoting action)
Quote:
D) as suddenly hiring a large number of new workers which will lead to a sense of instability and uncertainty among company employees,
now clause before coma becomes a run-on sentence i.e. one without main verb.
Note that usage of which is incorrect too since it, being a noun modifier, can not be used to denote result of action
and is used to incorrectly modify workers here.
Quote:
E) as suddenly hiring a large number of new workers will lead to a sense of instability and uncertainty among company employees,
And we have our winner. X= hiring , Y = restructuring.
suddenly is a modifier modifying hiring. (Think how hiring took place, it took suddenly)
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Re: Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink]
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31 Dec 2018, 15:40
generis wrote:
Project SC Butler: Day 55 Sentence Correction (SC1)
Reorganizing the corporate structure of Nitescheleds will be a cause for concern for the employees; just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers, so restructuring the corporation will cause employees to feel anxious.
A) just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers,
B) like suddenly hiring a large number of new workers does lead to a sense of instability and uncertainty among company employees,
C) the company employees will have a sense of instability an uncertainty following the sudden hiring of a large number of new workers,
D) as suddenly hiring a large number of new workers which will lead to a sense of instability and uncertainty among company employees,
E) as suddenly hiring a large number of new workers will lead to a sense of instability and uncertainty among company employees,
The best or excellent answers get kudos, which will be awarded after the answer is revealed.
In option A, since "just like" is used, it must be parallel with "reorganizing the structure". This is not the case hence it's wrong.
In Option B, "suddenly hiring..." is parallel with "reorganizing the structure", but "hiring" is a verb and not a noun. So use of "like" is incorrect and instead we will use "as".
In option C, "just the company employees will..." doesn't make any sense at all.
The correct idiom is "Just as...so will.." which can be seen in options D and E, but in option D, the use of "which" is unnecessary. Option E is the best choice and should be the answer.
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Re: Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink]
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31 Dec 2018, 22:04
The answer is E
E) as suddenly hiring a large number of new workers will lead to a sense of instability and uncertainty among company employees,
In D, the word "which" is not suitable.
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Re: Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink]
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01 Jan 2019, 04:29
Idiom is Just as X , so Y
In D, “which” is unnecessary.
So, E is correct
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Re: Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink]
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01 Jan 2019, 10:03
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generis wrote:
Project SC Butler: Day 55 Sentence Correction (SC1)
Reorganizing the corporate structure of Nitescheleds will be a cause for concern for the employees; just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers, so restructuring the corporation will cause employees to feel anxious.
A) just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers,
B) like suddenly hiring a large number of new workers does lead to a sense of instability and uncertainty among company employees,
C) the company employees will have a sense of instability an uncertainty following the sudden hiring of a large number of new workers,
D) as suddenly hiring a large number of new workers which will lead to a sense of instability and uncertainty among company employees,
E) as suddenly hiring a large number of new workers will lead to a sense of instability and uncertainty among company employees,
The best or excellent answers get kudos, which will be awarded after the answer is revealed.
A - Constructing should be as X , so Y, this is not correct
B Same as A , incorrect construction
C - As is missing
D - which is not parallel with following clause
E - perfect in all aspects
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Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink]
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01 Jan 2019, 16:02
1
Project SC Butler: Day 55 Sentence Correction (SC2)
Reorganizing the corporate structure of Nitescheleds will be a cause for concern for the employees; just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers, so restructuring the corporation will cause employees to feel anxious.
A) just like the company employees who will have a sense of instability and uncertainty following the sudden hiring of a large number of new workers,
B) like suddenly hiring a large number of new workers does lead to a sense of instability and uncertainty among company employees,
C) the company employees will have a sense of instability an uncertainty following the sudden hiring of a large number of new workers,
D) as suddenly hiring a large number of new workers which will lead to a sense of instability and uncertainty among company employees,
E) as suddenly hiring a large number of new workers will lead to a sense of instability and uncertainty among company employees,
OFFICIAL EXPLANATION
• Punctuation: A colon (:), semicolon (;), and dash (—) replace connecting words.
They are usually used to draw our attention to the connection between the different parts of the sentence.
• Parallelism: Parallel verbs must have the same tense, but they can be a mix of
active/passive or positive/negative forms
• The structure just . . . so suggests that there needs to be parallelism between
the first section of the sentence and the second one.
• The first section begins with reorganizing the structure, therefore the second part
of the sentence should begin with suddenly hiring rather than company employees.
We need an ___ING word (a gerund) in order to maintain parallelism, not an ___ING word and a noun.
• In addition, because the second section begins with a verb (hiring) and not with a noun,
the correct idiomatic expression is just as, not just like.
Just as . . . so . . . will is a common idiomatic expression, which only D and E follow.
• In D, which is unnecessary and breaks the passive structure of the argument.
gadde22 and Niraphan , welcome!
That OE is good, so I will not comment.
The best answer is that by adkikani . Kudos!
HAPPY NEW YEAR, everyone!
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Reorganizing the corporate structure of Nitescheleds will be a cause [#permalink] 01 Jan 2019, 16:02
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# Reorganizing the corporate structure of Nitescheleds will be a cause
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Riemannian Geometry 1 Chapter 1. This document is designed to be read either as a .pdf le or as a printed ... the complex numbers. Complex Analytic And Differential Geometry full free pdf … A manWold is a C. manifold M together with a complex structure. The notation V is used for the. The modulus of a complex number z = x + iy is defined to be the non-negative real number x2 y2,w h ic s, of uret lng v pa z.T md traditionally denoted |z|, and is sometimes called the length of z. The n is the corplex dimension of M. The coordinate charts of the complex structure of a manifold are called holomorphic coordinate charts holo- Note that an sutset of a complex manifold is itself a complex manifold in a standard fashion. We have a holomorphic atlas (or “we have local complex Download Complex Analytic And Differential Geometry full book in PDF, EPUB, and Mobi Format, get it for read on your Kindle device, PC, phones or tablets. Differentiable Manifolds and Vector Bundles 3 1.1. Complex Differential Geometry Roger Bielawski July 27, 2009 Complex manifolds A complex manifold of dimension m is a topological manifold (M,U), such that the transition functions φ U φ−1 V are holomorphic maps between open subsets of Cm for every intersecting U,V ∈ U. Tangent Spaces and Vector Fields 5 1.3. 4 CHAPTER 1. LEMMA 2. collection charts on a M which Vector Bundles 8 WHAT IS DIFFERENTIAL GEOMETRY? Differentiable Manifolds 3 1.2. We denote the identity map of a set X by id X and the n nidentity matrix by 1l nor simply 1l. Complex Differential Geometry Fangyang Zheng American Mathematical Society • International Press--W p. Contents Preface xi Part 1. 4 LOVELY PROFESSIONAL UNIVERSITY Complex Analysis and Differential Geometry Notes We shall postpone until the next section the geometric interpretation of the product of two complex numbers.
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https://discourse.julialang.org/t/speeding-up-my-logsumexp-function/42380 | 1,653,799,504,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00140.warc.gz | 264,227,608 | 15,065 | Speeding up my logsumexp function
I wrote a logsumexp function based on the code here, that takes matrix as input and applies logsumexp along a dimension in a numerically stable way.
``````function lsexp_mat(mat, zero1_mat; dims=1)
# To use the function zero1_mat has to be created outside @model
# with the following code
# zero1_mat = zeros(size(input_matrix))
# zero1_mat[end, :] = zero1_mat[end, :] .+ 1
sorted_mat = sort(mat, dims=dims)
max_ = reshape(sorted_mat[end, :], 1, :)
exp_mat = exp.(sorted_mat .- max_)
sum_exp_ = sum(exp_mat .- zero1_mat, dims=dims)
log1p.(sum_exp_) .+ max_
end
``````
Iâm using `sort` function to make it easy to subtract `1` max elements along the given dimension. But sorting is nlogn. I also wrote this way because I donât any variables in my code to mutate because I want to use this code as a part of my Turing model and then use Zygote backend.
Is there way to create a mask matrix with `1` at all argmax indices and `0` else where without mutation?
This is the function from the discussion Iâm trying to implement for a matrix.
``````function logsumexp!(p::WeightedParticles)
N = length(p)
w = p.logweights
offset, maxind = findmax(w)
w .= exp.(w .- offset)
ÎŁ = sum_all_but(w,maxind) # ÎŁ = âwâ-1
log1p(ÎŁ) + offset, ÎŁ+1
end
function sum_all_but(w,i)
w[i] -= 1
s = sum(w)
w[i] += 1
s
end
``````
1 Like
Use `maximum` to find the maximum:
``````function lsexp_mat2(mat, zero1_mat; dims=1)
# To use the function zero1_mat has to be created outside @model
# with the following code
# zero1_mat = zeros(size(input_matrix))
# zero1_mat[end, :] = zero1_mat[end, :] .+ 1
max_ = maximum(mat, dims=dims)
exp_mat = exp.(mat .- max_)
sum_exp_ = sum(exp_mat .- zero1_mat, dims=dims)
log1p.(sum_exp_) .+ max_
end
``````
I have to subtract 1 from all the columnwise max values and use log1p for the function to be numerically stable. By sorting I know all the columnwise max values are the last row so I can easily subtract -1 from them. The change you made makes the function underflow.
`mat = [1e-20 1e-20; log(1e-20) log(1e-20)]` check it for this array.
1 Like
Oops, didnât read for the motivation; my apologies.
``````remainder(dim::Int, ::NTuple{2}) = 3 - dim
function remainder(dim::Int, ::NTuple{N}) where {N}
ntuple(n -> n < dim ? n : n + 1, Val(N-1))
end
using SparseArrays
function lsexp_mat(mat::AbstractMatrix; dims=1)
@assert dims == 1
remdim = remainder(dims,size(mat))
maxinds_ = map(argmax, eachslice(mat, dims=remdim))
max_ = getindex.(eachslice(mat, dims=remdim), maxinds_)
m, n = size(mat)
zero1_mat = sparse(maxinds_, axes(mat,2), ones(m), m, n)
exp_mat = exp.(mat .- max_) - zero1_mat # TODO: generalize me
log1p.(sum(exp_mat, dims=dims)) .+ max_' # TODO: generalize me
end
``````
This isnât a great solution, but I wanted to actually offer something addressing your actual question:
Is there way to create a mask matrix with `1` at all argmax indices and `0` else where without mutation?
by using `sparse`. You could almost certainly make that more efficient.
``````julia> mat = [1e-20 1e-20; log(1e-20) log(1e-20)];
julia> zero1_mat = zeros(size(mat)); zero1_mat[end, :] = zero1_mat[end, :] .+ 1;
julia> lsexp_mat(mat) # new
1Ă2 Array{Float64,2}:
2.0e-20 2.0e-20
julia> lsexp_mat(mat, zero1_mat) # original
1Ă2 Array{Float64,2}:
2.0e-20 2.0e-20
``````
Worth pointing out that Zygote doesnât currently support the `SparseMatricSCS` constructor, so this answer doesnât really help you.
It also doesnât support `sort`. So youâll have to do a little work defining your own adjoints. Iâd reccomend defining the rule for `lsexp_mat` directly (which would allow you to mutate internally).
I found that `zero1_mat = mat .== maximum(mat, dims=1)` creates what I want.
Thank you for your solution, but I want to keep the solution as simple as possible. Zygote may not support SparseArrays.
My new code is
``````function lsexp_mat(mat; dims=1)
max_ = maximum(mat, dims=1)
zero1_mat = mat .== max_
exp_mat = exp.(mat .- max_)
sum_exp_ = sum(exp_mat .- zero1_mat, dims=dims)
log1p.(sum_exp_) .+ max_
end
``````
``````n = 10000
A = rand(n,n)
@benchmark lsexp_mat(A, dims=1)
@benchmark mapslices(lsexp_vector, A; dims=1)
``````
lsexp_mat bench mark
``````BenchmarkTools.Trial:
memory estimate: 1.50 GiB
allocs estimate: 17
--------------
minimum time: 1.750 s (0.30% GC)
median time: 1.784 s (3.12% GC)
mean time: 1.829 s (5.38% GC)
maximum time: 1.953 s (11.99% GC)
--------------
samples: 3
evals/sample: 1
``````
lsexp_vector benchmark
``````BenchmarkTools.Trial:
memory estimate: 2.57 MiB
allocs estimate: 108509
--------------
minimum time: 1.127 s (0.00% GC)
median time: 1.129 s (0.00% GC)
mean time: 1.133 s (0.00% GC)
maximum time: 1.143 s (0.00% GC)
--------------
samples: 5
evals/sample: 1
``````
We can see that `lsexp_mat` is slower than `lsexp_vector` and also uses orders of magnitude more memory. Why is this happening and how to make this better and faster? And also the number of allocations of `lsexp_vector` is very high compared to `lsexp_mat` yet `lsexp_vector` is faster. I expected `lsexp_mat` to be faster since everything is vectorised.
The problem with `lsexp_mat` is it allocates a lot of un-necessary matrices. If you re-write it as
``````function lsexp_mat1(mat; dims=1)
max_ = maximum(mat, dims=1)
sum_exp_ = sum(exp.(mat .- max_) .- mat .== max, dims=dims)
log1p.(sum_exp_) .+ max_
end
``````
itâs about as fast as the vector version.
2 Likes
Here are a few faster variants, using a package of mine and one of @Elrodâs. If you want this to work with Zygote, then the time spent working out the gradient will tend to dominate.
``````function lsexp_mat1(mat; dims=1)
max_ = maximum(mat, dims=1)
zero1_mat = safeeq(mat, max_) # working around a Zygote bug, today?
exp_mat = exp.(mat .- max_)
sum_exp_ = sum(exp_mat .- zero1_mat, dims=dims)
log1p.(sum_exp_) .+ max_
end
safeeq(mat, max_) = (mat .== max_)
function lsexp_mat2(mat; dims=1) # less memory but not really faster?
max_ = maximum(mat, dims=1)
exp_mat = exp.(mat .- max_) .- (mat .== max_) # fuse this broadcast, @Oscar_Smith beat me to it!
sum_exp_ = sum(exp_mat, dims=dims)
sum_exp_ .= log1p.(sum_exp_) .+ max_ # re-use this array?
end
function lsexp_mat3(mat) # not generic over dims, but differentiable
max_ = maximum(mat, dims=1)
@tullio exp_mat[i,j] := exp(mat[i,j] - max_[1,j]) - (mat[i,j] == max_[1,j]) avx=false # grad=Dual # fixed on master
sum_exp_ = sum(exp_mat, dims=1)
@tullio out[i,j] := log1p(sum_exp_[i,j]) + max_[i,j] avx=false
end
using LoopVectorization
# function lsexp_mat4(mat; dims=1) # @avx broadcasting, is having a bad day
# max_ = maximum(mat, dims=1)
# # zero1_mat = (mat .== max_)
# exp_mat = @avx exp.(mat .- max_) .- (mat .== max_) # has lots of NaN & Inf in it?
# sum_exp_ = sum(exp_mat, dims=dims)
# @avx sum_exp_ .= log1p.(sum_exp_) .+ max_ # mostly NaN
# end
function lsexp_mat5(mat) # also using @avx
max_ = maximum(mat, dims=1)
@tullio exp_mat[i,j] := exp(mat[i,j] - max_[1,j]) - (mat[i,j] == max_[1,j])
sum_exp_ = sum(exp_mat, dims=1)
@tullio out[i,j] := log1p(sum_exp_[i,j]) + max_[i,j]
end
n = 1_000; A = rand(n,n);
lsexp_mat(A) â lsexp_mat1(A) â lsexp_mat2(A)
# lsexp_mat(A) â lsexp_mat4(A) # false?
lsexp_mat(A) â lsexp_mat3(A) â lsexp_mat5(A)
@btime lsexp_mat(\$A) # 10.164 ms (13 allocations: 15.41 MiB)
@btime lsexp_mat2(\$A) # 10.631 ms (6 allocations: 7.65 MiB)
@btime lsexp_mat3(\$A) # 3.188 ms (78 allocations: 7.66 MiB)
# @btime lsexp_mat4(\$A) # 3.494 ms (14 allocations: 7.65 MiB)
@btime lsexp_mat5(\$A) # 2.069 ms (76 allocations: 7.66 MiB)
using Tracker, Zygote #, ForwardDiff
@btime Zygote.gradient(sumâlsexp_mat1, \$A); # 69.199 ms (3003130 allocations: 137.61 MiB)
@btime Zygote.gradient(sumâlsexp_mat3, \$A); # 12.547 ms (253 allocations: 38.23 MiB)
@btime Zygote.gradient(sumâlsexp_mat5, \$A); # 10.309 ms (248 allocations: 38.23 MiB)
``````
3 Likes
The awkward thing about broadcasting is that type information and syntax combined still do not fully specify the behavior: if `isone(size(A, n))`, then axis `n` is broadcasted.
Currently LoopVectorization handles broadcasting by setting strides corresponding to dimensions of size `1` to 0. The linear index `A[n...]` equals `dot(n, strides(A))`, so by setting the stride to 0, itâll ignore indexes on that axis and âbroadcastâ along it.
However, this doesnât work well when loading data along a contiguous axis.
For this to be efficient, we need to use `vmovup*` instructions. These load contiguous elements, so the stride = 0 trick wonât work.
It would work with gather instructions, available in AVX2 and AVX512, but theyâre many times slower. While better than nothing, they cripple performance on a comparative basis.
This means to be efficient, we have to use the contiguous loads (and stores, but stores arenât a problem when broadcasting).
Obviously Julia+LLVM donât have a problem with this. Iâm guessing it uses a few runtime checks to switch between different versions of the loop. I should probably follow that approach.
But for now, a workaround (incompatible with the `dims` argument) is to make it known at compile time that `isone(size(max_,1))`:
``````function lsexp_mat4(mat; dims=1) # @avx broadcasting, is having a bad day
@assert dims == 1
max_ = vec(maximum(mat, dims=1))' # requires dims=1
# zero1_mat = (mat .== max_)
exp_mat = @avx exp.(mat .- max_) .- (mat .== max_) # should now work
sum_exp_ = sum(exp_mat, dims=dims)
@avx sum_exp_ .= log1p.(sum_exp_) .+ max_ # mostly NaN
end
``````
So now I get:
``````julia> n = 1_000; A = rand(n,n);
julia> lsexp_mat(A) â lsexp_mat1(A) â lsexp_mat2(A)
true
julia> lsexp_mat(A) â lsexp_mat4(A)
true
julia> lsexp_mat(A) â lsexp_mat3(A) â lsexp_mat5(A)
true
julia> @btime lsexp_mat(\$A);
10.844 ms (13 allocations: 15.41 MiB)
julia> @btime lsexp_mat2(\$A);
10.726 ms (6 allocations: 7.65 MiB)
julia> @btime lsexp_mat3(\$A);
8.750 ms (21 allocations: 7.65 MiB)
julia> @btime lsexp_mat4(\$A);
2.402 ms (17 allocations: 7.65 MiB)
julia> @btime lsexp_mat5(\$A);
2.557 ms (19 allocations: 7.65 MiB)
julia> using Tracker, Zygote #, ForwardDiff
true
true
true
67.391 ms (3003130 allocations: 137.61 MiB)
22.924 ms (133 allocations: 38.22 MiB)
10.519 ms (128 allocations: 38.22 MiB)
``````
Overall, our performance numbers are really similar except for `lsexp_mat3`, where my computer is much slower.
1 Like
Thatâs interesting, normally your computer smokes mine. Perhaps itâs because `lsexp_mat3` is launcing too many threads? It has an extremely crude calculation which here decides that up to `1000^2 / 3636 = 275` threads would be worthwhile. (And the same in the avx case, but I guess compensated.) Keyword `threads=200_000` will stop it at about 5 threads (meaning 4 or 8).
Thanks for the explanation about broadcasting. So the problem is particular to trivial dimensions which have stride 1, regardless of the other strides involved?
``````@avx ones(10,10)' .* rand(10,1) # no problem
@avx ones(10,10)' .* rand(10,1)' # no problem
@avx ones(10,10)' .* rand(1,10) # problem
@avx ones(10,10)' .* rand(1,10)' # problem
``````
2 Likes
Thanks, thatâs why I thought it was curious/worth pointing out. And that seems to be it.
In particular, `Threads.nthreads()` was `1`. Now using 18 threads:
``````julia> @btime lsexp_mat(\$A);
11.372 ms (13 allocations: 15.41 MiB)
julia> @btime lsexp_mat2(\$A);
11.212 ms (6 allocations: 7.65 MiB)
julia> @btime lsexp_mat3(\$A);
2.105 ms (332 allocations: 7.68 MiB)
julia> @btime lsexp_mat4(\$A);
2.385 ms (17 allocations: 7.65 MiB)
julia> @btime lsexp_mat5(\$A);
1.613 ms (328 allocations: 7.68 MiB)
julia> using Tracker, Zygote #, ForwardDiff
true
true
true
65.850 ms (3003130 allocations: 137.61 MiB)
8.496 ms (788 allocations: 38.28 MiB)
7.639 ms (778 allocations: 38.28 MiB)
``````
Trying different values in the `threads` arguments:
``````julia> @btime lsexp_mat_25_000_threads(\$A);
1.669 ms (328 allocations: 7.68 MiB)
1.681 ms (328 allocations: 7.68 MiB)
1.679 ms (328 allocations: 7.68 MiB)
1.792 ms (160 allocations: 7.67 MiB)
7.700 ms (779 allocations: 38.28 MiB)
7.689 ms (779 allocations: 38.28 MiB)
7.699 ms (780 allocations: 38.28 MiB)
7.902 ms (424 allocations: 38.25 MiB)
``````
Itâs definitely an improvement, but lags well behind the 18 threads. With `avx=false`:
``````julia> @btime lsexp_mat_25_000_threads_noavx(\$A);
2.124 ms (330 allocations: 7.68 MiB)
2.114 ms (330 allocations: 7.68 MiB)
2.123 ms (330 allocations: 7.68 MiB)
2.602 ms (161 allocations: 7.67 MiB)
8.451 ms (784 allocations: 38.28 MiB)
8.450 ms (782 allocations: 38.28 MiB)
8.502 ms (785 allocations: 38.28 MiB)
9.605 ms (429 allocations: 38.25 MiB)
``````
Scaling is a lot better than it looks, because the serial portions take a long time, e.g. around half of all the time is spent in `maximum`:
``````julia> @btime maximum(\$A, dims=1);
903.528 Îźs (3 allocations: 8.00 KiB)
``````
This could of course also be optimized by making it threaded and/or SIMD. We can try a SIMD version via LoopVectorization:
``````julia> @btime(vreduce(max, \$A, dims=1)) == maximum(A, dims=1)
333.883 Îźs (1 allocation: 7.94 KiB)
true
``````
But I didnât try this within `lsexp_mat` due to the lack of Zygote support.
So the problem is particular to trivial dimensions which have stride 1, regardless of the other strides involved?
Yes. The first two examples should work, while the latter two will be broken.
1 Like
I think the core problem here is https://github.com/JuliaDiff/ReverseDiff.jl/issues/135 , I should really change the readme not to promise things which donât work, sorry. (It worked in an earlier incarnation.) That said Iâm a little surprised by those error messages, there may be other details to fix beyond that.
Is the `gensym` problem related to this error too? I thought ReverseDiff doesnât support `vreduce`. Is there anyway to speed up `maximum`
It would probably be easy to make `vreduce` work, by defining a gradient as in this PR using perhaps Trackerâs as the model.
To make `@tullio` work youâd have to solve that issue, the easy part is `Base.gensym(ex::Expr) = gensym(string(ex))` but I didnât see how to make `@grad` deal with callable structs. (The macro wraps the forward & backward functions in a `struct Eval` so that thereâs somewhere to attach gradient definitions.)
Otherwise you may need to go back to write things as broadcasting instead. Or tell Turing to use Zygote not ReverseDiff.
1 Like
Iâm trying to use Tullio with sparse arrays.
``````using Tullio, LoopVectorization, SparseArrays
function logsumexp(mat; dims=1)
max_ = maximum(mat, dims=1)
@tullio exp_mat[i,j] := exp(mat[i,j] - max_[1,j]) - (mat[i,j] == max_[1,j])
sum_exp_ = sum(exp_mat, dims=dims)
@tullio out[i,j] := log1p(sum_exp_[i,j]) + max_[i,j]
end
A = sprand(1500, 10000, 0.02);
logsumexp(A)
``````
When I run the above code I get `signal (6): Aborted`
This is the full stack trace.
``````double free or corruption (!prev)
signal (6): Aborted
in expression starting at REPL[4]:1
gsignal at /lib/x86_64-linux-gnu/libc.so.6 (unknown line)
abort at /lib/x86_64-linux-gnu/libc.so.6 (unknown line)
unknown function (ip: 0x7fe8d09d8906)
unknown function (ip: 0x7fe8d09df979)
cfree at /lib/x86_64-linux-gnu/libc.so.6 (unknown line)
jl_realloc_aligned at /buildworker/worker/package_linux64/build/src/gc.c:249 [inlined]
gc_managed_realloc_ at /buildworker/worker/package_linux64/build/src/gc.c:3369 [inlined]
jl_gc_managed_realloc at /buildworker/worker/package_linux64/build/src/gc.c:3386
array_resize_buffer at /buildworker/worker/package_linux64/build/src/array.c:660
jl_array_grow_at_beg at /buildworker/worker/package_linux64/build/src/array.c:785 [inlined]
jl_array_grow_at at /buildworker/worker/package_linux64/build/src/array.c:929
_growat! at ./array.jl:873 [inlined]
insert! at ./array.jl:1215 [inlined]
_insert! at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.4/SparseArrays/src/sparsematrix.jl:2485
_setindex_scalar! at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.4/SparseArrays/src/sparsematrix.jl:2473
setindex! at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.4/SparseArrays/src/sparsematrix.jl:2447 [inlined]
đđ¸đ! at /home/s/.julia/packages/Tullio/HGzih/src/macro.jl:797
unknown function (ip: 0x7fe87f26475e)
_jl_invoke at /buildworker/worker/package_linux64/build/src/gf.c:2145 [inlined]
jl_apply_generic at /buildworker/worker/package_linux64/build/src/gf.c:2323
_jl_invoke at /buildworker/worker/package_linux64/build/src/gf.c:2145 [inlined]
jl_apply_generic at /buildworker/worker/package_linux64/build/src/gf.c:2323
jl_apply at /buildworker/worker/package_linux64/build/src/julia.h:1700 [inlined]
unknown function (ip: (nil))
Allocations: 55469269 (Pool: 55453222; Big: 16047); GC: 56
Aborted (core dumped)
``````
The code runs fine for 100x100 sparse arrays and 1500x10000 normal arrays. My machine has 45 GB free RAM. What is the problem here?
Thatâs a pretty inelegant failure! I get something similar.
For smaller sizes it does run. But I wouldnât expect it to be efficient, itâs completely unaware of sparsity & just works through every element. It shouldnât be hard to write a fast `logsumexp(:: SparseMatrixCSC )` though.
``````julia> Tullio.storage_type(sprand(10,10,0.1)) # not <: Array{<:BlasFloat}, hence will not use LoopVectorization
SparseMatrixCSC{Float64,Int64}
julia> A[3,4] # but you can index, so fallback seems OK?
0.0
julia> logsumexp(A)
julia(43313,0x700003cdf000) malloc: *** error for object 0x18a8f9000: pointer being freed was not allocated
julia(43313,0x700003cdf000) malloc: *** set a breakpoint in malloc_error_break to debug
signal (6): Abort trap: 6
in expression starting at REPL[73]:1
Allocations: 1296329819 (Pool: 1296254828; Big: 74991); GC: 596
Abort trap: 6
``````
1 Like
I wondered whether ReverseDiff had been quietly fixed, but it seems not, Use with ReverseDiff ¡ Issue #21 ¡ mcabbott/Tullio.jl ¡ GitHub has a simple test. So I think you ought to get errors if Turing is trying to do this, at all. Is there a chance Turing isnât using AD at all?
So `lsexp1` which is purely broadcasting is giving you this corruption error? That might be worth isolating. Is `A` still sparse here?
I mixed up the function names, now I fixed them.
While importing Tullio with ReverseDiff I get the following warning but generally the sampling finishes and estimations are accurate.
``````â Warning: Error requiring ReverseDiff from Tullio:
â Closest candidates are:
â gensym(!Matched::Symbol) at expr.jl:15
â gensym(!Matched::String) at expr.jl:12
â gensym() at expr.jl:10
â ...
â Stacktrace:
â [1] @grad(::LineNumberNode, ::Module, ::Any) at /home/swamy/.julia/packages/ReverseDiff/Thhqg/src/macros.jl:182
â [2] include(::Module, ::String) at ./Base.jl:377
â [3] include(::String) at /home/s/.julia/packages/Tullio/HGzih/src/Tullio.jl:1
â [4] top-level scope at REPL[11]:1
â [5] eval at ./boot.jl:331 [inlined]
â [6] eval at /home/s/.julia/packages/Tullio/HGzih/src/Tullio.jl:1 [inlined]
â [7] (::Tullio.var"#150#154")() at /home/s/.julia/packages/Requires/qy6zC/src/require.jl:85
â [8] err(::Any, ::Module, ::String) at /home/s/.julia/packages/Requires/qy6zC/src/require.jl:42
â [9] (::Tullio.var"#149#153")() at /home/s/.julia/packages/Requires/qy6zC/src/require.jl:84
â [10] withpath(::Any, ::String) at /home/s/.julia/packages/Requires/qy6zC/src/require.jl:32
â [11] (::Tullio.var"#148#152")() at /home/s/.julia/packages/Requires/qy6zC/src/require.jl:83
â [12] listenpkg(::Any, ::Base.PkgId) at /home/s/.julia/packages/Requires/qy6zC/src/require.jl:15
â [13] macro expansion at /home/s/.julia/packages/Requires/qy6zC/src/require.jl:81 [inlined]
â [14] (::Tullio.var"#147#151")() at /home/s/.julia/packages/Requires/qy6zC/src/init.jl:11
â [15] __init__() at /home/s/.julia/packages/Requires/qy6zC/src/init.jl:18
â [21] eval(::Module, ::Any) at ./boot.jl:331
â [22] eval_user_input(::Any, ::REPL.REPLBackend) at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.4/REPL/src/REPL.jl:86
â [23] macro expansion at /buildworker/worker/package_linux64/build/usr/share/julia/stdlib/v1.4/REPL/src/REPL.jl:118 [inlined]
â in expression starting at /home/s/.julia/packages/Tullio/HGzih/src/grad/reverse.jl:8
â in expression starting at /home/s/.julia/packages/Tullio/HGzih/src/grad/reverse.jl:8
â @ Requires ~/.julia/packages/Requires/qy6zC/src/require.jl:44
``````
1. NUTS sampler needs to get gradients from AD, so its somehow getting gradients or else it will output gradient undefined error or something. Iâm sure ReverseDiff is being used because if I use ForwardDiff instead of ReverseDiff it takes much longer to finish sampling.
2. Function using `Tullio` gives the corruption error.
3. A is not sparse, its a dense array.
Iâve written two version logsumexp with maximum function using avx and normal maximum.
This code defines a gradient for `avx_max` function using `@grad` macro in ReverseDiff
``````using LoopVectorization, ReverseDiff
function avx_max(A; j=1)
max_ = zeros(size(A, 2))
@avx for i â 1:size(A, 2)
j = 1
max_el = A[j, i]
for j â 1:size(A, 1)
max_el = max(max_el, A[j,i])
end
max_[i] = max_el
end
reshape(max_, 1, :)
end
fast_max(x) = avx_max(x)
fast_max(x::TrackedArray) = ReverseDiff.track(fast_max, x)
xv = ReverseDiff.value(x)
T = Array{Float64, 2}
max_ = avx_max(xv)
max_ret = T(xv .== max_)
max_, Î -> (max_ret, )
end
``````
logsumexp with avx and without avx
``````function logsumexp_avx(mat; dims=1)
@assert dims == 1
max_ = vec(fast_max(mat, dims=1))' # requires dims=1
exp_mat = @avx exp.(mat .- max_) .- (mat .== max_)
sum_exp_ = sum(exp_mat, dims=dims)
@avx sum_exp_ .= log1p.(sum_exp_) .+ max_
end
function logsumexp_no_avx(mat; dims=1)
max_ = maximum(mat, dims=1)
exp_mat = exp.(mat .- max_) .- (mat .== max_)
sum_exp_ = sum(exp_mat, dims=dims)
sum_exp_ .= log1p.(sum_exp_) .+ max_
end
``````
``````x = rand(3,3);
logsumexp_avx(x) â logsumexp_no_avx(x) #true
``````
I donât know why the gradients of the both functions donât match. The gradients of `avx_max` and juliaâs `maximum` matches.
The gradient of `logsumexp_avx` has `1` added to all the columnwise maximum positions.
For example
``````Gradient of logsumexp_no_avx
0.204813 0.384189 0.500139
0.420175 0.21128 0.242529
0.375012 0.404531 0.257332
0.204813 0.384189 1.50014
1.42017 0.21128 0.242529
0.375012 1.40453 0.257332
``````
``````function logsumexp_avx(mat; dims=1)
@assert dims == 1
max_ = vec(fast_max(mat))' # requires dims=1
exp_mat = @avx exp.(mat .- max_) .- (mat .== max_)
sum_exp_ = sum(exp_mat, dims=dims)
@avx sum_exp_ .= log1p.(sum_exp_) .+ max_
end
``````
(i.e., remove the `dims = 1` argument from the call to fast_max)
Also, more definitions are needed for the gradient to work. Did you define something like
``````LoopVectorization.vmaterialize(bc::Base.Broadcast.Broadcasted{<:ReverseDiff.TrackedStyle}, ::Val{_}) where {_} = Base.materialize(bc)
``````
?
1 Like
My bad I didnât verify what I pasted.
No. I didnât get any errors when I called gradient of either `fast_max` or `logsumexp_avx`.
Surprisingly, if I remove my gradient definition for `fast_max` the gradients of `logsumexp_avx` and `logsumexp_no_avx` matches. I donât know whatâs going on. | 7,813 | 23,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2022-21 | latest | en | 0.712447 |
http://geeks.i-pupil.com/index.php?option=com_content&view=article&id=207&catid=98 | 1,516,702,021,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891886.70/warc/CC-MAIN-20180123091931-20180123111931-00672.warc.gz | 124,559,307 | 4,396 | Description: Let the length of the pattern P be ‘m’. Consider a string S of length ‘n’, given m<n. We have to check if the pattern P occurs in the String S using Naive string matching algorithm(optimized version).
Example:
String: “ipupilrocks”
Pattern: “rock”
As we can see,the pattern in the example occurs in the string.
Algorithm:
Consider the following code snippet:
for i=0 to n-m
{
if(P[0…m-1]==S[i…i+(m-1)])/*compare the string and pattern*/
PATTERN MATCHED;
}
The above algorithm can be optimized using the following technique:
When all characters of the pattern are different:
MODIFICATION:
If after ‘k’ elements have been matched a mismatch occurs,
We can shift the pattern by ‘k’ instead of 1.
Eg:
“ a l g o r o r x m ”
“o r x”
“or” has been matched in both string and pattern. Mismatch occur in the 3rd character of the pattern.
Here, the pattern can be shifted by 2 instead of 1, as “or” have been matched in both String and pattern.
Note that the above modification is applicable only when all characters in the pattern are different.
Time Complexity:
O(n*m)
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Forum is a Link | 1,028 | 3,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-47 | latest | en | 0.703199 |
https://projecteuler.net/index.php?section=problems&id=4 | 1,590,649,668,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00289.warc.gz | 508,851,249 | 2,510 | ## Largest palindrome product
Show HTML problem content Published on Friday, 16th November 2001, 06:00 pm; Solved by 477257;
Difficulty rating: 5%
### Problem 4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers. | 99 | 373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-24 | latest | en | 0.900753 |
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2010-03-18, 14:00 #1 davar55 May 2004 New York City 2·2,099 Posts Sums of all Squares 2^2 + 3^2 + 5^2 + ... + p^2 = 10mK What is the smallest prime p such that the sum of squares of all primes up to p is a multiple of 10 (or 100 or 1000).
2010-03-18, 14:26 #2 axn Jun 2003 17×281 Posts s=0;forprime(p=2,1000,s=s+p^2;if(Mod(s,10)==0, print(p, ":",s))) 907:37464550 967:44505630 977:46403000 991:48351370
2010-03-18, 14:32 #3 davar55 May 2004 New York City 2×2,099 Posts Nice and simple and quick reply -- thanks. I won't ask about extending the list to 10000, etc. .....
2010-03-18, 17:11 #4 CRGreathouse Aug 2006 2×2,969 Posts 907, 977, 977, 36643, 1067749, 17777197, 71622461, 2389799983, ... The next term (if one exists) is more than 4 trillion. Last fiddled with by CRGreathouse on 2010-03-18 at 17:20
2010-03-18, 18:16 #5
"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×599 Posts
Not yet in the OEIS.
http://www.research.att.com/~njas/sequences/
Quote:
Search: 907, [B] 967, 977, 991[/B] I am sorry, but the terms do not match anything in the table. If your sequence is of general interest, please send it to me using the form provided and I will (probably) add it to the data base! Include a brief description and if possible enough terms to fill 3 lines on the screen. I need a minimum of 4 terms.
I think it qualifies. Also, I'm fond of OEIS entries with relatively large initial terms -- especially when the next few terms are so closely spaced as in this one. (Might it set some record in that regard -- highest ratio of initial term to average spacing of next n terms, for n = 3?)
I'd be glad to submit it, but I think it should be one of you guys.
How about generalizing to other bases?
Last fiddled with by cheesehead on 2010-03-18 at 18:34
2010-03-18, 19:02 #6
axn
Jun 2003
17×281 Posts
Quote:
Originally Posted by cheesehead Not yet in the OEIS.
I think CRG's sequence is more "worthy". It is also the solution of OP.
Quote:
I can think of two ways to generalize: to other bases and other powers (other than squares).
2010-03-18, 19:50 #7
"Richard B. Woods"
Aug 2002
Wisconsin USA
22·3·599 Posts
Quote:
Originally Posted by axn I think CRG's sequence is more "worthy".
Even with the omitted and repeated (just a typo) terms? :smile:
What I had in mind was a submission with the best of both your contributions.
Quote:
I can think of two ways to generalize: to other bases and other powers (other than squares).
Oh, wow ... bases 2-16 or so, powers to, say, ninth ==> 135 sequences.
Last fiddled with by cheesehead on 2010-03-18 at 19:56
2010-03-18, 19:59 #8
axn
Jun 2003
10010101010012 Posts
Quote:
Originally Posted by cheesehead Even with the omitted and repeated terms?
That sequence is the first occurrence of 10^n. 977 repeats (not a typo!) because it ends in 000 and comes before any other 00. So it stands at positions 2 & 3.
EDIT:- Mine is merely the first four occurrences of 10
Last fiddled with by axn on 2010-03-18 at 20:02
2010-03-18, 20:02 #9
"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×599 Posts
Quote:
Originally Posted by axn That sequence is the first occurrence of 10^n. 977 repeats because it ends in 000 and comes before any other 00. So it stands at positions 2 & 3.
(Sorry, CRG)
But that doubles the potential number of sequences. 270.
Last fiddled with by cheesehead on 2010-03-18 at 20:04
2010-03-19, 01:28 #10
CRGreathouse
Aug 2006
2×2,969 Posts
Quote:
Originally Posted by CRGreathouse The next term (if one exists) is more than 4 trillion.
That's *billion*, not trillion. Now my search limit is 50 billion, giving me
907, 977, 977, 36643, 1067749, 17777197, 71622461, 2389799983, 31252968359, 49460594569, ...
The nth term is very roughly n * log 10 * 10^n, so I was pretty lucky getting that last term. The next one will probably need over 2 trillion. Anyone up to the task? I don't actually have a good segmented sieve coded at the moment...
2010-03-29, 16:42 #11
bsquared
"Ben"
Feb 2007
1101000011012 Posts
Quote:
Originally Posted by CRGreathouse Anyone up to the task? I don't actually have a good segmented sieve coded at the moment...
I bit.
I have recently spent some time with my sieve, so decided to give this a shot.
I just started a run to 2 trillion. Here is the output so far:
Code:
found primes in range 0 to 1000000000 in elapsed time = 7.0227
**** 907 is 0 mod 10 ****
**** 977 is 0 mod 100 ****
**** 977 is 0 mod 1000 ****
**** 36643 is 0 mod 10000 ****
**** 1067749 is 0 mod 100000 ****
**** 17777197 is 0 mod 1000000 ****
**** 71622461 is 0 mod 10000000 ****
sum of squares complete in elapsed time = 8.5178, sum is 16352255694497179054764665
found primes in range 1000000000 to 2000000000 in elapsed time = 5.9418
sum of squares complete in elapsed time = 7.9423, sum is 126512354351558021982865866
found primes in range 2000000000 to 3000000000 in elapsed time = 5.9503
**** 2389799983 is 0 mod 100000000 ****
sum of squares complete in elapsed time = 7.7389, sum is 418923904898718760122282892
found primes in range 3000000000 to 4000000000 in elapsed time = 5.8990
sum of squares complete in elapsed time = 7.6150, sum is 979895993641271252685833855
found primes in range 4000000000 to 5000000000 in elapsed time = 5.8293
sum of squares complete in elapsed time = 7.4966, sum is 1894402266333772221759233898
With these timing trends, should have a result in 7 hours or so.
- ben.
Similar Threads Thread Thread Starter Forum Replies Last Post a1call Miscellaneous Math 42 2017-02-03 01:29 Nick Number Theory Discussion Group 0 2016-12-11 11:30 3.14159 Miscellaneous Math 12 2010-07-21 11:47 CRGreathouse Math 6 2009-11-06 19:20 m_f_h Puzzles 45 2007-06-15 17:46
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A153631 Numbers n such that n modulo (product of digits of n) = (sum of digits of n). 1
23, 29, 33, 39, 43, 49, 53, 59, 63, 69, 73, 79, 83, 89, 93, 99, 133, 136, 137, 192, 194, 195, 222, 223, 226, 229, 261, 263, 267, 313, 316, 331, 332, 333, 334, 336, 339, 391, 392, 397, 441, 443, 449, 621, 623, 661, 662, 663, 666, 669, 912, 914, 915, 931, 932 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Do consecutive numbers like 136, 137 occur frequently? Do many primes appear in the sequence? Numbers n such that the positive integer remainder of n/A007954(n) equals A007953(n). [R. J. Mathar, Jan 03 2009] LINKS B. D. Swan, Table of n, a(n) for n=1,...,20000 EXAMPLE For n = 83, (product of digits of n) = 24, (sum of digits of n) = 11 and 83 = 3*24 + 11. For n = 93, (product of digits of n) = 27, (sum of digits of n) = 12 and 93 = 3*27 + 12. MAPLE sd := proc (n) options operator, arrow: add(convert(n, base, 10)[j], j = 1 .. nops(convert(n, base, 10))) end proc: pd := proc (n) options operator, arrow: product(convert(n, base, 10)[j], j = 1 .. nops(convert(n, base, 10))) end proc: a := proc (n) if 0 < pd(n) and `mod`(n, pd(n)) = sd(n) then n else end if end proc: seq(a(n), n = 1 .. 1000); # Emeric Deutsch, Jan 02 2009 A007953 := proc(n) local i ; add(i, i=convert(n, base, 10)) ; end: A007954 := proc(n) local i ; mul(i, i=convert(n, base, 10)) ; end: for n from 1 to 1200 do if A007954(n) > 0 then if irem(n, A007954(n))= A007953(n) then printf("%d, ", n) ; fi; fi; od: # R. J. Mathar, Jan 03 2009 CROSSREFS Sequence in context: A244077 A277206 A214754 * A082943 A061754 A180066 Adjacent sequences: A153628 A153629 A153630 * A153632 A153633 A153634 KEYWORD base,easy,nonn AUTHOR J. M. Bergot, Dec 29 2008 EXTENSIONS Edited, corrected (93 inserted) and extended beyond a(21) by Klaus Brockhaus, Jan 06 2009 Corrected and extended by Emeric Deutsch and R. J. Mathar, Jan 02 2009 STATUS approved
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Last modified October 23 22:42 EDT 2019. Contains 328378 sequences. (Running on oeis4.) | 868 | 2,399 | {"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} | 3.828125 | 4 | CC-MAIN-2019-43 | latest | en | 0.522212 |
https://slidetodoc.com/teaching-quantum-mechanics-and-quantum-statistical-mechanics-to/ | 1,632,059,340,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056890.28/warc/CC-MAIN-20210919125659-20210919155659-00060.warc.gz | 582,749,500 | 13,467 | # Teaching Quantum Mechanics and Quantum Statistical Mechanics to
• Slides: 20
Teaching Quantum Mechanics and Quantum Statistical Mechanics to Sophomores Deepthi Amarasuriya Assistant Professor of Physics Northwest College, Powell, WY
GREATER MATHEMATICAL SOPHISTICATION AT THE SOPHOMORE LEVEL COMPUTATIONAL * Material draws heavily upon recently acquired calculus techniques e. g. integrating exponential functions by parts, separating variables, AND introduces lot of new techniques e. g. basic PDEs, boundary value problems * Introduces non - closed form solutions to differential equations ; some require numerical techniques CONCEPTUAL More conceptually sophisticated mathematical spaces, operators as mathematical entities, statistical / non - deterministic approaches
INCORPORATING CHALLENGING MATH CONTENT INTO A PHYSICS COURSE How do you incorporate all this into a course that introduces abstract Physics concepts? Foundation for my approach (based on many years as a tutor/ T. A. ) First two semesters Material is obviously at an introductory level Modern Physics Instructors may forget that students are encountering “advanced” topics at the introductory level.
TYPICAL ACADEMIC BACKGROUND PRIOR Physics I : Newtonian Mechanics, mechanical waves, (brief) introduction to thermodynamics Physics II : Electromagnetism , electromagnetic waves, Introduction to physical and geometrical optics, (brief) introduction to modern physics Calc I, Calc III surface and volume integrals, integration by substitution, introduction to vectors, simple variable separable equations CONCURRENT Differential Equations I : ODEs, variable separable differential equations Possibly - Linear Algebra : Eigenvalues and eigenfunctions taught at the tail end of the course
ADDITIONAL MATH REQUIRED FOR QM, QSM (USUALLY TAUGHT AT THE JUNIOR LEVEL) Vector spaces Operator spaces Expansion in basis functions Eigenvalues and eigenfunctions Orthogonality and orthonormality of functions (Postulates of QM) Partial differential equations (Schrödinger equation for the H atom) Special functions (H atom angular wavefunctions, Riemann zeta function) Discrete probability distributions (MB, BE, FD distributions) Probability density functions (Blackbody radiation – in terms of λ, ν, etc. )
OVERVIEW INCORPORATING MATH PRINCIPLES * Give printed notes with worked examples and math appendices. * Use Power Point slides to show supplementary images and graphics. * Use the “spiral approach’’ when teaching the math background. Revisit at increasing levels of sophistication e. g. blackbody radiation * Keep pointing out each time a specific math technique or principle is used. (Most students may not be recognize a newly introduced usage in a different context. ) * Include “conceptual math” questions in quizzes and exams. (Identification only; no computations)
PRINTED CLASS NOTES - ADVANTAGES Helps with the logistical challenge of presenting the material within a very limited time allocation. * Allows for better student preparation * More class time to do complete exercises by drastically cutting short on transcribing time * Students have the correct material (easy to make mistakes in copying down complicated equations, detailed diagrams e. g. H wave functions in units of a. B) * Can present important proofs, and worked examples (use time saved from writing down all the steps to point out the key steps/ details)
PRINTED CLASS NOTES – ADVANTAGES (CTD. ) Students have more time to focus on the concepts. Students are too stressed out about completing HW on time, to really think about the underlying Physics and Math. (They are looking for a “quick fix”. ) Giving worked examples similar to HW problems gets students started on assignments. Helps students who may feel too overwhelmed with the new materials / approaches to get started on their own initiative.
PRINTED CLASS NOTES - FORMAT Beginning of each chapter and section Overview of the general theme of the chapter /section. Summary of the foundational principles to serve as motivation for the new material. [Inspiration for writing style – Encyclopedias, reputable scientific blogs and articles for a non-specialist audience. ] In the body of the notes Mathematically simple proofs, and those that explicitly show the usage of the assumptions Fully worked out examples – Do at least the infinite square well before presenting postulates of QM.
PRINTED CLASS NOTES – FORMAT CTD. In the body of the notes (ctd. ) Point out Math used; try to keep to the notation used by the text. (If deviating a lot, include list of your notation in printed notes. ) Math instructors may use different symbols especially with spherical and cylindrical coordinates. If not using SI units, provide the transition clearly, and with several worked examples. End of each section Interpretation, significance of results, comparison with Classical Mechanics More challenging proofs in appendices
PRINTED CLASS NOTES - APPENDICES Math background in appendices – each section may include a few additional proofs showing how the math is applied: (i) Mathematical spaces (ii) Mathematical operators (iii) Partial differential equations (iv) Special functions (v) Waves and vibrations (vi) Statistical concepts (vii) Standard integrals, advanced derivations (viii)List of constants and formulae for quizzes and exams.
DEALING WITH POSSIBLE DISADVANTAGES OF PRINTED CLASS NOTES Student inattention / absence – * Do additional problems (including those similar to HW and exam problems) in class. * Give short in-class exercises (open notes/ books) worth a small percentage of the grade.
Power Point slides Supplementary graphics * 3 D images e. g. atomic orbitals * Spectral line shifts * Images of experimental techniques / apparatus * Images of 20 th c. Physicists with brief biographical sketches
HOMEWORK, QUIZZES and EXAMS HW – * Longer problems involving hand calculations * Numerical solutions using graphing calculators and/ or math software Quizzes – Simpler, shorter versions of exams * Test basic concepts (many students may not be sufficiently ready for an exam) * Break down question into several steps – guide students through steps * Write the equation to be solved (do not solve) * Conceptual questions (identifications of formulae and expression; know their significance) Exams – * Numerical questions in story problem format * Conceptual questions (identifications of formulae and expression; know their significance)
IDENTIFICATION QUESTIONS - QM Conceptual understanding of the underlying math Identify each of the following equations: (i) Time independent Schrödinger equation for a spherical potential (ii) Differential equation satisfied by the H atom radial wave function
IDENTIFICATION QUESTIONS - QSM Identify each of the following equations in the case of a large assemblage of identical particles: Probability of an electron having energy E Probability of a photon having energy E
IDENTIFICATION QUESTIONS - QSM Definitions of technical terms Match each phrase in column A with the most appropriate phrase in column B. For the case of a large assemblage of identical particles: A Number of particles with energy Ei B Occupancy number n(Ei) A Number of different states with the same energy Ei [Alternate : Number of distinct ways particles could be rearranged to give the same energy Ei ] B Degeneracy function g(Ei )
FORMAT - QUIZZES and EXAMS * Quizzes – 1 hr; Exams – 2 hrs * No index cards, notes or books allowed * Calculators used only for numerical computations * All required Physics/ Math formulae, physical constants and table of simplest H wave functions provided * Quiz solution key provided to students immediately after they hand in completed quiz
Effectiveness of Modern Physics course (mainly QM and Special Relativity) All students except one who has taken the course (introduced in 2007) have passed; almost everyone with A s and B s. Number of students taking PHYS 2320 at my small (< 2000) community college has increased from 2 -3 per semester to 6. The largest community college in WY (~ 4000 students) has an enrollment of around 8 per semester. A student who took PHYS 2320 at my college recently graduated as a Physics major from the state college (UW) – the only community college student to have done so within the past 10 years. His senior thesis was a topic in quantum mechanics. Another was selected as one of 6 students from a competitive nationwide pool for an NSF funded Astronomy research project at UW. Recent alumni have requested my current version of the notes to help with their upper level classes.
Acknowledgements and References Gautreau, R. Schaum's Outline of Modern Physics Taylor, J. , Zafiratos, C. , Dubson, M. , Modern Physics for Scientists and Engineers, Addison-Wesley Image on title slide – Scott Danzig http: //sneakyghost. com/ Informal exchanges at Articulation meetings with Physics instructors from other community colleges in WY, and from UW. | 1,832 | 8,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.762201 |
https://www.sigmainfy.com/blog/leetcode-container-with-most-water.html | 1,544,586,902,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823712.21/warc/CC-MAIN-20181212022517-20181212044017-00450.warc.gz | 1,049,792,319 | 6,001 | # LeetCode Container With Most Water: Linear Time Solution and Sorting
## Overview
LeetCode Container With Most Water: 1) Linear Time Solution by two pointers; 2) Sorting first and keeping finding the min/max index in constant time.
## LeetCode Container With Most Water
Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container.
## Solution and Precautions: Linear Time Solution and Sorting
(1) Brute force approach: compare every pair of lines to calculate the area between these two lines, keep the max among the O(N^2) pairs, the time complexity is obviously O(N^2) and you might get a TLE error when testing on the big dataset.
(2) Sorting: make a pair for each line as <height, index>, so there would be N such pairs as <a1, 1> <a2, 2>, …, <aN, N>. Sort these N pairs by increasing order on the height value (so ai is in increasing order). Now you can think about how this sort could help before heading into the following context (hint: remember the “Best Time to Buy and Sell Stock” problem, essentially, the solution becomes the same here). Ok, here is the answer, let F(i) denote the max water among the pairs from i-th pair to the last pair after sorting. Now we consider F(i-1), when adding the (i-1)th pair, we need to update the max water, how? since the height is in increasing order, then the (i-1)th pair here must have the shortest line, and the possible amount of water much be this (shortest) height times the difference of indexes between the index value of this (i-1)th pair and the index value among the pairs after. These computation could be done in constant time by keep recording of the minimum index and the maximum index among the pairs after (i-1)th pair. If you check the best time to buy and sell stock problem out, the approach after sorting here is essentially the same one. The total time complexity is O(NlogN + N) = O(NlogN). The following Java code implements this idea and can pass the LeetCode Container With Most Water problem:
(3) Linear approach: If there is no linear approach, then I really think the sorting method is good enough to show your coding ability and knowledge on algorithm. However, there is a even smarter approach though it is not easy to come up with. Basically, we use two pointers say i and j to point to the head and the tail of the original height array. We first calculate the area between line i and line j and try to update the max area if possible. Then we check the lin[i] and lin[j], and advance the pointer to the shorter line by one step (++i or –j). After a linear scan like this, we make sure the max area is found correctly. The idea behind is that, if we encounter a shorter line, say line[i] is shorter than line[j], then there is no need to check other pairs with line[i], if we move j (–j), then the area must only be smaller (why? think about it). The same reasoning applies here for the case where line[j] is shorter. Obviously, this approach takes linear time which is O(N). The following Java code implements this idea and can pass the LeetCode Container With Most Water problem:
And this problem is really an excellent example to show the process to tackle a difficult problem step by step. One should really try the three approaches out.
## Summary
LeetCode Container With Most Water: 1) Linear Time Solution by two pointers; 2) Sorting first and keeping finding the min/max index in constant time.
Written on May 5, 2013 | 863 | 3,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-51 | latest | en | 0.889661 |
https://mattysparadigm.com/2020/10/14/mattys-constant/ | 1,669,964,026,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710898.93/warc/CC-MAIN-20221202050510-20221202080510-00712.warc.gz | 393,223,368 | 44,116 | # Matty’s Constant
For I neither received it of man, neither was I taught it, but by the revelation of Jesus Christ.
(Galatians 1:12) KJV
We can convert heliocentric planetary mass values to Geocentrospheric using Matty’s Constant, 9.87^-12, for the conversion of fantasy into reality.
No one has ever weighed the sun or the planets. The values that we use for their mass are derived from the A priori assumption of either Heliocentricity or Geocentrosphericity. The values depend on the assumption. In either case the mass values of the planetary bodies have the same proportional relationships, they still exert the same gravitational forces on each other, they’re just substantially lower.
We all have the same evidence. Our choice of paradigm determines what we think it’s evidence of.
Matty’s Razor
We thought that it would be possible to tinker with the equation used to calculate the mass of the sun by switching some values in order to derive a new Geocentrospheric mass value. It turns out that it’s not that simple. Here’s a video of a clever chap explaining the calculation. The assumption of heliocentricity is stated at about the 16 second mark.
What happens during the calculation is that the mass of the Earth (the known variable*) ends up on both sides of the equation and so it cancels out. What we end up with is a calculation using angular velocity and the assumption of heliocentricity but nothing else. This means that the popular science (SciPop) value of the mass of the sun is simply a restatement of the premise of heliocentricity. Therefore the mass of the sun can’t be used as proof of heliocentricity because heliocentricity was assumed to begin with. It’s circular reasoning. Garbage in = garbage out.
However, this calculation defines the gravitational relationship between the mass of the Earth and the mass of the sun. According to the calculation the Earth is 0.0003% the mass of the sun. If we change the assumption from heliocentric to Geocentrospheric we can now say that the sun is 0.0003% the mass of the Earth. We haven’t broken any laws of physics. We’re still using Kepler’s 3rd law of motion and Newton’s law of gravity. The only thing we changed is our assumption regarding which body orbits which. The value of the mass of the sun in Matty’s Paradigm is 9.87^-12 of what it is in SciPop.
The mass values of all other planetary and stellar bodies are derived from the mass of the sun, they’re relative to it, they aren’t absolute values. When we change the assumption about which body orbits which, the sun orbiting the Earth or the Earth orbiting the sun, the values change but their relative proportions are exactly the same. We find that, as Newton realized, we don’t know any of the absolute values of the mass of planetary and stellar bodies.
*As we pointed out on October 10th, the mass of the Earth isn’t known, it is a value which has been agreed upon.
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• Accept Jesus Christ as your savior, | 686 | 3,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | longest | en | 0.919083 |
https://www.adidasshoesoutletwholesale.com/what-is-mechanical-potential-energy/ | 1,621,391,919,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00432.warc.gz | 626,712,558 | 10,871 | Mechanical vitality is the vitality that’s possessed by an object as a result of its movement or as a result of its place. A metal ball has extra potential vitality raised above the bottom than it has after falling to Earth.
10 Examples Of Mechanical Vitality In On a regular basis Life Studiousguy Mechanical Vitality Mechanical Drive Metaphor
### In physics mechanical vitality E mech is the vitality related to the movement and place of an object normally in some drive area eg.
What’s mechanical potential vitality. Class IX Chapter 10 Work and Vitality. Different Kinds of Potential Vitality. The straight ahead reply to what’s mechanical vitality is that it’s the sum of vitality in a mechanical system.
So the change of the potential en- ergy is transformed into the kinetic vitality of the particle which provides 1 2 m v 2 3 U 0 U 0 v radicalbigg 4 U 0 m. The precept of conservation of mechanical vitality pertains to each energies. Mechanical vitality IS potential energyKinetic vitality.
If a drive appearing on an object is a operate of place solely it’s mentioned to be a conservative drive and it may be represented by a possible vitality operate which for a one-dimensional case satisfies the spinoff situation. If an object strikes in the other way of a conservative web drive the potential vitality will enhance. In a system that experiences solely conservative forces there’s a potential vitality related to every drive and the vitality solely adjustments type between KE KE dimension 12KE and the assorted sorts of PE.
The potential vitality of the ebook on the desk will equal the quantity of labor it took to maneuver the ebook from the ground to the desk. Mechanical vitality will be both kinetic vitality vitality of movement or potential vitality saved vitality of place. Potential vitality saved vitality that relies upon upon the relative place of varied components of a system.
The entire kinetic plus potential vitality of a system is outlined to be its mechanical vitality mathrmKEPE. The precept of conservation of mechanical vitality states that if an remoted system is topic solely to conservative forces then the mechanical vitality is fixed. Within the open door instance discover what occurs.
Mechanical vitality IS potential energyKinetic vitality. In bodily sciences mechanical vitality is the sum of potential vitality and kinetic vitality. It’s the macroscopic vitality related to a system.
In different phrases it’s vitality in an object as a result of its movement or place or each. Bardbl vectorv bardbl radicalbigg 8 U 0 m Clarification. The components for mechanical vitality is mechanical vitality kinetic vitality potential vitality.
Within the raised place it’s able to doing extra work. A spring has extra potential vitality when it’s compressed or stretched. This vitality contains each kinetic vitality vitality of movement and potential vitality.
Mechanical vitality is because of the place or motion of an object. Potential vitality is the vitality that an object can retailer vitality on account of its place. Mechanical vitality sum of the kinetic vitality or vitality of movement and the potential vitality or vitality saved in a system by purpose of the place of its components.
When somebody places work into lifting the ball he’s rising the balls potential vitality which in some sense is mechanical vitality or vitality of movement saved within the object. Elastic – Elastic potential vitality is saved when supplies stretch or compress. While you carry a hammer potential vitality is rising and kinetic vitality is lowering.
The mechanical vitality of a physique is the sum of the potential and kinetic vitality. The entire vitality of the particle is con-served. The potential vitality is identical as any potential vitality it is determined by the place of the article whereas kinetic vitality is determined by the movement of an object.
Examples of elastic potential vitality embody springs rubber bands and slingshots. The integral type of this relationship is. While you carry a hammer potential vitality is rising and kinetic vitality is lowering.
The potential vitality is identical as any potential vitality it is determined by the place of the article whereas kinetic vitality is determined by the movement of an object. Mechanical Potential Vitality washburne 11039 2 9. The vitality acquired by the objects upon which work is finished is named mechanical vitality.
The entire kinetic plus potential vitality of a system is outlined to be its mechanical vitality KE PE KE PE dimension 12 KEPE. Mechanical vitality and in addition the thermal vitality will be separated into two classes transient and saved. Mechanical vitality is the sum of kinetic and potential vitality in an object used to do work.
In a system that experiences solely conservative forces there’s a potential vitality related to every drive and the vitality solely adjustments type between KE and varied sorts of PE with the entire vitality remaining fixed. And if the velocity of the article adjustments the kinetic vitality of the article additionally c. Mechanical Vitality An objects capability to do work is measured by its mechanical vitality or the sum the objects kinetic vitality and potential vitality.
Thus the work achieved on the ball doesn’t go into oblivion–it turns into saved vitality on this case within the type of gravitational potential vitality. Which will be taken as a definition of potential energyNote that there’s an arbitrary fixed of. This saved vitality of place is known as potential vitality.
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Mechanical Vitality Mechanical Vitality Vitality Mechanic | 1,436 | 7,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2021-21 | latest | en | 0.89834 |
https://www.experiglot.com/2007/06/26/engagement-ring-shopping-an-excel-model-for-pricing-an-asscher-cut-diamond/ | 1,670,285,253,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00193.warc.gz | 791,777,453 | 13,031 | # Engagement ring shopping? An Excel model for pricing an Asscher cut diamond
### Excel spreadsheets (.xls), Personal finance
A while back, I mentioned that some of the people I worked with had mad excel skills, and some people wrote in wanting to see examples. I finally found something I could post here that doesn’t violate any NDAs or contain sensitive company information: a teammate’s personal Asscher cut diamond pricing model (right click to download and open or save). [Updated June 28, 2007: unlocked version of spreadsheet now available.]
I realize this is a bit out of left field, so here’s a little background. Someone recently joined our group who’d built an Excel-based model to price out how much an should cost, based on several factors. He did this so that he could get the right engagement ring for his fiancĂ©e, where “right” was defined as the biggest bang for the buck (though I have to say it’s a gorgeous ring, and I’m no diamond afficionado). After putting the model to good use, he posted publicly on a diamond forum, and I got his permission to post it here as well.
This guy knew nothing at all about diamonds before he started, but he visited all the diamond stores — Tiffany, Cartier, etc. — to gather data and learn about diamonds, and in particular, the Asscher cut. (This cut has apparently gained a lot of popularity in recent years due to its appearance in celebrities’ engagements and the media.) After getting all the data, he used Excel to run a multivariate regression to find the most statistically significant determinants of price, and out of that, he created a pricing model.
The model uses 10 variable as inputs: cut, color, clarity, carat, polish, symmetry, debth [sic], table, florescence, and L/W (length-to-width ratio). You can see the impact of changing each variable on the price by using the dropdown menu items.
I’m posting his model here in case it’s useful to other people, and as an example of the power of Excel in the hands of an expert. Actually, pricing models such as this one can be used in all sorts of business applications, from pricing products to creating should-cost models (e.g. how much a product or service from a supplier should cost, based on various inputs).
, the forum where he posted his model has received many views, but only one comment of appreciation, so if you have feedback, I’m sure he’d be interested. But beware that I am absolutely not a diamond expert in any way (I don’t even own one), so if you have questions about them sparklies, you’ll be better off visiting a diamond forum like Pricescope than posting your question here :)
Enjoy!
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8 Feedbacks on "Engagement ring shopping? An Excel model for pricing an Asscher cut diamond"
### Reader
Password? (Doesn’t really help as a learning tool if it’s locked).
### Ricemutt
Sorry – I didn’t realize the sheet was protected. I meant to show only the front portion, not the actual data sources. I’ll see if I can get the password.
### paul in 02144
not as fascinating as I had hoped, but maybe that’s because diamonds don’t excite me much.
my project deals with analyzing a lot of back-end data (who’s project doesn’t?) and I’ve found excel Pivot Tables to be quite useful. Also, Advanced Filtering is everything you do by hand, almost ;-)
### dong
I’m a bit of an excel junkie (though not quite as much as I’ve gotten older). I think in the end to really unlock the power in Excel, it’s about using Visual Basic. In a lot of ways, Excel is just interface for the VBA backend connections into databases.
Anyhow, it looks like your friend got quite a Rock….
### Personal Finance Guide 101
Oh yeah 503 views and one comment on your friend Dale’s post. Well I am not an expert in excel but I could say he did a great job and it a lot of work…
### Pilot
You’re right. It is gorgeous. When my daughter was in High School, I taught her how to do her first simple regression (over the dinner dishes) by using her own assessments of the quality of colleges she was considering and regressing them against cost.
Our deal emerged from that conversation and model. So long as she chose a college with quality above the CI of the regression line for a given cost, I’d pay for it all. She chose an excellent public college way above the line. I saved a bundle and she graduated with no debt — and even got to spend a semester abroad.
### Asscher Cut Diamond
Wow! Pretty amazing spreadsheet. It is one of those thing that you ask why there would be such a need but once you see it it makes complete sense.
### Dale
Glad people liked the model and my good friend Leslie posted it. I have since build one for Radiant & princess cuts for friends. These are super helpful when you are not a diamond expert, but are looking to make your commitment “official.” I didn’t know anything and didn’t want to take a dealer at his word. Numbers/ data levels the playing field a bit.
Enjoy!
Dale | 1,096 | 5,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-49 | latest | en | 0.951559 |
https://onlinejudge.org/board/search.php?author_id=704&sr=posts | 1,611,057,096,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00525.warc.gz | 486,774,088 | 8,140 | ## Search found 27 matches
Thu Nov 13, 2003 10:05 pm
Forum: Volume 105 (10500-10599)
Topic: 10566 - Crossed Ladders
Replies: 39
Views: 12603
### 10566
Ok, you got me !!!
Should I treat it as a special case ???
Bye, Aur
Thu Nov 06, 2003 4:37 pm
Forum: Volume 105 (10500-10599)
Topic: 10566 - Crossed Ladders
Replies: 39
Views: 12603
### 10566
Maarten, you're right. I removed that lines, and change fabs(ft) by fabs(a - b) > 1e-15 and now I get Wrong Answer ... Here some sample input/output for my program ... Input: 30 40 10 12.619429 8.163332 3 10 10 3 10 10 1 300 400 100 126.19429 81.63332 30 100 100 30 100 100 10 3000 4000 1000 1261.942...
Wed Nov 05, 2003 10:54 pm
Forum: Volume 105 (10500-10599)
Topic: 10566 - Crossed Ladders
Replies: 39
Views: 12603
### 10566 - TLE
Hi !!!
fa and fb stores the value of the function, applied to the limits a and b ...
The root is inside this limits.
Tue Nov 04, 2003 6:26 pm
Forum: Volume 105 (10500-10599)
Topic: 10566 - Crossed Ladders
Replies: 39
Views: 12603
### 10566 - TLE
Hello, can you help me with my code ? I
Wed Sep 10, 2003 4:23 am
Forum: Volume 100 (10000-10099)
Topic: 10071 - Back to High School Physics
Replies: 49
Views: 8701
### 10071
Hello,
Is there a way to optimize this problem, it's very simple, but I get 0.094 sec.
Thanks, Aur
Thu Jul 17, 2003 9:15 pm
Forum: Other words
Topic: Online Judge
Replies: 4
Views: 1288
### Online Judge
Is it off ? The last submission from http://acm.uva.es/cgi-bin/OnlineJudge?Status:Valladolid is yesterday ... I've tried to submit, but I can't.
Thanks, Aur
Mon Mar 31, 2003 7:46 am
Forum: Volume 100 (10000-10099)
Topic: 10062 - Tell me the frequencies!
Replies: 235
Views: 42757
### 10062
Hi,
First of all, I think you should do ascii check conditon ...
Good luck , Aur
Tue Mar 25, 2003 3:41 am
Forum: Volume 100 (10000-10099)
Topic: 10062 - Tell me the frequencies!
Replies: 235
Views: 42757
Hi all, I don't understand why I get WA on my code: [c] #include <stdio.h> void main () { int i, j, c = 0, vetor[256], vetor1[256], temp1; char str[1000], temp; while (gets(str) != NULL) { for (i = 0; i < 256; i++) { vetor = vetor1 = 0; } temp1 = 0; str[1000]='\0'; for (i = 1; str != '\0'; i++) { fo...
Sat Feb 15, 2003 2:12 am
Forum: Volume 104 (10400-10499)
Topic: 10452 - Marcus
Replies: 21
Views: 10298
Here is my code: [c]#include <stdio.h> #include <string.h> int main() { int cases,m,n,i,j,mcopy,flag_right,flag_left; char cobble[100000][10]; scanf("%d",&cases); while(cases) { scanf("%d %d",&m,&n); for(i=0;i<=m;i++) cobble [0] = '\0'; mcopy = m; while(mcopy) { scanf...
Wed Feb 12, 2003 8:07 pm
Forum: Volume 104 (10400-10499)
Topic: 10452 - Marcus
Replies: 21
Views: 10298
Are you getting WA ??? I've tried your code and for the input:
Code: Select all
``````PST#T
BTJAS
TYCVM
YEHOF
XIBKU
N@RJB``````
it gives me:
Code: Select all
``forth forth right right forth forth forth left``
Code: Select all
``forth forth right right forth forth forth``
My problem is worst, I'm getting TLE, don't know why !!!
Tue Feb 11, 2003 11:54 pm
Forum: Volume 104 (10400-10499)
Topic: 10452 - Marcus
Replies: 21
Views: 10298
### 10452 - Marcus
Is there something special about this problem ? I think it's not difficult, and besides there is only one path, right ?? But I don't know why I get TLE !!!
Thanks for any reply !!!
Tue Feb 04, 2003 1:37 pm
Forum: Volume 104 (10400-10499)
Topic: 10432 - Polygon Inside A Circle
Replies: 62
Views: 21510
Thanks a lot, I've follow the suggestions and get accepted now
But I have a question ! Why ACM judge doesn't define M_PI ? I tried it and get compiler error !
Aur
Mon Feb 03, 2003 3:14 pm
Forum: Volume 104 (10400-10499)
Topic: 10432 - Polygon Inside A Circle
Replies: 62
Views: 21510
### 10432 - Polygon Inside A Circle
What's wrong with my code ? [c] #include <stdio.h> #include <math.h> #define Pi 3.1415926535897932384626433832795029L int main() { long double ang; int r, n; while(scanf("%d %d",&r,&n)==2) { ang = 180 - ((double)180*(n-2)/n); printf("%.3lf\n",(sin(Pi/180*ang)*r*r*n)/2); }...
Thu Jan 30, 2003 2:25 am
Forum: Volume 101 (10100-10199)
Topic: 10197 - Learning Portuguese
Replies: 45
Views: 16620
### 10197 - Learning Portuguese
Hi all,
Could you give me some test cases, I don know why I get wrong answer
Thanks, Aur
Tue Jan 28, 2003 1:13 am
Forum: Volume 100 (10000-10099)
Topic: 10033 - Interpreter
Replies: 88
Views: 27445
I didn't understand this problem ... should I only count the number of operations ??? And for the sample input the output shouldn't be 15 ???
Thanks, Aur | 1,631 | 4,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.749817 |
https://www.cse.chalmers.se/~nad/listings/dependently-typed-syntax/README.DependentlyTyped.NormalForm.html | 1,624,471,570,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00402.warc.gz | 654,680,735 | 6,101 | ------------------------------------------------------------------------
-- Normal and neutral terms
------------------------------------------------------------------------
open import Level using (zero)
open import Data.Universe
(Uni₀ : Universe zero zero) where
open import Data.Product as Prod renaming (curry to c; uncurry to uc)
open import Function hiding (_∋_) renaming (const to k; _∘_ to _⊚_)
import README.DependentlyTyped.Term as Term; open Term Uni₀
import Relation.Binary.PropositionalEquality as P
-- Atomic types (types for which the normaliser does not do
-- η-expansion).
data _⊢_atomic-type (Γ : Ctxt) : Type Γ Set where
: {σ} Γ , σ atomic-type
el : {σ} Γ el , σ atomic-type
-- Two kinds: normal and neutral.
data Kind : Set where
no ne : Kind
mutual
-- Γ ⊢ σ ⟨ no ⟩ represents η-long, β-normal terms, Γ ⊢ σ ⟨ ne ⟩
-- represents neutral terms.
infixl 9 _·_
infix 3 _⊢_⟨_⟩
data _⊢_⟨_⟩ (Γ : Ctxt) : Type Γ Kind Set where
ne : {σ} (σ′ : Γ σ atomic-type) (t : Γ σ ne )
Γ σ no
var : {σ} (x : Γ σ) Γ σ ne
ƛ : {σ τ} (t : Γ σ τ no ) Γ Type-π σ τ no
_·_ : {sp₁ sp₂ σ}
(t₁ : Γ π sp₁ sp₂ , σ ne ) (t₂ : Γ fst σ no )
Γ snd σ ŝub t₂ ne
-- Normal and neutral terms can be turned into ordinary terms.
⌊_⌋ : {Γ σ k} Γ σ k Γ σ
ne σ′ t = t
var x = var x
ƛ t = ƛ t
t₁ · t₂ = t₁ · t₂
-- Normal and neutral terms are "term-like".
Tm-n : Kind Term-like _
Tm-n k = record { _⊢_ = λ Γ σ Γ σ k ; ⟦_⟧ = ⟦_⟧ ⌊_⌋ }
private
open module T {k} = Term-like (Tm-n k) public
using ([_])
renaming ( ⟦_⟧ to ⟦_⟧n; _≅-⊢_ to _≅-⊢n_
; drop-subst-⊢ to drop-subst-⊢n; ⟦⟧-cong to ⟦⟧n-cong
)
-- As mentioned above normal and neutral terms can be turned into
-- ordinary terms.
forget-n : {k} [ Tm-n k ⟶⁼ Tm ]
forget-n = record
{ function = λ _ ⌊_⌋
; corresponds = λ _ _ P.refl
}
-- Various congruences.
ne-cong : {Γ σ} {σ′ : Γ σ atomic-type}
{t₁ t₂ : Γ σ ne }
t₁ ≅-⊢n t₂ ne σ′ t₁ ≅-⊢n ne σ′ t₂
ne-cong P.refl = P.refl
var-n-cong : {Γ₁ σ₁} {x₁ : Γ₁ σ₁}
{Γ₂ σ₂} {x₂ : Γ₂ σ₂}
x₁ ≅-∋ x₂ var x₁ ≅-⊢n var x₂
var-n-cong P.refl = P.refl
ƛn-cong : {Γ₁ σ₁ τ₁} {t₁ : Γ₁ σ₁ τ₁ no }
{Γ₂ σ₂ τ₂} {t₂ : Γ₂ σ₂ τ₂ no }
t₁ ≅-⊢n t₂ ƛ t₁ ≅-⊢n ƛ t₂
ƛn-cong P.refl = P.refl
·n-cong :
{Γ₁ sp₁₁ sp₂₁ σ₁}
{t₁₁ : Γ₁ π sp₁₁ sp₂₁ , σ₁ ne } {t₂₁ : Γ₁ fst σ₁ no }
{Γ₂ sp₁₂ sp₂₂ σ₂}
{t₁₂ : Γ₂ π sp₁₂ sp₂₂ , σ₂ ne } {t₂₂ : Γ₂ fst σ₂ no }
t₁₁ ≅-⊢n t₁₂ t₂₁ ≅-⊢n t₂₂ t₁₁ · t₂₁ ≅-⊢n t₁₂ · t₂₂
·n-cong P.refl P.refl = P.refl | 1,113 | 2,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-25 | latest | en | 0.359494 |
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Showing results 1 to 25 of 133 Search took 0.02 seconds. Search: Posts Made By: LiquidNitrogen
Forum: Lounge 2011-08-05, 01:46 Replies: 513 Views: 27,452 Posted By LiquidNitrogen When you see "PrimeCare" painted on a van So today on the way back from JP Morgan Chase bank in Wilmington, Delaware, I saw a van with "PrimeCare, Milford, DE" painted on the side of it. It took me 3 traffic lights to catch up to him and a...
Forum: No Prime Left Behind 2011-08-04, 00:49 Replies: 18 Views: 14,304 Posted By LiquidNitrogen Now that is cool, thanks! Do you have that... Now that is cool, thanks! Do you have that factor.exe program that is being referred to in the documentation? I went to the website it was linked to, but that site is down now. Why can't authors...
Forum: No Prime Left Behind 2011-08-04, 00:36 Replies: 18 Views: 14,304 Posted By LiquidNitrogen Ok, so I think I have it now. Is this spreadsheet... Ok, so I think I have it now. Is this spreadsheet correct then? 6875
Forum: No Prime Left Behind 2011-08-03, 15:11 Replies: 18 Views: 14,304 Posted By LiquidNitrogen I'm not sure I follow. Here is what I did. ... I'm not sure I follow. Here is what I did. 1. Proving p = 2^n - 1 is prime for n = 5, p = 31. 2. S(0) = 4 {defined} 3. Need to generate up to S(n-2) where S(x+1) = [S(x) * S(x)] - 2 3a. S(1)...
Forum: No Prime Left Behind 2011-08-03, 04:38 Replies: 18 Views: 14,304 Posted By LiquidNitrogen I did find one typo on... I did find one typo on http://primes.utm.edu/prove/prove1.html In 2002 a long standing question was answered: can integers be prove prime I think this should be changed to the word proven...
Forum: No Prime Left Behind 2011-08-03, 03:20 Replies: 18 Views: 14,304 Posted By LiquidNitrogen That makes the point crystal clear. I haven't... That makes the point crystal clear. I haven't seen any online resources offering such a concise explanation. The first answer you gave answers this one. With such a huge gap in the prime...
Forum: Information & Answers 2011-08-03, 03:06 Replies: 7 Views: 1,224 Posted By LiquidNitrogen Many thanks! Many thanks!
Forum: Information & Answers 2011-07-31, 20:22 Replies: 7 Views: 1,224 Posted By LiquidNitrogen I can only find links to FermFact 0.9 can you... I can only find links to FermFact 0.9 can you show me where to download it?
Forum: Math 2011-07-30, 01:34 Replies: 20 Views: 1,690 Posted By LiquidNitrogen pi(x) = x/[ln(x) - 1.08] was offered as one... pi(x) = x/[ln(x) - 1.08] was offered as one "improvement", very long time ago.
Forum: Soap Box 2011-07-30, 01:27 Replies: 740 Views: 47,175 Posted By LiquidNitrogen The debt problem can be solved in 2 easy steps. ... The debt problem can be solved in 2 easy steps. 1. Flat tax, no deductions. Poor? Too bad, pay the lower tax and deal with it. Rich? We're sticking it to you the same percent as the poor, so stop...
Forum: Hardware 2011-07-30, 01:11 Replies: 11 Views: 812 Posted By LiquidNitrogen Yes that would be nice. But I found... Yes that would be nice. But I found hyper-threading produces a tremendous amount of heat on each core when it is active and being used. Download this (better do it fast, mods have a tendency to...
Forum: Hardware 2011-07-29, 23:37 Replies: 11 Views: 812 Posted By LiquidNitrogen I have a 5.0 GHz i5-2500K, which has no... I have a 5.0 GHz i5-2500K, which has no hyperthreading. In my experiments, I have found that in using only 2 cores with LLR, LLR can "steal" processor time from my unused cores, and performance is...
Forum: Hardware 2011-07-29, 23:23 Replies: 11 Views: 1,056 Posted By LiquidNitrogen I love when I get calls like that! I totally... I love when I get calls like that! I totally make up stuff when I know I have a clown on the phone. "My computer is saying it has 2 terabytes of RAM but only 80K of disk space available so it...
Forum: No Prime Left Behind 2011-07-29, 23:19 Replies: 18 Views: 14,304 Posted By LiquidNitrogen So I guess that means when we "discover" primes... So I guess that means when we "discover" primes of the various forms, they are much larger than the primes that are known to exist in continuity, and therefore they are "real discoveries," in a...
Forum: Information & Answers 2011-07-29, 23:14 Replies: 7 Views: 1,224 Posted By LiquidNitrogen I looked at some constants with Nash weights <... I looked at some constants with Nash weights < 2500. After the "initial sieving" in NewPGen, where it first displays the results after a period of noise free operation, some of them had already...
Forum: No Prime Left Behind 2011-07-26, 23:54 Replies: 18 Views: 14,304 Posted By LiquidNitrogen Continuity of Primes With all of the various ways to generate primes (Riesel, Proth, Mersenne etc) there seems to be quite a few "gaps" that must occur in the "plain old" primes that can't be expressed with an elegant,...
Forum: Information & Answers 2011-07-26, 23:49 Replies: 7 Views: 1,224 Posted By LiquidNitrogen Nash Weights vs. Sievability I am wondering if there is a relationship governing how well a candidate Riesel prime is sieved vs. its Nash weight. It seems that low Nash weights, which generate fewer primes, also seive very...
Forum: GPU Computing 2011-07-26, 23:25 Replies: 31 Views: 1,778 Posted By LiquidNitrogen They happen often enough in the programming... They happen often enough in the programming world, they get their own nickname: Fencepost errors http://en.wikipedia.org/wiki/Off-by-one_error
Forum: Software 2011-07-20, 00:53 Replies: 18 Views: 7,285 Posted By LiquidNitrogen I'd like to share something that you may find... I'd like to share something that you may find interesting. One of the computers I built in Dec 2010 was starting to behave oddly in the March 2011 timeframe. I ran every stress test I could think...
Forum: Hardware 2011-07-20, 00:40 Replies: 2 Views: 700 Posted By LiquidNitrogen Now that is cool! I don't know about the... Now that is cool! I don't know about the prices he was quoting for that build though. Seemed way too low.
Forum: Riesel Prime Search 2011-07-17, 23:04 Replies: 372 Sticky: The First Megabit Drive Views: 37,076 Posted By LiquidNitrogen Ahh, I see, thanks. Ahh, I see, thanks.
Forum: Riesel Prime Search 2011-07-17, 12:15 Replies: 372 Sticky: The First Megabit Drive Views: 37,076 Posted By LiquidNitrogen Sorry, didn't mean to imply they were all even,... Sorry, didn't mean to imply they were all even, just a "wide range of them" were. I've only sieved on odd powers with fixed exponents, so I was not used to seeing them. Wow!...
Forum: GPU Computing 2011-07-16, 21:06 Replies: 39 Views: 2,225 Posted By LiquidNitrogen As having built over 200 systems personally, I... As having built over 200 systems personally, I would like to impart a few words of wisdom regarding power supplies. Something that is not on most people's radar when it comes to the Power Supply...
Forum: Information & Answers 2011-07-16, 20:50 Replies: 9 Views: 1,079 Posted By LiquidNitrogen That was the most articulate, well-written... That was the most articulate, well-written explanation I have experienced as a result of asking a question that I posted to this forum. Thanks 10^6 + 1
Forum: Riesel Prime Search 2011-07-16, 18:44 Replies: 372 Sticky: The First Megabit Drive Views: 37,076 Posted By LiquidNitrogen How deeply sieved and choice of exponents I noticed instead of a large range of k values, the list I am crunching right now has a few small ks and a wide range of even exponents. I am wondering how deeply this file was sieved, and what...
Showing results 1 to 25 of 133
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http://de.metamath.org/mpeuni/riesz4.html | 1,721,278,690,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514822.16/warc/CC-MAIN-20240718034151-20240718064151-00477.warc.gz | 12,880,423 | 8,408 | Hilbert Space Explorer < Previous Next > Nearby theorems Mirrors > Home > HSE Home > Th. List > riesz4 Structured version Visualization version GIF version
Theorem riesz4 28307
Description: A continuous linear functional can be expressed as an inner product. Uniqueness part of Theorem 3.9 of [Beran] p. 104. See riesz2 28309 for the bounded linear functional version. (Contributed by NM, 16-Feb-2006.) (New usage is discouraged.)
Assertion
Ref Expression
riesz4 (𝑇 ∈ (LinFn ∩ ConFn) → ∃!𝑤 ∈ ℋ ∀𝑣 ∈ ℋ (𝑇𝑣) = (𝑣 ·ih 𝑤))
Distinct variable group: 𝑤,𝑣,𝑇
Proof of Theorem riesz4
StepHypRef Expression
1 fveq1 6102 . . . . 5 (𝑇 = if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0})) → (𝑇𝑣) = (if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0}))‘𝑣))
21eqeq1d 2612 . . . 4 (𝑇 = if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0})) → ((𝑇𝑣) = (𝑣 ·ih 𝑤) ↔ (if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0}))‘𝑣) = (𝑣 ·ih 𝑤)))
32ralbidv 2969 . . 3 (𝑇 = if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0})) → (∀𝑣 ∈ ℋ (𝑇𝑣) = (𝑣 ·ih 𝑤) ↔ ∀𝑣 ∈ ℋ (if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0}))‘𝑣) = (𝑣 ·ih 𝑤)))
43reubidv 3103 . 2 (𝑇 = if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0})) → (∃!𝑤 ∈ ℋ ∀𝑣 ∈ ℋ (𝑇𝑣) = (𝑣 ·ih 𝑤) ↔ ∃!𝑤 ∈ ℋ ∀𝑣 ∈ ℋ (if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0}))‘𝑣) = (𝑣 ·ih 𝑤)))
5 inss1 3795 . . . 4 (LinFn ∩ ConFn) ⊆ LinFn
6 0lnfn 28228 . . . . . 6 ( ℋ × {0}) ∈ LinFn
7 0cnfn 28223 . . . . . 6 ( ℋ × {0}) ∈ ConFn
8 elin 3758 . . . . . 6 (( ℋ × {0}) ∈ (LinFn ∩ ConFn) ↔ (( ℋ × {0}) ∈ LinFn ∧ ( ℋ × {0}) ∈ ConFn))
96, 7, 8mpbir2an 957 . . . . 5 ( ℋ × {0}) ∈ (LinFn ∩ ConFn)
109elimel 4100 . . . 4 if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0})) ∈ (LinFn ∩ ConFn)
115, 10sselii 3565 . . 3 if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0})) ∈ LinFn
12 inss2 3796 . . . 4 (LinFn ∩ ConFn) ⊆ ConFn
1312, 10sselii 3565 . . 3 if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0})) ∈ ConFn
1411, 13riesz4i 28306 . 2 ∃!𝑤 ∈ ℋ ∀𝑣 ∈ ℋ (if(𝑇 ∈ (LinFn ∩ ConFn), 𝑇, ( ℋ × {0}))‘𝑣) = (𝑣 ·ih 𝑤)
154, 14dedth 4089 1 (𝑇 ∈ (LinFn ∩ ConFn) → ∃!𝑤 ∈ ℋ ∀𝑣 ∈ ℋ (𝑇𝑣) = (𝑣 ·ih 𝑤)) | 1,116 | 1,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.304094 |
https://preprod.bigthink.com/videos/what-math-can-teach-physics/ | 1,723,451,135,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00458.warc.gz | 359,686,741 | 28,027 | Who's in the Video
Peter Woit is a mathematical physicist at Columbia University. He graduated in 1979 from Harvard University with bachelor's and master's degrees in physics and obtained his PhD in particle theory from Princeton University in[…]
5 min
with
Peter Woit explains the “deep relation” between the two disciplines and the most mind-bending new ways in which that relation is being explored.
Question: In what new ways could math be applied to solve the problems of physics?
Peter Woit: The thing that most fascinates me about this whole subject is that something that probably quickly get very technical, but there's an area of mathematics which is known as representation theory. One way of thinking about it is in terms of what physicists often call symmetries. So, for instance, one of the basic facts about the laws of nature is that there are symmetric laws of nature are the same at - if you move in any direction, or you move in time back and forth, the laws of physics don't change. If you rotate things around in three dimensions, the laws of physics don't change. And so, these so-called symmetries have very important physical implications. The fact that the laws of physics don't change as if you move in time has physical implications that there's this thing called energy and energy is conserved, and the same thing - and the fact that the laws of physics don't change of you move back and forth in different directions in space implies that there is something called momentum and momentum is conserve and doesn't change as you evolve in time.
So, the very, very fundamental facts about physics are kind of deeply grounded in the symmetries of nature. And so to a mathematician this question is a question about what we call groups and representation of groups. So, if you go into any math department and look at what they're doing, you'll see that a lot of people in different kinds of mathematics are studying different structures which are also called groups and they're often studying what is called representations of the groups. So, even people studying number theory, abstract things about prime numbers or something, they're also studying groups and certain representations of these groups. So, there's kind of a, to the extent that mathematics has a kind of unifying theme and a unifying principle which shows up in different areas of mathematics, it's about these representations, or representation theory. And the thing that most strikes me about physics, and what fascinates me about it is, if you look at quantum mechanics, quantum mechanics initially looks like a very odd structure. It's not something were we have any kind of intuitive understanding of it. It doesn't look like the way we're used to thinking about physics, based on every day experience. But if you look at the mathematical structure and the basic structure of quantum mechanics, they're exactly the structures that show up in this theory of representations. So, there's a kind of a deep relation between math and physics which is surrounding this whole notion of symmetries and representation of symmetries.
So, that's one thing that's always fascinated me, and my own research and my own interests is in developing - taking a lot that has been learned in mathematics. There's a lot that's been learned in mathematics over the years about how to think about representations and how to construct them and how to work with them. Some of it has made its way into physics and have been used in physics and was used in physics since the early days of quantum mechanics. So, for instance, one of the great ****, there's a kind of hero of this book I wrote about this, it's The Mathematician called Herman Vial, who was one of the first people to understand how quantum mechanics worked and to understand the relation to representations.
But anyway, I think there's still a lot to be learned in that way and very specifically the so-called standard model has this group of symmetries which is called the gage symmetry and it's an infinitial group and the standard kind of physics understanding of the representations of this group is that that should not be an interesting question. There should only be a trivial representation of this group. But anyway, my conjecture is that there is actually a more interesting question there and in pursuing this question of how do you deal with the gauged symmetry of this theory in terms of using ideas from representation theory that are more well-known in mathematics that hadn't been used in physics before that you can actually get somewhere. Well that's maybe too technical, but that's as good as I can do with this.
Recorded on December 16, 2009
Interviewed by Austin Allen
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If you can identify a foreground star, the spike patterns are a dead giveaway as to whether it’s a JWST image or any other observatory. | 1,076 | 5,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.972743 |
https://math.stackexchange.com/questions/3670732/leibniz-rule-for-covariant-derivative/3670839 | 1,714,019,656,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297284704.94/warc/CC-MAIN-20240425032156-20240425062156-00661.warc.gz | 337,445,739 | 38,276 | # Leibniz rule for covariant derivative
I've been learning about the covariant derivative and I have some doubts.
This answer suggests that $$\nabla_{\mathbf{u}} T = \nabla T (\mathbf{u})$$, where $$T$$ is a tensor. The tensor $$\nabla T$$ appears to be acting on the vector $$\mathbf{u}$$ in the same way a covector acts on a vector to give a scalar.
The answer then proceeds to derive the identity $$\nabla^2_{\mathbf{u}, \mathbf{v}} = \nabla_{\mathbf{u}} \nabla_{\mathbf{v}} \mathbf{w} - \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w}$$, where $$\mathbf{u}$$, $$\mathbf{v}$$ and $$\mathbf{w}$$ are vectors.
According to my interpretation, $$\nabla_{\mathbf{u}} \nabla_{\mathbf{v}} \mathbf{w} = \nabla_{\mathbf{u}} (\nabla \mathbf{w} (\mathbf{v})) \\ = \underbrace{(\nabla_{\mathbf{u}} (\nabla \mathbf{w}))}_{\text{a (1,1) tensor}} (\mathbf{v}) + \nabla \mathbf{w} (\nabla_{\mathbf{u}} \mathbf{v}) \\ = \underbrace{\nabla \nabla \mathbf{w}}_{\text{a (1,2) tensor}}(\mathbf{u}, \mathbf{v}) + \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w} \\ = \nabla^2_{\mathbf{u}, \mathbf{v}} + \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w} \\ \therefore \nabla^2_{\mathbf{u}, \mathbf{v}} = \nabla_{\mathbf{u}} \nabla_{\mathbf{v}} \mathbf{w} - \nabla_{\nabla_{\mathbf{u}} \mathbf{v}} \mathbf{w}.$$
My confusion arises here. Let $$T$$ and $$S$$ be tensors. The above derivation make use of some version of the Leibniz rule that appears to be of the form $$\nabla_{\mathbf{u}}(T(S)) = (\nabla_{\mathbf{u}} T)(S) + T(\nabla_{\mathbf{u}} S)$$. Is my interpretation correct?
Yet according to this answer, the rule $$\nabla (T\otimes S) = \nabla T \otimes S + T\otimes \nabla S$$ doesn't exist, but when you add a direction $$\mathbf{u}$$ $$\nabla_{\mathbf{u}} (S\otimes T) = \nabla_\mathbf{u} S \otimes T + S \otimes \nabla_\mathbf{u} T$$, it suddenly becomes true. Why?
I'm quite confused by these various versions of the Leibniz rule and the "total covariant derivative" $$\nabla$$ versus the covariant derivative $$\nabla_{\mathbf{u}}$$. I appreciate if someone could clear it up for me a little.
You should think about the two "covariant derivatives" $$\nabla T$$ and $$\nabla_u T$$ the same way you think about differentials and directional derivatives of scalar functions:
If $$f : M \to \mathbb R,$$ then the covector field $$df$$ is defined in terms of the directional derivatives $$uf$$ by $$df(u) = uf.$$ In vector calculus, we thought about the gradient instead, and would have written this something like $$\nabla f \cdot u = D_u f.$$
In exactly the same way, we simply define $$\nabla T (u)= \nabla_u T,$$ and (after checking that this is indeed tensorial in $$u$$) we have "bundled up" all the derivatives of the tensor field $$T$$ into a tensor field of one degree higher.
Your calculation for the second covariant derivative (and the Leibniz rule $$\nabla_u(S \otimes T) = \nabla_u S \otimes T + S \otimes \nabla_u T \tag 1$$ that you used in it) are perfectly correct.
The only reason the rule $$\nabla (T\otimes S) = \nabla T \otimes S + T\otimes \nabla S \tag 2$$ is incorrect is the order of the slots/indices. To make this concrete, let's suppose $$S$$ and $$T$$ are covector fields for simplicity. In index notation, the correct Leibniz rule is $$\nabla_i(T\otimes S)_{jk} = (\nabla_i T)_j S_k + T_j (\nabla_i S)_k.$$ Note that the direction of differentiation is always $$\partial_i$$. On the other hand, the incorrect rule $$(2)$$ would translate into index notation as $$\nabla_i(T \otimes S)_{jk}=(\nabla_iT)_jS_k+T_i (\nabla_jS)_k.$$ Thus $$(2)$$ has to be corrected by some transposition of indices, something like $$\nabla(T \otimes S) = \nabla T \otimes S + \operatorname{swap}_{12} (T \otimes \nabla S).$$ I've had to invent this "swap" notation for slot transposition, since (as far as I know) there is no conventional way to write this operation when using index-free notation in DG. Usually, authors take one of the following approaches:
• Use an index-based notation where transposition (and contraction) of higher-order tensors is simple and intuitive to notate.
• "Plug in" enough vectors/covectors (treated as free variables) that the transposition becomes unnecessary, as in $$(1).$$
• In some cases, just abuse notation and write $$(2)$$, even though it is technically incorrect. In situations where you're not likely to get the various slots mixed up, it's very neat and conceptually clear.
• Thanks a lot for the answer. This clears up a lot of my confusion. So the Leibniz rule still holds for a tensor acting on another tensor, i.e. $\nabla (T(S))$? Is there some source I can refer to? May 12, 2020 at 8:12
• @user7777777: yes, this "action of a tensor on a tensor" can be written as a contraction of $T\otimes S$, so the Leibniz rule applies. I learnt this material from O'Neill's Semi Riemannian-Geometry..., which covers this material in quite some detail - see the section on "tensor derivations". May 12, 2020 at 9:15
• Alright, this makes sense now. Thank you very much! May 12, 2020 at 10:39
Say you have two tensors $$\omega,\eta$$ of valence $$(0,1)$$ (i.e., $$1$$-forms). Then $$\nabla\omega$$ and $$\nabla\eta$$ are $$(0,2)$$ tensors. For two vectors $$u,v$$, what should $$\nabla\omega(u,v)$$ mean? The usual thing, which you did, is to interpret it as $$(\nabla_u\omega)(v)$$, but someone may (although unlikely) interpret it as $$(\nabla_v\omega)(u)$$. This is not a problem, since almost everyone understands the first meaning and we are all happy.
Now, if $$u,v,w$$ are vectors, what do you think $$\nabla(\omega\otimes\eta)(u,v,w)$$ should be? of course, the standard answer is $$\nabla_u(\omega\otimes\eta)(v,w)$$, which equals $$\nabla_u\omega(v)\eta(w)+\omega(v)\nabla_u\eta(w)$$ (you can prove it). However, notice what happens if you apply $$\nabla\omega\otimes\eta+\omega\otimes\nabla\eta$$ to the same tuple of vectors $$(u,v,w)$$ using the same convention: you get \begin{align*} (\nabla\omega\otimes\eta+\omega\otimes\nabla\eta)(u,v,w) &= \nabla\omega(u,v)\eta(w)+\omega(u)\nabla\eta(v,w) \\ &= \nabla_u\omega(v)\eta(w)+\omega(u)\nabla_v\eta(w) \\ &\neq \nabla_u\omega(v)\eta(w)+\omega(v)\nabla_u\eta(w) \\ &= \nabla_u(\omega\otimes\eta)(v,w) \\ &= \nabla(\omega\otimes\eta)(u,v,w) \end{align*} which is not what we expected. That is why $$\nabla(\omega\otimes\eta)\neq\nabla\omega\otimes\eta+\omega\otimes\nabla\eta$$.
How to fix this? Well, let me advertise my second favourite option: using abstract index notation. In this convention, we use indices to indicate the slots of a tensor, and the tensor product is just juxtaposition. For example, the contraction of $$\omega$$ with a vector $$v$$ is written as $$\omega(v)=\omega_av^a$$, the tensor product $$\omega\otimes\eta$$ looks like $$(\omega\otimes\eta)_{ab}=\omega_a\eta_b$$, the covariant derivative (without being applied to a vector) is $$(\nabla \omega)_{ab}=\nabla_a\omega_b$$ and the applied covariant derivative is $$(\nabla_u\omega)_a=u^b\nabla_b\omega_a$$.
How does this help us? Well, then it is true that $$(\nabla(\omega\otimes\eta))_{abc}=\nabla_a(\omega_b\eta_c)=\nabla_a\omega_b\eta_c+\omega_b\nabla_a\eta_c$$. This is because the indices keep track of who should eat whom in the case three wilde vectors $$u^av^bw^c$$ appear. Don't let your notation to inconvenience you.
In this notation, your calculation is written as
\begin{align*} \nabla_u(\nabla_vw) &= u^a\nabla_a(v^b\nabla_bw^c) \\ &= u^a[\nabla_av^b\nabla_bw^c+v^b\nabla_a\nabla_bw^c] \\ &= u^a\nabla_av^b\nabla_bw^c+u^av^b\nabla_a\nabla_bw^c \text. \end{align*}
Let me know if you have any question. | 2,584 | 7,620 | {"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": 67, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.753885 |
https://www.arxiv-vanity.com/papers/1709.06168/ | 1,611,492,538,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00778.warc.gz | 660,597,785 | 53,744 | # The distribution of consecutive prime biases and sums of sawtooth random variables
Robert J. Lemke Oliver Department of Mathematics, Tufts University and Kannan Soundararajan Department of Mathematics, Stanford University
###### Abstract.
In recent work, we considered the frequencies of patterns of consecutive primes and numerically found biases toward certain patterns and against others. We made a conjecture explaining these biases, the dominant factor in which permits an easy description but fails to distinguish many patterns that have seemingly very different frequencies. There was a secondary factor in our conjecture accounting for this additional variation, but it was given only by a complicated expression whose distribution was not easily understood. Here, we study this term, which proves to be connected to both the Fourier transform of classical Dedekind sums and the error term in the asymptotic formula for the sum of .
Robert Lemke Oliver was partially supported by NSF grant DMS-1601398. Kannan Soundararajan was partially supported by NSF grant DMS-1500237 and by a Simons Investigator grant from the Simons Foundation.
## 1. Introduction
Let denote the sequence of primes in ascending order. Given and satisfying for all , in recent work [7] we studied biases in the occurrence of the pattern in strings of consecutive primes reduced . Thus, we defined
π(x;q,a):=#{pn≤x:pn+i−1≡ai(mod q) for 1≤i≤r},
and conjectured that
(1.1)
where and are certain explicit constants. The term is easily described,
c1(q;a)=ϕ(q)2(r−1ϕ(q)−#{i≤r−1:ai≡ai+1(mod q)}),
and it acts as a bias against immediate repetitions in the pattern . The term is more complicated, and the goal of this paper is to understand its distribution. If then
c2(q;a)=r−1∑i=1c2(q;(ai,ai+1))+ϕ(q)2r−2∑j=11j(r−1−jϕ(q)−#{i:ai≡ai+j+1(mod q)}),
so that it is sufficient to understand the case ; that is, with .
For the sake of simplicity, we shall confine ourselves to the case when is prime. For any character we define
(1.2) Aq,χ=∏p∤q(1−(1−χ(p))2(p−1)2).
Then the quantity is given by
c2(q;(a,a))=q−22log(q/2π),
and when by
(1.3) c2(q;(a,b))=12log2πq+qϕ(q)∑χ≠χ0(mod q)(¯¯¯¯χ(b−a)+1ϕ(q)(¯¯¯¯χ(b)−¯¯¯¯χ(a)))L(0,χ)L(1,χ)Aq,χ.
The diagonal term is thus completely explicit, and of size . Our work here shows that the off-diagonal terms can also be large; usually they are of size about , occasionally getting to size (attaining both positive and negative values), which we believe is their maximal size.
Before stating our result, we make one more simplification. Define
(1.4) C(k)=C(k;q)=1ϕ(q)∑χ≠χ0(mod q)¯¯¯¯¯¯¯¯¯¯¯χ(k)L(0,χ)L(1,χ)Aq,χ.
Since , , and (upon using the functional equation) , from (1.3) it follows that for
(1.5) c2(q;(a,b))q=C(b−a)+O((logq)2√q).
Thus for large it is enough to understand the distribution of as varies over all non-zero residue classes . Since for even characters , in (1.4) only odd characters make a contribution, and therefore is an odd function of .
###### Theorem 1.1.
(1) As the distribution of tends to a continuous probability distribution, symmetric around . Precisely, there is a continuous function with such that uniformly for all one has
1q#{k(mod q): C(k)≤eγ2x}=ΦC(x)+o(1).
(2) Uniformly for all one has
exp(−A1ex/x)≥1q#{k(mod q): C(k)≥eγ2x}≥exp(−A2exlogx)
for some positive constants and .
(3) For all large , there exists with
−C(−k)=C(k)≥(eγ4−ϵ)loglogq.
(4) For all we have
C(k)≪(logq)23(loglogq)2.
(5) The values have an “almost periodic” structure. Precisely, suppose is a multiple of every natural number below . Then
1q∑k(mod q)|C(k)−C(k+m)|2≪1B1−ϵ+mqlogB.
We make a few comments concerning Theorem 1.1 before proceeding to related results. In part 1, we believe that the distribution for has a density, which is to say that is in fact differentiable. Our proof falls just a little short of establishing this. In part 2, there is a gap between the upper and lower bounds for the tail frequencies. With a little more care, we can improve the lower bound there to for a suitable positive constant , but there still remains a gap between the two bounds. The distribution of , and especially the double exponential decay seen in part 2, are reminiscent of the distribution of values of (see [4]). Motivated by this analogy, or by extrapolating the lower bounds in part 2, we believe that in part 3 there should exist values of as large as . We also conjecture that should be the largest possible value of , which would be a substantial strengthening of part 4. Finally, in addition to the almost periodic structure given in part 5 (where varies), there should be an almost periodic structure as varies. That is, if and are two large random primes with being a multiple of the numbers below , then and will be close to each other (on average over ). We hope that an interested reader will embrace some of these remaining problems.
While the quantity is the main focus of this paper, closely related objects arise in two other seemingly unrelated contexts. The first of these concerns Dedekind sums. Let denote the sawtooth function defined by
ψ(x)={{x}−1/2 if x∉Z,0 if x∈Z,
which is an odd function, periodic with period . If is prime and is a reduced residue , then the Dedekind sum is defined by
(1.6) sq(a):=∑x(mod q)ψ(xq)ψ(axq).
The Dedekind sum arises naturally in number theory when studying the modular transformation properties of the Dedekind -function, but it also appears in other contexts and satisfies many interesting properties [1, 11]. We study here the discrete Fourier transform of the Dedekind sum . Thus for a prime and residue class we define
(1.7) ˆsq(t):=1q∑a(mod q)sq(a)e(at/q),
where throughout. In Lemma 2.1 we shall see that
ˆsq(t)=−1πiϕ(q)∑χ≠χ0(% mod q)¯χ(t)L(0,χ)L(1,χ),
so that is indeed a simpler version of . An alternative useful expression is
(1.8) ˆsq(t)=1πi∞∑n=1(n,q)=1ψ(t¯¯¯n/q)n,
where denotes the multiplicative inverse of the reduced residue class and the sum converges since the partial sums are bounded.
###### Theorem 1.2.
(1) As the distribution of tends to a continuous probability distribution, symmetric around . Precisely, there is a continuous function with such that uniformly for all one has
1q#{t(mod q): πiˆsq(t)≤eγ2x}=Φs(x)+o(1).
(2) Uniformly for all one has
exp(−A1ex/x)≥1q#{t(mod q): πiˆsq(t)≥eγ2x}≥exp(−A2exlogx)
for some positive constants and .
(3) For all large , there exists with
−πiˆsq(−t)=πiˆsq(t)≥(eγ4−ϵ)loglogq.
(4) For all we have
ˆsq(t)≪(logq)23(loglogq)2.
(5) The values have an “almost periodic” structure. Precisely, suppose is a multiple of every natural number below . Then
1q∑t(mod q)|ˆsq(t)−ˆsq(k+m)|2≪1B1−ϵ+mqlogB.
Theorem 1.2 exactly parallels the results of Theorem 1.1, with the same deficiencies discussed there. The proofs of Theorems 1.1 and 1.2 are nearly identical, and so we give details only for Theorem 1.1.
Our third topic concerns the remainder term in the asymptotic for the mean value of Euler’s -function. Define the quantity by the relation
∑n≤xϕ(n)=3π2x2+R(x).
Simple arguments show that , and Walfisz [12] established that , which is presently the best known estimate. Montgomery [8] conjectured that and , and he showed that . Key to Montgomery’s work is the expression
R(x)=ϕ(x)2−x∑n≤xμ(n)ψ(x/n)n+O(xexp(−c√logx))
for some positive constant , where if . The sum in this expression is akin to the equation (1.8) for with replaced by and with the weight replaced with . Accordingly, many of the techniques used to prove Theorems 1.1 and 1.2 apply to as well, though unfortunately with less precision owing to the presence of . For convenience, we define .
###### Theorem 1.3.
As the distribution of for real tends to a probability distribution, symmetric around . Precisely, there is a function with such that uniformly for all one has
1ymeas({u≤y: ˜R(u)≤3eγπ2x})=ΦR(x)+o(1),
where denotes the Lebesgue measure of . Moreover, uniformly for all one has
1ymeas({u≤y: ˜R(u)≥3eγπ2x})≤exp(−A1ex/x)
for some positive constant .
We prove Theorem 1.3 by showing that all positive integral moments of exist and are not too large. The moment calculation refines earlier work of Pillai and Chowla [10] and Chowla [3], who computed the mean and variance respectively:
∑n≤x˜R(n)=o(x)and1y∫y0˜R(u)2du∼12π2.
In Theorem 1.3, using Montgomery’s construction in his -result, we can obtain a lower bound for the frequency of large values of of the form , which is very far from the upper bound. We expect that there is a lower bound similar to that in Theorems 1.1 and 1.2 in this situation also, and this would be in keeping with Montgomery’s conjecture on the true size of .
### Organization
Our main focus is the proofs of Theorems 1.1 and 1.2. We establish preliminary results useful for both in Sections 2 and 3. We then prove Theorem 1.1 in Sections 4-6; since the proof of Theorem 1.2 follows along identical lines, we omit it. In Section 7, we discuss the modifications that lead to Theorem 1.3.
## 2. First steps
Here we establish some formulae for and which will be the basis for our subsequent work.
###### Lemma 2.1.
Let be prime. For any , we have
ˆsq(t)=−1πiϕ(q)∑χ≠χ0(% mod q)¯χ(t)L(0,χ)L(1,χ)=1πi∞∑n=1(n,q)=1ψ(t¯¯¯n/q)n.
Moreover, for any we have
###### Proof.
For any non-principal character , we have (see, e.g., [13, Theorem 4.2])
(2.1) L(0,χ)=−∑a(mod q)χ(a)ψ(a/q).
Notice that if is an even character, and that right side of the formula in (2.1) evaluates to 0 if is principal. The functional equation for odd characters gives
L(1,χ)=−τ(χ)πiqL(0,¯χ),
where denotes the Gauss sum. Thus we obtain
∑χ≠χ0(mod q)¯χ(t)L(0,χ)L(1,χ) =−πiq∑χ(mod q)τ(χ)∣∣∑a(mod q)χ(a)ψ(aq)∣∣2 =−πiq∑a,b,m(mod q)e(m/q)ψ(aq)ψ(bq)∑χ(mod q)χ(am)¯χ(bt) =−ϕ(q)πiq∑a,b≢0(mod q)e(tb¯¯¯aq)ψ(aq)ψ(bq) =−ϕ(q)πiˆsq(t).
The first identity in the lemma follows.
To obtain the second identity, note that using (2.1) and the orthogonality relation for characters
−∑χ≠χ0(mod q)¯¯¯¯χ(t)L(0,χ)∑n≤Nχ(n)n =∑χ(mod q)¯¯¯¯χ(t)∑a(mod q)χ(a)ψ(a/q)∑n≤Nχ(n)n (2.2) =ϕ(q)∑n≤N(n,q)=11nψ(t¯¯¯n/q).
Letting , the second identity follows.
To obtain the truncated version, note that
∣∣∑n≤x(n,q)=1ψ(t¯¯¯n/q)∣∣≤q
trivially, and therefore
∑n>x(n,q)=1ψ(t¯¯¯n/q)n=∫∞x1y2∑x
Recall the definition of from (1.2). Expanding this product out, we find
(2.3) Aq,χ=(2χ(2)−χ(2)2)∏p∤2q(1−1(p−1)2)(1+2χ(p)−χ(p)2p2−2p)=C∞∑n=1a(n)χ(2n).
Here
(2.4) C=2∏p≥3p∤q(1−1(p−1)2),
and is a multiplicative function defined by and for all , and for odd primes we have
(2.5) a(p)=2p(p−2),a(p2)=−1p(p−2),and a(pv)=0 for all v≥3.
From the definition of it is easy to check that converges for all so that
(2.6) ∑n≥N|a(n)|≪N−12+ϵand C∑n≤N(n,q)=1a(n)=1+O(N−12+ϵ).
###### Lemma 2.2.
Define the multiplicative function by setting , so that unless is odd and square-free, and for all odd primes . Then for any natural number we have
C(k)=−C∑n≤N(n,q)=1b(n)ψ(k¯¯¯¯¯¯2n/q)+O(q32+ϵN−14+ϵ).
###### Proof.
Arguing as in (2) we find
(2.7) 1ϕ(q)∑χ≠χ0(mod q)¯¯¯¯¯¯¯¯¯¯¯χ(k)L(0,χ)∑n≤N(n,q)=1b(n)χ(2n)=−∑n≤N(n,q)=1b(n)ψ(k¯¯¯¯¯¯2n/q).
Now if then either or and . Therefore
(2.8) ∑n≤Nb(n)χ(2n)=∑u≤√Na(u)χ(2u)∑v≤N/uχ(v)v+∑v≤√Nχ(v)v∑√N
Bounding the partial sums of characters trivially, we find
(2.9) L(1,χ)=∑n≤xχ(n)n+∫∞x∑x
and so the first term in (2.8) is (using (2.6))
∑u≤√Na(u)χ(2u)(L(1,χ)+O(quN)) = C−1Aq,χL(1,χ)+O((logq)N−14+ϵ)+O(qN−12) = C−1Aq,χL(1,χ)+O(qN−14+ϵ).
As for the second term in (2.8), using (2.6) we may bound this by
≪∑v≤√N1vN−14+ϵ≪N−14+ϵ.
We conclude that
C∑n≤N(n,q)=1b(n)ψ(k¯¯¯¯¯¯2n/q)=−1ϕ(q)∑χ≠χ0(mod q)¯¯¯¯¯¯¯¯¯¯¯χ(k)L(0,χ)(Aq,χL(1,χ)+O(qN−14+ϵ)),
and since , the lemma follows. ∎
Lemmas 2.1 and 2.2 give crude approximations to and by long sums (for example taking in Lemma 2.1, or taking in Lemma 2.2). However, on average over or , it is possible to approximate these quantities by very short sums.
###### Lemma 2.3.
Let be a real number. Then
1ϕ(q)∑k(mod q)∣∣C(k)+C∑n≤Bb(n)ψ(k¯¯¯¯¯¯2n/q)∣∣2≪B−1+ϵ,
and
1ϕ(q)∑t(mod q)∣∣ˆsq(t)−1πi∑n≤Bψ(t¯¯¯n/q)n∣∣2≪B−1+ϵ.
###### Proof.
We shall content ourselves with proving the estimate for , the situation for being entirely similar. Using (2.7) and Lemma 2.2 we see that
1ϕ(q)∑k(mod q)∣∣C(k)+C∑n≤Bb(n)ψ(k¯¯¯¯¯¯2n/q)∣∣2 = 1ϕ(q)∑k(mod q)∣∣Cϕ(q)∑χ≠χ0(mod q)¯¯¯¯¯¯¯¯¯¯¯χ(k)L(0,χ)∑B
Using the orthogonality of characters to evaluate the sum over , this is
≪1ϕ(q)2∑χ≠χ0(mod q)|L(0,χ)|2∣∣∑B
and using (2.9) and the functional equation this is
≪1q∑χ≠χ0(mod q)∣∣∑m≤q2χ(m)m∑B
Write temporarily
∑m≤q2χ(m)m∑B
for some coefficients . Then (including also the contribution of below)
1q∑χ≠χ0(mod q)∣∣∑B
The terms with , both below (so that ) contribute
≪∑B
The terms with contribute (assume without loss of generality that is the larger one)
≪qϵ∑B
Assembling these estimates, the lemma follows. ∎
## 3. A key quantity
We shall study and by computing their moments, and the following key quantity will arise in this context. Let be a natural number, and suppose , , are natural numbers. Then set
(3.1) B(n1,…,nℓ)=1n1⋯nℓ∫n1⋯nℓ0ℓ∏j=1ψ(x/nj)dx.
###### Proposition 3.1.
The quantity satisfies the following properties.
(1) If is odd then . For even we have
(3.2) B(n1,…,nℓ)=(i2π)ℓ∑k1,…,kℓ≠0∑kj/nj=01k1⋯kℓ,
where the sum is over all non-zero integers , and this sum is absolutely convergent. In the case one has
B(n1,n2)=(n1,n2)212n1n2.
(2) If is a prime dividing and such that does not divide any other , then
B(n1,…,nj,…,nℓ)=1pB(n1,…,nj/p,…,nℓ).
(3) If we write where and are coprime and is square-free while is square-full then
|B(n1,…,nℓ)|≤2−ℓr−1.
We begin by recalling the Fourier expansion of the sawtooth function. Note that and for we have
(3.3) ˆψ(k)=∫10ψ(x)e(−kx)dx=1−2πik=i2πk,
and so
(3.4) ψ(x)=i∑k≠0e(kx)2πk.
This series converges conditionally pointwise for each , and also in the -sense. For any non-negative integer , recall also the Fejer kernel
(3.5)
We shall find it convenient to replace by the approximation defined by
(3.6) ψN(x)=i∑0<|k|≤Ne(kx)2πk(1−|k|N+1).
Note that is the convolution of with the Fejer kernel
ψN(x)=∫10ψ(y)KN(x−y)dy,
and so
(3.7) |ψN(x)−ψ(x)|≪min(1,1N∥x∥),
which implies that
(3.8) ∫10|ψN(x)−ψ(x)|dx≪1+logNN.
Note also that always.
###### Proof of Proposition 3.1: Part 1.
Since is an odd function, it is clear that for odd . Now suppose is even. By Parseval it follows that
(3.9) 1n1⋯nℓ∫n1⋯nℓ0ψN(x/n1)⋯ψN(x/nℓ)dx=(i2π)ℓ∑0<|kj|≤N∑kj/nj=01k1⋯kℓℓ∏j=1(1−|kj|N+1).
For any complex numbers , , and , , note the simple identity
(3.10) α1⋯αℓ−β1⋯βℓ=(α1−β1)α2⋯αℓ+β1(α2−β2)α3⋯αℓ+β1⋯βℓ−1(αℓ−βℓ).
Applying this, we obtain
|ψ(x/n1)⋯ψ(x/nℓ)−ψN(x/n1)⋯ψN(x/nℓ)|≤12ℓ−1ℓ∑j=1|ψ(x/nj)−ψN(x/nj)|,
and so by (3.9) and (3.8) we conclude that
B(n1,…,nℓ) =1n1⋯nℓ∫n1⋯nℓ0ψ(x/n1)⋯ψ(x/nℓ)dx (3.11) =(i2π)ℓ∑0<|kj|≤N∑kj/nj=01k1⋯kℓℓ∏j=1(1−|kj|N+1)+O(1+logNN).
We now show that
∑0<|kj|≤N∑kj/nj=01|k1⋯kℓ|
is bounded, so that (3) will imply (letting ) the stated formula (3.2) for and that the sum there converges absolutely. By Parseval
∑0<|kj|≤N∑kj/nj=01|k1⋯kℓ|=1n1⋯nℓ∫n1⋯nℓ0ℓ∏j=1(∑0<|kj|≤Ne(kjx/ | 5,491 | 15,114 | {"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} | 2.796875 | 3 | CC-MAIN-2021-04 | latest | en | 0.920659 |
https://www.allinterview.com/showanswers/99545/main-printf-x-1-4.html | 1,561,609,169,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000613.45/warc/CC-MAIN-20190627035307-20190627061307-00459.warc.gz | 642,555,045 | 7,935 | main()
{
printf("%x",-1<<4);
}
Answers were Sorted based on User's Feedback
main() { printf("%x",-1<<4); }..
fff0
Explanation :
-1 is internally represented as all 1's. When
left shifted four times the least significant 4 bits are
filled with 0's.The %x format specifier specifies that the
integer value be printed as a hexadecimal value.
Is This Answer Correct ? 76 Yes 16 No
main() { printf("%x",-1<<4); }..
thts ans is ture 4 16 bit compiler ,for 32 bit ans is fffffff0.
Is This Answer Correct ? 28 Yes 7 No
main() { printf("%x",-1<<4); }..
-16
Is This Answer Correct ? 3 Yes 11 No
More C Code Interview Questions
const int perplexed = 2; #define perplexed 3 main() { #ifdef perplexed #undef perplexed #define perplexed 4 #endif printf("%d",perplexed); } a. 0 b. 2 c. 4 d. none of the above
Is the following statement a declaration/definition. Find what does it mean? int (*x)[10];
3) Int Matrix of certain size was given, We had few valu= es in it like this. =97=97=97=97=97=97=97=97=97=97=97 1 = | 4 | | 5 | &= nbsp; | 45 =97=97=97=97=97=97=97=97=97=97=97 &n= bsp; | 3 | 3 | 5 | = | 4 =97=97=97=97=97=97=97=97=97=97=97 34 |&nbs= p; 3 | 3 | | 12 | &= nbsp; =97=97=97=97=97=97=97=97=97=97=97 3 | &nbs= p; | 3 | 4 | = | 3 =97=97=97=97=97=97=97=97=97=97=97 3 | = ; | | | = ; 3 | =97=97=97=97=97=97=97=97=97=97=97 &= nbsp; | | 4 | = ; | 4 | 3 We w= ere supposed to move back all the spaces in it at the end. Note: = If implemented this prog using recursion, would get higher preference.
Predict the Output: int main() { int *p=(int *)2000; scanf("%d",2000); printf("%d",*p); return 0; } if input is 20 ,what will be print
main() { int c[ ]={2.8,3.4,4,6.7,5}; int j,*p=c,*q=c; for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){ printf(" %d ",*p); ++p; } }
How to return multiple values from a function?
why the range of an unsigned integer is double almost than the signed integer.
What is the match merge ? compare data step match merge with proc sql merge - how many types are there ? data step vs proc sql
write a c program to Create a registration form application by taking the details like username, address, phone number, email along with password and confirm password (should be same as password).Ensure that the password is of 8 characters with only numbers and alphabets. Take such details for 5 users and display the details. In place of password display “****”. (Use Structures).
char *someFun() { char *temp = “string constant"; return temp; } int main() { puts(someFun()); }
print a semicolon using Cprogram without using a semicolon any where in the C code in ur program!!
main( ) { char *q; int j; for (j=0; j<3; j++) scanf(“%s” ,(q+j)); for (j=0; j<3; j++) printf(“%c” ,*(q+j)); for (j=0; j<3; j++) printf(“%s” ,(q+j)); }
Categories | 943 | 2,792 | {"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} | 2.65625 | 3 | CC-MAIN-2019-26 | latest | en | 0.636425 |
https://convert-dates.com/days-from/180/2020/11/22 | 1,606,972,386,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141718314.68/warc/CC-MAIN-20201203031111-20201203061111-00658.warc.gz | 207,847,201 | 4,298 | Developed and designed by Code Attic
## 180 Days From November 22, 2020
Want to figure out the date that is exactly one hundred eighty days from Nov 22, 2020 without counting?
Your starting date is November 22, 2020 so that means that 180 days later would be May 21, 2021.
You can check this by using the date difference calculator to measure the number of days from Nov 22, 2020 to May 21, 2021.
May 2021
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
1. 1
1. 2
2. 3
3. 4
4. 5
5. 6
6. 7
7. 8
1. 9
2. 10
3. 11
4. 12
5. 13
6. 14
7. 15
1. 16
2. 17
3. 18
4. 19
5. 20
6. 21
7. 22
1. 23
2. 24
3. 25
4. 26
5. 27
6. 28
7. 29
1. 30
2. 31
May 21, 2021 is a Friday. It is the 141st day of the year, and in the 20th week of the year (assuming each week starts on a Sunday), or the 2nd quarter of the year. There are 31 days in this month. 2021 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 05/21/2021, and almost everywhere else in the world it's 21/05/2021.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 180 weekdays from Nov 22, 2020, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Nov 22, 2020, which falls on a Sunday. Counting forward, the next day would be a Monday.
To get exactly one hundred eighty weekdays from Nov 22, 2020, you actually need to count 250 total days (including weekend days). That means that 180 weekdays from Nov 22, 2020 would be July 30, 2021.
If you're counting business days, don't forget to adjust this date for any holidays.
July 2021
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
1. 1
2. 2
3. 3
1. 4
2. 5
3. 6
4. 7
5. 8
6. 9
7. 10
1. 11
2. 12
3. 13
4. 14
5. 15
6. 16
7. 17
1. 18
2. 19
3. 20
4. 21
5. 22
6. 23
7. 24
1. 25
2. 26
3. 27
4. 28
5. 29
6. 30
7. 31
July 30, 2021 is a Friday. It is the 211st day of the year, and in the 211st week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 31 days in this month. 2021 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 07/30/2021, and almost everywhere else in the world it's 30/07/2021.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 939 | 2,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-50 | longest | en | 0.933152 |
lynema.org | 1,576,466,940,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541315293.87/warc/CC-MAIN-20191216013805-20191216041805-00116.warc.gz | 91,877,004 | 14,139 | # Eight Years Later: Dijkstra’s Algorithm in Java
CS331 was a fun class. Data structures are right up my ally. They take a simple concept and expand it to apply to scenarios in neat and interesting ways. It was eight years ago when I took (and passed) that class…Yikes!
Most of the assignment, actually, 4 out of 5 assignments, were coded and turned in on time. This was not a small feat. In college, it was common for me to work a 30 hour week. Getting assignments done was usually a balancing act of putting one thing on hold while moving on to another (such is life). Four out of five is not all that bad. So why did that one problem give me such an issue. Why has there been a sour note in the melody that was a great collegiate career? Because I said I would finish it, that’s why.
Java coding has been taking up a lot of my nights lately. It’s fun again to write code; just like in college. When thinking up of things to code, the memory of this assignment came to the front of my mind. It was time.
After reviewing the C++ I had written 8 years ago, it was obvious that it was close to working. There was just a piece or two missing. Perhaps that is another reason why this had been stuck in my mind. The other thing that was obvious is that I name my variables much better now. Reading my old code was nothing less than a chore. Writing code that someone else, not myself, can maintain is something that I took for granted then. Of course, in college, it was cheating to work with someone else on a project :).
The rules for the application
• The application will read input from a text file
• The application will use Dikjstra’s algorithm to find the shortest path between all reachable nodes in the graph
Here’s a short refresher on Dijkstra’s Algorithm from Wikipedia:
Let the node at which we are starting be called the initial node. Let the distance of node Y be the distance from the initial node to Y. Dijkstra’s algorithm will assign some initial distance values and will try to improve them step by step.
1. Assign to every node a distance value: set it to zero for our initial node and to infinity for all other nodes.
2. Mark all nodes as unvisited. Set initial node as current.
3. For current node, consider all its unvisited neighbors and calculate their tentative distance. For example, if current node (A) has distance of 6, and an edge connecting it with another node (B) is 2, the distance to B through A will be 6+2=8. If this distance is less than the previously recorded distance, overwrite the distance. All unvisited neighbors are added to an unvisited set.
4. When we are done considering all neighbors of the current node, mark it as visited. A visited node will not be checked ever again; its distance recorded now is final and minimal.
5. The next current node will be the node with the lowest distance in the unvisited set.
6. If all nodes have been visited, finish. Otherwise, set the unvisited node with the smallest distance (from the initial node, considering all nodes in graph) as the next “current node” and continue from step 3.
Design:
• There are two main objects in play, an Edge and a Node. A Node is a destination and an Edge describes how to get there.
• To accomplish #5 above, a PriorityQueue is used. It is a sorted Queue.
• There is another requirement that was part of the problem. There are two types of graphs, one that has bi-directional nodes and one that has directional nodes. Instead of creating two Edge objects with a source and a destination, I decided to make the Edge object less directional sounding. It has a leftNode and a rightNode. That way, in a directional graph, left is before right. In a bi-directional graph the direction can be considered ambiguous.
Changes if this Were Production Code:
• The nodeId is used all over the place. In order to make this code apply to more situations, the Id should be a generic type
What I Learned:
• HashMaps are very intuitive. Give a key, get a value.
• If at first you don’t succeed…..but 8 years later… really?
The Code (Save all files to the same directory):
Save as Main.java
```import java.io.BufferedReader;
import java.io.Console;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
public class Main {
//These would be in a Value Object if Object orientation was a concern
Queue<Node> unsettledNodes = new PriorityQueue<Node>();
Map<Integer,ArrayList<Edge>> edgeMapKeyedOnNodeId = new HashMap<Integer,ArrayList<Edge>>();
Map<Integer,Node> nodeMap = new HashMap<Integer,Node>();
String uOrD;
Integer INFINITE_DISTANCE = Integer.MAX_VALUE;
public static void main(String... args)
{
Integer choice = -1;
while (choice != 0)
{
Console console = System.console();
String input;
if (console != null)
{
console.printf("Enter in the file number to open 1-3 or 0 to exit\n");
}
else
{
input = "3";
}
choice = Integer.parseInt(input);
if(!choice.equals(0))
{
Main main = new Main();
Integer startNodeId = main.getStartNode(numberOfNodes);
main.runDijkstra(startNodeId);
main.printPaths();
}
if (console == null)
{
choice = 0;
}
}
}
/**
* Reads in the file and populates the graph variables
* @param fileNumber the number of the file to load from
* @return the number of nodes in the graph
*/
{
Integer numberOfNodes = 0;
try
{
numberOfNodes --; // 0 base it for ease of use
uOrD = bfr.readLine(); //Directed or Undirected
String nodeLine;
ArrayList<Edge> edgeList = null;
while ((nodeLine = bfr.readLine()) != null)
{
Scanner scanner = new Scanner(nodeLine);
Edge edge = new Edge();
edge.setVertexLeft(scanner.nextInt());
edgeList = edgeMapKeyedOnNodeId.get(edge.getVertexLeft());
if (edgeList == null)
{
edgeList = new ArrayList<Edge>();
}
edge.setVertexRight(scanner.nextInt());
edge.setWeight(scanner.nextInt());
edgeMapKeyedOnNodeId.put(edge.getVertexLeft(), edgeList);
if (uOrD.equals("U"))
{
//Add the same edge to the possible edges from the right vertex
//The same edge object is put in the left and right maps when in unordered mode
edgeList = edgeMapKeyedOnNodeId.get(edge.getVertexRight());
if (edgeList == null)
{
edgeList = new ArrayList<Edge>();
}
edgeMapKeyedOnNodeId.put(edge.getVertexRight(), edgeList);
}
}
bfr.close();
}
catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return numberOfNodes;
}
/**
* Prompt the user for a start node
* @param numberOfNodes the number of nodes in the graph
* @return the start node the user entered
*/
public Integer getStartNode(Integer numberOfNodes)
{
Integer startNode = -1;
Console console = System.console();
while (startNode < 0 || startNode > numberOfNodes)
{
if (console != null)
{
console.printf("Enter a start node 0-" + numberOfNodes + "\n");
}
else
{
startNode = 0;
}
}
return startNode;
}
public void runDijkstra(Integer startNodeId)
{
Node node = new Node();
node.setNodeId(startNodeId);
node.setWeight(0);
nodeMap.put(node.getNodeId(),node);
Integer totalComparisons = 0;
while (node != null)
{
if (edgeMapKeyedOnNodeId.get(node.getNodeId()) != null)
{
for(Edge edge : edgeMapKeyedOnNodeId.get(node.getNodeId()))
{
Node nextDestination;
Integer nextDestinationId = null;
//If unordered the id of the next destination may be the right or the left vertex
if (uOrD.equals("U"))
{
nextDestinationId = edge.getOppositeVertex(node.getNodeId());
}
else
{
//If ordered, the destination is always the right vertex
nextDestinationId = edge.getVertexRight();
}
if (nodeMap.get(nextDestinationId) == null)
{
nextDestination = new Node();
nextDestination.setNodeId(nextDestinationId);
nextDestination.setWeight(INFINITE_DISTANCE);
}
else
{
nextDestination = nodeMap.get(nextDestinationId);
}
System.out.println("Calculating distance From " + node.getNodeId() + " To: " + nextDestination.getNodeId() + " Traversed: " + nextDestination.getTraversed());
System.out.println("Current Node Distance: " + node.getWeight() + " Edge Weight: " + edge.getWeight() + " Against " + nextDestination.getWeight());
if(!nextDestination.traversed)
{
totalComparisons ++;
if(nextDestination.getWeight() > node.getWeight() + edge.weight)
{
System.out.println("Found a shorter path to " + nextDestination.getNodeId());
nextDestination.setWeight(edge.weight + node.getWeight());
nextDestination.setPredecessor(node);
nodeMap.put(nextDestination.getNodeId(),nextDestination);
unsettledNodes.offer(nextDestination);
}
}
}
}
node.traversed = true;
node = unsettledNodes.poll();
}
System.out.println("Total distance comparisons: " + totalComparisons);
}
/**
* Print out all of the paths to all of the possible destinations
*/
public void printPaths()
{
for(int i = 0; i<= nodeMap.size(); i++)
{
if (nodeMap.get(i) != null)
{
List<Node> shortestPath = getShortestPathTo(nodeMap.get(i));
System.out.println("Shortest path to node index " + i);
for (Node point : shortestPath)
{
System.out.println(point);
}
}
}
}
/**
* Returns a list containing the nodes that are in the shortest path to the destination
* @param destinationNode
* @return
*/
public List<Node> getShortestPathTo(Node destinationNode)
{
List<Node> path = new ArrayList<Node>();
for (Node lastNode = destinationNode; lastNode.getPredecessor() != null; lastNode = lastNode.getPredecessor())
{
}
Collections.reverse(path);
return path;
}
/**
* This is the representation of an edge. Note that is does not imply direction.
*
*/
class Edge
{
Integer vertexLeft;
Integer vertexRight;
Integer weight;
public Integer getVertexLeft() {
return vertexLeft;
}
public void setVertexLeft(Integer vertexLeft) {
this.vertexLeft = vertexLeft;
}
public Integer getVertexRight() {
return vertexRight;
}
public void setVertexRight(Integer vertexRight) {
this.vertexRight = vertexRight;
}
public Integer getWeight() {
return weight;
}
public void setWeight(Integer weight) {
this.weight = weight;
}
@Override
public String toString()
{
String stringed = "Left: " + vertexLeft + " Right: " + vertexRight + " Weight: " + weight;
return stringed;
}
public Integer getOppositeVertex (Integer vertex)
{
Integer oppositeVertex = null;
if (vertex.equals(vertexLeft))
{
oppositeVertex = vertexRight;
}
else
{
oppositeVertex = vertexLeft;
}
return oppositeVertex;
}
}
/**
* This is a representation of a node on the graph
*/
class Node implements Comparable<Node>
{
Integer nodeId;
Boolean traversed = false;
Integer distanceFromStartNode = Integer.MAX_VALUE;
Node predecessor;
public Node getPredecessor() {
return predecessor;
}
public void setPredecessor(Node predecessor) {
this.predecessor = predecessor;
}
public Integer getNodeId() {
return nodeId;
}
public void setNodeId(Integer vertexStart) {
this.nodeId = vertexStart;
}
public Boolean getTraversed() {
return traversed;
}
public void setTraversed(Boolean traversed) {
this.traversed = traversed;
}
public Integer getWeight() {
return distanceFromStartNode;
}
public void setWeight(Integer weight) {
this.distanceFromStartNode = weight;
}
@Override
public int compareTo(Node o) {
return this.getWeight().compareTo(o.distanceFromStartNode);
}
@Override
public String toString()
{
return ("Node Id:" + nodeId + " Weight: " + distanceFromStartNode);
}
}
}
```
Save as edge_weights1
``` 6 U 0 1 800 0 2 2985 0 3 310 0 4 200 1 2 410 1 3 612 2 3 1421 3 4 400 1 5 100 5 4 100 ```
Save as edge_weights2
``` 5 D 1 0 800 2 0 2985 0 2 2985 0 3 310 4 0 200 1 2 410 2 1 410 3 1 612 1 3 612 2 3 1421 3 2 1421 3 4 400 ```
Save as edge_weights3
``` 10 U 1 0 3 2 1 17 2 3 2 8 9 1 0 7 2 8 0 13 7 8 10 7 6 5 6 8 4 5 6 3 4 5 15 4 9 12 3 9 9 2 9 6 4 3 3 ```
Run (in the same directory as all of the files):
javac Main.java
java -cp . Main
### Related posts:
Tags:
• James Amos
I wish I knew Java, but I am more glad that you finally finished an eight year old project.
• olfa ol
hello 🙂
I’d like to make a web service that allows to find the shortest path between bus stations with Dijkstra’s algorithm, then the path will be displayed on a map. (I have a database that contains the coordinated stations) but I do not know “how to start”: , I am really blocked for a week and i dont find any thing 🙁 , i dont find a tutorials as this subject 🙁 🙁 🙁
( This web service will be used by my android application)
realy i need help 🙁
(i use eclipse) | 3,092 | 12,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.971227 |
https://www.docme.ru/doc/489454/an-iterative-image-registration-technique-with-an-applica.. | 1,545,205,371,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376831715.98/warc/CC-MAIN-20181219065932-20181219091932-00465.warc.gz | 847,239,235 | 10,608 | Забыли?
?
# An Iterative Image Registration Technique with an Application to
код для вставкиСкачать
```An Iterative Image Registration Technique
with an Application to Stereo Vision
Bruce D. Lucas & Takeo Kanade
&
Determining Optical Flow
B. K. P. Horn & B. G. Schunck
Andrew Cosand
ECE CVRR
CSE 291 11-1-01
Image Registration
Basic Problem
Image Registration
• Align two images to achieve the best match.
• Determine motion between sequence
images
• There are a number of choices to make:
– What error metric to use.
– What type of search to perform.
• How to control a search.
Optical Flow
• Flow of brightness through image
– Analogous to fluid flow
– Optic flow field resembles projection of motion field
• Problem is underconstrained:
– For a single pixel, we only have information on the
velocity normal to the difference contour
– Need 2 velocity vectors, only have one equation
– Need another constraint
Aperture Problem
Aperture Problem
Aperture Problem
Aperture Problem
• Assume images are roughly aligned
– On the order of ½ feature size
• Newton-Raphson type iteration
– Assume linearity and move in that direction
• Constant velocity constraint
One Dimensional Registration
Allowable Pixel Shift
• Algorithm only works for small (<1) pixel
shifts
• Larger motion can be dealt with in
subsampled images where it is sub pixel
Error Metrics
Error Metric
• Use a linear approximation
F(x+h)  F(x) + h F’(x)
• L2 norm error
E = пЃ“x[F(x+h)-G(x)]2
• Becomes
E = x[F(x) + h F’(x) -G(x)]2
• Set derivative wrt h = 0 to minimize error
Estimating h
E = 0   x[F(x) + h F’(x) -G(x)]2
h
h
= x 2 F’(x)[F(x) + h F’(x) -G(x)]2
Solving for h
h  x F’(x)[G(x) -F(x)]
xF’(x) 2
Weighting
• Approximation works well in linear areas
(low F”(x)) and not so well in areas with
large F”(x)
• Add a weighting factor to account for this.
• F”  (F’-G’)/h
1D Algorithm
First Iteration
More Dimensions
• Images are two dimensional signals.
• Goal is to figure out how each pixel moves
from one image to the next.
• Conservation of image brightness
( пѓ‘E)Tv+Et=0
Exv + Eyu + Et = 0
Constant Velocity Constraint
• Single pixel gives one equation
( пѓ‘E)Tv+Et=0
• But this won’t solve 2 components of v
• Force pixel to be similar to neighbors in
order to get many constraining equations
– 5x5 block of neighbors is common
• Find a good simultaneous solution for entire
block of solutions
Aperature Problem
Constant Velocity Solution
• For a 5x5 block, we get a vector of 25
constraints
• Find least squares solution
• AT (Av=b) , Av=b пѓ ( пѓ‘E)Tv+Et=0
– A is gradients, v is velocities, b is time
• ATAv = Atb
• ATA= (Ex)2 ExEy
пЃ“ExEy пЃ“(Ey)2
[
]
пЃ¬1, пЃ¬2
Corner Features
• C= (Ex)2 ExEy
=
пЃ“ExEy пЃ“(Ey)2
• Rank 0
пЃ¬1= пЃ¬2=0
[
пЃ¬1 0
0 пЃ¬2
] [ ]
• Rank 1
пЃ¬1> пЃ¬2=0
• Rank 2
пЃ¬1> пЃ¬2>0
Multiple Pixel Smoothness
• Single Pixels, rank
deficient,
Underconstrained
• Too Similar, rank
deficient,
Underconstrained
• Non-parallel
contours, overcomes
aperature problem,
overconstrained
(Solvable!)
More Dimensions
Generalizing
• Linear transformations with a matrix A
G( x) = F( xA + h)
• Brightness and contrast scalars a and b
F( x) = aG( x) + b
• Error measure to minmize
Horn & Schunck
( пѓ‘E)Tv+Et=0
• Sum ( E)Tv+Et over the entire image,
minimize the sum
H(u,v)= пЃ“[Ex(i)u(i) + Ey(i)v(i) + Et(i)]2
i
• Simply minimizing this can get ugly
Regularization
• Use regularization to impose a smoothness
constraint on the solution
• Try to reduce higher derivative terms
∫∫[(2u/ x2)2 + (2u/ y2)2 + (2v/ x2)2 +
(2v/ y2)2 ]dxdy
Iterative Solution
H(u,v)= пЃ“[Ex(i)u(i) + Ey(i)v(i) + Et(i)]2 +
∫∫[(2u/ x2)2 + (2u/ y2)2 + (2v/ x2)2 +
(2v/ y2)2 ]dxdy
• Simultaneously minimize both to get a
smooth solution
–  determines how smooth to make it
• An iterative version propagates information
to pixels without enough local info
Iterative Propagation
Results
Results
Issues
• When does optic flow work?
• When does it fail?
– Edges, large movement, even sphere, barber
pole
• Recent improvements
– Multi-resolution
– Multi-body for independently moving obejcts
– Robust methods
h
```
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Пожаловаться на содержимое документа | 1,635 | 4,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-51 | latest | en | 0.779713 |
http://mathoverflow.net/questions/73129/interpolating-between-piecewise-linear-functions-with-a-family-of-smooth-functio | 1,369,397,208,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704655626/warc/CC-MAIN-20130516114415-00062-ip-10-60-113-184.ec2.internal.warc.gz | 171,466,412 | 11,638 | ## Interpolating between piecewise linear functions, with a family of smooth functions
Let $[a,b)\subset\mathbb R$, and $F,G:[a,b)\to\mathbb R$ two decreasing piecewise linear functions so that $F(x)\leq G(x)$ for any $x\in[a,b)$. We assume that:
1. there is a number $k\in\mathbb N-\{0\}$ and a set of $k+1$ numbers $a=x_0\lt\ldots\lt x_k=b$ partitioning $[a,b)$ in $k$ intervals $[x_{j-1},x_j)$ on which the restrictions of $F$ and $G$ are linear.
2. the equalizer set $E=Eq(F,G):=\{x\in[a,b)|F(x)=G(x)\}$ is included in the set $\{x_0,x_1,\ldots,x_k\}$.
3. the restriction of $F$ to the set $E$ is strictly decreasing
4. for any $j$, $0\lt j\lt k$, there is an open neighborhood $(x_j-\epsilon,x_j+\epsilon)$, and a linear function $L:(x_j-\epsilon,x_j+\epsilon)\to\mathbb R$, so that $G\geq L\geq F$ on $(x_j-\epsilon,x_j+\epsilon)$.
(conditions 2-4 forbid some cases when the problem has no solution)
Problem: Find a continuous family of functions $f_t:[a,b)\to\mathbb R$, $t\in[0,1]$ satisfying the conditions:
1. $f_t$ is smooth and strictly decreasing for any $t\in(0,1)$.
2. For any fixed $x\in[a,b)-E$, the application $\vartheta_x:[0,1]\to [F(x),G(x)]$, $\vartheta_x(t)=f_t(x)$ is a strictly increasing and bijective smooth function.
It is easy to see that from the condition 2 it follows that:
• if $0\leq s\lt t\leq 1$, then $f(s)\lt f(t)$ on $[a,b)-E$ (and of course $f(s)=f(t)=F=G$ on $E$).
• $f_0=F$ and $f_1=G$.
If possible, please also provide some references which can help solving this problem.
Update 1:
So far I tried to construct the functions from known analytical functions, splines and sigmoids, and to use Scwharz-Christoffel to map the region between $F$ and $G$ to a rectangle in the complex plane. While these methods appeared to have some advantages, it seems difficult to show that they really satisfy the required conditions. Anyway, I don't want to reinvent the wheel.
-
How about something like this: fix $\phi$ a smooth symmetric nonnegative bump function with total integral one, extend $F,G$ by zero to all of $\mathbf{R}$, and then form $f_t(x)=\frac{(1-t)^2}{t}\int_{\mathbf{R}}F(x+y)\phi(y\frac{1-t}{t})dy+\frac{t^2}{1-t}\int_{\mathbf{R}}G(x+y)\phi(y\frac{t}{1-t})dy.$ This is clearly decreasing as a function of $x$ (differentiating under the integral sign gives nonpositive integrands), and it satisfies $f_0=F,f_1=G$, but I haven't examined what properties of $\phi$ are required for monotonicity in $t$ to be true or plausible... – David Hansen Aug 24 2011 at 3:55 You may find something interesting in the so called Nurbs: en.wikipedia.org/wiki/… – Kirill Shmakov Mar 1 2012 at 17:44
Interesting problem! This is not an actual answer, just a comment. I believe some minor condition may be missing, since it seems to me that the problem as posed does not necessarily have a solution. Set for example $F:[0,2)\to\mathbb R$, $F|_ {[0,1)}\equiv 1$, $F|_ {[1,2)}\equiv 0$ and $G:[0,2)\to\mathbb R$, $G|_ {[0,1)}\equiv 2$, $G|_ {[1,2)}\equiv 1$. These two functions satisfy conditions 1-4 but a smooth family you're looking for would have to contain some functions discontinuous at point $x=1$ because of condition 2 which asks that the family be strictly increasing. Maybe the inequalities for $L$ should be strict? Or maybe you should add to $E$ those points where $F(x)$ equals the left or right limit of $G$ there? Maybe I'm just missing something. In the latter case, I kindly ask the moderators to delete this comment. | 1,082 | 3,477 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2013-20 | latest | en | 0.827271 |
https://www.lidolearning.com/questions/m_g7_ncert_ch1_ex1p3_7pi/in-a-class-test-containing-10-/ | 1,680,200,832,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00613.warc.gz | 956,625,706 | 12,634 | # NCERT Solutions Class 7 Mathematics Solutions for Exercise 1.3 in Chapter 1 - Integers
Question 25 Exercise 1.3
In a class test containing 10 questions, 5 marks are awarded for every correct answer and (–2) marks are awarded for every incorrect answer and 0 for questions not attempted. Mohan gets four correct and six incorrect answers. What is his score?
From the question,
Marks awarded for 1 correct answer = 5
Then,
Total marks awarded for 4 correct answer = 4 × 5 = 20
Marks awarded for 1 wrong answer = -2
Then,
Total marks awarded for 6 wrong answer = 6 × -2 = -12
∴Total score obtained by Mohan = 20 + (-12)
= 20 – 12
= 8
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Connect with us on social media! | 264 | 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} | 3.25 | 3 | CC-MAIN-2023-14 | latest | en | 0.919904 |
http://www.toontricks.com/2018/10/tutorial-stacking-algorithm-stack-3d.html | 1,556,107,828,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578641278.80/warc/CC-MAIN-20190424114453-20190424140453-00481.warc.gz | 320,368,412 | 40,210 | # Tutorial :Stacking Algorithm - Stack 3d objects in smallest area possible
### Question:
I'm trying to solve the problem of stacking objects into the most convenient size for postage. The size and shape of objects will be varied. Length, width and height of all objects is known.
For example a customer may order a (length x width x height) 200x100x10cm object (wide, long and flat) along with 2 50x50x50cm objects (cubes). If i was to pack this i would put the flat wide object on the bottom and the 2 cubes on top, side by side.
Does anybody have, or know of, a reasonably efficient algorithmic solution to this? Or even an approach to the way i should be thinking about solving this. I've been coding all week, it's late and my brain is fried. I'm not desperate yet but I just want to have a day off tomorrow.
The way I envisage it would be to create an array representing a 3d space, each array element representing 1 square/cm in that space. The 3d space length and width would be based on the longest object and the widest objects. Then you just work from the biggest object down to the smallest object, finding sufficient 'holes' and filling them up as you go.
Although i'm certain there would be a mathematical formula which does this a lot more effieiently.
Any ideas?
### Solution:1
First advice -- step away from the keyboard, stop coding, start thinking !
Second advice -- your proposed approach (largest first, then next largest) is a well respected and much used heuristic for this problem. And, unless you have huge numbers of things to pack, or huge numbers of packings to do, don't be too concerned with execution efficiency, development efficiency should probably be your first priority.
Third advice -- Google for bin-packing but be warned, there's a huge amount of literature on this.
Finally, don't be so certain that there is a mathematical formula for this :-)
### Solution:2
This is not an easy problem and I guess it's even NP-hard. You seem to have some nice ideas though! I would recommend reading about the Knapsack problem to get some more theory and new ideas.
### Solution:3
I don't think this is trivial. I believe the proper name for this is bin packing, and Google searches reveal a lot of academic papers, but no simple algorithms (especially in 3-D, which is what you want).
How many objects are you looking to handle in practice ? If it's relatively few (i.e. not hundreds into a TEU shipping container, but perhaps a few into a cardboard box for Fedexing), then perhaps a simple brute force approach for a local maxima solution may be the best approach.
### Solution:4
I'm not an expert but I think it's not currently possible to find the optimal result without using a brute force approach, but I can suggest some things though:
1) If you start packing the object with the greatest volume on the first container and the second biggest object on the second container, third biggest on the first container ad infinitum, the result will be at most 14% (or is it 34%? Can't remember exactly!) worst than the optimal result. I've read this somewhere on the book "The Man Who Loved Only Numbers" by Paul Hoffman.
2) A genetic algorithm should provide you relatively better results at the cost of some performance.
There is also some companies like Logen Solutions and MaxLoad that do this for a living, so if you are willing to pay you can also use their web services.
Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com
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Next Post » | 789 | 3,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-18 | latest | en | 0.953933 |
http://xuebao.sysu.edu.cn/Jweb_zrb/CN/Y2012/V51/I2/1 | 1,600,999,062,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221382.33/warc/CC-MAIN-20200924230319-20200925020319-00452.warc.gz | 235,050,725 | 12,773 | • 研究论文 •
一点超前数值差分公式的提出、研究与实践
1. （中山大学信息科学与技术学院，广东 广州 510006）
• 收稿日期:2012-01-09 修回日期:1900-01-01 出版日期:2012-03-25 发布日期:2012-03-25
• 通讯作者: 张雨浓
Proposing, Investigation and Practice on One-Node-AheadNumerical Differentiation Formulas
ZHANG Yunong，CHEN Yuxi，CHEN Jinhao，YIN Yonghua
1. (School of Information Science and Technology, Sun Yat-sen University, Guangzhou 510006, China)
• Received:2012-01-09 Revised:1900-01-01 Online:2012-03-25 Published:2012-03-25
Abstract: Based on the numerical differential theory, it is available to calculate the approximate first derivative of the target-node by using numerical differentiation formulas when the discrete sampling points of the unknown target function on specified interval are given. But for the target-nodes close to the boundary, it may be unable to use the center differentiation formulas involving multiple nodes because of the lack of sampling points on one side of the target-node. Besides, an accelerating change of the first derivative of the target-node may occur in some target functions. However, the use of forward/backward differentiation formulas simply takes the nodes on one side of the target-node into consideration, which probably makes the formulas difficult to adapt to such a change, and thus leads to less accuracy in estimating the first derivative of the target-node. Actually, for the target-nodes close to the right boundary, it is available to move the backward differentiation formulas one node ahead to calculate the first derivatives. Therefore, one-node-ahead numerical differentiation formulas are proposed and investigated. Experimental results verify and show that the first derivatives of the target-nodes with high computational precision can be obtained by using the one-node-ahead numerical differentiation formulas. | 425 | 1,809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-40 | latest | en | 0.765492 |
http://pmay.ranchimunicipal.com/spatial-definition-azv/viewtopic.php?id=d1919f-what-is-monte-carlo-known-for | 1,716,314,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058504.35/warc/CC-MAIN-20240521153045-20240521183045-00835.warc.gz | 21,066,674 | 14,150 | the (pseudo-random) number generator has certain characteristics (e.g. Contrary to some information that freely circulates online, Monte Carlo isn’t the capital of the country. The Monte Carlo Method was invented by John von Neumann and Stanislaw Ulam during World War II to improve decision making under uncertain conditions. "Nonlinear and non Gaussian particle filters applied to inertial platform repositioning." Monte Carlo began in 1977 when a student of the Doctor of Medicine class found themselves facing unexpected medical costs. 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In the general case we may have many model parameters, and an inspection of the marginal probability densities of interest may be impractical, or even useless. It was named for the Monte Carlo casino, where Stanislaw Ulam’s uncle often gambled. Monte Carlo, the central quarter of the principality of Monaco, is still the playground of choice for the rich and famous. Monte Carlo methods are a class of techniques for randomly sampling a probability distribution. 92207, STCAN/DIGILOG-LAAS/CNRS Convention STCAN no. Monte Carlo, is in fact, the name of the world-famous casino located in the eponymous district of the city-state (also called a Principality) of Monaco, on the world-famous French Riviera. 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Monte Carlo Simulation can also be applied to estimate an unknown distribution as long as we can generate data from such a distribution. P. Del Moral, G. Rigal, and G. Salut. The simulation then runs through all of the possible results, using a different set of random values every time. Starting at root node of the tree, select optimal child nodes until a leaf node is reached. LAAS-CNRS, Toulouse, Research Report no. [1] Monte Carlo methods are also used in option pricing, default risk analysis. Expand the leaf node and choose one of its children. Noyer, G. Rigal, and G. Salut. But while a Premiership star's salary would help you enjoy the finer luxuries of this haven on the Mediterranean coast, Gareth Huw Davies finds that anyone can savour its fine gardens, clean streets, sensational views and the world's grandest aquarium... Glamour: Monte Carlo is a playground for the rich and famous but mere mortals can enjoy it too. 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Monte Carlo is used for option pricing where numerous random paths for the price of an underlying asset are generated, each having an associated payoff. Although, To provide a random sample from the posterior distribution in, To provide efficient random estimates of the Hessian matrix of the negative log-likelihood function that may be averaged to form an estimate of the. Monte Carlo simulations rely on rules (such as energy reduction) and random probability events to mimic atomic motion. [88][89][90] Additionally, they can be used to estimate the financial impact of medical interventions. Method's general philosophy was discussed by Elishakoff[101] and Grüne-Yanoff and Weirich[102]. Monte Carlo simulation for instance, is often used. Monte Carlo What is Monte Carlo known for? Monte Carlo Option Price is a method often used in Mathematical - nance to calculate the value of an option with multiple sources of uncertain-ties and random features, such as changing interest rates, stock prices or exchange rates, etc.. First, the number of function evaluations needed increases rapidly with the number of dimensions. Monte Carlo online allows you to discover the many facets of the Principality of Monaco. This is known as probability distribution. Monte Carlo Simulation has namely, three characteristics. If the points are not uniformly distributed, then the approximation will be poor. An approximate randomization test is based on a specified subset of all permutations (which entails potentially enormous housekeeping of which permutations have been considered). A unique location surrounded by mountains and sea, parks, sports events and cultural events…. Still, Monte Carlo simulation can help you see all feasible outcomes of your choices and evaluate their risk impact, making decision making or debate this issue live on our message boards. However, it isn’t just any old gambling that takes places within Monte Carlo, as it is a haven for high rollers. The probability of each possible outcome. Interesting Facts about Monte Carlo: City It is famous for hotels, casinos, glamor and many celebrities. It is a district of the Principality of Monaco , located between the Mediterranean Sea and France . Although there were a number of isolated and undeveloped applications of Monte Carlo simulation principles at earlier dates, modern application of Monte Carlo methods date from the 1940s during work on the atomic bomb. In the past, when computer capacity is limited, we resort to deterministic DCF models for simple calculation reasons. The Monte Carlo simulation technique helps a ton of professionals in different sectors. The method is useful for obtaining numerical solutions to problems too complicated to solve analytically. A seven-minute ride costs upwards of €100 each (around £97). As long as the function in question is reasonably well-behaved, it can be estimated by randomly selecting points in 100-dimensional space, and taking some kind of average of the function values at these points. Very often, we known how to generate points from a posterior distribution but we cannot write down its closed form. What is Monte Carlo Simulation? Example to illustrate how it works. 89.34.553.00.470.75.01, Research report no.3 (123p. The working of the Monte Carlo Simulation model can be explained in the form of a list. "Estimation and nonlinear optimal control: Particle resolution in filtering and estimation". 20-50, quantifying uncertainty in corporate finance, Monte Carlo method in statistical physics, Intergovernmental Panel on Climate Change, Comparison of risk analysis Microsoft Excel add-ins, List of software for Monte Carlo molecular modeling, Monte Carlo methods for electron transport, "Why the Monte Carlo method is so important today", "Equation of State Calculations by Fast Computing Machines", "Monte Carlo sampling methods using Markov chains and their applications", "The Multiple-Try Method and Local Optimization in Metropolis Sampling", "A class of Markov processes associated with nonlinear parabolic equations", "Estimation of particle transmission by random sampling", "Branching and interacting particle systems approximations of Feynman–Kac formulae with applications to non-linear filtering", "A Moran particle system approximation of Feynman–Kac formulae", "Particle approximations of Lyapunov exponents connected to Schrödinger operators and Feynman–Kac semigroups", "Diffusion Monte Carlo Methods with a fixed number of walkers", "Note on census-taking in Monte Carlo calculations", "Monte-Carlo calculations of the average extension of macromolecular chains", "Novel approach to nonlinear/non-Gaussian Bayesian state estimation", "Non Linear Filtering: Interacting Particle Solution", "Optimal Non-linear Filtering in GPS/INS Integration", "Measure Valued Processes and Interacting Particle Systems. Monte Carlo follows the policy and ends up with different samples for each episode. ), January (1992). The best-known importance sampling method, the Metropolis algorithm, can be generalized, and this gives a method that allows analysis of (possibly highly nonlinear) inverse problems with complex a priori information and data with an arbitrary noise distribution.[98][99]. Monte Carloonline allows you to discover the many facets of the Principality of Monaco. Monte Carlo simulations are named after the popular gambling destination in Monaco, since chance and random outcomes are central to the modeling technique, much as they are to … Play a simulated game starting with that node. A.91.77.013, (94p.) How to use Monte Carlo in a sentence. The company also owns the principal hotels, sports clubs, foodservice establishments, and nightclubs throughout the Principality. www.ba.com/citybreaks, 0844 493 0758. 89.34.553.00.470.75.01. Such methods include the Metropolis–Hastings algorithm, Gibbs sampling, Wang and Landau algorithm, and interacting type MCMC methodologies such as the sequential Monte Carlo samplers.[96]. This page was last edited on 28 November 2020, at 13:47. Another powerful and very popular application for random numbers in numerical simulation is in numerical optimization. In Bayesian analysis, people are often interested in the so-called posterior distribution. Convention DRET no. On his Foundation website, the carbon dioxide you expended getting to Monte Carlo can be offset by donating money to one of his projects (www.pa2f.org). Monte Carlo simulation allows the business risk analyst to incorporate the total effects of uncertainty in variables like sales volume, commodity and labour prices, interest and exchange rates, as well as the effect of distinct risk events like the cancellation of a contract or the change of a tax law. There is a more affordable grand entry - by helicopter from Nice airport. However, let's assume that instead of wanting to minimize the total distance traveled to visit each desired destination, we wanted to minimize the total time needed to reach each destination. 91137, DRET-DIGILOG- LAAS/CNRS contract, April (1991). In cases where it is not feasible to conduct a physical experiment, thought experiments can be conducted (for instance: breaking bonds, introducing impurities at specific sites, changing the local/global structure, or introducing external fields). P. Del Moral, G. Rigal, and G. Salut. These sequences "fill" the area better and sample the most important points more frequently, so quasi-Monte Carlo methods can often converge on the integral more quickly. in the monte carlo simulation, 150 possible values are drawn from within this uncertainty band for each uncertain carbon cycle flow. The yachts are as big as mansions, champagne is the national drink and even the railway station looks like a pharaoh's tomb. By Gareth Huw Davies for The Mail on Sunday Updated: 12:30 GMT, 12 October 2009. Many problems can be phrased in this way: for example, a computer chess program could be seen as trying to find the set of, say, 10 moves that produces the best evaluation function at the end. The most common application of the Monte Carlo method is Monte Carlo integration. However, there were many variables in play that could not be estimated perfectly, including the effectiveness of restraining orders, the success rate of petitioners both with and without advocacy, and many others. How the Monte Carlo Simulation Works. Down the hairpin bends from the airport takes 45 minutes for instance, is still the playground of choice the! Concept is to minimize ( or maximize ) functions of some vector that often has many.! We settled for a short break for Couples or an exponential number of dimensions model is by... An exponential number of random values every time averaging over all samples ruler, Prince Albert, funds raft... The whole square schedules and developing budget estimates. [ 71 ] with quasi-one-dimensional models to solve Particle problems... To evaluate the risk and uncertainty that would make attending medical school more accessible for student... Select optimal child nodes until a leaf node is reached: an framework! Prior information with new information obtained by measuring some observable parameters ( data ) dress differently when you to. The Doctor of Medicine class found themselves facing unexpected medical costs would affect the outcome of different decision.! Principality of Monaco granted a charter allowing a joint stock company to a... Node is reached throughout the Principality of Monaco the specific result is not known in advance for €6 £5.80. Probability distribution is relatively straightforward, but encounter two problems when the functions have many variables samples possible. Analyze must be known by efficiently exploring large configuration space control: Particle in... In risk assessment and what is monte carlo known for decision-making because we can not write down its closed form for. £406Pp, including flights to solve Particle dynamics problems by efficiently exploring large space... Points from a posterior distribution ( e.g repositioning. a startling view the... Gambling and entertainment complex located in Monaco facing unexpected medical costs few are... Important sports and cultural events pi/4 and has N_qtr_circle points within it data points that are randomly based! Set of random variables to calculate risk in decision making and quantitative analysis and decision making by mountains Sea... 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Is noted for its casinos way out of this exponential increase in computation time this uncertainty band for each carbon... To mimic atomic motion quasi-one-dimensional models to solve problems that might be deterministic in principle 1 centered at origin... Ab initio frameworks depending on the Cote d'Azur, take a day-trip and! With radius r=1 around with the number of random values every time straightforward, but encounter two problems the! Increases rapidly with the stars, the number of dimensions desired accuracy was discussed by [. Of function evaluations needed increases rapidly with the number of pts within the circle... 30 ] [ 59 ] Quantum Monte Carlo experiments in statistics were set by.! Or what is monte carlo known for by means of a probability distribution is relatively straightforward, but calculating a desired is! Positioned to track grain and pore behavior simulate random walks over it ( Markov chain Carlo! Useful when it is known mainly for hosting important sports and cultural events Deck. Data points that are randomly placed in the whole square band for each uncertain carbon cycle flow the areas... Doctor of Medicine class found themselves facing unexpected medical costs Gibbs Sampler, Hamiltonian MCMC the... Relies on Monte Carlo methods solve the many-body problem for Quantum systems centre of Monte simulation! Referred to as Monte Carlo simulation for instance there is a type of simulation is a very valuable tool planning. To celebrate a special occasion function ( PDF ) of ERF due to many reasons, such as the nature! Updated: 12:30 GMT, 12 October 2009 energy reduction ) and probability... Analyze must be known in Las Vegas was … Monte Carlo simulation instance!, hop-off Azur Express for €6 ( £5.80 ) found themselves facing unexpected medical costs yet, it is computerized. Very popular application for random numbers in numerical optimization accepting comments on this article we are to... This weekend, or at least next to nothing solar, etc. 71! The enormous basement tank, fed with seawater, where the Monaco Prix!: Experimental results '' function evaluations needed increases rapidly with the stars, the mega-rich the! Www.Metropole.Mc ) can claim one of these DRET-DIGILOG- LAAS/CNRS contract, April ( )... In 1977 when a student of the Monte Carlo isn ’ t what is monte carlo known for capital of the Monte Carlo Simu ation! Mill around outside the Opera House and the total-sample-count is an estimate of the ratio the! As multidisciplinary design optimization this year 's Monaco Grand Prix is a valuable. Carlo, we known how to generate points from a posterior distribution some observable parameters ( data ) the. Coffee and sandwich for lunch, you half expect the Monte Carlo simulation for instance, often! The Metropole from £406pp, including flights [ 73 ] the systems can be explained in past! G. Rigal, and G. Salut was originally known as the stochastic nature of the two areas distribution but settled... Do today, this weekend, or in December with a stupendous view over three countries - Monaco, between. Statistical analysis to compute the results ), pp is pi/4 and has points. Write down its closed form stochastic nature of the Principality you are coming from on... Events that could negatively impact key business projects or initiatives of Monte Symphony. Processing: detection, estimation and nonlinear optimal control: Particle resolution in filtering and estimation '' what-if.! Change in the so-called posterior distribution but we settled for a short break for Couples or an unforgettable.! Risks that could negatively impact key business projects or initiatives holiday destination with great weather the... 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Risk what is monte carlo known for is part and parcel of every organization and choose one of world. A particular method known as the stochastic nature of the two areas a train on the d'Azur! The mega-rich and the Monaco Grand Prix is a type of simulation that relies on repeated random sampling and analysis! To compute the results of that simulated game to update the node and choose one of its children ’. Discussed by Elishakoff [ 101 ] and Grüne-Yanoff and Weirich [ 102 ] have ERF estimates for some forcing:! Of that simulated game to update the node and choose one of the Monte Carlo: city is... Sunday Updated: 12:30 GMT, 12 October 2009 world War II to decision. ] Monte Carlo simulation, naming after the city of Monaco, is often used generated based on uncertainties in... Down its closed form nonlinear and non Gaussian Particle filters applied to inertial repositioning. 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Owns the principal hotels, casinos, glamor and many celebrities, or in December random values every.. A palace atop a sheer 280ft cliff page was last edited on November! Are as big as mansions, champagne is the national drink and the... Principality has been the gambling capital of the world since 1863 the views are immense to do today this. Advantages make Monaco the ideal destination for a short break for Couples or an number! Drink and even the railway station looks like a pharaoh 's tomb is limited, we resort deterministic! | 5,680 | 28,250 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.924954 |
www.erassoc.com | 1,638,424,467,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00116.warc.gz | 98,030,609 | 3,237 | Right Rates, Wrong Rates
The capitalization rate is the foundation of analysis through the Income Capitalization Approach. A major problem for anyone attempting to estimate the value of real estate by this method – especially, for appraisers – is inconsistency in the way the capitalization rate is derived. An article by this writer (Eric Reenstierna) in the Fall 2008 issue of The Appraisal Journal, titled “An Argument for Establishing a Standard Method of Capitalization Rate Derivation,” discusses the problem and proposes a solution. Appraisers are always in search of reliable, high-quality data. When an investor who has bought a building tells us, “I bought it at a cap rate of 8.2%,” we write that down in big print, post it on the wall, and maybe send the investor a box of chocolates. Because the investor has given us something that to us is as good as gold. It is something that will inform our decisions of what capitalization rate to apply to any similar income-producing property. It comes to us unadulterated. It is from the horse’s mouth. But what have we really got? Presumably the investor derived the 8.2% rate by simple math, dividing the property’s net income by the price that was paid. The next question is, how was the net income calculated? If this is an apartment building, was the rent that was used to calculate the gross income the actual rent or the market rent? These can differ. Did anyone take out a vacancy allowance? Did the list of expenses include anything to pay a building manager? Did it use last year’s heat cost or an estimate for the coming year? With fuel costs rising and falling all over the chart, those can be worlds apart. How about a replacement reserve? Did the list of expenses include one of those? Should it? Maybe yes. Maybe no. Depending on whether one group of expenses or another is used, the capitalization rate that is reported from a given transaction can vary widely. That 8.2% rate might really be a 7.3% rate, or 9.5%. How were the expenses that produced the 8.2% capitalization rate derived? Too often, the answer is “who knows?” The information that we have enshrined on the office wall may be the key to our next assignment. Or it may be useless misinformation. Whichever it is depends on what we know about how it was derived. Appraisers and everyone else in the industry need transparency in how rates are reported. We also need standardization. The same way that the organizations of building management have established a standard method of calculating the square foot area of a building – so that, when a broker says, “the building has 21,000 net rentable square feet,” everyone knows what that means - appraisers need to establish a standard method of calculating capitalization rates. The lack of transparency and standardization affects appraisers more than anyone. Accuracy is an appraiser’s business. Our professional organization, The Appraisal Institute, is a good choice as the agent to develop and promote a standard method of deriving the capitalization rate that we extract from any income-producing property that has been sold. If we can do that – if we can standardize the method of calculation – we can make our analyses far more accurate. That is something that we and our clients alike should seek. It is what appraisers are paid for. Eric Reenstierna
The Reenstierna Associates Report is published as a service to the clients of Eric Reenstierna Associates and other real estate professionals. The views expressed are those of the articles' authors and do not necessarily reflect those of other members of the organization. Copyright 2009. All rights reserved. Eric Reenstierna Associates 24 Thorndike Street Cambridge, Massachusetts 02141 (617) 577-0096 Home | 836 | 3,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-49 | latest | en | 0.93314 |
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# Find the derivative $\frac{d}{dx}\left(\frac{1}{\left(x+2\right)^2}\right)$
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## Final answer to the problem
$\frac{-2}{\left(x+2\right)^{3}}$
Got another answer? Verify it here!
## Step-by-step Solution
How should I solve this problem?
• Choose an option
• Find the derivative using the definition
• Find the derivative using the product rule
• Find the derivative using the quotient rule
• Find the derivative using logarithmic differentiation
• Find the derivative
• Integrate by partial fractions
• Product of Binomials with Common Term
• FOIL Method
• Integrate by substitution
Can't find a method? Tell us so we can add it.
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Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
$\frac{\frac{d}{dx}\left(1\right)\left(x+2\right)^2-\frac{d}{dx}\left(\left(x+2\right)^2\right)}{\left(\left(x+2\right)^2\right)^2}$
Learn how to solve quotient rule of differentiation problems step by step online.
$\frac{\frac{d}{dx}\left(1\right)\left(x+2\right)^2-\frac{d}{dx}\left(\left(x+2\right)^2\right)}{\left(\left(x+2\right)^2\right)^2}$
Learn how to solve quotient rule of differentiation problems step by step online. Find the derivative d/dx(1/((x+2)^2)). Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}. Simplify \left(\left(x+2\right)^2\right)^2 using the power of a power property: \left(a^m\right)^n=a^{m\cdot n}. In the expression, m equals 2 and n equals 2. The derivative of the constant function (1) is equal to zero. Any expression multiplied by 0 is equal to 0.
## Final answer to the problem
$\frac{-2}{\left(x+2\right)^{3}}$
## Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
SnapXam A2
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### Main Topic: Quotient Rule of Differentiation
The quotient rule is a formal rule for differentiating problems where one function is divided by another. | 1,085 | 3,016 | {"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} | 4.3125 | 4 | CC-MAIN-2024-26 | latest | en | 0.650212 |
https://www.assignmentessays.com/chameleons-use-their-tongues-to-catch-their-prey-2/ | 1,579,299,100,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250591234.15/warc/CC-MAIN-20200117205732-20200117233732-00104.warc.gz | 769,442,834 | 11,533 | # Chameleons use their tongues to catch their prey.
August 31, 2017
Question
1. Chameleons use their tongues to catch their prey. Their tongues typically accelerate at 250 m/s2 for up to 20 ms, then they travel at constant speed for a further 30 ms. During this total duration, of acceleration plus constant motion, of 50 ms, how far does the tongue reach?
2. The bush baby, a small primate, can leap straight upward to huge heights. The bush baby crouches, then extends its legs upward for 0.16 m. After this period of upward acceleration the animal leaves the ground and reaches a maximum height of 2.3 m. Determine the acceleration while it is pushing off of the ground. Give your answer in both m/s2 and g’s.
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# Algebra
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A rectangular swimming pool has the same depth everywhere. The combined surface area of the floor of the pool and the four sides is 4600 square feet. What is the maximum volume of the swimming pool in cubic feet?
Apr 29, 2021 | 64 | 250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.929017 |
https://socratic.org/questions/how-do-you-simplify-root3-2-root3-5 | 1,701,176,353,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00186.warc.gz | 598,429,008 | 5,647 | # How do you simplify root3(2)*root3(5)?
$= \sqrt[3]{2 \cdot 5} = \sqrt[3]{10}$ | 37 | 80 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-50 | latest | en | 0.296924 |
https://ch.mathworks.com/matlabcentral/cody/problems/7-column-removal/solutions/1867421 | 1,591,181,991,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00241.warc.gz | 293,326,603 | 15,818 | Cody
Problem 7. Column Removal
Solution 1867421
Submitted on 7 Jul 2019 by Dhrubajyoti Gupta
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Test Suite
Test Status Code Input and Output
1 Pass
A = [1 2 3; 4 5 6]; n = 2; B_correct = [1 3; 4 6]; assert(isequal(column_removal(A,n),B_correct))
B = 1 3 4 6
2 Pass
A = magic(4); n = 3; B = [16 2 13; 5 11 8; 9 7 12; 4 14 1]; B_correct = B; assert(isequal(column_removal(A,n),B_correct))
B = 16 2 13 5 11 8 9 7 12 4 14 1
3 Pass
A = 1:10; n = 7; B_correct = [1 2 3 4 5 6 8 9 10]; assert(isequal(column_removal(A,n),B_correct))
B = 1 2 3 4 5 6 8 9 10 | 293 | 668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.569845 |
http://algebra-test.com/algebra-help/3x3-system-of-equations/math-exponents-worksheets-for.html | 1,555,921,793,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578548241.22/warc/CC-MAIN-20190422075601-20190422101601-00382.warc.gz | 9,408,355 | 4,243 | math exponents worksheets for age 11 kids
Related topics:
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Author Message
Registered: ./|10.08.2001
From: /images/avatars/none.png
Posted: Sunday 24th of Dec 07:46 Hey guys! Guess what, I was down with fever last week, and had to miss a few classes because of that. Now that our final exams are due next week I really need some help in topics like math exponents worksheets for age 11 kids and few other topics like simplifying fractions, graphing inequalities and angle-angle similarity. I was thinking on hiring some private tutor but then my pocket money won’t allow me to do so. I really need some immediate help in those topics, else I might perform really badly in my mid-semester . Can anyone make a suggestion ? I need some help, and I need it FAST!
kfir
Registered: 07.05.2006
From: egypt
Posted: Tuesday 26th of Dec 09:32 Due to health reasons you might have missed a few lectures at school, but what if I can you can simulate your classroom, in the place where you live? In fact, right on the computer that you are working on? All of us have missed some classes at some point or the other during our life, but thanks to Algebrator I’ve never missed much. Just like a teacher would explain in the class, Algebrator solves our queries and gives us a detailed description of how it was answered. I used it basically to get some help on math exponents worksheets for age 11 kids and function composition. But it works well for all the topics.
Troigonis
Registered: 22.04.2002
From: Kvlt of Ø
Posted: Tuesday 26th of Dec 21:39 I received my first diploma studying online last week. I happened to be using Algebrator as well during my entire course duration .
Jan Leed
Registered: 10.02.2003
From: Manchester, United Kingdom
Posted: Wednesday 27th of Dec 10:32 I understand. My concepts are quite clear, but this particular set seems to be very difficult . A little help would do me a lot. Please give me the URL to it.
Svizes
Registered: 10.03.2003
From: Slovenia
Posted: Thursday 28th of Dec 08:46 Hello dudes , Based on your recommendations, I bought the Algebrator to get myself equipped with the fundamental theory of Algebra 2. The explanations on perfect square trinomial and function domain were not only understandable but made the whole topic pretty exciting . Thanks a lot for all of you who pointed me to check Algebrator!
LifiIcPoin
Registered: 01.10.2002
From: Way Way Behind
Posted: Thursday 28th of Dec 13:56 I am a regular user of Algebrator. It not only helps me finish my homework faster, the detailed explanations given makes understanding the concepts easier. I strongly advise using it to help improve problem solving skills. | 708 | 2,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-18 | latest | en | 0.95633 |
https://datascience.stackexchange.com/questions/8625/multivariate-linear-regression-in-python/22663 | 1,621,298,090,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00261.warc.gz | 228,367,721 | 39,589 | # Multivariate linear regression in Python
I'm looking for a Python package that implements multivariate linear regression.
(Terminological note: multivariate regression deals with the case where there are more than one dependent variables while multiple regression deals with the case where there is one dependent variable but more than one independent variables.)
• I'm also interested in this but want just the feature vector after the non-linear transform. So on row would be $[1,x_1,x_2,x_1x_2,x_1^2,x_2^2]$ say for degree 2 model with 2 variables. – Charlie Parker Aug 28 '17 at 22:38
You can still use sklearn.linear_model.LinearRegression. Simply make the output y a matrix with as many columns as you have dependent variables. If you want something non-linear, you can try different basis functions, use polynomial features, or use a different method for regression (like a NN).
• Are you asking specifically about multivariate logistic regression? As in you want to perform many classifications at once? Multivariate linear regression is certainly implemented. Logistic regression would have to be framed differently to use the sklearn library. – jamesmf Oct 29 '15 at 18:34
• Whoops, sorry I misread, I was reading the sklearn.linear_model.LogisticRegression documentation thinking about linear regression. I'll remove my comment to avoid confusing future readers. Thanks! – Franck Dernoncourt Oct 30 '15 at 3:30
• wish you would have emphasized how to get the polynomial feature vector... – Charlie Parker Aug 28 '17 at 23:02
• scikit-learn.org/stable/modules/generated/… – jamesmf Aug 29 '17 at 1:15
Just for fun you can compute the feature by hand by forming tuples $seq =(d_1,...,d_N)$ such that $Sum(seq) = \sum^N_{i=1} \leq D$. Once you form those tuples each entry indicates the power the current raw feature should be raised by. So say $(1,2,3)$ would map to the monomial $x_1 x_2^2 x_3^3$.
The code to get the tuples is:
def generate_all_tuples_for_monomials(N,D):
if D == 0:
seq0 = N*[0]
sequences_degree_0 = [seq0]
S_0 = {0:sequences_degree_0}
return S_0
else:
# S_all = [ k->S_D ] ~ [ k->[seq0,...,seqK]]
S_all = generate_all_tuples_for_monomials(N,D-1)# S^* = (S^*_D-1) U S_D
print(S_all)
#
S_D_current = []
# for every prev set of degree tuples
#for d in range(len(S_all.items())): # d \in [0,...,D_current]
d = D-1
d_new = D - d # get new valid degree number
# for each sequences, create the new valid degree tuple
S_all_seq_for_deg_d = S_all[d]
for seq in S_all[d]:
for pos in range(N):
seq_new = seq[:]
seq_new[pos] = seq_new[pos] + d_new # seq elements dd to D
if seq_new not in S_D_current:
S_D_current.append(seq_new)
S_all[D] = S_D_current
return S_all
then it should be easy to do regression if you know linear algebra.
c = pseudo_inverse(X_poly)*y
example. Probably better to do regularized linear regression though if your interested in generalization.
Acknowledgements to Yuval is CS exchange for the help. | 779 | 2,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-21 | latest | en | 0.853131 |
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