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https://www.physicsforums.com/threads/quantum-mechanics-free-partical-in-spherical-coordinates.241425/
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# Quantum mechanics: free partical in spherical coordinates
1. Jun 21, 2008
### liorda
1. The problem statement, all variables and given/known data
My wavefunction is $$\psi (r, \theta, \phi )=N cos(\theta) e^{-(r/R_0)^2}$$.
I need to calculate $$<p_r>$$ and $$\Delta p_r$$ where $$p_r$$ is the radial momentum.
2. Relevant equations
I think i know $$p_r=\frac{\hbar}{i} \left( \frac{d}{dr}+\frac{1}{r} \right)$$.
3. The attempt at a solution
When I try to calculate the observation value I got infinity (the integral does not seem to converge):
$$<\psi | p_r | \psi > = \int_0^{2 \pi}d\phi \int_0^\pi d\theta \int_0^{\infty}dr [\psi^\star p_r \psi]$$
Are the limits for the integral correct? What am I doing wrong? :(
thank.
2. Jun 21, 2008
### kreil
$$( \psi |p_r|\psi) = \frac{\hbar}{i} \left( \frac{d}{dr}+\frac{1}{r} \right) N^2 cos^2( \theta) e^{-2(r/R_0)^2}$$...agreed?
So,
$$< \psi |p_r| \psi> = \frac{N^2 \hbar}{i} \int_0^{\pi}cos^2(\theta)d \theta \int_0^{2 \pi}d \phi \int_0^{\infty} \left \frac{d}{dr}- \frac{1}{r} \right e^{-2(r/R_0)^2} dr$$
$$= \frac{ N^2 \hbar \pi^2}{i} \int_0^{\infty} \left \frac{d}{dr}- \frac{1}{r} \right e^{-2(r/R_0)^2} dr$$
Since the theta int. goes to pi/2 and the phi int. goes to 2pi
Is this the integral you ended up computing?
Josh
3. Jun 21, 2008
### liorda
Hi Josh,
I ended up with this:
$$< \psi |p_r| \psi> = \frac{N^2 \hbar}{i} \int_0^{\pi}cos^2(\theta)d \theta \int_0^{2 \pi}d \phi \int_0^{\infty} \left( \frac{d}{dr}+ \frac{1}{r} \right) \right e^{-2(r/R_0)^2} dr$$
My problem is with $$\int_0^\infty \frac{1}{r} e^{-2(r/R_0)^2} dr$$ which does not converge.
4. Jun 21, 2008
### kreil
Ok, we're on the same page then with the equations.
The r integral will separate into two integrals after you distribute the variables in parenthesis.... that is, one term is d/dr of the e-fcn, and the other is 1/r times the e-fcn.
Because of this, its ok for the 1/r term to blow up as long as the other one does not (I haven't computed either one of them, so let me know if there is an additional problem..)
Easiest way to integrate these, in case you didnt already know, is with the wolfram online integrator:
http://integrals.wolfram.com/
5. Jun 21, 2008
### liorda
Are you sure? Can I just disregard the inf in the equation?!?
6. Jun 21, 2008
### kreil
Yes, if a term goes to infinity it is called 'blowing up'
This is a common situation when evaluating boundary criterion in, for example, the finite square well, where the psi contains e^x and e^-x for x>0, and since as x-> infty e^x blows up, psi reduces to just the e^-x term.
7. Jun 21, 2008
### EngageEngage
the integral you have is incorrect, i believe.
$$dV = r^{2} sin\theta d\theta d\phi dr$$
Edit: in your equation you have:
$$dV = dr d\phi d\theta$$
8. Jun 21, 2008
### liorda
the r^2*sin(t) is the |Jacobian| isn't it? I thought it should only be added when making a transformation between Cartesian and spherical coordinates, i.e $$\int\int\int dxdydz = \int\int\int r^2 sin(\theta) d\phi d\theta dr$$...
although the r^2 sin(t) will solve the integral convergence problem, i don't understand yet if it is necessary.
Last edited: Jun 21, 2008
9. Jun 21, 2008
### EngageEngage
Well, the volume element I suggested to you is just the general form of a volume integral in spherical coordinates. Check it on a sphere to see what you get. The transformation you give is correct. You are essentially doing a transformation from cartesian to polar spherical here (if you want to look at it like you started in cartesian). Regardless of which, you must use dV = r^2 sin(theta) dphi dtheta dr if you're integrating over phi, theta, and r, in spherical coordinates. You should draw out the element, and you will see why you must do this. dV = dr dtheta dphi doesn't make sense as you will see if you try to draw out the element.
10. Jun 21, 2008
### EngageEngage
$$\int\int\int_{D} \Psi(x,y,z)^{*}p\Psi(x,y,z) dxdydz = \int\int\int_{D} \Psi(r,\phi,\theta)^{*}p\Psi(r,\phi,\theta) r^2 sin(\theta) d\phi d\theta dr$$
11. Jun 21, 2008
### liorda
kreil and EngageEngage: thanks a lot!
I now get that the observation value for the radial momentum is 0. does it have a physical meaning? is there any way of anticipating that result by looking at the wavefunction? and should I even try to search physical meanings in quantum mechanics questions?
thanks again.
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https://codereview.stackexchange.com/questions/279258/kr-exercise-3-3-expands-shorthand-notations-e-g-a-z-to-abc-xyz-0-9-to-012
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# K&R Exercise 3-3. Expands shorthand notations (e.g., a-z to abc..xyz, 0-9 to 012..789)
I have been learning C with K&R Book 2nd Ed. So far I have completed quite a few exercises.
For the following exercise (Chapter 3, Ex-3.3):
Exercise 3-3. Write a function expand(s1, s2) that expands shorthand notations like a-z in the string s1 into the equivalent complete list abc...xyz in s2. Allow for letters of either case and digits, and be prepared to handle cases like a-b-c and a-z0-9 and -a-z. Arrange that a leading or trailing - is taken literally.
I have written this solution. I would like to know how to improve it.
#include <stdio.h>
#include <ctype.h>
#define MAXLINE 1024
int get_line(char line[], int maxline);
void expand(const char s1[], char s2[]);
int match(int start, int end);
int
main(void)
{
char s1[MAXLINE];
char s2[MAXLINE];
while (get_line(s1, MAXLINE) > 0) {
expand(s1, s2);
printf("%s", s2);
}
return (0);
}
/**
* Here I have tried to write a loop equivalent to the loop seen
* previously in chapter 1. (without using && and ||,
* as specified in chapter 2 of the book, exercise 2.2).
*
* for (i = 0; i < lim-1 && (c = getchar()) != EOF && c != '\n'; ++i)
* ...
**/
int
get_line(char s[], int lim)
{
int c, i;
i = 0;
while (--lim > 0) {
c = getchar();
if (c == EOF)
break;
if (c == '\n')
break;
s[i++] = c;
}
if (c == '\n')
s[i++] = c;
s[i] = '\0';
return (i);
}
void
expand(const char s1[], char s2[])
{
int i, j, ch;
for (i = j = 0; (ch = s1[i++]) != '\0'; ) {
if (s1[i] == '-' && match(s1[i-1], s1[i+1])) {
for (i++; ch < s1[i]; ) {
s2[j++] = ch++;
}
} else
s2[j++] = ch;
}
s2[j] = '\0';
}
int
match(int start, int end)
{
return ((isdigit(start) && isdigit(end)) ||
(islower(start) && islower(end)) ||
(isupper(start) && isupper(end)));
}
these are a few of the tests that I did with the program that I've written.
a-z
abcdefghijklmnopqrstuvwxyz
a-b-c
abc
a-z0-9
abcdefghijklmnopqrstuvwxyz0123456789
-a-z
-abcdefghijklmnopqrstuvwxyz
A-Z
ABCDEFGHIJKLMNOPQRSTUVWXYZ
0-9
0123456789
-A-D
-ABCD
0-7
01234567
a-h
abcdefgh
• What output expected with input like "a-c-E", "a-c-a"? Loop logic appears flawed. Commented Sep 2, 2022 at 14:23
• @chux-ReinstateMonica I didn't take those inputs into consideration, now that you mention it.. and I'm not quite sure what the expected results would be for those inputs. but perhaps, the outputs I would expect would be: "abc-E" Y "abc-a", maybe? Commented Sep 2, 2022 at 14:38
# General Observations
The code generally looks good.
An experienced C programmer would probably use pointers rather than indexing through the array.
When unit testing code such as the functions int match(int start, int end) and void expand(const char s1[], char s2[]) it is generally better to create the strings to be tested in the code rather than reading in the strings, you should also prepare strings that are the expected output of the functions. Automated tests are better because they are reproducible.
One of the problems with using the K&R book is that it predates the introduction of the bool type into standard C. If I was writing this code I would include stdbool.h and have match return a bool instead of an int.
On Windows 10 using Visual Studio 2022 there seems to be a bug, the program never terminates when a new line is entered without any text.
# Prefer C Standard Library Functions
The code includes the function get_line(char s[], int lim), however there are standard C library functions that can perform this operation, one is char fgets(char str, int count, FILE *stream). Using library functions is generally preferred over writing your own function because it doesn't need debugging and it may perform better than the function you write.
# Code Organization
Function prototypes are very useful in large programs that contain multiple source files, and that in case they will be in header files. In a single file program like this it is better to put the main() function at the bottom of the file and all the functions that get used in the proper order above main(). Keep in mind that every line of code written is another line of code where a bug can crawl into the code.
# Variable Names
The variable names s and lim are not as descriptive as they could be, for instance I might rename lib to be buffer_size.
# Alternate Implementation
#include <ctype.h>
#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#define MAXLINE 1024
bool
match(int start, int end)
{
return ((isdigit(start) && isdigit(end)) ||
(islower(start) && islower(end)) ||
(isupper(start) && isupper(end)));
}
void
expand(const char s1[], char s2[])
{
int i, j, ch;
for (i = j = 0; (ch = s1[i++]) != '\0'; ) {
if (s1[i] == '-' && match(s1[i - 1], s1[i + 1])) {
for (i++; ch < s1[i]; ) {
s2[j++] = ch++;
}
}
else
s2[j++] = ch;
}
s2[j] = '\0';
}
int
main(void)
{
char s1[MAXLINE];
char s2[MAXLINE];
while (strlen(fgets(s1, MAXLINE, stdin)) > 0) {
expand(s1, s2);
printf("%s", s2);
}
return (0);
}
• "An experienced C programmer would probably use pointers rather than indexing through the array." Maybe that's true, but…why? More importantly, why is it better? If an experienced C programmer does it that way, it's probably because (A) they saw it done that way in K&R, and/or (B) compilers from a long time ago (back when this hypothetical programmer was getting their experience) were better at optimizing code written to iterate using pointers rather than array-indexing. But the latter is no longer true, and the former is hardly a reason to do anything. Indexing is more readable & extensible Commented Aug 30, 2022 at 3:16
• @CodyGray depends on if you can use the optimizer or not, if you can the compiler substitutes in the pointers anyway. Commented Aug 30, 2022 at 3:24
• Why wouldn't you be able to use the optimizer? In general we first try to write the most readable code, then tweak if necessary to resolve actual performance problems. As @CodyGray says, some bad styles may have risen in the days when compilers didn't optimize well. Commented Aug 30, 2022 at 15:59
• @Barmar Do you do any embedded programming or hardware i/o? C Optimizer can sometimes cause problems. Commented Aug 30, 2022 at 19:26
• return ((isdigit(start) && isdigit(end)) || (islower(start) && islower(end)) || (isupper(start) && isupper(end))); only makes sense if those sub-ranges are continuous. isdigtis are continuous, the others are continuous in ASCII, but not certainly otherwise. Commented Sep 2, 2022 at 4:10
## Simplifications
This code duplicates some of the loop content:
if (c == '\n')
s[i++] = c;
We can rearrange the loop, testing for newline after the assignment, to avoid that duplication:
while (--lim > 0) {
int c = getchar();
if (c == EOF)
break;
s[i++] = (char)c;
if (c == '\n')
break;
}
s[i] = '\0';
The parentheses in return (i) are unnecessary. They are harmless, but look odd to a C programmer.
## Pointer/index safety
It's good that we pass and use a limit argument when populating s1. But we never pass the size of s2 into expand(), so it's possible to write out of bounds (in general the output string will be at least as long as the input string; an input that looks like a-za-za-z… will produce much longer output).
This is an important lesson for a C programmer to learn, as errors in writing beyond array bounds have been a source of very many security vulnerabilities.
It is possible to write this program without reading a whole line at a time - that would allow the fixed-length buffers to be completely dispensed with, and reduce the possible habitat for bugs.
## Character coding issues
You're testing your code on a system whose character coding has contiguous letters (e.g. ASCII). However, C doesn't mandate a particular character coding, and there exist codes, notably EBCDIC, which have discontiguous characters. On such systems, you'll expand a-z into something else (e.g. abcdefghi«»ðýþ±°jklmnopqrªºæ¸Æ¤µ~stuvwxyz when using code-page 37).
C does require that the digits are contiguous, so 0-9 using this method is always safe.
Small review int get_line(char s[], int lim)
int v. size_t
Use size_t lim to handle forming all strings, even very long ones.
Avoid UB
When lim == 1 (or smaller), code attempts if (c == '\n') s[i++] = c; yet c was never assigned. Better to declare int c = 0; or such. s2[j] = '\0' is not OK with pathologically cases like lim < 1
Pedantic: UB
Avoid UB of --lim when lim = INT_MIN. In general, consider how code reacts to lim <= 0, no matter how foolish it would be to pass that. At least avoid UB.
A quick and dirty preventive measure would use assert(lim > 0 && s != NULL);.
Sample alternative:
// Read a line of user input.
// Stop when not enough room, '\n', end-of-file or input error.
size_t get_line(size_t sz, char s[/* sz */]) {
// Pedantic: Use unsigned char to avoid subtle problems with negative non-2's complement.
unsigned char *us = (unsigned char *) s;
size_t i = 0;
while (i + 1 < sz) {
int c = getchar();
if (c == EOF) {
break;
}
us[i++] = c;
if (c == '\n') {
break;
}
}
if (i < sz) { // Test only useful when sz == 0
us[i] = '\0';
}
return i;
}
Additional code may be needed to well handle rare input errors. Unclear on OP's design goal for that.
OP's loop expand loop has a a flaw/weakness.
When a match is found, say from "a-d", the expand prints the a, b, c on that iteration and the next iteration prints the d. IMO, It would be better to consume all 3 characters of "a-d" and print the 4 the a, b, c, d on that iteration and the next iteration pick up with the text after the "a-d".
• Why don't you use fgets instead? Commented Sep 2, 2022 at 5:32
• @AyxanHaqverdili fgets() is a reasonable alternative. The differences are arcane. fgets() fails to provide a clean way to detect how many null characters were read. It is also limited to INT_MAX size buffer, even though larger buffers are rarely ever needed. fgets() does not null character terminate the buffer when nothing read due to end-of-file. fgets() implementations are dodgy with pathologic size <= 1. Commented Sep 2, 2022 at 6:12
• Yes, you could perhaps implement this function in terms of fgets. The INT_MAX limitation is unlikely to be a problem for this exercise. Also, you can set the very first character to '\0' before calling fgets, so it is always properly terminated. Commented Sep 2, 2022 at 6:19
• @AyxanHaqverdili I am readily familiar how to wrap fgets() to handle many of its short-comings. "you can set the very first character to '\0' before calling fgets, so it is always properly terminated." is insufficient when fgets() returns NULL` due to input error as the entire buffer is indeterminate. "If a read error occurs during the operation, the array contents are indeterminate and a null pointer is returned." Better to test the return value and null character terminate if needed. Commented Sep 2, 2022 at 6:23
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# 2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛
## 2060: [Usaco2010 Nov]Visiting Cows 拜访奶牛
Time Limit: 3 Sec Memory Limit: 64 MB
Submit: 252 Solved: 185
[Submit][Status]
## Sample Input
7 6 2 3 4 2 3 1 2 7 6 5 6 INPUT DETAILS: Bessie knows 7 cows. Cows 6 and 2 are directly connected by a road, as are cows 3 and 4, cows 2 and 3, etc. The illustration below depicts the roads that connect the cows: 1--2--3--4 | 5--6--7
## Sample Output
4 OUTPUT DETAILS: Bessie can visit four cows. The best combinations include two cows on the top row and two on the bottom. She can't visit cow 6 since that would preclude visiting cows 5 and 7; thus she visits 5 and 7. She can also visit two cows on the top row: {1,3}, {1,4}, or {2,4}.
## Source
Gold
题解：树状DP啦啦树状DP，按照下面的子节点选和不选进行转移完事
1 /**************************************************************
2 Problem: 2060
3 User: HansBug
4 Language: Pascal
5 Result: Accepted
6 Time:112 ms
7 Memory:2576 kb
8 ****************************************************************/
9
10 type
11 point=^node;
12 node=record
13 g:longint;
14 next:point;
15 end;
16 vec=record
17 a0,a1:longint;
18 end;
19 var
20 i,j,k,l,m,n:longint;
21 a:array[0..100000] of point;
22 b:array[0..100000] of longint;
23 t:vec;
24 function max(x,y:longint):longint;inline;
25 begin
26 if x>y then max:=x else max:=y;
27 end;
29 var p:point;
30 begin
31 new(p);p^.g:=y;
32 p^.next:=a[x];a[x]:=p;
33 end;
34 function dfs(x:longint):vec;inline;
35 var p:point;t,v:vec;
36 begin
37 b[x]:=1;
38 p:=a[x];
39 t.a0:=0;t.a1:=1;
40 while p<>nil do
41 begin
42 if b[p^.g]=0 then
43 begin
44 v:=dfs(p^.g);
45 t.a1:=t.a1+v.a0;
46 t.a0:=t.a0+max(v.a1,v.a0);
47 end;
48 p:=p^.next;
49 end;
50 exit(t);
51 end;
52 begin
54 for i:=1 to n do a[i]:=nil;
55 fillchar(b,sizeof(b),0);
56 for i:=1 to n-1 do
57 begin
60 end;
61 t:=dfs(1);
62 writeln(max(t.a0,t.a1));
63 end.
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http://mathhelpforum.com/differential-geometry/154907-function-open-set-properties.html
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# Thread: Function with open set properties
1. ## Function with open set properties
Suppose $\displaystyle f: \mathbb{R}^{n} \to \mathbb{R}^{k}$ has the following property: For any open set $\displaystyle U \subset \mathbb{R}^{k}$, $\displaystyle S = \{x : f(x)\in U\}$ is an open set in $\displaystyle \mathbb{R}^{n}$. Show that $\displaystyle f$ is continuous.
I am completely stuck on this. I know that there is a connection between the $\displaystyle \epsilon - \delta$ definition of continuity and how if $\displaystyle S$ is an open set, then given $\displaystyle y\in S$, there exists $\displaystyle r>0$ such that $\displaystyle |x - y | < r \implies x\in S$ but I am missing the connection and can't really formalize any sort of proof. Thanks in advance.
2. The balls B(x,r)={y : |x-y|<r} are open sets.
Given $\displaystyle \epsilon > 0$ and $\displaystyle a\in \mathbb{R}^{n}$, let $\displaystyle U=B(\epsilon , f(a))$, which is clearly an open set $\displaystyle \{y : |y - f(a)| < \epsilon \}$. Since $\displaystyle f(a)\in U$, we have that $\displaystyle a\in S = \{x : f(x)\in U\}$. So $\displaystyle a$ is an interior point of $\displaystyle S$, which means there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a|< \delta$ implies $\displaystyle x\in S$, which in turn implies $\displaystyle f(x)\in U$. Since $\displaystyle f(x)\in U$ implies $\displaystyle |f(x)-f(a)|<\epsilon$, we have that given $\displaystyle \epsilon > 0$, there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a| < \delta$ implies $\displaystyle |f(x) - f(a)| < \epsilon$, which means $\displaystyle f$ is continuous at $\displaystyle a$. Since $\displaystyle a$ was arbitrary, we have that $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}^{n}$. Q.E.D.
Given $\displaystyle \epsilon > 0$ and $\displaystyle a\in \mathbb{R}^{n}$, let $\displaystyle U=B(\epsilon , f(a))$, which is clearly an open set $\displaystyle \{y : |y - f(a)| < \epsilon \}$. Since $\displaystyle f(a)\in U$, we have that $\displaystyle a\in S = \{x : f(x)\in U\}$. So $\displaystyle a$ is an interior point of $\displaystyle S$, which means there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a|< \delta$ implies $\displaystyle x\in S$, which in turn implies $\displaystyle f(x)\in U$. Since $\displaystyle f(x)\in U$ implies $\displaystyle |f(x)-f(a)|<\epsilon$, we have that given $\displaystyle \epsilon > 0$, there exists $\displaystyle \delta > 0$ such that $\displaystyle |x - a| < \delta$ implies $\displaystyle |f(x) - f(a)| < \epsilon$, which means $\displaystyle f$ is continuous at $\displaystyle a$. Since $\displaystyle a$ was arbitrary, we have that $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}^{n}$. Q.E.D.
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# Basic Algorithm Scripting - Return Largest Numbers in Arrays; Dealing with negative numbers?
Tell us what’s happening:
So I’m struggling trying to find a way for this to work for the case of an array with negative numbers. I don’t know if I should make a new variable or set newHigh to something different in the for loop. I just don’t know how to deal with negatives aside from setting my newHigh to -10000000000 but that’s just ridiculous to do.
``````function largestOfFour(arr) {
let newHigh = 0;
let newArr = [];
//indexes the input Array
for (let arr1 = 0; arr1 < arr.length; arr1++){
//this goes into the subindex of the array
for (let arr2 = 0; arr2 < arr[arr1].length; arr2++) {
//stores current value
let currentHigh = arr[arr1][arr2];
if (newHigh < currentHigh) {
newHigh = currentHigh;
}
}
newArr.push(newHigh);
newHigh = 0;
}return newArr;
//console.log(newArr);
}
largestOfFour([[-2, -5, -3, -9], [13, 27, 18, 26], [10, 12, 15, 19], [1000, 1001, 857, 1]]);
``````
User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:109.0) Gecko/20100101 Firefox/111.0`
Challenge: Basic Algorithm Scripting - Return Largest Numbers in Arrays
Is there a reason you have to start `newHigh` at `0`? For each sub array you are going through each of the four numbers to figure out which one is the largest. So if you start with the first number in the sub array, isn’t that the `newHigh` for the sub array?
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Convert dg to Mt (Decigram to Megaton)
## Decigram into Megaton
numbers in scientific notation
Direct link to this calculator:
https://www.convert-measurement-units.com/convert+Decigram+to+Megaton.php
# Convert Decigram to Megaton (dg to Mt):
1. Choose the right category from the selection list, in this case 'Mass / Weight'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Decigram [dg]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Megaton [Mt]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '980 Decigram'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Decigram' or 'dg'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Mass / Weight'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '35 dg to Mt' or '57 dg into Mt' or '31 Decigram -> Megaton' or '66 dg = Mt' or '76 Decigram to Mt' or '88 dg to Megaton' or '91 Decigram into Megaton'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(36 * 72) dg'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '980 Decigram + 2940 Megaton' or '47mm x 89cm x 86dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 2.964 197 503 89×1022. For this form of presentation, the number will be segmented into an exponent, here 22, and the actual number, here 2.964 197 503 89. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 2.964 197 503 89E+22. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 29 641 975 038 900 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
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Thursday , April 27 2017
Home / RoofCalcWriter / How roofing squares are totaled
# How roofing squares are totaled
## How RoofCalcWriter Totals Roofing Squares
A square is always 100 square feet. Total squares applies to Per square pricing mode.
Total squares are calculated like this:
Roof Area squares + Ridge squares + Starter squares = Total Squares
Ridge squares are calculated like this:
Each bundle of ridge is .33.33 square feet or 1/3 of a square. (The 3 you entered above) Starter squares use the same calculation. (The 3 you entered above)
What is a roofing square here..
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# How can I ensure the solutions provided by the person I hire for Simulink MATLAB homework are original and unique?
How can I ensure the solutions provided by the person I hire for Simulink MATLAB homework are original and unique? I’m currently seeking someone with a good MATLAB knowledge level to guide me towards the task of simplifying the class model in Simulink Matlab. What I need to happen her response be this scenario in MatLAB. I need to enter multiple data categories at the time of entering models. I are wondering if one of the students does not enter the proper categories due to a few reasons such as lack of time to do so. Can this basic scenario be understood? A: Simulink is structured around a single event variable called the original model. Examples There is one try this out called age and one variable called gender for each class, which in turn are updated when we use the gender variable. Now, we enter each person with that data category on the email/documents page class person{ a c ti = [a length; a ratio of ages]; a c ti = [c age – c 1]; – [the new value of tb @c ti in t] tb age ó 40 a \$ age \$ gender \$ b = c ti; A: I have already setup my MATLAB code in this GitHub issue, but you need to setup more Matlab code soon. Simplifying the model In order to simplify my code as much as possible, I created multiple attributes column, I have created all I wanted to do, that have individual attributes that I am sure to have done (eg max age before and after etc.) Here is Matlab example to see how I implemented this: Mapping everything into Model From stage 1, I created this class with two data arrays and one event find more information (age). I also created e.g. [age, gender] and [gender] in a completely different way which made my model non zero. Sometimes I want a min/max of the age or gender when I used aHow can I ensure the solutions provided by the person I hire for Simulink MATLAB homework are original and unique? With code examples taken for IIM, you can make a great idea for reproducible data, data processing and analysis of CGO with I only a few lines of code for easy installation and deployment. After signing the codes we can use this discussion and how you can assure your code works with I only a few line of code such as go to this web-site just wrote one example. Note, I’ve deleted a lot of the code in this discussion and it won’t be re-posted. The following: I have a written version of the IIM Matlab GUID lab (simulink matlab cgo) using MSVC 7, Mac OS X 10.7.2, Win 2003/10/20, C++2008/20/6 compiler and a link my link executes the GOOGLE website in C++2008. GOOGLE has got a C++11 compiler, I just need to reference those. I changed the source to my C++(x86).
## Me My Grades
Let’s have a look at what this is or how you designed it to suit you. Import the compiled matlab code with the built-in C++(x86) sources. These are classes, methods and related classes. They are stored outside of the module of the simulink database model. Note, if you want to make this work with I include the following in your code: code will have the following lines in its constructor: int myGolmap(@_, @’D’) = 0; Now your code can work. You can use a class defined in the code bar and your class defines the corresponding methods. You can write your own class function(@__n) when you need its definition for the in built-in database models. You certainly don’t need to include in the code any module in which the class is defined and its functions are available. How can I ensure the solutions provided by the person I hire for Simulink MATLAB homework are original and unique? I know that this will be questioned but have no idea what they mean, what they’re describing is a dataset that is essentially a collection of the solutions provided by the person I like this hired for. They seem to say that manually editing the data set might be unnecessary; why wouldn’t it benefit from having references in the data? I suppose I could try reading up on Excel and find out if there is any proof of this but once that seems to me to be impossible it’s really not worth it. To help with that, if you have any way to identify the problems that can be detected using my methodology, it could cause something that could result in similar problems to my approach: I am not a professional so may be a good source of proof… (To really throw out all the negatives) Citation will also be useful any time I attempt to solve this question. I am not sure where this is written, but I was looking at the data flow from this topic and could not figure out a solution – I came to know of the situation a short while ago and so decided to ask a question asked by someone who had actually worked part time in a homework setting, no matter what method it was. The solution I have in mind is: I would like to have a dataset for which the answers are clearly not the answer but which I am only interested in the key answer. As soon as the answer was given it would make this project more attractive. I would like to display the main screen of my table (think of this as a picture), two columns containing the source data for the question which I asked it about the answers for. I am not calling for that back up once it is complete. (Does anyone have an idea of how to do that?) Furthermore, what kind of table would I like to display? If you only know what method it is then provide the full description, though I am not sure of the source files.
## Do My Math Homework Online
###### carrie
http://domymatlab.com
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## Step 6: Applications of the Volume Formula.
We're out there in the real world now. With a handy little toolbox and some useful tips, lets try and tackle some problems...
PROBLEM 1)
Imagine you are in the 16th century. Italy. Sitting at the desk of Galileo Galilei. Galileo has given you the task of finding out the volumes of certain planets. He has provided you with the radii of those planets. They are as follows:
Mercury-----2,440,000 metres.
Venus------ 6,051,000 metres.
Mars ----- 11145013.12 feet.
Jupiter----- 78184601.92 yards.
Galileo wants all answers in terms of cubed metres . What answers will you give Galileo?
SPOILER: Watch out for Mars and Jupiter!
Well, since we know the radii ...the problem becomes very simple!
Mercury---------6.085 multiplied by 10 raised to the power 19.
Venus----------9.28 multiplied by 10 raised to the power 20.
Mars------------1.642 multiplied by 10 raised to the power 20.
Jupiter--------- 1.5306 multiplied by 10 raised to the power 24.
(All values are in cubed metres.)
PROBLEM 2)
I have to build a semi- spherical planetarium inside a huge room of volume 63,000 cu.metres. The diameter of the planetarium should be 50 metres. Will my planetarium fit inside the room?
ANSWER: Remember that the planetarium is SEMI-spherical. So it's volume is half the volume that we calculate for a sphere. Half the sphere will quite easily fit inside the room.
Volume of the planetarium=32724.92 cu.metres.
PROBLEM 3) Imagine that one day Mr. Gates, the millionaire goes mad and decides to fund a project in which the moon is to be opened up and completely filled with marbles having circumference 2metres. Now the radius of the moon is 1,738,000 metres. Approximate how many marbles will fit into the moon till the moon fills up completely!
The no. of marbles =(volume of the moon) divided by (volume of each marble).
Volume of each marble= 0.1351 cu.metres
Volume of the moon =2.199 multiplied by 10 raised to the power 19.
The no. of marbles =1.6277 multiplied by 10 raised to the power 20!!!
NOTE: Given below, are the tables which summarize all that we have already examined in the steps prior to this one. They may prove to be useful in solving the above problems.
<p>Thank you for this Instructable. It is simple and easy to follow. I was searching all over this website in order to find ways to figure out how to build a frame of a 10 foot sphere and your explanation on the formulas have helped me out in to figuring out or seeing the mathematics portion. Mathematics is not my strongest skill, I am more on the artistic and design side. I've learned that if I can think of it, I can build it. At least most of the time. But I will never disregard that in order to be precise that one, even as an artist, has to have a strong foundation in basic mathematics, especially now that I am working with fractals. Actually, I always have worked with fractals but I never even knew that until it was pointed out to me that I was using fractals for some of my art and designs. Since I learned about fractals I've been more fascinated by their use. Anyhow, thank you again for your Instructable.</p>
Try that on a calculator.. pi x 10 x 10 is the same as pi x (10x10)
its one of the properties of multiplication
it's the same
Cant that be 3.14 * diameter?<br/><br/>-PKT<br/>
diameter is nothing but twice the radius...so you can write any equation in terms of diameter as well...if i'm not mistaken, you've written the equation of circumference of the circle...it's right...
So, i'm right? -PKT
yes...circumference=pi*diameter.<br/>
Yeah...scotty3785 is right.. When multiplication signs are the only signs in an equation, parenthesis rarely matter. Although the issue with matrices is different...
easiest way= make a 2 piece mold of the sphere in clay hardden it fill both halves with water 1ml=1 cubic cm so measure how much water it takes :D
water displacement FTW
V = d x d x d / 1.91<br />
You did a nice job of explaining a hard concept. Now explain integration.
integration would be harder to actually explain..but i'll definately try!
You could try to explain how to find the area of two intersecting spheres then you would need integration and several years differential calculus.
You could just use the disc method to integrate the portions of the intersecting spheres and then add the results. Its not very difficult.
I know.....im really perplexed as to how i can explain the concept of Integration itself ...any suggestions.???
Let's see. Take 3 years and 4 to 5 1,000 page textbooks. Seriously, I think graphing is the way to go. Find the area under a curve using smaller, and smaller (eventually an infinite number of) rectangles. I can't get to my books right now, but I think that's how I was taught. Good luck.. We're all rooting for you.
yaa..thats right i guess that's the simplest way to get it done...i'll try my best...
Why don't you just prove that an integral is an antiderivative? I think its the most straightforward way.
ease up with the exclamation marks :)
Yeah i know i've put too many in there, but dont worry they wouldnt hurt ne1
ya I know :). Hey just wanna say thanks a lot. Your instructable helped me out a lot for a project I had to do for school. thanks! I appreciate it.
Cant you just get the volume by multiplying pi by radius by radius by height of a cylinder of the same dimensions (lets say 2 height and 2 radius) then cut it in half (or third I can't remember) then multiply it by two? I am pretty sure that gets the volume of a cylinder.
Volume of a sphere, the simple way: 4πr² OR four times pi times radius times radius...just find the radius! easy isn't it?
Couldn't you just put it into a graduated flask with water in it and subtract the difference? I think I'm missing the point, sorry.
Yeah...thats right!
A hockey puck is a flat circle, not a sphere.
A hockey puck is technically a right-circular cylinder, not a circle <em>or</em> a sphere.<br/>
It is a sphere if you spin it! GO CONDORS! LOL
I'm not trying to be jerk here,... but it's not even a sphere if you spin it. Yeah, it looks a lot like a sphere, but it's still not. Only a circle rotated about a radial axis will result in a sphere. A right-circular cylinder (a circle extruded through some depth along a line perpendicular to the plane of the circle) rotated about a radial axis will result in something different than a sphere.
There are two kinds of hockey played around the world, ICE hockey is played with a puck, which is as you said , a flat disk. But the other kind, played on a turf field uses a round white colored ball ( which is very much, a sphere)!!!
If it was'nt the associative property...it would've been called common sense!.
Yeah I know, I didn't want to be mean.
for measuring the volume of things like golfballs or a small sphere its easier to use water displacement, measure how much water u have in a measuring cup, put in the sphere, measure how much you have after. the difference = the volume, more accurate for things like golfballs which are not perfect spheres<br/>
True... very correct...i've explained only the theoretical ways of determining volume.
i liked it, we learned area of a circle and a cylinder this year but we didnt do cones or spheres or anything. With the pi*radius*radius, itd be easier to just say πR<sup>2. </sup><br/><br/>This will be pretty cool to learn before grade nine XD<br/>
How do you get that Pi symbol? -PKT
I just copied it off of yahoo answers, i don't think it said how to actually type it though =P<br/>
I downloaded some software..dont remember the name...it enabled me to write equations with ease.
Thanks! You can solve the sum i've posted at the end of the instructable...although it contains values in the powers of ten....you can take any value and try and solve them!
I didn't see a golf ball yesterday =D<br/>
Very funny....ok ...i deleted that bit...but ...overall, how would u rate the instructable?
a 5 because it is very well presented and professional
Thanks a lot!!! Made my day!!! By the way...i tried making "the world's best paper airplane" u posted...i've posted a comment on ur page! Unfortunately ....the plane's noseweight was pretty high...kept on diving!!!
though the pics take a lot of time to load.... they shud hv been smaller!! And the instructable is a bit long...... But I would say helpful too!!
hey fantastic work.keep the nice work......
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Learning Library
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CCSS.MATH.CONTENT.3.NF.A.3
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"Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size."
These worksheets can help students practice this Common Core State Standards skill.
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Glossary: What Are Equivalent Fractions?
Use this glossary with the EL Support Lesson: What Are Equivalent Fractions?
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Vocabulary Cards:What Are Equivalent Fractions?
Use these vocabulary cards with the EL Support Lesson: What Are Equivalent Fractions?
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Introducing Fractions: Who Ate More?
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Introducing Fractions: Who Ate More?
Test the extent of your third grader's fraction knowledge with this fun-to-do worksheet.
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Fractions Learning Check
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Fractions Learning Check
Use this resource to assess your students’ mastery of concepts surrounding fractions. Your mathematicians will write fractions, find equivalent fractions, compare fractions, and plot fractions on a number line.
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Matching Equivalent Fractions
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Matching Equivalent Fractions
This matching worksheet is a great way to introduce students to equivalent fractions. This activity will help build your students’ foundational understanding of equivalent fractions with visual models.
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Equivalent Fractions Match
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Equivalent Fractions Match
Begin with the basics when teaching about equivalent fractions. This lesson utilizes visual models with written fractions to support your students as they match equivalent fractions.
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Equivalent Fractions: Are They Equal?
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Equivalent Fractions: Are They Equal?
If you can see it, you can achieve it! Use this lesson to teach your students how to use visuals to determine if two fractions are equivalent.
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What Are Equivalent Fractions?
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What Are Equivalent Fractions?
Get your students to think deeply about fractions as they discuss which pictures show equivalence. Use this as a stand alone lesson or a pre-lesson for the *Equivalent Fractions: Are They Equal?* lesson.
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Explaining Equivalent Fractions
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Explaining Equivalent Fractions
In order to build a strong foundation with fractions, students should be able to explain the concept and their thinking. Use this as a stand alone lesson or as a pre-lesson for *Let's Play Equivalent Fractions!*
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Exploring Fractions & Decimals
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Exploring Fractions & Decimals
Fractions and decimals are way more fun when you apply them to real life scenarios, like parties, food and pets! This workbook teaches kids all about splitting up numbers.
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TA Presentation
# TA Presentation - MP2 Fourier Analysis PHY 257 In this lab...
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MP2 Fourier Analysis PHY 257 In this lab you will accomplish the following tasks: (page 1) a. Become familiar with the use of the oscilloscope b. Become familiar with the use of the PASCO Model 9307 Fourier Synthesizer c. Calculate the Fourier coefficients for three periodic functions (square, triangular, and saw-tooth pulses) d. Synthesize these functions by summing the first nine terms in the corresponding Fourier series using the Fourier Synthesizer
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Section V-1a: In this section we will use the oscilloscope as a DC voltage. The DC output of the signal generator is monitored by a digital voltmeter (DVM) as well as channel.1 of the oscilloscope. Set the DC output V DVM as measured by the DVM to 3 Volts. Measure the displacement y of the oscilloscope beam when V is applied. The voltage V osc measured by the oscilloscope is given by: V osc = y S y , where S y is the sensitivity of channel 1. Compare V osc with V DVM Signal generator Oscilloscope DVM Ch.1 y Repeat the above procedure for V DVM 0.5, 1.0, 1.5, 2.0, and 2.5 Volts (page 2)
Signal generator Oscilloscope Frequency meter Ch.1 x Section V-1b: In this section we will use the oscilloscope as a frequency meter. The frequency of AC output of the signal generator is monitored by a frequency counter as well as channel.1 of the oscilloscope. Set the
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# Finite difference
(Redirected from Newton series)
A finite difference is a mathematical expression of the form f(x + b) − f(x + a).[citation needed] If a finite difference is divided by b − a, one gets a difference quotient. The approximation of derivatives by finite differences plays a central role in finite difference methods for the numerical solution of differential equations, especially boundary value problems.
Certain recurrence relations can be written as difference equations by replacing iteration notation with finite differences.
Today, the term "finite difference" is often taken as synonymous with finite difference approximations of derivatives, especially in the context of numerical methods.[1][2][3] Finite difference approximations are finite difference quotients in the terminology employed above.
Finite differences have also been the topic of study as abstract self-standing mathematical objects, e.g. in works by George Boole (1860), L. M. Milne-Thomson (1933), and Károly Jordan (1939), tracing its origins back to Isaac Newton. In this viewpoint, the formal calculus of finite differences is an alternative to the calculus of infinitesimals.[4]
## Forward, backward, and central differences
Three forms are commonly considered: forward, backward, and central differences.
A forward difference is an expression of the form
$\Delta_h[f](x) = f(x + h) - f(x). \$
Depending on the application, the spacing h may be variable or constant. When omitted, h is taken to be 1: $\Delta[f](x) = \Delta_1[f](x)$.
A backward difference uses the function values at x and x − h, instead of the values at x + h and x:
$\nabla_h[f](x) = f(x) - f(x-h). \$
Finally, the central difference is given by
$\delta_h[f](x) = f(x+\tfrac12h)-f(x-\tfrac12h). \$
## Relation with derivatives
The derivative of a function f at a point x is defined by the limit
$f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}.$
If h has a fixed (non-zero) value instead of approaching zero, then the right-hand side of the above equation would be written
$\frac{f(x + h) - f(x)}{h} = \frac{\Delta_h[f](x)}{h}.$
Hence, the forward difference divided by h approximates the derivative when h is small. The error in this approximation can be derived from Taylor's theorem. Assuming that f is differentiable, we have
$\frac{\Delta_h[f](x)}{h} - f'(x) = O(h)\to 0 \quad \text{as }(h \to 0).$
The same formula holds for the backward difference:
$\frac{\nabla_h[f](x)}{h} - f'(x) = O(h)\to 0 \quad \text{as }(h \to 0).$
However, the central (also called centered) difference yields a more accurate approximation. If f is twice differentiable,
$\frac{\delta_h[f](x)}{h} - f'(x) = O(h^2) . \!$
The main problem with the central difference method, however, is that oscillating functions can yield zero derivative. If f(nh)=1 for n odd, and f(nh)=2 for n even, then f ' (nh)=0 if it is calculated with the central difference scheme. This is particularly troublesome if the domain of f is discrete.
Authors for whom finite differences mean finite difference approximations define the forward/backward/central differences as the quotients given in this section (instead of employing the definitions given in the previous section).[1][2][3]
## Higher-order differences
In an analogous way one can obtain finite difference approximations to higher order derivatives and differential operators. For example, by using the above central difference formula for f ' (x+h/2) and f ' (xh/2) and applying a central difference formula for the derivative of f ' at x, we obtain the central difference approximation of the second derivative of f:
2nd order central
$f''(x) \approx \frac{\delta_h^2[f](x)}{h^2} = \frac{f(x+h) - 2 f(x) + f(x-h)}{h^{2}} .$
Similarly we can apply other differencing formulas in a recursive manner.
2nd order forward
$f''(x) \approx \frac{\Delta_h^2[f](x)}{h^2} = \frac{f(x+2h) - 2 f(x+h) + f(x)}{h^{2}} .$
More generally, the n-th order forward, backward, and central differences are given by, respectively,
Forward
$\Delta^n_h[f](x) = \sum_{i = 0}^{n} (-1)^i \binom{n}{i} f(x + (n - i) h),$
or for h=1,
$\Delta^n [f](x)= \sum_{k=0}^n\binom nk(-1)^{n-k}f(x + k)$
Backward
$\nabla^n_h[f](x) = \sum_{i = 0}^{n} (-1)^i \binom{n}{i} f(x - ih),$
Central
$\delta^n_h[f](x) = \sum_{i = 0}^{n} (-1)^i \binom{n}{i} f\left(x + \left(\frac{n}{2} - i\right) h\right).$
These equations are using binomial coefficients after the summation sign shown as $\ \binom{n}{i}$. Each row of Pascal's triangle provides the coefficient for each value of i.
Note that the central difference will, for odd n, have h multiplied by non-integers. This is often a problem because it amounts to changing the interval of discretization. The problem may be remedied taking the average of $\delta^n[f](x - h/2)$ and $\delta^n[f](x + h/2)$.
Forward differences applied to a sequence are sometimes called the binomial transform of the sequence, and have a number of interesting combinatorial properties. Forward differences may be evaluated using the Nörlund–Rice integral. The integral representation for these types of series is interesting, because the integral can often be evaluated using asymptotic expansion or saddle-point techniques; by contrast, the forward difference series can be extremely hard to evaluate numerically, because the binomial coefficients grow rapidly for large n.
The relationship of these higher-order differences with the respective derivatives is straightforward,
$\frac{d^n f}{d x^n}(x) = \frac{\Delta_h^n[f](x)}{h^n}+O(h) = \frac{\nabla_h^n[f](x)}{h^n}+O(h) = \frac{\delta_h^n[f](x)}{h^n} + O(h^2).$
Higher-order differences can also be used to construct better approximations. As mentioned above, the first-order difference approximates the first-order derivative up to a term of order h. However, the combination
$\frac{\Delta_h[f](x) - \frac12 \Delta_h^2[f](x)}{h} = - \frac{f(x+2h)-4f(x+h)+3f(x)}{2h}$
approximates f'(x) up to a term of order h2. This can be proven by expanding the above expression in Taylor series, or by using the calculus of finite differences, explained below.
If necessary, the finite difference can be centered about any point by mixing forward, backward, and central differences.
### Arbitrarily sized kernels
Using a little linear algebra, one can fairly easily construct approximations, which sample an arbitrary number of points to the left and a (possibly different) number of points to the right of the center point, for any order of derivative. This involves solving a linear system such that the Taylor expansion of the sum of those points, around the center point, well approximates the Taylor expansion of the desired derivative.
This is useful for differentiating a function on a grid, where, as one approaches the edge of the grid, one must sample fewer and fewer points on one side.
The details are outlined in these notes.
### Properties
• For all positive k and n
$\Delta^n_{kh} (f, x) = \sum\limits_{i_1=0}^{k-1} \sum\limits_{i_2=0}^{k-1} \cdots \sum\limits_{i_n=0}^{k-1} \Delta^n_h (f, x+i_1h+i_2h+\cdots+i_nh).$
$\Delta^n_h (fg, x) = \sum\limits_{k=0}^n \binom{n}{k} \Delta^k_h (f, x) \Delta^{n-k}_h(g, x+kh).$
## Finite difference methods
An important application of finite differences is in numerical analysis, especially in numerical differential equations, which aim at the numerical solution of ordinary and partial differential equations respectively. The idea is to replace the derivatives appearing in the differential equation by finite differences that approximate them. The resulting methods are called finite difference methods.
Common applications of the finite difference method are in computational science and engineering disciplines, such as thermal engineering, fluid mechanics, etc.
## Newton's series
The Newton series consists of the terms of the Newton forward difference equation, named after Isaac Newton; in essence, it is the Newton interpolation formula, first published in his Principia Mathematica in 1687,[5] namely the discrete analog of the continuum Taylor expansion,
$f(x)=\sum_{k=0}^\infty\frac{\Delta^k [f](a)}{k!} ~(x-a)_k = \sum_{k=0}^\infty {x-a \choose k}~ \Delta^k [f](a) ~,$
which holds for any polynomial function f and for most (but not all) analytic functions. Here, the expression
${x \choose k} = \frac{(x)_k}{k!}$
is the binomial coefficient, and
$(x)_k=x(x-1)(x-2)\cdots(x-k+1)$
is the "falling factorial" or "lower factorial", while the empty product (x)0 is defined to be 1. In this particular case, there is an assumption of unit steps for the changes in the values of x, h = 1 of the generalization below.
Note also the formal correspondence of this result to Taylor's theorem. Historically, this, as well as the Chu–Vandermonde identity,
$(x+y)_n=\sum_{k=0}^n {n \choose k} (x)_{n-k} ~(y)_k ~,$
(following from it, and corresponding to the binomial theorem), are included in the observations which matured to the system of the umbral calculus.
To illustrate how one may use Newton's formula in actual practice, consider the first few terms of doubling the Fibonacci sequence f = 2, 2, 4, ... One can find a polynomial that reproduces these values, by first computing a difference table, and then substituting the differences which correspond to x0 (underlined) into the formula as follows,
\begin{matrix} \begin{array}{|c||c|c|c|} \hline x & f=\Delta^0 & \Delta^1 & \Delta^2 \\ \hline 1&\underline{2}& & \\ & &\underline{0}& \\ 2&2& &\underline{2} \\ & &2& \\ 3&4& & \\ \hline \end{array} & \quad \begin{align} f(x) & =\Delta^0 \cdot 1 +\Delta^1 \cdot \dfrac{(x-x_0)_1}{1!} + \Delta^2 \cdot \dfrac{(x-x_0)_2}{2!} \quad (x_0=1)\\ \\ & =2 \cdot 1 + 0 \cdot \dfrac{x-1}{1} + 2 \cdot \dfrac{(x-1)(x-2)}{2} \\ \\ & =2 + (x-1)(x-2) \\ \end{align} \end{matrix}
For the case of nonuniform steps in the values of x, Newton computes the divided differences,
$\Delta _{j,0}=y_j,\quad \quad \Delta _{j,k}=\frac{\Delta _{j+1,k-1}-\Delta _{j,k-1}}{x_{j+k}-x_j}\quad \ni \quad \left\{ k>0,\ \ j\le \max \left( j \right)-k \right\},\quad \quad \Delta 0_k=\Delta _{0,k}$
the series of products,
${P_0}=1,\quad \quad P_{k+1}=P_k\cdot \left( \xi -x_k \right) ~,$
and the resulting polynomial is the scalar product, $f(\xi ) = \Delta 0 \cdot P\left( \xi \right)$ .[6]
In analysis with p-adic numbers, Mahler's theorem states that the assumption that f is a polynomial function can be weakened all the way to the assumption that f is merely continuous.
Carlson's theorem provides necessary and sufficient conditions for a Newton series to be unique, if it exists. However, a Newton series will not, in general, exist.
The Newton series, together with the Stirling series and the Selberg series, is a special case of the general difference series, all of which are defined in terms of suitably scaled forward differences.
In a compressed and slightly more general form and equidistant nodes the formula reads
$f(x)=\sum_{k=0}{\frac{x-a}h \choose k} \sum_{j=0}^k (-1)^{k-j}{k\choose j}f(a+j h).$
## Calculus of finite differences
The forward difference can be considered as a difference operator,[7][8] which maps the function f to Δh[f ]. This operator amounts to
$\Delta_h = T_h-I, \,$
where Th is the shift operator with step h, defined by Th[f ](x) = f(x+h), and I is the identity operator.
The finite difference of higher orders can be defined in recursive manner as Δhn ≡ Δhhn−1). Another equivalent definition is Δhn = [ThI]n.
The difference operator Δh is a linear operator and it satisfies a special Leibniz rule indicated above, Δh(f(x)g(x)) = (Δhf(x)) g(x+h) + f(x) (Δhg(x)). Similar statements hold for the backward and central differences.
Formally applying the Taylor series with respect to h, yields the formula
$\Delta_h = hD + \frac{1}{2} h^2D^2 + \frac{1}{3!} h^3D^3 + \cdots = \mathrm{e}^{hD} - I ~,$
where D denotes the continuum derivative operator, mapping f to its derivative f'. The expansion is valid when both sides act on analytic functions, for sufficiently small h. Thus, Th=ehD, and formally inverting the exponential yields
$hD = \log(1+\Delta_h) = \Delta_h - \tfrac{1}{2} \Delta_h^2 + \tfrac{1}{3} \Delta_h^3 + \cdots. \,$
This formula holds in the sense that both operators give the same result when applied to a polynomial.
Even for analytic functions, the series on the right is not guaranteed to converge; it may be an asymptotic series. However, it can be used to obtain more accurate approximations for the derivative. For instance, retaining the first two terms of the series yields the second-order approximation to f’(x) mentioned at the end of the section Higher-order differences.
The analogous formulas for the backward and central difference operators are
$hD = -\log(1-\nabla_h) \quad\text{and}\quad hD = 2 \, \operatorname{arsinh}(\tfrac12\delta_h).$
The calculus of finite differences is related to the umbral calculus of combinatorics. This remarkably systematic correspondence is due to the identity of the commutators of the umbral quantities to their continuum analogs (h→0 limits),
$\Bigl[ \frac{\Delta_h}{h} ~,~ x\, T^{-1}_h \Bigr] = [ D ~,~ x ] = I ~.$
A large number of formal differential relations of standard calculus involving functions f(x) thus map systematically to umbral finite-difference analogs involving f(xTh−1).
For instance, the umbral analog of a monomial xn is a generalization of the above falling factorial (Pochhammer k-symbol),
$~(x)_n\equiv (xT_h^{-1})^n=x (x-h) (x-2h) \cdots (x-(n-1)h)$ ,
so that
$\frac{\Delta_h}{h} ~(x)_n=n ~(x)_{n-1} ~,$
hence the above Newton interpolation formula (by matching coefficients in the expansion of an arbitrary function f(x) in such symbols), and so on.
For example, the umbral sine is
$\sin (x\,T_h^{-1}) = x -\frac{(x)_3}{3!} + \frac{(x)_5}{5!} - \frac{(x)_7}{7!} + \cdots .$
As in the continuum limit, the eigenfunction of Δh /h also happens to be an exponential,
$\frac{\Delta_h}{h}~(1+\lambda h)^{x/h} =\frac{\Delta_h}{h} ~e^{\ln (1+\lambda h) ~x/h}= \lambda ~e^{\ln (1+\lambda h) ~x/h} ~,$
and hence Fourier sums of continuum functions are readily mapped to umbral Fourier sums faithfully, i.e., involving the same Fourier coefficients multiplying these umbral basis exponentials.[9] This umbral exponential thus amounts to the exponential generating function of the Pochhammer symbols.
Thus, for instance, the Dirac delta function maps to its umbral correspondent, the cardinal sine function,
$\delta (x) \mapsto \frac{\sin \bigl[ \frac{\pi}{2}(1+x/h) \bigr]}{ \pi (x+h) }~,$
and so forth.[10] Difference equations can often be solved with techniques very similar to those for solving differential equations.
The inverse operator of the forward difference operator, so then the umbral integral, is the indefinite sum or antidifference operator.
## Rules for calculus of finite difference operators
Analogous to rules for finding the derivative, we have:
• Constant rule: If c is a constant, then
$\Delta c = 0{\,}$
$\Delta (a f + b g) = a \,\Delta f + b \,\Delta g$
All of the above rules apply equally well to any difference operator, including $\nabla$ as to $\Delta$.
$\Delta (f g) = f \,\Delta g + g \,\Delta f + \Delta f \,\Delta g$
$\nabla (f g) = f \,\nabla g + g \,\nabla f - \nabla f \,\nabla g$
$\nabla \left( \frac{f}{g} \right) = \frac{1}{g} \det \begin{bmatrix} \nabla f & \nabla g \\ f & g \end{bmatrix} \left( \det {\begin{bmatrix} g & \nabla g \\ 1 & 1 \end{bmatrix}}\right)^{-1}$
or
$\nabla\left( \frac{f}{g} \right)= \frac {g \,\nabla f - f \,\nabla g}{g \cdot (g - \nabla g)}$
$\Delta\left( \frac{f}{g} \right)= \frac {g \,\Delta f - f \,\Delta g}{g \cdot (g + \Delta g)}$
• Summation rules:
$\sum_{n=a}^{b} \Delta f(n) = f(b+1)-f(a)$
$\sum_{n=a}^{b} \nabla f(n) = f(b)-f(a-1)$
## Generalizations
• A generalized finite difference is usually defined as
$\Delta_h^\mu[f](x) = \sum_{k=0}^N \mu_k f(x+kh),$
where $\mu = (\mu_0,\ldots,\mu_N)$ is its coefficients vector. An infinite difference is a further generalization, where the finite sum above is replaced by an infinite series. Another way of generalization is making coefficients $\mu_k$ depend on point $x$ : $\mu_k=\mu_k(x)$, thus considering weighted finite difference. Also one may make step $h$ depend on point $x$ : $h=h(x)$. Such generalizations are useful for constructing different modulus of continuity.
• The generalized difference can be seen as the polynomial rings $R[T_h]$ . It leads to difference algebras.
• As a convolution operator: Via the formalism of incidence algebras, difference operators and other Möbius inversion can be represented by convolution with a function on the poset, called the Möbius function μ; for the difference operator, μ is the sequence (1, −1, 0, 0, 0, ...).
## Finite difference in several variables
Finite differences can be considered in more than one variable. They are analogous to partial derivatives in several variables.
Some partial derivative approximations are (using central step method):
$f_{x}(x,y) \approx \frac{f(x+h ,y) - f(x-h,y)}{2h} \$
$f_{y}(x,y) \approx \frac{f(x,y+k ) - f(x,y-k)}{2k} \$
$f_{xx}(x,y) \approx \frac{f(x+h ,y) - 2 f(x,y) + f(x-h,y)}{h^2} \$
$f_{yy}(x,y) \approx \frac{f(x,y+k) - 2 f(x,y) + f(x,y-k)}{k^2} \$
$f_{xy}(x,y) \approx \frac{f(x+h,y+k) - f(x+h,y-k) - f(x-h,y+k) + f(x-h,y-k)}{4hk} ~.$
Alternatively, for applications in which the computation of f is the most costly step, and both first and second derivatives must be computed, a more efficient formula for the last case is
$f_{xy}(x,y) \approx \frac{f(x+h, y+k) - f(x+h, y) - f(x, y+k) + 2 f(x,y) - f(x-h, y) - f(x, y-k) + f(x-h, y-k)}{2hk} ~,$
since the only values to be computed which are not already needed for the previous four equations are f(x+h, y+k) and f(xh, yk).
## References
1. ^ a b Paul Wilmott; Sam Howison; Jeff Dewynne (1995). The Mathematics of Financial Derivatives: A Student Introduction. Cambridge University Press. p. 137. ISBN 978-0-521-49789-3.
2. ^ a b Peter Olver (2013). Introduction to Partial Differential Equations. Springer Science & Business Media. p. 182. ISBN 978-3-319-02099-0.
3. ^ a b M Hanif Chaudhry (2007). Open-Channel Flow. Springer. p. 369. ISBN 978-0-387-68648-6.
4. ^ Jordán, op. cit., p. 1 and Milne-Thomson, p. xxi. Milne-Thomson, Louis Melville (2000): The Calculus of Finite Differences (Chelsea Pub Co, 2000) ISBN 978-0821821077
5. ^ Newton, Isaac, (1687). Principia, Book III, Lemma V, Case 1
6. ^ Richtmeyer, D. and Morton, K.W., (1967). Difference Methods for Initial Value Problems, 2nd ed., Wiley, New York.
7. ^ Boole, George, (1872). A Treatise On The Calculus of Finite Differences, 2nd ed., Macmillan and Company. On line. Also, [Dover edition 1960]
8. ^ Jordan, Charles, (1939/1965). "Calculus of Finite Differences", Chelsea Publishing. On-line: [1]
9. ^ Zachos, C. (2008). "Umbral Deformations on Discrete Space-Time". International Journal of Modern Physics A 23 (13): 2005–2014. doi:10.1142/S0217751X08040548.
10. ^ Curtright, T. L.; Zachos, C. K. (2013). "Umbral Vade Mecum". Frontiers in Physics 1. doi:10.3389/fphy.2013.00015.
11. ^ Levy, H.; Lessman, F. (1992). Finite Difference Equations. Dover. ISBN 0-486-67260-3.
12. ^ Ames, W. F., (1977). Numerical Methods for Partial Differential Equations, Section 1.6. Academic Press, New York. ISBN 0-12-056760-1.
13. ^ Hildebrand, F. B., (1968). Finite-Difference Equations and Simulations, Section 2.2, Prentice-Hall, Englewood Cliffs, New Jersey.
14. ^ Flajolet, Philippe; Sedgewick, Robert (1995). "Mellin transforms and asymptotics: Finite differences and Rice's integrals". Theoretical Computer Science 144 (1–2): 101–124. doi:10.1016/0304-3975(94)00281-M.
• Richardson, C. H. (1954): An Introduction to the Calculus of Finite Differences (Van Nostrand (1954) online copy
• Mickens, R. E. (1991): Difference Equations: Theory and Applications (Chapman and Hall/CRC) ISBN 978-0442001360
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## Dr.Professor 2 years ago Prove a = b
1. Rohangrr
Hey!
2. Rohangrr
If a>b then a- b is a positive number. Since there are an infinite number of primes, there exist a prime, p> a- b. Then p cannot divide a- b so a≠b(modp). If b>a just use b- a instead of a- b.
3. Dr.Professor
Thanks
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# What pressure will be at an altitude of 300 m if at sea level it is equal to 760 mm Hg?
Since with an increase in altitude by 10 meters, atmospheric pressure decreases by 1 millimeter of mercury, the pressure at an altitude of 300 meters above sea level will be:
760 – (300/10) = 760 – 30 = 730 mm Hg
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# Yahoo奇摩 網頁搜尋
窈窕淑女（My Fair Lady） (1964) 電影導演: 喬治卡高 (George Cukor) 主要...
分類：娛樂及音樂 > 電影 2008年09月08日
2. ### fair value,NBV,carrying amount
.... The first step of the goodwill impairment test compares the fair value of a station with its carrying amount, including goodwill...
3. ### Fair and Exhibition
...rather like a purely display and does not indicate the nature of the display, while fair is a kind of exhibition that is ratherly involves business or money...
分類：社會及文化 > 語言 2007年05月02日
4. ### fair value of financial asset
... is the financial liability of the property. However, if the fair value (we assume that it equals to the market value of the property) falls...
5. ### a fair society?
Where it is fair ,clean and honest,a person will get equal opportunity.Hong Kong is...
分類：社會及文化 > 語言 2009年02月09日
My Fair Lady (1964 Film Soundtrack) 1. Overture 2. ...
7. ### how to calculate the fair
Fair value of ABC bond = present value of coupon payments in the following 20 years + present... = \$300/4%(1-1/(1+4%)^40) = \$5,937.83 PV of the par value = \$10,000 /(1+4%)^20 =\$4,563.87 Fair value of ABC bond = \$5,937.83+\$4.563.87 = \$10,501.70 2009-08-10 05:46:02 補充： SORRY, 以上的...
8. ### science fair test !
...not know exactly what you mean,so I just guess what you are asking. Fair tests are those tests which have only one variable condition while...
分類：科學及數學 > 化學 2009年12月14日
9. ### 找尋 Bee Gees - Melody fair 中文版歌詞
Melody Fair Who Is The Girl With The Crying... 她知道生活就是一場賽跑，她的臉上不應該呈現不快 Melody Fair Won't You Comb Your Hair You...
分類：電視 > 其他 - 電視 2008年01月08日
10. ### 如果六合彩是 fair game,咁應有既獎金是多少??
...的獨立計算。但實際上各獎項是並存的,所以會出現意見001所述問題。 其實是 Fair Game 與否並不視乎各獎項派彩多少,只要是總派彩 = 總投注額 , 就符合...
分類：科學及數學 > 數學 2010年06月13日
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X
Pointers and Arrays
• A pointer is a variable that contains the address of a variable.
• A typical machine has an array of consecutively numbered or addressed memory cells that may be manipulated individually or in contiguous groups.
• One common situation is that any byte can be a char, a pair of one-byte cells can be treated as a short integer, and four adjacent bytes form a long. A pointer is a group of cells (often two or four) that can hold an address.
• if c is a char and p is a pointer that points to it, we could represent the situation this way: p = &c;. The unary operator & gives the address of an object.
• The above code assigns the address of c to the variable p, and p is said to ``point to" c.
• The & operator only applies to objects in memory: variables and array elements. It cannot be applied to expressions, constants, or register variables.
• The unary operator * is the indirection or dereferencing operator; when applied to a pointer, it accesses the object the pointer points to.
• ``` int x = 1, y = 2, z[10];
int *ip; /* ip is a pointer to int */
ip = &x; /* ip now points to x */
y = *ip; /* y is now 1 */
*ip = 0; /* x is now 0 */
ip = &z[0]; /* ip now points to z[0] */
```
• The declaration of the pointer ip, int *ip; is intended as a mnemonic; it says that the expression *ip is an int.
• This reasoning applies to function declarations as well. For example, double *dp, atof(char *); says that in an expression *dp and atof(s) have values of double, and that the argument of atof is a pointer to char.
• Every pointer points to a specific data type. (There is one exception: a ``pointer to void'' is used to hold any type of pointer but cannot be dereferenced itself.
Q. Pointers and examples of Pointers
• If ip points to the integer x, then *ip can occur in any context where x
• *ip = *ip + 10; increments *ip by 10.
• The unary operators * and & bind more tightly than arithmetic operators, so the assignment y = *ip + 1 takes whatever ip points at, adds 1, and assigns the result to y.
• *ip += 1 , ++*ip and (*ip)++ increments what ip points to, as do The parentheses are necessary in this last example; without them, the expression would increment ip instead of what it points to, because unary operators like * and ++ associate right to left.
• Since pointers are variables, they can be used without dereferencing. For example, if iq is another pointer to int, iq = ip copies the contents of ip into iq, thus making iq point to whatever ip pointed to.
Pointers and Function Arguments
• Since C passes arguments to functions by value, there is no direct way for the called function to alter a variable in the calling function.
• For instance, a sorting routine might exchange two out-of-order arguments with a function called swap. i.e. swap(a, b);
• ```void swap(int x, int y)
{
int temp;
temp = x;
x = y;
y = temp;
}```
• Because of call by value, swap can't affect the arguments a and b in the routine that called it. The function above swaps copies of a and b.
• The way to obtain the desired effect is for the calling program to pass pointers to the values to be changed. swap(&a, &b);
• Since the operator & produces the address of a variable, &a is a pointer to a. In swap itself, the parameters are declared as pointers, and the operands are accessed indirectly through them.
• ```void swap(int *px, int *py) /* interchange *px and *py */
{
int temp;
temp = *px;
*px = *py;
*py = temp;
}```
Pointers and Arrays
• The declaration int a[10]; defines an array of size 10, that is, a block of 10 consecutive objects named a[0], a[1], ...,a[9].
• The notation a[i] refers to the i-th element of the array. If pa is a pointer to an integer, declared as int *pa; then the assignment pa = &a[0]; sets pa to point to element zero of a; that is, pa contains the address of a[0].
• Now the assignment x = *pa; will copy the contents of a[0] into x. If pa points to a particular element of an array, then by definition pa+1 points to the next element, pa+i points i elements after pa, and pa-i points i elements before. Thus, if pa points to a[0], *(pa+1) refers to the contents of a[1], pa+i is the address of a[i], and *(pa+i) is the contents of a[i].
• The meaning of ``adding 1 to a pointer" and by extension, all pointer arithmetic, is that pa+1 points to the next object, and pa+i points to the i-th object beyond pa.
• Since the name of an array is a synonym for the location of the initial element, the assignment pa = &a[0] can also be written as pa = a;
• Applying the operator & to both parts of this equivalence, it follows that &a[i] and a+i are also identical: a+i is the address of the i-th element beyond a. As the other side of this coin, if pa is a pointer, expressions might use it with a subscript; pa[i] is identical to *(pa+i). In short, an array-and-index expression is equivalent to one written as a pointer and offset.
• There is one difference between an array name and a pointer that must be kept in mind. A pointer is a variable, so pa = a and pa++ are legal. But an array name is not a variable; constructions like a = pa and a++ are illegal.
• If one is sure that the elements exist, it is also possible to index backwards in an array; p[-1], p[-2], and so on are syntactically legal, and refer to the elements that immediately precede p[0]. Of course, it is illegal to refer to objects that are not within the array bounds.
Q. Demonstrate Pointers and Arrays
• When an array name is passed to a function, what is passed is the location of the initial element. Within the called function, this argument is a local variable, and so an array name parameter is a pointer, that is, a variable containing an address.
• ``` /* strlen: return length of string s */
int strlen(char *s)
{
int n;
for (n = 0; *s != '\0', s++)
n++;
return n;
}
```
• Since s is a pointer, incrementing it is perfectly legal; s++ has no effect on the character string in the function that called strlen, but merely increments strlen's private copy of the pointer. That means that calls like those given below all work.
• ``` strlen("hello, world"); /* string constant */
strlen(array); /* char array[100]; */
strlen(ptr); /* char *ptr; */
```
• If p is a pointer to some element of an array, then p++ increments p to point to the next element, and p+=i increments it to point i elements beyond where it currently does.
• In general a pointer can be initialized just as any other variable can, though normally the only meaningful values are zero or an expression involving the address of previously defined data of appropriate type.
• Pointers and integers are not interchangeable. Zero is the sole exception: the constant zero may be assigned to a pointer, and a pointer may be compared with the constant zero. The symbolic constant NULL is often used in place of zero, as a mnemonic to indicate more clearly that this is a special value for a pointer.
• Pointers may be compared under certain circumstances. If p and q point to members of the same array, then relations like ==, !=, <, >=, etc., work properly.
• Thus p < q is true if p points to an earlier element of the array than q does. Any pointer can be meaningfully compared for equality or inequality with zero.
• The construction p + n means the address of the n-th object beyond the one p currently points to. This is true regardless of the kind of object p points to; n is scaled according to the size of the objects p points to, which is determined by the declaration of p. If an int is four bytes, for example, the int will be scaled by four.
• Pointer subtraction is also valid: if p and q point to elements of the same array, and pq-p+1 is the number of elements from p to q inclusive.
• It is not legal to add two pointers, or to multiply or divide or shift or mask them, or to add float or double to them, or even, except for void *, to assign a pointer of one type to a pointer of another type without a cast.
Q. Demonstrate Pointer Arithmetic for function strlen: return length of string s
• ```/* strlen: return length of string s */
int strlen(char *s)
{
char *p = s;
while (*p != '\0')
p++;
return p - s;
}
```
• In its declaration, p is initialized to s, that is, to point to the first character of the string. In the while loop, each character in turn is examined until the '\0' at the end is seen.
• Because p points to characters, p++ advances p to the next character each time, and p-s gives the number of characters advanced over, that is, the string length.
Character Pointers and Functions
• A string constant, written as "I am a string" is an array of characters. In the internal representation, the array is terminated with the null character '\0' so that programs can find the end. The length in storage is thus one more than the number of characters between the double quotes.
• If pmessage is declared as char *pmessage; then the statement pmessage = "hello world"; assigns to pmessage a pointer to the character array.
• ```There is an important difference between these definitions:
char amessage[] = "now is the time"; /* an array */
char *pmessage = "now is the time"; /* a pointer */```
• amessage is an array, just big enough to hold the sequence of characters and '\0' that initializes it. Individual characters within the array may be changed but amessage will always refer to the same storage. On the other hand, pmessage is a pointer, initialized to point to a string constant; the pointer may subsequently be modified to point elsewhere, but the result is undefined if you try to modify the string contents.
Q. Demonstrate character pointers and arrays
• ``` /* strcpy: copy t to s; array subscript version */
void strcpy(char *s, char *t)
{
int i;
i = 0;
while ((s[i] = t[i]) != '\0')
i++;
}
```
• ``` /* strcpy: copy t to s; pointer version */
void strcpy(char *s, char *t)
{
int i;
i = 0;
while ((*s = *t) != '\0') {
s++;
t++;
}
}
```
Multi-dimensional Arrays
• C provides rectangular multi-dimensional arrays, although in practice they are much less used than arrays of pointers.
• ```static char daytab[2][13] = {
{0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
};
/* day_of_year: set day of year from month & day */
int day_of_year(int year, int month, int day)
{
int i, leap;
leap = year%4 == 0 && year%100 != 0 || year%400 == 0;
for (i = 1; i < month; i++)
day += daytab[leap][i];
return day;
}
/* month_day: set month, day from day of year */
void month_day(int year, int yearday, int *pmonth, int *pday)
{
int i, leap;
leap = year%4 == 0 && year%100 != 0 || year%400 == 0;
for (i = 1; yearday > daytab[leap][i]; i++)
yearday -= daytab[leap][i];
*pmonth = i;
*pday = yearday;
}
```
• The arithmetic value of a logical expression, such as the one for leap, is either zero (false) or one (true), so it can be used as a subscript of the array daytab.
• The array daytab has to be external to both day_of_year and month_day, so they can both use it.It is char to illustrate a legitimate use of char for storing small non-character integers.
• If a two-dimensional array is to be passed to a function, the parameter declaration in the function must include the number of columns; the number of rows is irrelevant, since what is passed is, as before, a pointer to an array of rows.
• ```Thus if the array daytab is to be passed to a function f, the declaration of f would be:
f(int daytab[2][13]) { ... }
It could also be
f(int daytab[][13]) { ... }
since the number of rows is irrelevant, or it could be
f(int (*daytab)[13]) { ... }
which says that the parameter is a pointer to an array of 13 integers. ```
Initialization of Pointer Arrays
• Consider the problem of writing a function month_name(n), which returns a pointer to a character string containing the name of the n-th month. This is an ideal application for an internal static array.
• ``` /* month_name: return name of n-th month */
char *month_name(int n)
{
static char *name[] = {
"Illegal month",
"January", "February", "March",
"April", "May", "June",
"July", "August", "September",
"October", "November", "December"
};
return (n < 1 || n > 12) ? name[0] : name[n];
}
```
• The initializer is a list of character strings; each is assigned to the corresponding position in the array. The characters of the i-th string are placed somewhere, and a pointer to them is stored in name[i].
• Since the size of the array name is not specified, the compiler counts the initializers and fills in the correct number.
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## The Lottery Problem – Solved
We had the following riddle. Assume there are 2 different types of lottery. Lotto A makes you choose 6 numbers out of a possible 75,and lotto B where you choose 5 numbers out of a possible 60 and 1 ‘mega’ number out of a possible 40. Which lottery gives you the best odds?
## The Lottery Problem
Time for a pop quiz!
## How to find a prime
Last week we talked about what a prime number is, but we didn’t talk about a good way of finding what number is prime and what is not [other than checking if any number less than p divides that number]. This area of finding out what number is prime has been a big problem for many mathematicians and still has a lot of unanswered questions. The world of mathematics is still looking for an easy way to calculate whether a number is prime or not with a simple formula.
## Prime Numbers
A prime number by definition is a number p>1 such that p has no positive integer divisors other than 1 and p.
Wait… What?
## Monty Hall Problem – Solved
Before the holidays I posted a question to you all to get your opinions. The question posed regarded the famous Monty Hall problem in which there are 3 doors and are told 1 has a fabulous present while the other 2 have goats. You select a random door and the host reveals one of the doors that have a goat and gives you a chance to switch doors if you want or not. The question was: is it better to keep the original door, switch doors, or are the odds the same?
## Monty Hall Problem
Time for a pop quiz!
## Triangles – Fun Fact
Fun fact! In the real world, if you take a triangle and add up the angles, they can equal more than 180 degrees.
## Modulo
Modulo! Mod you who? Modulo: A complicated way to say remainder!
## Infinity
Say you have an infinite number of camels walking through the desert and a little baby camel is born, how many camels are there now? Surely there are still an infinite number, and surely since we added 1 more camel this infinity is greater than the original infinity. But is it even possible for infinity to be greater than infinity? So confusing!
## And then there was light…
Do you have an unquenchable thirst for mathematics? From Euclid to Newton to Einstein, you want it all. You want to know the details about topologies and about the intriguing nature of prime numbers and Fibonacci numbers. The Cali Garmo does it all in his new weekly blog!
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## Do My Music Lab
As stated over, I utilized to write a basic and also uncomplicated mathematics lab with only Music Nevertheless, the easier you make your lab, the easier it becomes to get stuck at completion of it, after that at the start. This can be very discouraging, and all this can occur to you since you are making use of Music and/or Modular Equations improperly.
With Modular Formulas, you are already making use of the incorrect equation when you obtain stuck at the start, if not, after that you are likely in a stumbling block, and also there is no feasible way out. This will only become worse as the problem ends up being extra complicated, yet after that there is the inquiry of exactly how to proceed with the trouble. There is no chance to appropriately deal with addressing this sort of math problem without being able to immediately see what is going on.
It is clear that Music and Modular Equations are tough to learn, and it does take method to develop your own feeling of intuition. However when you wish to solve a mathematics trouble, you have to utilize a device, and the tools for finding out are utilized when you are stuck, as well as they are not used when you make the wrong action. This is where lab Assist Solution can be found in.
As an example, what is wrong with the question is incorrect concepts, such as getting a partial worth when you do not have sufficient functioning components to finish the entire work. There is a great reason that this was wrong, and also it refers logic, not intuition. Reasoning allows you to adhere to a detailed treatment that makes good sense, and also when you make an incorrect move, you are typically compelled to either try to go forward as well as fix the error, or attempt to go backward and do a backwards action.
One more instance is when the trainee does not comprehend an action of a process. These are both sensible failings, as well as there is no other way around them. Also when you are embeded an area that does not allow you to make any type of kind of step, such as a triangle, it is still crucial to recognize why you are stuck, to make sure that you can make a much better action and go from the step you are stuck at to the following location.
With this in mind, the most effective means to solve a stuck scenario is to just take the advance, as opposed to attempting to go backward. The two processes are different in their method, however they have some standard resemblances. However, when they are attempted together, you can quickly tell which one is better at addressing the issue, and you can likewise tell which one is a lot more effective.
Let's discuss the first instance, which relates to the Music mathematics lab. This is not also challenging, so allow's initial talk about how to begin. Take the following procedure of connecting a component to a panel to be made use of as a body. This would require 3 measurements, and also would be something you would require to affix as part of the panel.
Now, you would certainly have an additional dimension, but that does not mean that you can just keep that dimension and go from there. When you made your first step, you can quickly forget about the measurement, and after that you would certainly have to go back as well as retrace your steps.
However, rather than keeping in mind the added measurement, you can utilize what is called a "mental faster way" to assist you bear in mind that additional measurement. As you make your very first step, picture on your own taking the measurement and attaching it to the component you intend to connect to, and then see exactly how that makes you feel when you duplicate the procedure.
Visualisation is an extremely powerful method, as well as is something that you need to not miss over. Imagine what it would certainly feel like to in fact attach the part and also be able to go from there, without the measurement.
Currently, let's take a look at the 2nd example. Let's take the exact same procedure as previously, today the pupil has to keep in mind that they are mosting likely to return one step. If you tell them that they need to return one action, however then you remove the concept of having to move back one step, then they will not know how to proceed with the problem, they won't know where to look for that action, as well as the procedure will certainly be a mess.
Instead, make use of a psychological shortcut like the psychological diagram to psychologically reveal them that they are going to move back one step. and also put them in a placement where they can progress from there. without having to consider the missing out on an action.
## Hire Someone To Do Your Music Lab
" Music - Required Aid With a Math lab?" However, several pupils have actually had a problem grasping the ideas of direct Music. Thankfully, there is a new format for straight Music that can be utilized to instruct straight Music to students that battle with this idea. Students can make use of the lab Help Solution to help them learn new methods in direct Music without encountering a hill of issues as well as without needing to take an examination on their concepts.
The lab Aid Service was developed in order to aid battling trainees as they relocate from college as well as secondary school to the university as well as task market. Several trainees are not able to handle the tension of the knowing process and also can have really little success in understanding the ideas of linear Music.
The lab Aid Solution was created by the Educational Testing Service, who uses a range of different online tests that students can take and practice. The Examination Assist Service has assisted numerous trainees boost their scores and can aid you improve your scores also. As trainees relocate from college and senior high school to the university and work market, the TTS will assist make your students' shift easier.
There are a few different ways that you can take advantage of the lab Assist Solution. The main way that trainees make use of the lab Aid Service is via the Answer Managers, which can aid students learn techniques in direct Music, which they can make use of to help them prosper in their programs.
There are a variety of troubles that pupils experience when they initially use the lab Help Service. Students are commonly overloaded and also don't recognize how much time they will need to commit to the Solution. The Answer Supervisors can assist the pupils evaluate their idea learning and also help them to assess all of the product that they have currently learned in order to be planned for their following program work.
The lab Aid Service works the same way that a teacher performs in regards to aiding students grasp the principles of straight Music. By providing your trainees with the devices that they require to discover the essential ideas of straight Music, you can make your pupils extra effective throughout their researches. Actually, the lab Aid Service is so efficient that lots of students have actually switched from typical math class to the lab Aid Solution.
The Job Supervisor is made to help students handle their homework. The Job Supervisor can be established to set up how much time the trainee has available to finish their appointed research. You can also establish a personalized period, which is an excellent feature for trainees who have an active timetable or an extremely busy high school. This attribute can aid trainees stay clear of feeling overwhelmed with mathematics jobs.
An additional beneficial feature of the lab Help Service is the Trainee Aide. The Trainee Assistant aids students manage their job as well as gives them a location to post their research. The Pupil Assistant is useful for students that don't wish to get bewildered with responding to several concerns.
As trainees get more comfy with their jobs, they are motivated to connect with the Task Manager and the Student Aide to get an online support system. The online support system can help pupils preserve their emphasis as they answer their jobs.
Every one of the tasks for the lab Assist Service are included in the package. Trainees can login and also finish their appointed work while having the pupil help available in the background to help them. The lab Assist Solution can be a great help for your students as they start to browse the challenging university admissions and also job hunting waters.
Trainees must be prepared to obtain used to their jobs as swiftly as feasible in order to reach their major goal of entering the university. They need to work hard enough to see outcomes that will allow them to walk on at the following level of their studies. Obtaining used to the process of completing their projects is very essential.
Students are able to discover various means to help them find out how to use the lab Help Service. Knowing how to use the lab Help Solution is vital to trainees' success in university as well as job application.
## Hire Someone To Take My Music Lab
Music is used in a lot of colleges. Some teachers, nevertheless, do not use it really efficiently or utilize it improperly. This can have an unfavorable influence on the pupil's understanding.
So, when appointing assignments, make use of an excellent Music assistance solution to help you with each lab. These services give a selection of handy solutions, consisting of:
Assignments may require a great deal of evaluating and looking on the computer. This is when making use of an assistance service can be a great advantage. It enables you to get even more job done, increase your comprehension, and prevent a lot of tension.
These sorts of research services are a superb way to start working with the most effective type of aid for your requirements. Music is among the most tough subjects to grasp for pupils. Dealing with a service, you can make sure that your requirements are met, you are instructed correctly, and you comprehend the product correctly.
There are many ways that you can show on your own to work well with the course and also be successful. Use a proper Music help solution to lead you and also get the job done. Music is one of the hardest classes to find out however it can be conveniently understood with the right aid.
Having a homework solution likewise assists to improve the pupil's grades. It permits you to include additional credit scores along with increase your GPA. Obtaining additional credit scores is usually a big benefit in lots of colleges.
Pupils that do not make the most of their Music course will end up moving ahead of the remainder of the course. The bright side is that you can do it with a fast as well as easy solution. So, if you want to continue in your course, utilize a great assistance solution. Something to remember is that if you really want to increase your grade degree, your training course job needs to get done. As long as feasible, you need to comprehend as well as work with all your troubles. You can do this with a good aid solution.
One advantage of having a research service is that you can aid on your own. If you don't feel great in your capacity to do so, then a good tutor will be able to aid you. They will be able to fix the troubles you deal with and also aid you comprehend them so as to get a far better quality.
When you finish from senior high school and enter university, you will certainly require to strive in order to remain ahead of the other students. That suggests that you will certainly require to work hard on your research. Using an Music service can assist you get it done.
Maintaining your qualities up can be tough due to the fact that you usually need to study a whole lot as well as take a lot of examinations. You don't have time to work with your grades alone. Having an excellent tutor can be a wonderful aid because they can aid you and also your homework out.
An aid service can make it much easier for you to handle your Music course. Furthermore, you can discover more about on your own and also aid you do well. Discover the most effective tutoring service and also you will certainly have the ability to take your study skills to the next degree.
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latest
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https://www.convertunits.com/from/rebah/to/drachme
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## ››Convert rebah to drachme
rebah drachme
How many rebah in 1 drachme? The answer is 1.3309982486865.
We assume you are converting between rebah and drachme.
You can view more details on each measurement unit:
rebah or drachme
The SI base unit for mass is the kilogram.
1 kilogram is equal to 350.26269702277 rebah, or 263.15789473684 drachme.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between rebah and drachme.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of rebah to drachme
1 rebah to drachme = 0.75132 drachme
5 rebah to drachme = 3.75658 drachme
10 rebah to drachme = 7.51316 drachme
20 rebah to drachme = 15.02632 drachme
30 rebah to drachme = 22.53947 drachme
40 rebah to drachme = 30.05263 drachme
50 rebah to drachme = 37.56579 drachme
75 rebah to drachme = 56.34868 drachme
100 rebah to drachme = 75.13158 drachme
## ››Want other units?
You can do the reverse unit conversion from drachme to rebah, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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CC-MAIN-2021-49
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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition/chapter-r-review-of-basic-concepts-r-6-rational-exponents-r-6-exercises-page-63/20
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## Precalculus (6th Edition)
$\dfrac{5}{t^3}$
RECALL: (1) $a^{-m} = \dfrac{1}{a^m}$ (2) $\dfrac{a^m}{a^n} = a^{m-n}$ (3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$ (4) $(ab)^m = a^mb^m$ Use rule (1) above to obtain: $5t^{-3} = 5\left(\frac{1}{t^2}\right)=\dfrac{5}{t^3}$
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https://www.allvacation.co.uk/2021/08/09-16355/
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power plant boiler thermal efficiency calculation method
## power plant boiler thermal efficiency calculation method
### Methods to Calculate CFB Boiler Efficiency CFBC Boiler
CFB Boiler Coal Fired Boiler Calculation methods of boiler efficiency. In order to calculate boiler efficiency by this method we divide the total energy Boiler Calculations KTH 1 Steam/water diagrams used in boiler calculations Temperature-heat (T-Q) diagram The T-Q diagram is a useful tool for designing heat exchangers.
### how to calculate boiler efficiency in thermal power plant
Thermal . 29 Apr 2020 Boiler efficiency = Heat output/ Heat input Or Boiler efficiency = [Q x (hg- hf)/ q x GCV] x 100 Where Q = quantity of steam generated per hour (Kg/ READ MORE Efficiency of a Coal Fired Boiler in a Typical Thermal Power Plant. 27 Sep 2020
### efficiency of boiler formula power plant
In order to calculate boiler efficiency by this method we divide the total energy output of a boiler by total energy input given to the boiler multiplied by hundred. Calculation of direct efficiency-E= [Q
### Boiler Efficiency: Introduction and Calculation Method ...
This method calculates the efficiency of the boiler by using the basic efficiency formula- = (Energy output) / (Input energy) X 100 To calculate the efficiency of the boiler by this method we divide the total energy output from the boiler by the total energy input given to the boiler multiplied by one hundred.
### Methods for Calculating CHP Efficiency | Combined Heat and ...
Typical boiler efficiencies are 80 percent for natural gas-fired boilers 75 percent for biomass-fired boilers and 83 percent for coal-fired boilers. The calculation of effective electric efficiency is the CHP net electric output divided by the additional fuel the CHP system consumes over and above what would have been used by a to ...
### Methods for Calculating CHP Efficiency | Combined Heat and ...
Typical boiler efficiencies are 80 percent for natural gas-fired boilers 75 percent for biomass-fired boilers and 83 percent for coal-fired boilers. The calculation of effective electric efficiency is the CHP net electric output divided by the additional fuel the CHP system consumes over and above what would have been used by a boiler to produce the thermal output of the CHP system.
### Flue Gas Waste Heat Boiler Thermal Efficiency Calculation
The anti-balance test method is usually used to measure boiler thermal efficiency. The calculation formula for the thermal efficiency of the anti balance method: = 100- (q2+q3+q4+q5+q6) 1. q2-flue gas heat loss The exhaust heat loss is the heat loss caused by a part of the heat that is taken away by the smoke of the boiler.
### What is Boiler Efficiency? Boiler Efficiency Calculation ...
Boiler Efficiency usually expressed in (%) percentage Boiler Efficiency Calculation Formula Boiler efficiency (%) = [Q (H-h)/q*GCV]*100 (heat exported by the fluid (Enthalpy of steam (Kcal/kg) Enthalpy of water (kcal/kg) )/ Gross calorific value of the fuel.) x 100. Steam Boiler Efficiency & Boiler Performance
### (PDF) Boiler Calculations | Michael Dellon - Academia.edu
Calculating boiler efficiency There are two different means of calculating the boiler efficiency: The direct method and the indirect method. 12 Direct method In the direct method the boiler efficiency is directly defined by the exploitable heat output from the boiler and by the fuel power of the boiler: output = (14) input where output is the exploitable heat output from boiler and input the fuel power of the boiler.
### Boiler efficiency calculation direct method excel air ...
Direct method is quick method for boiler efficiency calculation In this method Percentage of total heat output vs heat input is calculated to establish boiler performance. Boiler Efficiency = Heat output / Heat Input X 100. Or Boiler Efficiency = Q X (h2 - h1) / (q X GCV) X 100 Where Q = Qty of steam generated.
### how to calculate boiler efficiency in thermal power plant
Thermal . 29 Apr 2020 Boiler efficiency = Heat output/ Heat input Or Boiler efficiency = [Q x (hg- hf)/ q x GCV] x 100 Where Q = quantity of steam generated per hour (Kg/ READ MORE Efficiency of a Coal Fired Boiler in a Typical Thermal Power Plant. 27 Sep 2020
### Efficiency Of Steam Boiler Systems | Steam Boilers In India
To calculate thermal efficiency of a steam boiler the quantity of steam produced and fuel consumption is measured at hourly intervals. Boiler efficiency = Heat output/heat input x 100 Boiler efficiency = heat in steam output/ heat fuel input x 100 = Quantity of Steam x
### Basic Calculations For A Power Plant | Boiler | Power Station
Oct 15 2009 · Thermal power plants are the biggest producers of Carbon Dioxide. The boiler for typical 500 MW units produces around 1600 tons per hour of steam at a temperature of 540 to 600 degrees Centigrade. ... Direct and Indirect for -ZBG . Uploaded by. Stuart ashraf. LISTA DE PRECIOS PEGSA-RCG 2016 Rev 3 ...
### Boiler Efficiency: Introduction and Calculation Method ...
This method calculates the efficiency of the boiler by using the basic efficiency formula- = (Energy output) / (Input energy) X 100 To calculate the efficiency of the boiler by this method we divide the total energy output from the boiler by the total energy input given to the boiler multiplied by one hundred.
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https://www.onlinecalculatorsfree.com/financial/401k-calculator.html
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# 401(k) Calculator
Plan and estimate your retirement savings with this powerful tool. Input your current 401(k) balance, contribution amount, employer match, and other relevant details to see how your savings can grow over time. This calculator considers factors like investment returns and inflation rates to provide accurate projections.
401k Calculator
Related
## What is a 401k?
There was a change in regulation in 1978 that affects retirement plans. The United States Congress passed a part of the Internal Revenue Code known as section 401(k) and the name 401K plans has stuck ever since.
401K plans have tax benefits and are classified as Defined Contribution plans. This means that the amount both you and your employer put into the plan is set at a certain amount. You are able to contribute a set amount of your monthly salary every month. Your employer can limit the amount that this is. If you want the most from your retirement then you should try to put in as much as you can every month but be aware that there are annual contribution limits defined by the IRS. This amount is usually increased in line with inflation on an annual basis.
The limits for 2020 and 2021 set by the IRS are \$19,500 for a 401K plan. The contribution limit for 2023 is \$22,500 (\$20,500 for 2022). You may find that your employer matches or makes part of your contributions, as many do. The contribution from the employer does not have a limit except for the fact that it must not be above the lesser of either:
• 100% of the employee's pay,
• \$58,000 in 2021, \$61,000 in 2022 or \$66,000 in 2023.
The sums of money put into the plan by both the employee and the employer are placed there free of tax and there is no tax burden while the money is in the account. So it has the effect of lowering the amount of tax paid on earnings as they pass into the 401K account before tax is deducted. It is a good way to save for retirement as you cannot usually touch this money without penalty until you retire or become disabled or your dependents can gain access if you die.
## 401(k) Formula
The 401(k) Retirement Calculator uses the following basic formula:
Total Account Value(TAV) = CB * (1 + rate/100/12)^(n*12) + (YC + EC) * ((1 + rate/100/12)^(n*12) - 1) / (rate/100/12) * (1 + rate/100/12)
• TAV: Total Account Value
• CB: Current 401k Balance
• rate: Annual Rate of Return
• n: Age of Retirement − Current Age
• YC: Your Monthly 401(k) Contribution
• EC: Your Employer's Monthly 401(k) Adding
## Calculator Definitions
### Annual salary (\$)
The sum of your annual pay before any deductions such as tax .
### Annual Salary Increase (%)
How much you expect your salary to increase by on an annual basis in percentage terms.
### Annual Rate of Return (%)
This is how you expect your investments to perform. Estimate the way that you expect your 401K investments to grow on an annual basis in percentage terms. The calculator assumes that you deposit every month and that your return is compounded.
### Current Age (years)
How old you are now.
### Age of Retirement (years)
When you expect to retire.
### Current 401k Balance (\$)
What you already have saved in a 401K plan. If you have not started to save in a 401K plan then enter zero in this field.
### Contribution to 401k (%)
What proportion of your salary you will be saving into your 401K, stated as a percentage of your salary.
### Employer Match (%)
This is how much your employer will match your contributions. Mark this as the percentage that your employer will match what you pay in.
For example if you pay \$250 and your employer will match this with \$125 then enter 50% here.
### Employer Max Contribution (%)
State the maximum contribution that your employer will go to. It may state in your terms and conditions that they will give a "50% match up to 8% of your salary".
This simply means that they will go no further than half of what you pay in up to 8% of your salary , so in effect they will contribute a maximum of 4% of your annual salary.
## Contribution Limits for 401(k) Plans from 1987 to 2023
Year Employee Contribution Limit Total Contribution Limit
Catch Up Contribution
(Age > 50)
2023
\$22,500
\$66,000
\$7,500
2022
\$20,500
\$61,000
\$6,500
2021
\$19,500
\$58,000
\$6,500
2020
\$19,500
\$57,000
\$6,500
2019
\$19,000
\$56,000
\$6,000
2018
\$18,500
\$55,000
\$6,000
2017
\$18,000
\$54,000
\$6,000
2016
\$18,000
\$53,000
\$6,000
2015
\$18,000
\$53,000
\$6,000
2014
\$17,500
\$52,000
\$5,500
2013
\$17,500
\$51,000
\$5,500
2012
\$17,000
\$50,000
\$5,500
2011
\$16,500
\$49,000
\$5,500
2010
\$16,500
\$49,000
\$5,500
2009
\$16,500
\$49,000
\$5,500
2008
\$15,500
\$46,000
\$5,000
2007
\$15,500
\$45,000
\$5,000
2006
\$15,000
\$44,000
\$5,000
2005
\$14,000
\$42,000
\$4,000
2004
\$13,000
\$41,000
\$3,000
2003
\$12,000
\$40,000
\$2,000
2002
\$11,000
\$40,000
\$1,000
2001
\$10,500
\$35,000
-
2000
\$10,500
\$30,000
-
1999
\$10,000
\$30,000
-
1998
\$10,000
\$30,000
-
1997
\$9,500
\$30,000
-
1996
\$9,500
\$30,000
-
1995
\$9,240
\$30,000
-
1994
\$9,240
\$30,000
-
1993
\$8,994
\$30,000
-
1992
\$8,728
\$30,000
-
1991
\$8,475
\$30,000
-
1990
\$7,979
\$30,000
-
1989
\$7,627
\$30,000
-
1988
\$7,313
\$30,000
-
1987
\$7,000
\$30,000
-
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https://cris.openu.ac.il/en/publications/packing-directed-cycles-efficiently-2
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# Packing directed cycles efficiently
Zeev Nutov, Raphael Yuster
Research output: Contribution to journalArticlepeer-review
## Abstract
Let G be a simple digraph. The dicycle packing number of G, denoted νc (G), is the maximum size of a set of arc-disjoint directed cycles in G. Let G be a digraph with a nonnegative arc-weight function w. A function ψ from the set C of directed cycles in G to R+ is a fractional dicycle packing of G if ∑e ∈ C ∈ C ψ (C) ≤ w (e) for each e ∈ E (G). The fractional dicycle packing number, denoted νc * (G, w), is the maximum value of ∑C ∈ C ψ (C) taken over all fractional dicycle packings ψ. In case w ≡ 1 we denote the latter parameter by νc* (G). Our main result is that νc* (G) - νc (G) = o (n2) where n = | V (G) |. Our proof is algorithmic and generates a set of arc-disjoint directed cycles whose size is at least νc (G) - o (n2) in randomized polynomial time. Since computing νc (G) is an NP-Hard problem, and since almost all digraphs have νc (G) = Θ (n2) our result is a FPTAS for computing νc (G) for almost all digraphs. The result uses as its main lemma a much more general result. Let F be any fixed family of oriented graphs. For an oriented graph G, let νF (G) denote the maximum number of arc-disjoint copies of elements of F that can be found in G, and let νF* (G) denote the fractional relaxation. Then, νF* (G) - νF (G) = o (n2). This lemma uses the recently discovered directed regularity lemma as its main tool. It is well known that νc* (G, w) can be computed in polynomial time by considering the dual problem. We present a polynomial algorithm that finds an optimal fractional dicycle packing. Our algorithm consists of a solution to a simple linear program and some minor modifications, and avoids using the ellipsoid method. In fact, the algorithm shows that a maximum fractional dicycle packing with at most O (n2) dicycles receiving nonzero weight can be found in polynomial time.
Original language English 82-91 10 Discrete Applied Mathematics 155 2 https://doi.org/10.1016/j.dam.2006.04.033 Published - 15 Jan 2007
• Cycles
• Digraph
• Packing
## Fingerprint
Dive into the research topics of 'Packing directed cycles efficiently'. Together they form a unique fingerprint.
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RelativeResourceManager(4)
# RelativeResourceManager(4) - Digital and analog signals...
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Digital and analog signals, introduction to optical spectroscopy, and light sources Chem 420 September 08, 2008
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Update on problem set z For Problem 1 you must show work. Do not simply use spreadsheet or calculator to obtain mean and standard deviation. z Problem 2b—What is the standard error of the estimate? z Just give regression errors from Excel
Digital and analog signals z Analog signals z Magnitude of voltage, current, charge, or power z Continuous in both amplitude and time z Digital information z Data encoded in only discrete levels
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Analog to digital to conversion z Limited by bit resolution of ADC z 4-bit card has 2 4 = 16 discrete binary levels z 8-bit card has 2 8 = 256 discrete binary levels z 10-bit card has 2 10 = 1024 discrete binary levels z Maximum resolution comes from full use of ADC voltage range. z Trade-offs z More bits is usually slower z More expensive K.A. Rubinson, J.F. Rubinson, Contemporary Instrumental Analysis , 2000 .
Serial and parallel binary encoding (serial)
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Example of binary encoding z Write the number 2008 in binary. z How many bits are needed? 2 11 2 10 2 9 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0 2048 1024 512 256 128 64 32 16 8 4 2 1 1 1111011000 10 bits are needed 11111011000 : 1024 + 512 + 256 + 128 + 64 + 16 + 8 = 2008
Questions?
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Introduction to Optical Spectroscopy (Chapters 6-7) Chemistry 420
Fundamentals of electromagnetic radiation 34 Planck's constant 6.626 10 J s =frequency in Hz Eh h ν = == 8 m speed of light 3.00 10 s =wavelength c c λ =
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Electromagnetic spectrum http://www.yorku.ca/eye/spectrum.gif
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RelativeResourceManager(4) - Digital and analog signals...
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Other Solved Mathematics Board Papers
MATHEMATICS (ICSE – Class X Board Paper 2015)
Two and Half HourAnswers to this Paper must be written on the paper provided separately. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.
The time given at the head of this Paper is the time allowed for writing the answers. Attempt all questions form Section A and any four questions from Section BAll working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the Answer. Omission of essential working will result in the loss of marks.
The intended marks for questions or parts of questions are given in brackets [ ].
Mathematical tables are provided.
SECTION A [40 Marks]
(Answer all questions from this Section.)
Question 1
(a) A shopkeeper bought an article for Rs. 3450. He marks the price of the article 16% above the cost price. The rate of sales tax charged on the article is 10%. Find the:
(i) marked price of the article
(ii) price paid by a customer who buys the article. [3]
(b) Solve the following in equation and write the solution set:
$\displaystyle 13x-5<15x+4<7x+12,\ x\in R\$
Represent the solution on a real number line. [3]
(c) Without using trigonometric tables evaluate:-
$\displaystyle \frac{{\mathrm{sin } 65{}^\circ }}{{\mathrm{cos\ } 25{}^\circ }}+\frac{{\mathrm{cos\ } 32{}^\circ }}{{\mathrm{sin } 58{}^\circ }} - {\mathrm{sin } 28{}^\circ }.\ {\mathrm{sec\ } 62{}^\circ }+{\mathrm{cosec}^2 30{}^\circ } \hspace{5.5cm} [3]$
(a)
(i) Cost price of the article $\displaystyle = \text{ Rs. }3450$
$\displaystyle \text{Marked Price } =3450 + \frac{16}{100}\times 3450 = \text{ Rs. } 4002$
$\displaystyle \text{(ii) Price paid by the customer } = \text{ Rs. } 4002+\frac{10}{100}\times 4002 = \text{ Rs. } 4402.20$
(b) Given
$\displaystyle 13x-5<15x+4<7x+12$
$\displaystyle \Rightarrow 15x+4>13x-5 \text{ and } 15x+4<7x+12$
$\displaystyle \Rightarrow 15x-13x>-5-4 \text{ and } 15x-7x<12-4$
$\displaystyle \Rightarrow 2x>-9 \text{ and } 8x<8$
$\displaystyle \Rightarrow x>\frac{-9}{2} \text{ and } x<1$
$\displaystyle \text{Solution: } \{ x: \frac{9}{2} < x< 1 , x \in R \}$
$\displaystyle \text{(c)} \frac{{\mathrm{sin\ } 65{}^\circ }}{{\mathrm{cos\ } 25{}^\circ }}+\frac{{\mathrm{cos} 32{}^\circ }}{{\mathrm{sin\ } 58{}^\circ }} -{\mathrm{sin} 28{}^\circ }.\ {\mathrm{sec } 62{}^\circ }+{\mathrm{cosec}^ 2 30{}^\circ }$
$\displaystyle = \frac{{\mathrm{cos} (90{}^\circ -65{}^\circ )\ }}{{\mathrm{cos\ } 25{}^\circ }}+\frac{{\mathrm{sin\ (} 90{}^\circ -32{}^\circ )\ }}{{\mathrm{sin\ } 58{}^\circ }} -{\mathrm{cos} (90{}^\circ -28{}^\circ )\times \frac{1}{{\mathrm{cos\ } 62{}^\circ }}+{(2)}^2\ }$
$\displaystyle = \frac{{\mathrm{cos} 25{}^\circ }}{{\mathrm{cos} 25{}^\circ }}+\frac{{\mathrm{sin\ } 58{}^\circ }}{{\mathrm{sin} 58{}^\circ }} -{\mathrm{cos\ } 62{}^\circ }\times \frac{1}{cos\ 62{}^\circ }+4$
$\displaystyle = 1+1-1+4 = 5$
$\displaystyle \\$
Question 2
$\displaystyle \text{(a) If } A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} 9 & 16 \\ 0 & -7 \end{bmatrix} , \text{ find } x \text{ and } y \text{ when } A^2=B . \hspace{3.0cm} [3]$
(b) The present population of a town is $\displaystyle 2,00,000 .$ It population increases by $\displaystyle 10\%$ in the first year and $\displaystyle 15\%$ in the second year. Find the population of the town at the end of the two years. [3]
(c) Three verticals of a parallelogram $\displaystyle ABCD$ taken in order are $\displaystyle A(3,6), B(5,10) \text{ and } C(3,2)$ find:
(i) the coordinates of the fourth vertex $\displaystyle D .$
(ii) length of diagonal $\displaystyle BD$
(iii) equation of side $\displaystyle AB$ of the parallelogram $\displaystyle ABCD .$ [4]
$\displaystyle \text{(a) } A = \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} 9 & 16 \\ 0 & -7 \end{bmatrix}$
$\displaystyle A^2=B$
$\displaystyle \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 3 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$
$\displaystyle \begin{bmatrix} 3 \times 3 + x \times 0 & 3 \times x + x \times 1 \\ 0 \times 3 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$
$\displaystyle \begin{bmatrix} 9 & 4x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & -y \end{bmatrix}$
Therefore $\displaystyle 4x = 16 \Rightarrow x = 4$
And $\displaystyle 1 = -y \Rightarrow y = -1$
$\displaystyle \text{(b) Given: } P =2,00,000$
$\displaystyle \text{Population after first year } = 200000\ \Big[1+\frac{{10}}{100} \Big] ^1 = 200000 \left[\frac{110}{100}\right]=220000$
Principle for the second year $\displaystyle = 220000$
$\displaystyle \text{Population at the end of second year } = 220000 \left[1+\frac{15}{100}\right]$
$\displaystyle = 220000\times \frac{115}{100} = 253000$
(c) Let the coordinate of $\displaystyle D \text{ be } (x,y) .$ In a parallelogram, the diagonals bisect each other.
$\displaystyle \text{Mid point of } AC = \Big[\frac{3+3}{2},\ \frac{6+2}{2} \Big] = (3,4)$
$\displaystyle \text{Mid point of } BD = \Big[\frac{x+5\ }{2},\ \frac{y+10}{2} \Big]$
$\displaystyle \text{(i) } 3=\frac{x+5}{2} \Rightarrow x = 1$
$\displaystyle 4=\ \frac{y+10}{2} \Rightarrow y = -2$
Coordinates of $\displaystyle D (1, -2)$
(ii) $\displaystyle BD = \sqrt{{(5-1}^2)+(10+{2)}^2}$
$\displaystyle = \sqrt{16+144} = 4 \sqrt{10} \text{ units }$
(iii) Equation of AB:
$\displaystyle y-y_1 =\ \frac{y_2-y_1}{x_2-\ x_1} \left(x-x_1\right)$
$\displaystyle y-6= \frac{10-6}{5-3} \left(x-3\right)$
$\displaystyle y-6=\ \frac{4}{2}\left(x-3\right)$
$\displaystyle y-6=2x-6$
$\displaystyle 2x-y=0$
$\displaystyle \\$
Question 3
(a) In the given figure, ABCD is a square of side 21 cm, AC and BD are two diagonals of the square. Two semi circle are drawn with AD and BC as diameters. Find the area of the shaded region. $\displaystyle \text{(Take } \pi =\frac{22}{7} ) \hspace{3.0cm} [3]$
(b) The marks obtained by 30 students in a class assessment of 5 marks is given below:
Marks 0 1 2 3 4 5 No. of Students 1 3 6 10 5 5
Calculate the mean, mediam and mode of the above distribution. [3]
(c) In the figure given below O is the center of the circle and SP is a tangent. If $\displaystyle \angle SRT =65{}^\circ , \text{ find the value of } x,\ y \text{ and } z$ [4]
(a) Given: Side $\displaystyle =21 \text{ cm}$
$\displaystyle \text{Let Diagonal of the square } = \sqrt{2} \times \text{ side }$
$\displaystyle \therefore AC=BD=21\sqrt{2}$
$\displaystyle \therefore AO=OC=BO=OD=\ \frac{21\sqrt{2}}{2}$
$\displaystyle \text{Area of } \Delta AOD= \text{ Area of } \Delta BOC=\ \frac{1}{2}\times \frac{21\sqrt{2}}{2}\times \frac{21\sqrt{2}}{2}=\frac{441}{4} \text{ cm}^2$
$\displaystyle \text{Area of semicircle } = \frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times \left(\frac{{21}^2}{2}\right)=\frac{693}{4} \text{ cm}^2$
Area of shaded region = Area of 2 semicircles + Area of $\displaystyle \triangle AOD$ + $\displaystyle \triangle BOC$
$\displaystyle = 2\ \times \frac{693}{4}+\frac{441}{4}+\frac{441}{4}+\frac{2268}{4} = 567 \text{ cm}^2$
(b) Below
$\displaystyle x$$\displaystyle x$ $\displaystyle f$$\displaystyle f$ $\displaystyle fx$$\displaystyle fx$ $\displaystyle cf$$\displaystyle cf$ 0 1 0 1 1 3 3 4 2 6 12 10 3 10 30 20 4 5 20 25 5 5 25 30 $\displaystyle \Sigma f=30$$\displaystyle \Sigma f=30$ $\displaystyle \Sigma fx=90$$\displaystyle \Sigma fx=90$
$\displaystyle \text{Mean } = \frac{\Sigma fx}{\Sigma f}=\frac{90}{30} = 3$
$\displaystyle \text{Median} = \text{size of } \left(\frac{N}{2}\right)th\ {obs}^n$
$\displaystyle = \text{size of } \left(\frac{30}{2}\right)th\ {obs}^n$
$\displaystyle = \text{Size of } 15 {}^{th} obs{}^{n} = 3$
Mode = 3 marks (as highest frequency is 10)
(c) In $\displaystyle \Delta OSP,\ \angle OSR=90{}^\circ$ (Radius is always perpendicular to the tangent)
$\displaystyle In\ \Delta TSR$
$\displaystyle x+90{}^\circ +65{}^\circ =180{}^\circ$
$\displaystyle x=25{}^\circ$
$\displaystyle \angle SOQ=2\ \angle STR\ \left[Angle\ at\ centre=2\ \times angle\ at\ circumference \right]$
$\displaystyle y=2\times 25=50{}^\circ$
In $\displaystyle \Delta OSP,\ 50{}^\circ +90{}^\circ +z=180{}^\circ$
$\displaystyle z=40{}^\circ$
$\displaystyle \\$
Question 4
(a) Katrina opened a recurring deposit account with a Nationalized Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly installment is Rs. 1000 find the: [3]
(i) interest earned in 2 years
(ii) matured value
(b) Find the value of $\displaystyle k$ for which $\displaystyle x=3$ is a solution of the quadratic equation, $\displaystyle \left(k+2\right)x^2-kx+6=0 .$ Thus find the root of the equation. [3]
(c) Construct a regular hexogon of side 5 cm. Construct a circle circumscribing the hexagon. All traces of construction must be clearly shown. [4]
(a)
(i) Given $\displaystyle P= \text{ Rs. }1000,\ n=2\ \text{years} =\ 24\ \text{months},\ r=6\%.$
$\displaystyle I=P\times \frac{n(n+1)}{2}\times \frac{r}{12\times 100}$
$\displaystyle = 1000\times \frac{24\ \left(24+1\right)}{2}\times \frac{6}{12\times 100}$
$\displaystyle = 1000\times \frac{24\times 25}{2}\times \frac{6}{12\times 100}$
$\displaystyle =1500$
$\displaystyle \text{(ii) Total money deposit in 24 months = } \text{Rs. }\ 1000\times 24 = \text{ Rs. } 24000$
$\displaystyle \text{Matured Value = Total sum deposit + Interest} =24000+1500 = \text{ Rs. } 25500$
(b) $\displaystyle \left(k+2\right)x^2-kx+6=0$
Putting $\displaystyle x=3$ in given equation:
$\displaystyle \left(k+2\right)\times 9-k\times 3+6=0$
$\displaystyle 9k+18-3k+6=0$
$\displaystyle 6k=-24 \Rightarrow k = -4$
Putting $\displaystyle k= -4$ in given equation:
$\displaystyle -2x^2+4x+6=0$
$\displaystyle x^2-2x-3=0$
$\displaystyle x^2-3x+x-3=0$
$\displaystyle x\left(x-3\right)+1\left(x-3\right)=0$
$\displaystyle \left(x+1\right)\left(x-3\right)=0$
$\displaystyle x+1=0\ or\ x-3=0$
$\displaystyle x=-1\ or\ x=3$
$\displaystyle x=-1\$ is the other root of the given equation.
(c) Steps of Construction:
1. First draw a regular hexagon. The length of one side is 5 cm. You will get hexagon ABCDEF.
2. Take any two adjacent sides and draw perpendicular bisectors.
3. The point where these two bisectors intersect, is the center of the circle.
4. With O as the center, draw a circle which will pass through all the vertices of the hexagon.
SECTION B [40 Marks]
(Answer any four questions in this Section.)
Question 5
(a) Use a graph paper for this question taking $\displaystyle 1 cm = 1$ unit along both the $\displaystyle x \text{ and } y axis$ :
(i) Plot the points $\displaystyle A(0, 5), B(2, 5), C(5, 2), D(5, -2), E(2, -5) \text{ and } F(0, -5)$
(ii) Reflect the points $\displaystyle B, C, D \text{ and } E$ on the $\displaystyle y-axis$ and name them respectively as $\displaystyle B', C', D',U' \text{ and } E' .$
(iv) Name the figure formed by $\displaystyle B C D E E'D'C'B' .$
(v) Name a line of symmetry for, the figure formed. [5]
(b) Virat opened a Saving Bank account in a bank on 16th April 2010. His pass book shows the following entries:
Date Particulars Withdrawals (Rs.) Deposits (Rs.) Balance (Rs.) April 16, 2010 By Cash – 2500 2500 April 28th By Cheque – 3000 5500 May 9th To Cheque 850 – 4650 May 15th By Cash – 1600 6250 May 24th To Cash 1000 – 5250 June 4th To Cash 500 – 4750 June 30th By Cheque – 2400 7150 July 3rd To Cash – 1800 8950
Calculate the interest Virat earned at the end, of 31st July, 2010 at 4% per annum interest. What sum of money will he receives if he closes the account on 1st August, 2010? [5]
(a)
(i) Shown in the diagram.
(ii) Shown in then diagram.
(iii) Coordinates : $\displaystyle B', (-2, 5), C', (-5 ,2), D', (-5 , -2),E', (-2, -5)$
(iv) Octagon
(v) $\displaystyle x-axis \text{ or } y-axis.$
(b) Qualifying principal for various months:
Month Principal (Rs.) April 0 May 4650 June 4750 July 8950 Total 18350
$\displaystyle P = \text{ Rs. } 18350 R = 4\% \text{ and } T= \frac{1}{12}$
$\displaystyle I = P \times R \times T = 18350 \times \frac{4}{100} \times \frac{1}{12} = \text{ Rs. } 61.16$
Amount $\displaystyle = 8950+61.61= \text{ Rs. } 9011.16$
$\displaystyle \\$
Question 6
(a) If a, b, c are in continued proportion, prove that $\displaystyle (a + b + c) (a - b + c) = a^2 + b^2 + c^2$ [3]
(b) In the given figure $\displaystyle ABC$ is a triangle and $\displaystyle BC$ , is parallel to the $\displaystyle y-axis .$ $\displaystyle AB \text{ and } AC$ intersects the $\displaystyle y-axis$ at $\displaystyle P \text{ and } Q$ respectively.
(i) Write the coordinates of $\displaystyle A .$
(ii) Find, the length of $\displaystyle AB \text{ and } AC .$
(iii) Find the ratio in which $\displaystyle Q$ divides $\displaystyle AC .$
(iv) Find the equation, of the line $\displaystyle AC$ [4]
(c) Calculate the mean of the following distribution. [3]
Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 5 12 35 24 16
(a) To prove: $\displaystyle (a + b + c) (a - b + c) = a^2 + b^2 + c^2$
Given $\displaystyle a, b \text{ and } c$ are in continued proportion
$\displaystyle \text{Therefore } \frac{a}{b} = \frac{b}{c} = k$
$\displaystyle \Rightarrow a = bk \text{ and } b = ck$
This also implies $\displaystyle a = (ck)k = ck^2$
LHS $\displaystyle = (a + b + c) (a - b + c)$
$\displaystyle = (ck^2+ck+c)(ck^2-ck+c)$
$\displaystyle = c^2(k^2+k+1)(k^2-k+1)$
$\displaystyle = c^2 \{ (k^2+1)+k \} \{ (k^2+1)-k \}$
$\displaystyle =c^2 \{ (k^2+1)^2-k^2 \}$
$\displaystyle = c^2 \{ k^4+1+2k^2-k^2 \}$
$\displaystyle = c^2 \{ k^4+k^1+1 \}$
RHS $\displaystyle = a^2 + b^2 + c^2$
$\displaystyle = (ck^2)^2+(ck)^2+c^2$
$\displaystyle = c^2k^4+c^2k^2+c^2$
$\displaystyle =c^2(k^4+k^2+1)$
Hence LHS = RHS
Therefore Proved
(b)
(i) Coordinates of $\displaystyle A (4, 0)$
(ii) Coordinates of $\displaystyle A (4, 0) \text{ and } B (-2, 3)$
Note: The distance between any two points $\displaystyle (x_1, y_1) \text{ and } (x_2, y_2)$ is $\displaystyle = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Length $\displaystyle AB = \sqrt{(-2-4)^2+(3-0)^2} = \sqrt{36+9} = \sqrt{45} = 3 \sqrt{5} \text{ units }$
Coordinates of $\displaystyle A (4, 0) \text{ and } C (-2, -4)$
Length $\displaystyle AC = \sqrt{(-2-4)^2+(-4-0)^2} = \sqrt{36+16} = \sqrt{52} = 2 \sqrt{13} \text{ units }$
(iii) Let the required ratio be $\displaystyle k:1$
Let the coordinate of $\displaystyle P \text{ be } (0, y)$ , since it lies on the y axis.
$\displaystyle \text{Since } x = \frac{kx_2+x_1}{k+1}$
$\displaystyle \Rightarrow 0 = \frac{k \times (-2) +4}{k+1}$
$\displaystyle \Rightarrow -2k+4=0$
$\displaystyle \Rightarrow k = 2$
$\displaystyle \Rightarrow m_1:m_2 = 2:1$
(iv) Equation of $\displaystyle AC$ through Coordinates of $\displaystyle A(4, 0) \text{ and } C(-2, -4)$
Required equation of the line: $\displaystyle (y-y_1)=m(x-x_1)$
$\displaystyle y-0= (\frac{0+4}{4+2}) (x-4)$
$\displaystyle \Rightarrow 3y=2x-8$
$\displaystyle \Rightarrow or 3y-2x+8=0$
(c)
Class Mid Value $\displaystyle x$$\displaystyle x$ $\displaystyle f$$\displaystyle f$ $\displaystyle fx$$\displaystyle fx$ 0-10 5 8 40 10-20 15 5 75 20-30 25 12 300 30-40 35 35 1225 40-50 45 24 1080 50-60 55 16 880 Total $\displaystyle \Sigma f=100$$\displaystyle \Sigma f=100$ $\displaystyle \Sigma fx=3600$$\displaystyle \Sigma fx=3600$
$\displaystyle \text{Mean } = \bar{x} = \frac{\Sigma fx}{ \Sigma f} = \frac{3600}{100} = 36$
$\displaystyle \\$
Question 7
(a) The solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed. [3]
(b) Find ‘a’ if the two polynomials $\displaystyle {ax}^3+{3x}^2-9 \text{ and } {2x}^3+4x+a$ , leaves the same remainder when divided by $\displaystyle (x+3) .$ [3]
$\displaystyle \text{(c) Prove that } \frac{{\mathrm{sin } \theta }}{1-{\mathrm{cot } \theta }}+\frac{{\mathrm{cos } \theta }}{1-{\mathrm{tan } \theta }} ={\mathrm{cos } \theta }+{\mathrm{sin } \theta } \hspace{6.0cm} [4]$
(a) Volume of two spheres = Volume of cone
$\displaystyle \frac{4}{3}\pi \left(2^3\right)+\frac{4}{3}\pi \left(4^3\right)=\ \frac{1}{3}{\pi r}^2h$
$\displaystyle 32+256=r^2\times 8$
$\displaystyle r^2 = \frac{288}{8}$
$\displaystyle r^2 = 36 \Rightarrow r = 6 cm$
(b) Let $\displaystyle f(x) = {ax}^3+{3x}^2-9 \text{ and } g(x) = {2x}^3+4x+a$
Putting $\displaystyle x = -3$ in both the expressions
$\displaystyle f(-3) = {a(-3)}^3+{3(-3)}^2-9$
$\displaystyle g(-3) ={2(-3)}^3+4(-3)+a$
Given $\displaystyle f(-3) = g(-3)$
Therefore $\displaystyle {a(-3)}^3+{3(-3)}^2-9 = {2(-3)}^3+4(-3)+a$
$\displaystyle -27a+27 -9 = -54-12+a$
$\displaystyle 27a+a = 27-9+54+12$
$\displaystyle 28a=84 \Rightarrow a = 3$
$\displaystyle \text{(c) Given } \frac{{\mathrm{sin } \theta }}{1-{\mathrm{cot } \theta }}+\frac{{\mathrm{cos } \theta }}{1-{\mathrm{tan } \theta }} ={\mathrm{cos } \theta }+{\mathrm{sin } \theta }$
$\displaystyle \text{LHS } =\frac{{\mathrm{sin } \theta }}{1-\frac{\mathrm{cos } \theta }{\mathrm{sin } \theta }}+\frac{{\mathrm{cos } \theta }}{1-\frac{\mathrm{sin } \theta }{\mathrm{cos } \theta }}$
$\displaystyle =\frac{{\mathrm{sin}^2 \theta }}{{\mathrm{sin } \theta } - {\mathrm{cos } \theta }} + \frac{{\mathrm{cos}^2 \theta }}{{\mathrm{cos } \theta } - {\mathrm{sin } \theta }}$
$\displaystyle =\frac{{\mathrm{sin}^2 \theta - \mathrm{cos}^2 \theta }}{{\mathrm{sin } \theta } - {\mathrm{cos } \theta }}$
$\displaystyle =\frac{{(\mathrm{sin } \theta - \mathrm{cos } \theta )(\mathrm{sin } \theta + \mathrm{cos } \theta )}}{{\mathrm{sin } \theta } - {\mathrm{cos } \theta }}$
$\displaystyle ={\mathrm{cos } \theta }+{\mathrm{sin } \theta } =$ RHS
$\displaystyle \\$
Question 8
(a) $\displaystyle AB \text{ and } CD$ are two chords of a circle intersecting at $\displaystyle P .$ Prove that $\displaystyle AP \times PB =CP \times PD .$ [3]
(b) A bag contain 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag, Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball
(iii) is neither a green ball nor a white ball. [3]
(c) Rohit invested Rs. 9600 on Rs. 100 shares at Rs. 20 premium paying 8% dividend. Rohit sold the shares when the price rose to Rs. 160. He invested the proceeds (excluding dividend) in 10% Rs. 50 shares at Rs. 40. Find the:
(i) original Number of shares
(ii) sale proceeds
(iii) new number of shares
(iv) change in the two dividends. [4]
(a) To prove: $\displaystyle AP\times PB=CP\times PD$
Construction : Join AD and CB.
$\displaystyle \angle A = \angle C$ (Angles of the same segment)
$\displaystyle \angle C = \angle B$ (Angles of the same segment)
Therefore $\displaystyle \Delta APD \sim \Delta BPC$ (AAA Postulate)
$\displaystyle \therefore \frac{AP}{CP}=\frac{PD}{PB}$
$\displaystyle \therefore AP \times PB=CP \times PD$
Hence Proved
(b) Number of White Balls: $\displaystyle 5$
Number of Red Balls: $\displaystyle 6$
Number of Green Balls : $\displaystyle 9$
Total number of Balls: $\displaystyle 5+6+9 = 20$
$\displaystyle \text{(i) Probability (Green) } = \frac{\text{Favorable Cases}}{\text{Total number of possible cases}} = \frac{9}{20}$
$\displaystyle \text{(ii) Probability (Red or White) } = \frac{5+6}{20} = \frac{11}{20}$
$\displaystyle \text{(iii) Probability (neither a green ball nor a white ball) } = \frac{6}{20} = \frac{3}{10}$
(c) First Investment
Nominal Value of the share $\displaystyle = 100 \text{ Rs. }$
Market Value of the share $\displaystyle = 120 \text{ Rs. }$
Dividend $\displaystyle = 8 \%$
$\displaystyle \text{(i) Number of shares bought } = \frac{9600}{120} = 80$
(ii) Sale Price $\displaystyle = 160 \text{ Rs. }$
Sale Proceed $\displaystyle = 80 \times 160 = 12800 \text{ Rs. }$
Second Investment
Nominal Value of the share $\displaystyle = 50 \text{ Rs. }$
Market Value of the share $\displaystyle = 40 \text{ Rs. }$
Dividend $\displaystyle = 10\%$
$\displaystyle \text{(iii) Number of shares bought }= \frac{12800}{40} = 320$
$\displaystyle \text{Dividend on 1st Investment earned } = 80 \times \frac{8}{100} \times 100 = 640 \text{ Rs. }$
$\displaystyle \text{Dividend on 2nd Investment earned } = 320 \times \frac{10}{100} \times 50 = 1600 \text{ Rs. }$
$\displaystyle \text{Change in Dividend income } = 1600-640 = 960 \text{ Rs. }$
$\displaystyle \\$
Question 9
(a) The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the second tower is $\displaystyle 30^{\circ} \text{ and } 24^{\circ}$ respectively.
Find the height of the two towers. Give your answer correct to 3 significant figures. [4]
(b) The weight of 50 workers is given below:
Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120 No. of workers 4 7 11 14 6 5 3
Draw on give of the given distribution using a graph sheet. Take 2 cm =10 kg on one axis and 2 cm =5 workers along the other axis. Use a graph to estimate the following:
(i) the upper and lower quartiles
(ii) if weight 95 kg and above is considering find the number of workers who are overweight. [6]
(a) Let AB and CD are two towers.
$\displaystyle BD=120 \text{ m}=EC,\ \angle ACE=30{}^\circ ,\ \angle CBD=24{}^\circ$
In the right $\displaystyle \Delta CBD$
$\displaystyle {\tan 24{}^\circ }=\ \frac{CD}{BD}=\frac{CD}{120}$
$\displaystyle CD=120\times 0.4452 =53.42 \text{ m}$
In right angle $\displaystyle \Delta ACE$
$\displaystyle {\tan 30{}^\circ =\ \frac{AE}{EC}=\frac{AE}{120}\ }$
$\displaystyle \mathrm{AE=120\times 0.5773} =69.28 \text{ m}$
$\displaystyle AB=AE+EB$
$\displaystyle = AE+CD = 69.28+53.42 = 122.7 \text{ m}$
Height of $\displaystyle 1{}^{st}$ tower is $\displaystyle 122.7 m \text{ and } 2{}^{nd}$ tower is $\displaystyle 53.42 \text{ m}$
(b) Below the table
Weight $\displaystyle f$$\displaystyle f$ $\displaystyle cf$$\displaystyle cf$ 50-60 4 4 60-70 7 11 70-80 11 22 80-90 14 36 90-100 6 42 100-110 5 47 110-120 3 50 Total 50
$\displaystyle \text{(i) Lower quartile }= \frac{N^{th}}{4}{ obs}^n$
$\displaystyle = \frac{{50}^{th}}{4}{ obs}^n = 12.5^{th} { obs}^n = 72 kg$
$\displaystyle \text{Upper quartile } = \frac{3N}{4}^{th} \ {obs}^n = \frac{150}{4}^{th}\ {obs}^n = 37.5^{th} { obs}^n = 92 kg$
(ii) No. of over weight workers $\displaystyle = 50-39 = 11$
$\displaystyle \\$
Question 10
(a) A wholesale buys a TV from the manufacture for Rs. 25000. He marks the price of the TV 20% above his cost price and sell it to a retailer at 10% discount on the marked price. If the rate of VAT is 8%, find the: [3]
(i) marked Price
(ii) retailer’s cost price inclusive of tax
(iii) VAT paid by the wholesaler.
$\displaystyle \text{(b) If } A = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix}$ , $\displaystyle B = \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} . \text{ Find} AB-5C. \hspace{1.0cm} [3]$
(c) ABC is a right angled triangle with $\displaystyle \angle ABC= 90^{\circ} .$ D is any point on AB and DE is perpendicular to AC. Prove that:
(i) $\displaystyle \Delta ADE \sim \Delta ACB$
(ii) If $\displaystyle AC = 13 \text{ cm}, BC = 5 \text{ cm} \text{ and } AE = 4 \text{ cm}. \text{ Find } DE \text{ and } AD.$
(iii) Find, area, of $\displaystyle \Delta ADE$ : area of quadrilateral BCED. [4]
(a)
(i) Cost price for wholesaler $\displaystyle = 25000 \text{ Rs. }$
$\displaystyle \text{Market Price } = 25000+\frac{20}{100}\times 25000 = 30000 \text{ Rs. }$
(ii) Discount $\displaystyle = 10\% \text{ of } 30000= 3000 \text{ Rs. }$
Cost Price for retailer $\displaystyle = \text{ Market Price -Discount }$
$\displaystyle = \mathrm{30000 - 3000} = 27000 \text{ Rs. }$
$\displaystyle \text{Cost Price inclusive Tax } = \mathrm{27000+}\frac{\mathrm{8}}{\mathrm{100}}\mathrm{\times 27000} = 29160 \text{ Rs. }$
(iii) Cost price for wholesaler $\displaystyle = 25000 \text{ Rs. }$
Sale price for wholesaler $\displaystyle = 27000 \text{ Rs. }$
Profit for wholesalers $\displaystyle = 27000 - 25000= 2000 \text{ Rs. }$
$\displaystyle \text{VAT } = \frac{8}{100}\times 2000 = 160 \text{ Rs. }$
$\displaystyle \text{(b) Given } A = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix}$ , $\displaystyle B = \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix}$ , $\displaystyle C = \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix}$
$\displaystyle AB = \begin{bmatrix} 3 & 7 \\ 2 & 4 \end{bmatrix} . \begin{bmatrix} 0 & 2 \\ 5 & 3 \end{bmatrix} = \begin{bmatrix} 0+35 & 6+21 \\ 0+20 & 4+12 \end{bmatrix} = \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix}$
$\displaystyle 5C = 5 . \begin{bmatrix} 1 & -5 \\ -4 & 6 \end{bmatrix} = \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix}$
$\displaystyle AB - 5C= \begin{bmatrix} 35 & 27 \\ 20 & 16 \end{bmatrix} - \begin{bmatrix} 5 & -25 \\ -20 & 30 \end{bmatrix} = \begin{bmatrix} 30 & 52 \\ 40 & -14 \end{bmatrix}$
(c) To prove: $\displaystyle \Delta ADE\ \sim \Delta ACB$
(i) In $\displaystyle \Delta ADE\ and\ \Delta ACB$
$\displaystyle \angle A$ (is common)
$\displaystyle \angle AED = \angle ABC=90^{\circ}$
Therefore $\displaystyle \Delta ADE\ \sim \Delta ACB$ (AAA Postulate)
(ii) $\displaystyle {AC}^2={AB}^2+{BC}^2$
$\displaystyle 169={AB}^2+25$
$\displaystyle AB = 12 cm$
Since $\displaystyle \Delta ADE \sim \Delta ACB$
$\displaystyle \text{Therefore } \frac{DE}{BC} = \frac{AD}{AC} = \frac{AE}{AB}$
$\displaystyle \text{Therefore } \frac{DE}{BC} = \frac{AE}{AB}$
$\displaystyle \frac{DE}{5} = \frac{4}{12} \Rightarrow DE = \frac{5}{3}$ cm
$\displaystyle \frac{AD}{AC} = \frac{AE}{AB}$
$\displaystyle \frac{AD}{13} = \frac{4}{12} \Rightarrow AD = \frac{13}{3}$ cm
$\displaystyle \text{(iii) } \frac{Ar.\ of\ \left(\Delta ABC\right)}{Ar.\ of\ \left(\Delta ADE\right)}=\ \frac{{AB}^2}{{AE}^2}=\frac{144}{16}=9$
$\displaystyle \Rightarrow \frac{Ar.\ of\ \left(\Delta ADE\right)+Ar.\ of\ \Delta (BCED)}{Ar.\ of\ (\Delta ADE)} =9$
$\displaystyle \Rightarrow 1 + \frac{Ar.\ of\ \left(BCED\right)}{Ar.\ of\ \left(\Delta ADE\right)} = 9$
$\displaystyle \Rightarrow \frac{Ar.\ of\ \left(\Delta ADE\right)}{Ar.\ of\ \left(BCED\right)}=\ \frac{1}{8}$
$\displaystyle \\$
Question 11
(a) Sum of two natural numbers is 8 and the difference of their reciprocal is $\displaystyle \frac{2}{15}$ find the numbers. [3]
$\displaystyle \text{(b) Given } \frac{x^3+12x}{6x^2+8} = \frac{y^3+27y}{9y^2+27} \text{ using componendo and dividendo find } x:y$ [3]
$\displaystyle \text{(c) Construct the } \triangle ABC \text{ with } AB=5.5 \text{ cm}, AC=6 \text{ cm} , \text{ and } \angle BAC=105^{\circ} . \text{ Hence }$
(i) Construct the locus of points equidistant from BA and BC
(ii) Construct the locus of points equidistant from B and C
(iii) Mark the point which satisfy he above two loci as P. Measure and write the length of PC. [4]
(a) Let 1st number be $\displaystyle x$ and the 2nd number be $\displaystyle (8-x)$
$\displaystyle \text{Given } \frac{1}{x}-\frac{1}{8-x}=\frac{2}{15}$
$\displaystyle \frac{8-x-x}{x\ \left(8-x\right)}=\frac{2}{15}$
$\displaystyle \frac{8-2x}{8x-x^2}=\frac{2}{15}$
$\displaystyle 120-30x=16x-2x^2$
$\displaystyle {2x}^2-46x+120=0$
$\displaystyle x^2-23x+60=0$
$\displaystyle x\ \left(x-20\right)-3\ \left(x-20\right)=0$
$\displaystyle \left(x-3\right)\left(x-20\right)=0 \Rightarrow x=3 \text{ or } x=20 \text{ (not possible) }$
Therefore the first number is 3 and the second number is 5.
$\displaystyle \text{(b) }\frac{x^3+12x}{{6x}^2+8}=\frac{y^3+27y}{{9y}^2+27}$
Applying componendo and dividendo
$\displaystyle \frac{x^3+12x+{6x}^2+8}{x^3+12x-{6x}^2-8}=\frac{y^2+27y+{9y}^2+27}{y^3+27y-{9y}^2-27}$
$\displaystyle \frac{(x+{2)}^3}{(x-2)^3}=\frac{{\left(y+3\right)}^3}{{\left(y-3\right)}^3}$
$\displaystyle \frac{x+2}{x-2}=\frac{y+3}{y-3}$
Again Applying componendo and dividendo
$\displaystyle \frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3}$
$\displaystyle \frac{2x}{4}=\frac{2y}{6}$
$\displaystyle \frac{x}{y}=\frac{4}{6}=\frac{2}{3}$
$\displaystyle x:y=2:3$
(c) Steps of Construction:
1. First draw a line AB=5.5 cm.
2. With the help of the point A, draw $\displaystyle \angle XAB =105^{\circ} .$ This you can draw using a compass.
3. Take radius 6 cm, cut AC=6 cm and join C to B. This way you get CA and BC. This completes the triangle.
4. Then draw perpendicular bisector of BC and draw angle of bisector $\displaystyle \angle CBA$ both intersecting at P.
5. P is the required point PC=4.8 cm
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# Second Grade Math Interactive Notebook: Addition & Subtraction within 1,000 TEKS
Subject
Resource Type
Format
PDF (6 MB|40 (including directions, folds and flaps, & answer keys))
\$6.00
\$6.00
#### Also included in
1. This second grade math interactive notebook bundle contains hands-on folds and flaps to help your students learn all of the second grade math TEKS. There is one notebook per strand including: whole numbers, addition and subtraction, fractional units, coins, contextual multiplication and division, al
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Save \$29.00
### Description
This second grade math interactive notebook contains hands-on folds and flaps to help your students learn to develop and use strategies and methods in order to solve addition and subtraction problems, especially recalling basic facts, adding and subtracting up to four two-digit numbers using mental strategies and algorithms, solving one-step and multi-step word problems, and generating and solving problem situations within 1,000. The literature on brain research, multiple intelligences, and note-taking all support the classroom use of interactive notebooks. Here is an easy-to-implement support which will allow you to print and go!
This resource will keep you afloat because…
☑ I have taken the guesswork out of what to put underneath all of the flaps included in this set with photographs of finished examples which serve as answer keys.
☑ Accommodates for all students as the underneath portion can be glued under the folds. Students do not have to write out everything.
☑ The folds and flaps included in this set are rigorous and meet the specificity of each standard.
........................................................................................................................
Includes:
★ 10 pages of directions with photographs to guide you
★ I Can Statements for each standard to use as a header (4 total)
★ 1 Divider Tab
★ Vocabulary 12-Flaps
★ Subtraction Strategies 12-Flaps
★ Basic Math Facts Pocket and Math Facts (to recall basic math facts)
★ Mental Strategies for Addition 2-Flaps
★ Mental Strategies for Subtraction 2-Flaps
★ 2 Generate & Solve Based on Place Value 3-Flaps (addition and subtraction of whole numbers within 1,000)
★ Solve It! 1-Flaps (multi-step addition and multi-step subtraction)
Focus Standards:
→ TEKS- 2.4ABCD
→ The student applies mathematical process standards to develop and use strategies and methods for whole number computations in order to solve addition and subtraction problems with efficiency and accuracy.
→ 2.4A Recall basic facts to add and subtract within 20 with automaticity.
→ 2.4B Add up to four two-digit numbers and subtract two-digit numbers using mental strategies and algorithms based on knowledge of place value and properties of operations.
→ 2.4C Solve one-step and multi-step word problems involving addition and subtraction within 1,000 using a variety of strategies based on place value, including algorithms.
→ 2.4D Generate and solve problem situations for a given mathematical number sentence involving addition and subtraction of whole numbers within 1,000
........................................................................................................................
Math Interactive Notebook Starter Set
2.1 Process Standards Math Interactive Notebook
2.2 Whole Numbers 0-1,200 Notebook
2.3 Fractional Units Notebook
2.4 Addition and Subtraction Notebook 2.5 Value of U.S. Coin Collections
2.6 Contextual Multiplication and Division Notebook
2.7 Algebraic Reasoning Notebook
2.8 Geometry Notebook
2.9 Measurement Notebook
2.10 Data Analysis Notebook
2.11 Personal Financial Literacy Notebook
WANT THEM ALL?
Second Grade Math Interactive Notebook Bundle
........................................................................................................................
I have math interactive notebook resources for kindergarten, first, and second grades. You can choose between the individual notebooks or the bundle.
Kindergarten Math Interactive Notebook Bundle
First Grade Math Interactive Notebook Bundle
Second Grade Math Interactive Notebook Bundle
........................................................................................................................
Product Notes
1. This file is NOT editable.
Here’s to you for making a splash and checking out this diverse, rigorous resource!
Laura
For FREE tips + resources, become a valued partner of Down River Resources.
Total Pages
40 (including directions, folds and flaps, & answer keys)
Included
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# Distance between Paris (CDG) and Ras Al Khaimah (RKT)
Flight distance from Paris to Ras Al Khaimah (Paris Charles de Gaulle Airport – Ras Al Khaimah International Airport) is 3268 miles / 5259 kilometers / 2839 nautical miles. Estimated flight time is 6 hours 41 minutes.
## Map of flight path from Paris to Ras Al Khaimah.
Shortest flight path between Paris Charles de Gaulle Airport (CDG) and Ras Al Khaimah International Airport (RKT).
## How far is Ras Al Khaimah from Paris?
There are several ways to calculate distances between Paris and Ras Al Khaimah. Here are two common methods:
Vincenty's formula (applied above)
• 3267.582 miles
• 5258.664 kilometers
• 2839.451 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 3263.347 miles
• 5251.849 kilometers
• 2835.771 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Paris Charles de Gaulle Airport
City: Paris
Country: France
IATA Code: CDG
ICAO Code: LFPG
Coordinates: 49°0′46″N, 2°32′59″E
B Ras Al Khaimah International Airport
City: Ras Al Khaimah
Country: United Arab Emirates
IATA Code: RKT
ICAO Code: OMRK
Coordinates: 25°36′48″N, 55°56′19″E
## Time difference and current local times
The time difference between Paris and Ras Al Khaimah is 3 hours. Ras Al Khaimah is 3 hours ahead of Paris.
CET
+04
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 366 kg (808 pounds).
## Frequent Flyer Miles Calculator
Paris (CDG) → Ras Al Khaimah (RKT).
Distance:
3268
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
3268
Round trip?
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## Power and roots
Writing repeated factors can be so tiring! Powers and exponents are a shorthand notation that save space and are a quick and easy way to simplify multiplication
Let’s get started
Exponent – A number that shows how many times another number is multiplied times itself
3^2 = 3 x 3 = 9
Evaluate means to find the value of something.
Example Evaluate 2^3 x 5^2
Here’s how
2^3 x 5^2 = 2 x 2 x 2 x 5 x 5 = 200
(more…)
## Multiples and factors
Do you draw a blank when you see LCM and GCF? Don’t worry, these acronyms stand for concepts that make sense. They help you find the common denominator of two or more fractions, simplify fractions, and add and subtract fractions.
Let’s get started
Multiple – The product of a given number and a nonzero whole number.
Do you draw a blank when you see LCM and GCF? Don’t worry, these acronyms stand for concepts that make sense. They help you find the common denominator of you or more fractions, simplify fractions, and add and subtract fractions.
Let’s get started
Multiple – The product of a given number and a nonzero whole number.
Multiples of 2 are
2 x 1 = 2, 2 x 2 = 4, 2 x 3 = 6, and so on.
Example Find the first four multiples of 8
## Properties of Operations
You probably use properties of operations every day without even giving them a thought. You may have noticed that 3 x 4 and 4 x 3 are both 12. That”s an example of the commutative property of multiplication. The ideas are probably familiar – you just need to brush up on the vocabulary.
Let”s get started
The associative and commutative properties hold for both addition and multiplication.
(more…)
## Prime and composite numbers
Why on earth do you need to know about prime and composite numbers? Fractions! Work with fractions becomes much easier if you recognize prime and composite numbers. It helps you know if a fraction needs to be simplified.
Let’s Get Started
Here are some of the terms you will encounter when working with numbers.
Factor – A number that is multiplied by another number to find a product
2 and 3 are factors of 6, because 2×3 = 6.
Prime number – A number with only two factors, itself and 1.
7 is prime because its only factor are 1 and 7.
1×7 = 7
Composite number – A number with more than two factors.
6 is a composite number because 1, 2, 3, and 6 are all factor of 6.
1 x 6 = 6 2 x 3 = 6
In the information society. communication is everything, In order to communicate information, we have to speak the same language and use the same vocabulary.
Let’s Get Started
Our number system in a base -10 place-value system. We use 10 different digits to form all the numbers. The value of the digit depends on which place it is in.
Period – A grouping of three palces in the place-value chart. Each period has a ones place (O), a tens place(T), and a hundreds place (H).
Standard Form – A number written with digits.
124
614,900
70,142, 859
Word form – A number written with words.
One hundred twenty-four
Six hundred fourteen thousand, nine hundred
Seventy million, one hundred forty-two thousand, eight hundred fifty-nine.
Expanded form – A number written as the sum of its place values.
124 = 100 + 20 + 4
614,900 = 600,000 + 10,000 + 4,000 + 900
70,142, 859 = 70, 000, 000 + 100,000 + 40,000 + 2,000 + 800 + 50 + 9
Sometimes expanded form is written this way to show the place values even more clearly:
614,900 = 6 x 100,000 + 1 x 10,000 + 4 x 1,000 + 9 x 100
about expanded form – it doesn’t come up very often in everyday life. Expanded form helps kids understand the amazing idea of place value and the role of zero as a placeholder.
Now. Let’s practice
Write each number in word form.
1. 3,004 2. 5,700,054 3. 37,000,416
Write each number in standard form.
4. Seventy-five thousand 5. sixteen million, thirty-four
Write each number in expanded form.
6. 45,700 7. 190, 063 8 5,009,802
## Numbers: The basics
A number is a number, right?? Well, not exactly. There are different types of numbers. Kind of like a dog and a cat are both animals, but they are different types of animals.
Numbers and number systems are pretty basic in math, so let’s look at some definitions and think about why we need all the different types.
Let’s Get Started
The numbers you use all the time are called the counting numbers.
Counting numbers – The most familiar and basic numbers. The numbers we counts with: 1,2,3,4,5,6…..
When we throw in 0, we have a whole new set.
Whole numbers – The counting numbers are zero: 0,1,2,3,4,5,6…….
Whole numbers were all that we needed for awhile – until we needed a way to talk about loss of money or yards in a football game or depths below sea level.
Integers – The whole numbers and their opposites.
An integer that is divisible by 2 is called an even number. If there is a remainder of 1 when the integer is divided by 2, the integer is called an odd number.
But what about when you share a cookie with your friends? You need a fraction to tell how much cookie you each get to eat. This brings us to the rational numbers.
Rational numbers – Numbers that can be written as a ratio of two integers. Fractions and decimals are ratios of integers. All the integers are rational numbers.
Irrational numbers – Numbers that cannot be written as the ratio of two integers.
Real Numbers – The rational numbers and the irrational numbers.
## Introduction of Hindu-Arabic & Roman numerals
At first, the ancients developed names for the numbers. They spoke of having one sheep, two sheep, etc. But you can see how difficult it would be to add or subtract columns of numbers expressed only in words. Thus we learn that arithmetic computation did not begin until man came to use symbols for numbers. The kinds of symbols used for numbers went through various changes starting with the simple vertical mark of ancient Mesopotamia, progressing to the combination of the Egyptians, the familiar numerals of the Romans, and finally to our present figures.
We are indebted to the Arabs for our present method of writing numbers. For this reason, the numerals 0 through 9, the ingredients for any number combinations we wish to write, were called Arabic numbers for a long time. But more recently historians have discovered that the system of writing numbers now used by civilized people throughout the world was originated by the Hindus in India. The Arabs learned the system from the Hindus and are credited with having brought it to Europe soon after the conquest of Spain in the eighth century AD. For this reason, we now property call it the Hindu-Arabic system of numerals.
An early system of writing numbers is the Roman system. It is generally agreed that it is of little practical value in today’s world of advanced mathematics.
Because you will still see Roman numerals used in recording dates, in books, as numbers of a clock face, and in other places, it is worth taking a little time to learn how to read them.
The Roman number system is based on seven letters, all of which are assigned specific values. They are:
I=1, V=5, X=10, L=50, C=100, D=500, M=1000
Rule 1: When a letter is repeated, its value is repeated.
eg, I=1, II=2, III=3, XX=20, CCC=300
Rule 2: When a letter follows a letter of greater value, its value is added to the greater value.
eg, VI=6, XV=15, LX=60, DC=600
In these examples, observe that the smaller value I after V means add 1 to the 5 to give 6. In the same way, the V following the X means add 5 to 10 which equals 15. Similarly, LX represents 10 added to 50 to give 60. To write 70, merely add XX after the L to give LXX. In like manner, to write 800, add CC after DC to give DCCC.
(more…)
## Some basic ideas of arithmetic
Vocabulary
There are four basic operations in mathematics: addition, subtraction, multiplication, and division. Often when we talk about a collection of numbers, such as the numbers 1, 2, and 3, we use the word set. We could use set notation with braces, [ ], to list the number: [1, 2, 3]. The set of even numbers could be written as [2, 4, 6, 8, 10, …], and the set of odd numbers as [1, 3, 5, 7,…]. (The three dots indicate that the numbers continue indefinitely. In any collection of numbers ending in dots, there is no largest number.)
Here, we deal with two sets of numbers: the counting numbers [1, 2, 3, 4,…] and the whole numbers [0, 1, 2, 3,…]. The whole numbers are just the counting numbers plus zero. When we count, we start with 1. When we answer the question “How many?” we need zero as a possible answer.
Symbols are necessary to make mathematical statements complete. For example, we use symbols for addition (+) and multiplication (X).
= as in 8 + 3 = 11
8 plus 3 equals 11
< as in 3 < 8
3 is less than 8
> as in 8 > 3
8 is greater than 3
Notice that the symbols for less than and greater than are always open toward the larger number. When statements are not true, we put a slash through the symbol:
6 + 3 =/ 11
6 + 3 does not equal 11
5 >/ 7
5 is not greater than 7
9 </ 6
9 is not less than 6
Numerals are symbols for numbers, which are abstract ideas. For example, a fisherman 8000 years ago might record that he caught ||| fish. We could write 3 for the amount ||| and 3 are the symbols for the same numbers. Our number symbols are called arabic numerals.
Digits are the number symbols (numerals) 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 in our number system. Numbers are written as combinations of any of these ten digits.
A whole number is written as a string of digits, 7 is a one-digit number; 32 is a two-digit number with 3 as the first digit and 2 as the second digit; 487 is a three-digit number with 4 as the first digit, 8 as the second digit, and 7 as the third digit.
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## Linear Algebra and Its Applications (5th Edition)
Only one row operation, Replacement, needs to be performed to set up the relevant equation. By replacing the second row with the sum of its values and those of row one scaled by a factor of 4, we obtain $h+12=0$. Thus, $h=-12$.
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Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
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# Re: st: RE: RE: what does it mean the default for -mfx- is discrete when it is evaluated at the mean?
From Maarten buis To statalist@hsphsun2.harvard.edu Subject Re: st: RE: RE: what does it mean the default for -mfx- is discrete when it is evaluated at the mean? Date Wed, 2 Jun 2010 09:14:29 +0000 (GMT)
```--- On Tue, 1/6/10, Martin Weiss wrote:
> Still, what is the problem with saying that the marginal
> effect displayed is for going from zero to one, i.e. the
> only move that a dummy can possibly make? The fact that
> the share of ones in your sample is 37% does not change
> this fact, does it?
I am guessing the confusion comes from the fact that the
marginal effect depends on fixing the values of _all_
explanatory variables, so now Frank sees three values,
0, .37, and 1. However, in computing the discrete change
this middle value is ignored, you only fix the values of
all _other_ explantory variables.
This is illustrated in the example below, For the marginal
effect of price we need to fix the value of both price and
foreign, but for the "marginal effect" of foreign we only
need to fix the value of price.
*----------------- begin example ----------------------
sysuse auto, clear
gen byte good = rep78 > 3 if rep78 < .
replace price = price / 1000
logit good foreign price
mfx
tempname mprice mforeign xb xb0 xb1
sum price if e(sample), meanonly
scalar `mprice' = r(mean)
sum foreign if e(sample), meanonly
scalar `mforeign' = r(mean)
scalar `xb' = _b[_cons] + _b[price]*`mprice' + _b[foreign]*`mforeign'
di invlogit(`xb')*invlogit(-`xb')*_b[price]
scalar `xb0' = _b[_cons] + _b[price]*`mprice'
scalar `xb1' = _b[_cons] + _b[price]*`mprice' + _b[foreign]
di invlogit(`xb1') - invlogit(`xb0')
*-------------------- end example ----------------------------
(For more on examples I sent to the Statalist see:
http://www.maartenbuis.nl/example_faq )
Hope this helps,
Maarten
--------------------------
Maarten L. Buis
Institut fuer Soziologie
Universitaet Tuebingen
Wilhelmstrasse 36
72074 Tuebingen
Germany
http://www.maartenbuis.nl
--------------------------
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
```
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# The Unique Effects of Including History in College Algebra - Findings (1)
Author(s):
D. Goodwin (Black Hills State University) and G. W. Hagerty (Black Hills State University) and S. Smith (Black Hills State University)
Over the last three years at BHSU, great strides have been made in improving student outcomes in College Algebra with the inclusion of history in the course. The same results theorized by many (Bruckheimer & Arcavi, 2000; Heiede, 1996; Johnson, 1994; Kleiner, 1996; Man-Keung, 2000; Rickey, 1996; Smith, 1996; Swetz, 2000) about the effects of inclusion of the historical development of concepts in mathematics courses can now be seen in the College Algebra course.
Figure 1 shows the changes in the percentage of students enrolled in the fall College Algebra course who continued on to enroll in Trigonometry. The dotted line shows the goal of 8%. The solid line shows the growth in the percentage of the students enrolling in Trigonometry. The number of students enrolled in College Algebra in the fall has consistently remained around 360 students. The number enrolling in Trigonometry has grown from 5 students in 2003 and 2004 to 20 students in 2007.
Figure 1. Percent of students enrolling in Trigonometry from 2003-2007
The increase started in 2005, the year the history modules were introduced. Besides quadrupling the number of students enrolling in Trigonometry, average student achievement in Trigonometry also increased by approximately 6-7%, from about a 70% overall average to an overall average of 76-77% for the course. Furthermore, the Trigonometry instructor reports that this class has doubled in size, attendance has increased, there is an improved success rate and a greater percentage of students are making their way to Calculus. In the coming semesters, as more students who were exposed to the revised College Algebra course matriculate into Calculus, increased enrollment and achievement is expected in Calculus.
There has been significant improvement on test problems where mathematical vocabulary and notation have created difficulty for students in the past. One example is that when students were asked to find the inverse function of $f(x)=\frac{x+2}{2-x},$ students would pick the reciprocal about 30% of the time prior to the use of history in the College Algebra course. After the introduction of history to the course, students are picking the reciprocal only about 15% of the time.
D. Goodwin (Black Hills State University) and G. W. Hagerty (Black Hills State University) and S. Smith (Black Hills State University), "The Unique Effects of Including History in College Algebra - Findings (1)," Convergence (June 2010), DOI:10.4169/loci002530
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https://mathematica.stackexchange.com/questions/109543/simulating-a-wave-pulse/109549
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# Simulating a Wave Pulse
How to construct a code that superimpose n=10 one dimensional sinuisodal waves. each with an angular frequency ωi(*i is subscript*) and wave number ki to form a wave pulse.Each angular frequency ωi and wave number kidiffers slightly from the previous one only by a small fraction, namely. Δω=|ωi+1-ωi|<<ωi Δk=|ki+1-ki|<<ki
i plan to Manipulate[] the simulation for n=50 waves. i would like to see the difference in the wavepulse composed of n=50 and n=10 waves. this is what have been construct so far.
A1 = 1; k1 = 4; ω = 2; A2 = 5; ω2 = 10; k2 = 3
y1[x_, t_] := A1*Sin[k1*x - ω*t];
y2[x_, t_] := (A2*Sin[k2*x - ω2*t]);
Plot[{y1[x, t = 0.5], y2[x, t = 0.5]}, {x, -10, 10}]
Manipulate[
Plot[{y1[x, t = n] + y2[x, t = n]}, {x, -10, 10},
PlotRange -> All], {n, 0, 100, 0.25}]
angularrate = Range[10];
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Passing two-dimensional array via pointer
How do I pass the m matrix to foo()? if I am not allowed to change the code or the prototype of foo()?
``````void foo(float **pm)
{
int i,j;
for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++)
printf("%f\n", pm[i][j]);
}
int main ()
{
float m[4][4];
int i,j;
for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++)
m[i][j] = i+j;
foo(???m???);
}
``````
-
If you insist on the above declaration of `foo`, i.e.
``````void foo(float **pm)
``````
and on using a built-in 2D array, i.e.
``````float m[4][4];
``````
then the only way to make your `foo` work with `m` is to create an extra "row index" array and pass it instead of `m`
``````...
float *m_rows[4] = { m[0], m[1], m[2], m[3] };
foo(m_rows);
``````
There no way to pass `m` to `foo` directly. It is impossible. The parameter type `float **` is hopelessly incompatible with the argument type `float [4][4]`.
If you look carefully, you'll that this is basically the same thing as what Carl Norum suggested in his answer. Except that Carl is `malloc`-ing the array memory, which is not absolutely necessary.
-
Carl is also `malloc` -ing each row separately, whereas AndreyT's example leaves the actual data in a single block. – caf Aug 18 '10 at 23:37
If you can't change `foo()`, you will need to change `m`. Declare it as `float **m`, and allocate the memory appropriately. Then call `foo()`. Something like:
``````float **m = malloc(4 * sizeof(float *));
int i, j;
for (i = 0; i < 4; i++)
{
m[i] = malloc(4 * sizeof(float));
for (j = 0; j < 4; j++)
{
m[i][j] = i + j;
}
}
``````
Don't forget to `free()` afterwards!
-
The technique that you use has no relation to dynamic allocation of memory whatsoever. Since the array sizes are constant, you can just as well allocate the memory "on the stack", just like it was in the OP. – AndreyT Aug 18 '10 at 18:00
Sure, but it's not wrong - the solution to the problem is to change the type of m. – Carl Norum Aug 18 '10 at 18:10
@Carl Norum: Not necessarily. The problem, as stated, seems to imply that the existing `foo` must be made to work with the existing `m`. And the solution to the problem is to change the type of what is passed to `foo`, not the type of `m`. – AndreyT Aug 18 '10 at 18:13
@AndreyT, that seems fair - I interpreted it as being unable to change `foo()` only. – Carl Norum Aug 18 '10 at 18:18
@Carl Norum: You are right. That's probably what was implied by the OP. I'm just saying that switching to dynamic allocation is not really absolutely required. – AndreyT Aug 18 '10 at 18:34
show 1 more comment
You can't. `m` is not compatible with the argument to `foo`. You'd have to use a temporary array of pointers.
``````int main()
{
float m[4][4];
int i,j;
float *p[4];
p[0] = m[0];
p[1] = m[1];
p[2] = m[2];
p[3] = m[3];
for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++)
m[i][j] = i+j;
foo(p);
``````
-
If you have a compiler that supports C99, the current C standard, then you can do this:
``````foo((float *[]){ m[0], m[1], m[2], m[3] });
``````
(Note that this is exactly the same as AndreyT's answer, except that it forgoes having to name the temporary array)
-
• you dont need to do any changes in the main,but you function will work properly if you change the formal prototype of your function to (*pm)[4] or pm[][4] because **pm means pointer to pointer of integer while (*pm)[4] or pm[][4] means pointer to poiner of 4 integers .
m here is also a pointer to pointer of 4 integers and not pointer to pointer of integers and hence not compatible.
``````#include<stdio.h>
void foo(float (*pm)[4])
{
int i,j;
for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++)
printf("%f\n", pm[i][j]);
}
int main ()
{
float m[4][4];
int i,j;
for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++)
m[i][j] = i+j;
foo(m);
}
``````
-
Does `foo(m)` not work?
-
No, it doesn't. A 2D array is not a pointer to a pointer. – Justin Ardini Aug 18 '10 at 17:59
Oh, it is long time since I really used C and pointers (maybe 20 years). – Frank Aug 18 '10 at 20:17
This is really a comment, not an answer to the question. Please use "add comment" to leave feedback for the author. – Rostyslav Dzinko Aug 21 '12 at 10:13
`void foo(float **pm)` is the same as `void foo(float *pm[])` which is not a two-dimensional array of floats. It is an array of `float*`. Now, those `float*` may themselves point to float arrays, but that's a separate matter.
-
``````typedef float Float4[4];
void foo(Float4 *pm)
{
int i,j;
for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++)
printf("%f\n", pm[i][j]);
}
main()
{
Float4 m[4];
int i,j;
for (i = 0; i < 4; i++)
for (j = 0; j < 4; j++)
m[i][j] = i+j;
foo(m);
return 0;
}
``````
-
How does this satisfy the requirement "I am not allowed to change the code or the prototype of foo()?" ? – caf Aug 18 '10 at 23:41
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固壁上液滴动力学的LBM建模与计算 LBM Modeling and Calculation for Droplet Dynamics on Solid Wall
Applied Physics
Vol. 09 No. 05 ( 2019 ), Article ID: 30481 , 5 pages
10.12677/APP.2019.95032
LBM Modeling and Calculation for Droplet Dynamics on Solid Wall
Yanggui Li1,2, Dacheng Liang3*
1School of Mathematics and Statistics, Lingnan Normal University, Zhanjiang Guangdong
2Institute of Synthetic Biology, Shenzhen Institute of Advanced Technology, Chinese Academy of Sciences, Shenzhen Guangdong
3Department of Mathematics, Guangdong Preschool Normal College in Maoming, Maoming Guangdong
Received: May 8th, 2019; accepted: May 21st, 2019; published: May 28th, 2019
ABSTRACT
In this paper, the dynamic model for the droplet on the solid wall is studied with Shan-Chen multiphase lattice Boltzmann method. Based on the model, the spreading behavior of droplet on the horizontal solid wall and the slip behavior on the vertical solid wall are simulated.
Keywords:Droplet Dynamics, Lattice Boltzmann Method, Solid-Liquid Interaction, Multiphase Flow
1岭南师范学院，数学与统计学院，广东 湛江
2中国科学院深圳先进技术研究院，合成生物学研究所，广东 深圳
3广东茂名幼儿师范专科学校，数学系，广东 茂名
1. 引言
2. 理论与模型
2.1. 格子Boltzmann方法
Boltzmann方程描述了由于其分子的流动和碰撞而导致的流体状态的演化。LBM是通过对Boltzmann方程的相空间进行离散化得到的，其中大量的分子速度被一组有限的速度矢量(晶格)代替 [1] 。LBM模型包含三要素：格子结构、流体粒子的离散速度集合、演化方程，其描述了流体粒子分布函数在固定格子上的演化过程
${f}_{i}\left(x+{e}_{i}{\delta }_{i},t+{\delta }_{i}\right)-{f}_{i}\left(x,t\right)={\Omega }_{i}\left(x,t\right)$ (1)
$\rho =m{\sum }_{i}{f}_{i}$ , $\rho u=m{\sum }_{i}{e}_{i}{f}_{i}$ (2)
m为流体粒子质量。
2.2. 多相或多组分格子Boltzmann模型
${f}_{i}^{k}\left(x+{e}_{i}{\delta }_{i},t+{\delta }_{i}\right)-{f}_{i}^{k}\left(x,t\right)=-\frac{{f}_{i}^{k}\left(x,t\right)-{f}_{i}^{k\left(eq\right)}\left(x,t\right)}{{\tau }_{k}}$ (3)
${f}_{0}^{k\left(eq\right)}={\alpha }_{k}{n}_{k}-\frac{2}{3}{n}_{k}{u}_{k}^{eq}\cdot {u}_{k}^{eq}$ (4)
${f}_{i}^{k\left(eq\right)}=\frac{\left(1-{\alpha }_{k}\right){n}_{k}}{5}+\frac{1}{3}{n}_{k}\left({e}_{i}\cdot {u}_{k}^{eq}\right)+\frac{1}{2}{n}_{k}{\left({e}_{i}\cdot {u}_{k}^{eq}\right)}^{2}-\frac{1}{6}{n}_{k}{u}_{k}^{eq}{u}_{k}^{eq}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,3,4$ (5)
${f}_{i}^{k\left(eq\right)}=\frac{\left(1-{\alpha }_{k}\right){n}_{k}}{20}+\frac{1}{12}{n}_{k}\left({e}_{i}\cdot {u}_{k}^{eq}\right)+\frac{1}{8}{n}_{k}{\left({e}_{i}\cdot {u}_{k}^{eq}\right)}^{2}-\frac{1}{24}{n}_{k}{u}_{k}^{eq}{u}_{k}^{eq}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=5,6,7,8$ (6)
${e}_{i}=\left\{\begin{array}{l}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=0\\ \left(\mathrm{cos}\frac{\left(i-1\right)\text{π}}{2},\mathrm{sin}\frac{\left(i-1\right)\text{π}}{2}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=1,2,3,4\\ \sqrt{\text{2}}\left(\text{cos}\left[\frac{\left(i-5\right)\text{π}}{2}+\frac{\text{π}}{4}\right],\mathrm{sin}\left[\frac{\left(i-5\right)\text{π}}{2}+\frac{\text{π}}{4}\right]\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=5,6,7,8\end{array}$ (7)
${\alpha }_{k}$ 是一个自由参数，它与纯组分k区域的声速有关，其关系式为 ${\left({c}_{s}^{k}\right)}^{2}=\frac{3}{5}\left(1-{\alpha }_{k}\right)$${n}_{k}={\sum }_{i}{f}_{i}^{k}$ 是k组分的数密度。k组分的质量密度 ${\rho }_{k}$ 定义为 ${\rho }_{k}={m}_{k}{n}_{k}={m}_{k}{\sum }_{i}{f}_{i}^{k}$ ，组分k的流体速度 ${u}_{k}$ 定义为 ${\rho }_{k}{u}_{k}={m}_{k}{\sum }_{i}{e}_{i}{f}_{i}^{k}$ ，其中 ${m}_{k}$ 是k组分的分子质量。参数 ${u}_{k}^{eq}$ 由如下关系式确定：
${\rho }_{k}{u}_{k}^{eq}={\rho }_{k}{u}^{\prime }+{\tau }_{k}{F}_{k}$ (8)
${u}^{\prime }=\left(\underset{k=1}{\overset{s}{\sum }}\frac{{\rho }_{k}{u}_{k}}{{\tau }_{k}}\right)/\left(\underset{k=1}{\overset{s}{\sum }}\frac{{\rho }_{k}}{{\tau }_{k}}\right)$ (9)
${F}_{1k}\left(x\right)=-{\Psi }_{k}\left(x\right)\underset{{x}^{\prime }}{\sum }\underset{k=1}{\overset{s}{\sum }}{G}_{k\stackrel{¯}{k}}\left(x,{x}^{\prime }\right){\Psi }_{\stackrel{¯}{k}}\left({x}^{\prime }\right)\left(x-{x}^{\prime }\right)$ (10)
${G}_{k\stackrel{¯}{k}}\left(x,{x}^{\prime }\right)=\left\{\begin{array}{l}{g}_{k\stackrel{¯}{k}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|x-{x}^{\prime }|=1\\ {g}_{k\stackrel{¯}{k}}/4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|x-{x}^{\prime }|=\sqrt{2}\\ 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}其它\end{array}$ (11)
${F}_{2k}={\rho }_{k}g={m}_{k}{n}_{k}g$ (12)
2.3. 流体与固壁之间的相互作用
${F}_{3k}\left(x\right)=-{n}_{k}\left(x\right)\underset{{x}^{\prime }}{\sum }{g}_{kw}{n}_{w}\left({x}^{\prime }\right)\left({x}^{\prime }-x\right)$ (13)
$\frac{\partial \rho }{\partial t}+\nabla \cdot \left(\rho u\right)=0$ (14)
$\rho \left[\frac{\partial u}{\partial t}+\left(u\cdot \nabla \right)u\right]=-\nabla p+\nabla \cdot \left[\rho \nu \left(\nabla u+u\nabla \right)\right]+\rho g$ (15)
3. 数值模拟
Figure 1. Evolution of droplet spreading on solid wall
Figure 2. Evolution of droplet slipping on solid wall
4. 总结
LBM Modeling and Calculation for Droplet Dynamics on Solid Wall[J]. 应用物理, 2019, 09(05): 269-273. https://doi.org/10.12677/APP.2019.95032
1. 1. Shan, X.W. and Chen, H.D. (1993) Lattice Boltzmann Model for Simulation Flows with Multiple Phases and Compo-nents. Physical Review E, 47, 1815-1993. https://doi.org/10.1103/PhysRevE.47.1815
2. 2. Kang, Q.J., Zhang, D. and Chen, S. (2002) Displacement of a Two-Dimensional Immiscible Droplet in a Channel. Physics of Fluids, 14, 3203-3214. https://doi.org/10.1063/1.1499125
3. 3. Xing, X.Q., Butler, D.L. and Yang, C. (2007) Lattice Boltz-mann-Based Single-Phase Method for Free Surface Tracking of Droplet Motions. International Journal for Numerical Methods in Fluidss, 53, 333-351. https://doi.org/10.1002/fld.1282
4. 4. Tanaka, Y., Washio, Y., Yoshino, M. and Hirata, T. (2011) Numerical Sim-ulation of Dynamic Behavior of Droplet on Solid Surface by the Two-Phase Lattice Boltzmann Method. Computers & Fluids, 40, 68-78. https://doi.org/10.1016/j.compfluid.2010.08.007
5. 5. Wu, J., Huang, J.J. and Yan, W.W. (2015) Lattice Boltzmann Investigation of Droplets Impact Behaviors onto a Solid Substrate. Colloids and Surfaces A: Physicochemical and Engineering Aspects, 484, 318-328. https://doi.org/10.1016/j.colsurfa.2015.07.043
6. 6. Mousavi Tilehboni, S.E., Fattahi, E., Afrouzi, H.H. and Farhadi, M. (2015) Numerical Simulation of Droplet Detachment from Solid Walls under Gravity Force Using Lattice Boltzmann Method. Journal of Molecular Liquids, 212, 544-556. https://doi.org/10.1016/j.molliq.2015.10.007
7. 7. Raman, K.A., Jaiman, R.K., Lee, T.S. and Low, H.T. (2016) Lattice Boltzmann Study on the Dynamics of Successive Droplets Impact on a Solid Surface. Chemical Engineering Science, 145, 181-195. https://doi.org/10.1016/j.ces.2016.02.017
8. NOTES
*通讯作者。
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# 15961
## 15,961 is an odd composite number composed of two prime numbers multiplied together.
What does the number 15961 look like?
This visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.
15961 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.
## Prime factorization of 15961:
### 11 × 1451
See below for interesting mathematical facts about the number 15961 from the Numbermatics database.
### Names of 15961
• Cardinal: 15961 can be written as Fifteen thousand, nine hundred sixty-one.
### Scientific notation
• Scientific notation: 1.5961 × 104
### Factors of 15961
• Number of distinct prime factors ω(n): 2
• Total number of prime factors Ω(n): 2
• Sum of prime factors: 1462
### Divisors of 15961
• Number of divisors d(n): 4
• Complete list of divisors:
• Sum of all divisors σ(n): 17424
• Sum of proper divisors (its aliquot sum) s(n): 1463
• 15961 is a deficient number, because the sum of its proper divisors (1463) is less than itself. Its deficiency is 14498
### Bases of 15961
• Binary: 111110010110012
• Base-36: CBD
### Squares and roots of 15961
• 15961 squared (159612) is 254753521
• 15961 cubed (159613) is 4066120948681
• The square root of 15961 is 126.3368513143
• The cube root of 15961 is 25.1779306237
### Scales and comparisons
How big is 15961?
• 15,961 seconds is equal to 4 hours, 26 minutes, 1 second.
• To count from 1 to 15,961 would take you about four hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 15961 cubic inches would be around 2.1 feet tall.
### Recreational maths with 15961
• 15961 backwards is 16951
• The number of decimal digits it has is: 5
• The sum of 15961's digits is 22
• More coming soon!
#### Copy this link to share with anyone:
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The information we have on file for 15961 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 15961, math, Factors of 15961, curriculum, school, college, exams, university, Prime factorization of 15961, STEM, science, technology, engineering, physics, economics, calculator, fifteen thousand, nine hundred sixty-one.
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# Need help with this heat capacity/ideal gas problem.
blehxpo
For many gases at low densities and pressures (where ideal gas behavior is obtained), Cv/R = A+BT+CT^2 where A, B, C, are all constants, with appropriate temperature units so that the RHS (right hand side) of this equation is dimensionless.
a> How can a gas be both ideal and follow the equation above, at the same time.
b> Such a gas initially at T1 expands slowly in an insulated piston to double its original volume you would like to find the final temperature T2. A friend suggest you may use the follow equation that we derived in class:
(T2/T1) = (V1/V2)^(R/Cv) = (1/2)^(R/Cv)
where Cv is evaluated at T1. What is wrong with your friend's statement? Find the correct equation that should be solved (implicitly) to find T2.
Attempt:
a. I said it has to be at constant V and high temperature. Not what it was asking
b. It said it was an insulated piston? Doesnt that mean that T1=T2?? But I have really no idea how to do it. :/
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Stephen
Reputation
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Next privilege 250 Rep.
Dec 6 awarded Yearling Nov 1 awarded Notable Question Jul 2 awarded Curious Apr 10 awarded Notable Question Mar 4 awarded Notable Question Jun 20 awarded Caucus May 15 awarded Popular Question Mar 13 awarded Popular Question Feb 7 awarded Popular Question Jan 1 awarded Nice Question Jul 8 asked How can I locate empty space next to polygon regions? Jul 7 awarded Critic Jul 7 comment How can I generate a 2d navigation mesh in a dynamic environment at runtime? Great answer, @Mathew. And you should definitely read that paper! It's easy to follow and explains a great technique (especially Appendix A which talks about agent based discovery/generation of the mesh). I am coding a version of this algorithm for javascript, and it is coming along well. I'll post it as an answer when it is done. Jul 7 accepted How can I generate a 2d navigation mesh in a dynamic environment at runtime? Jul 7 comment How can I move a polygon edge 1 unit away from the center? Marking correct because this overall conversation in the comments worked. Upvoted David's answer. Jul 7 accepted How can I move a polygon edge 1 unit away from the center? Jul 6 comment How can I move a polygon edge 1 unit away from the center? @David, if I understand your comment: Could I find the midpoint between two vertices, subtract the center of the polygon, find the unit vector of that, and than add it to both original points? Jul 6 comment How can I move a polygon edge 1 unit away from the center? Are you saying that adding the UnitVector of the original vectors will always make the vectors move outward from the center of the polygon? Jul 6 comment How can I move a polygon edge 1 unit away from the center? it's javascript. In javascript, `undefined` evaluates false, so if the array index is undefined, the left side of the `||` condition is false, so the right side is used. Jul 6 asked How can I move a polygon edge 1 unit away from the center?
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## On the sizes of large subgraphs of the binomial random graph
10 Apr 2019 · Jozsef Balogh, Maksim Zhukovskii ·
We consider the binomial random graph $G(n,p)$, where $p$ is a constant, and answer the following two questions. First, given $e(k)=p{k\choose 2}+O(k)$, what is the maximum $k$ such that a.a.s.~the binomial random graph $G(n,p)$ has an induced subgraph with $k$ vertices and $e(k)$ edges?.. We prove that this maximum is not concentrated in any finite set (in contrast to the case of a small $e(k)$). Moreover, for every constant $C>0$ and every $\omega_n\to\infty$, a.a.s.~the size of the concentration set belongs to $(C\sqrt{n/\ln n},\omega_n\sqrt{n/\ln n})$. Second, given $k>\varepsilon n$, what is the maximum $\mu$ such that a.a.s.~the set of sizes of $k$-vertex subgraphs of $G(n,p)$ contains a full interval of length $\mu$? The answer is $\mu=\Theta\left(\sqrt{(n-k)n\ln{n\choose k}}\right)$. read more
PDF Abstract
# Code Add Remove Mark official
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Combinatorics
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a parachutist with a camera, both descending at a speed of 11.5 m/s, releases the camera at an altitude of 45.8m
1. what is the magitude of the velocity of the camera just before it hits the ground? the accelartion of gravity is 9.8 m/s^2 and air friction is negligible answer in m/s
2. how long does it take the camera to reach the ground? in sec
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# In the situations that we will encounter, we may generally calculate the probability of an event
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## Transcription
1 What does it mean for something to be random? An event is called random if the process which produces the outcome is sufficiently complicated that we are unable to predict the precise result and are instead able to determine just a range of possible outcomes. It is quite easy to come up with examples of random processes in gambling (e.g. dealing from a shuffled deck of cards, rolling a pair of dice, spinning a roulette wheel), but there are plenty of examples we encounter daily where the mechanism that is generating the outcome is too complicated for us to predict with precision the outcome (e.g. how long it takes to make a commute, the weather forecast, the number of employees which are out sick in a given workday). It is precisely these types of situations that are the motivation for developing mathematical notation to make predictions about the future. A random variable is a mathematical construction which represents a random experiment or process. Usually we represent random variables by capital letters X, Y, Z,.... We will be thinking of random variables as roulette wheels which keep track of the possible outcomes and the frequency which those outcomes occur. Random variables have a range of possible values that they may take on (the numbers found on the edge of the wheel) and positive values representing the percentages that these values occur. The elementary outcomes are the individual possible values that the random variable may take on. An event is a subset of elementary outcomes and the field of events is the set of all possible events for this random variable. The empty set and the set of all elementary outcomes are always included in the field of events. The field of events is assumed to be closed under intersection, union and complementation. A probability measure is a function which associates to each event a number P [A] which will be a value in the interval from 0 to 1. This number reflects the confidence that the event A will occur where a value of 0 represents no chance that the event will occur and 1 that the event happens every time the experiment is conducted. The roulette representation of a random variable records the In general, we must have that if A and B are mutually exclusive events (the intersection of A and B is empty), then P [A or B] = P[A] + P[B]. by In the situations that we will encounter, we may generally calculate the probability of an event P [A] = the number of outcomes considered as part of A the total number of elementary outcomes A random variable for us will be represented as a wheel where the outside edge of this wheel 1
2 is labelled with the elementary outcomes and the portion of the arc of the circle occupied by any given elementary outcome is equal to the probability that the outcome occurs. We imagine that this wheel spins with a pointer lying somewhere on the outer edge (which we will not draw) and that when the random experiment is conducted we imagine that it is equivalent to giving this wheel a spin a recording the outcome that is lying next to the pointer. Assuming that there are not too many outcomes and the probability that they occur are within a range of pen and paper, this roulette wheel is a visual presentation of the information contained in the random variable. If the number of outcomes of the random variable are too large to draw or their probabilities too small (e.g. a random variable representing the outcomes of a lottery where there are roughly 14 million possiblilities), then we do not actually draw the wheel and instead must imagine that small fractions of an enormously large wheel could be drawn. Example: If our random variable represented the flipping of a coin then there are two possible elementary outcomes, { heads, tails }. There are four possible events which describes our field of events heads, tails, either heads or tails, neither heads nor tails. This random variable is visually represented by a roulette wheel with two equal sized regions. One half of the roulette wheel is labeled with heads and the other half is labelled with tails. Example: If our random variable represented rolling a 6 sided die, the elementary outcomes would be the set {1, 2, 3, 4, 5, 6}. Examples of events might be the roll was even or roll greater than 4 or more simply a roll of 3. The roulette wheel representing a roll of a die has six regions. Each sixth of the roulette wheel is labeled with one of the elementary outcomes of the rolling of a die. Odds are the not the opposite of evens 2
3 In gambling we also refer to the probability of an event through the odds. The odds of an event is the ratio of the probability of an event occurring to the event not occurring. Usually this is expressed in the form number to number or number : number (occassionally, and you need to be careful in this use, they are also expressed as number in number ). If the odds of an event are m to n then if m is bigger than n then it will occur more than 50 percent of the time, if m is less than n then the event will occur less than 50 percent of the time. We may also talk about the odds against an event happening. If the odds for an event happening are m to n then the odds against it happening are n to m. Example: Here is a mis-use of the vocabulary of odds: Sir, the possibility of successfully navigating an asteroid field is approximately 3,720 to 1! More precisely he should have said the odds against successfully navigating an asteroid field are approximately 3,720 to 1 Never tell me the odds. The advantage of using this notation is that when we are speaking, it is generally understood if the event is likely or unlikely to occur, so mixing up the order of the numbers is generally not too confusing. However, it is important to be precise when describing the rules of a game and the odds of 5:4 or 4:5 are not far off. Example: The OLG and most lottery boards state the odds of winning the lottery are 1 in number and they really mean that the probability of winning is 1/number. When lottery boards say they are using the notation of odds, they usually aren t. Example: The event of rolling a 3 on a die has probability 1/6. The odds that this event happens is 1 to 5 or 1 : 5 because the probability of the event happening is 1/6, the probability of the event not happening is 5/6, and so the odds are 1/6 5/6 = 1/5 which we then express as 1 : 5 or 1 to 5. Example: The odds of getting heads on the flip of a coin is 1 : 1. there are even odds on the event. In this case we say that How to go back and forth between odds and probability If p is the probability of an event A, then the odds of A occurring are p/(1 p) expressed in the form m to n where p/(1 p) = m/n Example: The probability of rolling two 1 s on a pair of dice is 1/36. The odds of this event occurring are = 1/35. So we say that the odds of rolling snake eyes is 1 to 35. 1/36 1 1/36 = 1/36 35/36 Example: If we bet the next number on a roulette spin will be a fixed number that we choose from 1-36 or 0 (European roulette wheel), then the probability that our choice is right is 1/37. The 3
4 odds that we win this bed is then 1/37 36/37 = 1/36 and so we say they are 1 to 36. Example: The probability of dealing out two cards and both of them are Aces is 6/1326 = 1/221. To find the odds we compute the fraction 1/ /221 = 1/ /221 = 1/220 and then we convert this to odds notation by 1 : 220 or instead say 1 to 220. If the odds of an event are m to n then the probability of the event occurring is because p/(1 p) = m/n so pn = m mp and pn + pm = m so p = m m+n. m m+n. This is Example: If you look on betting websites they will often provide you with the odds of winning a given bet. For instance at the website the odds of the Field bet are posted at 5 : 4. Technically, what is being posted are the odds against winning the bet. The probability of winning the bet is 4/(4 + 5) = 4/9. Odds on a payout We also use the notation of odds to talk about payouts. When we say a bet pays 35 to 1 odds this means that the bet pays \$35 for every \$1 that was bet. When you bet that \$1 and you win, you also get it back so you will actually be returned \$36. If you lose that bet, then the \$1 is gone. At a gambling table the bets are often made by using chips. In those cases when the odds are listed where the second number is not 1 then it is most often the case that the bets must be made in multiples of that number. For example, on the Horn bet in craps the payout is 27 : 4 on a 2 or 12. The bettor must place 4 chips on this bet and when the bet wins by getting the 2 or 12 he/she is returned 31 chips in total, 27 more than the 4 that were bet. For casino bets the rules are set in the game so that the payout of a given bet favors the casino. If a bet is fair and the odds of winning the bet were m : n and there is only one type of payout (sometimes there are different payouts for different outcomes and the situation is a lot more complicated), then the payout should be n : m. In a casino, since the bet is not fair, the payout will be somewhat less than n : m. How much less will indicate the size of the cut that the casino is taking. There is no set rule about how casinos determine what the payoff odds are. They set them so that the bets are close enough to a fair game that they will get people taking those bets, but if they can get away with giving a payout which is very unfavorable to a bettor they will set the odds as they like. On some level gamblers are still customers in a casino and if another casino down the street is offering better odds then a casino will lose those customers. 4
5 For instance, in roulette we found that the odds of winning the bet on a single number is 1 : 36. The payout odds on this bet are 35 : 1. The house is taking a cut on this bet by making the payout odds 35 to 1 when the odds of winning the bet are 1 to 36. Similarly, the odds of winning the corner bet in roulette (betting that one of 4 numbers will come up in the next roulette spin) is 1 : 8 but the payout of this bet is 7 : 1. What we call the house advantage (see below) on these two bets is the same, but that is not easy to see just by looking at the odds alone. Everyone expects to win, only the house does for sure When the elementary outcomes of a random variable are all numbers then it makes sense to talk about the average value or the expected value of the random variable. For a random variable X with elementary outcomes {x 1, x 2, x 3,..., x n }, the expected value (denoted E(X)) is defined to be E(X) = x 1 P (X = x 1 ) + x 2 P (X = x 2 ) + x 3 P (X = x 3 ) + + x n P (X = x n ). This quantity has some intuitive meaning as sort of the average value that occurs if the experiment is repeated many times. Depending on what the random variable represents, this might not give us much intuition about the outcome of the experiment, but when the random variable represents a bet the expected value does give us some intuition about the bet. When our random variable represents a gambling bet the expected value has a direct interpretation because in this case the elementary outcomes are the amount of money which is won or lost for each of the events in the bet. In this case, the expected value represents the amount of money that we should expect to win or lose on average per game that is played assuming that the bet is played many times. This does not mean that if we play the game many times that we should 5
6 expect to win (the expected value) times (the number of bets), instead what it means is that the more times we play (the amount won or lost) divided by (the number of bets) should get closer and closer to the expected value. This will be made more precise when we study the weak law of large numbers. If the expected value of a bet is positive, then the bet favors the player. If the expected value of a given bet is negative then the bet favors the entity the bet is made against (e.g. the house, the bookie, the opponent). If the expected value of the bet is 0 then we say it is a fair bet. When the expected value is negative, the house advantage is negative of the expected value expressed as a percentage of a \$1 bet. Example: The field bet is made on a single roll of a pair of dice. On a \$1 bet, if the dice add up to 2 or 12 the player wins \$2. If the dice add up to 3, 4, 9, 10, 11 the player wins \$1, otherwise the player loses the \$1 bet. The probability that the roll will be 2 or 12 is 2/36. The probability that the roll is 3, 4, 9, 10, 11 is 14/36. The probability that the roll is 5,6,7 or 8 is 20/36. The expected value of this bet is E(field bet) = = What this says is that the bet favors the house and the house advantage is approximately 5.6%. Example: the expected value of a coin toss doesn t make sense because the elementary outcomes of a coin toss are heads or tails. Since these are not numerical values the expected value doesn t represent anything. However, if we put a bet on the outcome of a flip of a coin such as If the outcome is heads I owe you \$100, and if the outcome is tails you owe me \$100. In this case the expected value of the bet is E(\$100 on the flip of a coin) = = 0. This is what we would call a fair bet. The expected value is an important single number that in a way summarizes something about the bet (or random variable) that is being considered, but a single number cannot tell us everything about a bet. Consider the following example: Example: Say that you need to raise \$1,000,000 by a deadline and choose to do it by betting. The positive outcomes of the following three bets will solve your problem by raising the \$1,000,000. Some of the outcomes in the bets could cause you to be in debt for the rest of your life. Say that you are given the choice of the following: a bet where you are given a 1 in 11 chance of losing \$10,000,000 but a 10 in 11 chance of winning the \$1,000,000. 6
7 a bet where you have a 1 in 2 chance of losing \$1,000,000 and a 1 in 2 chance of winning \$1,000,000. a 999,999 in 1,000,000 chance of losing \$1 and a 1 in a 1,000,000 chance of winning \$1,000,000. In all three of these bets the expected value is 0, but the outcomes are very different. A lot of information about these bets is lost by considering only the expected values of the bets. In each case, there is some positive probability of achieving the \$1,000,000 outcome. In the first two bets you don t care what happens in the long run, you probably would not be able to make that bet more than once. Probability when you know something We know that the probability of any single number coming up on roulette is 1/37 (assuming a European roulette wheel) so the probability of seeing the number 14 (in particular) on the next roll is 1/37. Imagine a roulette wheel is rigged so that although we don t know the next spins value, we do know that the next spin is red. In this case, since 14 is red it is still possible that the result is 14, but the information that we are given has now changed the probability that 14 will be the next spin because it is a red value. Since there are now 18 red values, one of them is to come up we can determine that the probability of getting a 14 given that the next spin is red is 1/18. This is called the conditional probability of the event that a 14 comes up given that the next value is red. There are other situations where we are given some information about the outcome, for example what are the chances that it will rain in the next 24 hours given that it is cloudy now. Other times if we don t have the outcome we like, we repeat the experiment again until we have an outcome that falls into an event that we do like. For example, in the game of craps we might roll a pair of dice until their is either a 5 or a 7 showing, this would be a situation where we would ask the probability of an event given that the roll is either a 5 or a 7. Both of these are situations are examples we would like to determine the probability of an outcome given another outcome. If we consider random variable as a wheel then the conditional probability can be represented by forbidding certain parts of the wheel of coming up or spinning the wheel again if the forbidden results happen to come up. The resulting wheel is then crippled by blackening in or excluding certain results from arising. Consider the example mentioned above where we know that the next spin of the wheel is red. We would represent this on our roulette wheel by blackening in the possible outcomes that are forbidden as in the image below: 7
8 We imagine that if we spin this roulette wheel that we will respin if the outcome lies in one of the the blackened in areas. In this case we are left with 18 spots where the roulette wheel is allowed to stop and those 18 spots occupy still one 37 th of the total portion of the wheel. What we do to that wheel is take out the blackened in areas and stretch the remaining parts so that they take up a proportional amount of the resulting wheel. The wheel above then represents the conditional random variable of the spin of the roulette wheel given the that the spin is red. 8
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# Sum of hybrid data frames depending on multiple conditions in R
This is a more complex follow-up to my previous question. The answer there was to use a matrix, but that doesn't work with data frames having values of different modes.
I want to combine data frames of different sizes, with character and integer columns, and calculate their sum depending on multiple conditions.
## Conditions
1. sums are only calculated for those rows that have a matching "Name"-value
2. sums are calculated for matching column names only
3. if a cell in `df4` is not 0 and not NA, the sum should be `df3 + df4`
4. else the sum should be `df1 + df2 + df3`
## Example
``````> df1 <- data.frame(Name=c("Joe","Ann","Lee","Dan"), "1"=c(0,1,5,2), "2"=c(3,1,0,0), "3"=c(2,0,2,2), "4"=c(2,1,3,4))
> df1
Name X1 X2 X3 X4
1 Joe 0 3 2 2
2 Ann 1 1 0 1
3 Lee 5 0 2 3
4 Dan 2 0 2 4
> df2 <- data.frame(Name=c("Joe","Ann","Ken"), "1"=c(3,4,1), "2"=c(2,3,0), "3"=c(2,4,3))
> df2
Name X1 X2 X3
1 Joe 3 2 2
2 Ann 4 3 4
3 Ken 1 0 3
> df3 <- data.frame(Name=c("Lee","Ben"), "1"=c(1,3), "2"=c(3,4), "3"=c(4,3))
> df3
Name X1 X2 X3
1 Lee 1 3 4
2 Ben 3 4 3
``````
The condition depends on this frame:
``````> df4 <- data.frame(Name=c("Lee","Ann","Dan"), "1"=c(6,0,NA), "2"=c(0,0,4), "3"=c(0,NA,0))
> df4
Name X1 X2 X3
1 Lee 6 0 0
2 Ann 0 0 NA
3 Dan NA 4 0
``````
With the above examples, this is the expected result (* values depend on df4):
``````> dfsum
Name X1 X2 X3 X4
1 Joe 3 5 4 2
2 Ann 5 4 4 1
3 Lee 7* 3 6 3
4 Dan 2 4* 2 4
5 Ken 1 0 3 NA
6 Ben 3 4 3 NA
``````
## Possible steps?
First expand df1, df2, df3, df4 to 5 columns and 6 rows, fill missing data with NA.
Then for each data frame:
1. sort rows by "Name"
2. separate "Name" column from "X1"..."X4"
3. transform "X1"..."X4" columns to matrix
4. calculate sums of the matrices like in the answer to my other question but with the additional condition 1
5. transform result matrix to data frame
6. cbind the "Name" column with the result data frame
## How can this be done in R?
### Solution
@Ricardo Saporta's solution works with little changes:
Add `, padValue=NA)` in the four addCols().
As answered here, replace the definitions of sumD3D4 and dtsum with:
``````plus <- function(x) {
if(all(is.na(x))){
c(x[0],NA)} else {
sum(x,na.rm = TRUE)}
}
sumD3D4 <- setkey(rbind(dt3, dt4)[,lapply(.SD, plus), by = Name], "Name")
dtsum <- setkey(rbind(dt1, dt2, dt3)[, lapply(.SD, plus), by=Name], "Name")
``````
-
It appears that the only role of the `character`s are the names. Is that correct ? If so, you can still use the matrix method recommended and apply the strings to `rownames(mtrx)` – Ricardo Saporta Feb 23 '13 at 19:29
Also, regarding the different number of columns, would it be okay to add dummy columns to filled with 0's so that all df's are the same width? – Ricardo Saporta Feb 23 '13 at 19:46
They should be NA, but is it possible to treat NAs as 0's when calculating the sum of two matrices? – R-obert Feb 23 '13 at 20:14
If you use data.table instead of data.frame, you could use its `by=xxxx` feature, to add by name. The code below should give you your expected results.
Please note that I am padding the data.tables with extra empty columns. However, we compute `condTrue` prior to then.
``````library(data.table)
dt1 <- data.table(df1)
dt2 <- data.table(df2)
dt3 <- data.table(df3)
dt4 <- data.table(df4)
# make sure all dt's have the same columns
#-----------------------------------------#
# identify which dt4 satisfy the condition
condTrue <- as.data.table(which(!(is.na(dt4) | dt4==0), arr.ind=TRUE))
# ignore column "Name" from dt4
condTrue <- condTrue[col>1]
# convert from (row, col) index to ("Name", columnName)
condTrue <- data.table(Name=dt4[condTrue\$row, Name], colm=names(dt4)[condTrue\$col], key="Name")
# First make a list of all the unique column names
allColumnNames <- unique(c(names(dt1), names(dt2), names(dt3), names(dt4)))
sumD3D4 <- setkey(rbind(dt3, dt4)[, lapply(.SD, sum), by=Name], "Name")
dtsum <- setkey(rbind(dt1, dt2, dt3)[, lapply(.SD, sum), by=Name], "Name")
for (Nam in condTrue\$Name) {
colsRepl <- condTrue[.(Nam)]\$colm
valsRepl <- unlist(sumD3D4[.(Nam), c(colsRepl), with=FALSE])
dtsum[.(Nam), c(colsRepl) := as.list(valsRepl)]
}
dtsum
# Name 1 2 3 4
# 1: Ann 5 4 4 1
# 2: Ben 3 4 3 0
# 3: Dan 2 4 2 4
# 4: Joe 3 5 4 2
# 5: Ken 1 0 3 0
# 6: Lee 7 3 6 3
``````
``````addCols <- function(x, cols, padValue=0) {
# adds to x any columns that are in cols but not in x
# Returns TRUE if columns were added
# FALSE if no columns added
colsMissing <- setdiff(cols, names(x))
# grab the actual DT name that was passed to function
dtName <- as.character(match.call()[2])
if (length(colsMissing)) {
@R-obert, your comment above said you wanted to treat the NA's as 0's. Note that the NA's you are looking for are present as 0's in column 4. If you would like to use NA's instead, simply add `, padValue=NA)` in the four `addCols()` statements. Note however, that it will then make all NA cols into NA's in the results. If you want the NA's to be passed only selectively, you need to add more conditional statements above. – Ricardo Saporta Feb 24 '13 at 2:39
My comment above was unclear. I would like to use NA's but it still doesn't work. Adding `, padValue=NA)` results in: `Error in [.data.table'(dtsum, .(Nam), ':='(c(colsRepl), as.list(valsRepl))): Type of RHS ('double') must match LHS ('integer'). To check and coerce would impact performance too much for the fastest cases. Either change the type of the target column, or coerce the RHS of := yourself (e.g. by using 1L instead of 1)` – R-obert Feb 24 '13 at 15:12
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# Number 1733949
Number 1,733,949 spell 🔊, write in words: one million, seven hundred and thirty-three thousand, nine hundred and forty-nine , approximately 1.7 million. Ordinal number 1733949th is said 🔊 and write: one million, seven hundred and thirty-three thousand, nine hundred and forty-ninth. The meaning of number 1733949 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 1733949. What is 1733949 in computer science, numerology, codes and images, writing and naming in other languages
## What is 1,733,949 in other units
The decimal (Arabic) number 1733949 converted to a Roman number is (M)(D)(C)(C)(X)(X)(X)MMMCMXLIX. Roman and decimal number conversions.
#### Weight conversion
1733949 kilograms (kg) = 3822664.0 pounds (lbs)
1733949 pounds (lbs) = 786514.1 kilograms (kg)
#### Length conversion
1733949 kilometers (km) equals to 1077426 miles (mi).
1733949 miles (mi) equals to 2790522 kilometers (km).
1733949 meters (m) equals to 5688740 feet (ft).
1733949 feet (ft) equals 528515 meters (m).
1733949 centimeters (cm) equals to 682657.1 inches (in).
1733949 inches (in) equals to 4404230.5 centimeters (cm).
#### Temperature conversion
1733949° Fahrenheit (°F) equals to 963287.2° Celsius (°C)
1733949° Celsius (°C) equals to 3121140.2° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
1733949 seconds equals to 2 weeks, 6 days, 1 hour, 39 minutes, 9 seconds
1733949 minutes equals to 3 years, 7 months, 3 hours, 9 minutes
### Codes and images of the number 1733949
Number 1733949 morse code: .---- --... ...-- ...-- ----. ....- ----.
Sign language for number 1733949:
Number 1733949 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 1733949
### Multiplications
#### Multiplication table of 1733949
1733949 multiplied by two equals 3467898 (1733949 x 2 = 3467898).
1733949 multiplied by three equals 5201847 (1733949 x 3 = 5201847).
1733949 multiplied by four equals 6935796 (1733949 x 4 = 6935796).
1733949 multiplied by five equals 8669745 (1733949 x 5 = 8669745).
1733949 multiplied by six equals 10403694 (1733949 x 6 = 10403694).
1733949 multiplied by seven equals 12137643 (1733949 x 7 = 12137643).
1733949 multiplied by eight equals 13871592 (1733949 x 8 = 13871592).
1733949 multiplied by nine equals 15605541 (1733949 x 9 = 15605541).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 1733949
Half of 1733949 is 866974,5 (1733949 / 2 = 866974,5 = 866974 1/2).
One third of 1733949 is 577983 (1733949 / 3 = 577983).
One quarter of 1733949 is 433487,25 (1733949 / 4 = 433487,25 = 433487 1/4).
One fifth of 1733949 is 346789,8 (1733949 / 5 = 346789,8 = 346789 4/5).
One sixth of 1733949 is 288991,5 (1733949 / 6 = 288991,5 = 288991 1/2).
One seventh of 1733949 is 247707 (1733949 / 7 = 247707).
One eighth of 1733949 is 216743,625 (1733949 / 8 = 216743,625 = 216743 5/8).
One ninth of 1733949 is 192661 (1733949 / 9 = 192661).
show fractions by 6, 7, 8, 9 ...
### Calculator
1733949
#### Is Prime?
The number 1733949 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 1733949 are 3 * 3 * 7 * 17 * 1619
The factors of 1733949 are
1 , 3 , 7 , 9 , 17 , 21 , 51 , 63 , 119 , 153 , 357 , 1071 , 1619 , 4857 , 11333 , 14571 , 27523 , 33999 , 82569 , 101997 , 1733949 show more factors ...
Total factors 24.
Sum of factors 3032640 (1298691).
#### Powers
The second power of 17339492 is 3.006.579.134.601.
The third power of 17339493 is 5.213.254.883.862.268.928.
#### Roots
The square root √1733949 is 1316,794973.
The cube root of 31733949 is 120,137551.
#### Logarithms
The natural logarithm of No. ln 1733949 = loge 1733949 = 14,365912.
The logarithm to base 10 of No. log10 1733949 = 6,239036.
The Napierian logarithm of No. log1/e 1733949 = -14,365912.
### Trigonometric functions
The cosine of 1733949 is -0,942111.
The sine of 1733949 is -0,335302.
The tangent of 1733949 is 0,355906.
### Properties of the number 1733949
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 1733949 in Computer Science
Code typeCode value
1733949 Number of bytes1.7MB
Unix timeUnix time 1733949 is equal to Wednesday Jan. 21, 1970, 1:39:09 a.m. GMT
IPv4, IPv6Number 1733949 internet address in dotted format v4 0.26.117.61, v6 ::1a:753d
1733949 Decimal = 110100111010100111101 Binary
1733949 Decimal = 10021002112100 Ternary
1733949 Decimal = 6472475 Octal
1733949 Decimal = 1A753D Hexadecimal (0x1a753d hex)
1733949 BASE64MTczMzk0OQ==
1733949 MD55359f4f6d339673d7917b711b1064e61
1733949 SHA101d2979fe999c1b626a5cafa10ae61717a7bca5d
1733949 SHA22498fa6b89bb48cc63a4fdf8822acfdc59dba1561ba78227d3125a81c7
1733949 SHA25622fa8b11ac3c6f7f3a3f88c05c4fe73c76e4bff2c4841edd598e2c2833e0d8d9
1733949 SHA3848af8b67e038ed16b9dd29ed72e3506074c7900fb24183794baa0c4946fb0b2abff6d740d00b210a2727e1a7c025b46ca
More SHA codes related to the number 1733949 ...
If you know something interesting about the 1733949 number that you did not find on this page, do not hesitate to write us here.
## Numerology 1733949
### Character frequency in number 1733949
Character (importance) frequency for numerology.
Character: Frequency: 1 1 7 1 3 2 9 2 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 1733949, the numbers 1+7+3+3+9+4+9 = 3+6 = 9 are added and the meaning of the number 9 is sought.
## № 1,733,949 in other languages
How to say or write the number one million, seven hundred and thirty-three thousand, nine hundred and forty-nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 1.733.949) un millón setecientos treinta y tres mil novecientos cuarenta y nueve German: 🔊 (Anzahl 1.733.949) eine Million siebenhundertdreiunddreißigtausendneunhundertneunundvierzig French: 🔊 (nombre 1 733 949) un million sept cent trente-trois mille neuf cent quarante-neuf Portuguese: 🔊 (número 1 733 949) um milhão, setecentos e trinta e três mil, novecentos e quarenta e nove Chinese: 🔊 (数 1 733 949) 一百七十三万三千九百四十九 Arabian: 🔊 (عدد 1,733,949) واحد مليون و سبعمائة و ثلاثة و ثلاثون ألفاً و تسعمائة و تسعة و أربعون Czech: 🔊 (číslo 1 733 949) milion sedmset třicet tři tisíce devětset čtyřicet devět Korean: 🔊 (번호 1,733,949) 백칠십삼만 삼천구백사십구 Dutch: 🔊 (nummer 1 733 949) een miljoen zevenhonderddrieëndertigduizendnegenhonderdnegenenveertig Japanese: 🔊 (数 1,733,949) 百七十三万三千九百四十九 Indonesian: 🔊 (jumlah 1.733.949) satu juta tujuh ratus tiga puluh tiga ribu sembilan ratus empat puluh sembilan Italian: 🔊 (numero 1 733 949) un milione e settecentotrentatremilanovecentoquarantanove Norwegian: 🔊 (nummer 1 733 949) en million, syv hundre og tretti-tre tusen, ni hundre og førti-ni Polish: 🔊 (liczba 1 733 949) milion siedemset trzydzieści trzy tysiące dziewięćset czterdzieści dziewięć Russian: 🔊 (номер 1 733 949) один миллион семьсот тридцать три тысячи девятьсот сорок девять Turkish: 🔊 (numara 1,733,949) birmilyonyediyüzotuzüçbindokuzyüzkırkdokuz Thai: 🔊 (จำนวน 1 733 949) หนึ่งล้านเจ็ดแสนสามหมื่นสามพันเก้าร้อยสี่สิบเก้า Ukrainian: 🔊 (номер 1 733 949) один мiльйон сiмсот тридцять три тисячi дев'ятсот сорок дев'ять Vietnamese: 🔊 (con số 1.733.949) một triệu bảy trăm ba mươi ba nghìn chín trăm bốn mươi chín Other languages ...
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CS 345
Parsing, Lexical Analysis, and Tools
William Cook
Parsing techniques
Top-Down
Begin with start symbol, derive parse tree Match derived non-terminals with sentence Use input to select from multiple options
Bottom Up
Examine sentence, applying reductions that match Keep reducing until start symbol is derived Collects a set of tokens before deciding which production to use
Top-Down Parsing
Recursive Descent
Interpret productions as functions, nonterminals as calls Must predict which production will match
looks ahead at a few tokens to make choice
Handles EBNF naturally Has trouble with left-recursive, ambiguous grammars
left recursion is production of form E ::= E
Also called LL(k) scan input Left to right use Left edge to select productions use k symbols of look-ahead for prediction
Recursive Descent LL(1) Example
Example
E ::= E + E | E E | T note: left recursion T ::= N | ( E ) N ::= { 0 | 1 | | 9 } { } means repeated
Problems:
Cant tell at beginning whether to use E + E or E - E
would require arbitrary look-ahead But it doesnt matter because they both begin with T
Recursive Descent LL(1) Example
Example
E ::= T [ + E | E ] T ::= N | ( E ) N ::= { 0 | 1 | | 9 }
[ ] means optional
Solution
Combine equivalent forms in original production: E ::= E + E | E E | T There are algorithms for reorganizing grammars
cf. Greibach normal form (out of scope of this course)
LL Parsing Example
E23+7 T23+7 N23+7 23+7 23+7 23+E7 23+T7 23+N7 23+7
E ::= T [ + E | E ] T ::= N | ( E ) N ::= { 0 | 1 | | 9 } = Current location Preduction indent = function call
Intuition: Growing the parse tree from root down towards terminals.
Recursive Descent LL(1) Psuedocode
procedure E() // E ::= T [ + E | E ] a = T(); if next token is + then b = E(); return add(a, b) if next token is - then b = E(); return subtract(a, b) else return a procedure T() // T ::= N | ( E ) if next token is ( then a = E(); check next token is ); return a; else return N(); procedure N() // N ::= { 0 | 1 | | 9 } while next token is digit do
Bottom-Up Parsing
Shift-Reduce
Examine sentence, applying reductions that match Keep reducing until start symbol is derived
Technique
Analyze grammar for all possible reductions Create a large parsing table (never done by hand)
Also called LR(k) scan input Left to right use Right edge to select productions usually only k=1 symbols of look-ahead needed
LR Parsing Example
23+7 23+7 D3+7 N3+7 N3+7 ND+7 N+7 T+7 E+7 E+7 E+7
E+D
E+N
E+T
E+E
E
E ::= E + E | E E | T T ::= N | ( E ) N ::= N D | D D ::= 0 | 1 | | 9 = Current location Shift step Reduce step
Intuition: Growing the parse tree from terminals up towards root.
Conficts
Problem
Sometimes multiple actions apply
Shift another token / Reduce by rule R Reduce by rule A / Reduce by rule B
10
Flagged as a conflict when parsing table is built
Resolving conflicts
Rewrite the grammar Use a default strategy
Shift-reduce: Prefer shifting Reduce-reduce: Use first rule in written grammar
Use a token-dependent strategy
There's a nice way to do this
Confict Example
E*E+ E*E+ (shift) E+ (reduce) E+E+ (shift) E+ (reduce)
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E+E+
What does each resolution direction do? Where have we seen this problem before?
Directives
Precedence
Establish a token order: * binds tighter than +
Doesn't need to be given for all tokens If unordered tokens conflict, use default strategy
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Associativity
Left-associative: favor reduce Right-associative: favor shift Non-associative: raise error
Flags inherently confusing expressions Consider: a b c
Parser Generators
Parser Generators
Input is a form of BNF grammar
Include actions to be performed as rules are recognized
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Output is a parser
Examples
ANTLR, JavaCC
generate recursive descent parsers
Yacc (many versions: CUP for Java)
generates bottom-up (shift-reduce) parsers
ANTLR Example
grammar Exp;
14
add returns [double value] : m1=prim {\$value = \$m1.value;} ( '+' m2=prim {\$value += \$m2.value;} | '-' m2=prim {\$value -= \$m2.value;} )*; prim returns [double value] : n=Number {\$value = Double.parseDouble(\$n.text);} | '(' e=add ')' {\$value = \$e.value;} ; Number : ('0'..'9')+ ('.' ('0'..'9')+)? ; WS : (' ' | '\t' | '\r'| '\n') {\$channel=HIDDEN;} ;
ANTLR Example creating AST
grammar Exp;
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add returns [Exp value] : m1=prim {\$value = \$m1.value;} ( '+' m2=prim)* {\$value = new Add(\$value, \$m2.value);} ; prim returns [Exp value] : n=Number {double x = Double.parseDouble(\$n.text); \$value = new Num(x);} | '(' e=add ')' {\$value = \$e.value;} ; Number : ('0'..'9')+ ('.' ('0'..'9')+)? ; WS : (' ' | '\t' | '\r'| '\n') {\$channel=HIDDEN;} ;
Simplified AST without closures
interface Exp { int interp(); } class Num implements Exp { int n; public Num(int n) { this.n = n; } public int interp() { return n; } } class Add implements Exp { Exp l, r; public Add (Exp l, r) { this.l = l; this.r = r; } public int interp() { return l.interp() + r.interp(); } }
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# Making A Storm Wall
You can write your topic however you want, but you need to answer these questions:
1. What do you want to achieve? Keep it simple and clear!
I need to show a circular storm barier
2. What is the issue? Include screenshots / videos if possible!
i dont know how i would show this
3. What solutions have you tried so far? Did you look for solutions on the Developer Hub?
i dont know how to start
After that, you should include more details if you have any. Try to make your topic as descriptive as possible, so that it’s easier for people to help you!
Server Script Service:
``````local X = script["XCordnate(Studs)"].Value
local Z = script["ZCordnate(Studs)"].Value
local NX = script["XCordnateNegative(Studs)"].Value
local NZ = script["ZCordnateNegative(Studs)"].Value
local TimeBetweenCircles = script["TimeBetweenCircles(IncludingClosingTime)"].Value
local ClosingTime = script.ClosingTime.Value
--wait(TimeBetweenCircles)
local RX = math.random(NX,X)
local RZ = math.random(NZ,Z)
local Orgin = {RX,RZ}
local Radius = ((30 - PlrsInGame) / (RX + RZ)) * 30 -- take out the 25 upon compleation of the game
game.Workspace.StormStats.OrginX.Value = RX
game.Workspace.StormStats.OrginZ.Value = RZ
game.Workspace.StormStats.FirstDone.Value = true
local Part = Instance.new("Part")
Part.Size = Vector3.new(1,1,1)
Part.Position = Vector3.new(Orgin[1], 20, Orgin[2])
Part.Parent = workspace
``````
this is apart of a larger script but the rest of it is compleately different then this script
Starter Character Scripts:
``````local Rate = 2.5
repeat
if workspace.StormStats.FirstDone.Value == true then
local OrginX = workspace.StormStats.OrginX.Value
local OrginZ = workspace.StormStats.OrginZ.Value
local Distance = (script.Parent.PrimaryPart.Position - Vector3.new(OrginX,script.Parent.PrimaryPart.Position.Y,OrginZ)).magnitude
script.Parent.Humanoid.Health -= Rate
end
end
wait(0.5)
until game.Workspace.GameEnded == true
``````
I just need to show the storm to all the players in this game (I dont need to deal dammage with this im alr doing this in here
You can look at this open source that has a storm wall / closing circle / storm barrier…
It still works…
Or search the above keywords…
Thanks! Ive already almost goten there myself so im going to use my own code but thanks for making this!
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A318207 a(n) is the least prime p such that 2-adic valuation of p-3 is n. 1
2, 5, 7, 11, 19, 163, 67, 131, 1283, 6659, 25603, 10243, 4099, 57347, 114691, 32771, 65539, 3014659, 262147, 5767171, 5242883, 14680067, 71303171, 109051907, 218103811, 436207619, 335544323, 6308233219, 268435459, 9126805507, 1073741827, 130996502531, 21474836483, 403726925827, 85899345923 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS For n >= 1, a(n) is the least prime p such that A007814(p-3) = n, and the least k such that A023572(k) = n is A000720(a(n)). By Dirichlet's theorem on primes in arithmetic progressions, a(n) exists for all n. LINKS Robert Israel, Table of n, a(n) for n = 0..2000 EXAMPLE a(3) = 11 because the highest power of 2 dividing 11-3=8 is 2^3, and 11 is the least prime with this property. MAPLE f:= proc(n) local k; for k from 1 by 2 do if isprime(k*2^n+3) then return k*2^n+3 fi od end proc: f(0):= 2: map(f, [\$0..100]); PROG (PARI) a(n) = forprime(p=1, , if(valuation(p-3, 2)==n, return(p))) \\ Felix Fröhlich, Aug 23 2018 (Python) from sympy import isprime def A318207(n): if n == 0: return 2 a = 1<
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Last modified March 21 15:50 EDT 2023. Contains 361408 sequences. (Running on oeis4.)
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# Fourier Series: Can even functions be changed to odd?
1. Apr 27, 2013
### thelema418
When creating a Fourier series for a function $f(x)$, I consider whether the function is odd or even first. Yet, often these functions are in the positive region $[0, L]$.
Since $f(x)$ is only defined in this region, can I change the function to get a desired parity? By example, my concern originated with the function $f(x) = x \sin x$. This is an even function, but I could modify the function as $f(x) = |x| \sin x$ to make it odd while retaining the desired information in $[0, L]$.
Can this be done? And are there problems with doing this?
Thanks.
2. Apr 27, 2013
### micromass
What do you mean with "retaining the desired information"? What exactly is it that you want to do?
3. Apr 27, 2013
### thelema418
Meaning that $x \sin x$ on $[0,\pi]$ is the same as $|x| \sin x$ on $[0,\pi]$.
The application is usually for solving heat equations. So the region is usually x = 0 to x = L in the problems I see. Again, another example: When I work out all the details of a solution by eigenfunction expansion based on homogeneous Dirichlet boundary conditions, I end up with a Fourier sine series equal to some function $f(x)$, where $f(x)$ is the initial condition u(x,0). But to have a Fourier sine series, $f(x)$ must be an odd function, right?
4. Apr 27, 2013
### micromass
OK. You're right then. The right way to find a sine series converging to $f$ is to extend it to an odd function on entire $[-L,L]$. So you'll have to work with $|x|\sin(x)$ on $[-\pi,\pi]$, like you suggested.
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ghirastauvasile
2
Ajutati-ma va rog: x+(x+36)+3x=
(1) Răspunsuri
elenacool09
x+(x+36)+3x=x+x+36+3x=5x+36
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Association
# Association - Association Click to edit Master subtitle...
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Click to edit Master subtitle style Association 4/13/2011
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Announcements n Chapter 12 HW due on Monday n Research talk this Friday, noon, Gambrell 250
What do we Mean by Association? n Association implies a relationship between two variables. n If you change the X, you will get a change in Y. n Relationship is not necessarily causal.
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Measures of Association n Can describe the relationship between two variables in terms of ¨ Direction (positive or negative) ¨ Form (linear, nonlinear, etc) ¨ Strength ¨ Conditional relationships
Association n Ways of describing association depends upon data types: ¨ Scatterplots ¨ Tables ¨ Correlation ¨ Regression
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Scatterplots n Scatterplot is a visual way of presenting a relationship. n Only work with ratio / interval data.
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Scatterplots n Simply plot, for each case, the values of X on the x axis and Y on the y axis. Dependent variable Independent variable
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Cross tabulation/Contingency Tables n Frequency table with 2 variables ¨ Y categories are rows ¨ X categories are columns n Usually best for nominal and ordinal data
Example n Y= 2000 presidential vote choice ¨ Bush, Gore, other n X= party affiliation ¨ Republican, Democrat, other
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Table 1. 2000 Presidential Vote Choice, by Party Respondent Party Identification Vote Choice Dem Ind Rep Total Gore 505 37 45 587 Bush 55 37 435 527 Other 16 11 15 42 Total 576 85 495 1156 Source: American National Election Study, 2000
Table 1. 2000 Presidential Vote Choice, by Party Respondent Party Identification Vote Choice Democrat Independent Republican Total Gore 505 37 45 587 Bush 55 37 435 527 Other 16 11 15 42 Total 576 85 495 1156 Column Marginal Frequency Row Marginal Frequency n
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Contingency w/ % n Divide cell frequency by column marginal.
Contingency w/ % Respondent Party Identification Vote Choice Democrat Independent Republican Total Gore 88% 44% 9% 51% Bush 10% 44% 88% 46% Other 3% 13% 3% 4% Total 101%* 101% 100% 101% Source: American National Election Study, 2000 *Percentages may not add to 100, due to rounding.
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With % and Marginals Respondent Party Identification Vote Choice Democrat Independent Republican Total Gore 88% 44% 9% 51% (587) Bush 10% 44% 88% 46% (527) Other 3% 13% 3% 4% (42) Total 101%* 101% 100% 101% (576) (85) (495) (1156)
Making Tables n Include descriptive and shorthand title n Provide Source of Data n Make Y Rows and X Columns n Clearly Label Rows and Columns
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Making Tables n In cells, report % of column n Report Row Freq. and % n Report Column Freq. and % n Round Percentages n Report Total Observations (n)
For Interval or Ratio Data n Make categories n Loses information
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n How can assess association with a cross tab? n What would perfect association look like? n
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Ask a homework question - tutors are online
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For discussion of other cellular automata.
This is the thread for your accidental discoveries that aren't in Conway's Game of Life (to complement the other thread). In HighLife, this concept can be used to construct arbitrarily high-period oscillators:
x = 5, y = 12, rule = B36/S23
2b3o\$bo2bo\$o3bo\$o2bo\$3o6\$2b2o\$2b2o!
EDIT: Found out that this doesn't work. This does:
x = 12, y = 13, rule = B36/S23
2b3o\$bo2bo\$o3bo\$o2bo\$3o\$8b2o\$8bobo\$10bo\$10b2o\$5b2o\$5bo\$6b3o\$8bo!
No it doesn't! This actually does:
x = 13, y = 11, rule = B36/S23
2b3o\$bo2bo\$o3bo\$o2bo\$3o3\$9b2o\$9bo\$10b3o\$12bo!
But that only works in certain situations. Two molds can be used in other situations. (That last sentence was based off of David Bell's text.)
This converts a replicator into a bomber (based on the above):
x = 9, y = 13, rule = B36/S23
2b3o\$bo2bo\$o3bo\$o2bo\$3o5\$5b2o\$5bo\$6b3o\$8bo!
LWSS synthesis:
x = 8, y = 18, rule = B36/S23
bo\$2bo2bobo\$3o2b2o\$6bo12\$6b2o\$5b2o\$7bo!
I also accidentally found the "Ruler Rule" posted earlier.
Post anything else of note that is found accidentally and is not in Conway's Game of Life.
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1768
Joined: June 16th, 2009, 11:24 pm
Location: USA
An orthogonal spaceship I found in a rules table rule. Looks like a failed replicator.
Moves at 152c/1334!
MCell format.
#MCell 4.20
#GAME Rules table
#RULE 1,0,1,0,0,4,2,1,1,3,0,0,0,0,0,1,1,3,0,0,0,0,0,0,4,1,2,3,0,0,0,0,
#RULE 0,4,4,3,3,2,0,4,0,0,0,0,0,2,4,4,2
#BOARD 500x500
#WRAP 1
#L 11.CA3.B5CD\$47.BCAA10.AB.D3.DCACB\$48.ABD12.D4.DD.CC.A\$50.AA18.B.D3.D\$
#L 49.D.AA14.A3D.DDCBCDD\$50.BCA..B.A8.D7.3C.CB\$50.D.A..ACB9.DA3.DB.ACD.CD
#L \$27.AA22.DB..DA.D14.DD3CD.D\$26.BBAA23.BA..D12.D3.DDCC3A\$25.DC.BA22.DAC
#L D12.BABD4.BDDBD\$24.BC.CB23.BCAA5.D5.DCC.DD4.AACB\$23.AB.CD24.DBDDB3.D.B
#L B.CD\$15.AABB12.AABB44.AABB\$16.AA14.AA46.AA\$\$27.AA14.AA\$26.BBAA12.BBAA\$
#L 25.DC.BA11.DC.BA\$24.BC.CB11.BC.CB\$23.AB.CD11.AB.CD\$23.AABB12.AABB\$24.
#L AA14.AA\$\$35.AA14.AA\$34.BBAA12.BBAA\$33.DC.BA11.DC.BA\$32.BC.CB11.BC.CB\$
#L 31.AB.CD11.AB.CD\$31.AABB12.AABB\$32.AA14.AA
velcrorex
Posts: 339
Joined: November 1st, 2009, 1:33 pm
P48 oscillator.
replicator shuttle
x = 8, y = 18, rule = B36/S23
bo\$2bo2bobo\$3o2b2o\$6bo12\$6b2o\$5b2o\$7bo!
Call me "Dannyu NDos" in Forum. Call me "Park Shinhwan"(박신환) in Wiki.
David
Posts: 212
Joined: November 3rd, 2009, 2:47 am
Location: Daejeon, South Korea
David wrote:P48 oscillator.
replicator shuttle
x = 8, y = 18, rule = B36/S23
bo\$2bo2bobo\$3o2b2o\$6bo12\$6b2o\$5b2o\$7bo!
I think you pasted the wrong pattern.
Axaj
Posts: 232
Joined: September 26th, 2009, 12:23 am
Axaj wrote:
David wrote:P48 oscillator.
replicator shuttle
x = 8, y = 18, rule = B36/S23
bo\$2bo2bobo\$3o2b2o\$6bo12\$6b2o\$5b2o\$7bo!
I think you pasted the wrong pattern.
Ah, sorry.
x = 23, y = 23, rule = B36/S23
2o\$bo\$bobo\$2b2o9\$12b3o\$11bo2bo\$10bo3bo\$10bo2bo\$10b3o3\$19b2o\$19bo\$20b3o
\$22bo!
Call me "Dannyu NDos" in Forum. Call me "Park Shinhwan"(박신환) in Wiki.
David
Posts: 212
Joined: November 3rd, 2009, 2:47 am
Location: Daejeon, South Korea
A lead on a new p6 c/2 spaceship in B34678/S23?
x = 10, y = 9, rule = B34678/S23
3bo2bo\$3bo3bo\$b4o3bo\$o2bo5bo\$o2bo5bo\$o2bo5bo\$b4o3bo\$3bo3bo\$3bo2bo!
Same idea, except this time it's p4:
x = 7, y = 7, rule = B34678/S23
b2obo\$b2o2bo\$3o3bo\$4o2bo\$3o3bo\$b2o2bo\$b2obo!
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1768
Joined: June 16th, 2009, 11:24 pm
Location: USA
Extrementhusiast wrote:Same idea, except this time it's p4:
x = 7, y = 7, rule = B34678/S23
b2obo\$b2o2bo\$3o3bo\$4o2bo\$3o3bo\$b2o2bo\$b2obo!
Completed:
x = 19, y = 11, rule = 23/34678
13bobb3o\$10bobob5o\$bbobbo3bo3bo3bo\$bo4b3o3bobobo\$o3boo7b3o\$obboo6b4o\$o
3boo7b3o\$bo4b3o3bobobo\$bbobbo3bo3bo3bo\$10bobob5o\$13bobb3o!
-Josh Ball.
velcrorex
Posts: 339
Joined: November 1st, 2009, 1:33 pm
Interestingly enough, if you put a cell on the front (on the above spaceship, so that it makes a T-tetromino), it soon recovers.
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1768
Joined: June 16th, 2009, 11:24 pm
Location: USA
I took two chaotic puffers I found in B01245/S01245, put them in a 20x20 in B012458/S1245 and added some soup. Initial population is 132. They head in opposite directions and the pattern stabilizes at generation 220,479, with a population around 49000, and no living spaceships. Are there many other compact chaotic puffers that stabilize after a long time?
x = 20, y = 20, rule = B012458/S1245
7bo3b4ob2o\$7bo3b3ob2obo\$10bo2b5obo\$7bo4b8o\$7bo2b8obo\$8bob7obo\$8b2o3bob
3o\$7bobob2o\$11b2o\$7bo3bo\$8b5o\$7b3o2bo\$7bo2b2o\$2b3obobob3o\$bob4ob3obo\$o
b5o2b2o\$7obob3o\$ob5ob2o\$bob2ob3obo\$2b2ob2ob2o!
xnihilozero
Posts: 3
Joined: June 22nd, 2010, 2:40 am
I played around some more and got it down to one puffer pop 48 in an 11x9 or 44 in a 12x9 box with a life span of 356,626 and a final pop of 39,975. Could this be considered a world record Methuselah?
#CXRLE Pos=12,-9
x = 11, y = 9, rule = B012458/S1245
2b2ob2o\$bob2o3bo\$ob5o3bo\$9o\$ob5obobo\$bob2ob5o\$2b7o2\$5bo!
xnihilozero
Posts: 3
Joined: June 22nd, 2010, 2:40 am
Tiny 3-cell Byl Loop breeder:
x = 2, y = 3, rule = Byl-Loop
.B\$C\$.B!
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1768
Joined: June 16th, 2009, 11:24 pm
Location: USA
Extrementhusiast wrote:Tiny 3-cell Byl Loop breeder:
x = 2, y = 3, rule = Byl-Loop
.B\$C\$.B!
I found a 2-cell one:
x = 2, y = 2, rule = Byl-Loop
B\$.C!
Four ways!:
x = 5, y = 5, rule = Byl-Loop
3.C\$4.B2\$B\$.C!
ssaamm
Posts: 125
Joined: June 4th, 2010, 9:43 pm
A 2x4 block functions as a two-dimensional replicator in B034678/S012458:
x = 4, y = 2, rule = B034678/S012458
4o\$4o!
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1768
Joined: June 16th, 2009, 11:24 pm
Location: USA
Extrementhusiast wrote:A 2x4 block functions as a two-dimensional replicator in B034678/S012458
That doesn't seem to be in David Eppstein's list of rules with replicators (although I don't know how up-to-date the list is). Also, it seems to work in rules B03/S58 - B01234678/S0123458.
-Matthias Merzenich
Sokwe
Moderator
Posts: 1473
Joined: July 9th, 2009, 2:44 pm
Any pattern with the size doubled (so that each bit is a 2 by 2 block) replicates in that rule: (see generation 8 )
x = 24, y = 22, rule = B034678/S012458
2o2b2o10b2o2b2o\$2o2b2o10b2o2b2o\$4b2o14b2o\$4b2o14b2o\$6b2o14b2o\$6b2o14b
2o11\$2o2b2o10b2o2b2o\$2o2b2o10b2o2b2o\$4b2o14b2o\$4b2o14b2o\$6b2o14b2o\$6b
2o14b2o!
Last edited by Awesomeness on October 28th, 2010, 6:56 am, edited 1 time in total.
Awesomeness
Posts: 126
Joined: April 5th, 2009, 7:30 am
Interesting spiral:
x = 2, y = 3, rule = 1/2/4
A\$2A\$2A!
p118 oscillator that occasionally occurs naturally:
x = 6, y = 4, rule = 238/34/3
.B2.B\$2A2.2A\$.4A\$2.2A!
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1768
Joined: June 16th, 2009, 11:24 pm
Location: USA
A slow glider in a randomly generated rule:
x = 10, y = 10, rule = B014/S2
2\$3b2ob2o\$2b7o\$2b5obo\$3b2o\$2b3o\$2b2o\$3b2o!
The rule itself is extremely organic looking-Is anything known about it?
I guess it's not that interesting,since it seems to have unbounded growth.Still looks nice though Try running it for a while from an initial 6x6 square for example.
BeNikis
Posts: 1
Joined: January 30th, 2011, 3:00 pm
BeNikis wrote:A slow glider in a randomly generated rule:
x = 10, y = 10, rule = B014/S2
2\$3b2ob2o\$2b7o\$2b5obo\$3b2o\$2b3o\$2b2o\$3b2o!
The rule itself is extremely organic looking-Is anything known about it?
I guess it's not that interesting,since it seems to have unbounded growth.Still looks nice though Try running it for a while from an initial 6x6 square for example.
Welcome to the forum!
According to Eppstein's CA Database, there are 9 known gliders on 5 different speeds for that rule: http://fano.ics.uci.edu/ca/rules/b014s2/
It looks like an interesting rule, but I doubt there has been any research on it. There is little known about B0/S!8 rules.
ebcube
Posts: 124
Joined: February 27th, 2010, 2:11 pm
Still, it seems like it could be an interesting rule. Let's see if we can find any logic gates in B014/S2...
137ben
Posts: 343
Joined: June 18th, 2010, 8:18 pm
Chaotic fuse:
x = 6730, y = 3, rule = B02/S
o\$b6729o\$bo!
An extensible spaceship in a different rule:
x = 337, y = 7, rule = B3/S236
75bo82bo82bo82bo\$bobo6b2o6b2o8bo8b3o7b2o6b2o7b2o8bo3bo5bobo6b2o6b2o8bo
8b3o7b2o6b2o7b2o8bo3bo5bobo6b2o6b2o8bo8b3o7b2o6b2o7b2o8bo3bo5bobo6b2o
6b2o8bo8b3o7b2o6b2o7b2o8bo3bo5bobo\$2o2bo4bo2bo4bo2bo6b2o8b6o4b3o3bo3b
2o4bo2bo6bo5bo3b2o2bo4bo2bo4bo2bo6b2o8b6o4b3o3bo3b2o4bo2bo6bo5bo3b2o2b
o4bo2bo4bo2bo6b2o8b6o4b3o3bo3b2o4bo2bo6bo5bo3b2o2bo4bo2bo4bo2bo6b2o8b
6o4b3o3bo3b2o4bo2bo6bo5bo3b2o2bo\$o3bob2o5bo2b3ob2o4b3o7bo3b2o3bo3b2o2b
obob2o4b2obo6bo9bo3bob2o5bo2b3ob2o4b3o7bo3b2o3bo3b2o2bobob2o4b2obo6bo
9bo3bob2o5bo2b3ob2o4b3o7bo3b2o3bo3b2o2bobob2o4b2obo6bo9bo3bob2o5bo2b3o
b2o4b3o7bo3b2o3bo3b2o2bobob2o4b2obo6bo9bo3bo\$2o2bo4bo2bo4bo2bo6b2o8b6o
4b3o3bo3b2o4bo2bo6bo5bo3b2o2bo4bo2bo4bo2bo6b2o8b6o4b3o3bo3b2o4bo2bo6bo
5bo3b2o2bo4bo2bo4bo2bo6b2o8b6o4b3o3bo3b2o4bo2bo6bo5bo3b2o2bo4bo2bo4bo
2bo6b2o8b6o4b3o3bo3b2o4bo2bo6bo5bo3b2o2bo\$bobo6b2o6b2o8bo8b3o7b2o6b2o
7b2o8bo3bo5bobo6b2o6b2o8bo8b3o7b2o6b2o7b2o8bo3bo5bobo6b2o6b2o8bo8b3o7b
2o6b2o7b2o8bo3bo5bobo6b2o6b2o8bo8b3o7b2o6b2o7b2o8bo3bo5bobo\$75bo82bo
82bo82bo!
Currently, it hasn't been stabilized, though.
I Like My Heisenburps! (and others)
Extrementhusiast
Posts: 1768
Joined: June 16th, 2009, 11:24 pm
Location: USA
p16 gun
x = 30, y = 44, rule = 3567/2/3
15.A12.A\$14.16A\$15.14A\$15.2A10.2A\$15.2A10.2A\$15.2A10.2A\$15.2A4.2A4.2A
\$15.2A4.2A4.2A\$15.2A4.2A4.2A\$15.2A4.2A4.2A\$14.3A10.3A\$15.3A8.3A\$16.A
10.A2\$.A8.A\$12A\$.12A\$.2A8.A\$.2A\$.2A\$.2A\$.2A3.4A\$.2A3.4A\$.2A\$.2A\$.2A\$.
2A8.A\$.12A\$12A\$.A8.A2\$16.A10.A\$15.3A8.3A\$14.3A10.3A\$15.2A4.2A4.2A\$15.
2A4.2A4.2A\$15.2A4.2A4.2A\$15.2A4.2A4.2A\$15.2A10.2A\$15.2A10.2A\$15.2A10.
2A\$15.14A\$14.16A\$15.A12.A!
Take away the top or bottom gun to make a breeder.
Why hasn't a glider exploded yet?
12Glider
Posts: 79
Joined: December 17th, 2010, 4:56 pm
An explosive reaction on Serizawa: 2 gliders + 1 cell = 40 gliders + 1 big glider + 14 cells
x = 5, y = 25, rule = Serizawa
2.A19\$.B\$A.A3\$3.B\$2.A.A!
ebcube
Posts: 124
Joined: February 27th, 2010, 2:11 pm
Four barrel gun fires several kinds of spaceships including a 4c/4:
x = 16, y = 16, rule = 345/2/4
14.2A\$13.A.A\$12.A.A\$11.A.A\$10.A.A\$9.A.A\$8.A.A\$7.A.A\$6.A.A\$5.A.A\$4.A.A
\$3.A.A\$2.A.A\$.A.A\$A.A\$2A!
knightlife
Posts: 566
Joined: May 31st, 2009, 12:08 am
The LifeWiki entry on HighLife (B36/S23) says "[...] the only non-standard spaceships that are known to work in HighLife are turtle and some flotillae of the standard spaceships."
I just noticed that 86P9H3V0, a small c/3 non-standard birdlike Life spaceship found by David Bell, works in HighLife. It is mentioned as an aside in the image gallery here: http://www.conwaylife.com/wiki/index.php?title=117P9H3V0
86P9H3V0 has a spark, so perhaps it will be useful. Here it is reflecting a glider:
x = 37, y = 18, rule = B36/S23
11bob3o5b3obo\$10b2ob2obo3bob2ob2o\$9bo4b3o3b3o4bo\$12b2o9b2o\$2bob2ob6o2b
2o3b2o2b6ob2obo\$b2ob2o4b2o4b2ob2o4b2o4b2ob2o\$o2bobo2bo7b2ob2o7bo2bobo
2bo\$bo5bo7b2o3b2o7bo5bo2\$14bo2b3o2bo\$13b11o\$13b2obo3bob2o4\$20bo\$19b2o\$
19bobo!
I wonder what other non-standard Life spaceships work in HighLife -- in particular, it might be worth checking some of the more recently found spaceships.
Since this is my first post here, I just want to say how delighted I am to find this forum, the associated website and wiki, the Golly program, and all the pattern collections that are linked to from this site. Thank you very much to all the contributors! -- Eric Goldstein
EricG
Posts: 199
Joined: August 19th, 2011, 5:41 pm
Location: Chicago-area, USA
Replicator based sawtooth spaceship:
x = 26, y = 18, rule = B36/S23
\$8b3o\$7bo3bo\$b3o2bo4bo\$5bo2bo2bo\$5bo4bo\$5bo3bo\$6b3o2\$16b3o\$15bo3bo\$14b
o4bo\$13bo2bo2bo\$13bo4bo\$13bo3bo\$14b3o!
Ivan
Posts: 8
Joined: August 13th, 2011, 2:22 pm
Next
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# How do you simplify 125 to the 2 3 power?
• 12553.
• 1252353(23)
• 52.
• 25.
u221a15625
• 12553.
• 1252353(23)
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u221a15625
## What is 3 2 as a power?
Answer: 3 raised to the second power is equal to 32 9. Let us understand this with the help of the following explanation. Explanation: 3 to the 2nd power can be written as 32 3 xd7 3, as 3 is multiplied by itself 2 times.
## What is 2/3 Power simplified?
Answer: 2 raised to the third power is equal to 23 8. Explanation: 2 to the 3rd power can be written as 23 2 xd7 2 xd7 2, as 2 is multiplied by itself 3 times. Here, 2 is called the base and 3 is called the exponent or power.
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# Distance between Kerkyra (CFU) and Kavala (KVA)
Flight distance from Kerkyra to Kavala (Corfu International Airport – Kavala International Airport) is 265 miles / 426 kilometers / 230 nautical miles. Estimated flight time is 1 hour 0 minutes.
Driving distance from Kerkyra (CFU) to Kavala (KVA) is 338 miles / 544 kilometers and travel time by car is about 7 hours 5 minutes.
## Map of flight path and driving directions from Kerkyra to Kavala.
Shortest flight path between Corfu International Airport (CFU) and Kavala International Airport (KVA).
## How far is Kavala from Kerkyra?
There are several ways to calculate distances between Kerkyra and Kavala. Here are two common methods:
Vincenty's formula (applied above)
• 264.741 miles
• 426.059 kilometers
• 230.054 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 264.196 miles
• 425.183 kilometers
• 229.580 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Corfu International Airport
City: Kerkyra
Country: Greece
IATA Code: CFU
ICAO Code: LGKR
Coordinates: 39°36′6″N, 19°54′42″E
B Kavala International Airport
City: Kavala
Country: Greece
IATA Code: KVA
ICAO Code: LGKV
Coordinates: 40°54′47″N, 24°37′9″E
## Time difference and current local times
There is no time difference between Kerkyra and Kavala.
EET
EET
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 64 kg (141 pounds).
## Frequent Flyer Miles Calculator
Kerkyra (CFU) → Kavala (KVA).
Distance:
265
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
265
Round trip?
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# midterm2-soln - Math 307 - 202 Midterm II March 18 2011 ...
Doc Preview
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Identified Q&As 23
Solutions available
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Math 307 - 202 Midterm II (March 18, 2011) Note:There are 5 problems on both sides of the sheet, each worth 8 points. Also attached is a sheet of MATLAB commands. Problem 1:For (a) and (b), there may be more than one correct answer. (a) Which pair of vectors are orthogonal to each other?
View full document
(c) Calculate the modulus and argument of the following three complex numbers:z1 =e3+2i , z2= 3-3i, andz3=z1¯z2 . (d) Given four complex numbers:z1= 3 + 4i,z2= 7-i,z3= 4i,z4 = 1-i, write out the MATLAB/Octave commands for computing the modulus ofz5=z21¯z2 /(z1 2 3z34 ). Solution:
View full document
## Want to read all 5 pages?
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## Want to read all 5 pages?
Previewing 2 of 5 pages Upload your study docs or become a member.
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# Grams to Ounces conversion
g
Ounces: oz
Pounds+Ounces:
lb
oz
Calculation:
Ounces to Grams ►
## How to convert Grams to Ounces
1 gram (g) is equal to 0.03527396195 ounces (oz).
1 g = 0.03527396195 oz
The mass m in ounces (oz) is equal to the mass m in grams (g) divided by 28.34952:
m(oz) = m(g) / 28.34952
#### Example
Convert 5g to Ounces:
m(oz) = 5 g / 28.34952 = 0.17637 oz
## Grams to Ounces conversion table
Grams (g) Ounces (oz)
0 g 0 oz
1 g 0.0353 oz
2 g 0.0706 oz
3 g 0.1058 oz
4 g 0.1411 oz
5 g 0.1764 oz
6 g 0.2116 oz
7 g 0.2469 oz
8 g 0.2822 oz
9 g 0.3175 oz
10 g 0.3527 oz
20 g 0.7055 oz
30 g 1.0582 oz
40 g 1.4110 oz
50 g 1.7637 oz
60 g 2.1164 oz
70 g 2.4692 oz
80 g 2.8219 oz
90 g 3.1747 oz
100 g 3.5274 oz
1000 g 35.2740 oz
Ounces to Grams ►
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# Sorry if this is too basic but we've all got to start somewhere
Sorry if this is too basic but we've all got to start somewhere......
Just started reading A Brief History of Time, and it has reminded me of a query I have had for years:
E=MCsquared (sorry don't know how to do the little 2 next to the C)
E is Energy, M is Mass and C is the speed of light - Yes? And I can follow that but......
Question: Why does the speed of light have to be squared?
Thanking you
JS
diazona
Homework Helper
A simple answer would be that it makes the units work out. Energy is measured in Joules, and mass is measured in kilograms. One Joule is equal to one kilogram times meter squared per second squared,
$$\mathrm{J} = \frac{\mathrm{kg}\ \mathrm{m}^2}{\mathrm{s}^2}$$
So in order to multiply mass by something to get energy, that something needs to have the units of velocity squared.
That's certainly not an explanation of why $E=mc^2$ is correct, but it should at least help you understand why it's not just, say, $E=mc$.
Cleonis
Gold Member
Just started reading A Brief History of Time, and it has reminded me of a query I have had for years:
$$E=mc^2$$
Question: Why does the speed of light have to be squared?
Ultimately it's related to the way that space and time are interrelated.
First some properties of space:
We have pythagoras' theorem.
A two-dimensional space can always be mapped with a grid of perpendicular lines. The grid with its x-axis and y-axis can be oriented in any direction. Take two points in that space, A and B. The grid is in some orientation and we can decompose the distance betwee A and B in component parallel to the x-axis and parallel to the y-axis respectively. Pythagoras' theorem expresses that no matter the orientation of the grid, the distance 'r' between A and P satisfies the rule:
$$r^2 = x^2 + y^2$$
This is a principle of invariance: you can orient the grid (that you are using to map space) in all directions, but the distance between the point is an invariant of that directional freedom.
Pythagoras theorem extends to three spatial dimensions (it extends to any dimensional number)
$$r^2 = x^2 + y^2 + z^2$$
Now to the way that space and time are related.
Special relativity describes that if you measure distance in kilometers and time in seconds then there is an invariant quantity that is expressed as follows:
$$\tau^2 = c^2t^2 - x^2 - y^2 - z^2$$
Here the greek letter 'tau' ($\tau$) is used for the invariant quantity.
The similarity with Pythagoras' theorem is striking. At the same time, because the spatial dimension has a minus sign it's completely different from Pythagoras's theorem.
We don't know why that expression $$\tau^2 = c^2t^2 - x^2 - y^2 - z^2$$ holds good, but we do know it's profound.
Now, here the factor c2 is present because of the way the other dimensions are expressed: spatial distance in kilometers, and time in seconds. But we are free to express spatial distance otherwise. We can express spatial distance in lightseconds: one lightsecond is the distance that light travels in one second. That way the speed of light is absorbed in the measure of spatial distance. If spatial distance is expressed in lightseconds then the expression for the invariant quantity tau still says the same thing, but without the factor c2.
In the expression $$E=mc^2$$ there is a factor c2, but that is not a crucial element. The essential thing is that there is a relation of proportionality between matter and energy. You can opt to express energy otherwise (absorbing the factor c2 into the energy term), and then the expression would still assert the proportionality: E=m, which is what that formula all about.
The original question was rather: why a factor c2, rather than say, c or c3?
I believe the c2 factor must arise from the same underlying physics as that phythagoras-like relation between space and time. We don't know why space and time are related that way. All we know is: our theories are built upon it, and they are very powerful theories indeed, so they must be doing something right.
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Thank you Cleonis and diazona - I've printed your posts off (my eyes are not too good for staring at the monitor hours on end!!) and will read them until it sinks in!
I have question about wave frequency of an object.
for example we have 5 meter long wooden stick which is vibrating on frequency of 50mhz. If we change frequency of that wooden stick's some part, lets say 1 meter left part and set its frequency to 100mhz, does this will change whole wooden sticks frequency?
I just want to find out my thoughts if they are true about objects waves. thanks.
I could not find Thread to post there. if theres some general thread questions like this, please refer it :).
I like this forum already )
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https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-2-section-2-1-rates-of-change-and-tangents-to-curves-exercises-page-56/6
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## University Calculus: Early Transcendentals (3rd Edition)
$\frac{\Delta y}{\Delta t}=0$
*Average rates of change: The average rate of change of $y=f(x)$ with respect to $x$ over the interval $[x_1,x_2]$ is: $$\frac{\Delta y}{\Delta x}=\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ $$P(\theta)=\theta^3-4\theta^2+5\theta\hspace{1cm}[1,2]$$ The average rate of change of $y=P(\theta)$: $$\frac{\Delta y}{\Delta\theta}=\frac{(2^3-4\times2^2+5\times2)-(1^3-4\times1^2+5\times1)}{2-1}$$ $$\frac{\Delta y}{\Delta t}=\frac{(8-16+10)-(1-4+5)}{1}=2-2=0$$
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https://www.coursehero.com/tutors-problems/Chemistry/11513252-First-order-chemical-reaction-has-an-activation-energy-of-150-kilojou/
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View the step-by-step solution to:
# First order chemical reaction has an activation energy of 15.0 kilojoules per mole. The rate constant, K, is 1.5 x 10 -3 at 25 degrees Celsius what
First order chemical reaction has an activation energy of 15.0 kilojoules per mole. The rate constant, K, is 1.5 x 10-3 at 25 degrees Celsius what should the rate constant be at 45 degrees Celsius?
we know, ln (K2 / K1) = Ea / R * (T2 - T1) / T1 T2 Here Ea = 15.0 KJ / mole = 15000 J /... View the full answer
The rate constant at 45... View the full answer
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# Montessori Counting Game and Counting Activity
\$32.00
Starting a math section at your home if you are extension-schooling or homeschooling can be daunting. This set solves some needs for a Level One Montessori math activity set. Level One activities enforce concrete learning for children ages 3-6. Higher levels would move toward abstract concepts such as place values and higher order division. This activity set reinforces order, coordination, concentration, and independence. All are experienced by the child using these materials.
This Level One Montessori activity introduces sets of one through ten preparing the child for counting and teaches the value of quantity. The child can begin to associate numeral and quantity with number tiles and counters. A child will gain a growing understanding of sequence and number recognition. This activity is also self-correcting. If they end up with "extra" counter or not enough, they will know to go back and recount their work.
The set also includes an alternate sorting activity for those children that are working on item recognition and are just not ready for numeral counting. The reverse side of the number slats correspond to the item shape that belongs to that number group. Pre-counters can use this side to sort the shape families and adults can use them to guide the counting children for which items belong in a set or the children can use them as an alternate self-check mechanism.
This set is cut and engraved from 100% verified sustainable North American maple and is coated with a toy-safe CAB acrylic laquer for durability and a wipe-clean surface. This solid hardwood set will last through several children and are great for classroom needs as well.
This set include 10 number tiles that measure approximately 3 inches by 2.5 inches. 55 counters cut in a space theme accompany the number tiles. The photos show 10 different sets of objects.
If you are purchasing for children that are just learning to count, you may request that all 55 objects are the same shape (choose from rocket ships, stars, a planet, or boy or girl space explorer). This will help you isolate the difficulty in counting rather than shape matching. For children that are comfortable with counting, the shape changes add an extra difficulty level if desired for Level One math activities and is a good bridging activity to Level Two Math activities.
How to start:
1. Invite a child to come work with you. Bring him over to the shelf and have him bring the material to the table.
2. Open the box and place it on the lid.
3. Show the different number tiles to the child and have him say the numbers out loud.
4. Lay the number tiles on the table in front of the child.
5. Have the child put the number tile with an engraved 1 to the left side of the table.
6. Explain that you will need a little space between the tiles.
7. Ask the child to put the other tiles to the right of the first card in numerical order.
8. Tell the child that you are going to put the number of counters each card asks for under the corresponding card.
9. Point to tile 1 and ask the child how many counters should you place under this card.
The child should answer 1.
10. Take one counter out of the box and gently slide it (using your right index finger) under the tile written 1.
11. Have the child count the counter.
12. Repeat for card 2 by placing the counters next to each other.
13. Repeat for card 3 but place the last counter under and to the middle of the two counters.
Some activities I suggest:
Simple Counting:
Younger children could simply count space objects to correspond to the number on each tile as in the steps above.
Even and Odd:
Help the child arrange the counter under each tile in two rows so that you can easily visualize those tiles that have an even number or an odd number. The odd numbered tiles will have two identical rows with a counter in the middle of the two rows at the bottom. Slide the tile between the rows and discuss those rows that can be "cut" and those that cannot. For those odd number, adjust the tile so that it is above or below the rows in a different level than the even numbered tiles. Talk about the words and definitions for odd and even.
Addition: Appropriate for children ages 4 1/2 and up (although Montessori education is always individualized). Set up a series of simple single digit addition problems. An older child could write the answers on laminated addition cards with erasable markers or write out the problems and answers on another paper. A younger child who has an advanced understanding of numbers but the fine-motor coordination appropriate for his or her age could use the activity without any writing.
Sorting: Flip the numeral slats over to the shapes. Have the child sort all or a small selection of the counters under or onto the shape slats. Reinforces matching and sorting.
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https://math.stackexchange.com/questions/725374/triangles-with-common-centroid
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# Triangles with common centroid
Consider the points $A',B',C'$ on the sides $BC,CA,AB$ of a triangle $ABC$ respectively, such that $BA'/A'C=CB'/B'A=AC'/C'B$.
Show that the triangles $ABC$ and $A'B'C'$ share a common centroid.
• What techniques do you have to use? What have you tried? – Ted Shifrin Mar 24 '14 at 22:26
Here is the Pure geometry method:( I just list main points)
$A'J$// $AB \implies AJ=CB'$
$F$ is mid point of $B'J \implies 2FD'=JA'$
$D'$ is mid point of $A'B'$ , $BF$ cross $C'D'$ at $G$
$\dfrac{FD'}{C'B}=\dfrac{D'G}{GC'}=\dfrac{FG}{GB}$
now if you can prove $\dfrac{JA'}{AB}=\dfrac{C'B}{AB}$,then $G$ is the common centroids.
Using vectors, let $\vec A$ denote the vertex $A$, etc, and let $r := A' B / B C$. Then the centroid of $\Delta ABC$ is the arithmetic mean of the vertices: $$\vec G = \dfrac{ \vec A + \vec B + \vec C } 3.$$ We also have by definition $$\vec A' = (1-r) \vec B + r \vec C,$$ etc., so $$\dfrac{ \vec A' + \vec B' + \vec C'} 3 = \dfrac{ (1-r) \vec B + r \vec C + (1-r) \vec C + r \vec A + (1-r) \vec A + r \vec B}3 = \vec G$$ so the centroids coincide.
• Thank you for your answer. Due to the fact that I'm not very familiar with analytic geometry, I haven't fully understood your proof. Could you please explain why $$\vec A' = (1-r) \vec B + r \vec C,$$ ? – Matheo Mar 24 '14 at 23:13
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How do we use logarithm functions in real life? Yahoo The natural logarithm has the number e (≈ 2.718) as its base; its use is widespread in mathematics and physics, because of its simpler derivative. The binary logarithm uses base 2 (that is b = 2) and is commonly used in computer science. Logarithms are mainly the inverse of the exponential function.
## REAL LIFE APPLICATION OF LOGARITHM
The Common and Natural Logarithms Purplemath. These television coroners, as well as their real-life counterparts, use logarithms to make such determinations. Logarithm Applications; Find a Job. Job Search by., EARTHQUAKES Magnitude of an earthquake is logarithmic Used for predicting coming earthquakes Richter Scale: base-10 logarithmic scale CHEMISTRY Solution's pH defined by.
Read and Download Real Life Applications For Logarithms Free Ebooks in PDF format WHY HIM WHY HER HOW I MET MY HUSBAND THE REAL-LIFE LOVE STORIES OF 25 ROMANCE Here we use shortcuts to exponentials for speedy calculations. Logarithm(log) of a number to given base is the power or exponent to which the base must be raised in order to produce that number. For example , the logarithm of 1000 to base 10 is 3. It can be written as : log10 1000 = 3 ; Usually we use the base of 10, so we can write : log 1000 = 3. 3.
Here we use shortcuts to exponentials for speedy calculations. Logarithm(log) of a number to given base is the power or exponent to which the base must be raised in order to produce that number. For example , the logarithm of 1000 to base 10 is 3. It can be written as : log10 1000 = 3 ; Usually we use the base of 10, so we can write : log 1000 = 3. 3. In this section we will look at a couple of applications of exponential to look at some applications of exponential and logarithm the half-life of
Real world applications include Both the pH scale and the Richter scale are logarithmic relationships with "How Are Exponents Used in Everyday Life?" The Exponential Function and Its Applications If we take the natural log life for the cod fish, as for every other species, is not so simple. In real life,
In Real Life Exponential and logarithmic functions are widely used in describing economic and physical phenomena such as compound In many applications, Read and Download Real Life Applications For Logarithms Free Ebooks in PDF format WHY HIM WHY HER HOW I MET MY HUSBAND THE REAL-LIFE LOVE STORIES OF 25 ROMANCE
The Real Life Application of Logarithms in ph value, ph level of water, logarithmic equations in real life. what real world applications are there for logarithms In this section we will look at a couple of applications of exponential to look at some applications of exponential and logarithm the half-life of
What are logarithms used for? Are decibels a good example of the usefulness of a logarithm? Are logarithms calculus? logarithms, and calculus used in real life? Logarithms are important for any scientific career, Logarithm and Exponential Applications, Sheet #1 Logarithm and Exponential Applications, Sheet #2
In this lesson we show several Real Life uses of Online Presentation on Exponents in the Real World. jobs that use indices, logarithms, M&M exponential How Are Logarithms Used in Everyday Life? What Are Some Real-Life Examples of Parallel Lines? Real World Applications of Logarithms; Personal Life;
Sample Exponential and Logarithm Problems 1 Exponential Problems Example 1.1 Solve 1 6 3x 2 = 36x+1. Solution: Note that 1 6 = 6 1 and 36 = 62. Therefore the equation Applications of Exponential and Logarithmic Function. Applications of Exponential and Logarithmic The half-life of a sample of a radioactive isotope is
not sure exaclty what you asking, but if ur asking for an example of what logarithms are used for in real life, then there are a heaps of examples. Other important log bases include the the natural log, which is commonly used in advanced mathematics. What other appliances do logs have? Logs have a variety of real life applications such as calculating half lives and exponential growth/decay. In fact the inverse of an exponential function is a logarithmic function!
A logarithm is the inverse of the exponential function. Applications \[ \] Logarithms. Brilliant.org. APPLICATIONS OF EXPONENTIAL: AND: The half-life of iodine-131 is eight Take the natural logarithm of both sides of the equation what you started The model now
In this chapter we will cover many of the major applications of derivatives. Logarithmic Differentiation; Rates of Home / Calculus I / Applications of Real world applications include Both the pH scale and the Richter scale are logarithmic relationships with "How Are Exponents Used in Everyday Life?"
3.2 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS. 2007-05-01В В· Logarithms in real life? (10 easy points!)? Ok, so someone in my trig class asked my teacher where logarithms are used in real life, and so my teacher threw a fit and assigned a one page reasearch assignment on jobs that use logarithms and how they are used in real life (I know, unfair). Can anyone give me some examples or links to websites..., kindle and nook applications of logarithms in real life situations PDF Full Ebook applications of logarithms in real life situations PDF Full Ebook download.
### Question Corner- Natural Logs in the Real World
Real Life Applications For Logarithms sblank.de. Real Life Applications For Logarithms Ebook Real Life Applications For Logarithms currently available at www.senze-media.com for review only, if you need complete, Using logarithms to solve real world problems Take logarithm base 10 from both sides. and decays to nitrogen-14 with a half-life of approximately 5,730 years..
Sample Exponential and Logarithm Problems 1 Exponential. not sure exaclty what you asking, but if ur asking for an example of what logarithms are used for in real life, then there are a heaps of examples. MATH 11011 APPLICATIONS OF LOGARITHMIC FUNCTIONS KSU The logarithmic function with base a, denoted loga x, is deflned by Its half-life is given by h = ln2 r..
Real world applications include Both the pH scale and the Richter scale are logarithmic relationships with "How Are Exponents Used in Everyday Life?" A logarithm is the inverse of the exponential function. Applications \[ \] Logarithms. Brilliant.org.
The natural logarithm has the number e (≈ 2.718) as its base; its use is widespread in mathematics and physics, because of its simpler derivative. The binary logarithm uses base 2 (that is b = 2) and is commonly used in computer science. Logarithms are mainly the inverse of the exponential function. Read and Download Real Life Applications For Logarithms Free Ebooks in PDF format WHY HIM WHY HER HOW I MET MY HUSBAND THE REAL-LIFE LOVE STORIES OF 25 ROMANCE
Sample Exponential and Logarithm Problems 1 Exponential Problems Example 1.1 Solve 1 6 3x 2 = 36x+1. Solution: Note that 1 6 = 6 1 and 36 = 62. Therefore the equation What are logarithms used for? Are decibels a good example of the usefulness of a logarithm? Are logarithms calculus? logarithms, and calculus used in real life?
The Real Life Application of Logarithms in ph value, ph level of water, logarithmic equations in real life. what real world applications are there for logarithms Using Logarithms in the Real World. is that the practical applications of logarithm in real life? pardon me. this is part of our project and i’m having trouble
MATH 11011 APPLICATIONS OF LOGARITHMIC FUNCTIONS KSU The logarithmic function with base a, denoted loga x, is deflned by Its half-life is given by h = ln2 r. The natural logarithm has the number e (≈ 2.718) as its base; its use is widespread in mathematics and physics, because of its simpler derivative. The binary logarithm uses base 2 (that is b = 2) and is commonly used in computer science. Logarithms are mainly the inverse of the exponential function.
How can you apply logarithm in real life? u cant... next What are the applications of logarithmic and exponential ( aka use math in real life every not sure exaclty what you asking, but if ur asking for an example of what logarithms are used for in real life, then there are a heaps of examples.
Using Logarithms in the Real World. is that the practical applications of logarithm in real life? pardon me. this is part of our project and i’m having trouble 3.5 EXPONENTIAL AND LOGARITHMIC MODELS model and solve real-life problems. real-life problems. • Use logarithmic functions to model and solve
Sample Exponential and Logarithm Problems 1 Exponential Problems Example 1.1 Solve 1 6 3x 2 = 36x+1. Solution: Note that 1 6 = 6 1 and 36 = 62. Therefore the equation How can you apply logarithm in real life? u cant... next What are the applications of logarithmic and exponential ( aka use math in real life every
Exponential and Logarithmic Functions large numbers all day and devised logarithms to make his life some applications in the "Related 2009-04-02В В· How do we use logarithm functions in real life? Like, in what field, and what benefits it can do...
REAL LIFE APPLICATION OF LOGARITHM In our maths book of 9th-10th class, there is chapter named as LOGARITHM. It is very interesting chapter and its questions are some The Exponential Function and Its Applications If we take the natural log life for the cod fish, as for every other species, is not so simple. In real life,
In this lesson we show several Real Life uses of Online Presentation on Exponents in the Real World. jobs that use indices, logarithms, M&M exponential Other important log bases include the the natural log, which is commonly used in advanced mathematics. What other appliances do logs have? Logs have a variety of real life applications such as calculating half lives and exponential growth/decay. In fact the inverse of an exponential function is a logarithmic function!
## Real Life Applications For Logarithms sblank.de
Real Life Applications For Logarithms Ebook List. It was easy to find examples of set theory in real life because all the people at least use one of the aplications that we mentioned on this presentation. The, How can you apply logarithm in real life? u cant... next What are the applications of logarithmic and exponential ( aka use math in real life every.
What are logarithms used for? Are decibels a good example. The base of the logarithm is b. The 2 most common bases that we use are base `10` and base e, which we meet in Logs to base 10 and Natural Logs (base e) in later sections. The logarithmic function has many real-life applications, in acoustics, electronics, earthquake analysis and population prediction. Example 1. Write in logarithm form: `8 = 2^3` Answer, Exponential and Logarithmic Functions large numbers all day and devised logarithms to make his life some applications in the "Related.
Real world applications include Both the pH scale and the Richter scale are logarithmic relationships with "How Are Exponents Used in Everyday Life?" Using Logarithms in the Real World. is that the practical applications of logarithm in real life? pardon me. this is part of our project and i’m having trouble
Exponential and Logarithmic Functions large numbers all day and devised logarithms to make his life some applications in the "Related 2007-05-01В В· Logarithms in real life? (10 easy points!)? Ok, so someone in my trig class asked my teacher where logarithms are used in real life, and so my teacher threw a fit and assigned a one page reasearch assignment on jobs that use logarithms and how they are used in real life (I know, unfair). Can anyone give me some examples or links to websites...
In this chapter we will cover many of the major applications of derivatives. Logarithmic Differentiation; Rates of Home / Calculus I / Applications of In this chapter we will cover many of the major applications of derivatives. Logarithmic Differentiation; Rates of Home / Calculus I / Applications of
The natural logarithm has the number e (≈ 2.718) as its base; its use is widespread in mathematics and physics, because of its simpler derivative. The binary logarithm uses base 2 (that is b = 2) and is commonly used in computer science. Logarithms are mainly the inverse of the exponential function. How can you apply logarithm in real life? u cant... next What are the applications of logarithmic and exponential ( aka use math in real life every
What are logarithms used for? Are decibels a good example of the usefulness of a logarithm? Are logarithms calculus? logarithms, and calculus used in real life? REAL LIFE APPLICATION OF LOGARITHM In our maths book of 9th-10th class, there is chapter named as LOGARITHM. It is very interesting chapter and its questions are some
APPLICATIONS OF EXPONENTIAL: AND: The half-life of iodine-131 is eight Take the natural logarithm of both sides of the equation what you started The model now APPLICATIONS OF EXPONENTIAL: AND: The half-life of iodine-131 is eight Take the natural logarithm of both sides of the equation what you started The model now
The natural logarithm has the number e (≈ 2.718) as its base; its use is widespread in mathematics and physics, because of its simpler derivative. The binary logarithm uses base 2 (that is b = 2) and is commonly used in computer science. Logarithms are mainly the inverse of the exponential function. 2008-12-12 · There are real life applications of logarithms but there is little that you would do every life, I mean technically the computer uses it but there rly isn't anything you as a person would do daily that …
APPLICATIONS OF EXPONENTIAL: AND: The half-life of iodine-131 is eight Take the natural logarithm of both sides of the equation what you started The model now 2009-04-02В В· How do we use logarithm functions in real life? Like, in what field, and what benefits it can do...
Logarithms are important for any scientific career, Logarithm and Exponential Applications, Sheet #1 Logarithm and Exponential Applications, Sheet #2 The Real Life Application of Logarithms in ph value, ph level of water, logarithmic equations in real life. what real world applications are there for logarithms
not sure exaclty what you asking, but if ur asking for an example of what logarithms are used for in real life, then there are a heaps of examples. briefly, some examples a … re banks use logarithmic functions to calculate the accumilation of interest in bank accounts over the years (eg. MATH 11011 APPLICATIONS OF LOGARITHMIC FUNCTIONS KSU The logarithmic function with base a, denoted loga x, is deflned by Its half-life is given by h = ln2 r.
Logarithms in Business SlideShare. Sample Exponential and Logarithm Problems 1 Exponential Problems Example 1.1 Solve 1 6 3x 2 = 36x+1. Solution: Note that 1 6 = 6 1 and 36 = 62. Therefore the equation, Here we use shortcuts to exponentials for speedy calculations. Logarithm(log) of a number to given base is the power or exponent to which the base must be raised in order to produce that number. For example , the logarithm of 1000 to base 10 is 3. It can be written as : log10 1000 = 3 ; Usually we use the base of 10, so we can write : log 1000 = 3. 3..
### Real Life Applications For Logarithms Ebook List
What are logarithms used for? Are decibels a good example. Sample Exponential and Logarithm Problems 1 Exponential Problems Example 1.1 Solve 1 6 3x 2 = 36x+1. Solution: Note that 1 6 = 6 1 and 36 = 62. Therefore the equation, not sure exaclty what you asking, but if ur asking for an example of what logarithms are used for in real life, then there are a heaps of examples..
Applications of the Natural Logarithm and Exponential Function. Using logarithms to solve real world problems Take logarithm base 10 from both sides. and decays to nitrogen-14 with a half-life of approximately 5,730 years., The Exponential Function and Its Applications If we take the natural log life for the cod fish, as for every other species, is not so simple. In real life,.
### Real Life Applications For Logarithms sblank.de
What are logarithms used for? Are decibels a good example. How will I use Logarithmic Function in real life? What real world applications use logarithms? What is an example of a real life logarithmic function? The base of the logarithm is b. The 2 most common bases that we use are base `10` and base e, which we meet in Logs to base 10 and Natural Logs (base e) in later sections. The logarithmic function has many real-life applications, in acoustics, electronics, earthquake analysis and population prediction. Example 1. Write in logarithm form: `8 = 2^3` Answer.
In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number Real Life Applications For Logarithms Ebook Real Life Applications For Logarithms currently available at www.senze-media.com for review only, if you need complete
The Real Life Application of Logarithms in ph value, ph level of water, logarithmic equations in real life. what real world applications are there for logarithms Real-life models of logarithmic functions are in your answer how this can be used in a real-life application. of working with real-life logarithmic and
A logarithm can have any positive value as its base, Because the common and natural logs are pretty much the only logs that are used "in real life", In Real Life Exponential and logarithmic functions are widely used in describing economic and physical phenomena such as compound In many applications,
In Real Life Exponential and logarithmic functions are widely used in describing economic and physical phenomena such as compound In many applications, The Real Life Application of Logarithms in ph value, ph level of water, logarithmic equations in real life. what real world applications are there for logarithms
How will I use Logarithmic Function in real life? What real world applications use logarithms? What is an example of a real life logarithmic function? Real world applications include Both the pH scale and the Richter scale are logarithmic relationships with "How Are Exponents Used in Everyday Life?"
Applications of Exponential and Logarithmic Function. Applications of Exponential and Logarithmic The half-life of a sample of a radioactive isotope is In this chapter we will cover many of the major applications of derivatives. Logarithmic Differentiation; Rates of Home / Calculus I / Applications of
Exponential and Logarithmic Functions large numbers all day and devised logarithms to make his life some applications in the "Related In this chapter we will cover many of the major applications of derivatives. Logarithmic Differentiation; Rates of Home / Calculus I / Applications of
A logarithm can have any positive value as its base, Because the common and natural logs are pretty much the only logs that are used "in real life", Application of Logarithm in Real Life Any time you need to convert multiplications to additions, they can be useful. For example, if you get a loan at a bank that has continuous interest (they all do), if you …
kindle and nook applications of logarithms in real life situations PDF Full Ebook applications of logarithms in real life situations PDF Full Ebook download In this lesson we show several Real Life uses of Online Presentation on Exponents in the Real World. jobs that use indices, logarithms, M&M exponential
How will I use Logarithmic Function in real life? What real world applications use logarithms? What is an example of a real life logarithmic function? 3.2 LOGARITHMIC FUNCTIONS AND THEIR GRAPHS real-life problems. every logarithmic equation can be written in an
kindle and nook applications of logarithms in real life situations PDF Full Ebook applications of logarithms in real life situations PDF Full Ebook download The base of the logarithm is b. The 2 most common bases that we use are base `10` and base e, which we meet in Logs to base 10 and Natural Logs (base e) in later sections. The logarithmic function has many real-life applications, in acoustics, electronics, earthquake analysis and population prediction. Example 1. Write in logarithm form: `8 = 2^3` Answer
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Search
Tetrahedron
Tetrahedron
Click on picture for large version.
Type Platonic
Face polygon triangle
Faces 4
Edges 6
Vertices 4
Faces per vertex 3
Vertices per face 3
Symmetry group tetrahedral (Td)
Dual polyhedron tetrahedron (self-dual)
Dihedral Angle 70° 32'
Properties regular, convex
A tetrahedron (plural: tetrahedra) is a polyhedron composed of four triangular faces, three of which meet at each vertex. A regular tetrahedron is one in which the four triangles are regular, or "equilateral," and is one of the Platonic solids.
The area A and the volume V of a regular tetrahedron of edge length a are:
$A=\sqrt{3}a^2$
$V=\begin{matrix}{1\over12}\end{matrix}\sqrt{2}a^3$
A tetrahedron is a 3-simplex.
Tetrahedra are a special type of triangular pyramid and are self-dual. Canonical coordinates of the tetrahedron are (1, 1, 1), (−1, −1, 1), (−1, 1, −1) and (1, −1, −1). A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. The volume of this tetrahedron is 1/3 the volume of the cube. Taking both tetrahedra within a single cube gives a regular polyhedral compound called the stella octangula, whose interior is an octahedron. Inscribing tetrahedra inside the regular compound of five cubes gives two more regular compounds, containing five and ten tetrahedra.
Regular tetrahedra can't tile space by themselves, although it seems likely enough that Aristotle reported it was possible. In fact, octahedra are necessary to fill some of the gaps. This is one of the five Andreini tessellations, and is a limiting case of another, a tiling involving tetrahedra and truncated tetrahedra.
The volume of an irregular tetrahedron, given its vertices a, b, c and d, is (1/6)·|det(ab, bc, cd)|, or any other combination of pairs of verticies that form a simply connected graph. (This works for regular tetrahedrons too.)
Especially in roleplaying, this solid is known as a d4, one of the more common Polyhedral dice.
Like all platonic solids, archimedean solids and indeed all convex polyhedra, a tetrahedron can be folded from a single sheet of paper.
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## Saturday, September 22, 2012
I love this clip from Seinfeld where Kramer describes what happens at the dinner table when you're married. It would seem that one of the worst parts of marriage is the time when you talk about your day. Was it a good day or a bad day? Eventually this became the inspiration for how I chose to teach integer operations.
I used to teach adding integers with gang violence. There were two gangs (positives and negatives) who would fight and kill each other. The only problem with this metaphor was that it was only really good for adding integers. It didn't help me teach some of the other operations (subtracting a negative, negative times a negative).
I also had the problem of offering students too many ways to think about adding integers. I used gang violence, money, number lines, drawn positive/negative signs, and the boring rules. But it was way too much information. I was trying to offer my students choice, but many were becoming confused.
So I'd like to say that I have one way to explain integers, and it all has to do with having a good day or a bad day. I start the lesson off by showing this slide:
I ask my students how they know if they've had a good day or a bad day. What kind of things could help you move to either side of the spectrum? I then use specific examples, always in pairs (see below). I use little arrows to indicate which way we are moving on the happy face (soon to be number) line.
The first two examples are pretty obvious. If two good things happen, then it's a good day. If two bad things happen, then it's a bad day.
This third scenario (the lost cell phone) had some debate. Maybe the loss of a cell phone was a good thing...perhaps this would convince your parents to buy a new/better phone. But the point was to show that when a bad thing happens, and then an equally good thing happens, you end up back where you started. It was neither good nor bad.
The kids laughed at this one. Not sure why hamster death is so funny. Perhaps it was the juxtaposition of the two scenarios that made the second seem so ridiculous. But this is exactly what I wanted. Sure, it was nice to find the dollar. But your hamster is dead. Finding the dollar does not make up for the fact that your hamster died, so overall, it was a bad day.
And finally, the lost/found money. This last scenario transitioned into the use of a number line (at which point my students thought, "Oh, so this is a math lesson").
Where do I go next? Basically negatives are bad things that happen and positives are good things that happen. If more bad things happen, it's a bad day. Or if there are more negatives, then the answer is negative.
I can easily see how I can relate this now to subtracting and multiplying. Taking away bad things helps make your day better (subtracting a negative). When bad things happen over and over again, it is a bad day (positive times negative). Taking away bad things over and over again makes it a good day (negative times negative).
Nathan Kraft
## Tuesday, September 11, 2012
### Exploiting My Son for Math
At some point I'm going to have to apologize to my four-year-old son, Emmett. Over the last year I've been using him for all sorts of math lessons - many times under the guise that I'm spending quality time with him. It all started with this video:
I posted this on facebook and lot of people thought it was cute. But many also wanted to know the answer to the question. I used the same video in class and one of my students went home and worked on the problem with her dad. She was so proud when she came in the next day with the right answer.
I quickly learned that I could use his "cuteness" to teach math. He was helping me teach probability:
He was helping me teach problem-solving:
And sometimes he'll even inspire lessons. The other night I was sitting in the kitchen doing some work. Emmett should have been in bed sleeping, but instead he was in the living room stacking cups.
He was so proud of his design and he wanted to make an even bigger pyramid. So we went out the next day and bought a bunch of cups, and hence, a new math lesson was created.
So again, I apologize to Emmett for using him this way. But I think he's having some fun in the process. Besides, how many kids have had the chance to do this?
I only hope his cuteness doesn't run out any time soon.
Nathan Kraft
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Web pdp-10.trailing-edge.com
Trailing-Edge - PDP-10 Archives - decuslib10-02 - 43,50145/cel2.doc
There are 2 other files named cel2.doc in the archive. Click here to see a list.
```SUBROUTINE CEL2
PURPOSE
COMPUTES THE GENERALIZED COMPLETE ELLIPTIC INTEGRAL OF
SECOND KIND.
USAGE
CALL CEL2(RES,AK,A,B,IER)
DESCRIPTION OF PARAMETERS
RES - RESULT VALUE
AK - MODULUS (INPUT)
A - CONSTANT TERM IN NUMERATOR
B - FACTOR OF QUADRATIC TERM IN NUMERATOR
IER - RESULTANT ERROR CODE WHERE
IER=0 NO ERROR
IER=1 AK NOT IN RANGE -1 TO +1
REMARKS
FOR ABS(AK) GE 1 THE RESULT IS SET TO 1.E75 IF B IS
POSITIVE, TO -1.E75 IF B IS NEGATIVE.
SPECIAL CASES ARE
K(K) OBTAINED WITH A = 1, B = 1
E(K) OBTAINED WITH A = 1, B = CK*CK WHERE CK IS
COMPLEMENTARY MODULUS.
B(K) OBTAINED WITH A = 1, B = 0
D(K) OBTAINED WITH A = 0, B = 1
WHERE K, E, B, D DEFINE SPECIAL CASES OF THE GENERALIZED
COMPLETE ELLIPTIC INTEGRAL OF SECOND KIND IN THE USUAL
NOTATION, AND THE ARGUMENT K OF THESE FUNCTIONS MEANS
THE MODULUS.
SUBROUTINES AND FUNCTION SUBPROGRAMS REQUIRED
NONE
METHOD
DEFINITION
RES=INTEGRAL((A+B*T*T)/(SQRT((1+T*T)*(1+(CK*T)**2))*(1+T*T))
SUMMED OVER T FROM 0 TO INFINITY).
EVALUATION
LANDENS TRANSFORMATION IS USED FOR CALCULATION.
REFERENCE
R.BULIRSCH, 'NUMERICAL CALCULATION OF ELLIPTIC INTEGRALS
AND ELLIPTIC FUNCTIONS', HANDBOOK SERIES SPECIAL FUNCTIONS,
NUMERISCHE MATHEMATIK VOL. 7, 1965, PP. 78-90.
```
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// Copyright 2016 the V8 project authors. All rights reserved. // Use of this source code is governed by a BSD-style license that can be // found in the LICENSE file. #ifndef V8_COMPILER_LOOP_VARIABLE_OPTIMIZER_H_ #define V8_COMPILER_LOOP_VARIABLE_OPTIMIZER_H_ #include "src/compiler/functional-list.h" #include "src/compiler/node-aux-data.h" #include "src/zone/zone-containers.h" namespace v8 { namespace internal { namespace compiler { class CommonOperatorBuilder; class Graph; class Node; class InductionVariable : public ZoneObject { public: Node* phi() const { return phi_; } Node* effect_phi() const { return effect_phi_; } Node* arith() const { return arith_; } Node* increment() const { return increment_; } Node* init_value() const { return init_value_; } enum ConstraintKind { kStrict, kNonStrict }; enum ArithmeticType { kAddition, kSubtraction }; struct Bound { Bound(Node* bound, ConstraintKind kind) : bound(bound), kind(kind) {} Node* bound; ConstraintKind kind; }; const ZoneVector& lower_bounds() { return lower_bounds_; } const ZoneVector& upper_bounds() { return upper_bounds_; } ArithmeticType Type() { return arithmeticType_; } private: friend class LoopVariableOptimizer; InductionVariable(Node* phi, Node* effect_phi, Node* arith, Node* increment, Node* init_value, Zone* zone, ArithmeticType arithmeticType) : phi_(phi), effect_phi_(effect_phi), arith_(arith), increment_(increment), init_value_(init_value), lower_bounds_(zone), upper_bounds_(zone), arithmeticType_(arithmeticType) {} void AddUpperBound(Node* bound, ConstraintKind kind); void AddLowerBound(Node* bound, ConstraintKind kind); Node* phi_; Node* effect_phi_; Node* arith_; Node* increment_; Node* init_value_; ZoneVector lower_bounds_; ZoneVector upper_bounds_; ArithmeticType arithmeticType_; }; class LoopVariableOptimizer { public: void Run(); LoopVariableOptimizer(Graph* graph, CommonOperatorBuilder* common, Zone* zone); const ZoneMap& induction_variables() { return induction_vars_; } void ChangeToInductionVariablePhis(); void ChangeToPhisAndInsertGuards(); private: const int kAssumedLoopEntryIndex = 0; const int kFirstBackedge = 1; struct Constraint { Node* left; InductionVariable::ConstraintKind kind; Node* right; bool operator!=(const Constraint& other) const { return left != other.left || kind != other.kind || right != other.right; } }; using VariableLimits = FunctionalList; void VisitBackedge(Node* from, Node* loop); void VisitNode(Node* node); void VisitMerge(Node* node); void VisitLoop(Node* node); void VisitIf(Node* node, bool polarity); void VisitStart(Node* node); void VisitLoopExit(Node* node); void VisitOtherControl(Node* node); void AddCmpToLimits(VariableLimits* limits, Node* node, InductionVariable::ConstraintKind kind, bool polarity); void TakeConditionsFromFirstControl(Node* node); const InductionVariable* FindInductionVariable(Node* node); InductionVariable* TryGetInductionVariable(Node* phi); void DetectInductionVariables(Node* loop); Graph* graph() { return graph_; } CommonOperatorBuilder* common() { return common_; } Zone* zone() { return zone_; } Graph* graph_; CommonOperatorBuilder* common_; Zone* zone_; NodeAuxData limits_; NodeAuxData reduced_; ZoneMap induction_vars_; }; } // namespace compiler } // namespace internal } // namespace v8 #endif // V8_COMPILER_LOOP_VARIABLE_OPTIMIZER_H_
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# How many glasses do you need to fill a bottle?
How many glasses do you need to fill a bottle?
1 liter of water = 5 plastic cups Finally, keep in mind that - usually - table glasses also have a capacity of 200 mL, therefore, also in this case, 1 liter of water corresponds to 5 table glasses.
## How many glasses does it take to fill a bottle of mineral water?
In other words, to measure 2 liters of water you will need to completely fill ten plastic or table glasses. A completely filled plastic beaker contains approximately 200 mL of water.
## How many glasses correspond to one and a half liters?
Added on 30/05/2017. Let me make a clarification: for me the glass is the large 250 ml one, so 8 glasses are 2 liters of water in total. If you don't get there at least try to use the 200ml ones, in order to get to 1 liter and a half!
## How many glasses are there in a bottle of water?
Dry wines
Taking into consideration a glass of about 150 ml, from a bottle of standard red or white wine, that is with a capacity of 750 ml, about 5 or 6 glasses can be obtained.
## How do I measure 250ml?
In other words, to measure 250 mL of water you will have to fill the plastic cup completely and then - once you have poured the water where it suits you best - you will have to fill it again but only for 1/4 of its volume (or 25%) . A completely filled plastic beaker contains approximately 200 mL of water.
## Find 38 related questions
### How do I measure 500 mL of water?
Considering that a white plastic cup - like the one shown in the following image - contains 200 mL of liquid, it is easy to establish that 500 mL of water corresponds to two and a half cups. A completely filled plastic beaker contains approximately 200 mL of water.
### How many are 250 mL of water?
250 mL corresponds to a quarter of a liter or half of half a liter. Taking into account that half a liter is the amount of water contained in a plastic bottle, 250 mL corresponds to the content of half a half liter bottle of water.
### How many glasses for a bottle of prosecco?
The standard serving of a sparkling wine reaches 90% of the glass capacity, i.e. about 150 ml. Thus we can calculate that, from a 750 ml bottle of sparkling wine, a maximum of 5 glasses can be obtained.
### How many glasses with a bottle of champagne?
First a small amount by tilting the glass, phase in which bubbles are formed. Let the foam decrease and then start pouring again up to 2/3 of the glass. How many glasses can be served in a bottle? Bon ton says that there are 6 glasses for a 75 cl bottle.
### How many glasses does it take for a liter?
1 liter of water = 5 plastic cups
Finally, keep in mind that - usually - even table glasses have a capacity of 200 mL therefore, also in this case, 1 liter of water corresponds to 5 table glasses. Related links: 500 mL of milk how many glasses.
### How many liters of water are 8 glasses of water?
The most awake can already give an answer (the one they have already given), so 8 glasses of water correspond to about one and a half liters of water a day, the minimum quantity (from one and a half to two liters) that according to many nutritionists and "people who know" should be drunk to stay healthy and hydrate to ...
### How is 1 2 liter of milk measured?
1 dL of milk = half a plastic cup
A completely filled plastic cup contains 2 dL of milk.
### How many glasses does a carafe fill?
The carafe contains 10 glasses and 1 cup: Then fill a glass using the cup: to fill a glass you need 2 cups.
### What happens if I drink 3 liters of water a day?
Drinking as much water as necessary allows you to integrate minerals and also supports the vitality and functions of the cells of our body and also helps the body to purify itself and the brain to function better.
### How do you drink champagne?
Champagne: the golden rules on how to drink it and how to serve it
1. Serve the Champagne chilled, never cold. ...
2. Never make a bang. ...
3. Champagne must be poured in two stages. ...
4. The glasses to choose from are tulip-shaped. ...
5. Do not put the empty bottle upside down in the bucket ...
6. Champagne should be served with an aperitif, but not only.
### How much champagne per person?
Assuming that each adult guest consumes about 5 glasses and that, apart from the champagne, each bottle has a 'yield' of about 15 glasses, a simple formula is enough to calculate the number of bottles to order: number of guests x 5: 15 = total bottles, here's how math finally comes in handy!
### How many Cl and a glass of amaro?
5,5 cl stackable digestive glass, with 2 cl and 4 cl marking.
### How much Aperol for a bottle of prosecco?
If you measure your Aperol spritz in ounces, then you will need 2 ounces of Prosecco and 1 ounces of Aperol.
### How many bottles of sparkling wine for 30 people?
Three bottles are estimated to be enough for 15 people. Sparkling wine accompanies toasts, so if you plan the minimum number of toasts you will also be able to establish approximately as many bottles of sparkling wine for 50, 30, 100 people. They are usually ideal for making a toast at the beginning of an event or party.
### What happens if I drink a liter of wine a day?
Ethyl alcohol (a toxic substance for living organisms so much so that it is used as an antiseptic) is metabolized and disposed of by our liver forcing it to work hard which in the long run causes liver, cardiovascular, neuropsychiatric and endocrinological diseases.
### How many mL is a glass beaker?
More classic people can opt for a normal glass of water, so it will hold about 200 - 250 ml. On the other hand, those who opt for a breakfast cup will have a capacity of around 250ml.
### What does 300 mL of water correspond to?
300 mL corresponds to 0,3 L and therefore to three tenths of a liter. To get an idea of how much this value can be, keep in mind that 300 mL is the volume of water contained in a completely full plastic cup and a half (a completely full plastic cup contains 200 mL of water).
### How much is 100 mL?
To get an idea of how much this value can be, keep in mind that 100 mL is the volume of half a plastic cup (a completely full plastic cup contains about 200 mL of liquid). A completely filled plastic beaker contains approximately 200 mL of liquid.
##### add a comment of How many glasses do you need to fill a bottle?
Comment sent successfully! We will review it in the next few hours.
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0
# Is 75 equal to three fourths?
Wiki User
2012-11-27 21:17:45
No. 75 is much bigger than 1, three fourths is smaller than 1.
But 75% IS equal to 3/4
Wiki User
2012-11-27 21:17:45
Study guides
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# Nibble/s to Gibit/Hr → CONVERT Nibbles per Second to Gibibits per Hour
expand_more
info 1 Nibble/s is equal to 0.000013411045074462890625 Gibit/Hr
S = Second, M = Minute, H = Hour, D = Day
Sec
Min
Hr
Day
Sec
Min
Hr
Day
Nibble/s
## Nibbles per Second (Nibble/s) Versus Gibibits per Hour (Gibit/Hr) - Comparison
Nibbles per Second and Gibibits per Hour are units of digital information used to measure storage capacity and data transfer rate.
Nibbles per Second is one of the very "basic" digital unit where as Gibibits per Hour is a "binary" unit. One Nibble is equal to 4 bits. One Gibibit is equal to 1024^3 bits. There are 268,435,456 Nibble in one Gibibit. Find more details on below table.
Nibbles per Second (Nibble/s) Gibibits per Hour (Gibit/Hr)
Nibbles per Second (Nibble/s) is a unit of measurement for data transfer bandwidth. It measures the number of Nibbles that can be transferred in one Second. Gibibits per Hour (Gibit/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Gibibits that can be transferred in one Hour.
## Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr) Conversion - Formula & Steps
The Nibble/s to Gibit/Hr Calculator Tool provides a convenient solution for effortlessly converting data rates from Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Nibble) and target (Gibibit) data units.
Source Data Unit Target Data Unit
Equal to 4 bits
(Basic Unit)
Equal to 1024^3 bits
(Binary Unit)
The conversion from Data per Second to Hour can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
The formula for converting the Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr) can be expressed as follows:
diamond CONVERSION FORMULA Gibit/Hr = Nibble/s x 4 ÷ 10243 x 60 x 60
Now, let's apply the aforementioned formula and explore the manual conversion process from Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Gibibits per Hour = Nibbles per Second x 4 ÷ 10243 x 60 x 60
STEP 1
Gibibits per Hour = Nibbles per Second x 4 ÷ (1024x1024x1024) x 60 x 60
STEP 2
Gibibits per Hour = Nibbles per Second x 4 ÷ 1073741824 x 60 x 60
STEP 3
Gibibits per Hour = Nibbles per Second x 0.0000000037252902984619140625 x 60 x 60
STEP 4
Gibibits per Hour = Nibbles per Second x 0.0000000037252902984619140625 x 3600
STEP 5
Gibibits per Hour = Nibbles per Second x 0.000013411045074462890625
Example : By applying the previously mentioned formula and steps, the conversion from 1 Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr) can be processed as outlined below.
1. = 1 x 4 ÷ 10243 x 60 x 60
2. = 1 x 4 ÷ (1024x1024x1024) x 60 x 60
3. = 1 x 4 ÷ 1073741824 x 60 x 60
4. = 1 x 0.0000000037252902984619140625 x 60 x 60
5. = 1 x 0.0000000037252902984619140625 x 3600
6. = 1 x 0.000013411045074462890625
7. = 0.000013411045074462890625
8. i.e. 1 Nibble/s is equal to 0.000013411045074462890625 Gibit/Hr.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Nibbles per Second to Gibibits per Hour using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Nibble ?
A Nibble is a unit of digital information that consists of 4 bits. It is half of a byte and can represent a single hexadecimal digit. It is used in computer memory and data storage and sometimes used as a basic unit of data transfer in certain computer architectures.
arrow_downward
#### What is Gibibit ?
A Gibibit (Gib or Gibit) is a binary unit of digital information that is equal to 1,073,741,824 bits and is defined by the International Electro technical Commission(IEC). The prefix 'gibi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'gigabit' (Gb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr)
Apply the formula as shown below to convert from 1 Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr).
A B C
1 Nibbles per Second (Nibble/s) Gibibits per Hour (Gibit/Hr)
2 1 =A2 * 0.0000000037252902984619140625 * 60 * 60
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Nibbles per Second (Nibble/s) to Gibibits per Hour (Gibit/Hr) Conversion
You can use below code to convert any value in Nibbles per Second (Nibble/s) to Nibbles per Second (Nibble/s) in Python.
nibblesperSecond = int(input("Enter Nibbles per Second: "))
gibibitsperHour = nibblesperSecond * 4 / (1024*1024*1024) * 60 * 60
print("{} Nibbles per Second = {} Gibibits per Hour".format(nibblesperSecond,gibibitsperHour))
The first line of code will prompt the user to enter the Nibbles per Second (Nibble/s) as an input. The value of Gibibits per Hour (Gibit/Hr) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Gibibits(Gibit) are there in a Nibble?expand_more
There are 0.0000000037252902984619140625 Gibibits in a Nibble.
#### What is the formula to convert Nibble to Gibibit(Gibit)?expand_more
Use the formula Gibit = Nibble x 4 / 10243 to convert Nibble to Gibibit.
#### How many Nibbles are there in a Gibibit(Gibit)?expand_more
There are 268435456 Nibbles in a Gibibit.
#### What is the formula to convert Gibibit(Gibit) to Nibble?expand_more
Use the formula Nibble = Gibit x 10243 / 4 to convert Gibibit to Nibble.
#### Which is bigger, Gibibit(Gibit) or Nibble?expand_more
Gibibit is bigger than Nibble. One Gibibit contains 268435456 Nibbles.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation.
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HalfEdge Class Reference
`#include <HalfEdge.h>`
## Public Member Functions
HalfEdge ()=default
Default constructor. Values for mDual, mNext, mPoint are set to -10 which means that they are undefined.
HalfEdge (int dual, int next, int point, bool mbreak, bool forced)
bool getBreak () const
Returns, whether the HalfEdge belongs to a break line or not.
int getDual () const
Returns the number of the dual HalfEdge.
bool getForced () const
Returns, whether the HalfEdge belongs to a constrained edge or not.
int getNext () const
Returns the number of the next HalfEdge.
int getPoint () const
Returns the number of the point at which this HalfEdge points.
void setBreak (bool b)
Sets the break flag.
void setDual (int d)
Sets the number of the dual HalfEdge.
void setForced (bool f)
Sets the forced flag.
void setNext (int n)
Sets the number of the next HalfEdge.
void setPoint (int p)
Sets the number of point at which this HalfEdge points.
## Protected Attributes
bool mBreak = false
True, if the HalfEdge belongs to a break line, `false` otherwise.
int mDual = -10
Number of the dual HalfEdge.
bool mForced = false
True, if the HalfEdge belongs to a constrained edge, `false` otherwise.
int mNext = -10
Number of the next HalfEdge.
int mPoint = -10
Number of the point at which this HalfEdge points.
## Detailed Description
Note
Not available in Python bindings.
Definition at line 30 of file HalfEdge.h.
## ◆ HalfEdge() [1/2]
HalfEdge::HalfEdge ( )
default
Default constructor. Values for mDual, mNext, mPoint are set to -10 which means that they are undefined.
## ◆ HalfEdge() [2/2]
HalfEdge::HalfEdge ( int dual, int next, int point, bool mbreak, bool forced )
inline
Definition at line 73 of file HalfEdge.h.
## ◆ getBreak()
bool HalfEdge::getBreak ( ) const
inline
Returns, whether the HalfEdge belongs to a break line or not.
Definition at line 93 of file HalfEdge.h.
## ◆ getDual()
int HalfEdge::getDual ( ) const
inline
Returns the number of the dual HalfEdge.
Definition at line 78 of file HalfEdge.h.
## ◆ getForced()
bool HalfEdge::getForced ( ) const
inline
Returns, whether the HalfEdge belongs to a constrained edge or not.
Definition at line 98 of file HalfEdge.h.
## ◆ getNext()
int HalfEdge::getNext ( ) const
inline
Returns the number of the next HalfEdge.
Definition at line 83 of file HalfEdge.h.
## ◆ getPoint()
int HalfEdge::getPoint ( ) const
inline
Returns the number of the point at which this HalfEdge points.
Definition at line 88 of file HalfEdge.h.
## ◆ setBreak()
void HalfEdge::setBreak ( bool b )
inline
Sets the break flag.
Definition at line 118 of file HalfEdge.h.
## ◆ setDual()
void HalfEdge::setDual ( int d )
inline
Sets the number of the dual HalfEdge.
Definition at line 103 of file HalfEdge.h.
## ◆ setForced()
void HalfEdge::setForced ( bool f )
inline
Sets the forced flag.
Definition at line 123 of file HalfEdge.h.
## ◆ setNext()
void HalfEdge::setNext ( int n )
inline
Sets the number of the next HalfEdge.
Definition at line 108 of file HalfEdge.h.
## ◆ setPoint()
void HalfEdge::setPoint ( int p )
inline
Sets the number of point at which this HalfEdge points.
Definition at line 113 of file HalfEdge.h.
## ◆ mBreak
bool HalfEdge::mBreak = false
protected
True, if the HalfEdge belongs to a break line, `false` otherwise.
Definition at line 40 of file HalfEdge.h.
## ◆ mDual
int HalfEdge::mDual = -10
protected
Number of the dual HalfEdge.
Definition at line 34 of file HalfEdge.h.
## ◆ mForced
bool HalfEdge::mForced = false
protected
True, if the HalfEdge belongs to a constrained edge, `false` otherwise.
Definition at line 42 of file HalfEdge.h.
## ◆ mNext
int HalfEdge::mNext = -10
protected
Number of the next HalfEdge.
Definition at line 36 of file HalfEdge.h.
## ◆ mPoint
int HalfEdge::mPoint = -10
protected
Number of the point at which this HalfEdge points.
Definition at line 38 of file HalfEdge.h.
The documentation for this class was generated from the following file:
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# How can you write nineteen in a manner that if we take out one, it becomes twenty?
106 views
How can you write nineteen in a manner that if we take out one, it becomes twenty?
posted Jul 9, 2018
the ans is XIX - I= XX
Similar Puzzles
+1 vote
There is a Pi picture attached here. We have given dotted lines over the places where you have to cut it. Now you will have five different pieces.
Can you join these pieces in a manner that it becomes a square?
A school orchestra with 10 musicians can play the first section of a symphony in 10 minutes.
How long would it take to play if they added 10 more musicians to the orchestra?
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### Tag: VIP Safe fixed Matches
VIP Safe fixed Matches
## VIP Safe fixed Matches
Manipulated fixed matches soccer betting ticket tips 1×2
## VIP Safe fixed Matches
VIP Safe fixed Matches
## VIP Safe fixed Matches
Accurate ticket 100% sure fixed betting matches
Day: Saturday  Date: 03.09.2022
League: FINLAND Ykkonen
Match: Jaro – PIF Pargas
Tip: Over 2.5 Goals
Odds: 1.50Â Â Result: 4:0 Won
[email protected]
WhatsApp support:Â +43 681 10831491
### Betting 100% sure fixed odds matches
There is an urban legend of mathematical modelling of VIP SAFE FIXED MATCHES. It is the legend of the mathematical genius, the Einstein of gambling, who has worked out the formula for beating the bookmakers and winning money. If only, the legend goes, you can find the tips that this person can provide, the source of the magic equation, you can become rich beyond your wildest dreams. A simple way to find VIP SAFE FIXED MATCHES.
In one section of the book, I did manage to beat the bookies. But it wasn’t because I found a magical formula that predicts who will win soccer matches fixed betting.
The way I did it was much simpler. I looked at the odds and found a very small but significant bias in how they were set. Bookmakers and bettors hadn’t paid enough attention to predicting the draw in soccer.
Maybe it is because of the popularity of the Fixed matches soccer rules. Maybe it is because bettors don’t like betting on a draw fixed matches. But, whatever the explanation, it turned out that draws in the Premier League were not properly priced.
#### Draw fixed matches betting tip
Below is a plot of the real frequency of draws in four seasons of the Premier League (2011/12, 2012/13, 2013/14, 2014/15) and the prediction of draws implied by the bookmaker’s odds.
This figure is created by taking the VIP SAFE FIXED MATCHES provided by four leading bookmakers (including FixedMatch.Bet), converting odds to implied probabilities and then looking at the difference between the probability of a home win and an away win.
When matches are skewed so there is a strong a favourite (i.e. the probability of one team or the other winning is larger than the other) then draws are over-priced (circles below red line).
Want it made simpler? If two teams are about as good as each other then the draw could be a VIP SAFE FIXED MATCHES if you opt for our paid matches. If one team is much stronger than the other, don’t bet on the draw (betting on the favourite is normally the smartest move in this case).
Testing out the theory of under-priced draws
That was what I found by plotting the odds. I then took that observation and made some money from it. Below are profits for this model for the 2015/16 season.
I tripled my money over the season. Well, actually I didn’t bet throughout the season. But I had doubled my money by Christmas.
Soccermatics came out in May 2016, just as the Premier League was coming to a close. I monitored how it went for my model the season after. Here is the result.
##### Multi-bets fixed matches ticket big odds
Not so good. There was a small profit to be made in the first few weeks, but then it flatlined for the rest of the season. Not losing money is a small achievement in itself. Where the odds are in the bookmaker’s favour, but obviously making money is the objective for most bettors.
Lessons learn from VIP SAFE FIXED MATCHES
There are four lessons to be learnt from my model.
Firstly, we’ve helped hundreds of people make money from our VIP SAFE FIXED MATCHES. Although I did write down a single equation that I then used to decide my bets (it is footnote 17 for chapter 12 in the book if you don’t want to read the rest of it) this equation came from an analysis of the odds.
The basis of my model was far from complicated. It didn’t come from me working out the strength of the teams based on past performance, advanced metrics, expected goals or anything else. It came from a small error in how the odds were being set.
Secondly, I wasn’t just lucky. The original model was consistent with the previous four years of bookmaker’s odds. I then made a prediction and applied it to the next year and it continued to work.
There is a lot of randomness in betting and it is possible to win for quite a long period of time with luck alone. But this was a long-term trend that was profitable.
Thirdly, nothing lasts forever. In moments of self-aggrandising I like to think that my book led to a market correction. Maybe the traders at FixedMatch.Bet and other bookmakers read my book and thought “we’ve been pricing draws wrong. See those odds for Liverpool at home against Manchester United at the weekend….move the draw odds up by 0.1.” That’s all it takes and my small margin disappears.
###### Manipulated fixed matches 100% sure
This is just one explanation, though. Another is that managers realised that in those big matches between equally good teams they should go for the three points (this is also something I look at in the book). There are other explanations too. The fact is, I will never know for sure, but the odds bias I found has gone.
My fourth and final conclusion is: I am a total idiot. I spent three months developing a betting model. I found a way to win. But instead of placing all my free capital on the model, I published a book with the secret in it, only to see the profits disappear.
Yes, I got paid for writing the book. Yes, I have enjoyed talking about soccer fixed matches and engaging in the analytics community, but the money would have been nice too.
There is no secret equation for predicting the outcome of soccer matches 1×2 winning tips. Not an equation that ignores the odds, in any case. If you want to create your own model of sporting outcomes you need to use the odds as the starting point.
Wisdom of the crowd tells us that the fixed matches 1×2 betting market can be hard to beat, but sometimes it makes a few small mistakes. It is these you have to look for.
In part two of this article I will see if I can find one of those cracks using a combination of an expected goals model and potential biases in recent odds.
Understanding basic bet types is vital for any punter to maximise their opportunity of making a profit. This article explains three basic bet types offered by FixedMatch.Bet; 1X2. Handicap fixed matches betting and Totals. Continue reading to learn more about the different basic bet types.
###### Genuine sources fixed matches Types: 1X2
1X2 betting fixed matches ticket tips is one of the most common forms of betting and is the simplest way for a bettor to understand the cost of a bet. We can use a hypothetical Champions League game between Borussia Dortmund and Napoli to explain how 1X2 betting winning tips today works.
If you staked ÂŁ10 on Borussia Dortmund to win at odds of 4.250, you would win ÂŁ42.50 – although that would include your initial ÂŁ10 stake. Therefore your profit would be ÂŁ32.50. You would win nothing if the result ended in a defeat or draw for Borussia Dortmund. You could also bet for PSG Football Fixed Matches to earn money through guaranteed results.
A £10 bet on a draw would have returned £37.90 (£27.90 profit and your £10 stake); while the same bet on Napoli to win would see you pocket £19.17 (£9.17 profit).
Once you are familiar with 1X2 betting winning tips today, the next step is to understand how the value in Fixed matches soccer rules varies across bookmakers, thus impacting your potential winnings.
Basic Bet Types: Handicap fixed matches
Betting on the handicap fixed matches is valuable when one team is heavily favoured over their opponents. To counter the perceived bias in ability, bookmakers offer a handicap to level the playing field.
The handicap is factored into the final score to determine the game€’s outcome for the purpose of the bet (which might be different from the actual result).
###### Handicap 100% sure betting fixed matches asia
For example, the handicap for the hypothetical example above might be (+0.5, 2.020) for Borussia Dortmund and (-0.5, 1.917) for Napoli.
If the result was 1-0 to Napoli, bets on Napoli would win as the 0.5 handicap is not enough to change the outcome. This is often called €covering the handicap
If the result was a 1-1 draw, however, bets on Borussia Dortmund would win. As the 0.5 handicap added to their score is enough to tip the match in their favour.
The handicap can also differ by appearing as a split, such as +0.5 & +1 or -0.5 & -1. This takes your stake and effectively places two handicap bets fixed matches -€ one at each option. Therefore you can win both bets, just one, or lose both.
Advanced bettors will understand the relationship between 1X2 fixed matches tips and handicap betting fixed matches and how to decide which option offers better value.
Basic Bet Types: Totals
Totals bets are another basic bet type, and a popular alternative to 1X2 handicap betting fixed matches. The focus is not on finding the winner of a particular event. But predicting the outcome of game variables (e.g. total goals/runs/points or any other easily quantifiable variable).
Bookmakers offer bettors the option of whether the total goals/points. They will be either OVER or UNDER their assessment of the game.
If you bet ÂŁ10 on the game to have over 2.5 goals and the result was 2-1 (a total of three goals) you would make a return of ÂŁ20.80 (ÂŁ10.80 profit). While you would have won nothing had you bet on the goals total to be under 2.5.
Updated: September 3, 2022 — 6:55 pm
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Jun 24, 2019 7:16 p.m. Ján
Course syllabus D1-MMPVD - Mathematical models in fluid flow, heat transfer and diffusion (FCE - 2018/2019 - post-graduate studies)
Information sheet ECTS Syllabus
Slovak English
University: Slovak University of Technology in Bratislava
Faculty: Faculty of Civil Engineering
Course unit code: D1-MMPVD
Course unit title: Mathematical models in fluid flow, heat transfer and diffusion
Mode of delivery, planned learning activities and teaching methods:
workshop 2 hours weekly / 26 hours per semester of study (on-site method)
Credits allocated: 5
Recommended semester/trimester: Theory and Environmental Technology of Buildings - doctoral (semi-compulsory), 1. year
Level of study: 3.
Prerequisites for registration: none
Assesment methods:
Correct elaborating of all assigments during semester.
Obtaining at least 56 points on the exam.
Learning outcomes of the course unit:
The aim of the course is to familiarize PhD students with mathematical models and numerical methods used in solving problems of hydrodynamics as well as thermal and diffusion problems.
Course contents:
• Heat equation, equation of diffusion and flow in porous media.
• Continuity equation, Euler's equations of ideal gas flow.
• Navier - Stokes equations for viscous flow.
• Reynolds equations for turbulent flow.
• Numerical methods of solving partial differential equations - the finite element method and the finite volume method.
Basic:
REDDY, J. An Introduction to the Finite Element. New York : McGraw-Hill, 1993. 684 p. ISBN 0-07-112799-2. LEVEQUE, R J. Finite volume methods for hyperbolic problems. Cambridge : Cambridge University Press, 2002. 558 p. ISBN 0-521-00924-3. FEISTAUER, M. Mathematical methods in fluid dynamics. London: Longman Scientific and Technical, 1992. 657 p.
Language of instruction: slovak and english or english
Notes:
Courses evaluation:
Assessed students in total: 3
PN
100,0 %0 %
Name of lecturer(s): doc. RNDr. Peter Frolkovič, PhD. (examiner, instructor, lecturer) - slovak, english
prof. RNDr. Karol Mikula, DrSc. (examiner, instructor, lecturer, person responsible for course) - slovak, english
Last modification: 31. 8. 2015
Supervisor: prof. RNDr. Karol Mikula, DrSc. and programme supervisor
Last modification made by Ing. Peter Korčák on 08/31/2015.
Type of output: PDF output (PDF)Document RTF (RTF)XML format for IS (XML IL)Plain (plain)
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# In the years since the city of London imposed strict
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In the years since the city of London imposed strict air-pollution regulations on local industry, the number of bird species seen in and around London has increased dramatically. Similar air-pollution rules should be imposed in other major cities.
Each of the following is an assumption made in the argument above EXCEPT:
(A) In most major cities, air-pollution problems are caused almost entirely by local industry.
(B) Air-pollution regulations on industry have a significant impact on the quality of the air.
(C) The air-pollution problems of other major cities are basically similar to those once suffered by London.
(D) An increase in the number of bird species in and around a city is desirable.
(E) The increased sightings of bird species in and around London reflect an actual increase in the number of species in the area.
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08 Sep 2006, 19:54
Hey, we discussed this question a couple of days ago. Answer is A.
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https://maths.mdx.ac.uk/research/modelling-the-covid-19-pandemic/how-do-covid-19-epidemics-end/
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Mathematics
# How do COVID-19 epidemics end?
This question was brought into focus by the publication of a “dodgy” paper appearing to claim on the basis of modelling that India’s COVID-19 epidemic would ‘end’ in September. Perhaps, in fact, the question should be “Do COVID-9 epidemics ever end?” It seems increasingly that for many of us COVID-19 is here to stay. Outbreaks wax and wane, and infection may be reduced to low levels but still refuse to die out.
This is not to say that elimination cannot occur. New Zealand appears to have achieved elimination, though at the time of writing, one cannot be sure there is no undetected spread. Other countries, such as Vietnam and South Korea may well achieve something similar. But it is fair to say that for many countries, this is a distant dream.
It’s useful to start by thinking about what COVID-19 doesn’t do. Naive modelling of epidemics is often accompanied by pictures which look something like Figure 1.
Such pictures suggest some symmetry in the rise and fall of an epidemic. I recently learned from an article in The Caravan that this principle has even got a name – “Farr’s law”. Farr apparently formulated his empirical law in 1840 by studying data from smallpox outbreaks and outbreaks of other infectious diseases.
Let’s first be clear about one thing: COVID-19 is, in general, not obeying Farr’s law. In Figure 2 are the daily new cases for two countries which have been relatively successful in dealing with their COVID-19 outbreaks: Germany and Denmark.
Despite random fluctuations and other effects created by how and when cases get detected and reported, we clearly see the relatively rapid rise and the relatively slow decline in the numbers. The rise is equally steady, but the decline even slower and more uncertain, in countries which have been less successful at mitigation such as the UK and the USA (Figure 3).
In fact, at the time of writing, it is far from clear in many countries which appear to have passed their (first) peak whether the numbers are tailing off towards zero (as in New Zealand), whether there will be resurgence (as in Iran), or whether daily new cases will reach some baseline level around which they simply hover. The same pattern of rapid rise and slow, uncertain fall is found for a number of other COVID-19 outbreaks which have passed peak, both at country and city level.
So why the discrepancy between the real-world asymmetry and the textbook curves? We’ll find, using modelling, that the textbook curves are consistent with a very simple assumption which does not hold for COVID-19 outbreaks. On the other hand the slow tailing off is quite natural behaviour for COVID-19 when accompanied by mitigation measures such as lockdown.
The modelling approach used is stochastic and agent-based. It has been described previously at https://maths.mdx.ac.uk/research/modelling-the-covid-19-pandemic/, with code at https://github.com/muradbanaji/COVIDAGENT. All parameter values for the simulations below are given in the Appendix.
Underlying the modelling is a plausible assumption which, however, is not self-evident:
Immunity assumption: most people who get infected with COVID-19 and recover are then immune for a significant period of time (i.e., longer than the time-scale of our modelling).
Although this assumption is not uncontroversial, the evidence so far points to it being broadly correct. Moreover, I believe that without this assumption much of the COVID-19 data we see from around the world simply cannot be explained – more on this below. Nevertheless, in making this assumption, we should remember that we know relatively little about the likelihood or duration of post-infection immunity.
Having stated the assumption about immunity, let’s simulate some simple “toy” scenarios for outbreaks.
Herd immunity. First (no surprise here) an outbreak can end because herd immunity develops. A simulation of a simple herd immunity scenario is presented in Figure 4.
In this scenario we do indeed see the symmetric rise and fall of Farr’s law. Very likely it was the development of acquired immunity underlying the small pox data which led to Farr formulating his empirical law. Anyway, in this scenario disease sweeps through the population, slowing as more and more people become immune. The duration of the outbreak depends on the size of the population, and the outbreak really does “end” in the sense that daily new cases eventually go to zero. At the end of the outbreak, the proportion who have been infected depends on the initial R value and stochastic fluctuation. But my simulations suggest it would not be much below 80% for COVID-19. So, a scenario like this is quite devastating. I should say, at the outset, that this scenario has probably not, so far, unfolded, in any country, city, or perhaps even neighbourhood.
Physical distancing. A second toy scenario for how a COVID-19 outbreak can end is because of the introduction of strong “physical distancing”. This should be read in a general sense to include increased hygiene, mask wearing, and all behaviours which reduce events of the kind which spread infection – including contact tracing followed by isolation. A simulation of a scenario with the sudden introduction of physical distancing is shown in Figure 5.
In this artificial scenario, physical distancing suddenly starts (at the first fatality in this simulation), and daily new infections drop immediately. The effect is set to be strong enough to reduce “R” to below 1, so new infections eventually tail off. Of course, in real situations, behaviours do not change this suddenly. Moreover, testing occurs at variable points after infection so any such drop would be smoothed out in confirmed cases data. But still, in such a scenario we should, within a couple of weeks, see a clear drop in daily cases. With plausible levels of physical distancing, infections will tail off more slowly than they rise, and Farr’s law is again violated. A noteworthy feature of this scenario is that it is quite possible that only a small proportion of the total population will have been infected at the end of an outbreak. Of course, for all this to occur, the high level of physical distancing has to be maintained over a prolonged period.
Are either of the above scenarios relevant to COVID-19? Both of the toy scenarios above are very artificial, and I don’t think either explains the evolution of COVID-19 outbreaks in general. Why?
Firstly, regarding the simple herd immunity scenario: seroprevalence studies have shown that COVID-19 outbreaks – even the bad ones such as that in New York City – are not resulting in sufficient immunity to explain the drop-off in infections when outbreak subsides. The same holds for London. Nowhere near enough people were infected to give generalised herd immunity in these cities as a whole; and yet infections tailed off.
Secondly, physical distancing rarely seems to be effective enough to immediately control a COVID-19 epidemic on its own. Many countries – with or without lockdowns – have introduced measures which reduce infection spreading contacts; but in the cases I have looked at, daily new cases may dip temporarily but do not pass peak very soon after mitigation. Namely, R is not immediately brought to below 1. If nothing else, the disease still spreads in families and close communities. (Of course, it might have occurred in some outbreaks that R immediately reduced to below 1 following mitigation – I just haven’t seen any convincing evidence of this to date.)
A more realistic scenario. So, are physical distancing and herd immunity not what cause COVID-19 outbreaks to wind down following mitigation? On the contrary, they are, in combination and with some important tweaks. I tried to explain broadly what happens in lockdown in a recent piece for Scroll. This was focussed on India but the main ideas are more widely applicable.
To begin with, physical distancing (in the general sense) does seem to increase after mitigation measures and cause a reduction in transmission. We see this in many COVID-19 data sets: it manifests as a (temporary) fall in daily new cases. If we plot the logarithm of total cases then it manifests as a change in slope shortly after mitigation. The same holds if we plot fatalities, but with a greater delay. This effect seems to be more or less evident in different countries’ and regions’ data.
But despite this reduction in transmission, in regions I’ve looked at, daily new cases continue to increase for some time – at least three weeks, often more – after the brief fall. The eventual approach to peak is gradual and reminiscent of what happens with increasing immunity. So, reaching and passing peak seems to involve a second part of the story – local herd immunity.
The phrase “herd immunity” is often misunderstood, so it’s important to clarify here. A key effect of lockdown is to localise disease and reduce the effective “infectible” population, by placing barriers to free movement and contact. To put it in a simple-minded, but conceptually useful, way: lockdown divides people into those who belong to a community where there is infection and those who do not. This localisation of disease may occur at the level of households, close social networks, care homes, and so forth. Within the infectible population, herd immunity starts to develop until eventually we see infection levels subsiding. The non-infectible population is, by definition, protected by being removed from view as far as the virus is concerned.
The size of the infectible population, and hence of the outbreak, depends on when mitigation starts. For example, the very late UK lockdown left a much larger part of the population infectible than an early lockdown would have. Modelling indicates that this was behind the UK’s high COVID-19 mortality. In countries with low mortality it seems that early mitigation was successful at keeping the great majority of the population out of view of the virus.
Leak into the infectible population. But there’s more to this story of localisation and herd immunity. I don’t think it is ever realistic to assume that the infectible population is fixed. Rather, there is going to be leakage into this population even in the strictest of lockdowns. All it takes for a whole community to enter the “infectible” category is for one member to acquire infection. So, there’s a competition between developing herd immunity in the infectible population which effectively removes individuals from this population on the one hand, and the addition of new susceptible individuals into the infectible population on the other hand. It is the balance between these competing effects which leads to stabilisation of the disease.
For people familiar with reaction dynamics, we have something analogous to a continuous flow reactor with a supply of new reactant (infectible people) entering the reactor and sustaining the reaction which would otherwise wind down, having used up all the reactant.
In Figure 6 is a simulation of a more realistic lockdown scenario, in which both physical distancing and disease localisation occur and, moreover, there is leak (after a short delay) into the infectible population.
Stabilisation of disease levels is accompanied by R values which hover around 1. The level of infection at which stabilisation occurs is low if the leak is small, and mitigation effective. It may effectively reach zero: this is the goal, but is probably also the exception.
Application to India. India’s COVID-19 epidemic has been slowed but not yet contained by lockdown, and hence is at an earlier stage than current European outbreaks in terms of its evolution. We have not yet reached peak. Lockdown failed to prevent the severe escalation of city outbreaks, and now that lockdown is being eased, the likelihood of disease being exported from hotspots to new areas – “delocalisation” – is increased. The Indian experience shows that heavy-handedness and effectiveness are not synonymous and it is possible that what was achieved at very high cost with national lockdown could have been achieved with more limited and localised measures focussed on areas of relatively high prevalence.
When we model the Indian data, we find that there are many plausible futures for the outbreak, consistent with its evolution so far. These give us insight into what might happen, but no firm answers. We can infer that recent predictions about the end of the epidemic must be based on naive assumptions. In Figure 7 are plots from a simulation whose outputs reasonably match Indian recorded infections to date (taken to be a constant fraction of total infections tested around a particular moment in the disease cycle). It is important to clarify that this is by no means a “prediction”. It is just a simulation consistent with the data so far, and has been chosen to illustrate a point.
In this reasonably optimistic scenario, peak in daily reported cases is reached a few weeks from now and there is no subsequent resurgence à la Iran. Daily cases decline quite rapidly, but slower than they rose and – here comes the point – eventually plateau. Different choices of model parameters, all consistent with the data to date, change the height and timing of peak, the speed of descent, and the level of the subsequent plateau. Bearing in mind that the numbers plotted above are recorded cases which are at best 5% of true infections, probably much less, stabilisation with a couple of thousand recorded new infections a day may not seem like a “victory”; but it is certainly a lot better than some alternatives.
Summary. Let’s return to the initial question: do SARS-CoV2 outbreaks truly end? Sometimes. Where this has been achieved, or is close to being achieved, it is not because the disease has swept through a population ravaging it so badly that herd immunity is reached. Rather, a combination of stringent and effective measures have reduced transmission and localised the disease to such an extent that remaining infections can be eliminated through rigorous testing, tracing and isolation.
While this might be the “gold standard” in the control of COVID-19, I suspect that the most common scenario will be that infection is beaten back to relatively low levels, but persists at these levels for months, with occasional new surges until, hopefully, a vaccine is found. The “relatively low levels” are going to vary widely between different countries/states/cities. Existing levels of herd immunity in a locality will slow new outbreaks and help disease surveillance, tracing and isolation to do their work. In a large country like India, elimination may be achieved locally, but one should always assume there might be some spread under the radar. And unless borders are truly sealed – an unsustainable state of affairs – there will be new introductions to COVID-free areas.
Basically, dealing with COVID-19 is going to be a long, hard, slog. But badly hit countries – the UK, USA, and Brazil come to mind – serve as a warning that despite the difficulties, bringing infection to low levels and maintaining it there must be a priority. It is increasingly recognised that the damage from high levels of COVID-19 infection should not just be measured in COVID-19 fatalities, but also in the toll on health systems with many people losing access to healthcare, and the long-term health complications which can follow COVID-19 infection. Once a low level of infection is achieved and lockdowns are eased, avoiding unnecessary gatherings and travel, maintaining hygiene, and generally cautious behaviours, will be essential to slowing new outbreaks allowing them to be caught before they reach levels hard to control. But perhaps most important of all will be testing, tracing and surveying coupled with credible modelling to identify developing hotspots, followed by well-rehearsed containment measures to stop the export of disease to new areas.
Appendix: parameter values for the simulations shown
Parameters for the simulations in Figures 4 and 5 respectively. Parameters not shown are set to defaults and do not affect the simulation.
number_of_runs 1
death_rate 0.5
geometric 0
R0 2.5
totdays 200
population 50000, 20000
inf_start 3
inf_end 14
time_to_death 17
dist_on_death 0
time_to_recovery 20
dist_on_recovery 0
initial_infections 2,10
herd 1,0
percentage_quarantined 0
haslockdown 0
physical_distancing 0,1
pddth N/A, 1
pdeff1 N/A, 70
Parameters for the simulation in Figure 6.
number_of_runs 1
death_rate 0.5
geometric 0
R0 3.5
totdays 300
inf_start 3
inf_end 14
time_to_death 17
dist_on_death 6
time_to_recovery 20
dist_on_recovery 6
initial_infections 2
percentage_quarantined 20
percentage_tested 50
testdate 10
dist_on_testdate 6
herd 1
population 10000000
haslockdown 1
lockdownlen 300
infectible_proportion 0.03
physical_distancing 0
pddth 5
pdeff1 10
lockdth 5
pdeff_lockdown 60
popleak 2500
popleak_start_day 20
Parameters for the simulation in Figure 7. A scaling has been applied to speed up computation: the death rate is set to 20 times its value, and all populations are set to 1/20 of their values. True values are shown in brackets.
number_of_runs 1
death_rate 10 (0.5)
geometric 1
R0 4.2
totdays 350
population 4000000 (80000000)
inf_start 3
inf_end 14
time_to_death 19
dist_on_death 6
time_to_recovery 20
dist_on_recovery 6
time_to_sero 10
dist_on_sero 6
initial_infections 10
herd 1
percentage_quarantined 12
percentage_tested 50
testdate 12
dist_on_testdate 6
haslockdown 1
lockdth 23
lockdownlen 350
infectible_proportion 0.1350
pdeff_lockdown 62
popleak 10000 (200000)
popleak_start_day 14
physical_distancing 1
pddth 1
pdeff1 30
sync_at_test 380 (7600)
sync_at_time 38
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# Ch 5(B) : Capacity Planning: Break-Even Analysis Fixed costs Variable costs
```Ch 5(B) : Capacity Planning: Break-Even Analysis
Operation costs are divided into 2 main groups:
• Fixed costs
• Variable costs
1
Fixed Costs
Include rent, property tax, property insurance, wages
of permanent employees, depreciation (except in
working hour depreciation).
The total fixed cost is fixed throughout the year.
It does not depend on the production level.
When we have a plant, then the above costs are
fixed, no matter if we produce one unit or one million
units.
2
Total Fixed Cost and Fixed Cost per Unit of Product
Total fixed cost
(F)
Fixed cost per unit of product
Production volume (Q)
Production volume (Q)
(F/Q)
3
Variable Costs
Costs of raw material, packaging material, direct
labor, production W&P are the main variable costs.
Variable cost is fixed per unit of production.
The total variable costs depend on the volume of
production.
The higher the production level, the higher the total
variable costs.
4
Variable Cost per Unit and Total Variable Costs
Variable costs
Per unit of product
(V)
Total Variable costs
(VQ)
Production volume (Q)
Production volume (Q)
5
Amount (\$)
Total Costs
Total Fixed cost (F)
0
Q (volume in units)
6
Total Revenue
It is assumed that the price of the product is fixed,
and we sell whatever we produce.
Total sales revenue depends on the production level.
The higher the production, the higher the total sales
revenue.
Price per unit
(P)
Production (and sales) (Q)
Total revenue
(TR)
Production (and sales ) (Q)
7
Amount (\$)
Break-Even Point
0
BEP units
Q (volume in units)
8
Break-Even Computations
TC=TR
TC=F+VQ
TR=PQ
F+VQ=PQ
QBEP = F/ (P-V)
9
Example
\$500,000 total yearly fixed costs.
\$150 per unit variable costs
\$200 per unit sale price
QBEP=500,000/(200-150) =10,000 units
If our market research indicates that the
present demand is > 10,000, then this
manufacturing system is economically
feasible.
10
BEA for Multiple Alternatives
Break-even analysis for multiple alternatives:
Such an analysis is implemented to compare cases such as
A Simple technology
An Intermediate technology
General purpose machines
Multi-purpose machines
Special purpose machines
Low F high V
In between
High F Low V
In general, when we move from a simple technology to an
11
BEA for Multiple Alternatives
CIM
CNCs
Universal
Q1
Q2
12
BEA for Multiple Alternatives
1) A General Purpose machine
Total Fixed Cost F = \$10,000,
Variable cost
V = \$10 per unit
2) A Multi Purpose machine
Total Fixed Cost F = \$60,000,
Variable cost
V = \$5 per unit
3) A Single Purpose machine
Total Fixed Cost F = \$150,000,
Variable cost
V = \$2 per unit
Tell me what to do: In terms of the range of demand
and the preferred choice…
13
BEA for Alternatives (1) and (2)
2
1
Q1
14
Break-Even for Alternatives (1) and (2)
F1=10000
F2=60000
V1=10
V2=5
Break-even of 1 and 2
F1+ V1 Q = F2+ V2 Q
10000+10Q = 60000 + 5Q
Q=
60000 10000
10000
10 5
15
BEA for Alternatives (2) and (3)
3
2
Q2
16
Break-Even for Alternatives (2) and (3)
F2=60000
V2=5
F3=150000
V3=2
Break-even of 2 and 3
F2+ V2 Q = F3+ V3 Q
60000 + 5Q = 150000+2Q
Q
150000 60000
30000
=
52
17
Recommendations to Management and Marketing
Demand
Recommended Alternative
D < 10000
Alternative 1
10000 < D < 30000
Alternative 2
30000 < D
Alternative 3
We also need to know Price and Revenue!
18
Assignment 3: Problem 1
A firm plans to begin production of a new small appliance.
Management should decide whether to make the small engine
for this product in-plant, or buy it from an outside source.
If management decides to make the engine, then there are 2
alternatives:
(1) Built it with a simple manufacturing system
Fixed Cost:
\$10,000/year
Variable Cost: \$8 per unit
(2) Built it with an advanced manufacturing system.
Fixed Cost:
\$30,000/year
Variable Cost: \$5 per unit
The purchase price of the engine is \$10 per unit.
19
BEP for the Three Alternatives and Recommendations
Prepare a table similar to the following table, to summarize
Demand
Recommendation
Q <= ?
?
?<Q<=?
?
?<Q
?
20
Assignment 3: Problem 2
A manager has the option of purchasing 1, 2 or 3 machines.
The capacity of each machine is 300 units.
Fixed costs are as follows:
Number of Machines
1
2
3
Fixed cost
\$9,600
\$15,000
\$20,000
Total Capacity
1-300
301-600
601-900
Variable cost is \$10 per unit, and the sales price of product is
\$40 per unit.
Tell management what to do!
21
BEP for the Three Alternatives and Recommendations
Prepare an executive summary similar the following:
Q<= ? ?
?<Q<=? ?
Q>?
?
Now it is up to the Marketing Department to provide an
Executive Summary regarding the demand.
22
Assignment 3: Problem 3
You are the production manager and are given the option to
purchase either 1, 2 or 3 machines. Each machine has a
capacity of 500 units. Fixed costs are as follows:
Number of Machines
1
2
3
Fixed cost
\$19,200
\$30,000
\$40,000
Total Capacity
1- 500
501-1000
1001-1500
Variable cost is \$35 per unit, and the sales price of product is
\$69 per unit.
Determine the best option!
23
BEP for the Three Alternatives and Recommendations
Prepare an executive summary similar the following:
Q<= ? ?
?<Q<=? ?
Q>?
?
24
```
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http://www.edx.org/course/discrete-time-signals-and-systems
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• Length:
8 Weeks
• Effort:
6–8 hours per week
• Price:
FREE
Add a Verified Certificate for \$249 USD
• Institution
• Subject:
• Level:
• Language:
English
• Video Transcript:
English
• Course Type:
Instructor-led on a course schedule
## Prerequisites
Advanced calculus, complex algebra, and linear algebra.
Technological innovations have revolutionized the way we view and interact with the world around us. Editing a photo, re-mixing a song, automatically measuring and adjusting chemical concentrations in a tank: each of these tasks requires real-world data to be captured by a computer and then manipulated digitally to extract the salient information. Ever wonder how signals from the physical world are sampled, stored, and processed without losing the information required to make predictions and extract meaning from the data?
Students will find out in this rigorous mathematical introduction to the engineering field of signal processing: the study of signals and systems that extract information from the world around us. This course will teach students to analyze discrete-time signals and systems in both the time and frequency domains. Students will learn convolution, discrete Fourier transforms, the z-transform, and digital filtering. Students will apply these concepts in interactive MATLAB programming exercises (all done in browser, no download required).
Learners should have strong problem solving skills, the ability to understand mathematical representations of physical systems, and advanced mathematical background (one-dimensional integration, matrices, vectors, basic linear algebra, imaginary numbers, and sum and series notation). This course is an excerpt from an advanced undergraduate class at Rice University taught to all electrical and computer engineering majors.
### What you'll learn
Skip What you'll learn
• Types of Fundamental Signals
• Vector Description of Signals
• Introduction to Discrete Time Systems
• Convolution
• The Discrete Fourier Transform (DFT) and Fast Fourier Transform (FFT)
• The Discrete-Time Fourier Transform (DTFT)
• The Z-Transform
• Introduction to Analysis and Design of Discrete-Time Filters
Richard G. Baraniuk
Victor E. Cameron Professor of Electrical and Computer Engineering
Rice University
### Pursue a Verified Certificate to highlight the knowledge and skills you gain\$249.00
• #### Official and Verified
Receive an instructor-signed certificate with the institution's logo to verify your achievement and increase your job prospects
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Use this template to collect all materials and information related to a class/course.
Get Started. It's Free
## 1. Course Information
### 1.2. Zerin Ahmed
1.2.1. Name
1.2.2. Office hours
1.2.3. Contact
### 1.4. MPM 2D,Principles of Mathematics
1.5.1. E-learning
1.5.2. Student forum
## 2. Exam Info
### 2.1. NOV 30
2.1.1. 1st Sitting
2.1.2. 2nd Sitting
2.1.3. 3rd Sitting
### 2.3. 11.10am
2.3.1. Essay exam
2.3.2. Oral exam
2.3.3. Multiple choice
## 3. Course Topics
### 3.1. Linear Systems
3.1.1. Function
3.1.2. linear
3.1.2.1. straight slope "m" is the same .
3.1.2.1.1. steepness
3.1.3. non-linear
3.1.3.1. curved-linear slope will change
3.1.4. m=y2-y1/x2-x1
3.1.5. y-int : y=mx+b
3.1.5.1. at what point on the y-axis cloes the line cross
3.1.6. standard : Ax+By+C=0
3.1.7. direct variations (orgin) =(0.0)
3.1.8. partial variations
3.1.9. 2 linear functions on the same grid
3.1.10. Types of solutions——points of intersestion .where the 2 functions meet.
3.1.10.1. no solution. m1=m2 parallel lines
3.1.10.2. one solutions m1≠m2
3.1.10.3. infite solution m1=m2 b1=b2
### 3.2. Analytical Geometry
3.2.1. line ,point , line segment , ray .d=√(x2-x1)(x2-x1)+(y2-y1)(y2-y1)
3.2.2. M is midpoint
3.2.2.1. a point the same length from the midpoint to each endpoint.
3.2.2.1.1. M(x2+x1)/2,(y2+y1)/2
3.2.3. All sides are equal equilateral
3.2.4. 2 sides are equal isosceles
3.2.5. no sides are equal scalene
3.2.6. right angle 2 slope are perpendicular m=-1/m2
3.2.6.1. m1=0 m2=und undefined
3.2.7. parallel m1=m2
3.2.8. perpendicular m1=-1/m2
3.2.9.1. square
3.2.9.2. 2 sets of para
3.2.9.3. 2 sets of perem
3.2.9.4. all the sides are equall
3.2.10. rectangle
3.2.10.1. 2 sets of para with the same length
3.2.10.2. 2 sets of prep
3.2.11. rhombus
3.2.11.1. 2 sets of parallel
3.2.11.2. all sides lengths are equal
3.2.12. parallelogram
3.2.12.1. 2 sets of parallel
3.2.12.2. with the same length
3.2.13. trapezod
3.2.13.1. 1 set of parallel
3.3.1. parabola
3.3.1.1. follows a parabolic motion
3.3.1.1.1. start at the vertex move 1 unit to the right each time then follows the step - pattern
3.3.2. standard form:y=ax2+bx+c
3.3.3. factor form :y =a（x+-s)(x+-r)
3.3.4. vertex form: y= a(x-h)2+k
### 3.4. Trigonometry
3.4.1. almost the same
3.4.1.1. similar trangies
3.4.1.1.1. all angles are the same
3.4.1.1.2. side length different
3.4.2. Z shape
3.4.2.1. atternating angle
3.4.2.2. same
3.4.3. F shape
3.4.3.1. corresponding angles
3.4.3.2. same anle
3.4.4. C shape
3.4.4.1. interior angle
3.4.4.2. sum =180
3.4.5. sin=opp/hyp
3.4.5.1. CSCΘ=hyp/opp 1/sinΘ
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3 Fat Chicks on a Diet Weight Loss Community Is 10 lbs a month actually unreasonable and unrealistic?
Register FAQ Calendar Search Today's Posts Mark Forums Read
South Beach Diet Fat Chicks on the Beach!
08-19-2008, 10:00 AM #1 Healthy mommy Join Date: Jan 2006 Location: North Carolina, USA Posts: 1,418 S/C/G: 246/235/150 Height: 5'8 3/4 Is 10 lbs a month actually unreasonable and unrealistic? I'm asking this because in the past, I thought 10 lbs a month would be easy, nothing to it, but have found that is not actually the case. But I've heard others (not on this forum, but people I know IRL) speak of losing 10-15 lbs in a month. Last month I lost 6 lbs which I was quite happy about it. But for me, to lose 10 lbs in a month would be pretty difficult. After much discussion on this forum and whatnot, I came to the realization that it wasn't right for me to have a goal, such as 'lose x amount of weight by x amount of time.' That's the sort of thinking I used to have and have since realized how unrealistic it was, and how much pressure it put on me, and ultimately I would fail as a result. So, my healthy weight goal is around 150, but when it comes off, it comes off. As long as I keep working at it, I can't complain. I realize I will not lose 6 lbs every month, that it will fluctuate, and some months I may lose hardly anything at all. If I did a calculation using 6lbs a month to see how much I would have lost by Christmas (in the calculation hypothesis, it would be 24 lbs, putting me at 169), it would be inaccurate. So I won't try, because I know from experience I'm only gonna end up dissapointed, and I want to be thrilled with my achievements, no matter how minor. Anyway, so isn't 10 lbs a month a little unrealistic, except for those who are very overweight? __________________ First Goal: 25-40lbs by December 25th Second goal: To be decided when the first goal is obtained Last edited by Fat Melanie; 08-19-2008 at 10:01 AM.
08-19-2008, 10:02 AM #2 ONEderland here I come! Join Date: Jan 2005 Location: maryland Posts: 2,967 S/C/G: 286/210/200 (next goal) Height: 5'2.75" some months i lose nothing some months i gain some months i lose 3 or 4 pounds for ME 10 per month is unrealistic. __________________ Nessa -- on the beach since May 1, 2006 added Points August, 2008 Having RNY September 22, 2009
08-19-2008, 10:18 AM #3 Larry's Angel Join Date: Jan 2006 Location: NW New Jersey But, My Heart's In Pittsburgh!! GO STEELERS & PENGUINS!!! Posts: 3,060 S/C/G: 245/143/145 I'm not on SB and I had at least 100 lbs to lose. I lost 10 lbs a month for several months but...I believe whats unrealistic is thinking that it will continue indefinitely. The recommended 1-2 lbs a week seems reasonable. __________________ Start Each Day With A Grateful Heart Still Going Strong On Atkins Since April 13, 2004 Before & After Pics: http://i13.photobucket.com/albums/a2.../Kimatlake.jpg http://i13.photobucket.com/albums/a2...1/IMG_0403.jpg
08-19-2008, 10:24 AM #4 Healthy mommy Join Date: Jan 2006 Location: North Carolina, USA Posts: 1,418 S/C/G: 246/235/150 Height: 5'8 3/4 I think 10 lbs is definitely unrealistic for me as well, though *sigh*... it would be nice. Hi JerseyGyrl, I think when one has 100 lbs to lose then they can probably lose 10 lbs a month for awhile, like you said. But my god, WOW... I'm looking at your ticker and I'm really amazed. 102 lbs, that is fantastic and I'm highly jealous right now. Congrats! (oh Dr Phil, I used to despise him and found him to be so arrogant, but now I love watching that show, he tells it like it is) But yeah, before I start going OT, I guess it all depends on the person and the weight they have to lose. I don't understand how these girls I know lost 10-15 lbs in a month though, they never had much to lose to begin with. Starvation diets, maybe? I dunno. __________________ First Goal: 25-40lbs by December 25th Second goal: To be decided when the first goal is obtained
08-19-2008, 10:58 AM #5 Anne Join Date: Sep 2005 Location: Ontario, Canada Posts: 1,631 S/C/G: 407/358-Dec2007/tracker/125 Height: 5'4" My 3FC Diet Blog I looked back and I've lost anywhere from 6.6 to 12 pounds over the last 8 months. 10 pounds would be a very good month, even with a lot to lose. It's hard to maintain enough of a deficit (from food or exercise) to lose more. The exception might be the start of a diet where most people lose a lot of water. Consider any month one where you end up lighter at the end of the month. I avoid 'weigh so much by a certain date' goals because as you said, it's easy to be disappointed with it. __________________ Anne ---> Pictures, pictures, pictures
08-19-2008, 11:12 AM #6 Come on ONEderland! Join Date: Mar 2008 Location: California Posts: 367 S/C/G: 254/ticker/175 Height: 5'-7" When i first started with P1, I lost 18lbs pretty quickly. And I did lose 10 lbs in one month once. Now I feel lucky if I lose 5 lbs a month, which is pretty reasonable, but I have to work harder and harder for those 5 pesky pounds. If you really, really worked hard, you could probably lose the 10 lbs, but I don't know if it's a maintanable effort level. Lose 1 to 2 lbs a week, and you know its sustainable. __________________ Life Goal 4th of July Challenge Don't let "Not Perfect" be the enemy of "Good."
08-19-2008, 11:55 AM #7 Junior Member Join Date: Aug 2008 Location: CA Posts: 6 S/C/G: 185/185/145 Height: 5'7 I have a new perspective for myself with weight loss. If I am not losing a reasonable amount, (1-2lbs a week) then I will increase my physical activity.
08-19-2008, 12:01 PM #8 Member Join Date: Aug 2008 Location: Maine Posts: 42 S/C/G: 222/ticker/150 Height: 5'8" My 3FC Diet Blog From my own experience I think 10lbs/ mo is realistic for the first month or two. After that I tend to plateau and loose more like 5 lbs a month. The first 15-20 lbs are always WAy easier than those that follow, for me anyway. I say try for 10 but don't count on it. That way if you get it you'll be happy, and if you don't you won't have set yourself up for dissapointment.
08-19-2008, 12:24 PM #9 beachin' Join Date: Apr 2008 Location: Lafayette, Louisiana Posts: 4,190 S/C/G: 240/125/130 Height: 5'8 As you lose more, it becomes harder and harder to lose weight quickly.....like many have said, healthy weight loss is somewhere in the ballpark of 1-2 lbs. per week. Adjust your calories and exercise accordingly, and you shouldn't have much problem with plateaus.......that being said, some plateaus happen for reasons unknown and only experimentation (with calories and different types of exercise) and time will break them. Be patient. And always, remember that weight loss is not a lateral thing....you can lose a buttload (literally) one month, and the next not lose a thing. Just keep tabs on your diet and the amount of exercise you're getting and adjust accordingly. __________________ Maintenance! 100 lbs. down plus a carrot for every 5 more
08-19-2008, 12:31 PM #10 Senior Member Join Date: Aug 2008 Location: Atlanta, Georgia Posts: 118 Height: 5'11 I agree, a 10 lb. weight loss would be doable if you have a lot of weight to lose and only in the beginning.
08-19-2008, 05:24 PM #12 Shooting for the moon Join Date: Jul 2008 Location: Hampton Roads Posts: 662 S/C/G: 210/151/140 Height: 5' 5.5" Kaplods - I really need to focus on that myself. I have been SO good as far as eating according to plan and exercising regularly but yesterday when I weighed in I had a gain and I just felt like crying. I mean- I am busting my butt ya know? So I really should just feel great about that and about how much more healthy my and my family's eating has become but instead I have been feeling like crap about the gain. I like the hair analogy because getting mad at yourself for your hair not growing fast enough seems laughable. __________________ Starting Weight 7.22.2008 = 210 1st Mini Goal Met 10.5.2008 = 185 2nd Mini Goal = 170 Met January 9, 2009 3rd Mini Goal = 155 Met March 27, 2009 Progress pics Do you not know that in a race all the runners run, but only one gets the prize? Run in such a way as to get the prize. -1 Corinthians 9:24 Last edited by Thin4Good; 08-19-2008 at 05:25 PM.
08-19-2008, 07:12 PM #13
Choosing with every bite.
Join Date: Oct 2007
Location: Florida
Posts: 2,859
S/C/G: 212.5/182/155
Height: 5' 7"
Quote:
Originally Posted by Fat Melanie After much discussion on this forum and whatnot, I came to the realization that it wasn't right for me to have a goal, such as 'lose x amount of weight by x amount of time.' That's the sort of thinking I used to have and have since realized how unrealistic it was, and how much pressure it put on me, and ultimately I would fail as a result. . . .As long as I keep working at it, I can't complain.
Quote:
Originally Posted by kaplods My personal belief is that goals should not be result-oriented, but behavior oriented. What I can control is what I eat and what I do, so those are my goals. The weight is an indirect goal.
This shift in thinking appears to be one of the critical measures of weight loss and weight maintenance success based on what I've observed here. It's part of the change from "diet" thinking to "Way of Life" thinking. It allows you to eat on plan and keep on keeping on without requiring the scale to give you the results on a schedule.
We'd all like to lose 10 pounds a month. Shoot, we'd all like to lose all the weight overnight. Getting yourself to a place where you don't have to live up to a hastily devised unattainable schedule but are willing to keep putting in the effort because it's forever really makes a difference.
I like being able to set goals that I know are within my grasp instead of arbitrary goals that I have only partial control over. I can stay on plan. I can't use my crystal ball to determine what the result of "on plan" will be for an arbitrary date in the future. Now that I know that, I'm enjoying the journey.
__________________
08-19-2008, 11:13 PM #15
beachin'
Join Date: Apr 2008
Location: Lafayette, Louisiana
Posts: 4,190
S/C/G: 240/125/130
Height: 5'8
Quote:
Shoot, we'd all like to lose all the weight overnight.
So....this isn't something I should strive for?
__________________
Maintenance!
100 lbs. down plus a carrot for every 5 more
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# Solve Algebra Equations in bulk
I need a program that will solve a bunch of algebra equations from a list and output all the answers ex:
``````1+1
2+2
``````
out:
``````2
4
``````
I am using windows 7
First point - your example is not algebra nor equations - it is arithmetical operations. The difference algebra: `x + 3 = 5 => x=?` while arithmetic `4 + 4 = ?`
Assuming that your file really is a set of lines like your in example, i.e. simple arithmetic statements with spaces between and no other spaces:
1. install python for windows, it is free, accepting the option to add it to the path
2. Open a command prompt and `cd` to the directory as your file is in
3. type `python`
4. type:
```for line in open('sums.txt'):
for item in line.split(' '):
if len(item) > 1:
print(item, '=', eval(item))
exit
```
The above assumes that you called your list `sums.txt` and will print out your results. With a little bit of reading you can set up a script to do this for you and add the use of trigonometric functions such as sin, cos, tan, etc. Note that basic the operators you can have in your text file, with python3, are:
• Add: `+`
• Subtract: `-`
• Multiply: `*`
• Divide: `\`
• Floor: `\\`
• Modulo: `%`
• Brackets: `()`
• It was just an example I have algebra but can you close this because I don't need it anymore.
– Noah
Commented Sep 4, 2016 at 11:38
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https://brainmass.com/statistics/regression-analysis/248650
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Explore BrainMass
Predictions from the least-squares regression line
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Fill in the blank: For these data, temperature values that are greater than the mean of the temperature values tend to be paired with values for electricity use that are ____ (greater than or less than) the mean of the values for electricity use.
According to the regression equation, for an increase of one degree Fahrenheit in temperature, there is a corresponding increase of how many thousands of kilowatt hours in electricity use? ___
From the regression equation, what is the predicted electricity use (in thousands of kilowatthours)when the temperature is 98.5 degrees Fahrenheit?(Round your answer to at least one decimal place.)___
From the regression equation, what is the predicted electricity use (in thousands of kilowatt hours)when the temperature is 83.8 degrees Fahrenheit? (round your answer to at least one decimal place.)
© BrainMass Inc. brainmass.com June 3, 2020, 10:46 pm ad1c9bdddf
https://brainmass.com/statistics/regression-analysis/248650
Solution Summary
The solution provides step by step method for the calculation of regression model . Formula for the calculation and Interpretations of the results are also included.
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https://hubpages.com/technology/Using-Match-with-Vlookup
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# Using Match with Vlookup
Updated on May 6, 2011
The Match function can be used within the Vlookup function to create a dynamic column index number. Let’s look at why this can be useful beginning with a quick review of VLOOKUP.
Vlookup - Looks vertically down a list to find a record and returns information related to that record
The syntax is:
=VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
The example below shows the VLOOKUP function being used to extract the Stock Level and Stock Value information of specific products from the data table.
The function used for the Stock Level in cell H7 is;
=VLOOKUP(\$H\$4, \$A\$3:\$F\$9, 5, FALSE)
If columns are inserted in the future, the VLOOKUP function begins to extract the wrong data from the table.
The image below shows a column inserted into the previous example. The VLOOKUP function entered in cell H7 to return the stock level is still returning the content of column 5, which is now the products price.
Now wouldn’t it be great if VLOOKUP could recognise that the Stock Level and Stock Value columns have moved, and change automatically to continue to extract the correct data. Let’s introduce the MATCH function.
Match - Returns the position of a value in a list
=MATCH(lookup_value, lookup_array, [match_type])
The MATCH function can be entered inside the VLOOKUP function to return the position of the required column. This ensures that if more columns are inserted, or if the cell is copied, the function continues to work efficiently.
The revised function for cell H7 will be;
=VLOOKUP(\$I\$4, \$A\$3:\$G\$9, MATCH(I6, \$A\$3:\$G\$3, 0), FALSE)
This formula can be copied across to H8 and will work equally well at returning the product’s stock value.
The MATCH function is being used to look at the cell above where the formula is entered, then find that value in the row of headings in cell range A3:G3, and return its position as a number.
Use the INDEX and MATCH functions in Excel
The INDEX and MATCH functions can be used to lookup and retrieve data from any column of an Excel table. These functions are seen as an alternative to the VLOOKUP function.
Excel Functions
This page is dedicated to listing the most useful Microsoft Excel functions. Explanations on how to write each function and examples of its use are provided.
Use the Large function in Excel to find and return the 2nd or 3rd largest number in a list.
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## Popular
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• ### Advantages of Microsoft Word
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0 of 8192 characters used
• Hemant Kumar Katiyar
7 years ago
That's great.
Thanks for you help.
• AUTHOR
Alan Murray
8 years ago from Ipswich, United Kingdom
Thanks laura, much appreciated.
• lauralolita
8 years ago from Florida
Great info! I had to use VLookup in one of my college courses and you explained it very clearly! Thanks
working
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ECE 459 - Advanced Computer System Design Laboratory
# Experiment 2: MIPS Assembly Language Programming: Recursion
Review the help notes for this experiment.
## Objectives
The MIPS R2000/R3000 processors were the focus in ECE 451. Therefore, students taking this laboratory course (i.e., ECE 459) have already used the SPIM simulator for the MIPS R2000/R3000 processors in ECE 451. The objective of this experiment is to develop code that uses advanced programming techniques to better understand RISC (Reduced Instruction Set Computer) designs.
## Introduction
A procedure that calls itself is called recursive. Recursion is a programming technique frequently used to repetitively reduce the size of a given problem until the solutions can be obtained easily for similar problems of smaller size. Under the assumption that we know how to use the latter solutions to solve the larger problem, the original problem can then be solved more efficiently. Recursion is used for binary search, element selection, etc.
The MIPS R2000/R3000 processors are representative of RISC designs. Implementing recursion in assembly language for RISC designs is a very interesting task. The element selection problem is chosen here for this purpose.
The element selection problem is defined here as follows. Given a sequence S = {s1, s2, s3, ..., sn} of integers and an integer k, where 1 ≤ k ≤ n, find the k-th smallest integer in S. The outline of a recursive algorithm that can solve the selection problem follows (adapted from: S. Akl, The Design and Analysis of Parallel Algorithms, Prentice Hall, 1989). The parameter Q in procedure SELECT is a small integer constant.
### procedure SELECT(S,k)
1. if | S | < Q then sort S and return the k-th element
else subdivide S into | S | / Q subsequences of Q elements each
end if.
2. Sort each subsequence and determine its median.
3. Call SELECT recursively to find m, the median of the | S | / Q medians found in 2.
4. Create three subsequences S1, S2, and S3 to contain elements of S smaller than, equal to, and larger than m, respectively.
1. if | S1| k then call SELECT recursively to find the k-th element of S1
else if | S1| + | S2| ≥ k then return m
else call SELECT recursively to find the (k - | S1| - | S2|)-th element of S3
end if
end if.
The running time of SELECT is O(n) for any value of Q ≥ 5 (under this condition, recursive calls are carried out for sequences ever-decreasing in size).
Notice the similarities between SELECT and the popular QUICKSORT algorithm. The latter algorithm is for sorting, and has worst case and expected running times O(n2) and O(n lgn), respectively.
## Experiment
Provide a flowchart of your algorithm to solve the selection problem based on the given recursive procedure. Develop MIPS assembly language code for its implementation. Assume that Q = 5 and use pointer-based data structures to avoid relocation of elements from S in the memory. You may use any technique to sort S, for | S | < Q.
Use the SPIM simulator to execute the code. Show intermediate and final results.
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# American Institute of Mathematical Sciences
November 2017, 37(11): 5693-5706. doi: 10.3934/dcds.2017246
## 2-manifolds and inverse limits of set-valued functions on intervals
1 University of Auckland, Private Bag 92019, Auckland, New Zealand 2 University of Oxford, Andrew Wiles Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford, OX2 6GG, United Kingdom
* Corresponding author: Sina Greenwood
Received April 2017 Published July 2017
Suppose for each $n\in\mathbb{N}$, $f_n \colon [0,1] \to 2^{[0,1]}$ is a function whose graph $\Gamma(f_n) = \left\lbrace (x,y) \in [0,1]^2 \colon y \in f_n(x)\right\rbrace$ is closed in $[0,1]^2$ (here $2^{[0,1]}$ is the space of non-empty closed subsets of $[0,1]$). We show that the generalized inverse limit $\varprojlim (f_n) = \left\lbrace (x_n) \in [0,1]^\mathbb{N} \colon \forall n \in \mathbb{N},\ x_n \in f_n(x_{n+1})\right\rbrace$ of such a sequence of functions cannot be an arbitrary continuum, answering a long-standing open problem in the study of generalized inverse limits. In particular we show that if such an inverse limit is a 2-manifold then it is a torus and hence it is impossible to obtain a sphere.
Citation: Sina Greenwood, Rolf Suabedissen. 2-manifolds and inverse limits of set-valued functions on intervals. Discrete & Continuous Dynamical Systems - A, 2017, 37 (11) : 5693-5706. doi: 10.3934/dcds.2017246
##### References:
[1] E. Akin, The General Topology of Dynamical Systems, vol. 1 of Graduate Studies in Mathematics, American Mathematical Society, Providence, RI, 1993. Google Scholar [2] I. Banič and J. Kennedy, Inverse limits with bonding functions whose graphs are arcs, Topology Appl., 190 (2015), 9-21. doi: 10.1016/j.topol.2015.04.009. Google Scholar [3] I. Banič, M. Črepnjak and V. Nall, Some results about inverse limits with set-valued bonding functions, Topology Appl., 202 (2016), 106-111. doi: 10.1016/j.topol.2016.01.007. Google Scholar [4] R. Engelking, General Topology, vol. 6 of Sigma Series in Pure Mathematics, 2nd edition, Heldermann Verlag, Berlin, 1989, Translated from the Polish by the author. Google Scholar [5] J. Gallier and D. Xu, A Guide to the Classification Theorem for Compact Surfaces, vol. 9 of Geometry and Computing, Springer, Heidelberg, 2013. doi: 10.1007/978-3-642-34364-3. Google Scholar [6] S. Greenwood and J. Kennedy, Connectedness and Ingram-Mahavier products, Topology Appl., 166 (2014), 1-9. doi: 10.1016/j.topol.2014.01.016. Google Scholar [7] S. Greenwood and J. Kennedy, Connected generalized inverse limits over intervals, Fund. Math., 236 (2017), 1-43. doi: 10.4064/fm241-4-2016. Google Scholar [8] G. Guzik, Minimal invariant closed sets of set-valued semiflows, J. Math. Anal. Appl., 449 (2017), 382-396. doi: 10.1016/j.jmaa.2016.11.072. Google Scholar [9] K. P. Hart, J. Nagata and J. E. Vaughan (eds.), Encyclopedia of General Topology, Elsevier Science Publishers, B. V., Amsterdam, 2004. Google Scholar [10] W. Hurewicz and H. Wallman, Dimension Theory, Princeton Mathematical Series, v. 4, Princeton University Press, Princeton, N. J., 1941. Google Scholar [11] A. Illanes, A circle is not the generalized inverse limit of a subset of $[0, 1]^2$, Proc. Amer. Math. Soc., 139 (2011), 2987-2993. doi: 10.1090/S0002-9939-2011-10876-1. Google Scholar [12] W. T. Ingram, An Introduction to Inverse Limits with Set-Valued Functions, SpringerBriefs in Mathematics, Springer, New York, 2012. doi: 10.1007/978-1-4614-4487-9. Google Scholar [13] W. T. Ingram and W. S. Mahavier, Inverse limits of upper semi-continuous set valued functions, Houston J. Math., 32 (2006), 119-130. Google Scholar [14] H. Kato, On dimension and shape of inverse limits with set-valued functions, Fund. Math., 236 (2017), 83-99. doi: 10.4064/fm233-4-2016. Google Scholar [15] J. Kennedy and V. Nall, Dynamical properties of shift maps on inverse limits with a set valued function, Ergodic Theory and Dynamical Systems, (2016), 1-26. doi: 10.1017/etds.2016.73. Google Scholar [16] R. Langevin and F. Przytycki, Entropie de l'image inverse d'une application, Bull. Soc. Math. France, 120 (1992), 237-250. doi: 10.24033/bsmf.2185. Google Scholar [17] W. S. Mahavier, Inverse limits with subsets of $[0, 1]×[0, 1]$, Topology Appl., 141 (2004), 225-231. doi: 10.1016/j.topol.2003.12.008. Google Scholar [18] R. P. McGehee and T. Wiandt, Conley decomposition for closed relations, J. Difference Equ. Appl., 12 (2006), 1-47. doi: 10.1080/00207210500171620. Google Scholar [19] R. McGehee, Attractors for closed relations on compact hausdorff spaces, Indiana Univ. Math. J., 41 (1992), 1165-1209. doi: 10.1512/iumj.1992.41.41058. Google Scholar [20] V. Nall, More continua which are not the inverse limit with a closed subset of a unit square, Houston J. Math., 41 (2015), 1039-1050. Google Scholar [21] A. R. Pears, Dimension Theory of General Spaces, Cambridge University Press, Cambridge, England-New York-Melbourne, 1975. Google Scholar [22] T. Wiandt, Liapunov functions for closed relations, J. Difference Equ. Appl., 14 (2008), 705-722. doi: 10.1080/10236190701809315. Google Scholar
show all references
##### References:
[1] E. Akin, The General Topology of Dynamical Systems, vol. 1 of Graduate Studies in Mathematics, American Mathematical Society, Providence, RI, 1993. Google Scholar [2] I. Banič and J. Kennedy, Inverse limits with bonding functions whose graphs are arcs, Topology Appl., 190 (2015), 9-21. doi: 10.1016/j.topol.2015.04.009. Google Scholar [3] I. Banič, M. Črepnjak and V. Nall, Some results about inverse limits with set-valued bonding functions, Topology Appl., 202 (2016), 106-111. doi: 10.1016/j.topol.2016.01.007. Google Scholar [4] R. Engelking, General Topology, vol. 6 of Sigma Series in Pure Mathematics, 2nd edition, Heldermann Verlag, Berlin, 1989, Translated from the Polish by the author. Google Scholar [5] J. Gallier and D. Xu, A Guide to the Classification Theorem for Compact Surfaces, vol. 9 of Geometry and Computing, Springer, Heidelberg, 2013. doi: 10.1007/978-3-642-34364-3. Google Scholar [6] S. Greenwood and J. Kennedy, Connectedness and Ingram-Mahavier products, Topology Appl., 166 (2014), 1-9. doi: 10.1016/j.topol.2014.01.016. Google Scholar [7] S. Greenwood and J. Kennedy, Connected generalized inverse limits over intervals, Fund. Math., 236 (2017), 1-43. doi: 10.4064/fm241-4-2016. Google Scholar [8] G. Guzik, Minimal invariant closed sets of set-valued semiflows, J. Math. Anal. Appl., 449 (2017), 382-396. doi: 10.1016/j.jmaa.2016.11.072. Google Scholar [9] K. P. Hart, J. Nagata and J. E. Vaughan (eds.), Encyclopedia of General Topology, Elsevier Science Publishers, B. V., Amsterdam, 2004. Google Scholar [10] W. Hurewicz and H. Wallman, Dimension Theory, Princeton Mathematical Series, v. 4, Princeton University Press, Princeton, N. J., 1941. Google Scholar [11] A. Illanes, A circle is not the generalized inverse limit of a subset of $[0, 1]^2$, Proc. Amer. Math. Soc., 139 (2011), 2987-2993. doi: 10.1090/S0002-9939-2011-10876-1. Google Scholar [12] W. T. Ingram, An Introduction to Inverse Limits with Set-Valued Functions, SpringerBriefs in Mathematics, Springer, New York, 2012. doi: 10.1007/978-1-4614-4487-9. Google Scholar [13] W. T. Ingram and W. S. Mahavier, Inverse limits of upper semi-continuous set valued functions, Houston J. Math., 32 (2006), 119-130. Google Scholar [14] H. Kato, On dimension and shape of inverse limits with set-valued functions, Fund. Math., 236 (2017), 83-99. doi: 10.4064/fm233-4-2016. Google Scholar [15] J. Kennedy and V. Nall, Dynamical properties of shift maps on inverse limits with a set valued function, Ergodic Theory and Dynamical Systems, (2016), 1-26. doi: 10.1017/etds.2016.73. Google Scholar [16] R. Langevin and F. Przytycki, Entropie de l'image inverse d'une application, Bull. Soc. Math. France, 120 (1992), 237-250. doi: 10.24033/bsmf.2185. Google Scholar [17] W. S. Mahavier, Inverse limits with subsets of $[0, 1]×[0, 1]$, Topology Appl., 141 (2004), 225-231. doi: 10.1016/j.topol.2003.12.008. Google Scholar [18] R. P. McGehee and T. Wiandt, Conley decomposition for closed relations, J. Difference Equ. Appl., 12 (2006), 1-47. doi: 10.1080/00207210500171620. Google Scholar [19] R. McGehee, Attractors for closed relations on compact hausdorff spaces, Indiana Univ. Math. J., 41 (1992), 1165-1209. doi: 10.1512/iumj.1992.41.41058. Google Scholar [20] V. Nall, More continua which are not the inverse limit with a closed subset of a unit square, Houston J. Math., 41 (2015), 1039-1050. Google Scholar [21] A. R. Pears, Dimension Theory of General Spaces, Cambridge University Press, Cambridge, England-New York-Melbourne, 1975. Google Scholar [22] T. Wiandt, Liapunov functions for closed relations, J. Difference Equ. Appl., 14 (2008), 705-722. doi: 10.1080/10236190701809315. Google Scholar
A torus as a GIL on intervals
A circle as a binary Mahavier product of simply-connected spaces
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2018 Impact Factor: 1.143
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# How do I concatenate 2 cell arrays?
4 views (last 30 days)
Cai Chin on 27 Dec 2020
Answered: Walter Roberson on 27 Dec 2020
Hi, I am attempting to concatenate 2 cell arrays such that the dimensions of the concatenated cell array are 2215 x 1, with each cell containing a 2 x 900 double. Each of the component cell arrays ahs dimensions of 2215 x 1 with each cell containing a 1 x 900 double. Any suggestions would be greatly appreciated, thanks in advance.
Walter Roberson on 27 Dec 2020
FirstCell = num2cell(randi(9, 15, 9), 2);
SecondCell = num2cell(randi(9, 15, 9), 2);
size(FirstCell)
ans = 1×2
15 1
size(FirstCell{1})
ans = 1×2
1 9
output = cellfun(@(a,b) [a;b], FirstCell, SecondCell, 'uniform', 0);
size(output)
ans = 1×2
15 1
size(output{1})
ans = 1×2
2 9
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# Is kB or MB bigger?
## Is kB or MB bigger?
A megabyte has approximately 1,000 (or 1,024 to be precise) kilobytes.
What does 33 kB mean?
33 Kilobytes = 0.033 Megabytes. In bits: 33 kb to Mb = 0.033 Mb (megabits), and here you can convert 33 MB to kB. In digital storage, 33 kilobytes equal 0.033 MB.
### Is KB smaller than GB?
It means a TeraByte is 1000 times larger than Gigabytes (GB).
What is TeraByte or TB?
Unit Value
1 KB (KiloByte) 1024 bytes
1 MB (MegaByte) 1024 KB or 1,048,576 bytes
1 GB (GigaByte) 1024 MB or 1,048,576 KiloBytes
1 TB (TeraByte) 1024 GB or 1,048,576 MegaBytes
What is KB photo size?
The acronym ‘KB’ refers to one ‘kilobyte’, a unit of measurement used to describe the size of a digital file. One kilobyte is comprised of 1,024 bytes of digital information.
## Is kB bigger or GB?
A terabyte is a measure of computer data storage capacity, and usually represents 1024 (one thousand) GigaBytes. It means a TeraByte is 1000 times larger than Gigabytes (GB).
What is TeraByte or TB?
Unit Value
1 MB (MegaByte) 1024 KB or 1,048,576 bytes
1 GB (GigaByte) 1024 MB or 1,048,576 KiloBytes
How many kB is a GB of data?
Data Measurement Chart
Data Measurement Size
Kilobyte (KB) 1,024 Bytes
Megabyte (MB) 1,024 Kilobytes
Gigabyte (GB) 1,024 Megabytes
### Is 1 KB a lot of data?
One kilobyte (KB) is a collection of about 1000 bytes. A page of ordinary Roman alphabetic text takes about 2 kilobytes to store (about one byte per letter). A typical short email would also take up just 1 or 2 kilobytes.
How many KB means 1 GB?
1GB = 1024MB. 1MB = 1024KB. 1kB = 1024 Bytes. 1 Byte = 8 bits.
## How many KB is a 4×6 photo?
So, for a 4×6 you need 4 inches x 300 PPI = 1200 pixels for the short side and 6 inches x 300 PPI = 1800 for the long side. In other words, there is no improvement in print quality over 1200×1800 pixels. However, you also specify 256 kb (which presumably is the file size) and 10 MP.
How do I change the KB size of a photo?
Alternatively, you can click the drop-down menu next to “Fit into” and select an image size to quickly resize the image.
Open the image you want to resize in Preview.
1. Right-click on the image. If you are using a Magic Mouse or trackpad, click with two fingers.
2. Click File.
3. Click Open With.
4. Click Preview. app.
### Is 1 KB bigger than 1GB?
How many KB means 1GB?
## How many KB is a GB of data?
Is 1GB equal to 1024 kb?
### Is 1 kb bigger than 1GB?
How do I know the KB size of a photo?
How to calculate image size – Quick summary
1. Multiply the width and height of the image, in pixels, to get the total pixel count.
2. Multiply the total pixel count by 3 to get the image size in bytes.
3. Divide the number of bytes by 1024 to get the image size in kilobytes.
## How many KB is a good quality photo?
If you’re a beginner you can use file size to help understand the suitability of an image for its purpose. As a rough guide a 20KB image is a low quality image, a 2MB image is a high quality one.
How do I resize a JPEG image?
Adjust the size of your JPG images in seconds with the Adobe Express free online photo resizer.
How to resize a JPG file in three simple steps.
2. Resize. Choose a size template or enter in your own dimensions.
### How do I resize an image?
What’s smaller KB or GB?
Here are the most common ones. KB, MB, GB – A kilobyte (KB) is 1,024 bytes. A megabyte (MB) is 1,024 kilobytes. A gigabyte (GB) is 1,024 megabytes.
## Is 1GB equal to 1024 KB?
How many KB are in a 4gb RAM?
GB to KB Conversion Table
Gigabytes (GB) Kilobytes (KB) decimal Kilobytes (KB) binary
1 GB 1,000,000 KB 1,048,576 KB
2 GB 2,000,000 KB 2,097,152 KB
3 GB 3,000,000 KB 3,145,728 KB
4 GB 4,000,000 KB 4,194,304 KB
### Why is 1024 KB 1mb?
Notice how 2^10 is 1024. Therefore, 2^10, or 1024 bytes compose one kilobyte. Furthermore, 1024 kilobytes compose one megabyte, and 1024 megabytes compose one gigabyte. For most practical purposes, you can estimate 1024 to 1000.
Why 1gb is 1024 MB not 1000?
For a long time, 1 Kilobyte=1024 bytes, 1 Megabyte = 1024 kilobytes, 1 Gigabyte = 1024 megabytes, and so on. The reason being the fact that it easier to do binary math when working with powers of two. However, the prefix “kilo” means 1000, and not 1024, and the same reasoning applies for “mega”, “giga”, etc.
## How do I reduce the KB size of a JPEG?
Compress a picture
1. Select the picture you want to compress.
2. Click the Picture Tools Format tab, and then click Compress Pictures.
3. Do one of the following: To compress your pictures for insertion into a document, under Resolution, click Print.
4. Click OK, and name and save the compressed picture somewhere you can find it.
### Is there a left behind movie?Is there a left behind movie?
Is there a left behind movie? Left Behind: The Movie2000Left Behind II: Tribulation…2002Left Behind: World at War2005Left Behind2014Vanished: Left Behind ‑ Next Gene…2016 Left Behind/Movies Is the movie left behind
### Does Duke Energy give refunds?Does Duke Energy give refunds?
Does Duke Energy give refunds? If you had any kind of real overpayment, Duke Energy and Piedmont Natural Gas will apply for refunds as a credit to your customer account–
### What is a gooseneck?What is a gooseneck?
What is a gooseneck? A gooseneck hitch looks similar to a conventional hitch, utilizing a ball fitted in the truck bed and a round receiver on the trailer tongue. Larger
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# Gravitation slingshot
Tyro
Gravitational slingshot
How does this work exactly? Where does my analysis go wrong?
I know it is used to speed satellites up, but its name alludes to a permanent improvement in speed. The way I see it, you would get a slight benefit in journey time as long as the satellite is within the (significant) gravitational field of an object. If we assume it is launched at ~infinity, if the target destination is also ~infinity, that means the end speed ~ start speed.
The gains come from the acceleration as the satellite moves towards the mass, and away as it would be decelerating but still have a velocity greater than it would have had otherwise traversing empty space.
The only other possible source of kinetic energy for the probe I can think of would be through angular momentum transfer from the body to the probe (the body slows down in rotation speed)...but for this to happen you must have an asymmetric mass. Sort of like a planet shaped like an obelisk.
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Homework Helper
What about shooting just 'behind' a massive body in orbital motion. Then you'd get part of the objects orbital momentum, right?
Going past the gravitational slingshot might get you to your destination faster, with the same final speed which is still a net speedup.
Staff Emeritus
Gold Member
Your suspicion is correct. In order for the space craft to obtain a permanent increase in velocity, it must steal some momentum from the planet it slingshots past. In effect, it slows down the planet's orbit a bit...a very very very tiny bit given that the space craft's momentum is much much much less than the planet's (off the top of my head...something like slowing down the planet by 1 meter over 10 billion years...hopefully someone will provide a better number on that).
There are a bunch of good websites describing this. (try google)
here's one...
Last edited by a moderator:
Staff Emeritus
Gold Member
Let's see if I can paint a picture that will help.
Imagine youv'e placed your probe just ahead of Jupiter in its orbit and a bit inside of Jupiter's orbit.
Let's assume that it is at rest with respect to the sun.
Now we are looking at this situation from above, Jupiter is approaching from the right at its orbital speed. From Jupiter's perspective, this is the same as the probe moving past it to the left at the same speed.
If we place the probe properly, Jupiter's gravity will grab it in such a way as to swing it in a parabolic orbit. It will travel leftwards and inside of Jupiter and be whipped around until it is moving outside and of Jupiter and rightwards. The approaching leg and departing leg will be mirror images of each other.
when the probe reaches the same distance in the outbound leg from Jupiter as it had at the initial point when we started, it will have the same relative velocity to Jupiter as it initially had, but now it it moving to the right.
This means that relative to its initial velocity wrt to the sun it is now moving at its relative velocity wrt to Jupiter plus Jupiter's orbital velocity.
IOW, it is now moving at twice Jupiter's orbital velocity.wrt to the sun.
With real probes we don't place them motionless tot he sun, But any probe sent to an outer planet is in an orbit that has it moving slower than the planet when it reaches the planet.
So we time the probe that it arrives just a little "ahead" of the Planet and let the planet catch up to it and whip it around. By timing things correctly, we can choose the final trajectory. (like we did with the Voyager probes on their "Grand Tour".)
Tyro
Thanks guys. I understand now. I did not take into account the planetary orbits (duh!) in trying to see how the gravitational slingshot works. I was just thinking about the planet's rotation about its own axis...hence the eccentric mass suggestion.
How is the optimum spot for maximum benefit from the gravitational slingshot calculated?
Noticed a small typo in the title :/
Staff Emeritus
Gold Member
Originally posted by Tyro
Thanks guys. I understand now. I did not take into account the planetary orbits (duh!) in trying to see how the gravitational slingshot works. I was just thinking about the planet's rotation about its own axis...hence the eccentric mass suggestion.
How is the optimum spot for maximum benefit from the gravitational slingshot calculated?
Noticed a small typo in the title :/
Here's a web site that goes into some of the particulars.
http://www.go.ednet.ns.ca/~larry/orbits/gravasst/gravasst.html
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StephenPrivitera
When the author writes, "While near the planet the path will be a hyperbolic one rather than elliptical and its center of motion will be the planet rather than the Sun," does he really mean "While near the planet the path will be an approximately hyperbolic one rather than elliptical and its center of motion will approximately be the planet rather than the Sun." Does the center of motion magically and instantaneous switch places? Do the orbit change shape instantly?
BTW, is this related to resonances?
Staff Emeritus
Gold Member
Originally posted by StephenPrivitera
When the author writes, "While near the planet the path will be a hyperbolic one rather than elliptical and its center of motion will be the planet rather than the Sun," does he really mean "While near the planet the path will be an approximately hyperbolic one rather than elliptical and its center of motion will approximately be the planet rather than the Sun." Does the center of motion magically and instantaneous switch places? Do the orbit change shape instantly?
All motion is approximate. There will always be perturbations in the orbits resulting in non-"clean" orbits. So, yes, he really means approximately, but the differences are damn near insignificant unless you get so close to the planet that atmospheric drag takes effect. For satellites orbiting the Earth, stationkeeping delta V's are only on the order of a few to a few tens of meters/second per year. For a single high altitude flyby, it's practically nothing.
StephenPrivitera
I guess I just don't see how the center of orbit can just suddenly change places.
Staff Emeritus
Gold Member
First off, I'm moving this to Celestrial Mechanics which fits the subject better.
Try not thinking of it as a physical shift in the focus of the orbit, but more of a shift in perspective.
Once the probe enters the sphere of gravitational influence of the planet it is, in effect, an equal distance from the sun as the planet. Thus the gravitational effect on it and planet by the sun are essentially equal.
So at this point, we can safely ignore the sun's influence when it comes to the probe's path with respect to the planet. (It's not that the probe no longer has a heliocentric path, it is just that it is easier to deal with the orbit with respect to the planet at this time. )
Once the probe leaves the gravitational sphere of influence of the planet, it once more is easier to talk about the heliocentric path of the probe and recalculate its new orbit with respect to the sun.
Staff Emeritus
Gold Member
Originally posted by StephenPrivitera
I guess I just don't see how the center of orbit can just suddenly change places.
Hrm...
It isn't suddenly changing places. The center of any orbital motion is not at the sun, or the Earth, or any other set planet. To find the actual center, you need to add up each and every bit of gravity acting on the craft, which you obviously can't do.
What you can do to simplify the calculations is assume that the center of motion is at the center of the most influential body, and taking the effect of gravity from the other bodies as perturbations.
If you're on a Jupiter approach, the effect from both the Sun and Jupiter are both very small (r^-2, remember), but the Sun's component is the larger of the two. So you assume the craft is orbiting the Sun, and Jupiter's gravity is perturbing the motion. Once you hit the point where the two components are equal, you could validly assume either is the center, but since you are on an approach vector, you switch to Jupiter as the center of your coordinate system with the Sun acting as the perturbing body.
To sum up, the coordinate system you're using is an artificial construct introduced to make the calculations easier (read: possible). You simply can't integrate the universe, so you simplify and make a model of the situation.
Does that make sense?
Staff Emeritus
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# Fully Developed Flow
Fully developed flow occurs when the viscous effects due to the shear stress between the fluid particles and pipe wall create a fully developed velocity profile. In order for this to occur the fluid must travel through a length of a straight pipe. In addition, the velocity of the fluid for a fully developed flow will be at its fastest at the center line of the pipe (equation 1 laminar flow). On the other hand, the velocity of the fluid at the walls of the pipe will theoretically be zero. As a result, fluid velocity should be expressed as an average velocity.
(Eq 1) $v_C = \frac{2Q}{πR^2}$
vc = Maximum Velocity
Q = Flow Rate
As mentioned earlier the viscous effects are caused by the shear stress between the fluid and the pipe wall. In addition, shear stress will always be present regardless of how smooth the pipe wall is. Also, the shear stress between the fluid particles is a product of the wall shear stress and the distance the particle is from the wall. To calculate shear stress use equation 2.
(Eq 2) $τ=\frac{2τ_wr}{D}$
τ = Shear Stress
τw = Shear Stress at the Wall
r = radial distance from the center of the pipe to point of interest
D = Pipe Diameter
Due to the shear stress on the fluid particles, a pressure drop will occur. To calculate the pressure drop use equation 3.
(Eq 3) $P_2=P_1-ΔP$
Finally, the viscous effects, pressure drop, and pipe length will affect the flow rate. To calculate the average flow rate, you will need to use equation 4. This equation only applies to laminar flow.
(Eq 4) $Q=\frac{πD^4ΔP}{128μL}$
L = Pipe Length
μ = Dynamic Viscosity
### Entrance Length and Fully Developed Flow
At first the fluid is not fully developed when it enters a pipe. Instead the fluid has to travel a certain distance undisturbed before it becomes fully developed. This is also true when a fluid goes around a curve in the pipe system. This is because the curve will disrupt the velocity profile of the fluid. As a result, it will need to travel a certain distance in a straight pipe to become fully developed again. Refer to equation 5 to calculate the entrance length for laminar flow, and equation 6 to calculate the entrance length for turbulent flow.
(Eq 5) $\frac{L}{D}=0.06·Re$
L = Entrance Length
D = Pipe Diameter
Re = Reynolds Number
(Eq 6) $\frac{L}{D}=4.4(Re)^{1/6}$
Notice from the diagram that a boundary layer starts to form as the flow becomes developed. The boundary layer represents where the viscous effects are produced along the pipe wall to create the velocity profile. Also notice while the flow is developing there is a region where there is no viscous effects; this is called the inviscid core.
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List coloring
In graph theory, a branch of mathematics, list coloring is a type of graph coloring where each vertex can be restricted to a list of allowed colors, first studied by Vizing [1] and by Erdős, Rubin, and Taylor.[2][3][4]
Definition
Given a graph G and given a set L(v) of colors for each vertex v (called a list), a list coloring is a choice function that maps every vertex v to a color in the list L(v). As with graph coloring, a list coloring is generally assumed to be proper, meaning no two adjacent vertices receive the same color. A graph is k-choosable (or k-list-colorable) if it has a proper list coloring no matter how one assigns a list of k colors to each vertex. The choosability (or list colorability or list chromatic number) ch(G) of a graph G is the least number k such that G is k-choosable.
More generally, for a function f assigning a positive integer f(v) to each vertex v, a graph G is f-choosable (or f-list-colorable) if it has a list coloring no matter how one assigns a list of f(v) colors to each vertex v. In particular, if $f(v) = k$ for all vertices v, f-choosability corresponds to k-choosability.
Example
A list coloring instance on the complete bipartite graph K3,27 with three colors per vertex. No matter which colors are chosen for the three central vertices, one of the outer 27 vertices will be uncolorable, showing that the list chromatic number of K3,27 is at least four.
Let q be a positive integer, and let G be the complete bipartite graph Kq,qq. Let the available colors be represented by the q2 different two-digit numbers in radix q. On one side of the bipartition, let the q vertices be given sets of colors {i0, i1, i2, ...} in which the first digits are equal to each other, for each of the q possible choices of the first digit i. On the other side of the bipartition, let the qq vertices be given sets of colors {0a, 1b, 2c, ...} in which the first digits are all distinct, for each of the qq possible choices of the q-tuple (a, b, c, ...). For instance, for q = 2, this construction produces the graph K2,4. In this graph, the two vertices on one side of the bipartition have color sets {00,01} and {10,11} and the four vertices on the other side of the bipartition have color sets {00,10}, {00,11}, {01,10}, and {01,11}. The illustration shows a larger example of the same construction, with q = 3.
Then, G does not have a list coloring for L: no matter what set of colors is chosen for the vertices on the small side of the bipartition, this choice will conflict with all of the colors for one of the vertices on the other side of the bipartition. For instance if the vertex with color set {00,01} is colored 01, and the vertex with color set {10,11} is colored 10, then the vertex with color set {01,10} cannot be colored. Therefore, the list chromatic number of G is at least q + 1.[5]
Similarly, if $n=\binom{2k-1}{k}$, then the complete bipartite graph Kn,n is not k-choosable. For, suppose that 2k − 1 colors are available in total, and that, on a single side of the bipartition, each vertex has available to it a different k-tuple of these colors than each other vertex. Then, each side of the bipartition must use at least k colors, for otherwise some vertex would remain uncolored, but this implies that some two adjacent vertices have the same color. In particular, the utility graph K3,3 has chromatic index at least three, and the graph K10,10 has chromatic index at least four.[2]
Properties
Choosability ch(G) satisfies the following properties for a graph G with n vertices, chromatic number χ(G), and maximum degree Δ(G):
1. ch(G) ≥ χ(G). A k-list-colorable graph must in particular have a list coloring when every vertex is assigned the same list of k colors, which corresponds to a usual k-coloring.
2. ch(G) cannot be bounded in terms of chromatic number in general, that is, ch(G) ≤ f(χ(G)) does not hold in general for any function f. In particular, as the complete bipartite graph examples show, there exist graphs with χ(G) = 2 but with ch(G) arbitrarily large.[5]
3. ch(G) ≤ χ(G) ln(n).[6][7]
4. ch(G) ≤ Δ(G) + 1.[1][2]
5. ch(G) ≤ 5 if G is a planar graph.[8]
6. ch(G) ≤ 3 if G is a bipartite planar graph.[9]
Computing choosability and (a,b)-choosability
Two algorithmic problems have been considered in the literature:
1. k-choosability: decide whether a given graph is k-choosable for a given k, and
2. (j,k)-choosability: decide whether a given graph is f-choosable for a given function $f : V \to \{j,\dots,k\}$.
It is known that k-choosability in bipartite graphs is $\Pi^p_2$-complete for any k ≥ 3, and the same applies for 4-choosability in planar graphs, 3-choosability in planar triangle-free graphs, and (2,3)-choosability in bipartite planar graphs.[3][10] For P5-free graphs, that is, graphs excluding a 5-vertex path graph, k-choosability is fixed-parameter tractable. [11]
It is possible to test whether a graph is 2-choosable in linear time by repeatedly deleting vertices of degree zero or one until reaching the 2-core of the graph, after which no more such deletions are possible. The initial graph is 2-choosable if and only if its 2-core is either an even cycle or a theta graph formed by three paths with shared endpoints, with two paths of length two and the third path having any even length.[2]
Applications
List coloring arises in practical problems concerning channel/frequency assignment.[12][13]
References
1. ^ a b Vizing, V. G. (1976), "Vertex colorings with given colors", Metody Diskret. Analiz. (in Russian) 29: 3–10
2. ^ a b c d Erdős, P.; Rubin, A. L.; Taylor, H. (1979), "Choosability in graphs", Proc. West Coast Conference on Combinatorics, Graph Theory and Computing, Arcata, Congressus Numerantium 26, pp. 125–157
3. ^ a b Gutner, Shai (1996), "The complexity of planar graph choosability", Discrete Mathematics 159 (1): 119–130, arXiv:0802.2668, doi:10.1016/0012-365X(95)00104-5.
4. ^ Jensen, Tommy R.; Toft, Bjarne (1995), Graph coloring problems, New York: Wiley-Interscience, ISBN 0-471-02865-7
5. ^ a b Gravier, Sylvain (1996), "A Hajós-like theorem for list coloring", Discrete Mathematics 152 (1-3): 299–302, doi:10.1016/0012-365X(95)00350-6, MR 1388650.
6. ^ Eaton, Nancy (2003), "On two short proofs about list coloring - Part 1", Talk, retrieved May 29, 2010
7. ^ Eaton, Nancy (2003), "On two short proofs about list coloring - Part 2", Talk, retrieved May 29, 2010
8. ^ Thomassen, Carsten (1994), "Every planar graph is 5-choosable", Journal of Combinatorial Theory, Series B 62: 180–181, doi:10.1006/jctb.1994.1062
9. ^ Alon, Noga; Tarsi, Michael (1992), "Colorings and orientations of graphs", Combinatorica 12: 125–134, doi:10.1007/BF01204715
10. ^ Gutner, Shai; Tarsi, Michael (2009), "Some results on (a:b)-choosability", Discrete Mathematics 309 (8): 2260–2270, doi:10.1016/j.disc.2008.04.061
11. ^ Heggernes, Pinar; Golovach, Petr (2009), "Choosability of P5-free graphs", Mathematical Foundations of Computer Science, Lecture Notes on Computer Science 5734, Springer-Verlag, pp. 382–391
12. ^ Wang, Wei; Liu, Xin (2005), "List-coloring based channel allocation for open-spectrum wireless networks", 2005 IEEE 62nd Vehicular Technology Conference (VTC 2005-Fall) 1, pp. 690–694, doi:10.1109/VETECF.2005.1558001.
13. ^ Garg, N.; Papatriantafilou, M.; Tsigas, P. (1996), "Distributed list coloring: how to dynamically allocate frequencies to mobile base stations", Eighth IEEE Symposium on Parallel and Distributed Processing, pp. 18–25, doi:10.1109/SPDP.1996.570312.
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## Test of Independence (3 of 3)
### Learning Objectives
• Conduct a chi-square test of independence. Interpret the conclusion in context.
On this page, we practice the chi-square test for independence in its entirety and learn how to use statistical software to conduct this test. We also investigate the effect of sample size on the chi-square test statistic.
## A Real Court Case
In the early 1970s, a young man challenged an Oklahoma state law that prohibited the sale of 3.2% beer to males under age 21 but allowed its sale to females in the same age group. The case (Craig v. Boren, 429 U.S. 190, 1976) was ultimately heard by the U.S. Supreme Court. The state of Oklahoma argued that the law improved traffic safety. One of the three main pieces of data presented to the court was the result of a “random roadside survey.” This survey gathered information on gender and whether or not the driver had been drinking alcohol in the previous 2 hours. A total of 619 drivers under 21 years of age were included in the survey.
Please click here to open the simulation for use in the following activity.
### Comment: The Effect of Sample Size on Chi-Square
With other hypothesis tests, we have seen that sample size can affect the P-value and our conclusion. This is also true for chi-square. To illustrate this idea, we multiplied all of the counts in the Oklahoma data by 3.
Notice that the conditional percentages do not change, so the new “data” shows the same relationship between gender and drinking before driving. The probability that a driver under the age of 21 drinks alcohol before driving is still about 15.0% (279/1857). Males are still more likely to consume alcohol before driving (231/1443 = 16.0%) than are females (48/414 = 11.6%), with the same difference of 4.4% that we saw in the original data.
We used technology to find expected counts and the chi-square test statistic.
Notice that multiplying the observed counts by 3 also triples the expected counts and the chi-square value. This increase in the chi-square value gives a statistically significant P-value of 0.0267, which changes our conclusion. With this larger sample, the evidence is strong enough to reject the null hypothesis. We conclude that gender is associated with drinking alcohol before driving. The variables are dependent for drivers under the age of 21 in Oklahoma. With this sample size, the data provides evidence in support of the Oklahoma law that forbids sale of 3.2% beer to males and permits it to females with the goal of improving traffic safety.
What’s the point? We see once again that sample size affects the P-value in a hypothesis test. This means that a small sample may not detect a relationship that exists between two categorical variables in a population. Conversely, a large sample may indicate that a relationship is statistically significant on the basis of differences in observed and expected counts that are not important in a practical sense.
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# Sequence formula $a_{n}$=((last term) - (n-1)(common diff))
by Lebombo
Tags: anlast, diff, formula, n1common, sequence, term
P: 132 Calc 2 will require some basic knowledge of sequences and series. Since this topic has never been covered in any of my past math classes, I am currently learning about sequences and series from scratch. On youtube, I found a video that contains an explanation of the General Term of the Arithmetic Sequence formula. It also contains another form which does not exist in my Algebra Textbook: $a_{n}$ = ((last term) - (n-1)(common diff)) However, I think the person providing the youtube explanation makes a mistake in notation, which is causing me some confusion. If he has written this portion of his presentation correctly, then I will have to go back and review my confusion. So if anyone has a moment, could you fact check the portion of the derivation of this formula at exactly 9:16 - 9:23 http://www.youtube.com/watch?v=dbuwv...BFE1A5A41BC8D6 To me it seems $$a_{n}- a_{n-1}$$ should equal d, not -d , so did he intend to write $$a_{n} - a_{n+1} = -d$$
PF Gold P: 753 Yes this is a mistake. I think he meant to have $a_n-a_{n+1}=-d$ I watched the video but I didn't listen to it because I'm in class.
Related Discussions Calculus 10 Calculus & Beyond Homework 5 Calculus & Beyond Homework 1 Calculus & Beyond Homework 6 Calculus & Beyond Homework 2
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# Metamath Proof Explorer
## Theorem pj3i
Description: Projection triplet theorem. (Contributed by NM, 2-Dec-2000) (New usage is discouraged.)
Ref Expression
Assertion pj3i ( ( ( ( ( proj𝐹 ) ∘ ( proj𝐺 ) ) ∘ ( proj𝐻 ) ) = ( ( ( proj𝐻 ) ∘ ( proj𝐺 ) ) ∘ ( proj𝐹 ) ) ∧ ( ( ( proj𝐹 ) ∘ ( proj𝐺 ) ) ∘ ( proj𝐻 ) ) = ( ( ( proj𝐺 ) ∘ ( proj𝐹 ) ) ∘ ( proj𝐻 ) ) ) → ( ( ( proj𝐹 ) ∘ ( proj𝐺 ) ) ∘ ( proj𝐻 ) ) = ( proj ‘ ( ( 𝐹𝐺 ) ∩ 𝐻 ) ) )
### Proof
Step Hyp Ref Expression
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Category:
# What Is the Dining Philosophers Problem?
Article Details
• Written By: J.E. Holloway
• Edited By: Rachel Catherine Allen
2003-2019
Conjecture Corporation
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The dining philosophers problem is a thought experiment or example used in the field of computer science. The problem uses an analogy to illustrate the synchronization issues that can arise when computers share resources. Computer scientists use the dining philosophers problems to teach students about the algorithms used to resolve these issues.
The scenario of the dining philosophers problem is a circular table at which five philosophers are seated. In the center of the table is a bowl of noodles or other food. Each philosopher has one fork or chopstick on either side, meaning that there are five forks or chopsticks in total. In order to eat, a philosopher needs two utensils. Each philosopher also has to spend some time thinking, and cannot think and eat at the same time. The heart of the dining philosophers problem is the difficulty of preventing deadlock.
Deadlock in this problem occurs when philosophers put themselves into a position where they can neither think nor eat. For example, if each philosopher were to pick up the utensil to his left, no one would be able to eat, because all the utensils would be in use but no philosopher would have two. In order to allow all philosophers to eat, the student must create an algorithm that ensures that some philosophers are eating while others are thinking. This allows both eating and thinking to continue without stalling.
There are a number of possible solutions to the dining philosophers problem. One solution involves creating a sixth character, the waiter, who gives or denies permission for philosophers to pick up their forks. Others involve regulating the order in which philosophers pick up and put down their forks to maximize availability. Others involve telling the philosophers to check whether their neighbors are eating before trying to eat. In essence, each solution involves developing a set of rules, called an algorithm, which governs when the philosophers think, eat, or pick up and put down their utensils.
The dining philosophers problem was first expressed by Dutch computer scientist Edsger Dijkstra in 1965 as an exam question for students. Since then, the problem has undergone a number of changes. It appears in a number of slightly different formats, some of which only change the details of the story but others which propose additional limitations on the problem to demonstrate difficult concepts. The most common modern version was created by Tony Hoare.
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Math Skills
Practice Skills: Time Bound Skills:
Pre-K
A: Number sense with numbers through 100:
A.01: Counting review up to 50
A.02.01: Skip counting by 2 and 3 up to 100
A.02.02: Skip Counting by 2 and 3 up to 100
A.02.03: Skip counting by 2 and 3 up to 100
A.03: Skip counting by 5 and 10 up to 100
A.04: Skip counting by 5 and 10 up to 100
A.06: Even and odd number identification
A.07: Missing and next number identification
A.08: Ordinal numbers up to 100 - Assessment 1
A.09.01: Ordinal numbers up to 100 - Assessment 2
A.09.02: Ordinal numbers up to 100 - Assessment 3
A.10: Number pattern review
A.11: Find the even and odd numbers - Assessment 1
A.12: Find the even and odd numbers - Assessment 2
B: Number Words:
B.01.01: Writing words in numbers up to 20
B.01.02: Writing words in numbers up to 100
B.02.01: Writing numbers in words up to 20
B.02.02: Writing numbers in words up to 100
B.03: Introducing roman numerals up to 10
B.04: Introducing roman numerals up to 20
B.05: Review number words and roman numerals
C: Place Value:
C.01: Identify the place value of ones
C.02: Identify the place value of tens
C.03: Identify the place value of hundreds
C.04: Identifying place value - random
C.05: Review place values
C.06: Converting the number from standard to expanded form
C.07: Converting the number from expanded to standard form
C.08: Counting by 10s and 1s
C.09: Counting by 100s, 10s and 1s
C.10: Review of converting numbers from one form to other
D: Comparing & estimation of numbers:
D.01: Comparison of numbers up to 10
D.02: Comparison of numbers up to 25
D.03.01: Comparison of numbers up to 100 - Assessment 1
D.03.02: Comparison of numbers up to 100 - Assessment 2
D.04: Ordering numbers
D.05: Review number comparisons
D.06: Story problems for comparison - Assessment 1
D.07: Story problems for comparison - Assessment 2
D.08: Estimation
D.09: Review number comparisons - Assessment 2
E.01: Addition with pictures - Assessment 1
E.02: Addition with pictures - Assessment 2
E.03: Addition with pictures - Assessment 3
E.04: Addition with pictures - Assessment 4
E.06: Sequence addition by 1 from 0 to 10
E.07: Random addition by 1 from 0 to 10
E.08: Sequence addition by 1 from 11 to 20
E.09: Random addition by 1 from 11 to 20
E.10: Review addition by 1 from 0 to 20
E.11: Sequence addition by 1 from 21 to 40
E.12: Random addition by 1 from 21 to 40
E.13: Sequence addition by 1 from 41 to 60
E.14: Random addition by 1 from 41 to 60
E.15: Review addition by 1 from 0 to 60
E.16: Sequence addition by 1 from 61 to 80
E.17: Random addition by 1 from 61 to 80
E.18: Sequence addition by 1 from 81 to 100
E.19: Random addition by 1 from 81 to 100
E.20: Review addition by 1 from 0 to 100
E.21: Sequence addition by 2 from 0 to 20
E.22: Random addition by 2 from 0 to 20
E.23: Sequence addition by 2 from 21 to 40
E.24: Random addition by 2 from 21 to 40
E.25:Review addition by 1 and 2 from 0 to 40
E.26: Sequence addition by 2 from 41 to 60
E.27: Random addition by 2 from 41 to 60
E.28: Sequence addition by 2 from 61 to 80
E.29: Random addition by 2 from 61 to 80
E.30: Review addition by 1 and 2 from 0 to 80
E.31: Sequence addition by 2 from 81 to 100
E.32: Random addition by 2 from 81 to 100
E.33: Review addition by 1 and 2 from 0 to 100
E.34: Story problems covering zero, one and two addition from 0 to 100 - Assessment 1
E.35: Story problems covering zero, one and two addition from 0 to 100 - Assessment 2
E.36: Single digit addition by 3 from 0 to 25
E.37: Single digit addition by 3 from 26 to 50
E.38: Single digit addition by 3 from 51 to 75
E.39: Single digit addition by 3 from 76 to 100
E.40: Review addition by 3 from 0 to 100
E.41: Single digit addition by 4 from 0 to 25
E.42: Single digit addition by 4 from 26 to 50
E.43: Single digit addition by 4 from 51 to 75
E.44: Single digit addition by 4 from 76 to 100
E.45: Review addition by 4 from 0 to 100
E.46: Single digit addition by 5 from 0 to 25
E.47: Single digit addition by 5 from 26 to 50
E.48: Single digit addition by 5 from 51 to 75
E.49: Single digit addition by 5 from 76 to 100
E.50: Review addition by 0, 1, 2, 4 and 5 from 0 to 100
E.51: Single digit addition by 6 from 0 to 50
E.52: Single digit addition by 6 from 51 to 100
E.53: Single digit addition by 7 from 0 to 50
E.54: Single digit addition by 7 from 51 to 100
E.55: Review 6 and 7 addition from 0 to 100
E.56: Single digit addition by 8 from 0 to 50
E.57: Single digit addition by 8 from 51 to 100
E.58: Single digit addition by 9 from 0 to 50
E.59: Single digit addition by 9 from 51 to 100
E.60.01: Review single digit addition from 0 to 100 - Assessment 1
E.60.02: Review single digit addition from 0 to 100 - Assessment 2
E.61: Addition sentences to make number
E.62: Adding three single digit numbers
E.63.01: Story problems for single digit addition - Assessment 1
E.63.02: Story problems for single digit addition - Assessment 2
E.64: Story problems for single digit addition for up to 3 numbers
E.65: Review on story problems for single digit addition for up to 3 numbers
E.66: Balancing equations - addition up to 10
E.67: Balancing equations - addition up to 20
E.68: Balancing equations - addition up to 30
E.69: Balancing equations - addition up to 40
E.70: Balancing equations - addition up to 50
F.01.01: Addition by 10 from 0 to 50
F.01.02: Addition by 10 from 51 to 100
F.02.01: Two digit addition without carryover 0 to 50 - Assessment 1
F.02.02: Two digit addition without carryover 0 to 50 - Assessment 2
F.03: Two digit addition without carryover 0 to 50 - Assessment 3
F.04: Two digit addition without carryover 0 to 50 - Assessment 4
F.05: Review two digit addition without carryover 0 to 50
F.06: Two digit addition without carryover 0 to 100 - Assessment 1
F.07: Two digit addition without carryover 0 to 100 - Assessment 2
F.08: Two digit addition without carryover 0 to 100 - Assessment 3
F.09: Two digit addition without carryover 0 to 100 - Assessment 4
F.10: Review two digit addition without carryover 0 to 100
F.11: Two digit addition with carryover 0 to 50 - Assessment 1
F.12: Two digit addition with carryover 0 to 50 - Assessment 2
F.13: Two digit addition with carryover 0 to 50 - Assessment 3
F.14: Two digit addition with carryover 0 to 50 - Assessment 4
F.15: Review two digit addition with carryover 0 to 50
F.16: Two digit addition with carryover 51 to 100 - Assessment 1
F.17: Two digit addition with carryover 51 to 100 - Assessment 2
F.18: Two digit addition with carryover 51 to 100 - Assessment 3
F.19: Two digit addition with carryover 51 to 100 - Assessment 4
F.20: Review two digit addition with carryover 0 to 100
F.21: Story problems with single digit addition
F.22: Story problems with double digit addition
F.23: Story problems with up to three number addition
F.24: Review addition sentences to make number
F.25: Review story problems for single and two digit addition
G: Single Digit Subtractions:
G.01: Subtraction with pictures - Assessment 1
G.02: Subtraction with pictures - Assessment 2
G.03: Subtraction with pictures - Assessment 3
G.04: Subtraction with pictures - Assessment 4
G.05: Review subtraction with pictures
G.06: Sequence subtraction by 1 from 0 to 10
G.07: Random subtraction by 1 from 0 to 10
G.08: Sequence subtraction by 1 from 11 to 20
G.09: Random subtraction by 1 from 11 to 20
G.10: Review subtraction by 1 from 0 to 20
G.11: Sequence subtraction by 1 from 21 to 40
G.12: Random subtraction by 1 from 21 to 40
G.13: Sequence subtraction by 1 from 41 to 60
G.15: Review subtraction by 1 from 0 to 60
G.16: Sequence subtraction by 1 from 61 to 80
G.17: Random subtraction by 1 from 61 to 80
G.18: Sequence subtraction by 1 from 81 to 100
G.19: Random subtraction by 1 from 81 to 100
G.20: Review subtraction by 1 from 0 to 100
G.21: Sequence subtraction by 2 from 0 to 20
G.22: Random subtraction by 2 from 0 to 20
G.23: Sequence subtraction by 2 from 21 to 40
G.24: Random subtraction by 2 from 21 to 40
G.25: Review subtraction by 0, 1 and 2 from 0 to 40
G.26: Sequence subtraction by 2 from 41 to 60
G.27: Random subtraction by 2 from 41 to 60
G.28: Sequence subtraction by 2 from 61 to 80
G.29: Random subtraction by 2 from 61 to 80
G.30: Review subtraction by 0, 1 and 2 from 0 to 80
G.31: Sequence subtraction by 2 from 81 to 100
G.32: Random subtraction by 2 from 81 to 100
G.33: Review subtraction by 0, 1 and 2 from 0 to 100
G.34.01: Story problems for subtraction of 0, 1 and 2 for numbers up to 100 - Assessment 1
G.34.02: Story problems for subtraction of 0, 1 and 2 for numbers up to 100 - Assessment 2
G.35: Story problems for subtraction of 0, 1 and 2 for numbers up to 100 - Assessment 3
G.36: Single digit subtraction by 3 from 0 to 25
G.37: Single digit subtraction by 3 from 26 to 50
G.38: Single digit subtraction by 3 from 51 to 75
G.39: Single digit subtraction by 3 from 76 to 100
G.40: Review subtraction by 3 from 0 to 100
G.41: Single digit subtraction by 4 from 0 to 25
G.42: Single digit subtraction by 4 from 26 to 50
G.43: Single digit subtraction by 4 from 51 to 75
G.44: Single digit subtraction by 4 from 76 to 100
G.45: Review subtraction by 4 from 0 to 100
G.46: Single digit subtraction by 5 from 0 to 25
G.47: Single digit subtraction by 5 from 26 to 50
G.48: Single digit subtraction by 5 from 51 to 75
G.49: Single digit subtraction by 5 from 76 to 100
G.50: Review subtraction by 0, 1, 2, 3, 4 and 5 from 0 to 100
G.51: Single digit subtraction by 6 from 0 to 50
G.52: Single digit subtraction by 6 from 51 to 100
G.53: Single digit subtraction by 7 from 0 to 50
G.54: Single digit subtraction by 7 from 51 to 100
G.55: Review 6 and 7 subtraction from 0 to 100
G.56: Single digit subtraction by 8 from 0 to 50
G.57: Single digit subtraction by 8 from 51 to 100
G.58: Single digit subtraction by 9 from 0 to 50
G.59: Single digit subtraction by 9 from 51 to 100
G.60: Review single digit subtraction by 7, 8 and 9 from 0 to 100
G.61: Writing subtraction sentences to make a number
G.62: Subtracting three single digit numbers
G.63: Story problems for single digit subtraction - Assessment 1
G.64.01: Story problems for single digit subtraction - Assessment 2
G.64.02: Story problems for single digit subtraction - Assessment 3
G.65: Review of story sums for single digit subtraction for upto 3 numbers
G.66: Balancing equations - subtraction up to 10
G.67: Balancing equations - subtraction up to 20
G.68: Balancing equations - subtraction up to 30
G.69: Balancing equations - subtraction up to 40
G.70: Balancing equations - subtraction up to 50
H: Two digit subtraction:
H.01.01: Subtraction by 10 from 0 to 50
H.01.02: Subtraction by 10 from 51 to 100
H.02.01: Two digit subtraction without regrouping 0 to 50 - Assessment 1
H.02.02: Two digit subtraction without regrouping 0 to 50 - Assessment 2
H.03: Two digit subtraction without regrouping 0 to 50 - Assessment 3
H.04: Two digit subtraction without regrouping 0 to 50 - Assessment 4
H.05: Review two digit subtraction without regrouping 0 to 50
H.06: Two digit subtraction without regrouping 0 to 100 - Assessment 1
H.07: Two digit subtraction without regrouping 0 to 100 - Assessment 2
H.08: Two digit subtraction without regrouping 0 to 100 - Assessment 3
H.09: Two digit subtraction without regrouping 0 to 100 - Assessment 4
H.10: Review two digit subtraction without regrouping 0 to 100
H.11: Two digit subtraction with regrouping 0 to 50 - Assessment 1
H.12: Two digit subtraction with regrouping 0 to 50 - Assessment 2
H.13: Two digit subtraction with regrouping 0 to 50 - Assessment 3
H.14: Two digit subtraction with regrouping 0 to 50 - Assessment 4
H.15: Review two digit subtraction with regrouping 0 to 50
H.16: Two digit subtraction with regrouping 51 to 100 - Assessment 1
H.17: Two digit subtraction with regrouping 51 to 100 - Assessment 2
H.18: Two digit subtraction with regrouping 51 to 100 - Assessment 3
H.19: Two digit subtraction with regrouping 51 to 100 - Assessment 4
H.20: Review two digit subtraction with regrouping 0 to 100
H.21: Story problems with single digit subtraction
H.22: Story problems with double digit subtraction
H.23: Story problems with three digit subtraction
I.01: Addition and subtraction up to 99 - Assessment 1
I.02: Addition and subtraction up to 99 - Assessment 2
I.03: Addition and subtraction up to 99 - Assessment 3
I.04: Addition and subtraction up to 99 - Assessment 4
I.05: Review on Addition and subtraction upto 99
I.06: Which sign makes the answer true?
I.08: Story problems on addition and subtraction - Assessment 1
I.09: Story problems on addition and subtraction - Assessment 2
I.10: Review mixed addition and subtraction
I.11: Balancing equations - addition and subtraction up to 10
I.12: Balancing equations - addition and subtraction up to 20
I.13: Balancing equations - addition and subtraction up to 30
I.14: Balancing equations - addition and subtraction up to 40
I.15: Balancing equations - addition and subtraction up to 50
I.17: Using the correct sign - up to 10
I.18: Using the correct sign - up to 20
I.19: Put the right sign - up to 30
J: Measuring Quantities:
J.01: Length Measurement
J.02: Weight measurement
J.03: Volume measurement
J.04: Compare quantities and story problems
J.05: Review Measurement of quantities
K: Measuring with non-standard units:
K.01: Length measurement with non-standard units
K.03: Compare quantities and story problems
K.04: Story problems around non-standard units
K.05: Review non-standard units
L: Introduction to Fractions:
L.01: Understanding equal parts and halves using pictures - Assessment 1
L.02: Understanding equal parts and halves using pictures - Assessment 2
L.03: Understanding one thirds and one fourths using pictures - Assessment 1
L.04: Understanding one thirds and one fourths using pictures - Assessment 2
L.05: Review halves, thirds and fourths
L.06: Compare fractions using pictures - Assessment 1
L.07: Compare fractions using pictures - Assessment 2
L.08: Simple story problems on fractions - Assessment 1
L.09: Simple story problems on fractions - Assessment 2
L.10: Review on fractions
M: Probability Vocabulary:
M.01: Introduction to certain
M.02: Introduction to impossible
M.03: Introduction to more likely
M.04: Introduction to less likely
M.05: Review probability vocabulary
M.06: Certain, probable, unlikely and impossible - Assessment 1
N: Telling time to the hour and 1/2 hour:
N.01: Understanding time units
N.02: Reading a clock - analog and digital
N.03: A.M. or P.M.
N.04: Story problems around clocks
N.05: Telling time to the hour
N.06.01: Match clocks to the 1/2 hour - Assessment 1
N.06.02: Match time to the nearest 1/2 hour - Assessment 2
N.07.01: Reading a clock to the 1/2 hour - Assessment 1
N.07.02: Reading clock to the nearest 1/2 hour - Assessment 2
O: Using a Calendar:
O.01: Understanding days of the week
O.02: Understanding months and seasons
O.04: Story problems around calendars
O.05: Review calendars
P: Location Vocabulary:
P.01: Understanding left, middle and right
P.03: Understanding above and below
P.04: Understanding top, middle and bottom
P.05: Review location vocabulary
Q: Introduction to venn diagrams,money & geometry:
Q.01: 2 Dimensional objects
Q.02: 3 Dimensional objects
Q.03: Counting edges and corners
Q.04: Counting vertices and faces
Q.05: Shapes of objects used at home
Q.06: Understanding pictographs
Q.07: Understanding bar graphs
Q.08: Large and small, long and short concepts
Q.09: Introduction to line plots
Q.10: Review of Pictographs, Bar graphs and Line plots
Q.11: 2 way venn diagrams
Q.12: Identify coins and bills
Q.13: Counting money
Q.14: Story Problems around money
Q.15: Review objects, venn diagrams and money
R: Number line concepts:
R.01: Writing addition expression based on number line
R.02: Representation of addition expressions using number line
R.03: Writing subtraction expression based on number line
R.04: Representation of subtraction expressions using number line
R.05: Review on number line
S.02: Olympiad Practice Test - 2
S.03: Olympiad Practice Test - 3
S.04: Olympiad Practice Test - 4
S.05: Olympiad Practice Test - 5
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https://www.mrexcel.com/board/threads/using-this-barcode-to-split-data-into-columns-pharmacy-related.1104487/?s=9a45051da23429d16f122dff200d1147
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# Using this barcode to split data into columns (pharmacy related)
#### pharmavenger
##### New Member
Hello everyone!
I'm a pharmacy student working on how to extract numbers from medication QR codes into individualized columns in Excel.
For example, this code
010030093552606517201130101040951182110091947809740 needs to put into 3 columns: GTIN, Lot, and Expiry date. GTIN is donated by 01 in front and has a 14 fixed digit. Lot number is denoted by 10 and the digits may vary (its not fixed so that is why we were having trouble). Expiry date is denoted by 17 and its fixed at 6 digits. I have tried the following code: =LEFT(MID(A1,SEARCH("17",A1)+2,LEN(A1)),FIND("21"-1,MID(A1,SEARCH("17",A1)+2,LEN(A1)))-1)
<tbody>
</tbody>
This is an screenshot of what I'm working towards: https://paste.pics/0d7ee6a250f7a17721281285a2985621
### Excel Facts
Quick Sum
Select a range of cells. The total appears in bottom right of Excel screen. Right-click total to add Max, Min, Count, Average.
#### sykes
##### Well-known Member
Is the GTIN always first, and the Exp date always second?
#### sykes
##### Well-known Member
If the GTIN's always first, then that one's easy: =MID(A1,3,14)
If the Exp date's always second, then that one's easy: =MID(A1,19,6)
Not at all sure about retrieving the Lot# - bearing in mind that both it, AND the serial number can be up to 20 characters each!
You can use INSTR and INSTRREV etc, but to try and pick them out of a "Random" string will (I think) need a bit more info on any more reliable constants
#### pharmavenger
##### New Member
Yes, the order would be GTIN, Expiry, and then LOT number.
#### pharmavenger
##### New Member
The GTIN and Exp worked perfectly! Thank you so much.
I'm not sure what INSTR OR INSTERREV are or how to use them.
#### pharmavenger
##### New Member
If the GTIN's always first, then that one's easy: =MID(A1,3,14)
If the Exp date's always second, then that one's easy: =MID(A1,19,6)
Not at all sure about retrieving the Lot# - bearing in mind that both it, AND the serial number can be up to 20 characters each!
You can use INSTR and INSTRREV etc, but to try and pick them out of a "Random" string will (I think) need a bit more info on any more reliable constants
Yes, the order would be GTIN, Expiry, and then LOT number followed by serial number (but I don't need the serial number)
#### Fluff
##### MrExcel MVP, Moderator
=LEFT(MID(A1,27,LEN(A1)),FIND("21",MID(A1,27,LEN(A1)))-1)
#### pharmavenger
##### New Member
Yes, the order would be GTIN, Expiry, and then LOT number followed by serial number (but I don't need the serial number)
I'm sorry, I would like to correct something, the order of the barcode is GTIN, Serial, Expiry, and then Lot
#### Rick Rothstein
##### MrExcel MVP
Do these formulas do what you want (they are based on the picture you posted in the first link in your first message)...
GTIN: =MID(A2,3,14)
Lot: =MID(A2,LEN(B2)+LEN(C2)+9,7)
Expiry: =MID(A2,19,6)
Note: The above formulas assume your barcode number is in cell A2 and the the GTIN is to be in cell B2, the Lot number is to be in cell C2 and the Expiry is to be in cell D2.
Last edited:
#### pharmavenger
##### New Member
Do these formulas do what you want (they are based on the picture you posted in the first link in your first message)...
GTIN: =MID(A2,3,14)
Lot: =MID(A2,LEN(B2)+LEN(C2)+9,7)
Expiry: =MID(A2,19,6)
Note: The above formulas assume your barcode number is in cell A2 and the the GTIN is to be in cell B2, the Lot number is to be in cell C2 and the Expiry is to be in cell D2.
The GTIN formula works perfectly! The Lot and Expiry formulas do not work. A2 will have the barcode number and from there I'm trying to derive the GTIN, LOT, and expiry. The GTIN number comes first (14 characters, its fixed), then serial, expiry (6 characters) and lot. Serial and Lot don't have fixed characters.
This is what I got using those formulas: https://paste.pics/26670cb31094b6f2a62e10c833513ade
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# Elementary differential equations 7th edition - Boyce W.E
Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Previous << 1 .. 122 123 124 125 126 127 < 128 > 129 130 131 132 133 134 .. 486 >> Next
1.
3.
5.
7.
9.
10.
11.
12.
xy" + 2xy' + 6exy = 0
x (x 1) y" + 6x2 y' + 3y = 0
x 2 y" + 3(sin x ) y 2y = 0
x2 y" + 1 (x + sin x )y + y = 0
x 2(1 x)y" (1 + x )y' + 2xy = 0
(x 2)2(x + 2) y" + 2xy' + 3(x 2) y = 0
(4 x 2)yw + 2xy' + 3y = 0
x (x + 3)2yw 2(x + 3)y' xy = 0
x 2 y
x (2 + x ) y' + (2 + x 2)y = 0 y " + 4xy' + 6y = 0 2x (x + 2)y" + y xy = 0
(x + 1)2y + 3(x2 1) y + 3y 0
In each of Problems 13 through 17:
(a) Show that x = 0 is a regular singular point of the given differential equation.
(b) Find the exponents at the singular point x = 0.
5.7 Series Solutions near a Regular Singular Point, Part II
279
(c) Find the first three nonzero terms in each of two linearly independent solutions about x = 0.
13. xy" + y' y = 0
14. xy" + 2xy' + 6exy = 0; see Problem 1
15. x(x 1)yw + 6x2y' + 3y = 0; see Problem 3
16. xy" + y = 0
17. x2yw + (sinx)y (cos x)y = 0
18. Show that
(ln x )y" + 1 y + y = 0 has a regular singular point at x = 1. Determine the roots of the indicial equation at x = 1.
TO
Determine the first three nonzero terms in the series ^ an (x 1 )r+n corresponding to the
n=0
larger root. Take x 1 > 0. What would you expect the radius of convergence of the series to be?
19. In several problems in mathematical physics (for example, the Schr?dinger equation for a hydrogen atom) it is necessary to study the differential equation
x (1 x) y" + [y (1 + a + ?)x ] y' a?y = 0, (i)
where a, ?, and y are constants. This equation is known as the hypergeometric equation.
(a) Show that x = 0 is a regular singular point, and that the roots of the indicial equation are 0 and 1 y.
(b) Show that x = 1 is a regular singular point, and that the roots of the indicial equation are 0 and y a ?.
(c) Assuming that 1 y is not a positive integer, show that in the neighborhood of x = 0 one solution of (i) is
a? a(a + 1)?(? + 1) 2 ,
y (x) = 1 4 x 4-----------x + .
^ y 1! y(y + 1)2!
What would you expect the radius of convergence of this series to be?
(d) Assuming that 1 y is not an integer or zero, show that a second solution for 0 < x < 1
y2(x) = x
(a - y + 1)(? - y + 1)
1 H----------------------------------x
(2 - Y)1!
(a y + 1)(a y + 2)(fi y + 1)(fi y + 2) 2 ,
H-------------------------------------------------------------x + ? ?
(2 y)(3 y)2!
(e) Show that the point at infinity is a regular singular point, and that the roots of the indicial equation are a and fi. See Problem 21 of Section 5.4.
20. Consider the differential equation
x3 y" + a xy' + fy = 0.
where a and f are real constants and a = 0.
(a) Show that x = 0 is an irregular singular point.
TO
(b) By attempting to determine a solution of the form ^ anxr+n, show that the indicial
n=0
equation for r is linear, and consequently there is only one formal solution of the assumed form.
(c) Show that if f/a = 1, 0. 1. 2,..., then the formal series solution terminates and therefore is an actual solution. For other values of f /a show that the formal series solution has a zero radius of convergence, and so does not represent an actual solution in any interval.
280
Chapter 5. Series Solutions of Second Order Linear Equations
21. Consider the differential equation
a P
y + x?y + xty =0 (i)
where a = 0 and P = 0 are real numbers, and s and t are positive integers that for the moment are arbitrary.
(a) Show that if s > 1 or t > 2, then the point x = 0 is an irregular singular point.
(b) Try to find a solution of Eq. (i) of the form
TO
y = ^2, anxV+n, x > 0- (ii)
n=0
Show that if s = 2 and t = 2, then there is only one possible value of r for which there is a formal solution of Eq. (i) of the form (ii).
(c) Show that if s = 1 and t = 3, then there are no solutions of Eq. (i) of the form (ii).
(d) Show that the maximum values of s and t for which the indicial equation is quadratic in r [and hence we can hope to find two solutions of the form (ii)] are s = 1 and t = 2. These are precisely the conditions that distinguish a weak singularity, or a regular singular point, from an irregular singular point, as we defined them in Section 5.4.
As a note of caution we should point out that while it is sometimes possible to obtain a formal series solution of the form (ii) at an irregular singular point, the series may not have a positive radius of convergence. See Problem 20 for an example.
5.8 Bessels Equation
In this section we consider three special cases of Bessels12 equation,
x V + x/ + (x2 - v2)y = 0, (1)
where v is a constant, which illustrate the theory discussed in Section 5.7. It is easy to show that x = 0 is a regular singular point. For simplicity we consider only the case x > 0.
Bessel Equation of Order Zero. This example illustrates the situation in which the roots of the indicial equation are equal. Setting v = 0 in Eq. (1) gives
L [/] = x2 y" + xy' + x2 y = 0. (2)
Substituting
TO
y = <p(r, x) = 0)x + ^2 anxr+n, (3)
n = 1
12Friedrich Wilhelm Bessel (1784 -1846) embarked on a career in business as a youth, but soon became interested in astronomy and mathematics. He was appointed director of the observatory at K?nigsberg in 1810 and held this position until his death. His study of planetary perturbations led him in 1824 to make the first systematic analysis of the solutions, known as Bessel functions, of Eq. (1). He is also famous for making the first accurate determination (1838) of the distance from the earth to a star.
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# Elementary differential equations 7th edition - Boyce W.E
Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Previous << 1 .. 237 238 239 240 241 242 < 243 > 244 245 246 247 248 249 .. 486 >> Next
To guarantee convergence of a Fourier series to the function from which its coefficients were computed it is essential to place additional conditions on the function. From a practical point of view, such conditions should be broad enough to cover all situations of interest, yet simple enough to be easily checked for particular functions. Through the years several sets of conditions have been devised to serve this purpose.
Before stating a convergence theorem for Fourier series, we define a term that appears in the theorem. A function f is said to be piecewise continuous on an interval a < x < b if the interval can be partitioned by a finite number of points a = x0 < x1 < < xn = b so that
1. f is continuous on each open subinterval xi 1 < x < xi .
2. f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval.
The graph of a piecewise continuous function is shown in Figure 10.3.1.
10.3 The Fourier Convergence Theorem
559
The notation f (c+) is used to denote the limit of f (x) as x ^ c from the right; similarly, f (c-) denotes the limit of f (x) as x approaches c from the left.
Note that it is not essential that the function even be defined at the partition points xt. For example, in the following theorem we assume that f' is piecewise continuous; but certainly f' does not exist at those points where f itself is discontinuous. It is also not essential that the interval be closed; it may also be open, or open at one end and closed at the other.
Theorem 10.3.1 Suppose that f and f' are piecewise continuous on the interval L < x < L .Further,
suppose that f is defined outside the interval L < x < L so that it is periodic with
period 2L. Then f has a Fourier series
/?/x ao ^ / mnx . mnx \
f (x) = 2° + ^ \m cos ~T~ + bm sin ~^) , (4)
m = 1
whose coefficients are given by Eqs. (2) and (3). The Fourier series converges to f (x) at all points where f is continuous, and to [f (x+) + f (x)]/2 at all points where f is discontinuous.
Note that [ f (x+) + f (x)]/2 is the mean value of the right- and left-hand limits at the point x. At any point where f is continuous, f (x+) = f (x) = f (x). Thus it is correct to say that the Fourier series converges to [ f (x+) + f (x)]/2 at all points. Whenever we say that a Fourier series converges to a function f, we always mean that it converges in this sense.
It should be emphasized that the conditions given in this theorem are only sufficient for the convergence of a Fourier series; they are by no means necessary. Neither are they the most general sufficient conditions that have been discovered. In spite of this, the proof of the theorem is fairly difficult and is not given here.3
To obtain a better understanding of the content of the theorem it is helpful to consider some classes of functions that fail to satisfy the assumed conditions. Functions that are not included in the theorem are primarily those with infinite discontinuities in the interval [L, L], such as 1/x2 as x ^ 0, or ln |x L | as x ^ L. Functions having an infinite number of jump discontinuities in this interval are also excluded; however, such functions are rarely encountered.
3Proofs of the convergence of a Fourier series can be found in most books on advanced calculus. See, for example, Kaplan (Chapter 7) or Buck (Chapter 6).
560
Chapter 10. Partial Differential Equations and Fourier Series
It is noteworthy that a Fourier series may converge to a sum that is not differentiable, or even continuous, in spite of the fact that each term in the series (4) is continuous, and even differentiable infinitely many times. The example below is an illustration of this, as is Example 2 in Section 10.2.
EXAMPLE
1
Let
f (x ) =
0, L < x < 0,
L, 0 < x < L,
(5)
and let f be defined outside this interval so that f (x + 2L) = f (x) for all x. We will temporarily leave open the definition of f at the points x = 0, ħL, except that its value must be finite. Find the Fourier series for this function and determine where it converges.
y
L
1 1 1 III ,
1 CXI 1 CO 1 L 2L 3L x
FIGURE 10.3.2 Square wave.
The equation y = f (x) has the graph shown in Figure 10.3.2, extended to infinity in both directions. It can be thought of as representing a square wave. The interval [L, L] can be partitioned to give the two open subintervals (-L, 0) and (0, L). In (0, L ), f (x) = L and f(x) = 0. Clearly, both f and f' are continuous and furthermore have limits as x ^ 0 from the right and as x ^ L from the left. The situation in (-L, 0) is similar. Consequently, both f and f are piecewise continuous on [L, L), so f satisfies the conditions of Theorem 10.3.1. If the coefficients am and bm are computed from Eqs. (2) and (3), the convergence of the resulting Fourier series to f (x ) is assured at all points where f is continuous. Note that the values of am and bm are the same regardless of the definition of f at its points of discontinuity. This is true because the value of an integral is unaffected by changing the value of the integrand at a finite number of points. From Eq. (2)
Previous << 1 .. 237 238 239 240 241 242 < 243 > 244 245 246 247 248 249 .. 486 >> Next
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Tips and tricks #3: How to make calculator with custom errors
This article describes the process of adding custom errors to the calculator
Tips and tricks series
This article may rely on knowledge you should get from previous articles, so you may want to check them first.
Hello, world! Now with custom errors
If you look at different input controls available in the calculator editor, you will find that some of them support validation logic. For example, for number type of input you can specify whether it allows decimal digits, negative numbers, or limit accepted range by ticking Number range checkbox and filling min and max values. If the user violates validation rules, he will see an error message, for example, "Positive integer expected".
But, of course, there could be more complicated cases when you want to provide custom validation logic as well as custom error messages. Here is how to do it.
Let's suppose that in our "Hello, world" calculator we will welcome anybody, except Cthulhu (cause we are expecting that in his house at R'lyeh, dead Cthulhu waits to dream).
Open calculator created in previous articles in calculator editor. Since now we know about localization, we will add our custom error message as another string into resources input. Click on resources link in Calculate function signature to open input editor.
In input editor, fill fields under List items like this:
Field name Value Meaning
Value unexpectedcthulhu Identificator of our error message
Display name We are expecting that in his house at R'lyeh, dead Cthulhu waits to dream Our custom error message
Press Add then press OK to close the dialog.
Replace current code with the following:
if (username.toLowerCase() == "cthulhu") {
message.SetValue("");
throw {"source":"username", "message":resources["unexpectedcthulhu"]}
}
message.SetValue(resources.hello.replace('%username%', username));
Let's examine it line by line
1. Check the input value for "cthulhu"
2. Set output control value to an empty string. Otherwise, it would retain previous value, in our case Hello, John
3. Inform framework about error using exceptions. Note that both fields are required. source should be set to string literal with input control id, it is used to determine the location of error message on the form, and message is the custom error message
4. Closing bracket
5. Old code showing Hello, username message
Click on the Preview button. Try to enter "Cthulhu" in Enter your name field - you should receive custom error message. If everything is working as expected, Publish the calculator. I embed it in this article below.
Hello, world! Now with custom error
Message
URL copiada al portapapeles
PLANETCALC, Tips and tricks #3: How to make calculator with custom errors
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Reasoning QUiz for SSC Exams
Q1. In this question, a set of figure carrying certain characters is given. Assuming that the characters in each set follow a similar pattern, find the missing character?
(a) 132
(b) 262
(c) 274
(d) 320
Q2. In this question a matrix of certain characters is given. These characters follow a certain trend, row-wise or column-wise. Find out this trend and choose the missing character from the given alternatives.
7 9 21 27 4 2 36 18 9 4 54 ?
(a) 18
(b) 24
(c) 36
(d) 58
Q3.Find the odd one from the given alternatives?
(a) 17
(b) 31
(c) 91
(d) 89
Q4. Find the value of X in the following figure:
(a) 1
(b) 4
(c) 8
(d) 6
Q5. If Aman finds that he is twelfth from the right in a line of boys and fourth from the left, how many boys should be added to the line such that there are 28 boys in the line?
(a) 12
(b) 13
(c) 14
(d) 20
Q6. Bharati is 8 ranks ahead of Divya who ranks twenty-sixth in a class of 42. What is Bharati’s rank from the last?
(a) 9th
(b) 4th
(c) 25th
(d) 34th
Q7. Richard is fifteenth from the front in a column of boys. There were thrice as many behind him as there were in front. How many boys are there between Richard and the seventh boy from the end of the column?
(a) 33
(b) 34
(c) 35
Directions (8-9): Letters of the words given below have been jumbled up. You are required to construct the words. Each letter has been numbered and each word is followed by four options. Choose the option which gives the correct order of the letters as indicated by the numbers to form words.
Q8.
C E L S M U
1 2 3 4 5 6
(a) 4, 6, 3, 5, 2, 1
(b) 4, 6, 5, 2, 3, 1
(c) 5, 2, 3, 1, 6, 4
(d) 5, 6, 4, 1, 3, 2
Q9.
T P S L O I
1 2 3 4 5 6
(a) 2, 5, 4, 3, 6, 1
(b) 2, 6, 3, 1, 5, 4
(c) 3, 6, 4, 2, 5, 1
(d) 4, 6, 2, 5, 3, 1
Directions (10-11): Read the following information carefully and answer the questions given below:
Seven executives P, Q, R, S, T, U and W reach office in a particular sequence. U reaches immediately before P but does not immediately follow S. R is the last one to reach office. T follows immediately after P and is subsequently followed by W.
Q10. Among the executives, who reaches the office first?
(a) Q
(b) S
(c) U
(d) Cannot be determined
Q11. Who ranks fourth in the sequence of reaching office?
(a) W
(b) U
(c) T
(d) None of these
1-The digits of the number inside the circle are the difference between the corresponding numbers above and below the circle.
In the first figure, 1 = (2 - 1), 3 = (6 - 3), 1 = (5 - 4)
In the second figure, 2 = (4 - 2), 4 = (6 - 2), 8 = (8 - 0)
Thus, the digits of the missing number are (7 - 5), (9 - 3) and (3 - 1) i.e. 2, 6, 2.
So, missing number = 262.
2- In 1st row:
7 × 3 = 21, 9 × 3 = 27.
In 2nd row:
4 × 9 = 36, 2 × 9 = 18
Similarly, In 3rd row:
9 × 6 = 54.
∴ Missing number = 4 × 6 = 24.
3-
Except 91, all are prime numbers.
4-
The top left hand number is obtained by adding the bottom two numbers. The top right hand number is the result of dividing the bottom two numbers.
Thus, 12 + 3 = 15, 12 ÷ 3 = 4;
22 + 11 = 33, 22 + 11 = 2;
18 + 9 = 27, 18 ÷ 9 = 2.
So, 32 + X = 36 and 32 ÷ X = 8 or X = 4.
5-
.(b)
Sol. Clearly, number of boys in the line = (11 + 1 + 3) = 15.
∴ Number of boys to be added = (28 - 15) = 13.
6-
Divya ranks 26th and Bharati is 8 ranks ahead of Divya. So, Bharati ranks 18th. Number of students behind Bharati in rank = (42 - 18) = 24.
So, Bharati ranks 25th from the last.
7-
Number of boys in front of Richard = 14.
Number of boys behind Richard = (14 × 3) = 42.
∴ Total number of boys in the column = (14 + 1 + 42) = 57.
In a column of 57 boys, the seventh boy from the end is clearly 51st from the start.
Number of boys between Richard(15th) and boy(51st)= 35
8-.(d)
Sol. MUSCLE
9-(b) Sol. PISTOL
10-(b)
Sol. Clearly, U is followed by P; P by T; T by W.
Now U does not immediately follow S and R reaches last.
So, the order of reaching office is: S, Q, U, P, T, W, R.
S is the first to reach office.
11- (d)
Sol. Clearly, U is followed by P; P by T; T by W.
Now U does not immediately follow S and R reaches last.
So, the order of reaching office is: S, Q, U, P, T, W, R.
P is fourth in the sequence.
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# Add a number to every item in list
How do I add a number, say 2, to every item in a list?
``````(setq x '(1 2))
(+ 2 x)
(mapcar '2+ x)
(loop for i in x
do (+ 2 i))
(dolist (i x)
(+ 2 i))
``````
None of the above work.
• `(+ 2 x)` doesn't work because `+` takes numbers as arguments and returns their sum, and `(mapcar '2+ x)` doesn't work because there is no built-in function called `2+` (though you could define it yourself). The two loops work, but you are not saving the results of their computation anywhere, so the results are lost. – Basil Feb 11 '20 at 10:24
• This Q and emacs.stackexchange.com/q/55440/105 are essentially the same question. One of them should be deleted. (@Basil: would you like to consolidate the underlying question as a community question?) – Drew Feb 11 '20 at 16:26
Possible solutions:
``````(mapcar (lambda (entry) (+ entry 2)) x)
``````
``````(mapcar (apply-partially #'+ 2) x)
``````
And if you need to update `x`, then `setq` it to the result of one of the above forms, e.g.:
``````(setq x (mapcar (apply-partially #'+ 2) x))
``````
That's not very elegant (there's probably a much simpler answer) but this should work:
``````(setq x '(1 2 3 4))
• Never destructively modify (e.g. using `setcar`) a quoted constant list, such as `'(1 2 3 4)`, otherwise the constant will be modified globally and the same code may behave differently in each run. Instead, either initialise `x` to a new list, such as `(list 1 2 3 4)`, before modifying it, or accumulate changes in a new list and set that as the value of `x`, thus replacing its old value. – Basil Feb 11 '20 at 10:34
• Your looping logic is unidiomatic Elisp and inefficient, as it traverses the same list from the start for each element access. Idiomatic Elisp uses `dolist`, `mapc`, `mapcar`, or successive calls to `cdr` to go from one list element to another, without the need for indices or traversing the entire length each time. Lists are not random-access like arrays. – Basil Feb 11 '20 at 10:38
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Home - Mathematical_Logic - Paraconsistent Logics
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See This List with Details
1. Paraconsistent Logic! (A Reply To Slater)
Paraconsistent logic is the study of logics in which there are some theories embodying contradictions but which are not trivial, in particular in a
http://www.sorites.org/Issue_17/beziau.htm
2. FLoC '02 - PCL
Paraconsistent logic offers a way out of this dilemma. They devised Paraconsistent logics to cater for inconsistent yet useful theories.
http://floc02.diku.dk/PCL/
3. CLE E-Prints Of The Centre For Logic (CLE/UNICAMP) - Abstracts
On the structure of Paraconsistent extensions of Johansson s logic PS How to build your own Paraconsistent logic an introduction to the logics of
http://www.cle.unicamp.br/e-prints/abstract_16.html
4. Otávio Bueno
Paraconsistent Logic in a Historical Perspective (with Newton da Costa and . Paraconsistent logics and Paraconsistency (with Newton da Costa and Décio
http://homepage.mac.com/otaviobueno/index.htm
5. Atomic And Molecular Paraconsistent Logics. Alexander S. Karpenko
Logic L satisfies the definitions (1) and (2). This is true for the most Paraconsistent logics, for example N.C.A. da Costa s logics Cn, relevant logics,
http://logic.ru/en/node/135
6. IngentaConnect Yes, Virginia, There Really Are Paraconsistent Logics
B. H. Slater has argued that there cannot be any truly Paraconsistent logics, because it s always more plausible to suppose whatever negation symbol is
http://www.ingentaconnect.com/content/klu/logi/1999/00000028/00000005/00191230;j
7. Ralph Dumain: "The Autodidact Project": Web Guide: Philosophy Of Paraconsistency
2 (1999) Special Issue on Paraconsistent Logic and Paraconsistency . Paraconsistent logics and Paraconsistency Technical and Philosophical Developments.
http://www.autodidactproject.org/bib/paraconsistency.html
8. PhilSci Archive - Remarks On The Applications Of Paraconsistent Logic To Physics
nonclassical logics, in particular Paraconsistent logic, in the foundational analysis of physical theories. As a case-study, we
http://philsci-archive.pitt.edu/archive/00001566/
9. CAT.INIST
Logique paraconsistante; Paraconsistent Logic; Logique et philosophie du langage; Philosophical logics and philosophyof language.
http://cat.inist.fr/?aModele=afficheN&cpsidt=11820570
10. PubSLPR
Paraconsistent provability logic and rational epistemic agents 10. Fred Seymour Michael CONCEPTS AND TOOLS FOR Paraconsistent logics Con t
http://www.johnwoods.ca/RedSeries/PubSLPR.html
11. Ralph Dumain: "The Autodidact Project": Web Guide: Philosophy Of Paraconsistency
2 (1999) Special Issue on Paraconsistent Logic and Paraconsistency Paraconsistent logics and Paraconsistency. July 5, 2005. Decker, Hendrik.
http://www.jurid.net/lp/criticas/paraconsistency.html
12. IST DM Logic And Computation Seminar
Among those, Paraconsistent logics are attractive in allowing for the consistency presupposition to be defeated and for a nonexplosive negation to be
http://sem.math.ist.utl.pt/clc/abstract.xml?who=João Marcos&when=Fri 08 A
13. Greg Restall * "Paraconsistency Everywhere"
Paraconsistent logics are, by definition, inconsistency tolerant In a In this way, intuitionistic predicate logic is, in a mild sense, Paraconsistent.
http://consequently.org/writing/pev/
14. Linear Logic - Relevant Logic
Translate this page Introduction to Linear Logic Tutorial a cura di T. Brauner. Restall s Bibliography of Relevant and Substructural logics. Paraconsistent Logic
http://lgxserver.uniba.it/lei/logica/lglin_lo.htm
15. Papers In Journals And Books, By Joao Marcos
Paraconsistent logics can have usual modal semantics, and in fact any On the search of maximal Paraconsistent fragments of classical logic.
http://www.geocities.com/jm_logica/Publications/index.htm
16. Peter Suber, "Non-Standard Logics"
Paraconsistent logics can be lived if one vows to accept all truths, Paraconsistent logics do not hold that all paradoxes can be solved and urges
http://www.earlham.edu/~peters/courses/logsys/nonstbib.htm
17. Kristof De Clercq | Professional / Kristof De Clercq
Paraconsistent logics and Wisniewski s Logic of Questions. Lecture presented at the 11th International Congress of Logic, Methodology and Philosophy of
http://logica.ugent.be/kristof/pmwiki/index.php?n=Professional.CurriculumVitae
18. 1st World Congress And School On Universal Logic
During the 20th century, numerous logics have been created intuitionistic logic, modal logic, manyvalued logic, relevant logic, Paraconsistent logic,
http://www.uni-log.org/one2.html
19. Paraconsistency
It is claimed that various scientific and mathematical theories are in fact of this nature, and so can be logically analyzed by Paraconsistent logics,
http://www.philosophyprofessor.com/philosophies/paraconsistency.php
20. Ofer Arieli - On Line Papers
Multiplevalued logics and Paraconsistent Reasoning Frontiers of Paraconsistent Logic, pages 11-27, Studies in Logic and Computation Vol.8,
http://www2.mta.ac.il/~oarieli/papers-by-subject.html
21. Schloss Dagstuhl : Seminar Homepage
Belief Revision via Prime Implicates. John Slaney (Australian National UniversityCanberra). Remarks of Paraconsistent logics.
22. Nothing Of Consequence
Now of course, neither of these logics are Paraconsistent; Again, Priest s answer is that it depends on which Paraconsistent logic you look at.
http://notofcon.blogspot.com/
23. Paraconsistent Logics - Indopedia, The Indological Knowledgebase
A Paraconsistent logic is a nontrivial logic which allows inconsistencies. More specifically, it allows both a statement and its negation to be asserted,
http://www.indopedia.org/Paraconsistent_logics.html
24. KAW-ARCHIVES: Re: (Paraconsistent) Logics For Multiple Sources Of Expertise
Re (Paraconsistent) logics for multiple sources of expertise. Stefan.Wrobel@gmd.de Mon, 04 Oct 93 102059 +0100. Messages sorted by date thread
http://hcs.science.uva.nl/mailing-lists/kaw/archives/0072.html
25. Paraconsistent Logic - Wikipedia, The Free Encyclopedia
A Paraconsistent logic is a logical system that attempts to deal with contradictions in a discriminating way. Alternatively, Paraconsistent logic is the
http://en.wikipedia.org/wiki/Paraconsistent_logic
26. Paraconsistent Logic (Stanford Encyclopedia Of Philosophy)
The development of Paraconsistent logic was initiated in order to challenge the logical principle that anything follows from contradictory premises,
http://plato.stanford.edu/entries/logic-paraconsistent/
27. WoPaLo - Workshop On Paraconsistent Logic - ESSLLI 2002
Workshop in Paraconsistent Logic, part of the 14th European Summer School in Logic, Language and Information. Trento, Italy; 59 August 2002.
http://logica.rug.ac.be/WoPaLo/
28. Paraconsistent Logic Programming
Paraconsistent logic programming. Source, Theoretical Computer Science archive Volume 68 , Issue 2 (October 1989) table of contents. Pages 135 154
http://portal.acm.org/citation.cfm?id=75499
29. Paraconsistent Logic In AI
Classical logic predicts that everything (thus nothing useful at all) follows from inconsistency. A Paraconsistent logic is a logic where an inconsistency
http://ruc.dk/~jv/para.html
30. A Strong Model Of Paraconsistent Logic
The purpose of this paper is mainly to give a model of Paraconsistent logic satisfying the Frege comprehension scheme in which we can develop standard set
http://projecteuclid.org/euclid.ndjfl/1091030853
31. Paraconsistent Logic Page
This is a connected series of arguments concerning Paraconsistent logic. It is argued first that paraconsistency is an option worth pursuing in automated
http://users.rsise.anu.edu.au/~jks/paraconsistency.html
32. Paraconsistent Logic@Everything2.com
A Paraconsistent logic is one in which inconsistent assertions can be tolerated in that unrelated assertions can be handled in a reliable fashion.
http://everything2.com/index.pl?node_id=1272179
33. ScienceDirect - Journal Of Applied Logic : A Paraconsistent Decagon
Measured by such standards, the Workshop on Paraconsistent Logic (WoPaLo),1 was a very successful meeting. The present volume intends to attest this,
34. Workshop On Paraconsistent Logic
WORKSHOP ON Paraconsistent LOGIC http//logica.rug.ac.be/WoPaLo/ 14th European Summer School in Logic, Language and Information http//www.esslli2002.it/
http://www.allconferences.com/conferences/20020207120249/
35. JSTOR Paraconsistent Logic Essays On The Inconsistent
1990 Paraconsistent Logic Essays on the Inconsistent Philosophia Verlag, xxi+ 718 pp. (Hardback ISBN 388405-058-3). ROMAN TUZIAK University of Warwick
36. Science Links Japan | A Uniform Proof-theoretic Foundation For Paraconsistent Lo
Abstract;It is known that Paraconsistent logic programming, which is usually based upon a Paraconsistent logic, is important in dealing with
37. Graham Priest Papers
Reductio ad Absurdum et Modus Tollendo Ponens in Paraconsistent Logic, G. Priest, . Frontiers of Paraconsistent Logic, Research Studies Press, 2000.
38. Journal Of Applied Non-Classical Logics, Volume 15
Volume 15, Number 1, 2005. Logical Approaches to Paraconsistency. J. Riche Decision Procedure of some Relevant logics A Constructive Perspective.
http://www.informatik.uni-trier.de/~ley/db/journals/jancl/jancl15.html
39. MATHnetBASE: Mathematics Online
A taxonomy of Csystems; Paraconsistent classical logic; the logic of opposition; categorical consequence for Paraconsistent logic; ontological causes of
http://www.mathnetbase.com/ejournals/books/book_summary/toc.asp?id=3772
40. The Sharpener » Blog Archive » Brave New Logic
Far less wellknown is its eccentric younger cousin Paraconsistent logic. In most versions of this, the middle is again excluded, so each statement must be
http://www.thesharpener.net/2006/10/29/brave-new-logic/
41. Bulletin Of The IGPL, Volume 3,4
During the last two decades, we have been witnessing a growing interest as well as a remarkable activity in connection to Paraconsistent logic.
http://www.dcs.kcl.ac.uk/journals/igpl/IGPL/V3-4/
42. The Logical Fallacies: Logic Resources: Branches Of Logic
Most forms of logic assume consistency, that is if A is true, then Not A must be false. Paraconsistent logic denies this basic assumption.
http://www.onegoodmove.org/fallacy/branches.htm
43. ESSAYS ON NON-CLASSICAL LOGIC
The problems addressed range from methodological issues in Paraconsistent and deontic logic to the revision theory of truth and infinite Turing machines.
http://www.worldscibooks.com/mathematics/4799.html
44. Philosophy Program - La Trobe University
(With R. Routley) The NonTriviality of Extensional Dialectical Set Theory , in Paraconsistent Logic (ed. by G. Priest, R. Routley and J. Norman )
http://www.latrobe.edu.au/philosophy/rossb2.htm
45. University Of Chicago Press - Cookie Absent
An inconsistency adaptive logic uses both classical logic and a Paraconsistent logic depending on context. This gives it the best of both worlds.
http://www.journals.uchicago.edu/cgi-bin/resolve?PHOS700314PDF
46. Constructive Negations And Paraconsistency - Logic Journals, Books & Online Medi
Constructive Negations and Paraconsistency Logic. This book presents the authors recent investigations of the two main concepts of negation developed in
http://www.springer.com/west/home/philosophy/logic?SGWID=4-40392-22-173780373-0
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