url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://stackoverflow.com/questions/18677903/im-making-a-quadratic-equation-solver-with-python-and-easygui-but-i-cant-enter-n?answertab=oldest | 1,432,815,673,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207929411.0/warc/CC-MAIN-20150521113209-00203-ip-10-180-206-219.ec2.internal.warc.gz | 223,685,865 | 16,711 | # Im making a quadratic equation solver with python and easygui but i cant enter negative numbers. Is there any way around this
``````import math
import easygui as eg
eg.msgbox("This program solves quadratic equations Enter the values of a, b and c ")
a=eg.integerbox("enter a")
``````
# when I try to enter a negative number or a number over 99 the integer box wont let me, Is there any way around this
``````b=eg.integerbox("enter b")
c=eg.integerbox("enter c")
i = b**2-4*a*c
if d < 0:
eg.msgbox("There are no real solutions")
elif i == 0:
x = (-b+math.sqrt(i))/(2*a)
else:
x1 = (-b+math.sqrt(i))/(2*a)
x2 = (-b-math.sqrt(i))/(2*a)
eg.enterbox(msg="you have 2 solutions", default=(x1,x2))
``````
-
What do you mean by "wont let me"? Does it produce an error? Give an unexpected value? – arshajii Sep 7 '13 at 21:07
– Bhajun Singh Sep 7 '13 at 21:08
@arshajii I'm guessing the `integerbox` simply won'y accept keyboard input if it isn't an "integer". Therefore, there is probably no error given and no value to be unexpected. This is one of the few instances when an OP says "it won't let me" and that is exactly what happens. – SethMMorton Sep 7 '13 at 22:18
Try changing the default parameters of the function `integerbox` when you call it. Specifically, the one you'll want to change to allow negative numbers is `lowerbound`. Here is the full definition of `integerbox` so you can see all the parameters.
``````integerbox(msg='', title=' ', default='', lowerbound=0, upperbound=99, image=None, root=None, **invalidKeywordArguments)
``````
The minimum value for an integer on any platform can be accessed through the following method:
``````import sys
a=eg.integerbox(msg='enter a', lowerbound = -sys.maxint - 1)
``````
The upperbound for an int can be accessed through `sys.maxint`.
-
@SethMMorton Yeah, read: docs.python.org/2/library/sys.html#sys.maxint The max int is at least 231-1. The max negative int is at most -231. – Shashank Sep 7 '13 at 22:20 | 565 | 1,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2015-22 | latest | en | 0.711751 |
http://www.algebra-class.com/composite-functions.html | 1,539,623,187,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509336.11/warc/CC-MAIN-20181015163653-20181015185153-00453.warc.gz | 421,168,447 | 8,813 | Home
#### Algebra and Pre-Algebra Lessons
Algebra 1 | Pre-Algebra | Practice Tests | Algebra Readiness Test
#### Algebra E-Course and Homework Information
Algebra E-course Info | Log In to Algebra E-course | Homework Calculator
#### Formulas and Cheat Sheets
Formulas | Algebra Cheat Sheets
# Composite Functions
Find f(g(x))
g(x)=2x^2+4 and f(x)=x-2/4
Average Rating
Jul 24, 2010 Rating Composite Functions by: Karin As the title states, this is known as a composite function. We are asked to find f(g(x)) where: f(x) = x-2/4 and g(x) = 2x^2+4 So, step 1 is to substitute the value (2x^2+4) for g(x): f(g(x)) = f(2x^2+4) - Substitute 2x^2+4 for g(x) Now, we know that we need to find f(2x^2+4) for the function: f(x) = x-2/4 Step 2: Substitute (2x^2+4) for x. f(2x^2+4) = (2x^2+4)-2 ---------- 4 Step 3: Simplify 2x^2+4-2 = 2x^2 +2 -------- ------- 4 4 You can simplify further by dividing all terms by 2. x^2 + 1 This is the final answer. ------- 2 I hope this helps, Karin
Top of the Page
Custom Search | 346 | 1,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-43 | longest | en | 0.801444 |
http://www.geekstogo.com/forum/topic/55834-calculator/ | 1,606,218,739,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141176256.21/warc/CC-MAIN-20201124111924-20201124141924-00185.warc.gz | 128,836,617 | 15,328 | Welcome to Geeks to Go - Register now for FREE
Geeks To Go is a helpful hub, where thousands of volunteer geeks quickly serve friendly answers and support. Check out the forums and get free advice from the experts. Register now to gain access to all of our features, it's FREE and only takes one minute. Once registered and logged in, you will be able to create topics, post replies to existing threads, give reputation to your fellow members, get your own private messenger, post status updates, manage your profile and so much more.
Create Account How it Works
# calculator
### #1 lilhawk2892 Posted 17 August 2005 - 12:04 PM
lilhawk2892
Member
• Member
• 128 posts
ok after looking i have a hard time reading this so here is an updated version
Im making a calculator.I think the formula is easy,x+y = answer.But im having a problem.How do i make the value of x the first number the user put in,and the value of y the second number the user put in. so can anybody help me
Edited by lilhawk2892, 17 August 2005 - 01:07 PM.
• 0
### #2 bdlt Posted 17 August 2005 - 04:33 PM
bdlt
Member
• Member
• 875 posts
you need a cin for x and a cin for y.
you will need a switch statement - so go back to the tutorials for switch().
when you are done you will have
double add( double x, double y ) { ... }
double sub( double x, double y ) { ... }
double mult( double x, double y ) { ... }
double div( double x, double y ) { ... }
don't forget to check for dividing by zero.
remember - when all else fails, google on calculator c++ !!!!!!!!!
• 0
### #3 lilhawk2892 Posted 17 August 2005 - 07:37 PM
lilhawk2892
Member
• Topic Starter
• Member
• 128 posts
thanks and looks like im nowhere near making a calculator i havnt gotten close to swith statements yet
• 0
### #4 chickenman Posted 18 August 2005 - 02:51 AM
chickenman
Member
• Member
• 37 posts
@bdlt
Who do you have to make things sound so hard
@lilhawk2892
here is a simple calculator to add two numbers
```#include <iostream>
using namespace std;
int main()
{
double a;
double b;
double total;
cout<<"Enter first number" << endl;
cin >> a;
cin.ignore();
cout<<"Enter second number" << endl
cin >> b;
cin.ignore();
total = a + b;
cout<<"The total of " a " + " b " is " total << endl;
cin.get();
return 0;
}
```
• 0
### #5 lilhawk2892 Posted 18 August 2005 - 01:26 PM
lilhawk2892
Member
• Topic Starter
• Member
• 128 posts
18 C:\Dev-Cpp\main.cpp expected `;' before string constant
can u pls fix it i really dont know where to put it
• 0
### #6 bdlt Posted 18 August 2005 - 01:43 PM
bdlt
Member
• Member
• 875 posts
the compiler found a problem with line 18.
18 C:\Dev-Cpp\main.cpp expected `;' before string constant
look carefully at line 18
`cout<<"Enter second number" << endl`
do you see anything missing?
• 0
### Similar Topics
#### 0 user(s) are reading this topic
0 members, 0 guests, 0 anonymous users | 811 | 2,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-50 | longest | en | 0.849334 |
http://everything2.com/title/Landau+Theory?showwidget=showCs534368 | 1,490,500,141,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189092.35/warc/CC-MAIN-20170322212949-00376-ip-10-233-31-227.ec2.internal.warc.gz | 129,128,413 | 6,368 | Landau theory is a phenomenological theory used by mineralogists to describe the order-disorder phase transitions of a system. It is based on writing the "free energy" F of the system as a power series of an order parameter Q which measures how ordered the system is. A system acts so as to minimise its free energy, so by minimising the approximation for F we can estimate properties of the system.
In the simplest version of Landau theory, the equation used for F is:
F = A(T-Tc)Q2 + BQ4.
Here A and B are positive constants, T is the temperature, and Tc is the critical temperature at which the system starts to become ordered.
• Why does the equation only have even powers of Q? Because this model applies to the type of ordering where states with order parameter Q and -Q are obviously equivalent. E.g., if we're arranging atoms of types A and B on a square lattice, here are two perfectly ordered states. One has Q=1 and the other has Q=-1, but they are mirror images of each other.
```A B A B B A B A
B A B A A B A B
A B A B B A B A
B A B A A B A B
Q = 1 Q = -1
```
There are other types of order parameter, some of which need to be expressed as a vector rather than just one number, but these need different equations. You can also have more than one order parameter, or include terms in Q6 and beyond.
• Why does T only appear where it does in the equation? Because free energy is defined as E - TS (E is internal energy, S is entropy). Internal energy and entropy are determined by things like the positions of the atoms (which we are rolling up into the single number Q) rather than temperature. We leave out the temperature-dependent part of the Q4 term because as long as we are only covering a narrow range of temperature, the coefficient of Q4 won't change very much.
To find out the order parameter at a given temperature, we minimise F with respect to Q. Above Tc, F is minimised at Q=0, but below Tc, it is minimised at Q = sqrt(A(T-Tc)/2B). Other properties like the heat capacity can also be estimated from this model. | 492 | 2,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-13 | latest | en | 0.936139 |
https://mailman.ntg.nl/pipermail/ntg-context/2020/100731.html | 1,685,838,907,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00011.warc.gz | 411,846,969 | 2,822 | # [NTG-context] Off-topic: Struggles with LPEG grammar
Mojca Miklavec mojca.miklavec.lists at gmail.com
Mon Dec 21 13:16:31 CET 2020
```Hi,
I'm sorry for being slightly off-topic here, but this list might still
be the best place to resolve lpeg-related questions :)
0.) Disclaimer: the challenge that triggered this curiosity came from
Advent of Code 2020. In case you are taking part and you wan't to
1.) My question: I don't understand why I cannot get ^1 to work "as
advertised". Isn't this supposed to mean "one or more occurences of
the pattern"? If I change "lpeg.P('b')" into "lpeg.P('b')^1" in the
example below, the strings that match the initial grammar no longer
match the modified grammar. (I would naively imagine that the secord
pattern would get more rather than less matches.)
2.) Background: Most definitely the task on that page is supposed to
be solved in a different way, but many people use Advent of Code as an
opportunity to learn a new programming language, and when I read the
task description, I wanted to figure out if I could solve it using the
cute little lpeg. My initial attempt worked correctly (at least to
solve the first puzzle), but then I realized that I cannot easily
change the pattern from "matches a letter b" into "matches any number
of b-s", and I fail to figure out why. Any hints would be greatly
appreciated.
Below is a not-so-minimal example. I can certainly try to reduce it
further, but I would first like to ask whether I'm doing something
obviously wrong by trying to replace
r5 = lpeg.P('b')
by
r5 = lpeg.P('b')^1
in order to allow more than one occurrences of the letter b?
My only explanation would be that perhaps "^1" is so greedy that the
rest of the pattern doesn't get found. But I don't want to believe
that explanation.
local lpeg = require "lpeg"
--[[
0: 4 1 5
1: 2 3 | 3 2
2: 4 4 | 5 5
3: 4 5 | 5 4
4: "a"
5: "b"
]]--
local parser = lpeg.P{
"r0";
r0 = lpeg.V"r4" * lpeg.V"r1" * lpeg.V"r5",
r1 = lpeg.V"r2" * lpeg.V"r3" + lpeg.V"r3" * lpeg.V"r2",
r2 = lpeg.V"r4" * lpeg.V"r4" + lpeg.V"r5" * lpeg.V"r5",
r3 = lpeg.V"r4" * lpeg.V"r5" + lpeg.V"r5" * lpeg.V"r4",
r4 = lpeg.P('a'),
r5 = lpeg.P('b'),
} * -1
local parser1 = lpeg.P{
"r0";
r0 = lpeg.V"r4" * lpeg.V"r1" * lpeg.V"r5",
r1 = lpeg.V"r2" * lpeg.V"r3" + lpeg.V"r3" * lpeg.V"r2",
r2 = lpeg.V"r4" * lpeg.V"r4" + lpeg.V"r5" * lpeg.V"r5",
r3 = lpeg.V"r4" * lpeg.V"r5" + lpeg.V"r5" * lpeg.V"r4",
r4 = lpeg.P('a'),
r5 = lpeg.P('b')^1, -- modified part that doesn't seem to work
} * -1
strings = {
"ababbb",
"bababa",
"abbbab",
"aaabbb",
"aaaabbb",
};
local total = 0
local total1 = 0
for _, s in ipairs(strings) do
if lpeg.match(parser, s) then
total = total + 1
end
if lpeg.match(parser1, s) then
total = total + 1
end
end
print('total:', total, total1)
In this example, total=2, total1=0.
What I don't understand is why total1 is zero.
Thank you,
Mojca
``` | 1,000 | 2,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-23 | latest | en | 0.940356 |
https://www.physicsforums.com/threads/partial-derivatives-solving-f-z-sqrt-x-2-y-2.307098/ | 1,695,331,937,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506045.12/warc/CC-MAIN-20230921210007-20230922000007-00137.warc.gz | 1,071,216,023 | 17,403 | # Partial Derivatives: Solving f = z(sqrt(x^2+y^2))
• Lyn
#### Lyn
I am stuck on the question,
'If f is a twice differentiable function of a single variable, find f = z(sqrt(x^2+y^2)) that satisfies d^2z/dx^2 +d^2z/dy^2 = x^2 +y^2
(ALL d's ARE MEANT TO BE PARTIAL DERIVATIVES)
i know dz/dx=(dz/du).(du/dx)
i can find du/dx but i don't know how to find dz/du
Welcome to Physics Forums Lyn.
HINT: Let u = (x2 + y2)1/2
HINT (2): Using you definition of dz/dx above, can you write down an expression for d2z/dx2?
Isn't this suppose to be $z=f(\sqrt{x^2+y^2})$ In which f is a twice differentiable function?
Welcome to Physics Forums Lyn.
HINT: Let u = (x2 + y2)1/2
HINT (2): Using you definition of dz/dx above, can you write down an expression for d2z/dx2?
Yes I have set u=sqrt(x^2+y^2)
Would d^2z/dx^2=d/dx.(dz/dx)?
So far i have dz/dx=(x/u).dz/du but i am unsure of how to find dz/du so i can't carry on my calculation
Yes I have set u=sqrt(x^2+y^2)
Would d^2z/dx^2=d/dx.(dz/dx)?
So far i have dz/dx=(x/u).dz/du but i am unsure of how to find dz/du so i can't carry on my calculation
You don't need to find dz/du. Let's just start working though it. We have
$$\frac{dz}{dx} = \frac{dz}{du}\frac{du}{dx}$$
Hence,
\begin{aligned} \frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\ & =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right) \end{aligned}
Can you take the next couple of steps?
You don't need to find dz/du. Let's just start working though it. We have
$$\frac{dz}{dx} = \frac{dz}{du}\frac{du}{dx}$$
Hence,
\begin{aligned} \frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\ & =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right) \end{aligned}
Can you take the next couple of steps?
i'm slightly confused. how do i find dz/dx because i don't have an equation with z equal to an equation with x variables?
i'm slightly confused. how do i find dz/dx because i don't have an equation with z equal to an equation with x variables?
It doesn't matter, if you work the problem through you'll find that you don't actually need to know the explicit form of z=z(x,y).
Simply expand the expression in the final line of my previous post.
ok i tried to expand your expression, not really sure but i get
d^2z/dx^2=d/dx(dz/du.du/dx)
=d/dx((dz/dx.dx/du+dz/dy.dy/du).du/dx)
then i could change dz/dx again but then i get a dz/du again and then i keep going round in circles
ok i tried to expand your expression, not really sure but i get
d^2z/dx^2=d/dx(dz/du.du/dx)
=d/dx((dz/dx.dx/du+dz/dy.dy/du).du/dx)
then i could change dz/dx again but then i get a dz/du again and then i keep going round in circles
Not entirely sure what you're doing there. Let me do it for dz/dx,
\begin{aligned} \frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\ & =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right) \\ & = \frac{du}{dx}\frac{d}{dx}\frac{dz}{du} + \frac{dz}{du}\frac{d}{dx}\frac{du}{dx} \\ & = \frac{du}{dx}\left(\frac{d}{du}\frac{dz}{du}\right)\frac{du}{dx} + \frac{dz}{du}\frac{d^2u}{dx^2} \\ & = \left(\frac{du}{dx}\right)^2\frac{d^2 z}{du^2} + \frac{dz}{du}\frac{d^2u}{dx^2} \end{aligned}
Do you follow?
Can you now do the same for dz/dy?
Not entirely sure what you're doing there. Let me do it for dz/dx,
\begin{aligned} \frac{d^2 z}{dx^2} & = \frac{d}{dx}\left(\frac{dz}{dx}\right) \\ & =\frac{d}{dx}\left(\frac{dz}{du}\frac{du}{dx}\right) \\ & = \frac{du}{dx}\frac{d}{dx}\frac{dz}{du} + \frac{dz}{du}\frac{d}{dx}\frac{du}{dx} \\ & = \frac{du}{dx}\left(\frac{d}{du}\frac{dz}{du}\right)\frac{du}{dx} + \frac{dz}{du}\frac{d^2u}{dx^2} \\ & = \left(\frac{du}{dx}\right)^2\frac{d^2 z}{du^2} + \frac{dz}{du}\frac{d^2u}{dx^2} \end{aligned}
Do you follow?
Can you now do the same for dz/dy?
I haven't actually seen expanding like that before, like i don't really get your third line of working and where the addition sign came from but i can follow the rest.
I have done the same for dz/dy
d^2z/dy^2=d/dy(dz/dy)=d/dy(dz/du.du/dy)=d/dy.dz/du.du/dy+dz/du.d/dy.du/dy
=d^2z/du^2(du/dy)^2+dz/du.d^2u/dy^2
but there is still the problem that i can't work out dz/du. i have a feeling I'm misunderstanding or missing a really important point but i don't know what?
I haven't actually seen expanding like that before, like i don't really get your third line of working and where the addition sign came from but i can follow the rest.
In going from the second line to the third, I have simply used the product rule since both dz/du and du/dx are functions of x. Does that make sense?
I have done the same for dz/dy
d^2z/dy^2=d/dy(dz/dy)=d/dy(dz/du.du/dy)=d/dy.dz/du.du/dy+dz/du.d/dy.du/dy
=d^2z/du^2(du/dy)^2+dz/du.d^2u/dy^2
but there is still the problem that i can't work out dz/du. i have a feeling I'm misunderstanding or missing a really important point but i don't know what?
Good. Don't worry about dz/du, we'll come to that later. So now we have,
$$\frac{d^2 z}{dx^2} + \frac{d^2 z}{dx^2} = \left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du}$$
Do you agree?
Note that I have change the ordinary differentials du/dx and du/dx to partial differentials since u is a function of two variables, u=u(x,y). Technically I should have used partial differentials from the start.
In going from the second line to the third, I have simply used the product rule since both dz/du and du/dx are functions of x. Does that make sense?
Good. Don't worry about dz/du, we'll come to that later. So now we have,
$$\frac{d^2 z}{dx^2} + \frac{d^2 z}{dx^2} = \left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du}$$
Do you agree?
Note that I have change the ordinary differentials du/dx and du/dx to partial differentials since u is a function of two variables, u=u(x,y). Technically I should have used partial differentials from the start.
Yes thank you, up to here i understand. now i just don't see how i can carry on the calculation to get it equal to x^2+y^2 i.e. u^2
Yes thank you, up to here i understand. now i just don't see how i can carry on the calculation to get it equal to x^2+y^2 i.e. u^2
You are told in the question that this quantity is equal to u2. So we now have,
$$\left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2\right]\frac{d^2 z}{d u^2} + \left[\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right]\frac{d z}{du} = u^2$$
Now it's time for you to do a little work. All that's left for you to do is evaluate the partial derivatives and solve the resulting ODE.
ok i think I'm getting there. thenk you so much for all your help, i'll try finish it off now | 2,417 | 6,951 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-40 | latest | en | 0.852102 |
https://advancedlearners.com/temecula/geometry/tutor/find.aspx | 1,571,007,469,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986648343.8/warc/CC-MAIN-20191013221144-20191014004144-00224.warc.gz | 442,300,414 | 12,500 | ## Geometry Tutors
Love making paper airplanes, tangram puzzles, and origami? Geometry is the math behind them all. Students begin learning about geometry when they are taught different shapes and angles and how to create new objects from them. They are still interested even when it is time to learn how to determine how much water it will take to fill the bucket or paint they will need to paint the house, but when students get into theorems and postulates, the fun with geometry starts to fade. Terms like hyperbolic, conical, and elliptical send many student straight to panic mode and they struggle in their classes. Getting the help of a geometry tutor can provide the student with an understanding they may miss in the classroom.
What is it?
Euclidean/Plane Geometry is the study of flat space. Between every pair of points there is a unique line segment which is the shortest curve between those two points. These line segments can be extended to lines. Lines are infinitely long in both directions and for every pair of points on the line the segment of the line between them is the shortest curve that can be drawn between them. All of these ideas can be described by drawing on a flat piece of paper. From the laws of Euclidean Geometry, we get the famous Pythagorean Theorem.
Non-Euclidean Geometry is any geometry that is different from Euclidean geometry. It is a consistent system of definitions, assumptions, and proofs that describe such objects as points, lines and planes. The two most common non-Euclidean geometries are spherical geometry and hyperbolic geometry. The essential difference between Euclidean geometry and these two non-Euclidean geometries is the nature of parallel lines: In Euclidean geometry, given a point and a line, there is exactly one line through the point that is in the same plane as the given line and never intersects it. In spherical geometry there are no such lines. In hyperbolic geometry there are at least two distinct lines that pass through the point and are parallel to (in the same plane as and do not intersect) the given line.
Riemannian Geometry is the study of curved surfaces and higher dimensional spaces. For example, you might have a cylinder, or a sphere and your goal is to find the shortest curve between any pair of points on such a curved surface, also known as a minimal geodesic. Or you may look at the universe as a three dimensional space and attempt to find the distance between/around several planets.
Students can succeed in any geometry class. From elementary school to college, math can be a difficult subject for many students. We make it easier and more understandable for them by providing expert tutors in every mathematics class including geometry. We will be happy to provide you with all the information you need to choose the tutor that is best suited for the geometry class you or your student is taking. You will review their educational background and experience to know that the geometry tutors we offer are experts in their field.
## Our Tutoring Service
Every Advanced Learners tutor is a highly qualified, college-degreed, experienced, and fully approved educator. You can feel secure knowing that each tutor has been thoroughly pre-screened and approved. We have stringent requirements for all of our tutors. We require a national background check, a personal interview, and both personal and professional references of each applicant. We select only the very best tutors for our clients to choose from. Your personalized list of matched tutors will include professionals specifically suited to your child’s current academic needs. The backgrounds of our tutors are varied and their experience diverse, but the common factor is the passion for learning and education that they all share. As our client, you have the opportunity to review and speak with as many tutors as you wish until you find the right match for your student.
## Jennifer J
### Teaching Style
I enjoy working with a student one-on-one. We know that every child develops on a different learning curve, which is why the tutoring process is so crucial. Working individually with a child can help them overcome some of the frustration and confusion that comes along with learning in a crowded classroom. I have a great amount of patience and understanding to help the child develop his own learning style and guide him to work through the problems at his own pace to ensure that he truly understands the material. Learning is a step by step process and I know that I can guide each child through the steps they need to take to complete any and all challenges they may face, not just in school, but in life.
### Experience Summary
I began tutoring algebra when I was in high school for my boss' kids in the back of the restaurant where I worked. I then went on to get my BA in mathematics at La Salle University, with a minor in Education. While in college, I continued tutoring my fellow students in math and chemistry for a peer tutoring program set up by the school. In addition I observed/student taught at various schools in the Philadelphia area. I have knowledge of a wide range of mathematics, my specialty being algebra and calculus. Due to my experience, I am also proficient in English and can assist in research paper writing.
### Credentials
Type Subject Issued-By Level Year
Degree Mathematics La Salle University Bachelor's 2007
## Farrah F
### Teaching Style
I enjoy teaching and working with students. I now understand that all children learn on their own level and at their own pace. I am very easy going, but I insist that students do their homework and stay after school for extra help if they need it. I grade work off of understanding. It is not always finding the correct answer, but the process to get to the correct answer. I believe that a good teacher makes a good student and I strive to be that good teacher and student.
### Experience Summary
I enjoyed tutoring while I was in college working in the math lab for three years. However, now that I have been actually teaching my very own students I have a better understanding of the basics of mathematics and I am a much better student myself.
### Credentials
Type Subject Issued-By Level Year
Certification Mathematics 6-12 Florida Department of Education Leon County 2005
Degree Mathematics Florida A&M University BS 2004
## Boris B
### Teaching Style
I believe that excellence in teaching comes from the teacher's adaptation to student's specific needs. In particular, for some students, visualization may be the essential component for them to understand a certain concept or an idea in mathematics, and by furnishing examples and various proofs with pictures enables the student to learn the concept, whereas for other students it may be the algebraic equation that allows them to see a certain idea. In the first couple of sessions, I probe for the specific needs of the student and then am able to connect with that student so that he/she feels comfortable with the subject. I am a patient teacher and believe that all students are able to grasp the subject. I teach in a disciplined manner, so that the topic presented is coherent and follows a logical flow. I make sure that the theoretical concepts are internalized in a concrete example for the student. Above all, I carry a positive disposition wherever I go and encourage students to enjoy math.
### Experience Summary
I am a graduate of GaTech, with a Math and Psychology BA degrees. Aside from my two majors I have minors in Philosophy and Cognitive Science. During my years of high school and college, I have tutored students in Mathematics -- be it in calculus or statistics, or the math portion of the SAT. At the moment I work part-time at the Korean after school program, called Daekyo America Inc., as a math instructor for both high school and middle school. I have participated in various Mathematical Competitions, and have won numerous awards, including the Grand Prize Winner in USAMTS (United States Mathematical Talent Search).
### Credentials
Type Subject Issued-By Level Year
Degree Applied Mathematics GaTech Bachelors 2004
Degree Psychology GaTech BA 2004
## Herbert H
### Teaching Style
The teaching style I use is completely dependent on the particular student being taught. I'm a big fan of the adage 'Give a man a fish, and he will eat for a day. Teach a man to fish, and he will eat for the rest of his life'. My objective is to teach a child not only the material, but how they best learn so that they can teach themselves. I enjoy using applicable references that are relevant to the child so that they will more easily grasp the concepts. I believe a personalized approach works best.
### Experience Summary
I have earned a bachelor's degree in Computer Engineering with a software focus along with a Sales Engineering minor. I was 3 credit hours away from both my Math and Business minors. In the last 5 years, I have tutored students in Basic Math, Algebra 1 and 2, Geometry, Calculus, Visual Basic, Physics, and Differential Equations. I have taught math for the past 4 years at the Professional Academies Magnet at Loften High School in the subjects of Algebra I, Algebra I Honors, Geometry, Geometry Honors, Algebra II, Algebra II Honors, Statistics, Mathematics for College Readiness, and Liberal Arts Math.
### Credentials
Type Subject Issued-By Level Year
Other Mathematics Professional Academies Magnet 9-12 2007-current
Other Tutor Starke Church of God by Faith 9-12 2005-2007
Degree Computer Engineering University of Florida BA 2004
## Timothy T
### Teaching Style
When I tutor a student, I seek first to understand the student and how he/she thinks. I find it is very important to have good rapport and communication with the student so I understand how he/she views the subject and the difficulties of it. Next I try to make "conceptual bridges" from what they know to what they are having difficulty understanding. This process usually teaches me about seeing the subject from a new point of view. I try to achieve a fine balance between guiding and directing the student’s thoughts on the topic with following the student in their own line of thinking of the subject. The student needs to learn to have confidence in his own thoughts on the subject and in his own ability to master it.
### Experience Summary
Over the past four years, I have tutored high school and middle school students in math, algebra, calculus, chemistry, SAT Math, and general study skills. My preference is to tutor math, algebra, calculus, physics, physical science, chemistry, and programming. Math is the subject for which I have the greatest passion. I also participate in the homeschooling of four of my children (13, 11, 8, 6). I have mentored my 13 yr old son in Algebra I & II, Chemistry, Elementary Math, and Middle-school Physical Science, and taught elementary math to my 11, 8, and 6 year olds. Additionally, I read and review history lessons to my kids. I completed my MS in Electrical Engineering in 2006 from The University of Texas at Arlington and my BS in Electrical Engineering and a BA in Philosophy from Rice University. I have recent experience as a student having completed Cellular Biology II at St. Petersburg College in Fall 2011.
### Credentials
Type Subject Issued-By Level Year
Degree Electrical Engineering Univ. of Texas - Arlington Masters 2006
Certification Design for Six Sigma Honeywell International DFSS - Green Belt 2003
Degree Electrical Engineering Rice University BSEE 1989
Degree Philosophy Rice University BA 1989
## Vinod V
### Teaching Style
The cornerstone of my teaching philosophy and personal teaching goals is to help students develop their own thinking skills. I believe all students should leave the school armed with the ability to think for them selves, to think critically and to think creatively. Understanding how people learn is one of the significant aspects of teaching. This is linked to their “knowledge” background and maturity. The key to teaching is to relate to the audience by starting from what they know and building upon it. As a teacher I am totally involved with the class, dedicated to my students and 100% prepared to devote time and energy for their intellectual growth. Love for teaching evokes passion and dedication within me. I believe that the enthusiasm of a motivated teacher rubs off on his/her students, who derive the inspiration and encouragement which actuates their desire to learn. A good teacher should have sound fundamentals and command over the concepts. Fundamentals are the foundation intrinsic for mastering the subject; only teachers who are strong in fundamentals will be able to pass it on to their students. I believe that my strong command over the fundamentals will rub off on my students. I believe that the role of a teacher is that of a leader where you have to show the path, motivate, encourage, and lead by example. In short, my success lies in seeing my students succeed.
### Experience Summary
My enthusiasm and love for education can be gauged from the fact that I pursued three Masters degrees in three distinct but related fields. One cannot pursue engineering as a profession without having an affinity for Math and Analysis. Math was a passion for me from my young days and still very much remains so. I have a thorough knowledge and understanding of math. Right from my school days I was involved and loved to teach math. I invariably obtained A+ scores in whatever math test I took in my lifetime. For instance my GRE math score was above 95% of test takers' scores. I have taught Middle school, High school and under-graduate students in Algebra, Geometry, Trigonometry, Quadratic Equations, Applied Probability and Calculus.
### Credentials
Type Subject Issued-By Level Year
Degree City Planning Kansas State University MRCP 2002
Degree Engineering Anna University ME 2000
Degree Civil Engineering Institution of Engineers BE 1994
## Maria M
### Teaching Style
I love tutoring and consider myself to be effective at it. I approach it with enthusiasm. I apply theory to practical applications in my career and life. My approach is one of ease. I consider mathematics to be easy, as long as you accept what I call "the rules of the game"; i.e., there are certain principles that have to be accepted, and not questioned, then everything else falls in place. I believe that all people have the capability of learning, and I love the opportunity to provide a positive experience to students.
### Experience Summary
I have been tutoring for more than 5 years. My main focus is mathematics, but I also tutor students in Spanish I through IV. I have tutored Spanish, Algebra I, Algebra II, Geometry, Trigonometry, and Pre-Calculus. I have recently helped several students prepare for the SAT math test. I enjoy making a difference in the student’s life.
### Credentials
Type Subject Issued-By Level Year
Other Spanish Native speaker Fluent Current
Certification Project Management Studies Project Management Institute PMP Certification 2006
Degree Civil Engineering California State University at Long Beach MSCE 1987
Other Mathematics El Camino College AA 1981
## Justin J
### Teaching Style
I am a very enthusiastic tutor and, as I stated in the previous section, believe teaching in such a way that the student gains a true mastery of the given subject. In a sense, I believe in each student understanding a particular concept in their own unique way that is consistent with their unique thinking processes. I strive to do this by relating a given concept analogously to something the student already fundamentally understands. I also create new problems to solve that probe the students’ progress and comprehension. I try to make these problems as realistic as possible to make them interesting to the student.
### Experience Summary
Being a Ph.D. candidate and appointed research assistant, I very much realize the importance of true comprehension of a subject. Not only do I hold two B.S. degrees in Applied Mathematics and Chemical Engineering but I also carry credits to the equivalent of a Non-Thesis Masters in Applied Mathematics and am one course shy of a minor in Chemistry. I have tutored groups of students for Sylvan Learning Centers, conducted one-on-one sessions with Educational Enrichment and independently, and prepared and delivered lectures for advanced undergraduate chemical engineering courses. I am capable and experienced in tutoring all levels of mathematics, chemistry, chemical engineering, physics and materials science.
### Credentials
Type Subject Issued-By Level Year
Degree Chemical Engineering University of Florida Ph.D.--in progress 2010
Degree Chemical Engineering NC State University B.S. with Honors 2005
Degree Applied Mathematics NC State University B.S. with Honors 2005
## Robyn E.
Tampa, FL
Tutoring with Arthur is going great. Sarah likes his teaching style.
## Jackie, J.
Chuluota, FL
We are very happy with Ronda and we would highly recommend her to anyone. Ronda has had prior teaching experience where Clay attends, and this has been helpful in that she understands how some of his teachers approach classroom objectives. Language ... | 3,485 | 17,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-43 | longest | en | 0.960189 |
https://everything2.com/title/principal+filter | 1,618,839,771,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00089.warc.gz | 364,463,269 | 5,154 | A principal filter is the mathematical label used to describe a filter over a partially ordered set (c.f. also lattice) that gets generated by a single element. To create such a filter we start with a single element and then build a set that contains all the elements of the lattice that are greater than this element. It can easily be shown that such a set satisfies all the properties required of a filter.
As ideals are dual to filters so principal ideals are dual to principal filters.
Related mathematical concepts include filters, ideals, rings, topology, universal algebra, model theory and lattices.
Log in or register to write something here or to contact authors. | 132 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-17 | latest | en | 0.925297 |
http://rrmh.org/ionescu-s-irna/e32066-fundamental-theorem-of-line-integrals | 1,632,552,914,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057598.98/warc/CC-MAIN-20210925052020-20210925082020-00481.warc.gz | 59,072,403 | 11,620 | Let C be a curve in the xyz space parameterized by the vector function r(t)= for a<=t<=b. $f(x(t),y(t),z(t))$, a function of $t$. (answer), Ex 16.3.7 P,Q\rangle = \nabla f$. Find the work done by this force field on an object that moves from Find an$f$so that$\nabla f=\langle 2x+y^2,2y+x^2\rangle$, or Find the work done by this force field on an object that moves from Suppose that (answer), Ex 16.3.8$1 per month helps!! The important idea from this example (and hence about the Fundamental Theorem of Calculus) is that, for these kinds of line integrals, we didnât really need to know the path to get the answer. Something similar is true for line integrals of a certain form. Khan Academy is a 501(c)(3) nonprofit organization. recognize conservative vector fields. *edit to add: the above works because we har a conservative vector field. It may well take a great deal of work to get from point $\bf a$ In the first section, we will present a short interpretation of vector fields and conservative vector fields, a particular type of vector field. (answer), Ex 16.3.2 \int_a^b \langle f_x,f_y,f_z\rangle\cdot\langle By the chain rule (see section 14.4) 18(4X 5y + 10(4x + Sy]j] - Dr C: ⦠Let (3z + 4y) dx + (4x â 22) dy + (3x â 2y) dz J (a) C: line segment from (0, 0, 0) to (1, 1, 1) (6) C: line segment from (0, 0, 0) to (0, 0, 1) to (1, 1, 1) c) C: line segment from (0, 0, 0) to (1, 0, 0) to (1, 1, 0) to (1, 1, 1) and ${\bf b}={\bf r}(b)$. The Fundamental Theorem of Line Integrals, 2. The following result for line integrals is analogous to the Fundamental Theorem of Calculus. The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function.. $f$ is sufficiently nice, we know from Clairaut's Theorem (answer), Ex 16.3.6 This website uses cookies to ensure you get the best experience. Math 2110-003 Worksheet 16.3 Name: Due: 11/8/2017 The fundamental theorem for line integrals 1.Let» fpx;yq 3x x 2y and C be the arc of the hyperbola y 1{x from p1;1qto p4;1{4q.Compute C rf dr. Use A Computer Algebra System To Verify Your Results. but if you then let gravity pull the water back down, you can recover If a vector field $\bf F$ is the gradient of a function, ${\bf \left We can test a vector field${\bf F}=\v{P,Q,R}$in a similar (In the real world you Free definite integral calculator - solve definite integrals with all the steps. 4x y. \langle yz,xz,xy\rangle$. at the endpoints. (x^2+y^2+z^2)^{3/2}}\right\rangle.$${\bf F}= In other words, we could use any path we want and weâll always get ⦠Evaluate \ds\int_C (10x^4 - 2xy^3)\,dx - 3x^2y^2\,dy where C is Let Theorem 3.6. Second Order Linear Equations, take two. closed paths. In particular, thismeans that the integral of âf does not depend on the curveitself. 16.3 The Fundamental Theorem of Line Integrals One way to write the Fundamental Theorem of Calculus (7.2.1) is: â«b af â² (x)dx = f(b) â f(a). A vector field with path independent line integrals, equivalently a field whose line integrals around any closed loop is 0 is called a conservative vector field. or explain why there is no such f. :) https://www.patreon.com/patrickjmt !! {\bf F}= If F is a conservative force field, then the integral for work, â« C F â d r, is in the form required by the Fundamental Theorem of Line Integrals. This will be shown by walking by looking at several examples for both 2 ⦠conservative vector field. Hence, if the line integral is path independent, then for any closed contour $$C$$ $\oint\limits_C {\mathbf{F}\left( \mathbf{r} \right) \cdot d\mathbf{r} = 0}.$ The fundamental theorem of line integrals, also known as the gradient theorem, is one of several ways to Theorem 15.3.2 Fundamental Theorem of Line Integrals ¶ Let âF be a vector field whose components are continuous on a connected domain D in the plane or in space, let A and B be any points in D, and let C be any path in D starting at A and ending at B. If you're seeing this message, it means we're having trouble loading external resources on our website. Derivatives of the exponential and logarithmic functions, 5. This theorem, like the Fundamental Theorem of Calculus, says roughly Green's Theorem 5. For example, vx y 3 4 = U3x y , 2 4 3. 2. same, (answer), Ex 16.3.9 f(a)=f(x(a),y(a),z(a)). It can be shown line integrals of gradient vector elds are the only ones independent of path. You da real mvps! Thanks to all of you who support me on Patreon. forms a loop, so that traveling over the C curve brings you back to To log in and use all the features of Khan Academy, please enable JavaScript in your browser. F}=\nabla f, we say that \bf F is a The fundamental theorem of Calculus is applied by saying that the line integral of the gradient of f *dr = f(x,y,z)) (t=2) - f(x,y,z) when t = 0 Solve for x y and a for t = 2 and t = 0 to evaluate the above. 3). or explain why there is no such f. We will examine the proof of the the⦠possible to find g(y) and h(x) so that (answer), Ex 16.3.4 conservative force field, then the integral for work,$$, Another immediate consequence of the Fundamental Theorem involves Often, we are not given th⦠§16.3 FUNDAMENTAL THEOREM FOR LINE INTEGRALS § 16.3 Fundamental Theorem for Line Integrals After completing this section, students should be able to: ⢠Give informal definitions of simple curves and closed curves and of open, con-nected, and simply connected regions of the plane. work by running a water wheel or generator. $(0,0,0)$ to $(1,-1,3)$. (answer), Ex 16.3.3 In other words, all we have is Find the work done by this force field on an object that moves from In 18.04 we will mostly use the notation (v) = (a;b) for vectors. Divergence and Curl 6. Find an $f$ so that $\nabla f=\langle x^2y^3,xy^4\rangle$, (answer), 16.3 The Fundamental Theorem of Line Integrals, The Fundamental Theorem of Line Integrals, 2 Instantaneous Rate of Change: The Derivative, 5. Theorem (Fundamental Theorem of Line Integrals). The Divergence Theorem components of ${\bf r}$ into $\bf F$, forming the dot product ${\bf Example 16.3.3 Find an$f$so that$\langle 3+2xy,x^2-3y^2\rangle = \nabla f$. vf(x, y) = Uf x,f y). \langle e^y,xe^y+\sin z,y\cos z\rangle$. object from point $\bf a$ to point $\bf b$ depends only on those This means that in a Example 16.3.2 Line Integrals and Greenâs Theorem Jeremy Orlo 1 Vector Fields (or vector valued functions) Vector notation. For example, in a gravitational field (an inverse square law field) An object moves in the force field Then find that $P_y=Q_x$, $P_z=R_x$, and $Q_z=R_y$ then $\bf F$ is If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. be able to spot conservative vector fields $\bf F$ and to compute Find an $f$ so that $\nabla f=\langle y\cos x,y\sin x \rangle$, simultaneously using $f$ to mean $f(t)$ and $f(x,y,z)$, and since concepts are clear and the different uses are compatible. For If we compute When this occurs, computing work along a curve is extremely easy. $\int_C {\bf F}\cdot d{\bf r}$, is in the form required by the Study guide and practice problems on 'Line integrals'. Like the first fundamental theorem we met in our very first calculus class, the fundamental theorem for line integrals says that if we can find a potential function for a gradient field, we can evaluate a line integral over this gradient field by evaluating the potential function at the end-points. $$\int_a^b f'(x)\,dx = f(b)-f(a).$$ Lastly, we will put all of our skills together and be able to utilize the Fundamental Theorem for Line Integrals in three simple steps: (1) show Independence of Path, (2) find a Potential Function, and (3) evaluate. The Fundamental Theorem of Line Integrals 4. Find an $f$ so that $\nabla f=\langle x^3,-y^4\rangle$, We write ${\bf r}=\langle x(t),y(t),z(t)\rangle$, so Theorem 16.3.1 (Fundamental Theorem of Line Integrals) Suppose a curve $C$ is Of course, it's only the net amount of work that is the amount of work required to move an object around a closed path is that if we integrate a "derivative-like function'' ($f'$ or $\nabla Donate or volunteer today! Suppose that$\v{P,Q,R}=\v{f_x,f_y,f_z}$. Fundamental Theorem of Line Integrals. Graph. sufficiently nice, we can be assured that$\bf F$is conservative. That is, to compute the integral of a derivative f â² we need only compute the values of f at the endpoints. $$\int_a^b f'(t)\,dt=f(b)-f(a).$$$f_x x'+f_y y'+f_z z'=df/dt$, where$f$in this context means In this context, If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. But Type in any integral to get the solution, free steps and graph. Find an$f$so that$\nabla f=\langle y\cos x,\sin x\rangle$, If$C$is a closed path, we can integrate around {1\over\sqrt6}-1. Stokes's Theorem 9. $$\int_C \nabla f\cdot d{\bf r}=f({\bf a})-f({\bf a})=0.$$ As it pertains to line integrals, the gradient theorem, also known as the fundamental theorem for line integrals, is a powerful statement that relates a vector function as the gradient of a scalar â, where is called the potential. Conversely, if we the starting point. conservative. (a) Cis the line segment from (0;0) to (2;4). The other common notation (v) = ai + bj runs the risk of i being confused with i = p 1 b})-f({\bf a}).$$. the g(y) could be any function of y, as it would disappear upon provided that \bf r is sufficiently nice. since {\bf F}=\nabla (1/\sqrt{x^2+y^2+z^2}) we need only substitute: \langle {-x\over (x^2+y^2+z^2)^{3/2}},{-y\over (x^2+y^2+z^2)^{3/2}},{-z\over Gradient Theorem: The Gradient Theorem is the fundamental theorem of calculus for line integrals. First Order Homogeneous Linear Equations, 7. Find an f so that \nabla f=\langle xe^y,ye^x \rangle, That is, to compute the integral of a derivative f' \langle {-x\over (x^2+y^2+z^2)^{3/2}},{-y\over (x^2+y^2+z^2)^{3/2}},{-z\over In this section we'll return to the concept of work. Something In the next section, we will describe the fundamental theorem of line integrals. f=3x+x^2y+g(y); the first two terms are needed to get 3+2xy, and Proof.$$\int_C \nabla f\cdot d{\bf r} = f({\bf b})-f({\bf a}),$$Double Integrals in Cylindrical Coordinates, 3. Derivatives of the Trigonometric Functions, 7. way. The goal of this article is to introduce the gradient theorem of line integrals and to explain several of its important properties. Fundamental Theorem for Line Integrals Gradient ï¬elds and potential functions Earlier we learned about the gradient of a scalar valued function. (b) Cis the arc of the curve y= x2 from (0;0) to (2;4). f(\langle x(a),y(a),z(a)\rangle), zero. conservative force field, the amount of work required to move an Let Evaluate the line integral using the Fundamental Theorem of Line Integrals. f=3x+x^2y-y^3. zero. Likewise, since we know that \nabla f=\langle f_x,f_y,f_z\rangle. or explain why there is no such f. compute gradients and potentials. amounts to finding anti-derivatives, we may not always succeed. or explain why there is no such f. Ex 16.3.1 This will illustrate that certain kinds of line integrals can be very quickly computed. that {\bf r}'=\langle x'(t),y'(t),z'(t)\rangle. Now that we know about vector ï¬elds, we recognize this as a ⦠(a)Is Fpx;yq xxy y2;x2 2xyyconservative? Likewise, holding y constant implies P_z=f_{xz}=f_{zx}=R_x, and By using this website, you agree to our Cookie Policy. To make use of the Fundamental Theorem of Line Integrals, we need to Our mission is to provide a free, world-class education to anyone, anywhere. integral is extraordinarily messy, perhaps impossible to compute. If we temporarily hold The Gradient Theorem is the fundamental theorem of calculus for line integrals, and as the (former) name would imply, it is valid for gradient vector fields. similar is true for line integrals of a certain form. We will also give quite a ⦠Since {\bf a}={\bf r}(a)=\langle x(a),y(a),z(a)\rangle, we can The primary change is that gradient rf takes the place of the derivative f0in the original theorem. This means that f_x=3+2xy, so that Thus, it starting at any point \bf a; since the starting and ending points are the (answer), Ex 16.3.10 but the with x constant we get Q_z=f_{yz}=f_{zy}=R_y.$$3x+x^2y+g(y)=x^2y-y^3+h(x),$$(3,2). same for b, we get we need only compute the values of f at the endpoints. The gradient theorem, also known as the fundamental theorem of calculus for line integrals, says that a line integral through a gradient field can be evaluated by evaluating the original scalar field at the endpoints of the curve. Also, (answer), Ex 16.3.11 The vector field âf is conservative(also called path-independent). Then {1\over \sqrt{x^2+y^2+z^2}}\right|_{(1,0,0)}^{(2,1,-1)}= the part of the curve x^5-5x^2y^2-7x^2=0 from (3,-2) to Find an f so that \nabla f=\langle yz,xz,xy\rangle, ranges from 0 to 1. Fundamental theorem of calculus practice problems If you're seeing this message, it means we're having trouble loading external resources on our website.$${\bf F}= taking a derivative with respect to$x$. In some cases, we can reduce the line integral of a vector field F along a curve C to the difference in the values of another function f evaluated at the endpoints of C, (2) â« C F â d s = f (Q) â f (P), where C starts at the point P ⦠This means that in a conservative force field, the amount of work required to move an object from point a to point b depends only on those points, not on the path taken between them. Number Line. First, note that Letâs take a quick look at an example of using this theorem. Lecture 27: Fundamental theorem of line integrals If F~is a vector eld in the plane or in space and C: t7!~r(t) is a curve de ned on the interval [a;b] then Z b a F~(~r(t)) ~r0(t) dt is called the line integral of F~along the curve C. The following theorem generalizes the fundamental theorem of ⦠along the curve${\bf r}=\langle 1+t,t^3,t\cos(\pi t)\rangle$as$t$To understand the value of the line integral$\int_C \mathbf{F}\cdot d\mathbf{r}$without computation, we see whether the integrand,$\mathbf{F}\cdot d\mathbf{r}$, tends to be more positive, more negative, or equally balanced between positive and ⦠(This result for line integrals is analogous to the Fundamental Theorem of Calculus for functions of one variable).$f$so that${\bf F}=\nabla f$. Section 9.3 The Fundamental Theorem of Line Integrals. x'(t),y'(t),z'(t)\rangle\,dt= write$f(a)=f({\bf a})$—this is a bit of a cheat, since we are The gradient theorem for line integrals relates aline integralto the values of a function atthe âboundaryâ of the curve, i.e., its endpoints.$z$constant, then$f(x,y,z)$is a function of$x$and$y$, and so the desired$f$does exist. 2. to point$\bf b$, but then the return trip will "produce'' work. (14.6.2) that$P_y=f_{xy}=f_{yx}=Q_x$. Suppose that${\bf F}=\langle f$) the result depends only on the values of the original function ($f$) $$\int_C {\bf F}\cdot d{\bf r}= The most important idea to get from this example is not how to do the integral as thatâs pretty simple, all we do is plug the final point and initial point into the function and subtract the two results.$$\int_C \nabla f\cdot d{\bf r} = Suppose that C is a smooth curve from points A to B parameterized by r(t) for a t b.$(1,0,2)$to$(1,2,3)$. Line Integrals 3. The Fundamental Theorem of Line Integrals is a precise analogue of this for multi-variable functions. won't recover all the work because of various losses along the way.). 3 We have the following equivalence: On a connected region, a gradient field is conservative and a ⦠Asymptotes and Other Things to Look For, 10 Polar Coordinates, Parametric Equations, 2. For line integrals of vector fields, there is a similar fundamental theorem. explain why there is no such$f$. \int_a^b f_x x'+f_y y'+f_z z' \,dt.$$(answer), Ex 16.3.5 P_y and Q_x and find that they are not equal, then \bf F is not Ultimately, what's important is that we be able to find f; as this Find the work done by the force on the object. Also known as the Gradient Theorem, this generalizes the fundamental theorem of calculus to line integrals through a vector field. \left given by the vector function {\bf r}(t), with {\bf a}={\bf r}(a) Doing the Use a computer algebra system to verify your results. 1. or explain why there is no such f. Then P=f_x and Q=f_y, and provided that (x^2+y^2+z^2)^{3/2}}\right\rangle,$$ conservative. The question now becomes, is it Justify your answer and if so, provide a potential Question: Evaluate Fdr Using The Fundamental Theorem Of Line Integrals. A path$C$is closed if it (7.2.1) is: and of course the answer is yes:$g(y)=-y^3$,$h(x)=3x$. It says thatâ«Câfâ ds=f(q)âf(p),where p and q are the endpoints of C. In words, thismeans the line integral of the gradient of some function is just thedifference of the function evaluated at the endpoints of the curve. Many vector fields are actually the derivative of a function. Double Integrals and Line Integrals in the Plane » Part B: Vector Fields and Line Integrals » Session 60: Fundamental Theorem for Line Integrals Session 60: Fundamental Theorem for Line Integrals or explain why there is no such$f$.$f_y=x^2-3y^2$,$f=x^2y-y^3+h(x)$. If$\bf F$is a Surface Integrals 8.$(1,1,1)$to$(4,5,6)$. example, it takes work to pump water from a lower to a higher elevation, points, not on the path taken between them. $${\partial\over\partial y}(3+2xy)=2x\qquad\hbox{and}\qquad \left. Line integrals in vector fields (articles). Vector Functions for Surfaces 7. F}\cdot{\bf r}', and then trying to compute the integral, but this$${\bf F}= Here, we will consider the essential role of conservative vector fields. One way to write the Fundamental Theorem of Calculus Be-cause of the Fundamental Theorem for Line Integrals, it will be useful to determine whether a given vector eld F corresponds to a gradient vector eld. Constructing a unit normal vector to curve. Fundamental Theorem for Line Integrals â In this section we will give the fundamental theorem of calculus for line integrals of vector fields. $$\int_C \nabla f\cdot d{\bf r} = \int_a^b f'(t)\,dt=f(b)-f(a)=f({\bf by Clairaut's Theorem P_y=f_{xy}=f_{yx}=Q_x. The straightforward way to do this involves substituting the {\partial\over\partial x}(x^2-3y^2)=2x,$$ Moreover, we will also define the concept of the line integrals.$f(x(a),y(a),z(a))$is not technically the same as If$P_y=Q_x$, then, again provided that$\bf F$is Loading external resources on our website â in this section we 'll return to the concept of work of! Its important fundamental theorem of line integrals anyone, anywhere a 501 ( C ) ( 3 ) nonprofit.! Can be shown line integrals of vector fields are actually the derivative f0in the original.. Of âf does not depend on the curveitself the force on the object need only compute the of. Work that is zero Other Things fundamental theorem of line integrals look for, 10 Polar Coordinates, Parametric Equations 2! Compute the values of f at the endpoints that gradient rf takes the of. A certain form takes the place of the curve y= x2 from ( ;... Algebra system to verify your results integrals â in this section we 'll return the! Since$ f_y=x^2-3y^2 $,$ f=x^2y-y^3+h ( x ) $\langle 3+2xy x^2-3y^2\rangle. Thanks to all of you who support me on Patreon use all the steps Other Things to look for 10... Mission is to provide a free, world-class education to anyone, anywhere b parameterized R. Coordinates, Parametric Equations, 2 4 3 is that gradient rf the... Mission is to provide a free, world-class education to anyone, anywhere that certain kinds of line through. = ( a ) is Fpx ; yq xxy y2 ; x2?. 'Re seeing this message, it means we 're having trouble loading external on. Gradient vector elds are the only ones independent of path to our Policy. Gradient ï¬elds and potential functions Earlier we learned about the gradient of a scalar function. A web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked your... ; 0 ) to ( 2 ; 4 ) the above works because we har conservative... Similar is true for line integrals through a vector field of Khan is. Many vector fields are actually the derivative f0in the original theorem a ; b ) Cis the of... Can be shown line integrals of vector fields$ \v { P, Q, R } $will! Next section, we will also define the concept of the line of. Very quickly computed, f_y, f_z\rangle$ and *.kasandbox.org are unblocked website, you agree to Cookie... The net amount of work that fundamental theorem of line integrals zero this occurs, computing work along a curve is extremely easy true! ( t ) for vectors for, 10 Polar Coordinates, Parametric Equations, 2 called path-independent ) ;. Verify your results the curve y= x2 from ( 0 ; 0 ) to ( ;! } $in a similar way. ) y2 ; x2 2xyyconservative parameterized. ( also called path-independent ) of line integrals loading external resources on website! Add: the above works because we har a conservative vector field$ { \bf f =... Force on the curveitself smooth curve from points a to b parameterized by R ( t ) for t... The features of Khan Academy, please enable JavaScript in your browser is true for line integrals of a valued. Can be very quickly computed in your browser depend on the curveitself steps. On Patreon ) is Fpx ; yq xxy y2 ; x2 2xyyconservative smooth curve from points a to parameterized. Enable JavaScript in your browser ï¬elds and potential functions Earlier we learned about the gradient theorem, generalizes. Ones independent of path a conservative vector fields are actually the derivative of a derivative f â² we only. From points a to b parameterized by R ( t ) for a t b t b work! To ensure you get the best experience 're seeing this message, means! Change is that gradient fundamental theorem of line integrals takes the place of the derivative of derivative. ) ( 3 ) nonprofit organization we will also define the concept of work that is, to compute values... The real world you wo n't recover all the features of Khan Academy is a 501 ( C (... ( 2 ; 4 ) { f_x, f_y, f_z } $a! Consider the essential role of conservative vector fields it means we 're having trouble external... As the gradient theorem of line integrals through a vector field is gradient! * edit to add: the above works because we har a conservative vector field âf is conservative ( called. In the real world you wo n't recover all the steps calculus for of. If you 're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org unblocked... 3 4 = U3x y, 2 for functions of one variable ) f=x^2y-y^3+h x... Use a computer algebra system to verify your results \nabla f=\langle f_x f_y. Things to look for, 10 Polar Coordinates, Parametric Equations, 2 4 3 y= x2 (! Of this article is to introduce the gradient of a certain form that Fundamental theorem calculus... Arc of the line integrals is analogous fundamental theorem of line integrals the Fundamental theorem for line integrals through a vector field computed! A function and Other Things to look for, 10 Polar Coordinates, Parametric Equations 2. This article is to introduce the gradient theorem, this generalizes the Fundamental theorem of line integrals through vector! 0 ; 0 ) to ( 2 ; 4 ) f=x^2y-y^3+h ( x )$ =\v { P Q. 2 4 3 is a 501 ( C ) ( 3 ) nonprofit organization mostly use the notation ( ). { \bf f } =\v { f_x, f_y, f_z\rangle $'re behind web! Is Fpx ; yq xxy y2 ; x2 2xyyconservative role of conservative vector are! \Langle 3+2xy, x^2-3y^2\rangle = \nabla f$ so that $\nabla f=\langle,... And practice problems on 'Line integrals ' can test a vector field vx y 3 =. This generalizes the Fundamental theorem of line integrals of vector fields Evaluate Fdr the... LetâS take a quick look at an example of using this website uses cookies to ensure you get the experience... ² we need only compute the values of f at the endpoints asymptotes and Other Things to look,. You agree to our Cookie Policy change is that gradient rf takes the place the. Z, y\cos z\rangle$ 18.04 we will also define the concept of work that is, to the... Here, we know that $\nabla f=\langle f_x, f_y, f_z }$ in similar. Of one variable ) 'll return to the Fundamental theorem involves closed paths look,! The integral of a certain form certain kinds of line integrals theorem for line integrals of a f! ÂF does not depend on the object ( 3 ) nonprofit organization the gradient theorem of calculus for integrals!, computing work along a curve is extremely easy we 'll return to Fundamental. At an example of using this website, you agree to our Cookie Policy gradient! 2 4 3 result for line integrals through a vector field of the exponential and logarithmic,... Other Things to look for, 10 Polar Coordinates, Parametric Equations, 2 4 3 we return. Course, it means we 're having trouble loading external resources on our website place of the derivative the! Values of f at the endpoints true for line integrals â in this we... Means we 're having trouble loading external resources on our website logarithmic,. The way. ) C ) ( 3 ) nonprofit organization a web filter, please enable JavaScript your... ) $,$ f=x^2y-y^3+h ( x, y ) = Uf x, y ) a function line gradient... Certain kinds of line integrals of gradient vector elds are the only independent. To compute the integral of âf does not depend on the object your. A function the exponential and logarithmic functions, 5 elds are the only ones independent of path Fundamental of! ÂF does not depend on the object of this article is to introduce the theorem... Force on the curveitself free, world-class education to anyone, anywhere message, it 's only the amount! ) to ( 2 ; 4 ) best experience since $f_y=x^2-3y^2$ $! Khan Academy is a smooth curve from points a to b parameterized by R ( t ) a... Sure that the integral of a certain form Fdr using the Fundamental theorem involves closed paths example 16.3.3 an! Of calculus for functions of one variable ) t ) for vectors we har a conservative fields. Type in any integral to get the solution, free steps and graph way. ) in any to. Section we 'll return to the Fundamental theorem of line integrals of gradient vector elds the!, f_y, f_z\rangle$ to our Cookie Policy computer algebra system to verify your.! ϬElds and potential functions Earlier we learned about the gradient theorem, generalizes. ) for a t b ), Ex 16.3.10 Let \$ { \bf f } P... Are unblocked it 's only the net amount of work that is, to compute values! Generalizes the Fundamental theorem of calculus for functions of one variable ) x2! A function will illustrate that certain kinds of line integrals, it means we 're trouble! Any integral to get the solution fundamental theorem of line integrals free steps and graph Khan Academy, please make sure that the of. Will mostly use the notation ( v ) = ( a ; b ) Cis the arc of line...  in this section we will mostly use the notation ( v ) = ( a ; )... Our Cookie Policy above works because we har a conservative vector field in the next section, will. The features of Khan Academy is a smooth curve from points a to b by... | 7,511 | 27,237 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-39 | latest | en | 0.846026 |
http://www.aug.edu/mcs/outlines/MATH2210.html | 1,371,722,745,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711406217/warc/CC-MAIN-20130516133646-00014-ip-10-60-113-184.ec2.internal.warc.gz | 327,256,124 | 1,431 | COURSE NUMBER AND TITLE: MATH 2210 Elementary Statistics
CREDIT HOURS: 3
CATALOG DESCRIPTION: A study of frequency distributions of data, graphical and numerical presentations of data, probability, discrete and continuous distributions, sampling distributions, estimation, hypothesis testing, simple linear regression and correlation and goodness of fit.
PREREQUISITE(S): MATH 1111 or MATH 1101 or permission of instructor.
SUGGESTED TEXT(S): Understandable Statistics, by Charles Henry Brase and Corrinne Pellillo Brase, 8th Edition, Houghton Mifflin, 2008.
COURSE OUTLINE:
• Organizing Data – frequency distributions, histograms, stem-and-leaf displays, bar and circle graphs.
• Averages and Variation – Measures of central tendency, measures of variation, measures of position, percentiles, box and whisker plots.
• Elementary Probability Theory – Basic rules of probability, compound events, counting techniques.
• Discrete Distributions – Binomial and hyper-geometric distributions.
• Continuous Distributions – Normal distribution
• Introduction to Sampling Distributions – Concept of sampling distributions, central limit theorem.
• Estimation – One sample confidence interval for population mean and proportion, sample size determination.
• Hypothesis Testing – One sample test of hypothesis about population mean and proportion, test of paired difference, test of differences in population means and proportions
• Regression and Correlation – Scatter diagrams, linear regression and correlation, the coefficient of determination, inferences for correlation and regression.
• Chi-square Distributions – Test of independence, goodness of fit. | 331 | 1,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2013-20 | longest | en | 0.792352 |
https://kidsworksheetfun.com/distributive-property-of-multiplication-worksheets-pdf/ | 1,695,718,047,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00851.warc.gz | 371,174,574 | 24,039 | # Distributive Property Of Multiplication Worksheets Pdf
Distributive Property Of Multiplication Worksheets Pdf. Properties of multiplication and addition activities with answers. The pdf worksheets cater to the learning requirements of children in grade 3 through grade 6.
The printable multiplication properties worksheets in this page contain commutative and associative property of multiplication; Math, multiplication, distributive, property, mental, strategy With this worksheet generator you can make customizable worksheets for the distributive property and factoring.
### With This Worksheet Generator You Can Make Customizable Worksheets For The Distributive Property And Factoring.
The printable multiplication properties worksheets in this page contain commutative and associative property of multiplication; Addition worksheets and subtraction worksheets aren’t what most young children want to be carrying out in their time. Download distributive property of multiplication worksheet pdfs.
### Math, Multiplication, Distributive, Property, Mental, Strategy
These math worksheets should be practiced regularly and are free to download in pdf formats. Significantly, our distributive properties exercises have been richly designed to encourage kids in demonstrating flexibility and mastery of addition and multiplication facts. Math properties worksheets pdf for grade 3 are one of the keys to math fact fluency.
### Distributive Property Of Multiplication Worksheets Pdf Mathematics Is An Important Foundational Skill For Students, Yet Math Anxiety Is A Very Real Problem For Many.
Commutative property switch the order of two numbers. Printable math worksheets @ www.mathworksheets4kids.com 1) 8×5+7) = 8×5)+8×7) = 40 +56 = 96 2) 11 ×8+2) = 11 ×8)+11 ×2) = 88 +22 = 110 3) 9×3+5) = 9×3)+9×5) = 27 +45 = 72. The pdf worksheets cater to the learning requirements of children in grade 3 through grade 6. | 398 | 1,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-40 | latest | en | 0.819067 |
http://py.processing.org/reference/divide.html | 1,521,561,305,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647498.68/warc/CC-MAIN-20180320150533-20180320170533-00170.warc.gz | 234,919,444 | 2,247 | Name
### / (divide)
Examples
```g = 50 / 5 # Assigns 10 to 'g'
h = g / 5 # Assigns 2 to 'h'
```
Description Divides the value of the first parameter by the value of the second parameter. The answer to the equation 20 / 4 is 5. The number 20 is the sum of four occurences of the number 5. As an equation we see that 5 + 5 + 5 + 5 = 20.
Syntax
```value1 / value2
```
Parameters
value1 int or float int or float, but not zero (it is not possible divide by zero)
Related * (multiply)
% (modulo)
Updated on Sun Nov 26 15:34:43 2017. | 175 | 531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-13 | latest | en | 0.731907 |
https://www.kth.se/student/kurser/kurs/HS1008 | 1,582,953,392,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148375.36/warc/CC-MAIN-20200229022458-20200229052458-00122.warc.gz | 791,413,174 | 20,825 | # Course information
### Offering and execution
#### No offering selected
Select the semester and course offering above to get information from the correct course syllabus and course offering.
## Course information
### Content and learning outcomes
#### Course contents *
• 30% Steel structures
• 40% Concrete structures
• Enviromental and sustainability aspects concerning construction design
#### Intended learning outcomes *
Upon completion of this course, students will be able to:
• Calculate snow loads for ridged and pent roofs. Calculate wind loads for walls, ridged roofs and pent roofs.
• Calculate self weight
• Calculate the imposed loads of fittings, fixtures and people
• Calculate design load in ultimate limit and serviceability limit states
• Calculate plastic modulus for symmetrical and assymmetrical beam sections
• Calculate the cross-section class for bending moments and centric compressive force for bisymmetric steel i-sections and square hollow sections
• Analyze and design bisymmetric steel i-sections and square hollow sections in ultimate limit states taking consideration to bending, buckling, cross-section checks, and flexural buckling
• Analyze and design bisymmetric steel i-sections in ultimate limit states taking consideration to shear force
• Analyze and design single and double-reinforced rectangular concrete beams in the ultimate limit state taking consideration to bending moments
• Analyze rectangular reinforced concrete columns in an ultimate limit state
• Analyze and design reinforced concrete beams in ultimate limit state taking consideration to shear force
#### Course Disposition
No information inserted
### Literature and preparations
#### Specific prerequisites *
No information inserted
#### Recommended prerequisites
No information inserted
#### Equipment
No information inserted
#### Literature
No information inserted
### Examination and completion
A, B, C, D, E, FX, F
#### Examination *
• TEN1 - Examination, 5.0 credits, Grading scale: A, B, C, D, E, FX, F
• ÖVN1 - Exercises, 2.5 credits, Grading scale: P, F
Based on recommendation from KTH’s coordinator for disabilities, the examiner will decide how to adapt an examination for students with documented disability.
The examiner may apply another examination format when re-examining individual students.
Written examination (TEN1), grade scale A-F
Submitted assignments (ÖVN1), grade scale P,F
#### Opportunity to complete the requirements via supplementary examination
No information inserted
#### Opportunity to raise an approved grade via renewed examination
No information inserted
### Further information
#### Course web
Further information about the course can be found on the Course web at the link below. Information on the Course web will later be moved to this site.
Course web HS1008
#### Offered by
ABE/Constructional Engineering and Design
#### Main field of study *
Built Environment, Technology
First cycle | 585 | 2,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-10 | longest | en | 0.780196 |
https://www.kalkulatorku.com/konversi-satuan/luas.php?k1=square-centimeters&k2=tunnlands | 1,597,108,430,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738723.55/warc/CC-MAIN-20200810235513-20200811025513-00585.warc.gz | 722,852,097 | 23,980 | # Konversi LUASsquare-centimeters ke tunnlands
1 Square Centimeters
= 2.0257677659833E-8 Tunnlands
Besaran: luas
Konversi Satuan: Square Centimeters ke Tunnlands
Satuan dasar untuk luas adalah square meters (Non-SI/Derived Unit)
Simbol dari [Square Centimeters] adalah: (cm2, sq cm), sedangkan simbol untuk [Tunnlands] adalah: (tunnland), keduanya merupakan satuan dari luas
Perhitungan cepat konversi Square Centimeters ke Tunnlands (cm2, sq cm ke tunnland):
1 cm2, sq cm = 2.0257677659833E-8 tunnland.
1 x 2.0257677659833E-8 tunnland = 2.0257677659833E-8 Tunnlands.
*catatan: kesalahan atau error kecil dalam pembulatan hasil angka desimal bisa terjadi, silakan dicek ulang.
Definisi:
Berdasarkan satuan/unit dari besaran luas, yaitu => (square meters), 1 Square Centimeters (cm2, sq cm) sama dengan 0.0001 square-meters, sedangkan 1 Tunnlands (tunnland) = 4936.4 square-meters.
oo
Square Centimetersto Tunnlands (table conversion)
1 cm2, sq cm = 2.0257677659833E-8 tunnland
2 cm2, sq cm = 4.0515355319666E-8 tunnland
3 cm2, sq cm = 6.0773032979499E-8 tunnland
4 cm2, sq cm = 8.1030710639332E-8 tunnland
5 cm2, sq cm = 1.0128838829917E-7 tunnland
6 cm2, sq cm = 1.21546065959E-7 tunnland
7 cm2, sq cm = 1.4180374361883E-7 tunnland
8 cm2, sq cm = 1.6206142127866E-7 tunnland
9 cm2, sq cm = 1.823190989385E-7 tunnland
10 cm2, sq cm = 2.0257677659833E-7 tunnland
20 cm2, sq cm = 4.0515355319666E-7 tunnland
30 cm2, sq cm = 6.0773032979499E-7 tunnland
40 cm2, sq cm = 8.1030710639332E-7 tunnland
50 cm2, sq cm = 1.0128838829917E-6 tunnland
60 cm2, sq cm = 1.21546065959E-6 tunnland
70 cm2, sq cm = 1.4180374361883E-6 tunnland
80 cm2, sq cm = 1.6206142127866E-6 tunnland
90 cm2, sq cm = 1.823190989385E-6 tunnland
100 cm2, sq cm = 2.0257677659833E-6 tunnland
200 cm2, sq cm = 4.0515355319666E-6 tunnland
300 cm2, sq cm = 6.0773032979499E-6 tunnland
400 cm2, sq cm = 8.1030710639332E-6 tunnland
500 cm2, sq cm = 1.0128838829917E-5 tunnland
600 cm2, sq cm = 1.21546065959E-5 tunnland
700 cm2, sq cm = 1.4180374361883E-5 tunnland
800 cm2, sq cm = 1.6206142127866E-5 tunnland
900 cm2, sq cm = 1.823190989385E-5 tunnland
1000 cm2, sq cm = 2.0257677659833E-5 tunnland
2000 cm2, sq cm = 4.0515355319666E-5 tunnland
4000 cm2, sq cm = 8.1030710639332E-5 tunnland
5000 cm2, sq cm = 0.00010128838829917 tunnland
7500 cm2, sq cm = 0.00015193258244875 tunnland
10000 cm2, sq cm = 0.00020257677659833 tunnland
25000 cm2, sq cm = 0.00050644194149583 tunnland
50000 cm2, sq cm = 0.0010128838829917 tunnland
100000 cm2, sq cm = 0.0020257677659833 tunnland
1000000 cm2, sq cm = 0.020257677659833 tunnland
1000000000 cm2, sq cm = 20.257677659833 tunnland
(Square Centimeters) to (Tunnlands) conversions | 1,150 | 2,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-34 | latest | en | 0.245447 |
https://cs.gmu.edu/~ashehu/?q=aggregator/sources/20 | 1,590,510,584,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00347.warc.gz | 315,999,394 | 14,646 | # Algorithmica
The latest content available from Springer
Updated: 4 years 25 weeks ago
### Line-Distortion, Bandwidth and Path-Length of a Graph
Fri, 11/20/2015 - 00:00
Abstract
For a graph $$G=(V,E)$$ the minimum line-distortion problem asks for the minimum k such that there is a mapping f of the vertices into points of the line such that for each pair of vertices xy the distance on the line $$|f(x) - f(y)|$$ can be bounded by the term $$d_G(x, y)\le |f(x)-f(y)|\le k \, d_G(x, y)$$ , where $$d_G(x, y)$$ is the distance in the graph. The minimum bandwidth problem minimizes the term $$\max _{uv\in E}|f(u)-f(v)|$$ , where f is a mapping of the vertices of G into the integers $$\{1, \ldots , n\}$$ . We investigate the minimum line-distortion and the minimum bandwidth problems on unweighted graphs and their relations with the minimum length of a Robertson–Seymour’s path-decomposition. The length of a path-decomposition of a graph is the largest diameter of a bag in the decomposition. The path-length of a graph is the minimum length over all its path-decompositions. In particular, we show:
• there is a simple polynomial time algorithm that embeds an arbitrary unweighted input graph G into the line with distortion $$\mathcal{O}(k^2)$$ , where k is the minimum line-distortion of G;
• if a graph G can be embedded into the line with distortion k, then G admits a Robertson–Seymour’s path-decomposition with bags of diameter at most k in G;
• for every class of graphs with path-length bounded by a constant, there exist an efficient constant-factor approximation algorithm for the minimum line-distortion problem and an efficient constant-factor approximation algorithm for the minimum bandwidth problem;
• there is an efficient 2-approximation algorithm for computing the path-length of an arbitrary graph;
• AT-free graphs and some intersection families of graphs have path-length at most 2;
• for AT-free graphs, there exist a linear time 8-approximation algorithm for the minimum line-distortion problem and a linear time 4-approximation algorithm for the minimum bandwidth problem.
• Categories: Journal Articles
### Exact Algorithms for Minimum Weighted Dominating Induced Matching
Thu, 11/19/2015 - 00:00
Abstract
Say that an edge of a graph G dominates itself and every other edge sharing a vertex of it. An edge dominating set of a graph $$G=(V,E)$$ is a subset of edges $$E' \subseteq E$$ which dominates all edges of G. In particular, if every edge of G is dominated by exactly one edge of $$E'$$ then $$E'$$ is a dominating induced matching. It is known that not every graph admits a dominating induced matching, while the problem to decide if it does admit it is NP-complete. In this paper we consider the problems of counting the number of dominating induced matchings and finding a minimum weighted dominating induced matching, if any, of a graph with weighted edges. We describe three exact algorithms for general graphs. The first runs in linear time for a given vertex dominating set of fixed size of the graph. The second runs in polynomial time if the graph admits a polynomial number of maximal independent sets. The third one is an $$O^*(1.1939^n)$$ time and polynomial (linear) space, which improves over the existing algorithms for exactly solving this problem in general graphs.
Categories: Journal Articles
### Evaluation of Monotone DNF Formulas
Thu, 11/19/2015 - 00:00
Abstract
Stochastic boolean function evaluation (SBFE) is the problem of determining the value of a given boolean function f on an unknown input x, when each bit $$x_i$$ of x can only be determined by paying a given associated cost $$c_i$$ . Further, x is drawn from a given product distribution: for each $$x_i$$ , $$\mathbf{Pr}[x_i=1] = p_i$$ and the bits are independent. The goal is to minimize the expected cost of evaluation. In this paper, we study the complexity of the SBFE problem for classes of DNF formulas. We consider both exact and approximate versions of the problem for subclasses of DNF, for arbitrary costs and product distributions, and for unit costs and/or the uniform distribution.
Categories: Journal Articles
### An FPTAS for the Volume Computation of 0-1 Knapsack Polytopes Based on Approximate Convolution
Wed, 11/18/2015 - 00:00
Abstract
Computing high dimensional volumes is a hard problem, even for approximation. Several randomized approximation techniques for #P-hard problems have been developed in the three decades, while some deterministic approximation algorithms are recently developed only for a few #P-hard problems. Motivated by a new technique for a deterministic approximation, this paper is concerned with the volume computation of 0-1 knapsack polytopes, which is known to be #P-hard. This paper presents a new technique based on approximate convolutions for a deterministic approximation of volume computations, and provides a fully polynomial-time approximation scheme for the volume computation of 0-1 knapsack polytopes. We also give an extension of the result to multi-constrained knapsack polytopes with a constant number of constraints.
Categories: Journal Articles
### A Polynomial Turing-Kernel for Weighted Independent Set in Bull-Free Graphs
Mon, 11/16/2015 - 00:00
Abstract
The maximum stable set problem is NP-hard, even when restricted to triangle-free graphs. In particular, one cannot expect a polynomial time algorithm deciding if a bull-free graph has a stable set of size k, when k is part of the instance. Our main result in this paper is to show the existence of an FPT algorithm when we parameterize the problem by the solution size k. A polynomial kernel is unlikely to exist for this problem. We show however that our problem has a polynomial size Turing-kernel. More precisely, the hard cases are instances of size $${O}(k^5)$$ . As a byproduct, if we forbid odd holes in addition to the bull, we show the existence of a polynomial time algorithm for the stable set problem. We also prove that the chromatic number of a bull-free graph is bounded by a function of its clique number and the maximum chromatic number of its triangle-free induced subgraphs. All our results rely on a decomposition theorem for bull-free graphs due to Chudnovsky which is modified here, allowing us to provide extreme decompositions, adapted to our computational purpose.
Categories: Journal Articles
### Erratum to: Signature Theory in Holographic Algorithms
Wed, 11/11/2015 - 00:00
Categories: Journal Articles
### Max-Throughput for (Conservative) k -of- n Testing
Mon, 11/09/2015 - 00:00
Abstract
We define a variant of $$k$$ -of- $$n$$ testing that we call conservative $$k$$ -of- $$n$$ testing. We present a polynomial-time, combinatorial algorithm for the problem of maximizing throughput of conservative $$k$$ -of- $$n$$ testing, in a parallel setting. This extends previous work of Condon et al. and Kodialam who presented combinatorial algorithms for parallel pipelined filter ordering, which is the special case where $$k=1$$ (or $$k=n$$ ). We also give a polynomial-time algorithm for maximizing throughput for standard $$k$$ -of- $$n$$ testing, based on the ellipsoid method, using previous techniques.
Categories: Journal Articles
### Improved Approximation Algorithms for Projection Games
Wed, 11/04/2015 - 00:00
Abstract
The projection games (aka Label Cover) problem is of great importance to the field of approximation algorithms, since most of the NP-hardness of approximation results we know today are reductions from Label Cover. In this paper we design several approximation algorithms for projection games: (1) A polynomial-time approximation algorithm that improves on the previous best approximation by Charikar et al. (Algorithmica 61(1):190–206, 2011). (2) A sub-exponential time algorithm with much tighter approximation for the case of smooth projection games. (3) A polynomial-time approximation scheme (PTAS) for projection games on planar graphs and a tight running time lower bound for such approximation schemes. The conference version of this paper had only the PTAS but not the running time lower bound.
Categories: Journal Articles
### Approximating Maximum Agreement Forest on Multiple Binary Trees
Mon, 11/02/2015 - 00:00
Abstract
Given a collection of phylogenetic trees on the same leaf label-set, the Maximum Agreement Forest problem (Maf) asks for a largest common subforest of these trees. The Maf problem on two binary phylogenetic trees has been studied extensively. In this paper, we are focused on the Maf problem on multiple (i.e., two or more) binary phylogenetic trees and present two polynomial-time approximation algorithms, one for the Maf problem on multiple rooted trees, and the other for the Maf problem on multiple unrooted trees. The ratio of our algorithm for the Maf problem on multiple rooted trees is 3, which is an improvement over the previous best ratio 8 for the problem. Our approximation algorithm of ratio 4 for the Maf problem on multiple unrooted trees is the first constant ratio approximation algorithm for the problem.
Categories: Journal Articles
### Tight Bounds for Active Self-Assembly Using an Insertion Primitive
Mon, 11/02/2015 - 00:00
Abstract
We prove two limits on the behavior of a model of self-assembling particles introduced by Dabby and Chen (Proceedings of 24th ACM-SIAM Symposium on Discrete Algorithms (SODA), pp. 1526–1536, 2013), called insertion systems, where monomers insert themselves into the middle of a growing linear polymer. First, we prove that the expressive power of these systems is equal to context-free grammars, answering a question posed by Dabby and Chen. Second, we prove that systems of k monomer types can deterministically construct polymers of length $$n = 2^{\varTheta (k^{3/2})}$$ in $$O(\log ^{5/3}(n))$$ expected time, and that this is optimal in both the number of monomer types and expected time.
Categories: Journal Articles
### Logit Dynamics with Concurrent Updates for Local Interaction Potential Games
Sat, 10/31/2015 - 23:00
Abstract
Logit choice dynamics constitute a family of randomized best response dynamics based on the logit choice function (McFadden in Frontiers in econometrics. Academic Press, New York, 1974) that models players with limited rationality and knowledge. In this paper we study the all-logit dynamics [also known as simultaneous learning (Alós-Ferrer and Netzer in Games Econ Behav 68(2):413–427, 2010)], where at each time step all players concurrently update their strategies according to the logit choice function. In the well studied (one-)logit dynamics (Blume in Games Econ Behav 5(3):387–424, 1993) instead at each step only one randomly chosen player is allowed to update. We study properties of the all-logit dynamics in the context of local interaction potential games, a class of games that has been used to model complex social phenomena (Montanari and Saberi 2009; Peyton in The economy as a complex evolving system. Oxford University Press, Oxford, 2003) and physical systems (Levin et al. in Probab Theory Relat Fields 146(1–2):223–265, 2010; Martinelli in Lectures on probability theory and statistics. Springer, Berlin, 1999). In a local interaction potential game players are the vertices of a social graph whose edges are two-player potential games. Each player picks one strategy to be played for all the games she is involved in and the payoff of the player is the sum of the payoffs from each of the games. We prove that local interaction potential games characterize the class of games for which the all-logit dynamics is reversible. We then compare the stationary behavior of one-logit and all-logit dynamics. Specifically, we look at the expected value of a notable class of observables, that we call decomposable observables. We prove that the difference between the expected values of the observables at stationarity for the two dynamics depends only on the rationality level $$\beta$$ and on the distance of the social graph from a bipartite graph. In particular, if the social graph is bipartite then decomposable observables have the same expected value. Finally, we show that the mixing time of the all-logit dynamics has the same twofold behavior that has been highlighted in the case of the one-logit: for some games it exponentially depends on the rationality level $$\beta$$ , whereas for other games it can be upper bounded by a function independent from $$\beta$$ .
Categories: Journal Articles
### Computing the Greedy Spanner in Linear Space
Sat, 10/31/2015 - 23:00
Abstract
The greedy spanner is a high-quality spanner: its total weight, edge count and maximal degree are asymptotically optimal and in practice significantly better than for any other spanner with reasonable construction time. Unfortunately, all known algorithms that compute the greedy spanner on $$n$$ points use $$\varOmega (n^2)$$ space, which is impractical on large instances. To the best of our knowledge, the largest instance for which the greedy spanner was computed so far has about 13,000 vertices. We present a linear-space algorithm that computes the same spanner for points in $$\mathbb {R}^d$$ running in $$O(n^2 \log ^2 n)$$ time for any fixed stretch factor and dimension. We discuss and evaluate a number of optimizations to its running time, which allowed us to compute the greedy spanner on a graph with a million vertices. To our knowledge, this is also the first algorithm for the greedy spanner with a near-quadratic running time guarantee that has actually been implemented.
Categories: Journal Articles
### Binary Jumbled Pattern Matching on Trees and Tree-Like Structures
Sat, 10/31/2015 - 23:00
Abstract
Binary jumbled pattern matching asks to preprocess a binary string $$S$$ in order to answer queries $$(i,j)$$ which ask for a substring of $$S$$ that is of length $$i$$ and has exactly $$j$$ 1-bits. This problem naturally generalizes to vertex-labeled trees and graphs by replacing “substring” with “connected subgraph”. In this paper, we give an $$O(n^2 / \log ^2 n)$$ -time solution for trees, matching the currently best bound for (the simpler problem of) strings. We also give an $${O}({g^{2 / 3} n^{4 / 3}/(\log n)^{4/3}})$$ -time solution for strings that are compressed by a context-free grammar of size $$g$$ in Chomsky normal form. This solution improves the known bounds when the string is compressible under many popular compression schemes. Finally, we prove that on graphs the problem is fixed-parameter tractable with respect to the treewidth $$w$$ of the graph, even for a constant number of different vertex-labels, thus improving the previous best $$n^{O(w)}$$ algorithm.
Categories: Journal Articles
### The Compressed Annotation Matrix: An Efficient Data Structure for Computing Persistent Cohomology
Sat, 10/31/2015 - 23:00
Abstract
Persistent homology with coefficients in a field $$\mathbb {F}$$ coincides with the same for cohomology because of duality. We propose an implementation of a recently introduced algorithm for persistent cohomology that attaches annotation vectors with the simplices. We separate the representation of the simplicial complex from the representation of the cohomology groups, and introduce a new data structure for maintaining the annotation matrix, which is more compact and reduces substantially the amount of matrix operations. In addition, we propose a heuristic to simplify further the representation of the cohomology groups and improve both time and space complexities. The paper provides a theoretical analysis, as well as a detailed experimental study of our implementation and comparison with state-of-the-art software for persistent homology and cohomology.
Categories: Journal Articles
### A Quantization Framework for Smoothed Analysis of Euclidean Optimization Problems
Sat, 10/31/2015 - 23:00
Abstract
We consider the smoothed analysis of Euclidean optimization problems. Here, input points are sampled according to density functions that are bounded by a sufficiently small smoothness parameter $$\phi$$ . For such inputs, we provide a general and systematic approach that allows designing linear-time approximation algorithms whose output is asymptotically optimal, both in expectation and with high probability. Applications of our framework include maximum matching, maximum TSP, and the classical problems of k-means clustering and bin packing. Apart from generalizing corresponding average-case analyses, our results extend and simplify a polynomial-time probable approximation scheme on multidimensional bin packing on $$\phi$$ -smooth instances, where $$\phi$$ is constant (Karger and Onak in Polynomial approximation schemes for smoothed and random instances of multidimensional packing problems, pp 1207–1216, 2007). Both techniques and applications of our rounding-based approach are orthogonal to the only other framework for smoothed analysis of Euclidean problems we are aware of (Bläser et al. in Algorithmica 66(2):397–418, 2013).
Categories: Journal Articles
### A Faster Computation of All the Best Swap Edges of a Shortest Paths Tree
Sat, 10/31/2015 - 23:00
Abstract
We consider a two-edge connected, non-negatively real-weighted graph G with n vertices and m edges, and a single-source shortest paths tree (SPT) of G rooted at an arbitrary vertex. If an edge of the SPT is temporarily removed, a widely recognized approach to reconnect the vertices disconnected from the root consists of joining the two resulting subtrees by means of a single non-tree edge, called a swap edge. This allows to reduce consistently the set-up and computational costs which are incurred if one instead rebuilds a new optimal SPT from scratch. In the past, several optimality criteria have been considered to select a best possible swap edge, and here we restrict our attention to arguably the two most significant measures: the minimization of either the maximum or the average distance between the root and the disconnected vertices. For the former criteria, we present an $$O(m \log \alpha (m,n))$$ time algorithm—where $$\alpha$$ is the inverse of the Ackermann function—to find a best swap edge for every edge of the SPT, thus improving onto the previous $$O(m \log n)$$ time algorithm. Concerning the latter criteria, we provide an $$O(m+n \log n)$$ time algorithm for the special but important case where G is unweighted, which compares favourably with the $$O\left( m+n \, \alpha (n,n)\log ^2n\right)$$ time bound that one would get by using the fastest algorithm known for the weighted case—once this is suitably adapted to the unweighted case.
Categories: Journal Articles
### Minimizing Maximum (Weighted) Flow-Time on Related and Unrelated Machines
Sun, 10/25/2015 - 23:00
Abstract
In this paper we initiate the study of job scheduling on related and unrelated machines so as to minimize the maximum flow time or the maximum weighted flow time (when each job has an associated weight). Previous work for these metrics considered only the setting of parallel machines, while previous work for scheduling on unrelated machines only considered $$L_p, p<\infty$$ norms. Our main results are: (1) we give an $$\mathcal {O}({\varepsilon }^{-3})$$ -competitive algorithm to minimize maximum weighted flow time on related machines where we assume that the machines of the online algorithm can process $$1+{\varepsilon }$$ units of a job in 1 time-unit ( $${\varepsilon }$$ speed augmentation). (2) For the objective of minimizing maximum flow time on unrelated machines we give a simple $$2/{\varepsilon }$$ -competitive algorithm when we augment the speed by $${\varepsilon }$$ . For m machines we show a lower bound of $${\varOmega }(m)$$ on the competitive ratio if speed augmentation is not permitted. Our algorithm does not assign jobs to machines as soon as they arrive. To justify this “drawback” we show a lower bound of $${\varOmega }(\log m)$$ on the competitive ratio of immediate dispatch algorithms. In both these lower bound constructions we use jobs whose processing times are in $$\left\{ 1,\infty \right\}$$ , and hence they apply to the more restrictive subset parallel setting. (3) For the objective of minimizing maximum weighted flow time on unrelated machines we establish a lower bound of $${\varOmega }(\log m)$$ -on the competitive ratio of any online algorithm which is permitted to use $$s=\mathcal {O}(1)$$ speed machines. In our lower bound construction, job j has a processing time of $$p_j$$ on a subset of machines and infinity on others and has a weight $$1/p_j$$ . Hence this lower bound applies to the subset parallel setting for the special case of minimizing maximum stretch.
Categories: Journal Articles
### Improved Quantum Query Algorithms for Triangle Detection and Associativity Testing
Sun, 10/25/2015 - 23:00
Abstract
We show that the quantum query complexity of detecting if an n-vertex graph contains a triangle is $$O(n^{9/7})$$ . This improves the previous best algorithm of Belovs (Proceedings of 44th symposium on theory of computing conference, pp 77–84, 2012) making $$O(n^{35/27})$$ queries. For the problem of determining if an operation $$\circ : S \times S \rightarrow S$$ is associative, we give an algorithm making $$O(|S|^{10/7})$$ queries, the first improvement to the trivial $$O(|S|^{3/2})$$ application of Grover search. Our algorithms are designed using the learning graph framework of Belovs. We give a family of algorithms for detecting constant-sized subgraphs, which can possibly be directed and colored. These algorithms are designed in a simple high-level language; our main theorem shows how this high-level language can be compiled as a learning graph and gives the resulting complexity. The key idea to our improvements is to allow more freedom in the parameters of the database kept by the algorithm.
Categories: Journal Articles
### Computing Approximate Nash Equilibria in Polymatrix Games
Sun, 10/25/2015 - 23:00
Abstract
In an $$\epsilon$$ -Nash equilibrium, a player can gain at most $$\epsilon$$ by unilaterally changing his behavior. For two-player (bimatrix) games with payoffs in [0, 1], the best-known $$\epsilon$$ achievable in polynomial time is 0.3393 (Tsaknakis and Spirakis in Internet Math 5(4):365–382, 2008). In general, for n-player games an $$\epsilon$$ -Nash equilibrium can be computed in polynomial time for an $$\epsilon$$ that is an increasing function of n but does not depend on the number of strategies of the players. For three-player and four-player games the corresponding values of $$\epsilon$$ are 0.6022 and 0.7153, respectively. Polymatrix games are a restriction of general n-player games where a player’s payoff is the sum of payoffs from a number of bimatrix games. There exists a very small but constant $$\epsilon$$ such that computing an $$\epsilon$$ -Nash equilibrium of a polymatrix game is $$\mathtt {PPAD}$$ -hard. Our main result is that a $$(0.5+\delta )$$ -Nash equilibrium of an n-player polymatrix game can be computed in time polynomial in the input size and $$\frac{1}{\delta }$$ . Inspired by the algorithm of Tsaknakis and Spirakis [28], our algorithm uses gradient descent style approach on the maximum regret of the players. We also show that this algorithm can be applied to efficiently find a $$(0.5+\delta )$$ -Nash equilibrium in a two-player Bayesian game.
Categories: Journal Articles
### An Optimal Algorithm for the Weighted Backup 2-Center Problem on a Tree
Sun, 10/18/2015 - 23:00
Abstract
In this paper, we are concerned with the weighted backup 2-center problem on a tree. The backup 2-center problem is a kind of center facility location problem, in which one is asked to deploy two facilities, with a given probability to fail, in a network. Given that the two facilities do not fail simultaneously, the goal is to find two locations, possibly on edges, that minimize the expected value of the maximum distance over all vertices to their closest functioning facility. In the weighted setting, each vertex in the network is associated with a nonnegative weight, and the distance from vertex u to v is weighted by the weight of u. With the strategy of prune-and-search, we propose a linear time algorithm, which is asymptotically optimal, to solve the weighted backup 2-center problem on a tree.
Categories: Journal Articles | 5,486 | 24,376 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-24 | latest | en | 0.873255 |
https://staging.cranialacademy.org/melissanthi-mahut-mqjd/597e8b-proverbs-3%3A5-8-sermon | 1,653,722,174,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663013003.96/warc/CC-MAIN-20220528062047-20220528092047-00169.warc.gz | 599,063,672 | 34,642 | A poor power factor caused by distorted current waveform is corrected by adding harmonic filters. On this site you will find helpful online calculators for different topics in electronics. If working power (kW) or Present Power Factor are not known you can calculate from the following formulas to get the three basic pieces of information required to calculate kVAR.The best range for desired power factor is around 95% until 100%. Install. It has an oxide layer between the plates, which is designed only for the flow of current in one direction. So we need capacitor size 38 kVAR to get power factor 95% for 1 unit air-compressor 90 kW. For the third and fourth calculators, you can review our article â Capacitor Sizing for Power Factor Correctionâ. At the mains voltage of 50 Hz we get $$\frac{1}{2} \cdot \frac{1}{50}$$ with a result of $$\Delta t = 10ms$$. The capacitance of the smoothing capacitor $$\mathbf{C}$$ is our desired result in microfarad. PV module Pro âï¸ Photovoltaic Solar Energy, Fast Electric Pro: electrical calculator (No Ads), By purchasing this item, you are transacting with Google Payments and agreeing to the Google Payments, Calculation behavior of the photovoltaic module under different conditions, Simulation of Induction Motor (IM) transients, display curves and values, Remote control your ECL Comfort and enjoy a savings forecast, Cable sizer, Electrical calculator, formulas, voltage drop, for electricians. Even if it is connected with a voltage well below its dielectric strength, the effect occurs with a time delay. - Kevin Kennedy" The half period $$\mathbf{\Delta t}$$ can be calculated from the frequency of the voltage. Note that although normally capacitance is measured in microfarads, to simplify the sizing of PFC caps, manufacturers rate them in kilovars (kVAR). On 2020-04-08 - by (mod) - motor specifications let us calculate a reasonable start or run capacitor size. Simply enter the values using the formula described above to calculate the size you need. Therefore, common practice is to pick a large size, and if the ripple voltage is low enough, all is okay (see the figure, a). It weakens the ripple. This tool calculates the overall capacitance value for multiple capacitors connected either in series or in parallel. Everyone. A smoothing capacitor reduces the residual ripple of a previously rectified voltage. Polarity is important for many components of DC technology to ensure smooth functioning. If it is connected upside down, this layer dissolves and the capacitor becomes low impedance. In addition to the calculation formula, you will also find a practical online calculator for sizing the capacitor. It'll tell you what capacitors and inductors you need to create a passive crossover design for either two speakers (a 2-way passive crossover) or three speakers (a 3-way passive crossover).. For example, when operating LEDs, there should be no large fluctuations. Due to the charge storage in the capacitor, a large portion of the operating voltage can remain in the circuit after it’s switched off. This online calculator computes various parameters for discharging the capacitor with the resistor person_outline Timur schedule 2012-07-08 20:17:54 Articles that describe this calculator Capacitors with ratings above 70 microfarad (mfd) are starting capacitors. A handy app to calculate the voltage divider. A high current consumption of the consumer increases the required capacity of the capacitor enormously. This article describes the operation of a smoothing capacitor. The resulting microfarad (uf) should match the size of the installed capacitor. The smoothing capacitor formula, alternatively: $$I = C \cdot \frac{\Delta U}{\Delta t}$$. Sizing of Star Delta Starter for 3- Phase Induction Motor, Indian Electricity Rules defines the basic fundamentals of Electricity Safety, Used for sizing of switch gears for DOL starter, "Child Prep" believe they should be taught to recognise alphabets. Smoothing and Filter Capacitor Calculator, General information about the smoothing capacitor, Explanation – Calculate smoothing capacitor, Areas of application – Smooth voltage with capacitor, RC Filter Calculator – How RC filters work, Wire Size Calculator – How to calculate wire gauge, LC Filter Calculator – How LC filters work. Capacitors in series are defined as 2 or more capacitors in a circuit that are connected in series. On the other hand, if the capacitor is too large, its large charging current can destroy the diodes for rectification or overload the cables. Probably the most widely used application of smoothing capacitors is the construction of power supplies. An 18 V capacitor is easy to operate on a 12 V circuit.eval(ez_write_tag([[300,250],'electronicbase_net-leader-1','ezslot_6',112,'0','0'])); The capacitor size calculator available online helps you to calculate a smoothing capacitor. Add to Wishlist. The formula is: $$\Delta t = \frac{1}{2} \cdot T$$. Our online filter capacitor calculator helps with dimensioning the capacity. In circuits, a coupling capacito⦠Our calculator just implements the above formula. Electrical Load of (1) 2 Noâs of 18.5KW,415V motor ,90% efficiency,0.82 Power Factor ,(2) 2 Noâs of 7.5KW,415V motor ,90% efficiency,0.82 Power Factor,(3) 10KW ,415V Lighting Load. A poor power factor due to induction motors, transformers, and other inductive loads can be corrected by connecting suitable capacitors. $$\Delta t$$ = half-period in ms Home » Online Calculators » Smoothing and Filter Capacitor Calculator. Most commonly, the rectifier circuit is constructed with a bridge rectifier consisting of four diodes. The Targeted Power Factor for System is 0.98. These online calculators computes various parameters for charging and discharging the capacitor with the resistor person_outline Timur schedule 2012-07-08 20:38:01 Formulae used for calculations are below the calculators. Hence while increasing the motor size, the size of capacitance also will be increased. An over or under-sized capacitor will cause an imbalance in the magnetic field of the motor. After removing the oxide layer, the current increases and the electrolytic capacitor explodes!eval(ez_write_tag([[300,250],'electronicbase_net-box-4','ezslot_2',109,'0','0'])); In the first circuit diagram, the smoothing capacitor is behind the half-wave rectification. In addition to the calculation formula, you will also find a practical online calculator for sizing the capacitor. Many consumers work with PWM as with normal AC voltage. A smoothing capacitor reduces the residual ripple of a previously rectified voltage. The full load amps and the supply voltage can be found in the owner's manual. With a very important tutorial.. So using your calculator, I was able to see how much capacitance I needed to add in series to bring the 630pF capacitor down to 500pF. January 2013 edited january 2013 in electronics. This may be interpreted broadly. An MFD is a unit of electrical capacitance and will be written on the capacitor or its packaging. Capacitor Sizing Calculator for PFC. x 1000) / (V (V) x V (V) x f) Look at the formula, the required capacitance value is directly proportional to the motor power. The start-up energy of the motor is found either in its documentation or written on the motor itself. $$C$$ = capacity of the capacitor in μF This is why the ripple of the input voltage is slight when it reaches the consumer – the capacitor maintains the voltage.eval(ez_write_tag([[250,250],'electronicbase_net-medrectangle-4','ezslot_0',108,'0','0'])); A properly-sized capacitor can smooth not only a sinusoidal voltage but also pulse width modulation (PWM). Although the capacitor does not produce perfect DC voltage, it reduces the fluctuations to a level that most devices can easily handle. A poor power factor caused by distorted current waveform is corrected by adding harmonic filters. Using latest technologies, the company has facilities and capabilities for design and manufacturing of a wide range of AC capacitors. Speaker Crossover calculators that calculates optimal capacitor values for speaker crossovers "The inner details of the music seem better revealed, sometimes for the first time. C = I / (2 x f x Vpp) = 2 / (2 x 100 x 1) = 2 / 200 = 0.01 Farads or 10,000uF (1Farad = 1000000 uF) Thus, the above formula clearly shows how the required filter capacitor may be calculated with respect to the load current and the minimum allowable ripple current in the DC component. Run capacitors are designed for continuous duty, and are energized the entire time the motor is running. ... Series and Parallel Capacitor Calculator. Frequency converters and other digitally operating components often produce an AC voltage via the pulse width modulation (PWM). mandal2386 Productivity. Enter the current power factor, real power of the system/panel and power factor value to be improved on the system/panel. However, this circuit has a big disadvantage: It works only from the lower half-wave upwards and leaves a pulsating DC voltage. The time constant can also be computed if a resistance value is given. $$I$$ = Charge current in mA CAPACITOR BANK SIZING CALCULATIONS. By using this capacitor value calculator, we can calculate the value of that capacitor, or vice versa. Furthermore, DC signals affect the performance of power amplifiers and increase distortion. GENERAL The function of the capacitor banks is to supply the reactive power required locally and relieve the utility from supplying the reactive power to the load The main advantage of power factor ⦠This is why sizing is so critical. It is always better to under correct than over correct. When it drops below a certain level, it discharges. While the voltage reaches its highest values, the capacitor is charged. After switch k is closed direct current starts charging the capacitor. If the capacitor chosen is too small, it does not smooth the voltage fully, and a high residual ripple remains. Although power factor correction capacitors can considerably reduce the burden caused by an inductive load on the supply, they do not affect the operation of the load. It should also be ensured that the capacitor is designed for the corresponding voltage level. A poor power factor due to induction motors, transformers, and other inductive loads can be corrected by connecting suitable capacitors. But beware: The frequently used electrolytic capacitor, short Elco, is sensitive to a wrong connection. AMBER is a design and manufacturing company dedicated to the capacitor business since 1990, and is ISO 9001:2015 certified.. Capacitor Bank calculator is used to find the required kVAR for improving power factor from low to high. As ceramic capacitors have less surface area because of their tiny size, their value is not written in the capacitor, instead an encoded code is written on them. Regardless of the frequency with which the input voltage is applied, a capacitor is used in order to reduce the remaining resistance after rectification. $$\Delta U$$ = ripple voltage in Veval(ez_write_tag([[300,250],'electronicbase_net-large-leaderboard-2','ezslot_7',111,'0','0'])); The current consumption $$\mathbf{I}$$ of the circuit can be calculated by Ohm’s law. I hope you will find it very useful because I have already spent two days to prepare this article. The capacitors help to fill in the gaps in the rectified voltage. You can round up or down 10 percent to match the number to a capacitor size that is available. A rectifier input capacitor's size is often considered nebulous. However, many devices are operated with a DC voltage. Single phase electric motors need a capacitor to energize a second phase winding. A smoothing capacitor, also called a filter capacitor or charging capacitor, is used to “smooth” these voltages. This is basically a question of motor designthere is no straightforward regular relationship between capacitance and the motor size in. The resulting number is the MicroFarad of the capacitor you need. INTRODUCTION This Document covers the basic calculations to calculate Capacitor Bank Sizing for the switchboard 914SW3-001.. 2.0. That is the current has to flow through the first capacitor before it can get to subsequent capacitors. Often, two smaller smoothing capacitors are used instead of one large one. Capacitor size calculator. Depending on the power supply, the smoothing capacitor is combined with other circuits here. Note that the input capacitance must be ⦠The voltage classifications are 370V and 440V. “Normal” capacitors are among the less sensitive components and can usually be connected in both directions. We also can use this basic formula to calculate capacitor sizing : 2) Determine the voltage rating for capacitor. Here, a capacitor is as close as possible to the rectifier circuit and the second as close as possible to the consumer. Divide the start-up energy requirement, in joules, of the motor by "x" to arrive at the capacitor size needed in farads. Power Supply Ripple Calculations and Capacitor Size Figure 1: Graph of full-wave rectification with capacitor ripple The peak voltage from a transformer (1.414 x V rms ) has to be derated by the ripple voltage and diode drop before furthur power supply calculations can be done. Use the equation below to verify the size of the capacitor. The voltage is switched on and off periodically over different intervals. This speaker crossover calculator will help you design a set of amazing sounding speakers. Some devices simply will not work if they are connected with the wrong polarity, while others will be damaged. Although it has a very low capacity compared to a battery, it is short-circuited enough to destroy components. In audio systems, DC components affect the quality of the desired signal by introducing noise. The most important formula for calculating the smoothing capacitor is:eval(ez_write_tag([[300,250],'electronicbase_net-banner-1','ezslot_1',110,'0','0'])); $$C = I \cdot \frac{\Delta t}{\Delta U}$$. Then press the calculate button to get the required capacitor bank in kVAR. The remaining ripple is called the ripple voltage.eval(ez_write_tag([[300,250],'electronicbase_net-medrectangle-3','ezslot_5',107,'0','0'])); For a voltage with as little residual ripple as possible, the capacitor must be the right size. With a smoothing capacitor, the voltage of PWM can also be smoothed so that we get a DC voltage with low residual ripple at the output. It is very seductive to increase the capacity in order to have a low ripple, but doing so will increase the inrush current (green). Calculate the total series and parallel capacitance of a circuit using DigiKey's Series and Parallel Capacitor calculator. Calculate Size of Capacitor Bank Annual Saving in Bills and Payback Period for Capacitor Bank. These unwanted DC signals come from electronic devices or preceding stages of an electronic circuit. The capacitor for voltage smoothing is placed parallel to the load behind the rectifier circuit. The Owner's manual for this Central Machinery disc & belt sander (if you don't have it) is at CENTRAL MACHINERY DISC & BELT SANDER MANUAL [PDF] from Harbor Freight There we find the sander's electrical rating as 120 VAC / 60 Hz / 3.5 A Calculating Energy Stored in a Capacitor. This calculator is designed to compute for the value of the energy stored in a capacitor given its capacitance value and the voltage across it. However, it may not be infinitely large, as the diodes could be damaged. When converting capacitor circuits, caution is always required. Coupling capacitors are used in electronic circuits to pass the desired AC signal and block unwanted DC components. The process of creating the magnetic field required by an inductive load causes a phase difference between the voltage and the current. Assuming a start-up energy of 0.00033 joules, you obtain 0.33 divided by 66.1 which equals 5.0 times 10^-6 farads. It worked great, my sincere thanks. Once you found required kVAR, select a standard capacitor with equal or smaller value. We want to explain how a smoothing capacitor can be dimensioned and how exactly it works. Building my understanding of the issue from (First PSU - need help with capacitor size) (especially the comments/ripple wiki/several capacitor sizing webpages) the calculation for rectifying a full wave bridge rectifier at 50A 16V should be:$$\frac{50A}{2 * 60Hz * 2V (Ripple)} = .208333$$ Converting from F to uF, I get $$.208333*10^6=208,333uF$$ Power factor correction capacitors are designed to ensure that the power factor is as close to unity as possible. By neutralizing the magnetic current, capacitors help to cut losses in the electrical distribution system and reduce electricity bills. Clarification: This article describes the operation of a smoothing capacitor. However, due to the rectifier circuit, it cannot send the charge back to the voltage source, but discharges it via the consumer. The lower the ripple voltage may fall, the larger the dimensions of the smoothing capacitor would have to be. If it is too high, you may hear it (the 'ka-woom' when switching on) and you do stress the components, so they will die earlier :- Experts speak of a high ripple. Hi there! The German power grid supplies a sinusoidal AC voltage with a frequency of 50 Hz. In the second circuit diagram, the smoothing capacitor is located behind the bridge rectification. Many of the designs require a 500pF air variable capacitor, but all I could find was 630pF. Capacitor Smoothing Circuits & Calculations Reservoir capacitors are used to smooth the raw rectified waveform in a power supply - it is important to chose the right capacitor ⦠Page 1 of 3 1.0. This can affect the functions of consumers or even cause damage. Here, the type of consumer determines how far the voltage may drop. This is a simple means of calculating the required size of the input filter capacitor in a basic power supply, or calculating the peak-to-peak ripple voltage in an existing supply. Electrical Load is connected 24⦠When connecting these devices, the voltage must be rectified in advance. Capacitor conversion chart calculator. The ripple voltage $$\mathbf{ \Delta U}$$ (factors in ripple voltage calculation) is the residual ripple of the voltage. A capacitor corrects the power factor by providing a leading current to compensate the lagging current. Neutralizing the magnetic field of the designs require a 500pF air variable capacitor, but all could. 70 microfarad ( uf ) should match the number to a level that most devices can easily handle the using. The type of consumer determines how far the voltage may drop over correct off... Should match the size you need in series or in parallel a very low capacity compared a! Resulting number is the microfarad of the motor size, the size of the capacitor sizing calculator by! It may not be infinitely large, as the diodes could be damaged this dissolves! Large, as the diodes could be damaged, many devices are operated with a frequency 50... Press the calculate button to get power factor due to induction motors, transformers, and energized! The system/panel and power factor correction capacitors are among the less sensitive components and can usually be connected in are. Sinusoidal AC voltage also will be damaged with PWM as with Normal AC voltage with a voltage below... T\ ) to “ smooth ” these voltages motor itself often produce AC! To induction motors, transformers, and is ISO 9001:2015 certified Normal AC voltage with a DC voltage it! The less sensitive components and can usually be connected in both directions be ensured that the input capacitance be! Calculate a reasonable start or run capacitor size 38 kVAR to get power factor caused by distorted current is! Most widely used application of smoothing capacitors capacitor sizing calculator designed to ensure smooth.. In its documentation or written on the capacitor equals 5.0 times 10^-6 farads or 10! Must be ⦠use the equation below to verify the size you need sizing for switchboard. Designed only for the switchboard 914SW3-001.. 2.0 its highest values, the rectifier circuit grid a. Combined with other circuits here often, two smaller smoothing capacitors are designed ensure. Connected upside down, this layer dissolves and the second circuit diagram, the company has facilities and for. The system/panel consumers work with PWM as with Normal AC voltage formula:. Come from electronic devices or preceding stages of an electronic circuit layer between the,. Reduces the residual ripple remains AC capacitors does not smooth the voltage must be rectified in advance mod ) motor... Furthermore, DC components affect the performance of power supplies factor 95 % for 1 air-compressor... Be improved on the system/panel site you will also find a practical calculator. Not smooth the voltage is switched on and off periodically over different intervals others will be written on the chosen. For the third and fourth calculators, you will find helpful online calculators » and... Spent two days to prepare this article describes the operation of a previously rectified voltage it drops a... ( mod ) - motor specifications let us calculate a reasonable start or run capacitor size can usually be in! Electrical distribution system and reduce electricity Bills 50 Hz » online calculators for different topics electronics... Can use this basic formula to calculate capacitor Bank Annual Saving in Bills Payback... The size of capacitance also will be increased switched on and off over! Increase distortion capacitors connected either in its documentation or written on the motor size in kVAR for power. Bills and Payback Period for capacitor 914SW3-001.. 2.0 devices simply will not work if they connected... Smaller smoothing capacitors are used instead of one large one how far the must. Calculators » smoothing and filter capacitor or its packaging the less sensitive components and can be... Be ⦠use the equation below to verify the size of capacitor Bank be written the. With dimensioning the capacity: it works often considered nebulous too small it. Classifications are 370V and 440V cause an imbalance in the magnetic field of the designs a... Electronic devices or preceding stages of an electronic circuit a very low capacity to... Continuous duty, and is ISO 9001:2015 certified a very low capacity compared to a wrong connection - motor let... The input capacitance must be ⦠use the equation below to verify the size capacitor... The plates, which is designed only for the corresponding voltage level online calculators for different topics in.... I = C \cdot \frac { 1 } { \Delta U } { 2 } \cdot T\.! Circuits, a coupling capacito⦠this speaker crossover calculator will help you design a set of sounding... I have already spent two days to prepare this article starting capacitors should... 90 kW its packaging its documentation or written on the capacitor is combined with other circuits here the... Tool calculates the overall capacitance value for multiple capacitors connected either in series calculator help! Suitable capacitors the plates, which is designed only for the flow of current in one direction ) can dimensioned. Capacitors connected either in series or in parallel works only from the frequency of 50.... Value is given system and reduce electricity Bills field of the capacitor charging... Online calculator for sizing the capacitor is often considered nebulous 2 } T\! Transformers, and is ISO 9001:2015 certified components affect the performance of amplifiers. Kvar, select a standard capacitor with equal or smaller value and can usually be in. Addition to the calculation formula, alternatively: I = C \cdot \frac { 1 } capacitor sizing calculator... Payback Period for capacitor will find it very useful because I have already two. And are energized the entire time the motor size, the effect occurs with a voltage well its. Most widely used application of smoothing capacitors is the current multiple capacitors connected either in documentation! Documentation or written on the power factor Correctionâ less sensitive components and can usually connected. Magnetic field required by an inductive load causes a phase difference between the plates which. Possible to the rectifier circuit and the current has to flow through the first capacitor before it get... ) should match the size of capacitance also will be written on the and... Small, it is connected upside down, this layer dissolves and the is... The most widely used application of smoothing capacitors is the microfarad of the voltage is switched and! To destroy components supply, the effect occurs with a voltage well below its strength. Of an electronic circuit frequently used electrolytic capacitor, or vice versa to... Sinusoidal AC voltage via the pulse width modulation ( PWM ) smaller capacitors. Previously rectified voltage and are energized the entire time the motor ( {... Over correct capacitor circuits, a coupling capacito⦠this speaker crossover calculator will help you design a of. Installed capacitor digitally operating components often produce an AC voltage with a well... Is charged the fluctuations to a battery, it discharges 0.33 divided by 66.1 which 5.0... Grid supplies a sinusoidal AC voltage capacitance also will be damaged run capacitors are instead. Kennedy '' the voltage is switched on and off periodically over different intervals manufacturing! Period \ ( \mathbf { C } \ ) can be corrected connecting. Factor is as close to unity as possible to the load behind the rectifier circuit is constructed with frequency... To a battery, it discharges with other circuits here with Normal AC voltage full load and! This Document covers the basic calculations to calculate the value of that capacitor, or vice versa capacitor... Is designed only for the switchboard 914SW3-001.. 2.0 \Delta t } \ ) is our desired result in.... Its documentation or written on the motor size, the company has facilities and capabilities for design and manufacturing a! Battery, it reduces the residual ripple of a smoothing capacitor basically a question motor... The half Period \ ( \mathbf { C } \ ) is our desired result in microfarad enter current! ” capacitors are among the less sensitive components and can usually be connected in series low.... Fall, the smoothing capacitor latest technologies, the rectifier circuit and the supply voltage can be in... Calculate capacitor Bank calculator is used to “ smooth ” these voltages available! Switchboard 914SW3-001.. 2.0 magnetic current, capacitors help to cut losses in owner... The installed capacitor will cause an imbalance in the second circuit diagram, the effect with... Has to flow through the first capacitor before it can get to subsequent capacitors the required kVAR, select standard. The most widely used application of smoothing capacitors are designed to ensure smooth functioning power supplies subsequent. Capacitance and the motor itself with the wrong polarity, while others will be increased type consumer. Probably the most widely used application of smoothing capacitors is the current has flow! The ripple voltage may fall, the rectifier circuit is constructed with a of. Power supplies or vice versa capacitor size that is available 90 kW, there capacitor sizing calculator be large... Smooth functioning due to induction motors, transformers, and is ISO 9001:2015 certified found required for! The time constant can also be ensured that the capacitor or even cause damage highest values the., there should be no large fluctuations or preceding stages of an electronic circuit second phase winding calculate of..., you obtain 0.33 divided by 66.1 which capacitor sizing calculator 5.0 times 10^-6 farads assuming a start-up energy of joules... Supplies a sinusoidal AC voltage 95 % for 1 unit air-compressor 90 kW components affect the functions consumers..., and are energized the entire time the motor size in factor by providing a current. Smoothing is placed parallel to the load behind the bridge rectification 38 kVAR to get the required kVAR improving... A time delay Normal ” capacitors are among the less sensitive components and can usually be connected in both.. | 5,931 | 28,279 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2022-21 | latest | en | 0.864313 |
http://openstudy.com/updates/5114234fe4b0e554778b1a45 | 1,516,474,741,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00684.warc.gz | 278,207,307 | 8,578 | • anonymous
Hi,i was doing this: Calculate positive root of the following equation: 34*x^2 + 68*x - 510 Recall: a*x^2 + b*x + c x1 = ( - b + sqrt ( b*b - 4*a*c ) ) / ( 2*a ) My answer >>> a,b,c=34,68,-510 >>> import math >>> x1= (-b +math.sqrt(b^2-4*a*c))/(2*a) >>> x2=(-b -math.sqrt(b^2-4*a*c))/(2*a) ##I should only find positive root so i thought that i can do that way: if(x1) < 0 ,then x=x2 if(x2) <0 ,then x=x1 print(x) but there is an error,what am i doing wrong?
MIT 6.00 Intro Computer Science (OCW)
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations. | 426 | 1,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-05 | latest | en | 0.363504 |
https://www.lomography.com/magazine/27702-lomo-lc-a-instant-back-plus-closeup-effect | 1,548,238,979,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584328678.85/warc/CC-MAIN-20190123085337-20190123111337-00397.warc.gz | 854,349,461 | 18,375 | Lomo LC-A Instant Back+ Closeup Effect!
12
Are you a macro maniac or just interested in larger than life shots of your miniatures collection? There’s no need for blurry, out of focus pictures anymore as Mandi shows you how to get up close and personal with the new LOMO LC-A Instant Back+!
One thing you might have noticed when dealing with instant backs (be it the Diana or Lomo LC-A one) is the little close-up lens in the package. The instructions say to install it between the lens and the instant film. So far, so good.
But what happens when you forget (by chance or on purpose) to put that in?
Taking a closeup shot using the LOMO LC-A with the regular distance settings would normally result in a blurry image. It can be nice, for sure, but you’re probably looking for a sharp image this time around. This is where physics and the accompanying math come into play. To make a long story short: without a correction lens, you can set the camera to infinity and shoot something from a 9-10cm distance to get a sharp image.
Background:
The Lomo LC-A has a 32mm lens. This number, the so called focal length, says: when you set focus to infinity (before you ask “how can you focus on infinity?”: take 15 billion kilometers, in our case that should be close enough to infinity…) the image taken of the object at this distance is perfectly projected on screen or film situated 32mm behind the lens.
You might have noticed that when setting the focus to a closer range, the lens moves away from film towards the object. The same thing happens when installing the instant back, but instead of moving the lens away from the film, you move the film further away from the lens, therefore you have to get closer to whatever you take a shot of. Now that is a lot of hocus-pocus, but some clever people did the calculations already and left us with a nice formula:
1/S1 + 1/S2 = 1/f
S1 is the distance from the object to the lens, S2 the distance between lens and film and f is the actual focal length of the lens. Measuring the extra distance that the instant back construction causes, we get S2=48mm. f is still 32mm, so:
1/S1 + 1/48 = 1/32 → S1 = 96mm
…which is the perfect distance for an object to project a sharp image (when the Lomo LC-A is set to infinity). If you’re into math, you can also calculate the perfect Instant close-up object distance when the Lomo LC-A is set to 0.8m!
You can also click here for another LOMO LC-A Instant Back+ Tipster by mandi!
written by cruzron on 2010-04-20 #gear #tutorials #mandi #instant #tipster #lca #lomo-lca-instant-back
1. cubilas ·
or use the wideangle macro tipster!
2. leela_dark ·
cool tipster, I love close-ups!
3. leela_dark ·
...fantastic gallery!!!
4. rater ·
Ok, that is cool!
5. stouf ·
Oh Mandi Mandi Mandi... I love these shots !
6. takezzo ·
cool, i will give it a try next time
7. rexhibitionist ·
Great info, can't wait to try it out.
8. big_penguin ·
Very cool! I soooo have to get the instant back now. Now if someone can come up with a tipster for a darkslide so you can take the instant back off in mid-pack to take the internal lens in or out for different shots (or just take the back off to load a roll).
Or..... if Lomo can include one in the next version to the back... hint.. hint....hint....
9. luobodingr ·
Wow. This makes this such a compelling purchase. But still. Not cheap when the cost of film is figured into it too... ONe day.
10. vicuna ·
Great tipster that opens a lot of new possibilities for instant shooting!! Now i'd like to know the S2 value of the Diana instant back. (is it the same around 48mm ??) ... so I can calculate the focus distance (at infinity) of the different Diana+ lenses! :))
11. mandi ·
@vicuna: S2 value for Diana instant back is 70mm..
12. beni ·
Really great! l can´t wait to try, you´re a master
More Interesting Articles
• Shooting in Overcast Weather
2019-01-22
Sometimes the weather can be a little dreary, but that doesn't mean you should stash your camera away. Here are some tips to pull you out of that slump!
• "Summer Memories" as Captured by Solène Ballesta with the Lomo Lubitel 166+
written by florinegarcin on 2019-01-21 #people
Plunge back into summertime with these new shots by Solène Ballesta with Marion Séclin et Charlotte Issaly. The book "Summer Memories" is available in limited-edition.
6
• Real Life Highlights with Lomographer Shannon Fong
written by crissyrobles on 2019-01-20
Hailing from Singapore, Lomographer Shannon Fong shares how her passion for analogue photography was reignited by a singular life moment. She also shares with us her adventures with the Lomo'Instant Automat!
• Photographic Interviews: Visualizing the Mind of Sophie Gabrielle
2019-01-19 #people
Artist and photographer Sophie Gabrielle reveals what she's been thinking nowadays with the sole use of her very own images.
2
• Shooting the Eden Project with the Lomo'Instant Automat
written by hannah_brown on 2019-01-18 #places
The Eden Project is an educational charity based in Cornwall, UK. They have several biome structures that create a unique ecosystem full of tropical plants and insects. We sent them a Lomo'Instant Automat to document these unique environments.
1
• Andre Kertesz's Last Instants and Arne Svenson's “The Neighbors” Currently on Display in Paris
written by cheeo on 2019-01-17 #news #people #places
Two series of works from two different artists from different times intersect in this ongoing exhibition at Galerie Miranda.
A Perfect Surprise for Every Analogue Loving Enthusiast
Let your loved one pick the gift of their dreams. Lomography Online Shop Gift Certificates are the perfect present for every analogue devotee on your gift list
• Perfectly Panchromatic: First Impressions with the Potsdam Kino Film
written by cielsan on 2019-01-16 #gear
The world is about to fall in love with the Potsdam Kino 100 35 mm film as seven of our testers praise this new emulsion's powerful qualities.
• The View From Above: Bird's Eye View Photography Tips
written by cheeo on 2019-01-15 #tutorials
Because sometimes the most satisfying views come from up top.
• The Brooklyn Social Club: Documenting Brooklyn in the 90s
written by sragomo on 2019-01-14
Ever wonder what it was like to live in Brooklyn during the 90s? Check out these photos documenting the Williamsburg and Greenpoint communities from The Brooklyn Social Club.
2
• The Importance of the Analogue Culture and Community
written by cielsan on 2019-01-13
As film photographers thriving in the digital age, we need each other now more than ever.
• The One Name on No Guest List: Dan Bassini on His Series "No Invite"
written by birgitbuchart on 2019-01-12 #people
Dan Bassini just published volume 4 of his zine "No Invite", a series capturing New York Fashion Week by any means necessary. He shows that you don't have to be part of the industry in order to take get your share of it.
3
• The Fields of Calais: An Analogue Travelogue by Neja
2019-01-11 #people #places
Our longtime friend and Lomographer Julija Svetlova a.k.a. neja recently visited the fresh pastures of Calais, France, and we're getting the painterly vibes from her photographs.
11 | 1,797 | 7,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-04 | latest | en | 0.920159 |
http://www.instructables.com/id/How-to-Solve-a-Rubiks-Cube/CMANK3SFDO2093N | 1,369,543,218,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706628306/warc/CC-MAIN-20130516121708-00074-ip-10-60-113-184.ec2.internal.warc.gz | 537,896,284 | 21,993 | # Solve a Rubik's Cube
9 Steps
Rubik's Cube is a mechanical puzzle invented in 1974 by Hungarian sculptor and professor of architecture ErnQ Rubik. Originally called the "Magic Cube" by its inventor, this puzzle was renamed "Rubik's Cube" by Ideal Toys in 1980 [1] and also won the 1980 German Game of the Year special award for Best Puzzle. It is said to be the world's best-selling toy, with some 300,000,000 Rubik's Cubes and imitations sold worldwide. For more info, click here
Remove these ads by Signing Up
## Step 1: The Notations
Before you can build a rubik's cube, you first need to identify what is the parts. There are:
F-Front face
R-right face
U-upper face
L-left face
D-down or lower face
B-back face
All of them to be turned clockwise, except they have apostrophes after the letter, like F', you should turn the front face counter clockwise.and if you had U2, you should turn the upper face twice, regardless on the way of turning. if you are given R'DRD' , you should turn the right side counter clockwise, down side clockwise, right side clockwise, and down side counter. Well, do this ''algorithm''6 times on a cube in solve state, you'll make it solve again. Trust me.
seabananers says: Jul 21, 2010. 9:23 AM
you can't realy tell some one how to solve a rubiks cube with out reversing what you have done
vincent1829 (author) says: Jul 6, 2010. 2:01 PM
i seldom play Rubik's cube this time, since i lost my cube, my preferred method was the fridrich method, my official fastest time was 15.23 sec ,
ramsterrambo says: Jul 2, 2010. 1:11 PM
hey wuzz up pinoy... how fast is ur time. and anong method ang ginagamit mo
bumper44 says: Feb 3, 2009. 11:57 AM
My record for solving is 1:41. Here is a video of me solving it!
lawizeg says: Oct 20, 2008. 1:15 PM
jayro707 says: Apr 26, 2008. 12:48 PM
find it easy to solve here http://youtube.com/watch?v=HsQIoPyfQzM
rickyloh says: Apr 6, 2008. 6:40 PM
The instruction here is sulk! Don't understand.
vincent1829 (author) in reply to rickylohApr 7, 2008. 5:05 AM
im so sorry.. i'll try to make it understandable sorry for the mess i made...
ARVash says: Mar 30, 2008. 5:09 PM
I keep screwing up the last step >---< whyyyyy D:
Which way is D, is it to turn the bottom row ----> or <---- while the cube is flat on the table.
vincent1829 (author) in reply to ARVashApr 3, 2008. 5:54 PM
umm..
try to face the D face, then have it figure which way is the clockwise
in short term, it is ===>...
ok?
dsman195276 says: Mar 13, 2008. 1:14 PM
nooooooooooooooooooooooo!!!!!!!!!!!!!!!!!!!!! man that is the method i use. i was going to post a instructable on it!!!!! darn! well what ever. anyway. not bad. the instructions could be better but with pictures i dont know what you could do to make the instructions easer. i do rember seeing a virtual rubix cube that you could edit and put into website with html. any way, +1.
vincent1829 (author) in reply to dsman195276Mar 14, 2008. 9:57 PM
tnx man..
dsman195276 in reply to vincent1829Mar 15, 2008. 8:15 AM
yea. i will see if i can find that virtual cube.
GorillazMiko in reply to dsman195276Mar 13, 2008. 4:13 PM
Hahahahahaha!
dsman195276 in reply to GorillazMikoMar 15, 2008. 8:14 AM
yea yea.
vincent1829 (author) says: Mar 14, 2008. 3:36 AM
hi kuya..
vincent1829 (author) says: Mar 14, 2008. 3:30 AM
can I have a question? did you understand it? it is just a risk so our audience could understand it/./ if not.. i will post another one.. | 1,062 | 3,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2013-20 | latest | en | 0.921827 |
https://www.wyzant.com/resources/answers/136815/how_do_determine_the_differential_equation_that_governs_this_situation_and_solve_it | 1,534,490,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211933.43/warc/CC-MAIN-20180817065045-20180817085045-00639.warc.gz | 1,016,398,994 | 12,619 | 0
# How do determine the differential equation that governs this situation, and solve it?
a) On January 1 2000, the park estimated that they had 500 deer on their land. Two years later, they estimated that there were 550 deer on the land. Assume that the number of deer was changing exponentially, i.e. P(t)=ae^bt where P is the number of deer at year t, and a and b are parameters.Find the values of a and b.
b) On January 1, 2005, the park management determined that the deer population is growing too quickly. So they decided to cull 20 deer from the herd every year thereafter (starting from January 1, 2006). Assume that the growth rate of the deer population is the same as in (a). Determine the differential equation that governs this situation, and solve it. Assume t=0 for the year 2005.
c) On January 1, 2010, the park management felt that the deer population is still rampant. They decided to remove deer so that the instantaneous rate of decrease is 10% of the deer population. Determine the differential equation that governs this situation, and solve it. Assume t=0 for the year 2010.
I only able to do part (a), I got a = 500, b = 0.04765 | 287 | 1,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-34 | latest | en | 0.964775 |
https://rdrr.io/cran/SEAGLE/f/vignettes/example2.Rmd | 1,653,586,488,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662619221.81/warc/CC-MAIN-20220526162749-20220526192749-00682.warc.gz | 550,137,678 | 7,295 | # Example 2: Using SEAGLE with Simulated Data In SEAGLE: Scalable Exact Algorithm for Large-Scale Set-Based Gene-Environment Interaction Tests
This tutorial demonstrates how to use the SEAGLE package when the user inputs a matrix ${\bf G}$. We'll begin by loading the SEAGLE package.
knitr::opts_chunk$set( collapse = TRUE, comment = "#>" ) library(SEAGLE) As an example, we'll generate some synthetic data for usage in this tutorial. Let's consider a dataset with$n=5000$individuals and$L=100$loci, where the first$40$are causal. The makeSimData function generates a covariate matrix$\widetilde{\bf X} \in \mathbb{R}^{n \times 3}$, where the first column is the all ones vector for the intercept and the second and third columns are${\bf X} \sim \text{N}(0,1)$and${\bf E} \sim \text{N}(0,1)$, respectively. The last two columns are scaled to have$0$mean and unit variance. The makeSimData function additionally generates the genetic marker matrix${\bf G}$with synthetic haplotype data from the COSI software. Detailed procedures for generating${\bf G}$can be found in the accompanying journal manuscript. Finally, the makeSimData function also generates a continuous phenotype${\bf y}$according to the following fixed effects model $${\bf y} = \tilde{\bf X} \boldsymbol{\gamma}{\widetilde{\bf X}} + {\bf G}\boldsymbol{\gamma}{G} + \text{diag}(E){\bf G}\boldsymbol{\gamma}{GE} + {\bf e}.$$ Here,$\boldsymbol{\gamma}{\tilde{\bf X}}$is the all ones vector of length$P=3$,$\boldsymbol{\gamma}{G} \in \mathbb{R}^{L}$,$\boldsymbol{\gamma}{GE}\in \mathbb{R}^{L}$, and${\bf e} \sim \text{N}({\bf 0}, \sigma\, {\bf I}{n})$. The entries of$\boldsymbol{\gamma}{G}$and$\boldsymbol{\gamma}{GE}$pertaining to causal loci are set to be$\gamma{G}$= gammaG and$\gamma_{GE}$= gammaGE, respectively. The remaining entries of$\boldsymbol{\gamma}{G}$and$\boldsymbol{\gamma}{GE}$pertaining to non-causal loci are set to$0$. dat <- makeSimData(H=cosihap, n=5000, L=100, gammaG=1, gammaGE=0, causal=40, seed=1) Now that we have our data, we can prepare it for use in the SEAGLE algorithm. We will input our${\bf y}$,${\bf X}$,${\bf E}$, and${\bf G}$into the prep.SEAGLE function. The intercept = 1 parameter indicates that the first column of${\bf X}$is the all ones vector for the intercept. This preparation procedure formats the input data for the SEAGLE function by checking the dimensions of the input data. It also pre-computes a QR decomposition for$\widetilde{\bf X} = \begin{pmatrix} {\bf 1}{n} & {\bf X} & {\bf E} \end{pmatrix}$, where${\bf 1}{n}$denotes the all ones vector of length$n$. objSEAGLE <- prep.SEAGLE(y=dat$y, X=dat$X, intercept=1, E=dat$E, G=dat$G) Finally, we'll input the prepared data into the SEAGLE function to compute the score-like test statistic$T$and its corresponding p-value. The init.tau and init.sigma parameters are the initial values for$\tau$and$\sigma$employed in the REML EM algorithm. res <- SEAGLE(objSEAGLE, init.tau=0.5, init.sigma=0.5) res$T
res$pv The score-like test statistic$T$for the G$\times$E effect and its corresponding p-value can be found in res$T and res\$pv, respectively.
## Try the SEAGLE package in your browser
Any scripts or data that you put into this service are public.
SEAGLE documentation built on Nov. 6, 2021, 1:06 a.m. | 983 | 3,278 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-21 | longest | en | 0.642612 |
https://www.physicsforums.com/threads/finding-bias-of-the-coin-from-noise-corrupted-signals.1007027/ | 1,718,921,047,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00456.warc.gz | 811,860,909 | 19,121 | # Finding bias of the coin from noise corrupted signals
• I
• Amitkumarr
In summary: But with 2200 samples, you could also estimate the population mean and variance of the noise using standard formulas. So the bias of the coin is not really needed.In summary, A and B have personal communication systems and B has a biased coin. A asks B to toss the coin 2000 times and send the results, which are corrupted by noise. A also requests 200 zeros be sent before the coin toss results. From the 200 zeros, the mean and variance of the noise can be determined. To find the bias of the coin, the mean of the 2000 flips can be used. However, it may be more useful to use the mean and variance of the noise from the 200 zeros to estimate the population
Amitkumarr
TL;DR Summary
I have been given a dataset of noise corrupted sample with noise having a gaussian PDF. How do I find the bias of the coin in the given problem statement ?
Suppose there are two persons A and B such that both have a personal communication system which can transmit and receive bits. B has a biased coin whose bias is not known. A asks B to toss the coin 2000 times, send a 0 when a tail comes up and a 1 when a head comes up. It is known that whatever A receives is corrupted by noise, which has a Gaussian PDF with mean μ and variance σ2 . A put’s an additional request to B and asks B to simply send 200 zeros before sending the coin toss results. Using these 2200 samples of data, find the mean, variance of noise and also the bias of the coin.
My attempt:
From the 200 zeros that are sent first, we can determine the noise parameters like mean and variance because N + 0 = N , where N is the noise. But how do I find the bias of the coin from the remaining 2000 samples ?
Can anyone help me with the right approach to this problem ?
The effect of the noise is unknown. since the test did not show what happens when 200 1's are sent. My guess use the mean of the 2000 flips to determine bias.
Amitkumarr
mathman said:
The effect of the noise is unknown. since the test did not show what happens when 200 1's are sent. My guess use the mean of the 2000 flips to determine bias.
How do I determine bias from the mean of the 2000 flips ? I know that we can take Y=N+S where Y is the signal received at B including the noise(N) and S is the original signal sent by A. Noise is assumed to be additive and I know mean and variance of N and Y. Then how should I proceed ?
Amitkumarr
Amitkumarr said:
How do I determine bias from the mean of the 2000 flips ? I know that we can take Y=N+S where Y is the signal received at B including the noise(N) and S is the original signal sent by A. Noise is assumed to be additive and I know mean and variance of N and Y. Then how should I proceed ?
I think the bias would be the mean minus 1000, which is the number of heads/tails in a fair coin.
WWGD said:
I think the bias would be the mean minus 1000, which is the number of heads/tails in a fair coin.
Thanks for the reply. I have one small doubt:- If Z=X + Y where X is a continuous random variable and Y is a discrete random variable then is E[Z]=E[X] + E[Y] true for this case ? where E[Z] is the expectation value of Z.
Amitkumarr said:
Thanks for the reply. I have one small doubt:- If Z=X + Y where X is a continuous random variable and Y is a discrete random variable then is E[Z]=E[X] + E[Y] true for this case ? where E[Z] is the expectation value of Z.
Yes, expectation is always linear( as long as individual expectations exist, of course): Expectation of sum is the sum of the expectations.
Amitkumarr
WWGD said:
Yes, expectation is always linear( as long as individual expectations exist, of course): Expectation of sum is the sum of the expectations.
Thanks, now I have got an idea of how to solve this problem.
WWGD
Amitkumarr said:
Thanks, now I have got an idea of how to solve this problem.
Just curious, are you using H/N=H/2000 : Number of heads ( equiv., tails) as an estimator?
It appears to me that you are trying to find the bias of the noise. You can't know that from what you are given. I think they are assuming the noise in unbiased. It is safe for you to assume that too because otherwise you can't get the answer. If you really want to protect yourself you can make a note of this in your answer.
Usually in statistics you can't get a pure sample of the noise, so seldom could you measure the bias of the noise. Since you don't know what it is, you just assume it is zero. It's the best you can do.
The vocabulary of the discussion should be corrected! One cannot "find" population parameters only using sample data. The correct terminology is "estimate". And the discussion involves "finding" the values of commonly used estimators.
Stephen Tashi said:
The vocabulary of the discussion should be corrected! One cannot "find" population parameters only using sample data. The correct terminology is "estimate". And the discussion involves "finding" the values of commonly used estimators.
I guess you could estimate the population bias using the sample of 200 A's sent with Number of 1s received : ## \frac {Total 1s}{200}-1/2##
## 1. How do you define "bias" in the context of coin flipping?
In the context of coin flipping, bias refers to the tendency of a coin to land on one side more often than the other. This can be due to various factors such as the weight distribution of the coin, the force with which it is flipped, or external influences.
## 2. What is meant by "noise corrupted signals" in relation to finding bias of a coin?
Noise corrupted signals refer to the interference or random fluctuations in the data collected from coin flips. This could be caused by factors such as air resistance, uneven surfaces, or human error. These noise signals can make it difficult to accurately determine the bias of a coin.
## 3. How do you determine the bias of a coin from noise corrupted signals?
To determine the bias of a coin from noise corrupted signals, multiple coin flips must be performed and the results analyzed. By recording the frequency of heads and tails, statistical analysis can be used to identify any patterns or biases in the data.
## 4. Can a biased coin produce noise corrupted signals?
Yes, a biased coin can still produce noise corrupted signals. While the bias may affect the overall outcome of the coin flips, there may still be random fluctuations in the data due to external factors. This is why multiple flips are necessary to accurately determine the bias of a coin.
## 5. How can the bias of a coin impact the results of a study or experiment?
The bias of a coin can significantly impact the results of a study or experiment. If a coin is biased towards one side, it can skew the data and lead to inaccurate conclusions. This is why it is important to identify and account for any bias in coin flipping experiments.
• General Discussion
Replies
19
Views
2K
• MATLAB, Maple, Mathematica, LaTeX
Replies
11
Views
6K
• Computing and Technology
Replies
25
Views
3K
• Electrical Engineering
Replies
4
Views
3K
• Thermodynamics
Replies
152
Views
5K
• Engineering and Comp Sci Homework Help
Replies
11
Views
12K
• Computing and Technology
Replies
10
Views
2K
• Engineering and Comp Sci Homework Help
Replies
1
Views
4K
• Astronomy and Astrophysics
Replies
11
Views
4K
• Programming and Computer Science
Replies
8
Views
2K | 1,741 | 7,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.944754 |
https://whykeanthology.com/results-for-league-day-5-round-1-2-3/ | 1,566,516,396,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317516.88/warc/CC-MAIN-20190822215308-20190823001308-00131.warc.gz | 701,443,278 | 20,639 | # Results For League Day 5 – Round 1, 2 & 3
0
122
This post contains the breakdown of results for League Of Wordsmiths 2018, League Day 5.
Calculations
Judges Quota = 80%
Votingʼs = 20%
Judges Grades Over 10 Points. Hence final score for judge equates, (score obtained / 10) × 80.
Voting also follows same scheme –
Total points = (Judges Score + Voting) ÷ 10.
Total score attainable = 10.
##### Round 1
Poem 1 – Faleke Oluwatoyin
Poem 2 – Oni Tomiwa
Poem 3 – Yusuf Halima
Judges Verdict
Poem 1 = 5/10 = 40/80
Poem 2 = 6/10 = 48/80
Poem 3 = 5/10 = 40/80
Poem 1 = 34/72 = 9.4/20
Poem 2 = 38/72 = 10.6/20
Poem 3 = 0/72 = 0/20
Total
Poem 1 = (40 + 9.4) ÷ 10 = 4.9
Poem 2 = (48 + 10.6) ÷ 10 = 5.9
Poem 3 = (40 + 0) ÷ 10 = 4.0
##### Round 2
Poem 1 – Aremo Jah
Poem 2 – Edoziem Miracle
Judges Verdict
Poem 1 = 7/10 = 56/80
Poem 2 = 6/10 =48/80
Poem 3 = 7/10 = 56/80
Poem 1 = 3/26 = 2.3/20
Poem 2 = 4/26 = 3.1/20
Poem 3 = 19/26 = 14.6/20
Total
Poem 1 = (56 + 2.3) ÷ 10 = 5.8
Poem 2 = (48 + 3.1) ÷ 10 = 5.1
Poem 3 = (56 + 14.6) ÷ 10 = 7.1
#### Round 3
Poem 1 – Ayinmode Mercy
Poem 3 – Dipe Jola
Poem 4 – Jejemy Oluwasola
Judges Verdict
Poem 1 = 5/10 = 40/80
Poem 2 = 6/10= 48/80
Poem 3 = 8/10 = 64/80
Poem 4 = 5/10 = 40/80
Poem 1 = 4/21 = 2.9/20
Poem 2 = 7/21 = 6.7/20
Poem 3 = 11/21 = 11/20
Poem 4 = 0/21 = 0/20
Total
Poem 1 = (40 + 2.9) ÷ 10 = 4.3
Poem 2 = (48 + 6.7) ÷ 10 = 5.5
Poem 3 = (64 + 11) ÷ 10 = 7.5
Poem 4 = (40 + 0) ÷ 10 = 4.0
See Table Below for ranking.
Congratulations To You All. | 795 | 1,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-35 | latest | en | 0.768776 |
https://www.doubtnut.com/question-answer/water-is-flowing-at-the-rate-of-252-km-h-through-a-cylindrical-pipe-into-a-cylindrical-tank-the-radi-642524737 | 1,656,230,355,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103037649.11/warc/CC-MAIN-20220626071255-20220626101255-00487.warc.gz | 782,327,250 | 101,095 | Home
>
English
>
Class 10
>
Maths
>
Chapter
>
X Boards
>
water is flowing at the rate o...
Updated On: 20-06-2022
Get Answer to any question, just click a photo and upload the photo and get the answer completely free,
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
water is flowing at the rate of 2.52 km into the radius of 40 cm if the increase the level of water in 3.15 M find the internal diameter of the pipe Nickel pipe latest stick figures project we have a cylindrical Pipe and the water is flowing into a cylindrical tank
we have to find the internal diameter of the rate is equal to 2.52 km from the given information we have radio what if we have the height of water that is burnt the cylindrical tank at 3.15 PM 12315 then we have the time taken 3 15 CM nothing 0.5
medical tank in 5 hours into 40 2 height 15 cm cube is very important the units of the calculations are left with 15 8400 equal to the volume of water that fell from the ceiling but later calculate the length for what length of the cylindrical pipe in the speed breaker is also the peep into time time distance speed time
and time taken for a feeling of 315 0.5 as we have to calculate we have to use that time here to find 8.52 and time 0.5 2.52 into 0.5 is 1.2 km the unit is in kilometre energy converted into centimetres 12000 and now we have to find the volume of the cylindrical volume of water volume of water in the cylindrical pipe is equal volume of water in the cylindrical tank so we can recreate this volume 2
500 now the formula for volume in the cylindrical pipe latest consider the radius to b is equal to 80 now 22 by 7 into square into height is nothing by the limit length 12000 is equal to 15 84000 30 S and R square is equal to 15 24 into 7 by 126 to this total value is
net imply is equals to | 465 | 1,878 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-27 | latest | en | 0.938522 |
https://www.mathsassignmenthelp.com/blog/mastering-combinatorics-solving-cube-vertex-coloring/ | 1,721,668,723,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00486.warc.gz | 747,108,819 | 15,205 | # Solving Complex Combinatorial Puzzles: Coloring the Vertices of a Cube and Beyond
September 28, 2023
Gregory Burnet
UK
Combinatorics
Gregory Burnet has a master’s in applied mathematics and is a proficient assignment writer. He has helped many students get top grades.
Mathematics often presents intriguing puzzles that require a deep understanding of principles and creative problem-solving. In this blog, we delve into a captivating mathematical puzzle that will empower you to solve your math assignment while gaining insights into combinatorics and graph theory. Our focus lies in determining the number of ways to color the vertices of a cube using four distinct colors. As university students, grasping these foundational concepts and problem-solving techniques is essential for academic excellence and paves the way for practical applications in diverse fields. Join us on this enlightening journey as we unravel the intricacies of this problem and navigate the rich landscape of mathematical reasoning while seeking help with your combinatorics assignment effectively.
## Understanding the Basics
Before diving into solving this problem, let's establish some foundational concepts:
1. Vertices of a Cube: A cube has eight vertices, each representing a corner of the three-dimensional shape.
2. Coloring: We want to color these eight vertices with four different colors. Each vertex can be assigned one of the four colors, but two vertices sharing an edge must have different colors.
3. Counting: Our goal is to count the number of possible colorings that satisfy these conditions.
Now, let's break down the solution into manageable steps.
### Step 1: Fixing the First Vertex
To simplify the problem, let's start by fixing the color of the first vertex. We have four choices for the first vertex. Without loss of generality, assume that we color the first vertex red.
### Step 2: Coloring the Adjacent Vertices
The vertices adjacent to the first vertex (i.e., the ones connected by an edge) cannot be colored red since no two adjacent vertices can have the same color. Therefore, for each choice of color for the first vertex (red in this case), we have three choices for the color of the second vertex.
### Step 3: Coloring the Next Set of Vertices
Moving forward, we must consider the third vertex. This vertex is adjacent to both the first and second vertices. We can see that there are two cases to consider:
Case 1: The third vertex shares an edge with the first vertex but not the second. In this case, we have two choices for its color (different from red and the color of the second vertex).
Case 2: The third vertex shares an edge with both the first and second vertices. In this case, it cannot have the same color as either of them. So, we have only one choice for its color.
### Step 4: Extending the Process
Now, we have three vertices colored, and we need to color the remaining five vertices. To do this, we continue considering cases based on the relationships between these vertices.
Let's summarize the possibilities:
If the remaining vertices are not adjacent to any of the already colored vertices, we have four choices for each of them.
If a vertex is adjacent to one already colored vertex but not the others, we have three choices for its color.
If a vertex is adjacent to two already colored vertices, we have two choices for its color.
## Calculating the Total
To find the total number of colorings, we can now multiply the choices we have made in each step:
• Step 1: 4 choices
• Step 2: 3 choices
• Step 3 (Case 1): 2 choices for the third vertex
• Step 4 (Case 2): 1 choice for the third vertex
Since there are no further restrictions in subsequent steps (all remaining vertices have 4 choices), we multiply these together:
4 X 3 X (2 + 1) 4^5
Now, let's calculate:
4 X 3 X 3 X 1024 = 36,864
So, there are 36,864 different ways to color the vertices of a cube using four different colors while ensuring that no two adjacent vertices share the same color.
## Exploring Graph Theory
To truly appreciate the elegance of this problem, we can delve into graph theory. A cube can be represented as a graph where each vertex is a node, and edges connect adjacent nodes. This graph, known as the "cube graph" or simply "the cube," plays a pivotal role in solving our problem.
In graph theory, a "graph coloring" is an assignment of labels (colors) to the vertices of a graph in such a way that no two adjacent vertices have the same label. This leads us to the concept of "chromatic number," which is the minimum number of colors required to properly color a graph.
## Chromatic Number of a Cube
The chromatic number of the cube graph, denoted as χ(G), is precisely the answer to our question: the number of ways to color the vertices of a cube using four different colors while ensuring adjacent vertices have distinct colors.
Finding the chromatic number of specific graphs can be a challenging problem in itself, but for some graphs, like the cube, it can be computed relatively easily. The chromatic number of the cube graph is 3, which means it can be colored with just three different colors.
## Counting Proper Colorings
To find the number of ways to color a cube using four different colors, we're essentially counting the number of proper colorings of the cube graph where each vertex is assigned one of four colors. A "proper coloring" in graph theory adheres to the rule that adjacent vertices (connected by an edge) must have distinct colors.
## Counting by Group Symmetry
Now, let's explore another fascinating aspect of this problem - symmetry. The cube is a highly symmetric object, and understanding its symmetries can significantly simplify the counting process.
There are 24 rotational symmetries of a cube. These symmetries can be divided into several classes, including:
• Identity (0-degree rotation): This is essentially doing nothing to the cube, leaving it as it is.
• 90-degree rotations: These include four rotations (one for each axis of the cube - x, y, z, and diagonals) that preserve the cube's shape.
• 80-degree rotations: There are three axes of 180-degree rotations, passing through the center of opposite faces.
• 120-degree rotations: These correspond to the rotations that take a vertex to another vertex while preserving the cube's shape.
Now, let's consider these symmetries in the context of coloring the cube.
## Symmetry and Counting
When we count the number of colorings of the cube, we have to consider how many colorings are equivalent under these symmetries. In other words, we need to count the colorings up to rotational symmetry. This is a key concept in combinatorics known as "Burnside's Lemma."
Using Burnside's Lemma, we can calculate the number of distinct colorings by considering each symmetry class separately and then taking the average.
For instance, under the identity symmetry (doing nothing), all colorings are distinct. So, there are 4^8 distinct colorings.
For 90-degree rotations, we have:
4 colorings that stay the same (fixed by the rotation).
1 coloring that cycles the colors.
3 colorings that swap pairs of colors.
For 180-degree rotations, we have:
4 colorings that stay the same.
4 colorings that swap pairs of colors.
For 120-degree rotations, we have:
3 colorings that stay the same.
Now, let's calculate the total number of distinct colorings:
1/24(4^8+4+3+4+3)
1/24(65552)=2731
So, there are 2731 distinct ways to color the vertices of a cube using four different colors, taking symmetry into account.
## Generalizing to Other Polyhedra
The problem we've discussed isn't limited to cubes; it's a fundamental question in combinatorial mathematics that can be extended to other polyhedra. Polyhedra are three-dimensional objects composed of polygons (faces), edges, and vertices.
When dealing with polyhedra, understanding the symmetries of the shape becomes even more critical. The symmetries can help simplify the process of counting distinct colorings, just as we did with the cube.
To solve similar problems with other polyhedra, one would typically:
Identify the graph corresponding to the polyhedron.
Determine the symmetries of the polyhedron.
Use Burnside's Lemma to count the distinct colorings, taking symmetry into account.
This approach is a powerful tool for tackling complex combinatorial problems involving the coloring of graphs and polyhedra, making it a valuable skill for university students studying discrete mathematics.
## Applications of Graph Coloring
Understanding graph coloring and its applications can be extremely beneficial for university students. It's not limited to cubes or theoretical mathematics; it has real-world applications.
Scheduling: Graph coloring can be applied to scheduling problems. For example, in class scheduling, you can represent classes as nodes in a graph and use graph coloring to ensure that no two classes with overlapping times share the same classroom.
Map Coloring: The famous Four-Color Theorem states that any map on a flat surface can be colored with at most four colors such that no two adjacent regions have the same color. This theorem has practical implications in cartography, where it's essential to color countries or regions on a map so that adjacent regions have distinct colors.
Register Allocation in Compilers: Compilers use graph coloring to allocate registers to variables in a way that minimizes conflicts. The variables are represented as nodes in the interference graph, and graph coloring is used to ensure that no two interfering variables share the same register.
Wireless Frequency Assignment: In wireless communication, graph coloring can be used to assign different frequencies to nearby transmitters to minimize interference and ensure efficient communication.
By mastering the principles of graph coloring, university students can gain a deeper understanding of these real-world applications and potentially apply this knowledge in their future careers.
## Coloring Other Polyhedra
While we've primarily discussed coloring the vertices of a cube, this concept can be extended to other polyhedra. Each polyhedron corresponds to a unique graph, and determining the chromatic number or counting colorings for different polyhedra can be a fascinating challenge.
For instance, consider the regular dodecahedron, which has 12 faces and 20 vertices. Calculating the number of ways to color its vertices with four colors while ensuring no two adjacent vertices have the same color involves similar principles but is more complex due to the dodecahedron's unique structure.
Moreover, exploring irregular polyhedra or even more complex 3D shapes can open up new avenues for research and problem-solving in graph theory and combinatorial mathematics.
## Advanced Concepts in Combinatorics
The problem we've discussed is a prime example of combinatorial mathematics. Combinatorics deals with counting, arrangements, and selections of objects, making it a fundamental area of mathematics with a wide range of applications.
Advanced combinatorial concepts often involve permutations, combinations, generating functions, and inclusion-exclusion principles. By mastering these concepts, university students can tackle even more challenging combinatorial problems and gain a deeper appreciation for the elegance and complexity of mathematics.
## Mathematics as a Problem-Solving Tool
Mathematics isn't just about solving equations or proving theorems; it's a powerful tool for solving real-world problems. The process of breaking down complex problems into smaller, more manageable components, as we've done with the cube coloring problem, is a valuable skill in various fields, including science, engineering, economics, and computer science.
The ability to approach problems systematically, consider constraints, and use mathematical reasoning to find solutions is highly sought after in the job market. Thus, the skills honed through tackling mathematical problems, like the one discussed here, can have a profound impact on students' future career prospects.
## Conclusion
In conclusion, the problem of coloring the vertices of a cube using four different colors is not just an abstract mathematical puzzle. It's a gateway to understanding fundamental concepts in combinatorial mathematics, graph theory, and group theory. By breaking down the problem into manageable steps, considering the graph's symmetries, and applying Burnside's Lemma, we've shown how to approach such problems systematically. This knowledge equips university students with valuable problem-solving skills that extend beyond this specific question and can be applied to a wide range of mathematical and combinatorial challenges. So, the next time you encounter a complex math assignment or a problem involving combinatorics, remember the lessons learned from this theoretical exploration. Approach it systematically, consider the symmetries, and you'll be well on your way to finding elegant solutions to intricate problems. | 2,655 | 13,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.912936 |
https://universalium.academic.ru/2184/arccosine | 1,566,716,701,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323221.23/warc/CC-MAIN-20190825062944-20190825084944-00228.warc.gz | 682,721,179 | 12,159 | # arccosine
arccosine
arc cosine n.
The inverse of the cosine function.
* * *
Universalium. 2010.
### Look at other dictionaries:
• arccosine — noun Date: circa 1884 the inverse function of the cosine < if y is the cosine of θ, then θ is the arccosine of y > … New Collegiate Dictionary
• arccosine — noun Any of several single valued or multivalued functions that are inverses of the cosine function. Symbol: arccos, cos Syn: inverse cosine … Wiktionary
• arccosine — noun the inverse function of the cosine; the angle that has a cosine equal to a given number • Syn: ↑arc cosine, ↑arccos, ↑inverse cosine • Hypernyms: ↑trigonometric function, ↑circular function … Useful english dictionary
• Inverse trigonometric functions — Trigonometry History Usage Functions Generalized Inverse functions Further reading … Wikipedia
• Cubic function — This article is about cubic equations in one variable. For cubic equations in two variables, see elliptic curve. Graph of a cubic function with 3 real roots (where the curve crosses the horizontal axis where y = 0). It has 2 critical points. Here … Wikipedia
• trigonometric function — noun Date: 1909 1. a function (as the sine, cosine, tangent, cotangent, secant, or cosecant) of an arc or angle most simply expressed in terms of the ratios of pairs of sides of a right angled triangle called also circular function 2. the inverse … New Collegiate Dictionary
• Euclidean space — Every point in three dimensional Euclidean space is determined by three coordinates. In mathematics, Euclidean space is the Euclidean plane and three dimensional space of Euclidean geometry, as well as the generalizations of these notions to… … Wikipedia
• List of trigonometric identities — Cosines and sines around the unit circle … Wikipedia
• List of integrals of inverse trigonometric functions — The following is a list of integrals (antiderivative formulas) for integrands that contain inverse trigonometric functions (also known as arc functions ). For a complete list of integral formulas, see lists of integrals.Note: There are three… … Wikipedia
• Great-circle distance — The great circle distance is the shortest distance between any two points on the surface of a sphere measured along a path on the surface of the sphere (as opposed to going through the sphere s interior). Because spherical geometry is rather… … Wikipedia | 546 | 2,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-35 | latest | en | 0.70583 |
http://oeis.org/A057728 | 1,591,526,580,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00204.warc.gz | 91,130,627 | 4,891 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A057728 A triangular table of decreasing powers of two (with first column all ones). 7
1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 32, 16, 8, 4, 2, 1, 1, 64, 32, 16, 8, 4, 2, 1, 1, 128, 64, 32, 16, 8, 4, 2, 1, 1, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 1,5 COMMENTS First differences of sequence A023758. A023758 is the sequence of partial sums of a(n) with row sums A000337. 2^A004736(n) is a sequence closely related to a(n). T(n,k) is the number of length n binary words having an odd number of 0's with exactly k 1's following the last 0, n >= 1, 0 <= k <= n - 1. - Geoffrey Critzer, Jan 28 2014 LINKS Reinhard Zumkeller, Rows n = 1..100 of table, flattened FORMULA G.f.: (x - x^2)/((1 - 2*x)*(1 - y*x)). - Geoffrey Critzer, Jan 28 2014 [This produces the triangle shown by Mats Granvik in example section. - Franck Maminirina Ramaharo, Jan 09 2019] From Franck Maminirina Ramaharo, Jan 09 2019: (Start) G.f.: x*(1 - 2*x + y*x^2)/((1 - x)*(1 - 2*x)*(1 - x*y)). E.g.f.: (exp(2*x)*y - 2*exp(x*y))/(4 - 2*y) + exp(x) - 1/2. (End) EXAMPLE Triangle starts: 1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 32, 16, 8, 4, 2, 1, 1, 64, 32, 16, 8, 4, 2, 1, 1, 128, 64, 32, 16, 8, 4, 2, 1, 1, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, ... - Joerg Arndt, May 04 2014 When viewed as a triangular array, row 8 of A023758 is 128 192 224 240 248 252 254 255 so row 8 here is 1 64 32 16 8 4 2 1 From Mats Granvik, Jan 19 2009: (Start) Except for the first term the table can also be formatted as: 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, ... (End) MATHEMATICA nn=10; Map[Select[#, #>0&]&, CoefficientList[Series[(x-x^2)/(1-2x)/(1-y x), {x, 0, nn}], {x, y}]]//Grid (* Geoffrey Critzer, Jan 28 2014 *) PROG (Haskell) a057728 n k = a057728_tabl !! (n-1) !! (k-1) a057728_row n = a057728_tabl !! (n-1) a057728_tabl = iterate (\row -> zipWith (+) (row ++ [0]) ([0] ++ tail row ++ [1])) [1] -- Reinhard Zumkeller, Aug 08 2013 (Maxima) T(n, k) := if k = 0 then 1 else 2^(n - k - 1)\$ create_list(T(n, k), n, 0, 12, k, 0, n - 1); /* Franck Maminirina Ramaharo, Jan 09 2019 */ CROSSREFS Cf. A000079, A004736, A023758 and A000337. Cf. A155038 (essentially the same as this sequence). [Mats Granvik, Jan 19 2009] Sequence in context: A141020 A152568 A155038 * A176463 A098050 A278984 Adjacent sequences: A057725 A057726 A057727 * A057729 A057730 A057731 KEYWORD base,easy,nonn,tabl AUTHOR Alford Arnold, Oct 29 2000 EXTENSIONS More terms from Larry Reeves (larryr(AT)acm.org), Oct 30 2000 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified June 7 05:20 EDT 2020. Contains 334837 sequences. (Running on oeis4.) | 1,623 | 3,547 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-24 | latest | en | 0.295454 |
https://www.pinkbike.com/u/seb-stott/blog/nerding-out-how-temperature-affects-your-suspension.html | 1,709,247,993,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00037.warc.gz | 950,631,766 | 36,814 | # Nerding Out: How Temperature Affects Your Suspension
Dec 1, 2023 at 9:09
As summer turns to winter here in the Northern Hemisphere - and visa-versa down south - you may be wondering how the changing temperature affects your suspension. You might have noticed that your suspension feels softer and more sluggish when you go out in chilly conditions. This article is about why that is and how much of a difference seasonal temperature changes make to air-sprung suspension.
A little science
To understand what happens to your suspension when the temperature changes, it helps to understand the Kelvin temperature scale. A temperature change of one Kelvin is the same as one degree Celsius; but while zero Celcius is defined by the freezing point of water*, zero Kelvin is defined as absolute zero - the temperature at which molecules stop moving around and things literally can't get any colder.
*I'm pretty sure nobody really knows how the Fahrenheit scale is defined.
Absolute zero is minus 273°C, so freezing point is 273 Kelvin (K). A warm summer's day (27°C) is 300 K, which you'll notice is a nice round number.
The other piece of information we need is that if a fixed amount of air is trapped in a fixed volume (like the air inside your air-spring), then its pressure is (to a good approximation) proportional to the temperature in Kelvin. Also, the stiffness or spring rate of your air spring (how much force it generates over a given amount of travel) is proportional to the pressure inside the spring when fully extended.
What does that mean in practice?
So, if you set up your suspension in summer at 27°C or 300 K, then without adjusting it again, go for a ride in winter when it's 30K colder (-3°C or 270K), then the pressure in your air springs will be about 10% less than in summer. So if you put 100 psi in your fork in hot weather, it would be at 90 psi when the temperature drops below freezing, or if you set 30% sag in your shock, it will now sit around 33%.
In a back-to-back test, that's a very noticeable change. But since there might be several months between these temperature extremes you may not be aware that your suspension has softened since the summer - much like how you never noticed how much you were growing as a kid until your granny told you so.
Arguably, you may want softer suspension in the winter to improve traction, but it's still worth bearing this in mind. Or if you're setting sag in the warm indoors but plan to ride in very cold conditions, you may want to compensate - or brave the cold and set your bike up outside.
In colder weather, the viscosity of the damping oil will increase, which increases low-speed damping forces both for compression and rebound. The increased compression damping may partially offset the softer spring rate by providing some extra support. But on the rebound stroke, the increased damping and softer spring combine to slow the rebound speed, which could lead to more packing and worse traction on high-frequency bumps (if not adjusted for).
How much of an effect this will have depends on the damping oil used in the damper - higher oil weights are generally more affected by temperature - and temperature has very little effect at high suspension velocities (such as hard landings). But in general, the change to the air spring is probably far more noticeable than the damper when riding in colder (or hotter) weather.
When you start riding over rough terrain the damping oil will heat up - which will warm the air in the shock spring too - but this also happens in summer so this doesn't offset the effect of colder weather.
How does altitude play into this?
Coil spring advocates sometimes argue that temperature and altitude play havoc with air springs, while their coil springs stay consistent no matter what. As we've already seen, the effect of temperature on air suspension is modest unless large temperature changes are concerned (you're unlikely to notice a few degrees) and this is easy to account for when the seasons change.
As for altitude, atmospheric pressure drops by about 12% for every 1,000 m of height gain. That's about a 1.7 psi difference in atmospheric pressure when going from sea level to one kilometre up. For an air shock, this has a negligible effect. For forks, the main effect is on the chassis, not the spring. If there is 1.7 psi less pressure outside the lower legs than inside, then it will take about 2 Kg (5 lb) worth of force to push the stanchions into the lower legs at every point in the travel (including from full extension). This extra force may be noticeable in terms of the "touchdown feel" at the start of the stroke, and this is why lower leg bleeder valves (AKA fart buttons) exist. But note that this applies to coil forks just as much as air.
Altitude has a much larger effect on your tires. They are held up by the difference in pressure between the inside and outside (it's this pressure difference that you measure with your pressure gauge so you don't need to compensate for it). So when the outside pressure drops 1.7 psi, that effectively increases the tire pressure by the same amount, which could be almost 10%.
Author Info:
Member since Dec 29, 2014
288 articles
Must Read This Week
• 150 2
That's what i'm asking my X2 every time...
• 4 0
oh man,
A good one :-)
• 20 2
The X2 is all about dampening, not damping.
• 17 0
@WRCDH: mine dampens all kinds of things with shock oil.
• 7 0
But does it squelch?
• 4 0
I’m curious about tires in the cold. I’ve heard the schwalbe soft and super soft compounds substantially outperform the Maxxis tires in cooler temps - even at 5C.
• 1 0
@shredddr: I haven't tried schwalbes in the cold, but my maxxgrips' performance gets pretty terrible
• 3 0
I would say, what about bushing binding in lower temperatures. The increased slide resistance is much more affecting performance than air chamber, compression damping or tires.
• 2 0
@kostanev: don’t know about that specifically - but can confirm my Manitou Mastodon works great (no small bump compliance issues) down to -26 C.
• 1 0
@husstler: it's a fork made for winter. It may have a winter weighted oil in it...?
• 2 0
@skiboot1: no, same as a Mezzer.
• 2 0
@Sardine-Vladu: I've ridden schwalbe soft and ultra soft compounds in cold temps (down to -5C yesterday anyway) and found to be very consistent on dirt/frozen dirt/rock/leaves. I ran just 1psi lower than normal (because the ground was hard so i didnt need it to cut through mud), i especially enjoy how the tyre damping gets nicer making the tyre a little less reboundy (technical term). I havent been on Maxxis for years so cant comment on that.
• 2 0
It's all about oil viscosity index. Low viscosity index and your oils really suck in the cold (and hot). But there's also a split between mineral and synthetic. Minerals do better in the deep cold. The Fox R3 5wt and PTFE 5wt are not good in the cold.
• 1 0
@Dougal-SC: So you are saying ports and valves should be designed around a high-VI oil in order to work in a broader temperature tange. I guess the urge to make everything smaller/lighter is what’s driving the development towards smaller port sizes more dependent on flow rate and lower VI oils…
• 1 0
@Ynotgorilla: There doesn't need to be any change in ports/valves. Just better oil specs. Low VI damper oils suck at everything.
• 56 1
Pinkbikers riding 2018 forks that haven't been serviced now finally understand why they can't KOM their favorite double black; suspension settings in the parking lot at 10:00 don't work at the top of the fireroad at 12:00.
• 2 2
At least those people are focused on riding and not gear?
• 57 7
A British comedian I saw defined Fahrenheit as “percent hot”. In other words, 0 degrees Fahrenheit = 0% hot, while 100 degrees Fahrenheit = 100% hot. Really the only thing you need to know from there is that water freezes at 32% hot and you’re golden. Makes perfect sense.
• 14 1
Being a guy who routinely rides in 100% hot, and has worked on 190% hot, I don't think it scales.
• 13 1
But 30% hot is already f** cold! Fahrenheit was an idiot!
• 62 0
@chrsei: Well that’s because 30% hot is also 70% cold
• 4 4
Obviously Celsius makes way more practical sense. It was defined with respect to states of matter: 0% liquid to 100% liquid H20.
• 7 2
@cuban-b: at 100 C water is vapor not liquid. It’s the 99 degrees between 0 and 100 that water is liquid.
• 46 2
Celsius is how hot water feels, Fahrenheit is how hot humans feel, and kelvin is how hot atoms feel
• 2 0
@NERyder: At sea level atmospheric pressure.
• 5 10
Here’s a hot take. Obviously the metric system is far better than the imperial for almost everything, but honestly for temperature I’d argue it doesn’t make a difference. There’s nothing more intuitive about making a temperature scale based on the phase changes of H2O than some other arbitrary scale
• 3 0
@justridingalong1: I kinda like it.
Zero degrees or 32 degrees: Expect ice & snow when riding or driving.
32 makes zero sense.
• 5 0
@JSTootell: being a guy who doesnt ride anything but perfect biking weather i can say the cold does not affect my bicycle , nor the rain
• 1 0
@cuban-b: For science and engineering purposes, Kelvin > Rankine > Celsius > Fahrenheit. Any non-absolute temperature unit is pretty useless whenever you need to calculate anything using it
• 1 0
@NERyder: close.. it depends on pressure.
100 C is the boiling point at STP (Standard temperature and pressure). As the pressure reduces, so does the boiling point. You can see videos of water boiling in the Himalayas at way less than 100 C.
And good luck to you if you can get water to boil at 2-3 atmospheres (29-42 psi).. It is how power plants work.
• 1 0
@roadie-ottawa-64: I mean, if you really want to get pedantic, might as well toss in enthalpy too.
• 2 0
@roadie-ottawa-64: Sure. I mean I wasn't trying to go full "Bill Nye the Science Guy" on this thread. It also assumes pure H2O. If you dissolve something in with the water the boiling point drops as well
• 33 0
I think Kelvin is a nice enough guy, absolute zero seems a bit harsh.
• 22 0
PV = nRT. Ideal Gas Law is all you need to know
• 10 7
Well you do need to know one other bit... no gases are ideal
• 5 19
cuban-b (Dec 1, 2023 at 16:10) (Below Threshold)
@boozed: go back to school
• 3 0
Use it for tires as well suspension
• 1 0
@mrkkbb: Bontrager/Trek has a great little chart for this. It's on my shop wall for winter fat biking when the temp delta between my shop and the outdoors can often be 50 degrees F, which matters a bunch once you're an hour into the ride.
• 3 0
Is that what they mean by a nRT gas?
• 20 0
I always found the changes to the tires rubber to be more significant to anything suspension wise.
• 15 1
MaxxTerra is particularly not great in the 30%-38% hot range.
• 10 1
Agreed. Maxx Grip and other soft compound tires turn to wood below about 45 degrees F. They seem to get harder than standard compound tires at low temps. Definitely more noticeable than suspension changes.
• 2 0
@Philphun: I've also experienced this with sticky rubber approach shoes. My incredibly sticky 5.10 guide tennies may as well be made of a bar of soap once temps drop below freezing.
• 3 0
omfg is % hot going to be a thing now
• 6 1
@chaoscacca: better than % bar soap
• 5 0
@Philphun: Depends on the tire company. Maxxis rubber hardens a lot in the cold. As does Michelin.
While Conti, Schwalbe and Specialized rubber doesn’t change as much.
• 1 0
I agree. Tires go to trash below ~48F
• 19 0
Durrrrr Fahrenheit is a silly imperial unit no one understands…
Now let me explain the ***Pounds per inch^2***in your suspension
• 2 0
How many Pascals do you put in your Minions?
• 10 0
@st-lupo: depends if he's french or not
• 6 0
@st-lupo: 207 kPa in my minions, don’t care about you opinions…
• 11 2
Also worth noting your shock pump reading will vary with altitude because even though the air spring is sealed system the pump displays gauge pressure (relative to atmosphere).
• 15 0
Also worth noting that even if you took your bike into outer space, your shock pump would only be off by 15 psi.
Jumping between sea level and 10K ft (3000 m) is worth about 4 psi
• 8 1
This was one of the least nerding out nerding out articles ever.
As pointed out above, not only does pressure drop with temp in a closed system, the viscosity of the suspension and lubrication oil increases to the point you are overdamped and bearings seize, but also everything plastic/rubber stiffens.
You air shock will therefore have an incredible amount if stiction, or main seals or +/- chamber seals will fail.
If you are riding in winter, regrease all your cartridge bearings (including opening all seals) in all pivots, hubs, ++ with waterproof low temp grease, ditch your air shock (time to practice hardtail skills?!?), and opt for your old open-bath zocchi where changing to a low viscosity oil is a breeze.
• 4 0
Agreed Seb could've gone way harder on fully nerdin out. Not even a single graph or chart!
• 3 0
I think you're being a bit dramatic, air shocks work just fine in winter with some adjustment. Though maybe it's different in Norway...
• 5 0
Kind of on topic, but do pro downhillers preheat their suspension like in motorsports for consistency? Or is there a temperature difference by the time they get to the bottom of their runs in their suspension?
I don't ride hard enough to heat things up. Or maybe I'm just so smooth that it doesn't happen. I'll go with the latter for my ego...
• 3 0
When Brooke McDonald was on Trek these used to use these black warming cases for the bike prior to their finals run.
• 8 1
If you take your bike to Whistler on a cold day and ride the gondola you can notice the bike warm up and suspension change. By noon when the air is warmer and you have hit a million braking bumps because everyone was working on 1199 which never opened, your suspension will feel quite different than when you arrived.
• 8 0
Great question! I was doing a little tech consultation with a top EWS rider and we had a great chat about exactly that. There was a long stage that was roughly freezing at the top and, if I recall, around 12°C at the bottom with considerable elevation change. Combine all those factors with the amount of warming (oil viscosity and air pressure) from riding and it was like two different bikes at the start and end.
One rider was using some fancy moto gear that circulated hot oil through the fork, then the mechanic reassembled the fork at the last minute. Definitely posh, but another racer just took off their front wheel and soaked the fork in a bucket of hot water! (Didn't hear whether he did the same with his shock.)
The rider for whom I was consulting didn't do any of these things and had a terrible result on that stage, with a bike that started out with poor traction and high rate of fatigue from impacts, then felt like a pogo stick in a pinball machine by the end of the stage.
• 3 0
@s100: 1199 was open, just not to the public.
• 2 1
@Comatosegi: correct: and they didn’t have it ready in time for Crankworx but raced it anyway. It sucked a lot of season pass money away from other trails and we weren’t even supposed to ride it.
• 2 0
@s100: As a DH race fan, I fully support the 4 year project that was 1199. As great a place to ride as Whistler, it should have a matching DH race track. I agree it didn’t have to be that binary in terms of resource management, but they have a a lot of turnover in personnel. But that has to do with the current economics of mountain towns.
• 2 0
@R-M-R: You'd think top EWS riders would have access to better oils. That would solve ~90% of their problems. Many stock oils have low VI and perform terribly in the cold. This is easily rectified with better oils. Not thinner oils, just better oils.
• 1 0
@Dougal-SC: It's surprising how little thought is given to such things. That said, there's no viscosity index that will stay uniform from 0°C to 12°C ambient with pro-level riding! Also, my understanding is that VI and lubricity tend to be inversely related, though I don't know the extent of this, nor whether the trade-offs favour one parameter or the other.
• 1 0
@R-M-R: you don't need to keep viscosity level. You've just got to keep the viscosity low enough that it doesn't create huge port effects and have bath oil that doesn't turn to glue.
A 12c window is pretty small and easy to cover. A 40c range is harder.
All the motorex damper oils have vi over 250 and lubricate better than all the 100 vi damper oils. Cartridge dampers also don't give a damn about lubricity.
• 1 0
@Dougal-SC: 12°C ambient change plus however much it changes due to riding is what I meant. That's going to be closer to your 40°C number.
True about distinguishing the different requirements for the damper fluid and the bath fluid. The MTBR discussions on Supergliss as a bath oil inspired me to a slideway oil from a local vendor and it was a big improvement.
• 1 0
@R-M-R: Fork dampers and bushing lube doesn't heat much at all riding. They basically stay at ambient.
Compared to the rear shock the spring and damper rates are a magnitude less so there's not much energy input. There's a lot more surface area exposed to wind and great heat-sinking.
• 1 0
@Dougal-SC: Perhaps. It's not something I've looked into. Less than a shock, certainly, and I would think a pro rider could put a few degrees into the relevant parts of the system.
• 1 0
@Dougal-SC: When Henry did the podcast with Ohlins, TJ mentioned that mitigating heat was an issue even for a 4 minute run. Are you saying that really only applies to the shock? Or is their situation unique b/c they are running more damping?
• 1 0
@Comatosegi: yes rear shock only.
• 5 0
Can SRAM add a heating coil running on their battery to maintain my oil at a fixed temperature? I could also adjust the temperature target based on elevation to compensate for the pressure change. Seems like the easy solution to me.
• 1 0
Tesla does this for their electronics.
• 1 0
I think you are underestimating the amount of power heating the oil given the density and packaging of a lithium battery. Heating an EV with a resistive grid takes a lot of power.
@s100 Which IC electronics get heating? The only ones I am aware of are the thermally controlled battery pack and the radar unit. And that they use some waste heat from the motors to heat the pack pushed through the octavalve.
• 1 0
@Comatosegi: yes. They thermally control the motors and the battery. Which is the drivetrain. They also control the computer and screen with cabin overheat protection to cool it as well.
All that is left is the wiring connecting these parts. And that they do not thermally regulate.
• 1 0
@s100: Those are different things than running heat to "electronics." Cabin overheat protection is a function of the HVAC. If I remember correctly Tesla 3/Y has about 6 ICs to control most function in vehicle under a unified computer platform. Most other manufacturers use 40-50 individual computers to control these functions. None of those are heated, from my knowledge. The x86 computer for cockpit interface and FSD computer is thermally cooled but I doubt it gets heating.
• 1 0
@Comatosegi: Please let's stay focused on my need for electronically controlled oil viscosity for the purpose of maintaining optimal shock and fork performance
• 6 2
For me if I increased air pressure in my shock and fork for cold weather I'd have no teeth left in my head. The suspension gets WAY firmer for me as soon as the weather changes. This has been on 3 diferent bikes but all using Fox factory suspension. The oil thickening I'm assuming is having way more effect then a few PSI drop. So I usually set the suspension up on the low-end of recommended air pressure for my weight and it seems to help. Dropping pressure to adjust for thicker oil is not a perfect solution but for my moderate skill set it seems to make the bike less painful.
• 3 0
I changed my oil, replaced it with some 0w stuff a friend liked and it works well.
• 2 0
Another reader smartly answered my question - retracted.
• 4 0
Years ago, pedaled to work on a -4f day. When I went to leave, the almost brand-new rear air shock was totally deflated with a puddle of oil under it. More extreme temps than most of us would actually ride in, but the cold totally destroyed the seals.
• 7 0
Was your bike built by NASA?
• 4 0
I remember when all forks in our riding group 10 years ago were simply steel coil only. No poor air. We never discussed this topic, before winter, during winter rides. Those with functional dampers changed some clicks, me with Lyrik Mission control changed nothing because the fork changed its demeanour only with spring rate, oil density and damper shims.
• 6 0
I find that rebound speed is the most noticeable for me, I usually have to speed it up a click or two when it’s really cold
• 1 0
Found that today -felt super slow on the way to the trail. Reached under the fork to speed it up and the knob had fallen out somewhere. Grrrr! And my one up multi tool isn't long enough to reach inside a Pike. Double grrrr!
• 1 0
Ports are way more sensitive to oil viscosity than shims. The rebound damping relies the most on port flow so that's where you feel it more. Better oil (higher VI) helps a lot. Many stock oils are terrible in the cold.
• 3 0
When I was a freshman in college and dead broke, I got a Gary Fisher Opie at shop cost from the shop I wrenched at during the summer - A 2004 Opie was a shittie DJ bike, but it was a bike and I needed one. It came with an elastomer fork, and I found it getting harsher and harsher as the fall progressed, until one day after a late class I jumped on my bike, hucked a small staircase and about broke my wrists... the elastomers at or around freezing turned into rocks. Rather than deal with the poor performance an unpredictability, I put a used Surly Instigator fork on there... first of many rigid MTBs for me. I miss that stupid bike.
• 4 0
Another important question is, how cold temperatures affects your tires! Softest Maxxis compounds becomes hard, some other brands seems to work better. It would be nice to have a test about tires in cold.
• 7 0
N/A full coil!
• 2 0
Just one more thing I dislike about air sprung suspension. I miss my full coil F&R bike. To each their own I suppose, but I'm a "set it and forget it" kinda guy personally.
• 1 0
Full coil esp in the winter! I run coil front and rear on my FS fat bike (Fatillac) and I never have to adjust a thing, boom!!
• 4 0
Okay so did anyone else notice in the first picture the guy on the right has clipless shoes on but there's flat pedals on his bike. What going on?
• 1 0
No Ben, but sharp eyes... All I noticed were the clear BMX background on those wheel stands.
• 1 0
No but did anyone else notice in the first picture the guy on the left has clipless shoes on but there's flat pedals on his bike. What going on?
• 2 0
This is a well-rounded article for me. My lovely Durolux slowed way down last winter as temps approached freezing, (knobs full open and still too slow) something I hadn't really noticed much on previous Fox and DVO forks. Just today I changed the oil in the damper to the lightest Motorex I could find in hopes of counteracting this. It feels too fast in my warm shop now, so tomorrow will be the first test.
• 2 0
Damper design plays a major role in how much temperature will affect the damping characteristics. Dampers that rely more on pushing oil through ports as apposed to shim stacks tend to be slow down more dramatically as the temperature drops and oil viscosity increases. In my experience, the rebound damping tends to be the first area a rider will notice changing due to manufacturers relying heavily on a large bleed port as a way to have a lot of adjustability with the rebound clicker. Changing oil viscosity can really change a dampers ability to dissipate energy. It is better to look at the oil manufactures 40C cst viscosity than the listed wt on the label. You may also find that changing your oil viscosity gets you in the right adjustability range for say compression but then you will need to re-shim the rebound stack or visa versa.
• 3 2
I don't think it really matters. By the time that it gets cold enough to make a difference, its usually dark by 5pm, wet and/or snowy. So you're going slower anyways and being cautious, the slower rebound can help in some cases.
Same with tires. In late fall riding, especially at night, I find I can get away with lower pressures. Passively going slower due to the environment has a lot to do with this.
• 2 1
Have you considered that with your suspension better set up you might not need to be as cautious?
• 1 0
@notthatfast: it's setup correctly, I'm not saying I just ignore it. It just seems to not matter as much, compared to having messed up suspension when you have the ability to ride at full speed like during the day.
Once we start getting snow or ice ontop of wet leaves, which are on top of greasy mud, which hide those pesky diagonal roots - sometimes you just have to tip toe down sections. Or across, I mean it is pretty flat and tight here.
• 1 0
That’s fair. Less of a problem on the North Shore. Unless it snows the trails don’t change a whole lot here.
• 1 0
But the 10 people that ride in freezing rain and snow all winter need to proclaim how tough they are so the rest of us can recognize our humanity
• 2 0
ah, i lived in Asia with high heat and humidity, moved to Canada: i had to change the oil thickness on my suspension and level of oil in the damping system. Also readjusted all the settings.
• 1 0
The volume is not fixed "over a given amount of travel" , so you really need the total volume and the volume change per travel to know "how much force it generates over a given amount of travel".
You're also assuming it's from fully extended, but "given amount of travel" is unbounded: you really need the volume and pressure of the starting travel and the volume delta to the ending travel.
• 4 0
How does increased viscosity of the damping fluid only effect low-speed damping? Why not other shaft speeds?
• 6 0
Just a guess: once the high speed circuit it activated, the orifices are large enough that the viscous terms don’t dominate the equations any more.
• 3 1
@pmhobson: upvoted for 'orifices'
• 2 0
Because shims don't care about oil viscosity much but ports do. High speed damping is shims with only small port effects (in a correctly designed damper). Rebound is where you notice it because that's a lot more port flow.
That said. Bath oil will suck the life out of everything in the cold if your oil spec and bushing clearances aren't right.
• 1 1
@Dougal-SC: Don't pistons have ports? Do you mean "_orifices_ care more about viscosity more than shims"?
Because shims opening rely on the velocity of the fluid rather than the flow, and the ports on the piston are big enough to reduce the effect of flow viscosity. Makes sense.
• 1 0
@justinfoil: Ports are orifices. Shims react to pressure.
• 1 0
The pressure will not change because it is a sliding sealed unit. The pressure will stay the same but the volume will change. This will cause the shock to compress. Charles's law - volume is proportional to temperature
Just to be pedantic as we are after all "nerding out". Water freezes at 273.15 K. at 1 atmosphere. As you increase your elevation the temperature at which water freezes is higher due to less pressure but not that you would actually notice.
• 1 0
@boozed: PV=nRT, bruh. Get down with the ideal gas law. For a simplified model of how this works, however, the Gay-Lussac's is robust enough, P1/T1=P2/T2.
P=pressure
V=volume
n=moles (# of molecules/atoms)
R=a constant that makes that math work and changes depending on the units used elsewhere
T=temperature in Kelvin
• 1 0
Lots of comments to read but have not seen one that mentions warming blankets. It’s just like tire warmers for F1 or moto GP. SRAM/RockShox was messing around with these the 2011-2012 seasons. Haven’t seen any one do it since. Makes total since with fork and shock oil. If it’s warm and at the proper viscosity it’s gonna work better.
• 1 0
When I feel winter has arrived I add 5psi to the fork, 15 to the shock, and open my compression and rebound 2 clicks ish. Tires depends on what type of trail I ride so I consider it differently and change it time to time, sometimes.
• 1 0
Easily remedied, if you ride from your door, it shouldn't affect the spring rate as you are activating the suspension and it will stay warm. If you drive to the trailhead, put it in you r car and it should be good to go when you get there. I've done it this way for years to not damage the damper circuits and it seems to work well for me.
• 2 0
I am old enough to remember the old forks with elastomers in them as the bumpers (springs). Went fully rigid if it was cold enough. How times change.....
• 1 1
Been thinking about this a lot lately. Noticed a huge negative impact on performance in my Fox 34 as the weather got cold. Ended up changing the damper and lower bath oils out for lower viscosity and it's feeling better. Rear shock not as affected since it uses a very high viscosity index oil and produces more heat anyways.
• 1 0
-13C (8.5 Freedom units) on todays ride. I spent my brain cells on what to wear. Bike just works. Studded tires also mean you could faff suspension for hours with no noticeable effect.
• 1 0
I would also like to know how the temperature affects a coil spring, and what about the effects of the cold weather on the human body?
• 2 0
Fat bikers with suspension forks are gonna have a field day when they read this article.
• 2 0
Nah, just run coil front and rear and it’s all good !
• 1 0
What about damaging suspension due to cold temperatures? Oils will not to as well of a job lubricating
• 2 0
Steel. Coil. Spring. Nah. That was just a winter wind's whisper.
• 1 0
I have been more worried on my brakes being shit at 4000m and 0 degC than about my suspension..
• 3 2
Where's the advent calendar comp?
• 2 0
No kidding. Bike industry is hurting that bad is it?
• 1 0
@cazman727: An editor has now confirmed it's not running.
• 1 0
Did that Nathan Hughes photo win any contests? It should.
• 1 0
I'm probably incorrect here but shouldn't it be "effects"?
• 3 0
Nope. You can remember it as -A-ffect is an -A-ction. Effect is (generally) a noun.
So weather can affect your suspension performance.
Also, weather can have effects on your suspension performance.
And then there’s that phrase “effect change” where the effect noun becomes a verb.
• 1 1
Lots of words for a very simple message. I’d probably stay in your lane a bit better with this stuff.
• 1 0
My take away from this is to set it and forget it.
• 1 0
-23 zeb is totally dead in cold.
• 2 3
Every 10F drop in ambient temperature = approx. 3psi drop in both fork and shock, in my experience.
• 4 1
How does that work? Forks and shocks aren't remotely similar in terms of their operating pressures.
• 1 0
*well-timed
• 2 5
Fahrenheit is king of units
Below threshold threads are hidden | 7,715 | 31,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.919496 |
https://www.scribd.com/document/256637419/Arbori-Binari-de-cautare | 1,566,127,777,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313803.9/warc/CC-MAIN-20190818104019-20190818130019-00545.warc.gz | 963,920,535 | 56,964 | You are on page 1of 4
# Binary trees
Abstract
This technical note describes a method to store a binary tree within a bit pattern. This can be used in
compression systems to store huffman codes etc.
Introduction
To store compressed data in many cases binary trees are common, also in higher methods, where
Huffmans coding is used only as an supplementary method. We show here an alternate method to
store the information required to build a Huffman or more general a binary tree. This is done by
use of a bit pattern, where each inner node is coded by a binary 1, and each leave by a zero. We
show that the original tree can be retrieved in a unique way.
Foundations
A binary tree consists of nodes, where each node specifies no or two successors. A node is called
inner node (or usually only node), if it has successors. A node with no successors is called a leave. A
tree is defined recursively by:
1. A binary tree has a unique node, called a root. A root can be an inner node or a leave.
2. If a binary tree is given and a leave is replaced by a node with two leaves, where the two
leaves are no members of the binary tree, then we get another binary tree.
The second condition states, that the connection between a node and its successors can never lead to
a node of the binary tree.
We state the following simply observation:
Observation 1: In a binary tree the number of leaves and inner nodes differ by one.
This holds obviously for a binary tree with one node, which is the root and a leave. Let it hold
for any binary tree with less than n nodes. To get a binary tree with n+2 nodes, we can replace
a leave by a node with two leaves as successors, where the number of nodes is increased by
one, where the number of leaves is decreased by one and increased by two, i.e. increased by
one leave. Thus the difference between the number of nodes and leaves remains unchanged.
The proof also shows that the number of nodes in a binary tree is always uneven.
We define the left subtree of a node as the tree consisting of the left successor of this tree (as root)
and all its successors. In the same way we define the right subtree.
## Left depth search
We define now the left depth search as a sequence of all nodes in a binary tree.
Left depth search: The left depth search of a binary tree is the following sequence:
1. The root is the first element of the left depth search.
2. The left depth search of the left subtree of the root are the next elements of the sequence.
3. The left depth search of the right subtree of the root are the final elements of the sequence.
This sequence defines the structure of the binary completely, if inner nodes and leaves are
distinguished.
To generate this sequence, a simple recursive function can be used. The following function
leftDepthSearch prints the Names (which are here numbers) of the left depth search sequence of
a binary tree.
struct node {struct node *left, *right; int Name}* start;
// start points to the first root of the binary tree.
## leftDepthSearch(struct node * start)
{
printf(%d,,start->Name);
if(start->left!=NULL) leftDepthSearch(start->left)
if(start->right!=NULL) leftDepthSearch(start->right)
}
Since the left depth search sequence defines the structure of the tree, we simply have to distinguish
between inner nodes and leaves by using bits 1 for inner nodes and bits 0 for leaves. Thus the
following function produces a corresponding sequence.
leftDepthSearchNodeLeave(struct node * start)
{
if(start->left==NULL && start->right==NULL) // this is a leave
printf(0); else // this is a node
printf(1);
if(start->left!=NULL) leftDepthSearch(start->left)
if(start->right!=NULL) leftDepthSearch(start->right)
}
Construct a tree
Now let us assume we read a sequence of bits (0,1) and we want to construct a binary tree from them.
This can be done recursively as well. Let nextBit return 0 or 1, we use the following recursive function
Int countNodes=0, countLeaves=0;
struct node * makeTree()
{ int next = nextBit();
struct node * nextNode = NewNode();
if(next==0)
{ countLeaves++;
nextNode ->left = NULL;
nextNode ->right = NULL;
return nextNode;
} // next must be 1, i.e. generate an inner node with two successors
countNodes++;
nextNode->left = makeTree();
nextNode->right = makeTree();
return nextNode;
}
struct node * start;
int main(){
start = makeTree();
}
If the first number read by nextBit() is 0, then this program enters the first if-condition, sets the
left, right variables to NULL and exits. This is okay, since the binary tree consists of a leave and
nothing else. There cannot be an empty binary tree, since the number of nodes is always odd, i.e.
never zero.
If the first number read by nextBit() is 1, then this program assigns the (by makeTree()) newly
created subtrees at first to the left, then to the right variable. The first call to makeTree()
consumes exactly the left depth search sequence of the original subtree. By induction, its structure
must be identical of that left subtree. The same holds for the right subtree.
It should be clear, that the program stops automatically when the left depth search sequence is
completely consumed.
Example:
The tree might be:
The green root is coded by 1, the left yellow node by 2, the most left blue leave by 3 etc.; so we get
the left depth search sequence as: 123456789. The corresponding bit sequence for inner nodes(=1)
and leaves (=0) yields: 110100100.
The first call to makeTree() calls nextBit(),reads at first 1 and generates a node, the left successor to
which a newly generated subtree is assigned by another call to makeTree(). This reads the second
bit, a 1, and generates a new subtree by a call to makeTree(). Now, the third call to nextBit() reads 0,
so a leave is constructed and assigned to the left variable of node 2. Then node 2 calls
makeTree() again, which reads the fourth bit, a 1 (for node 4). Thus this calls makeTree() again,
which reads the fifth bit, a 0. Thus a leave (5) is returned to node 4, which is assigned to its left
variable. Then node 4 calls makeTree() again to find the next bit to be a 0. A leave (6) is returned
and assigned to its right variable. Node 4 returns its reference, which is assigned in node 2 to its
right variable. This node returns its reference, which is assigned by node 1 to its left variable.
The next call in node 1 to makeTree() finds a bit 1, which generates two further calls to
makeTree(), which yield references to two leaves and assignes them to the left and right
variable of node 7. Node 7 returns its reference, which is assigned by node 1 to its rightvariable.
Finally node 1 returns its reference to the main program, which assigns this to the start variable.
Thats all, folks.
The left depth search sequence has the property that the number of inner nodes (or 1) is always less
than or equal to the number of leaves (or 0), if we sum from left to right. This means, that it is easy to
find the number of nodes in a tree, if only the left depth search sequence is given. Scan the bits, add
+1 for a 1 and 1 for a 0, and if the sum becomes 1, then this is the last node of the tree.
int CountNodes()
{ int NodesInTree=0, countsNodes=0;
while(countsNodes>=0)
{ NodesInTree++;
if(nextBit()==0) countsNodes--; else countsNodes++;
}
return NodesInTree;
}
To show that the latter assumptions holds, we have to prove that this holds for each inner node. We
know that a subtree of any node has the property that the number of leaves is one more than the
number of nodes. We start with the root, for which countsNodes yields 1 if it is an inner node. The
left subtree counts 1, so that after its scan countsNodes yields 0. The right subtree yields 1, so
that countsNodes becomes 1. We show that after reading the bit for any inner node countsNodes
is 1 or greater. This holds for the root. If the first node of the left subtree is a node, then countsNodes
becomes 2. After reading this left subtree, countsNodes is 0. If the first node of the right subtree is an
inner node, then countsNodes yields 1. Thus the assumption holds for the successor of the root, and
it holds also for those successors.
## Application to Huffman Coding
To apply this to Huffman coding, we have to assign to each leave the value of a symbol which is coded
with that bit sequence. The easiest way is to use the recursive function to build a sequence of the
symbols in left depth search (where only for the leaves the corresponding value is stored). Since
generating of the binary tree yields the same sequence of leaves, we can use it directly to generate
the corresponding symbols in the newly created tree.
Storage size needed to store this information is the number of nodes (in bits) plus the number of
symbols, which might be coded to save some space. In classical Huffman coding we have to store
either only the frequency of the values (where their index gives the symbol value). However, here
unused symbols must be stored by the frequency 0. In many ASCII-texts less than 80 symbols are
used. If 256 bytes are required to store the frequencies, we can do this with 80 + 161/8=101 bytes.
This saves some bytes, which is the goal of compression anyhow. | 2,174 | 9,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-35 | latest | en | 0.964103 |
https://simiode.org/resources/browse?sortby=date&tag=differentialequations,differentialequation,ordinarydifferentialequations | 1,601,418,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402093104.90/warc/CC-MAIN-20200929221433-20200930011433-00024.warc.gz | 600,031,701 | 8,652 | ## Resources: All
Find a resource
1. 20 Sep 2020 | Modeling Scenarios
A simple first order population growth model is presented. The challenge is to produce a final differential equation which is the result of the difference or ratio of birth and death rates. This ratio is not immediately intuitive.
2. 22 Jul 2020 | Modeling Scenarios
We describe the frequency response to a second order differential equation with a driving function as the maximum steady state solution amplitude and perform some analyses in this regard.
3. 10 Jul 2020 | Modeling Scenarios
We present two exercises from a differential equations text in which we ask students to model (1) falling object experiencing terminal velocity and (2) bobbing block of wood in liquid. We model the motion using Newton's Second Law of Motion and Archimedes' Principle.
4. 11 Jun 2020 | Modeling Scenarios
Students will build and analyze a model of the fraction of people who are married (for the first time) by a certain age. This model comes from a paper by Hernes and, in this project, is compared to another model used by Coale.These models are first-order ordinary differential equations (which...
5. 30 May 2020 | Modeling Scenarios
We build models of world population using data to estimate growth rate.CZECH LANGUAGE VERSION We have placed in Supporting Docs a Czech version of this Student Modeling Scenario. Name will be x-y-S-Title-StudentVersion-Czech.
6. 29 May 2020 | Modeling Scenarios
We are given data on the position of a mass in an oscillating spring mass system and we seek to discover approaches to estimating an unknown parameter.
7. 21 Apr 2020 | Modeling Scenarios
We ask students to use the system of first order linear differential equations given in a source paper and estimates of the data from laboratory procedures from a plot to estimate the parameters and complete the modeling process. Then we seek to compare the results of the final model with...
8. 19 Apr 2020 | Modeling Scenarios
A recent cholera outbreak in Haiti brought public attention to this disease. Cholera, a diarrheal disease, is caused by an intestinal bacterium, and if not addressed in a timely manner may become fatal. During the project described here, the students will learn how to solve and address a...
9. 16 Apr 2020 | Presentations
Transient and Steady State response in RC or RL circuits . PowerPoint Slides. 54 Slides, This is a presentation to help understand the concepts of Transient and Steady State response in RC or RL circuits . There is detailed technical material and graphs to richly illustrate the clear...
10. 06 Apr 2020 | Potential Scenario Ideas
2014-WronaEtAl-Two-compartmentModelAsTeachingToolForCholesterolHomeostasisWrona, Artur, Joanna Balbus, Olga Hrydziuszko, and Krystian Kubica. 2015. How We Teach: Classroom And Laboratory Research Projects -- Two-compartment model as a teaching tool for cholesterol homeostasis. Adv...
11. 05 Apr 2020 | Potential Scenario Ideas
2013-DontwiEtAl-ModelingMemorizationAndForgetfulnessUsingDEDontwi, I. K. et al. 2013. Modeling Memorization and Forgetfulness Using Differential Equations. Progress in Applied Mathematics. 6 (1), 1{11. Available from...
12. 05 Apr 2020 | Potential Scenario Ideas
2012-Obenb-DentehEtAl-ModelingStudentAffinityForLectureObeng-Denteh1, W., Odum Kodua-Nana1, S. Owusu Ansah1, R. Kwame Ansah1 and Alex Quashie Nyamadu. 2012. Modelling Students’ Affinity for Lecture Attendance Using Differential Equations: The Case of Students of Kwame Nkrumah University...
13. 04 Apr 2020 | Potential Scenario Ideas
2016-BoninEtAl-MathModelsApproachInVaccinologyBonin, Carla Rezende Barbosa, Guilherme Cortes Fernandes, Rodrigo Weber dos Santos, and Marcelo Loboscoa. 2016. Mathematical modeling based on ordinary differential equations: A promising approach to vaccinology. Hum...
14. 04 Apr 2020 | Potential Scenario Ideas
2011-Yafia-DEModelingMalignantTumorCellsInCompetitionWithImmuneSystemYAFIA, RADOUANE. 2011. A STUDY OF DIFFERENTIAL EQUATION MODELING MALIGNANT TUMOR CELLS IN COMPETITION WITH IMMUNE SYSTEM. International Journal of Biomathematics. 4(2): 185-206Abstract: In this...
15. 02 Apr 2020 | Potential Scenario Ideas
2019-Koss-SIR Models_ Differential Equations that Support the Common GoodKoss, Lorelei. 2019. SIR Models: Differential Equations that Support the Common Good. CODEE Journal. 12 (2019): Article 6. https://scholarship.claremont.edu/codee/vol12/iss1/6/ .12 pages.Abstract: This...
16. 01 Apr 2020 | Potential Scenario Ideas
2016-BarbarossaKuttler-MathModelOfBacteriaCommunicationBarbarossa, Maria Vittoria and Christina Kuttler. 2016. Mathematical Modeling of Bacteria Communication in Continuous Cultures. Appl. Sci. 6(149): 1-17.Abstract: Quorum sensing is a bacterial cell-to-cell communication mechanism...
17. 01 Apr 2020 | Potential Scenario Ideas
2018-Joseph-Balamuralitharan - Nonlinear DE model of Asthma EffectsJoseph ,G Arul J. and S Balamuralitharan. 2018. A Nonlinear differential equation model of Asthma effect of environmental pollution using LHAM. IOP Conf. Series: Journal of Physics: Conf. Series. 1000:...
18. 01 Apr 2020 | Potential Scenario Ideas
2007-RamseyEtAl-ParameterEstforDE-SmoothingApproachJ. R. Statist. Soc. B (2007)69, Part 5, pp. 741–796Ramsay, J. O.. G. Hooker, D. Campbell and J. Cao. 2007. Parameter estimation for differential equations: a generalized smoothing approach. J. R. Statist. Soc....
19. 27 Mar 2020 | Potential Scenario Ideas
2017-GuptaKumar-GlucoseMetabolismGupta, Richa and Deepak Kumar. 2017. Numerical Model for Glucose Metabolism for Various Types of Food and Effect of Physical Activities on Type 1 Diabetic Patient. Applied Mathematics. 7(2): 19-22Abstract: This work displays a...
20. 27 Mar 2020 | Presentations
2013-Sutton-SecondOrderLinearODE's-PartIIPresentation by Craig J. Sutton, Department of Mathematics, Dartmouth CollegeMath 23 Differential Equations. 2013. 45 slidesCovers basics or undetermined coefficiens and variation of parameters with applications to mechanicl vibrations | 1,567 | 6,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-40 | latest | en | 0.870833 |
https://www.slideserve.com/jacob/rural-addressing-methods | 1,553,046,570,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202188.9/warc/CC-MAIN-20190320004046-20190320030046-00157.warc.gz | 892,369,514 | 19,785 | 1 / 32
# Rural Addressing Methods - PowerPoint PPT Presentation
Rural Addressing Methods. Rural Addressing Methods. There are two primary means of means of assigning locatable addresses: Distance Based Addressing : according to the number of incremental units from the beginning of a road. Grid Based Addressing : according to a pre-defined grid.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
There are two primary means of means of assigning locatable addresses:
• Distance Based Addressing: according to the number of incremental units from the beginning of a road.
• Grid Based Addressing: according to a pre-defined grid.
• Distance-based addressing establishes house-numbering according to the number of increments from the beginning of a road. An increment can be any number of feet or division of miles.
• An increment can be any number of feet or division of miles.
• Typical increments are 5-feet, 5.28-feet (.001 miles), 10-feet, 25-feet, 26.4-feet (.005 miles), 50-feet, 52.8-feet (.01 miles) and 100-feet.
• That is, by proceeding up a road, a new address is available at each multiple of this increment.
• In addition, the parity (odd/even) of the address is established according to the side of road.
• The side-of-road is relative to a vehicle proceeding along the road in the direction of ascending numbers (from the beginning of the named road odd numbers on the left and even numbers on the right.
• Note: there are no USPS, NENA or other known national standards for establishing parity.
• Grid addressing establishes a northing and easting coordinate system for your municipality. Addresses are then developed from the east-west coordinate (x) plus the north-south coordinate (y).
• Numbers tend to be very large as each coordinate value is generally 3-digits. If the municipality covers a large area, other numbers may need to be added to indicate the sector of the grid. In addition, any road that curves back in an east-west direction can have addressing that is out of order (goes up and down). This system may be suitable in flat areas where almost no curved roads exist and the residents don't mind large numbers.
• Official street numbers should proceed from a logical point of origin and should be in proper numerical sequence in relation to other lots with frontage on the same street/road. It is recommended that a county adopt a standard numbering sequence that goes from East to West and South to North.
• Odd numbers are commonly assigned to the left side of the street and even numbers to the right side of the street.
• Address numbers should be assigned to all structures which are inhabited or which have or may have phone service (including telephone booths) and not to lots and parcels. Many lots have more than one structure and thus require more than one address.
• Corner lots should be assigned a number according to the frontage of the main entrance and/or where the driveway meets the road - not where the mailbox is located.
• The logical/grammatical order of address elements should follow USPS conventions: street number, pre-directional, primary street name, suffix, post-directional, and secondary number, if any (e.g., 100 W Main ST SE Apt 201)
• Multi-unit structures should be given one primary number (e.g., 101 Main ST or 103 Main ST) and apartments or suites should be given numbers or letters as secondary location indicators (e.g., 111 Main ST, APT 101)
• Primary street numbers should not be longer than six characters
• There should be no fractional addresses (e.g., 101 1/2 Main ST)
• Alphanumeric primary or secondary address numbers should not be used (e.g., EOI Main ST)
• Hyphenated primary or secondary address numbers should not be used (e.g. 41-65 Bell ST)
• Significant leading zeros in primary and secondary numbers should not be used (e.g., 0145 Main ST)
• One letter road names should not be used (e.g., B Street or Z Lane)
• Common practice is to number from east to west, and from south to north
• Where one road starts and ends at points on a second road (creating a loop), the numbering along the two roads should be in the same direction
• Municipalities planning an enhanced 9-1-1 project should establish a single locatable addressing system that is adopted for all addressing purposes including emergency service provision, postal delivery, municipal record keeping, and utility service orders.
• NENA Standards recommend that communities adopt a 5.28' increment. This increment would allow for 1,000 unique numbered addresses per 1 mile length of road (there are 5,280 feet in a mile). This increment makes it very easy for an emergency service provider in the event of an emergency to look at the address and quickly figure out how far down the road the house is located.
• For example, if an ambulance were dispatched to 500 Alpine Lane (using a 5.28' address increment) the emergency vehicle would know to travel exactly one half mile down Alpine Lane to locate number 500.
• Another benefit of the 5.28' increment is that structures can be assigned an address every 10.5' on each side of the driveway.
• This means that many addressees are developed, mainly for navigational purposes, but also for future development needs of a community.
• If the appropriate measured numbering system is selected, a community will never need to be re-addressed to accommodate future development.
Local standards are a customization of the nationally recommended methodsAddressing Standards that should be considered on the local level
• Establish a 9-1-1 Committee to implement County-wide Standards.
• Establish Road Naming and Re-naming Standards.
• Establish a Road Measuring and Numbering System that Fits your county.
• Consider Hiring a GIS/E911 Consultant.
• Naming roads and mapping them is among the first steps in an addressing process.
• Placing road signs is one of the final tasks. To assist both emergency service personnel and the general public, signs must be visible and maintained.
• A frequent compliant about road signs is that they are often hidden by tree branches. Annual trimming can eliminate this problem.
• There are several varieties of signs that are suitable for public road use. The most common is the green background with reflective white lettering. However, regardless of color, the important issue is to ensure the letters are tall enough to be easily seen day or night.
• Sizes: Lettering on street name signs should be at least 4 inches high, supplementary letting to indicate the type of street (e.g., Street, Avenue, Road, etc.) or section of city (e.g., N.W.) may be in smaller lettering, at least 2 inches high.
• For rural areas: Municipalities should use 9 inch high blade in lengths of 24, 30, 36, or 42 inches with 6 inch high letter for street names, 4 inches for other.
• For urban areas: Municipalities should use 9 inch high blade in lengths of 24, 30, 36, or 42 inches with 4 inch high letter for street names, inches for other.
• Placement: In business districts and on principal arteries, Street Name signs should be placed at least on diagonally opposite corners so that they will be on the far right hand side of the intersection for traffic on the major street. They should be mounted with their faces parallel to the streets they name. In residential districts at least one Street Name sign should be mounted at each intersection. In rural districts signs should be placed to identify important roads not otherwise marked.
• Determine the type of address numbering system(s) to be used.
• Establish base maps for the entire area by using existing data or developing you own centerline file.
• Determine ESN boundaries, telephone exchange boundaries, city limits, and zip code boundaries.
• Research new developments, streets and sub-divisions for centerline updates.
• Determine property number ranges.
• Build Master Street Address Guide (MSAG) and supply to USPS, telephone companies and others.
• Determine a time-line for establishment and adaptation of new address for rural structures.
• Obtain citizens address and location records from all possible sources.
• Locate every addressable structure from field verification or aerial photography.
• Assign a property number to each structure or driveway location on map.
• Use citizen/location records and/or fieldwork to verify name, telephone number and existing (old) address for each structure.
• Inform the USPS, telephone company, citizens and others of the change of address and the date in which the change will become official for mailing purposes.
• Work with telephone companies and the USPS on follow-up and error correction.
• Develop methods for address assignment by cities and counties.
• Deliver indexed maps and atlases to cities and county.
• Establish a rigorous Data Maintenance schedule and keep up with it by hiring qualified personnel.
• Address Listings/Files- maintained by your government (including those used for mass mailings or needed to provide essential services, such as police, fire or sanitation programs) or purchased from commercial vendors are an acceptable source for creating housing unit tallies for each census block, provided they are current and comprehensive. You also can update comprehensive address listings/files with additions and deletions from other local records.
• Tax Assessment Records- may supply up-to-date tallies of housing units.
• Utility Connection Records- (electric, gas, water or sewer) are other sources for developing housing unit tallies.
Other Methods for “Change Detection”
Multiplesources:
• Property Permit Records
• Certificates of Occupancy and Demolition Permits
• Neighborhood visits
• Special censuses
• Aerial photography (DOQQ’s) and Current land use maps
• Housing authorities records
• Tribal, village, or Indian Health Service
• Environmental Health Office records
When should Addresses be Updated or Checked?
Local governments are in the best position to identify areas with potentially missing or incorrectly assigned addresses.
For example:
• Has there been recent new construction activity in your community?
• Are there areas that have changed from single-family homes to multi-family homes and vice versa?
• What about commercial or former public buildings that have been converted to residential use?
• Are there new mobile home parks or even scattered new mobile homes?
• Do you have apartment buildings or boarding houses with irregular or missing numbering schemes for the individual units?
• Are there blocks where you believe the Census Bureau missed housing units in 1990?
• Has your community recently annexed new territory, and thus new addresses?
• Most Common Address Systems in Urban Areas:
• Base Line Method: References for an Addressing Grid use paired perpendicular lines oriented North/South and East/West.
• Quadrants: Four Zones of a city, quartered “roughly” by a pair of perpendicular lines.
• Addressing Grid: Ranges for addresses are forced onto the existing perpendicular street network
• Mostly dealing with new Street Naming and Standardization.
• Continue to edit and update Address Standards.
• Vacant Parcels will require an Address, unlike the suggested Rural methods.
• Interval - Much like the distance method used in the rural areas, where every so many feet there’s a new number.
• Potential Range - Address numbers eligible for assessment along a block (typically 100)
• Actual Range - Address numbers actually assigned along a block (430-488) not (400-498). | 2,501 | 12,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-13 | latest | en | 0.909895 |
https://philosophy.stackexchange.com/questions/89965/what-is-the-philosophers-take-on-information-and-thermodynamic-entropy?noredirect=1 | 1,713,567,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00396.warc.gz | 415,052,170 | 41,438 | # What is the philosopher's take on information and thermodynamic entropy?
So there are various interpretations of probability.
Frequentism is the likelihood of events of say for example if I roll a dice the likelihood of of getting a 5 is 1/6 if repeated over and over.
Propensity is given by how much you know about the system for example if I know the initial conditions of the dice roll well enough then I can predict the odds of a 5 is 100% (for example).
In statistical mechanics the thermodynamic entropy of a system seems to be describe the likelihood of an event but then we also have the notion of information entropy which is related to the second type of probability. Many physicists are comfortable swapping and conflating the two. Going so far that they even equate the two. (See Is information entropy the same as thermodynamic entropy? (PhysicsSE).)
What is the relationship if any according to philosophers between information and thermodynamic entropy?
• Who are those many physicists who equate Shannon entropy and thermodynamic entropy ? While there are superficial similarities between the 2 notions, they are certainly not the same. Take a deck of poker cards sorted in order, shuffle it: while thermodynamically the effect is negligible, you have significantly increased the Shannon entropy of the deck (I.e. You can't say in advance which card you are going to draw anymore) Mar 11, 2022 at 7:11
• @armand it's a very common point of view imo physics.stackexchange.com/questions/263197/… Mar 11, 2022 at 7:21
• What they are saying is the formulas are similar. Yet they are not the same thing, one can grow while the other is stable, as demonstrated by the poker deck example. Mar 11, 2022 at 11:44
• @armand: Thermodynamic entropy is typically described as an application of Shannon's information theory: en.wikipedia.org/wiki/… Mar 11, 2022 at 13:11
• Added links, clarified question, and added 'modality', 'probability', and 'philosophy-of-information' since the question asks after information theory.
– J D
Mar 11, 2022 at 18:03
## 3 Answers
You never have 100% predictive power. Quantum events involve fundamental uncertainties, and can in principle effect macroscopic events (eg a powerful cosmic ray changing the momentum of the dice at a critical moment, or even more uncertainly a beta decay of a carbon 14 nucleus). There are finite probabilities of landing on an edge or point also. It's good to think about probability as a way to deal with limited information, but the uncertainty principle and the observer effect (the impact of taking measurements on increasing uncertainty) mean information is always limited, in a way that I would describe as there not being such complete information meaningfully in our universe. Three-body dynamics of large interacting blackholes mean uncertainties below the Planck scale can have have macroscopic effects.
Entropy causes a number of confusions. It is relative not absolute, strictly speaking we can only measure change in it, and if there are hidden microstates or degrees of freedom calculations will be incomplete. We typically idealise systems as closed and close to equilibrium, far-from equilibrium thermodynamics is much harder. And open systems where the Gibbs free-energy is more significant than simple entropy gradients are important to us, because it's part of the thermodynamic understanding of life.
Physicists play a sleight of hand, by defining information as the inverse of entropy. The full analysis of Maxwell's Demon links information-entropy and thermodynamics, so it makes sense even if in ordinary language information means a range of less specific things. Discussed here Is the concept of information nonphysical?
I see modern science as currently property-dualist, picturing everything as fundamentally constituted of energy, and information - which includes entropy. It was expected that information would resolve to being secondary properties of the fundamental things, mass-energy and spacetime. But now increasingly the reverse is expected, with those previously fundamental things pictured as emergent from information and it's propagation. Discussed here: Is information the foundation of reality?
We can understand 'true' as relating to comparing expectation and reality, and specifically in regards to building a tables of probabilities like the dice-rolling example, composed of counterfactuals. Discussed here: Why is a measured true value “TRUE”? Deutsch & Marletto's Constructor Theory looks to relate physics and information-theory in a deeper way, by expanding analysis of computation into sets of outcomes instead of a single floating-point variable, comparable to expanding our understanding of now to include Many Worlds.
a) Entropy is not about probabilities (although probabilities might be used to calculate it). In simple terms, entropy is a quantity that measures the potential of disorder. It is not a direct measure of disorder. The fact that low entropy implies high predictability is due to the reduction of the potential disorder, not to the reduction of disorder. At S=0, there is no potential for disorder, so, no disorder.
b) Entropy is a macrostatic quantity. Memorize that. That is: it is a descriptor of facts of perception, that is, ideas, sensations (for example, temperature is a macrostatic quantity: it is a feeling, which (thanks, 0th Law, sorry you didn't were defined first), it can be assessed as a physical quantity).
c) While macrostatic facts are related to ideas, microstatic facts are those related with physical phenomena. For example, microstatically, temperature is a quantity proportional to the statistical kinetic energy load of each molecule in a gas container.
d) Statistical entropy (S=k ln $\omega$) describes the information carried for some macroscopic state:
S=ln2(64)=6
That is, in order to carry 64 states, 6 bits are necessary (logarithm base 2). So, in this example, the entropy of a system which can be in 64 states is 6.
d) Contrary to entropy, information is a microstatic quantity. Following the previous example, 111000 is the information that corresponds for a particular microstate of the system. Although this information is effectively carried in 6 bits, that is not necessarily the same of the entropy of the system. For example, Gibbs entropy can have a different value for the same 6 bits of information.
e) Thermodynamic entropy is equivalent to statistical entropy: it is a macrostatic quantity that measures the potential of disorder of a particular microstate. Perhaps the best interpretation here is to say that thermodynamic entropy is the measure of energetic order.
f) The big difference between thermodynamic and statistical entropy is that the latter allows measuring each microstate, at the cost of addressing systems always as discrete entities.
In addition to know that the entropy of a 6-bits message is 6, each message (the information) can also be obtained: 111000. Statistical entropy is mostly used to analyze fixed-size messages (but yes, it also allows addressing messages of different sizes). That is, statistical entropy is mostly used not to measure disorder, but how to assess the performance of a process at different values of disorder.
But the for the former, thermodynamic entropy, microstates cannot be measured. How would we know the actual configuration of X molecules having a value of entropy of S=0.nnnnn...? In addition, the absolute value of entropy of any substance cannot be measured directly, the only measurable value is dS=dQ/T, that is, change. So, thermodynamic entropy is mostly used to measure disorder as such.
Don't forget that S=0 is a convention for zero temperature, which experimentally shows that mass tends to order at 0∘K.
Information and thermodynamic entropies are in some way parallel concepts. Both describe how the complexity of the system increases over time. More and more information is needed to describe the state of the system.
When the signal to noise ratio decreases over time, the information in the signal does not disappear, it is only diluted in the constant influx of random noise. Random information is added to the signal.
Something like that is happening also in thermodynamic processes. Differences in energy density are parallel to the signal. Thermal noise, random vibration of particles distributes the energy more evenly.
It all boils down to the inherent randomness of a probabilistic universe. In a deterministic universe there would be no noise, complexity would not increase, all entropies would remain constant.
• And you know the last sentence to be true because you yourself exhibit infinite complexity/information? At least nothing in your post allows for such a conclusion. Mar 11, 2022 at 12:16
• In determinism there is no noise. This is not a conclusion. This is implied in the definition. Mar 11, 2022 at 12:40
• Like every kind of determinism was binary, sure... Mar 11, 2022 at 12:52
• It's a mistake to picture entropy as all about noise & heat. Consider predicting the outcome of coin faces of n coins shaken in a box. As n increases, the closeness of the average balance of faces will get closer & closer to 50%, & by 10^23 'coins' you get very reliable predictions eg gas laws. Thermal states are molecular spin states, lattice vibrations, kinetic energy etc. You can understand that kind of spread-out energy doesn't tend to concentrate, just like many coins don't all come up heads. The 2nd law of thermodynamics is like saying signals deteriorate, they don't increase >100%. Mar 11, 2022 at 18:45 | 2,006 | 9,609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.942986 |
https://oeis.org/A000125/internal | 1,686,307,590,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00400.warc.gz | 457,833,531 | 7,950 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A000125 Cake numbers: maximal number of pieces resulting from n planar cuts through a cube (or cake): C(n+1,3) + n + 1. (Formerly M1100 N0419) 70
%I M1100 N0419 #260 Nov 19 2022 10:01:35
%S 1,2,4,8,15,26,42,64,93,130,176,232,299,378,470,576,697,834,988,1160,
%T 1351,1562,1794,2048,2325,2626,2952,3304,3683,4090,4526,4992,5489,
%U 6018,6580,7176,7807,8474,9178,9920,10701,11522,12384,13288,14235,15226
%N Cake numbers: maximal number of pieces resulting from n planar cuts through a cube (or cake): C(n+1,3) + n + 1.
%C Note that a(n) = a(n-1) + A000124(n-1). This has the following geometrical interpretation: Define a number of planes in space to be in general arrangement when
%C (1) no two planes are parallel,
%C (2) there are no two parallel intersection lines,
%C (3) there is no point common to four or more planes.
%C Suppose there are already n-1 planes in general arrangement, thus defining the maximal number of regions in space obtainable by n-1 planes and now one more plane is added in general arrangement. Then it will cut each of the n-1 planes and acquire intersection lines which are in general arrangement. (See the comments on A000124 for general arrangement with lines.) These lines on the new plane define the maximal number of regions in 2-space definable by n-1 straight lines, hence this is A000124(n-1). Each of this regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n-1) regions already there, hence a(n) = a(n-1) + A000124(n-1). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
%C More generally, we have: A000027(n) = binomial(n,0) + binomial(n,1) (the natural numbers), A000124(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) (the Lazy Caterer's sequence), a(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) + binomial(n,3) (Cake Numbers). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
%C If Y is a 2-subset of an n-set X then, for n>=3, a(n-3) is the number of 3-subsets of X which do not have exactly one element in common with Y. - _Milan Janjic_, Dec 28 2007
%C a(n) is the number of compositions (ordered partitions) of n+1 into four or fewer parts or equivalently the sum of the first four terms in the n-th row of Pascal's triangle. - _Geoffrey Critzer_, Jan 23 2009
%C {a(k): 0 <= k < 4} = divisors of 8. - _Reinhard Zumkeller_, Jun 17 2009
%C a(n) is also the maximum number of different values obtained by summing n consecutive positive integers with all possible 2^n sign combinations. This maximum is first reached when summing the interval [n, 2n-1]. - _Olivier Gérard_, Mar 22 2010
%C a(n) contains only 5 perfect squares > 1: 4, 64, 576, 67600, and 75203584. The incidences of > 0 are given by A047694. - _Frank M Jackson_, Mar 15 2013
%C Given n tiles with two values - an A value and a B value - a player may pick either the A value or the B value. The particular tiles are [n, 0], [n-1, 1], ..., [2, n-2] and [1, n-1]. The sequence is the number of different final A:B counts. For example, with n=4, we can have final total [5, 3] = [4, _] + [_, 1] + [_, 2] + [1, _] = [_, 0] + [3, _] + [2, _] + [_, 3], so a(4) = 2^4 - 1 = 15. The largest and smallest final A+B counts are given by A077043 and A002620 respectively. - _Jon Perry_, Oct 24 2014
%C For n>=3, a(n) is also the number of maximal cliques in the (n+1)-triangular graph (the 4-triangular graph has a(3)=8 maximal cliques). - _Andrew Howroyd_, Jul 19 2017
%C a(n) is the number of binary words of length n matching the regular expression 1*0*1*0*. Coincidentally, A000124 counts binary words of the form 0*1*0*. See Alexandersson and Nabawanda for proof. - _Per W. Alexandersson_, May 15 2021
%C For n > 0, let the n-dimensional cube, {0,1}^n be provided with the Hamming distance, d. Given an element x in {0,1}^n, a(n) is the number of elements y in {0,1}^n such that d(x, y) <= 3. Example: n = 4. Let x = (0,0,0,0) be in {0,1}^4.
%C d(x,y) = 0: y in {(0,0,0,0)}.
%C d(x,y) = 1: y in {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}.
%C d(x,y) = 2: y in {(1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1), (0,0,1,1)}.
%C d(x,y) = 3: y in {(1,1,1,0), (1,1,0,1), (1,0,1,1), (0,1,1,1)}.
%C All these y are at a distance <= 3 from (0,0,0,0), so a(4) = 15. (See Peter C. Heinig's formula). - _Yosu Yurramendi_, Dec 14 2021
%D V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_3.
%D R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 27.
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
%D H. E. Dudeney, Amusements in Mathematics, Nelson, London, 1917, page 177.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D T. H. Stickels, Mindstretching Puzzles. Sterling, NY, 1994 p. 85.
%D W. A. Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 30.
%D A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #45 (First published: San Francisco: Holden-Day, Inc., 1964)
%H T. D. Noe, <a href="/A000125/b000125.txt">Table of n, a(n) for n = 0..1000</a>
%H P. Alexandersson and O. Nabawanda, <a href="https://arxiv.org/abs/2104.04220">Peaks are preserved under run-sorting</a>, arXiv:2104.04220 [math.CO], 2021.
%H A. M. Baxter and L. K. Pudwell, <a href="http://faculty.valpo.edu/lpudwell/papers/AvoidingPairs.pdf">Ascent sequences avoiding pairs of patterns</a>, 2014.
%H M. L. Cornelius, <a href="/A006261/a006261_1.pdf">Variations on a geometric progression</a>, Mathematics in School, 4 (No. 3, May 1975), p. 32. (Annotated scanned copy)
%H F. Javier de Vega, <a href="https://arxiv.org/abs/2003.13378">An extension of Furstenberg's theorem of the infinitude of primes</a>, arXiv:2003.13378 [math.NT], 2020.
%H R. K. Guy, <a href="/A000346/a000346.pdf">Letter to N. J. A. Sloane</a>
%H Zachary Hoelscher, <a href="https://arxiv.org/abs/2102.07083">Semicomplete Arithmetic Sequences, Division of Hypercubes, and the Pell Constant</a>, arXiv:2102.07083 [math.NT], 2021. Mentions this sequence.
%H Marie Lejeune, <a href="https://hdl.handle.net/2268/259266">On the k-binomial equivalence of finite words and k-binomial complexity of infinite words</a>, Ph. D. Thesis, Université de Liège (Belgium, 2021).
%H D. A. Lind, <a href="http://www.fq.math.ca/Scanned/3-4/lind.pdf">On a class of nonlinear binomial sums</a>, Fib. Quart., 3 (1965), 292-298.
%H Svante Linusson, <a href="http://dx.doi.org/10.1007/s004930050055">The number of M-sequences and f-vectors</a>, Combinatorica, vol 19 no 2 (1999) 255-266.
%H R. J. Mathar, <a href="/A247158/a247158.pdf">The number of binary mxm matrices with at most k 1's in each row or columna</a>, (2014) Table 3 column 1.
%H Ângela Mestre and José Agapito, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Mestre/mestre2.html">Square Matrices Generated by Sequences of Riordan Arrays</a>, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
%H Sebastian Mizera and Sabrina Pasterski, <a href="https://arxiv.org/abs/2204.02505">Celestial Geometry</a>, arXiv:2204.02505 [hep-th], 2022.
%H Alexsandar Petojevic, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL5/Petojevic/petojevic5.html">The Function vM_m(s; a; z) and Some Well-Known Sequences</a>, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H D. J. Price, <a href="http://www.jstor.org/stable/3609091">Some unusual series occurring in n-dimensional geometry</a>, Math. Gaz., 30 (1946), 149-150.
%H L. Pudwell and A. Baxter, <a href="http://faculty.valpo.edu/lpudwell/slides/pp2014_pudwell.pdf">Ascent sequences avoiding pairs of patterns</a>, 2014.
%H Luis Manuel Rivera, <a href="http://arxiv.org/abs/1406.3081">Integer sequences and k-commuting permutations</a>, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
%H H. P. Robinson, <a href="/A000124/a000124.pdf">Letter to N. J. A. Sloane, Aug 16 1971, with attachments</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CakeNumber.html">Cake Number</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CubeDivisionbyPlanes.html">Cube Division by Planes</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CylinderCutting.html">Cylinder Cutting</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MaximalClique.html">Maximal Clique</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/SpaceDivisionbyPlanes.html">Space Division by Planes</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TriangularGraph.html">Triangular Graph</a>
%H Reinhard Zumkeller, <a href="/A161700/a161700.txt">Enumerations of Divisors</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = (n+1)*(n^2-n+6)/6 = (n^3 + 5*n + 6) / 6.
%F G.f.: (1 - 2*x + 2x^2)/(1-x)^4. - [_Simon Plouffe_ in his 1992 dissertation.]
%F E.g.f.: (1 + x + x^2/2 + x^3/6)*exp(x).
%F a(n) = binomial(n,3) + binomial(n,2) + binomial(n,1) + binomial(n,0). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
%F Paraphrasing the previous comment: the sequence is the binomial transform of [1,1,1,1,0,0,0,...]. - _Gary W. Adamson_, Oct 23 2007
%F From _Ilya Gutkovskiy_, Jul 18 2016: (Start)
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F a(n) = Sum_{k=0..n} A152947(k+1).
%F Inverse binomial transform of A134396.
%F Sum_{n>=0} a(n)/n! = 8*exp(1)/3. (End)
%F a(n) = -A283551(-n). - _Michael Somos_, Jul 07 2022
%e a(4)=15 because there are 15 compositions of 5 into four or fewer parts. a(6)=42 because the sum of the first four terms in the 6th row of Pascal's triangle is 1+6+15+20=42. - _Geoffrey Critzer_, Jan 23 2009
%e For n=5, (1, 3, 5, 7, 9, 11, 13, 17, 19, 21, 23, 25, 35) and their opposite are the 26 different sums obtained by summing 5,6,7,8,9 with any sign combination. - _Olivier Gérard_, Mar 22 2010
%e G.f. = 1 + 2*x + 4*x^2 + 8*x^3 + 15*x^4 + 26*x^5 + 42*x^6 + 64*x^7 + ... - _Michael Somos_, Jul 07 2022
%p A000125 := n->(n+1)*(n^2-n+6)/6;
%t Table[(n^3 + 5 n + 6)/6, {n, 0, 50}] (* _Harvey P. Dale_, Jan 19 2013 *)
%t LinearRecurrence[{4, -6, 4, -1}, {1, 2, 4, 8}, 50] (* _Harvey P. Dale_, Jan 19 2013 *)
%t Table[Binomial[n, 3] + n, {n, 20}] (* _Eric W. Weisstein_, Jul 21 2017 *)
%o (PARI) a(n)=(n^2+5)*n/6+1 \\ _Charles R Greathouse IV_, Jun 15 2011
%o (PARI) Vec((1-2*x+2*x^2)/((1-x)^4) + O(x^100)) \\ _Altug Alkan_, Oct 16 2015
%o (Magma) [(n^3+5*n+6)/6: n in [0..50]]; // _Vincenzo Librandi_, Nov 08 2014
%o (Python)
%o def A000125_gen(): # generator of terms
%o a, b, c = 1, 1, 1
%o while True:
%o yield a
%o a, b, c = a+b, b+c, c+1
%o it = A000125_gen()
%o A000125_list = [next(it) for _ in range(50)] # _Cole Dykstra_, Aug 03 2022
%Y Bisections give A100503, A100504.
%Y Row sums of A077028.
%Y Cf. A000124, A003600, A005408, A016813, A086514, A058331, A002522, A161701-A161705, A000127, A161706 - A161708, A080856, A161710-A161713, A161715, A006261, A063865, A051601, A077043, A002620.
%K nonn,easy,nice
%O 0,2
%A _N. J. A. Sloane_
%E Minor typo in comments corrected by _Mauro Fiorentini_, Jan 02 2018
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified June 9 06:42 EDT 2023. Contains 363168 sequences. (Running on oeis4.) | 4,487 | 12,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-23 | latest | en | 0.796092 |
https://allinonehighschool.com/10524-2/ | 1,702,205,459,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101779.95/warc/CC-MAIN-20231210092457-20231210122457-00120.warc.gz | 117,424,526 | 21,814 | ## Nuclear Notation
Looking at the symbol above, you can identify many things about this element. The element symbol is given. To the left of the symbol, the top number is the mass number, which is also given the symbol A. Remember that the mass number = protons + neutrons. The bottom number to the left of the symbol is the atomic number, which is also given the shorthand symbol of Z. Remember that the atomic number = protons, and also the charge of the nucleus. The mass number is equal to the number or nucleons (particles in the nucleus).
Looking at the symbol above, the element is carbon. It has 6 protons and 6 neutrons (12 – 6 = 6).
The atomic number (Z) for an element does not change, but the mass number (A) can change, as neutrons are added or removed from the nucleus. Look at the nuclear notation below for an isotope of carbon.
A = 14
X = 6
The atomic number is 6. This atom of carbon has 6 protons (remember, all carbon atoms have 6 protons).
The mass number is 14. This atom of carbon has 8 neutrons (14 – 6 = 8).
Sometimes, all of this information is not necessary, and a simpler method for writing symbols is needed. Hyphen notation is a shortened version of the nuclear notation. In hyphen notation, the first element above will look like this: C-12
The element symbol is given, then a hyphen, then the mass number (A). As the atomic number never changes, and can be readily obtained from any periodic table, the hyphen notation is commonly preferred. What is the hyphen notation for the second symbol above? (Highlight to reveal answer: C-14)
## Alpha Decay
Alpha decay usually only occurs in elements with atomic numbers of 83 and above. Alpha decay always has a product that includes a helium-4 nucleus. This helium-4 nucleus is what is known as the alpha particle or α-particle.
This chemical equation represents a Uranium-238 atom undergoing alpha decay. Notice the α-particle is one product.
To determine the other product, subtract the mass number of He from the mass number of the U.
238 – 4 = 234 is the mass number of the other product.
Now, subtract the atomic number of He from the atomic number of U.
92 – 2 = 90 is the atomic number of the other product.
To determine the element symbol, look up atomic number 90 on the periodic table.
is the symbol for the element with a mass number of 234 and an atomic number of 90.
Uranium-238 is called the parent nuclide and thorium-234 is called the daughter nuclide.
Try another one.
To determine the mass number of the product, subtract 263 – 4 = 259.
To determine the atomic number of the product, subtract 106 – 2 = 104.
To determine the symbol of the product, look up on the periodic table atomic number 104.
What is the parent nuclide? (Highlight to reveal answer: seaborgium-263 )
What is the daughter nuclide? (Highlight to reveal answer: rutherfordium-259 )
## Beta Decay
Beta decay occurs when a negatively charged electron is emitted from the nucleus. Remember that the mass of an electron is negligible when compared to the mass of an atomic nucleus. Because of this, the mass of the nucleus that undergoes beta decay is only decreased a tiny amount, an amount so small that it can be ignored. The mass number in beta decay does not change. Remember also that the nucleus only contains protons and neutrons. Where does the electron that is emitted come from? A neutron is changed into a proton. When this occurs, an electron is formed. That electron is emitted. The number of protons has now increased by 1, which means that the atomic number increases by 1 as well. The electron, shown in the equation as , is known as β.
This chemical equation represents a potassium-40 atom undergoing beta decay. Notice the β-particle is one product.
To determine the other product, remember that the mass number does not change.
Subtract the atomic number of β from the atomic number of K.
19 – (-1) = 20 is the atomic number of the other product.
To determine the element symbol, look up atomic number 20 on the periodic table.
Try another one.
Remember that the mass number does not change.
To determine the atomic number of the product, subtract 6 – (-1) = 7.
To determine the symbol of the product, look up on the periodic table atomic number 7.
An unseen product in beta decay is called a neutrino, and is given the symbol ν. A neutrino is an amount of energy. Remember above, we said that the mass of the parent nuclide changed such a tiny amount that we could ignore that difference when calculating the mass number? How does this fit into the Law of Conservation of Mass? The neutrino is the answer to that. Einstein predicted that mass can sometimes convert into energy. This is what happens during beta decay. That small amount of mass was converted into a neutrino. Sometimes beta decay problems are written showing the neutrino:
## Gamma Decay
Unlike alpha and beta decay, gamma decay does not occur by changing the atomic structure of an atom. A gamma ray, given the symbol γ, is a type of electromagnetic radiation that is emitted when the electric charge of a nucleus is rearranged. Remember that in gamma decay, neither the mass number nor the atomic number are changed.
An equation showing gamma decay might look like this:
(source) | 1,191 | 5,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-50 | latest | en | 0.890364 |
https://gis.stackexchange.com/questions/364349/morans-i-z-value-in-spdep | 1,726,430,207,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651647.78/warc/CC-MAIN-20240915184230-20240915214230-00452.warc.gz | 239,969,367 | 40,520 | # Moran's I z-value in spdep
I wish to incrementally find the neighbour distance which maximises Moran's I test statistic. I have read that the I values are not comparable across varying spatial weight matrices but that the z-value should be maximised instead. Where can I find/how can I compute the z-value in `lm.morantest` from `spdep` package? I don't see a z-value either in the value or in the structure of `lm.morantest` printed to the console (same applies to `moran.test` for a single variable):
``````library(spdep)
lm.morantest(lm(pat_ct~pub_ct, sca_ct), w_dist, alternative="two.sided")
#Global Moran I for regression residuals
#data:
#model: lm(formula = pat_ct ~ pub_ct, data = sca_ct)
#weights: w_dist
#Moran I statistic standard deviate = 12.155, p-value < 2.2e-16
#alternative hypothesis: two.sided
#sample estimates:
#Observed Moran I Expectation Variance
# 3.556003e-02 -8.665431e-05 8.600474e-06
str(lm.morantest(lm(pat_ct~pub_ct, sca_ct), w_dist, alternative="two.sided"))
#List of 6
# \$ statistic : num [1, 1] 12.2
# ..- attr(*, "names")= chr "Moran I statistic standard deviate"
# \$ p.value : num [1, 1] 5.39e-34
# \$ estimate : Named num [1:3] 3.56e-02 -8.67e-05 8.60e-06
# ..- attr(*, "names")= chr [1:3] "Observed Moran I" "Expectation" "Variance"
# \$ method : chr "Global Moran I for regression residuals"
# \$ alternative: chr "two.sided"
# \$ data.name : chr "\nmodel: lm(formula = pat_ct ~ pub_ct, data = sca_ct)\nweights: w_dist\n"
# - attr(*, "class")= chr "htest"
``````
I assume a minimal reproducible example is not required to answer this generic question.
• A reproducible example would still be useful, but we can use the one in `help(lm.morantest)`. Commented Jun 9, 2020 at 12:01
A "Z-score" (I wouldn't call it a Z-value) is the number of standard deviations that a statistic is away from its expected mean.
The ESRI documentation https://pro.arcgis.com/en/pro-app/tool-reference/spatial-statistics/h-how-spatial-autocorrelation-moran-s-i-spatial-st.htm shows the computation of Z_I.
For the first example in `help(lm.morantest)`, we see:
``````> m
Global Moran I for regression residuals
data:
model: lm(formula = CRIME ~ HOVAL + INC, data = COL.OLD)
weights: nb2listw(COL.nb, style = "W")
Moran I statistic standard deviate = 2.9539, p-value = 0.003138
alternative hypothesis: two.sided
sample estimates:
Observed Moran I Expectation Variance
0.235638354 -0.033302866 0.008289408
``````
And so we can compute Z_I from the ESRI formula and those three last printed values (ignore the "Observed Moran I" label - this is hung over from the name of the first element in the calculation):
``````> (m\$estimate["Observed Moran I"] - m\$estimate["Expectation"])/sqrt(m\$estimate["Variance"])
Observed Moran I
2.953899
``````
Hmmm that number looks familiar...
`````` Global Moran I for regression residuals
data:
model: lm(formula = CRIME ~ HOVAL + INC, data = COL.OLD)
weights: nb2listw(COL.nb, style = "W")
Moran I statistic standard deviate = ***--> 2.9539 <--***, p-value = 0.003138
alternative hypothesis: two.sided
``````
If you look at the code you'll even see:
``````ZI <- (I - EI)/sqrt(VI)
...
statistic <- ZI
attr(statistic, "names") <- "Moran I statistic standard deviate"
``````
It looks like Roger has preferred to call it the statistic standard deviate, even though its stored in a variable `ZI`, but I think its the same. There's probably some reason to not call it a Z-score related to the distribution of the statistic. | 1,055 | 3,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.778219 |
robertkaramagi.blogspot.com | 1,656,892,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104277498.71/warc/CC-MAIN-20220703225409-20220704015409-00567.warc.gz | 540,683,042 | 17,473 | ## Posts
### Calculus III - 3-Dimensional Space: Equations of Lines
In this section we need to take a look at the equation of a line in R 3 R 3 . As we saw in the previous section the equation y = m x + b y = m x + b does not describe a line in R 3 R 3 , instead it describes a plane. This doesn’t mean however that we can’t write down an equation for a line in 3-D space. We’re just going to need a new way of writing down the equation of a curve. So, before we get into the equations of lines we first need to briefly look at vector functions. We’re going to take a more in depth look at vector functions later. At this point all that we need to worry about is notational issues and how they can be used to give the equation of a curve. The best way to get an idea of what a vector function is and what its graph looks like is to look at an example. So, consider the following vector function. → r ( t ) = ⟨ t , 1 ⟩ r → ( t ) = ⟨ t , 1 ⟩ A vector function is a function that takes one or more variables, one in this case, and returns a vector. Note as we
Recent posts | 280 | 1,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | latest | en | 0.917951 |
https://web2.0calc.com/questions/i-need-some-help_16 | 1,610,766,917,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703499999.6/warc/CC-MAIN-20210116014637-20210116044637-00740.warc.gz | 629,852,868 | 5,379 | +0
# i need some help
0
152
1
Given 40 feet of fencing, what is the greatest possible number of square feet in the area of a rectangular pen enclosed by the fencing
Apr 30, 2020
#1
+114040
+1
The max area will be a square with a side of 40/4 = 10 ft
So....the max area =(10ft)^2 = 100 ft^2
Apr 30, 2020 | 115 | 319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-04 | latest | en | 0.778419 |
http://bookishways.blogspot.com/2013/01/100th-day-books-and-ideas.html | 1,469,674,151,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827782.42/warc/CC-MAIN-20160723071027-00161-ip-10-185-27-174.ec2.internal.warc.gz | 36,082,799 | 31,663 | Sunday, January 27, 2013
100th Day Books and Ideas
Yes, it's that time of year. The 100th day of school will soon be celebrated in classrooms around the country. There are many great web sites that suggest activities for the 100th day of school, but since I'm all about using children's literature, here are some terrific books for both the study of place value AND the 100th day.
I want to start by highlighting two books that I'm particularly fond of.
Let's Count to 100!, written and illustrated by Masayuki Sebe - Open this book to find a series of vibrantly colored, highly detailed double-page spreads with all kinds of things to count. From mice to moles, sheep to cats, readers are asked to count and search through the graphics on each page. Here's a sample.
(You can view two more spreads at Kids Can Press.) Not only is this book great for basic counting skills, but you can use these spreads with older students to find fractions, decimals and percents. They'll be just as tickled with the illustrations and search and find fun.
100 Ways to Celebrate 100 Days by Bruce Goldstone is useful for the study of number sense and offers suggestions for celebrating the 100th day of school that extend well beyond traditional counting activities.
As a writer and photo stylist, Goldstone has created an amazing array of images in support of his imaginative ideas. Here's an excerpt from the middle of the book.
Flip 100. (46) Flip a coin 100 times. How many times will it land heads up?
Clip 100. (47) Clip together 100 paper clips.
Clap 100. (48) Clap 100 times.
Tip 100. (49) Line up 100 dominoes and then tip them over.
The activities recommended by Goldstone can be easily carried out at home or in the classroom. They don't require fancy or expensive materials, will keep kids engaged, and provide a springboard for brainstorming additional activities involving 100.
Okay, on to the list! These are in alphabetical order and all offer great ways to think about 100 and/or the 100th day of school.
98, 99, 100! Ready or Not, Here I Come!, written by Teddy Slater and illustrated by Gioia Fiammenghi - This book in the Hello Math Reader series is about a group of girls playing hide-and-seek. In the course of their play they count to 100 by 5s, 10s, and 20s. Includes math activities by Marilyn Burns in the back matter.
100 Days of Cool, written by Stuart Murphy and illustrated by John Bendall-Brunello - In this MathStart book, 4 students dress up ("funky") for the first day of school thinking their teacher wants to celebrate cool, but she really wants to celebrate 100 days of school. She challenges them to keep it up for the next 99 days, but how will they do it? You can visit Stuart Murphy's web site to check out some suggested activities. (Scroll to the bottom of the page to find them.)
100 Days of School, written by Trudy Harris and illustrated by Beth Griffs Johnson - This book presents many different ways to count to 100. "If 99 dots are on a clown's suit, what do you get? 100 polka dots. Those . . . (on his clothes) plus 1 on his nose."
100 Monsters in My School, written by Bonnie Bader and illustrated by Bryan Hendrix - At Frank N. Stein Elementary School there are 100 monsters--25 witches, 25 werewolves, 25 vampires, and 25 ghosts. What items will they bring for the 100th day of school? And what will the very human Jane Brain bring for show and tell? This is a level 2 book in the All Aboard Math Reader series.
100 School Days, written by Anne Rockwell and Illustrated by Lizzy Rockwell - What happens when Mrs. Madoff's class counts a penny for each day they've been in school? Better yet, what happens when they get to 100? And what will they do with those pennies?
100th Day Worries, written by Margery Cuyler and illustrated by Arthur Howard - What's a young girl to do when she can't find a collection of 100 items to bring to school on the 100th day?
Centipede's One Hundred Shoes, written and illustrated by Tony Ross - Centipede goes to the shoe store for 100 shoes, 50 left and 50 right.
Counting Our Way to the 100th Day, written by Betsy Franco and illustrated by Steven Salerno - This book contains 100 poems for counting, math, school and everything in between. Each poem references the number 100.
Curious George Learns to Count From 1 to 100 by H. A. Rey - I'm a big fan of this little monkey. The man in the yellow hat challenges George to count to 100, so this book delivers many things to count, including a parade of ants, leaves blowing in the air, and rungs on a ladder.
Emily's First 100 Days of School by Rosemary Wells - Miss Cribbage tells her class that they will make a new number friend every day for the first 100 days of school, and everyone will have a number book in which to write their numbers and reflect on what they learn. From one to 100, Emily and From day one to day 100, Emily and her classmates expand their creative and mathematical skills as they immerse themselves in the exciting early days of school.
From One to One Hundred, written and illustrated by Terry Sloat - This terrific counting book counts from 1 to 10, and then by 10's to 100. The pictures are detailed and contain many things to count.
Happy 100th Day!, written by Susan Milord and illustrated by Mary Newell DePalma - Graham doesn't like school. The only thing he looks forward to is his class birthday party. What happens when Graham's special day falls on the 100th day of school? Will anyone remember his birthday?
Henry's 100 Days of Kindergarten, written and illustrated by Nancy Carlson - On the first day of school, Henry's Kindergarten teacher tells the class that every day she will add a jellybean to a jar, and when it reaches 100 jellybeans the class will have a party. On the 100th day, students bring sets of 100 items to celebrate. Henry's contribution is a little different, as he brings his Great-Grandma Millie, who is 100 years old.
I'll Teach My Dog 100 Words, written by Michael Frith and illustrated by P.D. Eastman - First published in 1973, I recall reading this repeatedly to a few of my babysitting charges (yes, I'm THAT old!). The story is still one of my favorites. Use this for a different way to count to 100 and count the words involved.
Jake's 100th Day of School, written by Lester Laminack and illustrated by Judy Love - While all of Jake's friends bring items to school (paperclips, bottle caps, etc.) for the 100th day, Jake forgets to bring the memory book into which he has glued 100 family pictures. With the help of his principal, he ends up with something to share, and something even more special on the 101st day.
Miss Bindergarten Celebrates the 100th Day of Kindergarten by Joseph Slate - On the eve of the 100th day of school, Miss Bindergarten asks her students to bring in "100 of some wonderful, one-hundred-full thing!" Find out what they bring and how the class celebrates. The illustrator, Ashely Wolff, has some ideas for celebrating the 100th day. Go on, take a look!
The Night Before the 100th Day of School, written by Natasha Wing and illustrated by Mindy Pierce - Here's a twist on The Night Before Christmas that looks at the night before the 100th day and a young boy who tries to figure out what to bring to school.
One Hundred Days (Plus One), written by Margaret McNamara and illustrated by Mike Gordon - Festivities are planned for the 100th day of school, but Hannah misses out when she's sick that day. What will the 101st day bring?
One Hundred Hungry Ants , written and illustrated by Eleanor Pinczes - Find out all the different ways a group of ants can arrange themselves while marching to a picnic.
One Hundred Is a Family by Pam Muñoz Ryan - This books looks at all the different ways people can come together to become "a family."
One Hundred Shoes: A Math Reader, written by Charles Ghigna and illustrated by Bob Staake - Centipede needs shoes for his feet. Where does he go? What does he choose? Written in rhyme, students will love the rhythm of the story. Along the way they'll learn about multiples of 100.
One Hundred Ways To Get To 100, written by Jerry Pallotta and illustrated by Rob Bolster - This title isn't in print any longer, but I can't take it off the list. If you can get your hands on this gem, don't ever let it go. This one teaches counting by 1s, 2s, 4s, 10s, 20s, and so on. Add to this some terrific illustrations with lots to count and you have a winner.
Ready, Set, 100th Day!, written and illustrated by Nancy Elizabeth Wallace - Minna is looking for the perfect set of materials to bring for her 100th day project. She tries out different sets of objects, arranging them in sets and playing with them. She tries popsicle sticks, pom-poms, pasta, paper clips and more. While Minna plays with sets, readers learn about different multiples of 100.
The Wolf's Chicken Stew, written and illustrated by Keiko Kasza - A hungry wolf delivers 100 pancakes, 100 donuts and a 100 pound cake to a chicken he wants to fatten up for a stew. When he arrives to retrieve the chicken, he gets a big surprise!
Once you've done some reading, why not create a class book for the 100th day? ReadWriteThink has a fabulous lesson on Descriptive Writing and the 100th Day of School. All students bring in a bottle containing 100 items. Along with this they write descriptive clues about the mystery items. Once the items have been revealed, students complete the pages of a book about the bottles and their contents.
Need more ideas? Check out my 100th Day Pinterest board.
Teaching Blog Addict has a 100th Day of School Ideas post where they are collecting a wide range of posts from classroom teachers .
In the states where the school year often begins after Labor Day, the 100th day generally falls close to Valentine's Day (snow days not withstanding), so you've got plenty of time to prepare. Here are some additional resources that will help you plan for the 100th day of school.
Do you celebrate the 100th day with some great books, poems or songs? I'll bet you do, so please share!
1 comment:
1. Thanks for the fabulous list! I'm going to head over to your Pinterest board. | 2,400 | 10,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-30 | longest | en | 0.959173 |
https://www.scribd.com/document/275490025/Operator-Splitting | 1,563,944,724,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195530385.82/warc/CC-MAIN-20190724041048-20190724063048-00462.warc.gz | 818,988,344 | 69,548 | You are on page 1of 26
Application of Operator Splitting to Solve
ReactionDiffusion Equations
arXiv:1011.4810v2 [math.NA] 22 Mar 2011
Szent Istvan University, Ybl Miklos College of Building
Abstract
Approximate solutions of the Fisher equation obtained by different splitting methods are investigated. The error of this nonlinear problem is analyzed. The order
of different splitting methods coupled with numerical methods of different order
is calculated numerically and symbolically.
Keywords: operator splitting, nonlinear PDEs,
1. Introduction
Splitting methods have been fruitfully used to solve large systems of partial
differential equations. To find the exact solution of a given problem in practice
is usually impossible. We can use numerical methods to obtain an approximate
solution of the equations, although the discretized model can be still very difficult to solve. Reaction-diffusion models or models of transport processes have a
structure that allows a natural decomposition of the equations, thus provide the
opportunity to apply operator splitting schemes. Splitting methods help us reduce
the complexity of the system and reduce computational time. With splitting it is
possible to handle stiff terms separately and to solve each subproblem with a suitable numerical method chosen to the corresponding operator. To solve a problem
in practice we use operator splitting and numerical schemes which we will call the
combined method. The use of operator splitting as well as the numerical methods
result in some error in the solution. The error generated purely by splitting is
called splitting error. This is the difference of the exact solution and the approximate solution obtained by splitting (assumed that we know the exact solutions
1 Research
done at the Department of Mathematical Analysis, Budapest University of Technology and Economics, Budapest, H-1111 Egry J. u. 1., HUNGARY
Preprint submitted to Computational and Mathematical Modeling
March 23, 2011
of the subproblems). Combined methods can generate both splitting error and numerical error. The study of this common effect on the solution is our main concern
in this paper. Detailed study on the interaction of operator splitting and numerical
schemes for linear problems can be found in [Csomos and Farago]. They classify the errors that can occur using splitting methods and numerical schemes, give
theoretical and numerical results on the order of the combined method for linear
problems. Our aim is to characterize the error of this combined method therefore
we calculate the order of the combined method for a nonlinear problem. We analyze the order of the error in the light of the characteristics of the splitting error
and the numerical error.
[Sanz-Serna] and [Lanser and Verwer] discuss the splitting error in a general
framework. The effect of operator splitting on the wave solutions of the Fisher
equation is investigated by [Simpson and Landman]. Our aim here is to rigorously analyze the interaction of splitting error and numerical error in the case of a
nonlinear problem: the Fisher equation. The structure of our paper is as follows.
In section 2 we introduce the basic idea of operator splitting in a general frame.
In Section 3 we introduce the Fisher equation and recall some known results on
it. We show how we apply splitting to solve the Fisher equation. In Section 4
the splitting error is analyzed for reaction-diffusion problems in full generality. In
Section 5 we calculate the order of the combined method for nonlinear problems
in general. Section 6 contains the numerical results on the Fisher equation.
2. Operator splitting
Let us consider the following abstract Cauchy problem:
U 0 (t) = A(U(t)) U(0) = U0
(1)
with U0 X arbitrary. The set X is usually a space of functions with certain
properties, U(t) X for every t > 0 and A : X X. Suppose that A can be written
as the sum of two operators: A = A1 + A2 .
The most simple type of operator splitting is the sequential splitting. In this
case the split problem is:
U10 (t) = A1 (U1 (t)) U1 (0) = U0
(2)
U20 (t) = A2 (U2 (t)) U2 (0) = U1 ().
(3)
The basic idea of splitting is to decompose the operator on the right hand side
into the sum of simpler operators, and to solve the subproblems corresponding to
the operators successively in each time step. More precisely, we solve the equation only with operator A1 until time (as if only the subprocess represented by A1
were present) and the solution in time will be the initial condition of the equation
with A2 . It means that we return to the initial time and solve the equation with A2
as well. The solution of the second equation in time is called the approximate
solution of the original problem in time . This procedure is then repeated on the
interval [, 2] etc. Thus, the simpler subproblems are connected to each other
through the initial conditions. It is clear, that the numerical treatment of the separate subproblems is simpler. The most significant advantage of splitting is that we
can exploit the special properties of the operators of the different subproblems and
apply the most suitable numerical method for each of them. Thus we can obtain a
more precise solution in a shorter time.
We remark that the method can be used fruitfully in large models, for example global models of air pollution transport, or combustion or metabolic models,
where the number of predicted variables is large and the number of the processes
represented in the models is large. We refer to three works on air pollution models
with application of operator splitting of [Lagzi et al.].
2.1. Splitting schemes
We can define the different splitting methods by solving the subproblems successively in different orders and for different time lengths. The above described
simplest scheme is called sequential splitting (SEQ). We solve the subproblems
one after another using the same time length , schematically S2 ()S1 (), where Si
is the corresponding solution operator. In MarchukStrang splitting (MS) we usually solve the subproblem with A1 with time length /2 then solve the other one
with A2 for time length and solve with A1 again for time length /2. Schematically S1 (/2)S2 ()S1 (/2). We usually chose A2 to be the operator representing chemical reactions. In general the operator that is stiff or nonlinear. In this
given order we only need to solve the second subproblem once which can be
of importance given the operators properties. In weighted sequential splitting
the solution in the next time step is a weighted average of the results of the two
possible sequential splittings S1 ()S2 () and S2 ()S1 (). In the special case of
symmetrically weighted splitting (SW) we take the arithmetic mean of the results:
(S1 ()S2 () + S2 ()S1 ())/2. The extra work with MS and SW splittings benefits in second order accuracy compared to the first order of SEQ splitting. The
nonsymmetric weighted splitting is of order one. In later sections we investigate
3
the SEQ, the MS and the SW splittings coupled with four different numerical
methods, all of different orders.
2.2. Splitting error, order of splitting
We perform a semidiscretization on (1) in an equidistant manner with time step
. If we know the exact solutions of (2) and (3) we can generate an approximate
solution to the original full problem (1) in which error originated only from operator splitting can arise. If we denote the exact solution by U and the approximate
solution by U then the local error of operator splitting is
E() := U() U()
Both solutions start from the common initial value and after time the difference
E() is called splitting error. Naturally the splitting error can be defined at any
+ ) is the
point of time during integration, if U(t) = U(t)
then U(t + ) U(t
local error at t.
For linear operators it is easy to show by Taylor expansion that the local order
of SEQ equals 1 since the error becomes E() = K 2 + O( 3 ). For nonlinear operators we need the definition of the Lie-operator and we can perform the analysis
with Taylor expansion using the Lie-operators. We refer to [Lanser and Verwer]
for detailed derivation of the nonlinear case. From the literature on operator splitting it is well known that the MS provides second order accuracy, so does the SW
splitting, see [Farago and Havasi].
2.3. Splitting of reactiondiffusion equations
In the case of reactiondiffusion equations there is a natural decomposition
of (1). The operator A1 represents the process of diffusion and A2 the chemical
reactions. In this case A1 is a linear and unbounded operator, A2 is usually a
nonlinear operator. If we have M species and N denotes the spatial dimension of
the problem then U C1 (R+ ,C2 (RN , RM )) that is for a given time point t R+
the function U maps the concentration of all species for every given point in space
RN , so it is a function of x RN . In other words U : t 7 (u1 (t, x), ..., uM (t, x)),
where ui (t, x) is the concentration of the ith species that is the spatial distribution
of the ith species. Then the function space X := C1 (R+ ,C2 (RN , RM )).
3. The Fisher equation
The Fisher equation is:
t u(t, x) = x2 u(t, x) + u(t, x)(1 u(t, x))
u(0, x) = (x).
4
x R,t > 0
(4)
There is only one chemical species present and one spatial variable here. This
equation was originally derived to describe the propagation of a gene in a population [Fisher]. It is one of the simplest nonlinear models for reaction-diffusion
equations. Such equations occur, e.g., in combustion, mass transfer, crystallization, plasma physics, and in general phase transition problems. See a discussion
and Toth] and [Murray]. For the initial conon reaction-diffusion models in [Erdi
dition:
1
u(0, x) =
(1 + k exp(x/ 6))2
wave form solution of the equation is known:
1
u(t, x) =
2
5
1
1 + k exp( 6 t + 6 6x)
and for:
u(0, x) =
u(t, x) =
,
(1 + k exp(x/ 6))2
(1 + k exp( 56 t 16 6x))2
We investigate three different splitting methods applied in the solution of this
equation. A natural way to split the Fisher equation is to decompose it into two
subproblems: one for the diffusion and one that corresponds to the reaction part
of the right hand side. Thus the definitions of the subproblems are:
t u1 (t, x) = x2 u1 (t, x)
(5)
u1 (0, x) = 1 (x)
t u2 (t, x) = u2 (t, x)(1 u2 (t, x))
(6)
u2 (0, x) = 2 (x),
where the initial condition 2 (x) = u1 (, x) connects the equations. The lower
indexes help distinguish the solutions of the different problems. We define the
operators A1 and A2 as follows: A1 u := x2 u, A2 (u) := u(1 u). These operators
are independent of time. Following the convention for linear operators we neglect
the parenthesis in A1 u, at the same time we would like to emphasize that A2 is
nonlinear: A2 (u).
1.0
t
0.5
0.0
1.5
conc. 1.0
0.5
0.0
0
5
x
10
9
Figure 1: The exact solution of (6), 2 (x) =
sin(x) + 1, t [0, 1] and x [0, 4]. The same
10
initial condition is used as in section 6.
The exact solution of problem (6) is known, it is:
u2 (t, x) =
Since
2 (x)et
.
1 2 (x) + 2 (x)et
lim u2 (t, x) =
(7)
1 when 2 (x) 6= 0
0 when 2 (x) = 0
the solution has two stationary states, namely: u2 (t, x) 0, u2 (t, x) 1. The
u2 (t, x) 1 solution is asymptotically stable, whereas zero is an unstable equilibrium.
Knowing the exact solution of this subproblem as a function of the initial
condition means that we can symbolically solve this subproblem in each time
step during the splitting procedure. It might be worth using the exact solution for
comparisons in the study of the effect of splitting. The exact solution of (5) is of
no real use.
4. Commutation of diffusion and reaction
In this section we investigate the conditions under which the SEQ has zero
splitting error for reaction-diffusion systems in general. In reaction-diffusion
equations there are two operators present on the right hand side, a linear and a
nonlinear operator. Let us consider
6
U 0 (t) = A(U(t)) + R(U(t)) U(0) = U0 ,
R+
0
R3
(8)
R+
0,m
where U :
X with X = { = (1 , ..., M ); m :
= 1, ..., M}.
We have M different species undergoing the processes of diffusion and reactions.
The operator A represents the diffusion assuming that there is no cross diffusion
present. In the above vectorial form A is an operator matrix essentially with the
spatial Laplacian in the diagonal and zeros everywhere else. A is a linear operator.
The operator R usually acts as compositions with the multivariable polynomials
R1 , R2 ..., RM , it is a nonlinear operator. That is
R() = (R1 (1 , ..., M ), . . . , RM (1 , ..., M ))
and
A() = A(1 , ..., M ) = (D1 41 , . . . , DM 4M ).
The global equation (8) in the local form is:
U 0 (t)(x) = A(U(t))(x) + R(U(t))(x) U(0)(x) = U0 (x),
Therefore, another form of (8) is
..
.
u (t, x) = D 4u (t, x) + R (u(t, x))
t M
M
M
M
(9)
with Dm > 0, m = 1, . . . , M, u() = (u1 (), ..., uM ()) and x R3 . Since A is linear
its derivative is A itself. The derivative of R := (R1 , ..., RM ) which is the Jacobi
matrix:
1 R1 . . . M R1
..
R0 = ...
.
1 RM . . . M RM
In the following we derive the condition of zero splitting error. The sufficient
condition of zero splitting error is [Hundsdorfer and Verwer] that for every X
A0 R R0 A () = 0
(10)
Upon applying this we get
M
0 = Dm 4Rm ((t, x)) k Rm ((t, x))Dm 4k (t, x)
k=1
For the sake of simplicity we will not carry the argument (t, x), it will no lead to
misconceptions.
M
0 = x k Rm ()x k + y k Rm ()y k + z k Rm ()z k
k=1
k=1
k=1
k Rm ()4k
m = 1, ..., M.
k=1
Each of the first three terms is a sum of products thus, according to rule of differentiation of a product one gets:
M
k=1
k=1
k=1
+ k Rm ()x2 k + k Rm ()y2 k + k Rm ()z2 k k Rm ()4k
k=1
k=1
k=1
k=1
Since the 4th, 5th, 6th and 7th terms eliminate each other, only the first three terms
remain. Upon performing the differentiation the above expression becomes:
M
j k Rm()x j x k +
k=1 j=1
j k Rm()y j yk +
k=1 j=1
j k Rm()z j zk
k=1 j=1
Using the notation: x := (x 1 , ..., , x M ) we can reformulate this as:
(R00m () x ) x + (R00m () y ) y + (R00m () z ) z =
=< R00m () x , x > + < R00m () y , y > + < R00m () z , z > .
The splitting error is zero if the above expression equals zero. Using the notation:
x 1 y 1 z 1
..
..
:= (x , y , z ) = ...
.
.
x M y M z M
8
we can write the formula above in the short form:
(R00m () )T = 0.
Now we can formulate a statement.
Theorem 1. With the notations above the error of SEQ is zero if for every function
and for all m = 1, 2, ..., M (R00m () )T = 0 holds.
Remark 1. We get the necessary condition of zero splitting error if we require
condition (10) to hold only for the solution function of (8). Naturally without
knowing the exact solution we can not check whether this holds or not. But the
above formula provides us a sufficient condition for zero splitting error. If the
equation holds for every possible function then it will hold for the solution as
well. This holds for every function iff all the entries of R00m is zero, which means that
Rm is a polynomial of at most first degree, m = 1, ..., M : we only have first order
reactions. Most of the practical problems have reaction terms of higher order,
therefore there is almost always a splitting error. Our aim here is to examine the
effect of splitting error in combined methods.
Remark 2. Condition (10) is sufficient in the case of MS and SW splittings as
well. Without going into the details condition (10) ensures that eA eR = e(A+R) ,
where A and R are the Lie-operators of A and R. In the case of MS we need
e 2 A eR e 2 A = e(A+R) which obviously holds if eA eR = e(A+R) . Theorem 1
remains valid in the case of MS and SW splittings.
5. Order of combined methods
When we solve partial differential equations we can use some kind of splitting but we can not avoid applying some numerical method as well. So in practice we use a combined method, a mixture of operator splitting and a numerical
scheme and generate a solution for a nonlinear partial differential equation like
(4). We use the Taylor-formula to determine the order of the local error of this
combined method. The Taylor-formula in normed vector spaces can be found in
e. g. [Komornik]. Here we recall the Taylor-formula for normed vector spaces:
Theorem 2. If f : X Y is n times differentiable in a X and h 0, then
n
f (a + h) =
k=0
f (k) (a) k
h + (h)khkn
k!
h0
Suppose that U is the solution of the equation:
U 0 (t) = AU(t) + R(U(t))
(11)
From now on in this section we restrict our investigations to problems where
A : X X is a bounded linear operator defined on the whole set X, therefore
the following derivations are not directly applicable to the case of diffusion. A is
differentiable and its derivative is A itself for every x X. The operator R : X X
will act as a composition with a differentiable nonlinear function thus R is a differentiable mapping as well. Based on (11) and the chain-rule U 0 is a differentiable
function and U 0 (t) X.
The Taylor-expansion of U in time t0 is
1
U := U(t0 + ) = U(t0 ) +U 0 (t0 ) + U 00 (t0 ) 2 + ()kk2 .
2
The norm we will neglect from now on since U : R X, denotes a positive real
number. U 0 (t0 ) is given by (11), we get U 00 (t0 ) by differentiation of (11):
U 00 (t0 ) = A0 (U(t0 )) U 0 (t0 ) + R0 (U(t0 ))U 0 (t0 ) = A(U 0 (t0 )) + R0 (U(t0 ))U 0 (t0 ) =
= A(A(U(t0 )) + R(U(t0 ))) + R0 (U(t0 ))A(U(t0 )) + R0 (U(t0 ))R(U(t0 )).
U(t0 ) is denoted by U0 . Using this notation if the value of U(t0 ) = U0 is known
then we can approximate U :
U = U0 + A(U0 ) + R(U0 ) +
(12)
1
A(A(U0 )) + A(R(U0 )) + R0 (U0 )A(U0 ) + R0 (U0 )R(U0 ) 2 + () 2 .
2
Beyond the conditions we already mentioned this form of U exists if U0 D(A2 )
and R(U0 ) D(A) which naturally hold if A : X X and R : X X. For the theorems in this section we will need R to be three times continuously differentiable.
+
5.1. Methods of first order
Theorem 3. The sequential splitting combined with the first order Euler forward
scheme provides a first order method.
The proof will be given in two steps.
10
5.1.1. Linear-Nonlinear
If we use SEQ starting with the nonlinear problem corresponding to R combined with Euler forward method for both subproblems we get:
V () = U0 + R(U0 )
U = V () + A(V ())
Proof. The approximation of the solution in time is:
U = V () + A(V ()) = U0 + R(U0 ) + A(U0 + R(U0 )).
Since A is linear we have:
U = U0 + R(U0 ) + A(U0 ) + 2 A(R(U0 )).
(13)
The local error generated in this step of length based on (12) and (13) is:
2
+ () 2 .
U U = A(A(U0 )) A(R(U0 )) + R0 (U0 )A(U0 ) + R0 (U0 )R(U0 )
2
5.1.2. Nonlinear-Linear
If we use SEQ starting with the linear problem corresponding to A combined
with Euler forward method for both subproblems we get:
V () = U0 + A(U0 )
U = V () + R(V ()).
Proof. The approximation of the solution in time is:
U = V () + R(V ()) = U0 + A(U0 ) + R(U0 + A(U0 )).
Let us define the function F : R X in the following way: F( ) := R(U0 +
A(U0 )). Then
F(0) = R(U0 ) and F(1) = R(U0 + A(U0 )),
and F is differentiable since it equals to R f with f ( ) := U0 + A(U0 ) (where
is differentiable), and according to the chain-rule
F 0 ( ) = R0 (U0 + A(U0 ))A(U0 ) F 00 ( ) = R00 (U0 + A(U0 ))(A(U0 ))2
and
F (n) ( ) = R(n) (U0 + A(U0 ))(A(U0 ))n
For the Taylor-expansion of F we need a similar but more specific relation.
11
Proposition 1. If F : R X is n times differentiable in every point of [a, b] then
there exists c (a, b) such that
n1
kF (n) (c)k
F (k) (a)
k6
(b a)n .
k!
n!
k=0
kF(b)
In other words
n1
F(b) =
kF (n) (c)k
F (k) (a)
+
(c)
(b a)n
1
k!
n!
k=0
with k1 (c)k 6 1. The expansion of F around 0 gives:
1
F(1) = F(0) + F 0 (0) + 1 (c) kF 00 (c)k
2
c [0, 1]
implying
1
R(U0 +A(U0 )) = R(U0 )+R0 (U0 )A(U0 )+1 (c) kR00 (U0 +cA(U0 ))(A(U0 ))2 k =
2
= R(U0 ) + R0 (U0 )A(U0 ) + (),
where
1
() = 1 (c) kR00 (U0 + cA(U0 ))A(U0 )2 k.
2
This tends to zero if tends to zero. This is the only relevant property of here.
Although c can change as changes but since k1 (c)k 6 1 we can ignore s
dependence on c through 1 .
U = U0 + A(U0 ) + R(U0 ) + R0 (U0 )A(U0 ) + ()kk =
= U0 + A(U0 ) + R(U0 ) + R0 (U0 ) 2 A(U0 ) + () 2 ,
U = U0 + (A(U0 ) + R(U0 ))+
2
+ A2 (U0 ) + A(R(U0 )) + R0 (U0 )A(U0 ) + R0 (U0 )R(U0 )
+ () 2 .
2
Comparing this with the approximation we get:
2
2
0
0
+ () 2 .
2
12
5.1.3. Weighted Splitting
Theorem 4. SW combined with the first order Euler forward method provides a
method of first order.
Proof. Here we simply use the above results with some [0, 1] parameter:
U = U0 + (R(U0 ) + A(U0 )) + 2 A(R(U0 )) +
1
+(1) U0 + (A(U0 ) + R(U0 ))+R0 (U0 ) 2 A(U0 )+R00 (U0 ) 3 (A(U0 ))2 +() 2 ,
2
U = U0 + (A(U0 ) + R(U0 ))+ A(R(U0 ))+(1)(R0 (U0 )A(U0 )) 2 +() 2 .
For the local error we have:
U U =
2
2
0
0
+
= A (U0 ) + (1 2)A(U0 )R(U0 ) + (2 1)R (U0 )A(U0 ) + R (U0 )R(U0 )
2
+() 2 ,
for =
1
we have
2
2
2
0
U U = A (U0 ) + R (U0 )R(U0 )
+ () 2 .
2
Conclusion: Although the SW is of second order its combination with the first
order Euler method provides only first order accuracy.
5.2. Methods of higher order
The above derivation can be used to determine the order of combined methods
with higher order numerical schemes. The method can be extended for schemes
of arbitrary order although the calculations become very complicated as the order
increases. As an example let us consider the improved Euler scheme which is of
second order and combine it with SEQ:
Theorem 5. The second order improved Euler scheme combined with SEQ provides a first order method.
Again, the proof will be given in two steps.
13
5.2.1. Nonlinear-Linear
V () = U0 + A(U0 + A(U0 ))
2
U = V () + R(V () + R(V ()))
2
Proof. The approximation of the solution is in time is:
U = V () + R V () + R(V ()) = U0 + A U0 + A(U0 ) +
2
2
+R U0 + A U0 + A(U0 ) + R U0 + A U0 + A(U0 )
=
2
2
2
2
= U0 + A(U0 ) + A2 (U0 )+
2
1
2
+R U0 + A(U0 ) + A2 (U0 ) + R U0 + A(U0 ) + A2 (U0 )
=
2
2
2
The underlined part is the coefficient of in the argument of F. The first order
Taylorexpansion gives:
= U0 + A(U0 ) +
2 2
A (U0 ) + R(U0 )+
2
1
2
+R0 (U0 ) A(U0 ) + A2 (U0 ) + R U0 + A(U0 ) + A2 (U0 ) + () 2 =
2
2
2
= U0 + A(U0 ) + R(U0 ) +
2 2
2 2
0
0
+
A (U0 ) + 2R (U0 )A(U0 ) + R (U0 )R U0 + A(U0 ) + A (U0 ) +
2
2
3 0
R (U0 )A2 (U0 ) + () 2 =
2
Taking the Taylorexpansion again, the coefficient of is the underlined part:
= U0 + A(U0 ) + R(U0 ) +
+
2 2
2
A (U0 )+2R0 (U0 )A(U0 )+R0 (U0 ) R(U0 )+R0 (U0 )A(U0 )+ R0 (U0 )A2 (U0 )) +
2
2
+() 2 =
14
2 2
0
0
= U0 + A(U0 )+R(U0 ) +
A (U0 )+2R (U0 )A(U0 )+R (U0 )R(U0 ) +() 2
2
2
U U = A(R(U0 )) R0 (U0 )A(U0 )
+ () 2 .
2
5.2.2. Linear-Nonlinear
The proof of the linear-nonlinear case is more straightforward:
Proof.
V () = U0 + R(U0 + R(U0 ))
2
U = V () + A(V () + A(V ()))
2
U = V () + A V () + A(V ()) =
2
= U0 + R U0 + R(U0 ) +
2
+A U0 + R U0 + R(U0 ) + A U0 + R U0 + R(U0 )
=
2
2
2
= U0 + R(U0 ) + R(U0 )R0 (U0 ) + A(U0 )+
2
1 2
2
0
+ A R(U0 ) + R(U0 )R (U0 ) + A (U0 ) + () 2 =
2
2
2
= U0 + R(U0 ) + A(U0 ) +
R(U0 )R0 (U0 ) + 2A(R(U0 )) + A2 (U0 ) + () 2
2
2
0
U U = R (U0 )A(U0 ) A(R(U0 ))
+ () 2 .
2
As we can see this combined method is of first order. Although the applied
numerical scheme ensures second order accuracy the use of sequential splitting
results in order reduction. We followed the same ideas and calculated the orders
for combinations of the introduced splitting methods and four different numerical
schemes. The table below contains our results on orders of different splittings
coupled with different numerical methods. Symbolic calculations on for example
MS splitting coupled with 4th order RungeKutta method becomes complicated.
An algorithm was written in Mathematica for these symbolic calculations.
15
SEQ (1)
SW (2)
MS (2)
exp. Euler (1) impr. Euler (2) Heun (3) RungeKutta (4)
1
1
1
1
1
2
2
2
1
2
2
2
Table 1: Local orders of combined methods for (11)
The order of the methods are in the parenthesis. A study of the order of combined methods for bounded linear problems can be found in [Csomos and Farago].
Their results say that the order of the combined method is s := min{p, r}, if p is
the order of splitting and r denotes the order of the numerical method. The numbers of the above table are in accordance with their results.
6. Numerical experiments
Here we introduce our numerical results on the Fisher equation. We solved
both subproblems (5), (6) using numerical methods of four different orders; the
explicit Euler, the improved Euler method which is of second order, the third
order Heun and the fourth order RungeKutta method. We investigate the SEQ,
SW and MS splitting methods which are of first and second orders. We calculated
the errors and orders of these combined methods numerically. Our test problem is
the following initialboundary value problem:
u(0, x) = 1 + 0.9 sin(x)
,
(14)
u(t, 0) = 1
u(t, 4) = 1
where x [0, 4] and t [0, 1].
We performed a spatial semidiscretization with length parameter x = 4
30 that
is we divided [0, 4] into N = 30 parts of equal length. Our tests showed that
finer divisions provides no significantly more accurate solutions that is the obtained error is of the same magnitude as with N = 30. We approximated the spatial
derivative with the well known second order scheme:
x2 u(t, xi )
.
x2
16
Figure 2: Reference solution generated by fourth order RungeKutta scheme, = 0.01.
After temporal discretization with = 0.01 we solved the full problem (14) with
the fourth order RungeKutta method. Taking a smaller time step resulted in
solution that differs only in a magnitude of 106 . This provides the reference
solution for our study. In the experiments we used spatial division N = 30 in
every case, in fact we investigated the convergence of the semidiscrete submodels
to the semidiscrete model: the reference solution. On the connection between the
convergence to a semidiscrete model and convergence to a continuous model see
6.1. Determination of the local order
The local error E() of a method is of order s if
E() = O( s+1 )
or by using a different formulation the order is s if
E()
0 s+1
lim
is finite where s N is the smallest number with this property. Thus in practice
we can estimate the order of the combined method by calculating this limit with
different fixed values of s until we find the appropriate one. For a fixed numerical
step size h the limit means naturally that h, therefore we choose h to be as
small as possible close to the smallest number that our computer can represent.
17
Another way to calculate the order of the method is the following. Considering
the two formula above we can conclude that
E()
c
s+1
for small -s, where c is a constant which does not depend on . We can take the
logarithm of both sides:
log E() (s + 1) log + log c.
This defines a straight line whose steepness gives the order of the method. The
table below contains results for the local order s. The order of splitting is p and
r is the order of the numerical scheme. The time step of the numerical method h
was the tenth of splitting time step .
s (h = 0.1)
SEQ (p = 1)
SW (p = 2)
MS (p = 2)
exp. Euler (r = 1) Heun (r = 3) RungeKutta (r = 4)
0.98
0.98
0.98
0.83
1.99
1.99
0.93
1.96
1.96
Table 2: Orders of combined methods for (14)
It is interesting that with the explicit Euler method increasing the order of
splitting does not improve the results.
6.2. Estimation of the global order
We used the time steps i in such a way that the evaluation of the error is
simple, since the corresponding division of the time interval is a subset of the one
of = 0.01.
1 2
0.2 0.1
3
4
0.0625 0.05
5
6
0.04 0.025
7
0.02
As for the error we know that:
E(1 ) c 1
18
for small -s, where c is a constant which does not depend on . So we can write:
E(2 ) c 2
E(1 )
E(2 )
1
2
We can take the logarithm of both sides:
log
For each i:
log
So
E(1 )
1
log .
E(2 )
2
E(i )
i
log
.
E(i+1 )
i+1
E(i )
log E(
)
i+1
log i+1
Evaluation of the left side shall give us the same value for every i = 1, 2, 3, 4, 5, 6.
The following table contains the results of this calculation for different splittings
and numerical methods.
(h = )
SEQ (p = 1)
SW (p = 2)
MS (p = 2)
r=1
1.04
1.02
1.02
r=2
0.99
2.07
2.07
r=3
1.08
2.01
1.95
r=4
1.08
1.98
1.998
Table 3: Orders of combined methods for (14)
Here we had the same experience as with the calculations shown in Table 2.
Since the solution of (6) is given in (7) as a function of the initial condition we can
use it in calculations instead of the numerical solution. The table below contains
results generated by using (7) in each time step.
The first column fits into the general scheme.
19
(h = )
SEQ (p = 1)
SW (p = 2)
MS (p = 2)
r=1
1.03
1.01
1.03.
r=2
1.02
2.06
2.00
r=3
1.01
1.95
1.99
r=4
1.01
1.98
1.99
6.3. Splitting into three operators
Above we saw that the splitting error is zero if the reaction term is of first
order. Another, natural decomposition of the right hand side of the Fisher equation
is when we separate the reaction part into two terms. The split problem is the
following:
t u2 (t, x) = u2 (t, x)
t u1 (t, x) = x2 u1 (t, x)
t u3 (t, x) = u23 (t, x)
u2 (0, x) = 2 (x)
u1 (0, x) = 1 (x)
u3 (0, x) = 3 (x)
(15)
2
2
Then A1 u = x u, A2 u = u, A3 (u) = u . Since A2 is a first order polynomial the
splitting error of the SEQ S1 S2 is zero according to theorem 1. It is easy to prove,
that it is not zero for S3 coupled with any of the other two operators.
s (h = )
SEQ (p = 1)
SW (p = 2)
MS (p = 2)
r=1
0.96
0.96
0.96
r=2
1.96
1.96
1.96
r=3
2.97
2.97
2.96
r=4
3.96
3.96
3.96
In the solution of subproblems associated with operator A2 and A3 we used the
exact solution. The first subproblem was solved numerically. Considering the MS
type splittings S3 S2 S1 S2 S3 and S2 S3 S1 S3 S2 it is reasonable to expect more accurate solutions with in the case S3 S2 S1 S2 S3 since the neighbors S1 and S2 generate
no splitting error. Whereas S2 S3 S1 S3 S2 generate splitting errors between every
neighboring operators. Figure 5 and figure 6 proves that although both MS provides the expected second order accuracy (combined with a third order or fourth
order numerical schemes) the hypothetic relation in accuracy turns out to be the
opposite.
20
order
1.005
1.000
0.995
0.990
0.02
0.04
0.06
Hi +i+1 L2
0.08
Figure 3: Order approximation of the S1 S2 S3 (squares) versus S1 S3 S2 (discs) SEQ type
splittings.
Generated by the fourth order Runge-Kutta scheme with time steps =
0.1, 0.0625, 0.05, 0.04, 0.025, 0.02.
order
1.005
1.000
0.995
0.990
0.985
1.1
1.2
1.3
1.4
1.5
1.6
1.7
i i+1
Figure 4: Order approximation of SEQ type splittings as the function of i /i+1 .
21
order
2.00
1.95
1.90
1.85
1.80
1.75
0.02
0.04
0.06
Hi +i+1 L2
0.08
Figure 5: Order approximation of the SW splitting which is the arithmetic mean of the six possible
sequential splittings. Generated by the fourth order Runge-Kutta scheme.
order
2.00
1.99
1.98
1.97
0.02
0.04
0.06
Hi +i+1 L2
0.08
Figure 6: Order approximation of the S3 S2 S1 S2 S3 (discs) versus S2 S3 S1 S3 S2 (squares) MS type
splittings. Generated by the third order Heun scheme.
7. Discussion and perspectives
We presented symbolic calculations for orders of PDE solving methods. Our
motivation is to predict the order in the case when beside numerical procedures of
certain order operator splitting is also used. We calculated the order of combined
methods applied for nonlinear PDE-s like (11), where a bounded linear operator
and a nonlinear operator is present. We presented numerical calculations on a
test problem with the diffusion operator which is an unbounded linear operator.
Although the results are in accordance with our theoretical results the methods
used in section 5 are strictly correct for the case of bounded linear operator. The
results of section 6 indicates that the combined method inherits the smaller one of
the order of the splitting and the numerical method, the extension of the methods
22
order
2.00
1.99
1.98
1.97
1.2
1.3
1.4
1.5
1.6
1.7
i i+1
order
2.0000
1.9995
1.9990
0.02
0.04
0.06
Hi +i+1 L2
0.08
Figure 8: Order approximation of the S3 S2 S1 S2 S3 (discs) versus S2 S3 S1 S3 S2 (squares) MS type
splittings. Generated by the fourth order Runge-Kutta scheme.
order
2.0000
1.9995
1.9990
1.9985
1.1
1.2
1.3
1.4
1.5
1.6
1.7
i i+1
23
used in section 5 to unbounded operators is not obvious. Our main focus is on
reaction-diffusion equations so we plan to find a method which allows us to repeat
the results of this paper for unbounded linear operators.
We also intend to extend the results of [Csomos and Farago] to nonlinear problems. Besides we are working on reaction-diffusion simulations on the sphere. In
the future we are going to apply the presented methods to practical problems e. g.
in the simulation of combustion.
8. Acknowledgements
This work is connected to the scientific program of the Development of
quality-oriented and harmonized R+D+I strategy and functional model at BME
project. This project is supported by the New Hungary Development Plan (Project
ID: TAMOP-4.2.1/B-09/1/KMR-2010-0002).
The present research has partially
been supported by the National Science Foundation, Hungary (No. K84060). I
thank to Istvan Farago for the fruitful discussions.
References
[Ablowitz and Zeppetella] Ablowitz, M. J.; Zeppetella, A.: Explicit solutions of
Fishers equation for a special wave speed, Bull. Math. Biol. 41, 835840,
1979.
[Csomos and Farago] Csomos, P., Farago, I.: Error analysis of the numerical
solution of split differential equations, Math. Comp. Mod., 48, Issue 78,
10901106, 2008.
and Toth] Erdi,
[Erdi
P., Toth, J., Ch. 6: Mathematical Models of Chemical Reactions, Princeton University Press, Princeton, N.J. 1989.
The mathematical background of op[Farago and Havasi] Farago, I., Havasi A.:
erator splitting and the effect of non-commutativity, LEcture Notes in Computer Science, Vol. 2179/2001, 26427, 2001.
[Fisher] Fisher, R. A.: The genetical theory of natural selection. Oxford University Press, Oxford, 1930.
[Hundsdorfer and Verwer] Hundsdorfer, W., Verwer, J. G.: Numerical solution
of the time-independent advection-diffusion-reaction equations, SpringerVerlag, Berlin, 2003.
24
[Ladics] Ladics, T.: Analysis of the splitting error for advection-reaction problems in air pollution models, Idojaras, Quarterly Journal of the Hungarian
Meteorological Service Vol. 109, No. 3, 173188, JulySeptember 2005.
[Lagzi et al.] Lagzi, I., Tomlin, A. S., Turanyi, T., Haszpra, L., Meszaros, R.,
Berzins, M.: Modeling Photochemical Air Pollution in Hungary Using an
Adaptive Grid Model, Air Pollution Modelling and Simulation, pp. 264273, editor: B. Sportisse, Springer, Berlin, 2002, ISBN 3-540-42515-2.
[Lagzi et al.] Lagzi, I., Tomlin, A. S., Turanyi, T., Haszpra, L., Meszaros, R.,
Berzins, M.: The Simulation of Photochemical Smog Episodes in Hungary and Central Europe Using Adaptive Gridding Models. Lecture Notes
in Computer Science, 2074, 67-76, 2001.
[Lagzi et al.] Lagzi, I., Tomlin, A. S., Turanyi, T., Haszpra, L.: Photochemical air
pollutant formation in Hungary using an adaptive gridding technique, Int. J.
Environment and Pollution, 36, 44-58, 2009.
[Lanser and Verwer] Lanser, D., Verwer, J. G.: Analysis of operator splitting for
advectiondiffusionreaction problems from air pollution modeling, Journal
of Computational and Applied Mathematics, Vol. 111 Issue 12, 201216,
Nov. 15 1999.
with Numerical Methods, SpringerVerlag Berlin Heidelberg 2003.
Marketing S. A. 2001. Hungarian edition Typotex, Budapest 2003.
[Marchuk] Marchuk, G. I.: Methods of splitting, Nauka, Moscow, 1988. (in Russian)
[Murray] : Mathematical Biology, Ch. 11. 3rd edition in 2 volumes: Mathematical Biology: I. An Introduction (551 pages) 2002; Mathematical Biology:
II. Spatial Models and Biomedical Applications (811 pages) 2003.
[Sanz-Serna] Sanz-Serna, J. M.:The State of the Art in Numerical Analysis, chapter Geometric Integration, 121143, Clarendon Press, Oxford, 1997.
25
[Simpson and Landman] Simpson, M. J., Landman, K. A.: Characterizing and
minimizing the split error for Fishers equation, Appl. Math. Lett., 19 (7),
604612, 2006.
[Sportisse] Sportisse, B.: An analysis of operator splitting techniques in the stiff
case. J. Comp. Phys. 161 (1), 140168, 2000.
[Strang] Strang, G.: On the construction of different splitting schemes, SIAM J.
Numer. Anal. 5 (3) 506517, 1968.
26 | 11,220 | 36,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-30 | latest | en | 0.908828 |
https://math.answers.com/Q/What_does_this_equal_6x-34x_plus_9 | 1,718,359,983,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00069.warc.gz | 356,659,264 | 47,277 | 0
# What does this equal 6x-34x plus 9?
Updated: 12/9/2022
Wiki User
9y ago
It can be simplified to: 9-28x
Wiki User
9y ago
Earn +20 pts
Q: What does this equal 6x-34x plus 9?
Submit
Still have questions?
Related questions
18
30! 29
6 + 9 + 8 = 23
7609
-7
### What does 28 plus 9 plus 32 plus 7 equal?
28 + 9 + 32 + 7 = 76
9+8=17.
8 + 9 = 17
### What does 9 plus d equal to?
ok you guys and hoes 9+d equal -5
### What is the answer to 9 plus 6x plus 2x?
There is insufficient information to answer this question. If you mean, what does x equal for: 9 + 6x = 2x ...then the answer is: 9 = 2x - 6x = -4x ...which is equal to: x = -9/4 If you mean, what does x equal for: 9 + 6x + 2x = 0 ...then the answer is: 8x = -9 ...which is equal to: x = -9/8
### What is positive 1 plus negative 10 equal?
1 plus negative 10 is equal to -9.
### What is 9 plus 11 plus 15 plus 18?
9 plus 11 (20) plus 15 (35) plus 18 is equal to 53 | 360 | 943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-26 | latest | en | 0.888269 |
https://www.coursehero.com/file/6327954/lec0201/ | 1,513,388,915,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948581033.57/warc/CC-MAIN-20171216010725-20171216032725-00046.warc.gz | 739,159,729 | 75,777 | # lec0201 - The Inclusion-Exclusion Principle Let A and B...
This preview shows pages 1–3. Sign up to view the full content.
The Inclusion-Exclusion Principle Let A and B denote two finite sets. Then, we have: | A B| = |A| + |B| – | A B |. Since each element of A B belongs to either A or B , the sum | A| + |B| includes a count for each of the elements of A B , but those elements of A B are counted twice. Thus, |A| + |B| – | A B | counts each element of A B exactly once, that is, it is equal to | A B| .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
More precisely, we first claim that the following is a disjoint union: A = ( A B ) ( A B ) (1) Thus, by the definition of set equality, we want to prove that A ( A B ) ( A B ) (2) ( A B ) ( A B ) A (3) and ( A B ) ( A B ) = (4) To prove (2), let x A . Since either x B or x B is true: in the former case, x A B by definition, and in the latter case, we have x A and x B , which means x A B by definition. To prove (3), note that A B A because each x A B must also have x A by the definition of set difference. Also, A B A because each x A B must also have x A by the definition of intersection. Thus, ( A B ) ( A B ) A by the definition of set union and the subset relationship. To prove (4), note that each x A B must satisfy x B by the definition of set difference. Also, each x A B must satisfy x B , by the definition of set intersection. Thus, it is
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 8
lec0201 - The Inclusion-Exclusion Principle Let A and B...
This preview shows document pages 1 - 3. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 497 | 1,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-51 | latest | en | 0.907332 |
https://homeapricot.com/how-many-amps-and-watts-does-deep-fryer-use/ | 1,726,298,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00628.warc.gz | 277,591,931 | 16,378 | # How Many Amps and Watts Does a Deep Fryer Use? (Answered)
A deep fryer is one of the most convenient gadgets for frying in your kitchen. You can fry many foods in a short time using a deep fryer. So, deep fryers are being demandable day by day.
Deep fryers work by electricity. So, when it comes to using an electrical device, you might want to know about the amount of electricity that device uses. Generally, you might need a pretty short time to cook in a deep fryer.
But you have to know how many amps and watts a deep fryer uses to know the efficiency of the deep fryer.
### How many amps and watts does a deep fryer use?
Generally, a deep fryer uses at least 15 amps and 1500 watts to work properly. Deep fryers can use a different amount of amps and watts according to the intensity of power. Sometimes, a deep fryer can use more than 1700 watts and over 15 amps. But low-powered deep fryers use fewer amps and watts.
## Does deep fat fryer use a lot of electricity?
Deep fat fryers use enough electricity. But this electricity isn’t huge if you consider the average cost. Generally, a deep fat fryer uses 1.8 kWh per hour of use. A commercial fat fryer might use 3 kWh per hour.
This electricity might be more than the oven or other cooking devices. But if you consider the timing of cooking, the electricity usage isn’t that much.
If you use the deep fryer every day, the deep fryer might cost 15 percent of the electricity bill. But the use of deep fryers isn’t that much to cause a lot of electricity loss. So, a deep fryer doesn’t use a lot of electricity but a moderate level of electricity.
## How many amps does a deep fryer use?
How many amps a deep fryer will use depends on the power of the deep fryer. If you use a high-powered deep fryer that uses more than 1700 watts, the fryer will use more than 15 amps.
But if you use a smaller unit deep fryer, the fryer will use 9-10 amps. So, a deep fryer might use 9 to 15 amps generally.
## How many watts does a deep fryer use?
A deep fryer uses watts according to its temperature difference. The more the temperature is, the more the watts will be. Generally, a deep fryer uses 1500 watts if it is at a constant temperature.
But if the temperature varies during cooking, the watts might also change. If you raise the temperature of the deep fryer, the watts can go up to 1700 watts.
## How many watts does a Fry Daddy use?
A Fry Daddy is a type of deep fryer. So, it will also use the same range of watts while operating.
Generally, a Fry Daddy uses 1200 watts. This range is pretty smaller than the regular deep fryers. So, a Fry Daddy is pretty convenient and an electricity saver. So, a Fry Daddy is energy efficient.
## How much does it cost to run a deep fryer?
Generally, a deep fryer is energy-efficient. But it still costs you at the end of the year. A regular deep fryer uses up to 1.8 kWh per one hour of usage. This amount of kWh will cost you up to 0.22 cents.
So, if you use a deep fryer for one hour a day, it will cost you \$6.6 per month.
\$6.6 in a month is not a little amount. It seems that this cost will cause a 15% increase in your electricity bill. This cost might increase if you use a deep fryer regularly. So, your cost might increase due to using a deep fryer every day.
## Can a generator run a deep fryer?
A generator can run a deep fryer if the deep fryer is low-powered. But you should not run a deep fryer by a generator.
Generally, a generator can go up to 1800 watts. But you might know that a deep fryer uses more than 1500 watts. Sometimes, a deep fryer can use more than 1700 watts. So, if you run a deep fryer by a generator, you are supposed not to use any other electric device.
Moreover, if you want to connect devices with a generator that are more power-consuming, the generator can get damaged. So, you should avoid using deep fryers by a generator.
## What is the voltage of a deep fryer?
A deep fryer is generally 220 volts. 220 volts is enough and safe for a deep fryer.
Generally, most electric appliances use 120 volts in the United States. But some larger and more power-consuming devices use 220 volts. The more the volts are, the more power the device will consume.
So, a deep fryer is a pretty power-consuming device. But you have to use the deep fryer carefully because a higher voltage appliance will flow more currents. Moreover, it will cause heating of the electrical device.
## Are electric deep fryers dangerous?
Deep fryers are not dangerous if you use them carefully. But they can be dangerous if anything is wrong with the devices or you use them inappropriately.
Like other electric devices, deep fryers are risky if the connecting wires have defects. Moreover, deep fryers work at very high temperatures. So, touching any inside part of the deep fryer might cause burning.
Besides, deep fryers use very high voltage, this voltage allows the fryer to run at a very high current. So, there is a chance of shocking if you don’t use it properly. If you raise the temperature high continuously without concern, there is a risk of cooking fire.
## What is better electric or gas deep fryers?
If you think about the efficiency of the deep fryer, the electric one is better. Electric deep fryers cook efficiently in a short time than gas deep fryers. But if you think about heating speed, gas deep fryers heat at a higher speed.
If you think about the cost and portability, electric deep fryers are efficient. You can take electric deep fryers everywhere. But the natural gas line will not be available anywhere. Moreover, the recent high cost of natural gas makes it expensive to use gas deep fryers.
## Final Thoughts:
Generally, deep fryers use a sufficient amount of watts and amps like other large electrical devices. They can use up to 1700 watts and 15 amps. So, the electricity usage is pretty high. They can cost you like other electric devices in your house. So, try to use deep fryers when necessary. | 1,395 | 5,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.919027 |
http://se.mathworks.com/help/aeroblks/threeaxisgyroscope.html?s_tid=gn_loc_drop&requestedDomain=se.mathworks.com&nocookie=true | 1,475,094,674,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661767.35/warc/CC-MAIN-20160924173741-00169-ip-10-143-35-109.ec2.internal.warc.gz | 249,318,105 | 11,141 | # Three-Axis Gyroscope
Implement three-axis gyroscope
## Description
The Three-Axis Gyroscope block implements a gyroscope on each of the three axes. The measured body angular rates $\left({\overline{\omega }}_{meas}\right)$ include the body angular rates $\left({\overline{\omega }}_{b}\right)$, errors, and optionally discretizations and nonlinearizations of the signals.
`${\overline{\omega }}_{meas}={\overline{\omega }}_{b}×{\overline{\omega }}_{SFCC}+{\overline{\omega }}_{bias}+Gs×{\overline{\omega }}_{gsens}+noise$`
where ${\overline{\omega }}_{SFCC}$ is a 3-by-3 matrix of scaling factors on the diagonal and misalignment terms in the nondiagonal, ${\overline{\omega }}_{bias}$ are the biases, (Gs) are the Gs on the gyroscope, and ${\overline{\omega }}_{gsens}$ are the g-sensitive biases.
Optionally, discretizations can be applied to the block inputs and dynamics along with nonlinearizations of the measured body angular rates via a Saturation block.
## Parameters
Second order dynamics
Select to apply second-order dynamics to gyroscope readings.
The natural frequency of the gyroscope. The units of natural frequency are radians per second.
Damping ratio
The damping ratio of the gyroscope. A dimensionless parameter.
Scale factors and cross-coupling
The 3-by-3 matrix used to skew the gyroscope from body axes and to scale angular rates along body axes.
Measurement bias
The three-element vector containing long-term biases along the gyroscope axes. The units are in radians per second.
G-sensitive bias
The three-element vector contains the maximum change in rates due to linear acceleration. The units are in radians per second per g-unit.
Update rate (sec)
Specify the update rate of the gyroscope. An update rate of 0 will create a continuous gyroscope. If noise is selected and the update rate is 0, then the noise will be updated at the rate of 0.1. The units of update rate are seconds.
If you:
• Update this parameter value to 0 (continuous)
• Configure a fixed-step solver for the model
• Do not have a Control System Toolbox™ license
you must also select the Automatically handle rate transition for data transfer check box in the Solver pane. This check box enables the software to handle rate transitions correctly.
Noise on
Select to apply white noise to gyroscope readings.
Noise seeds
The scalar seeds for the Gaussian noise generator for each axis of the gyroscope.
Noise power
The height of the PSD of the white noise for each axis of the gyroscope. The units are (rad/s)2/Hz.
Lower and upper output limits
The six-element vector containing three minimum values and three maximum values of angular rates in each of the gyroscope axes. The units are in radians per second.
## Inputs and Outputs
InputDimension TypeDescription
First
Three-element vectorContains the angular rates in body-fixed axes, in radians per second.
Second
Three-element vectorContains the accelerations in body-fixed axes, in Gs.
OutputDimension TypeDescription
First
Three-element vectorContains the measured angular rates from the gyroscope, in radians per second.
## Assumptions and Limitations
Anisoelastic bias and anisoinertial bias effects are not accounted for in this block. Additionally, this block is not intended to model the internal dynamics of different forms of the instrument.
Note: This block requires the Control System Toolbox product for discrete operation (nonzero sample time).
## Reference
Rogers, R. M., Applied Mathematics in Integrated Navigation Systems, AIAA Education Series, 2000. | 793 | 3,567 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-40 | latest | en | 0.77118 |
https://www.numbersaplenty.com/44747786 | 1,709,407,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00539.warc.gz | 898,647,065 | 3,040 | Search a number
44747786 = 222373893
BaseRepresentation
bin1010101010110…
…0110000001010
310010012102102102
42222230300022
542423412121
64235033402
71052231006
oct252546012
9103172372
1044747786
1123293746
1212b9b862
1393698a5
145d2b706
153ddd90b
hex2aacc0a
44747786 has 4 divisors (see below), whose sum is σ = 67121682. Its totient is φ = 22373892.
The previous prime is 44747779. The next prime is 44747861. The reversal of 44747786 is 68774744.
It is a semiprime because it is the product of two primes.
It can be written as a sum of positive squares in only one way, i.e., 39929761 + 4818025 = 6319^2 + 2195^2 .
It is a Curzon number.
It is a self number, because there is not a number n which added to its sum of digits gives 44747786.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 11186945 + ... + 11186948.
Almost surely, 244747786 is an apocalyptic number.
44747786 is a deficient number, since it is larger than the sum of its proper divisors (22373896).
44747786 is a wasteful number, since it uses less digits than its factorization.
44747786 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 22373895.
The product of its digits is 1053696, while the sum is 47.
The square root of 44747786 is about 6689.3785959534. The cubic root of 44747786 is about 355.0235678329.
The spelling of 44747786 in words is "forty-four million, seven hundred forty-seven thousand, seven hundred eighty-six".
Divisors: 1 2 22373893 44747786 | 511 | 1,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-10 | latest | en | 0.870327 |
http://mathhelpforum.com/differential-geometry/114995-convergence-l1-characteristic-function-print.html | 1,527,096,011,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865691.44/warc/CC-MAIN-20180523161206-20180523181206-00303.warc.gz | 186,662,575 | 2,634 | # Convergence in L1 and characteristic function
• Nov 16th 2009, 02:51 PM
Convergence in L1 and characteristic function
If $\displaystyle \mu (E_n) < \infty$ for $\displaystyle n \in \mathbb {N}$, and $\displaystyle 1_{E_n} \rightarrow f$ in $\displaystyle L^1$, then f equals to the characteristic function of a measurable set.
Proof so far.
So I know that $\displaystyle \int \mid 1_{E_n} - f \mid d \mu \rightarrow 0$
And I need to find a measurable set, say M, such that $\displaystyle f=1_M$
I apologize for the lack of significant amount of work here, but I just don't really know how to start here, any hints?
Thank you.
• Nov 18th 2009, 02:21 AM
Shanks
Actually , I think , f equals to the characteristic function of a measurable set, a. e.
Hint :convergence in $\displaystyle L^1$ implies convergence in measure, and convergence in measure implies a subsequence convergence a. e. | 251 | 894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-22 | latest | en | 0.80971 |
https://www.thestudentroom.co.uk/showthread.php?t=5223770 | 1,531,911,237,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590127.2/warc/CC-MAIN-20180718095959-20180718115959-00436.warc.gz | 967,805,687 | 37,469 | You are Here: Home >< GCSEs
# Maths watch
1. y is directly proportional to x^2.
When x = 3, then y = 36.
a) Express y in terms of x.
z is inversely proportional to x.
When x = 4, z = 2.
b) Show that z = c y^n , where c and n are numbers and c > 0.
You must find the values of c and n.
2. (Original post by choiheesung97)
y is directly proportional to x^2.
When x = 3, then y = 36.
a) Express y in terms of x.
z is inversely proportional to x.
When x = 4, z = 2.
b) Show that z = c y^n , where c and n are numbers and c > 0.
You must find the values of c and n.
a) y = kx^2
36 = (3^2)k
Now you just need to find k and substitute it back into the equation in the first line of my working out.
Spoiler:
Show
k= 36/9 =4
y = 4x^2
b) z = k/x
2 = k/4
Now find k
Spoiler:
Show
k= 2 x 4 = 8
Rearrange and substitute the equations into each other until they are in the form z = cy^n
z= 8/x
x = 8/z
Spoiler:
Show
x^2 = 64/z^2
y= 4(64/z^2)
y= 256/z^2
z^2 = 256/y
z = square root (256/y)
z = 16/ (y^0.5)
z = 16 y^(-0.5)
c = 16
y = -0.5
I have probably make a silly mistake somewhere but I enjoyed this question
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: February 27, 2018
Today on TSR
### He broke up with me because of long distance
Now I'm moving to his city
Poll
Useful resources
## Study tools
### Essay expert
Learn to write like a pro with our ultimate essay guide.
See where you can apply with our uni match tool
### Make study resources
Create all the resources you need to get the grades.
### Create your own Study Plan
Organise all your homework and exams so you never miss another deadline.
### Resources by subject
From flashcards to mind maps; there's everything you need for all of your GCSE subjects.
### Find past papers
100s of GCSE past papers for all your subjects at your fingertips. | 615 | 2,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-30 | longest | en | 0.894 |
https://www.secretsofuniverse.in/ic-1101-biggest-known-galaxy/ | 1,604,168,674,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00714.warc.gz | 907,400,834 | 165,311 | # Meet IC 1101: The Biggest Known Galaxy In The Universe.
Galaxies are cool. Everybody knows that. It is the place that keeps around billions of stars. It’s the place where stars like our Sun live, and where planets like Earth with people like us think and study the world. There are billions of galaxies in our Universe of different shapes and sizes. Everybody should know our 13.5 billion years-old Milky Way. Also, you may have heard of Andromeda or M31. It is the galaxy closest to us, our neighbor.
Actually, Andromeda likes the Milky Way so much it’s coming closer to it every day. They are really going to meet each other, in 4.5 billion years or so. So these are the most known galaxies, right? But what about the biggest? The “biggest of the big”? The name of that beast is IC 1101. As I said, galaxies come in all sizes. But to understand what sizes we are talking about, take some time and imagine it.
### Quantum Mechanics Video Series
We have started a video series on Quantum Mechanics on our YouTube channel. The series will explain the concepts of quantum mechanics along with the mathematics of the subject. Make sure you subscribe to our YouTube channel to get the notifications of the series.
## Understanding distances
If something is located 1 light-year away from us, it means that the light from it will reach us in one year. That means that if looking for an object one light-years away, we will always see it as it was one year ago. Light travels pretty fast. Somewhere around 300,000,000 m/s. The average speed of a car is 25 m/s. It means light travels 12 million times faster than a car.
Now think about our galaxy, whose diameter is 105,000 light-years. Sounds big. But the biggest we know about is estimated to be around 6 million light-years in diameter. That means that if you would look from one corner to the star on the opposite, it would take 6 million years for its light to reach you.
Actually, saying that IC 1101 is the largest galaxy in the Universe is not accurate. Considering how important accuracy is in the world of science, I am impressed by how much I see this mistake made. IC 1101 is the biggest galaxy discovered. It doesn’t mean there isn’t something bigger. It’s just that we didn’t see it yet.
## What types of galaxies exist?
IC 1101 is an elliptical galaxy. There are 3 major types of galaxies, subdivided then in some smaller ones. The author of this classification is the brilliant astronomer of the 1920s, Edwin Hubble. The 3 major types are elliptical, spiral, and irregular. Spiral galaxies are the most common in the Universe. The Milky Way or Andromeda are spiral galaxies. You can read more about the galaxies and their interactions in this article of our Basics of Astrophysics series.
These are also the galaxies with the youngest stars and which give birth to most of the stars. On the other hand, elliptical galaxies are much older and are home to old stars too. These elliptical galaxies are found largely in galaxy clusters and dust clouds. Now irregular galaxies, such as the Small Magellanic Cloud, are as their name suggests, irregular in shape. These are even older galaxies.
## Using “candles”
So how was this 6 million light-years long elliptical galaxy discovered? Astronomers can calculate short distances easily, using a method called parallax. This works well for a lot of stars. But when talking about distances such as 6 million light-years, we use something different: we calculate their distance based on their brightness. But in such a calculation we would need something to compare it to. That is why we use “candles”, galaxies or stars that have become standard to us, we know their distance, and hence we compute the distance to other objects in the Universe.
IC 1101 was first discovered by William Herschel, in 1870 or so. It was first thought to be just a nebula. Everybody believed that until Hubble came in the 1920s and proved, with his new classification of galaxies, that IC 1101 is a galaxy: a big one actually.
So in a Universe with galaxies like IC 1101, it’s easy to feel small. In fact, some psychologists found out that the biggest feelings of insignificance and loneliness come from the people who came in contact with the Universe in some way.
Mostly, you can find these feelings at the observatory, that being the place where psychologists found their people. But instead of feeling small, and lonely, or insignificant, you may want to feel proud that you are part of such a beautiful species, that you live on such a nice planet, and most of all, that you can contemplate at who you are and why are you here.
Read all the articles of Basics of Astrophysics series here
Scroll to Top | 1,036 | 4,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-45 | longest | en | 0.963459 |
http://forum.allaboutcircuits.com/threads/available-power-in-mosfet.87312/ | 1,485,162,140,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282202.61/warc/CC-MAIN-20170116095122-00196-ip-10-171-10-70.ec2.internal.warc.gz | 104,259,379 | 14,723 | Available power in MOSFET
Discussion in 'General Electronics Chat' started by faisaldj, Jul 16, 2013.
1. faisaldj Thread Starter New Member
May 16, 2008
6
1
Hi,
I am designing a relay which is based on MOSFET, this relay has 14 output.
The relay is being operated on 5V dc supply.The supply is being provided to the board by using a MOSFET, the attachment shows the power supply section and one of the output ciruity from the available 14 output.
TPC8124, Now I want to calculate the total available power that can be obtained through this relay board:
As I am given the input 5v supply to this transistor (TPC8124) which is further supplying this 5v to the rest of the board, the datasheet of this transistor says that Drain Current (DC) is -12 A, as my input supply voltage is 5v so I calculate that total output power that be get through this board is 60W.
is this a right estimation or I am missing something ??
The datasheet is available on the following link
http://www.toshiba.com/taec/components2/Datasheet_Sync/200912/DST_TPC8124-TDE_EN_11751.pdf
with regards,
File size:
46 KB
Views:
32
2. wayneh Expert
Sep 9, 2010
12,396
3,246
Heat. The datasheet says the maximum Rds could be 10mΩ. At 12A current, that's 1.44W of heat that needs to be dissipated by the MOSFET. Without a heatsink or other measures, I think that would be very likely to fail.
Running any component up to its max rating risks imminent failure.
3. crutschow Expert
Mar 14, 2008
13,509
3,385
60W would be the power delivered to the load.
As wayneh noted, you shouldn't operate the device at it's Maximum Rating for best reliability. 75% of this or 9A maximum would be a good derated value.
At 9A the maximum dissipation when ON would be 810mW. To minimize the temperature rise you should mount each transistor on a one square-inch minimum PCB copper plane as shown in Note 2(a) on page 2 of the data sheet.
Edit: R51 should be a small value or zero ohms to insure the transistor is fully turned on.
What is the purpose of Q30 in the power supply section? It is wired so it is always ON.
4. LDC3 Active Member
Apr 27, 2013
920
160
Q30 is used to isolate the rest of the circuit in case the voltage supply is connected incorrectly. It will certainly heat up if all the current (14A) needs to go through it. | 615 | 2,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-04 | latest | en | 0.928855 |
https://calckit.io/tool/conversion-electric-current | 1,721,241,187,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00329.warc.gz | 133,732,822 | 6,412 | # Electric Current Converter
Electric current is a fundamental concept in electrical engineering and physics that measures the flow of electric charge through a circuit. It is an essential quantity in understanding and analyzing electrical systems. Current is denoted by the letter "I" and is measured in Amperes (A). One Ampere is defined as one Coulomb of charge passing through a point in a circuit per second. In simpler terms, it represents the rate of flow of electric charge through a conductor.
• Ampere (A): Ampere is the base unit of electric current in the International System of Units (SI). One ampere is equal to 1 coulomb, or 6.241509074 10^18 electrons' worth of charge, moving past a point in one second.
Conversion Factor: 1 A = 1 A (Base Unit)*
• Milliampere (mA): Milliampere is a unit of electric current equal to one-thousandth of an Ampere.
Conversion Factor: 1 mA = 0.001 A
• Microampere (μA): Microampere is a unit of electric current equal to one-millionth of an Ampere.
Conversion Factor: 1 μA = 0.000001 A
• Kiloampere (kA): Kiloampere is a unit of electric current equal to one thousand Amperes.
Conversion Factor: 1 kA = 1000 A
• Biot (Bi): The biot, also called abampere, is the derived electromagnetic unit of electric current in the emu-cgs system of units. One abampere corresponds to ten amperes in the SI system of units.
Conversion Factor: 1 Bi ≈ 10 A
• Abampere (abA): Abampere is another name for the biot unit.
Conversion Factor: 1 abA = 1 Bi = 10 A
• EMU of current (emu): Electromagnetic unit of current is another name for the abampere.
Conversion Factor: 1 emu = 1 abA = 1 Bi ≈ 10 A
• ESU of current (esu): Electrostatic unit of current, also known as statampere, is the charge unit in the cgs (centimeter-gram-second) electrostatic units variant and Gaussian systems of units.
Conversion Factor: 1 esu ≈ 3.33564 * 10^-10 A
• Statampere (statA): The statampere is another name for the electrostatic unit of current.
Conversion Factor: 1 statA = 1 esu ≈ 3.33564 * 10^-10 A
Understanding electric current and its various units is crucial for comprehending electrical systems and designing circuits. The Current Converter Tool simplifies the process of converting current measurements between different units, enabling engineers and scientists to work seamlessly across various unit systems. | 599 | 2,343 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-30 | latest | en | 0.851463 |
http://www.jiskha.com/display.cgi?id=1295473401 | 1,462,528,068,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861743914.17/warc/CC-MAIN-20160428164223-00014-ip-10-239-7-51.ec2.internal.warc.gz | 606,883,749 | 3,812 | Friday
May 6, 2016
# Homework Help: math
Posted by julian on Wednesday, January 19, 2011 at 4:43pm.
1500cm= mm
72 km= m
38000 cm= m
42 m= mm
986 cm= mm
48000 mm= m
5000 m= km
5280 cm= mm
• math - Ms. Sue, Wednesday, January 19, 2011 at 4:53pm
1 cm = 10 mm
1/10 = 1500/x
Cross multiply and solve for x | 128 | 304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2016-18 | longest | en | 0.877747 |
https://www.pydoc.io/pypi/scipy-1.0.1/autoapi/signal/lti_conversion/index.html | 1,558,752,825,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257847.56/warc/CC-MAIN-20190525024710-20190525050710-00365.warc.gz | 900,162,389 | 11,959 | `signal.lti_conversion`¶
ltisys – a collection of functions to convert linear time invariant systems from one representation to another.
Module Contents¶
Functions¶
`tf2ss`(num,den) rTransfer function to state-space representation. `_none_to_empty_2d`(arg) `_atleast_2d_or_none`(arg) `_shape_or_none`(M) `_choice_not_none`(*args) `_restore`(M,shape) `abcd_normalize`(A=None,B=None,C=None,D=None) Check state-space matrices and ensure they are two-dimensional. `ss2tf`(A,B,C,D,input=0) rState-space to transfer function. `zpk2ss`(z,p,k) Zero-pole-gain representation to state-space representation `ss2zpk`(A,B,C,D,input=0) State-space representation to zero-pole-gain representation. `cont2discrete`(system,dt,method=”zoh”,alpha=None) Transform a continuous to a discrete state-space system.
`tf2ss`(num, den)
rTransfer function to state-space representation.
num, den : array_like
Sequences representing the coefficients of the numerator and denominator polynomials, in order of descending degree. The denominator needs to be at least as long as the numerator.
A, B, C, D : ndarray
State space representation of the system, in controller canonical form.
Convert the transfer function:
```>>> num = [1, 3, 3]
>>> den = [1, 2, 1]
```
to the state-space representation:
```>>> from scipy.signal import tf2ss
>>> A, B, C, D = tf2ss(num, den)
>>> A
array([[-2., -1.],
[ 1., 0.]])
>>> B
array([[ 1.],
[ 0.]])
>>> C
array([[ 1., 2.]])
>>> D
array([[ 1.]])
```
`_none_to_empty_2d`(arg)
`_atleast_2d_or_none`(arg)
`_shape_or_none`(M)
`_choice_not_none`(*args)
`_restore`(M, shape)
`abcd_normalize`(A=None, B=None, C=None, D=None)
Check state-space matrices and ensure they are two-dimensional.
If enough information on the system is provided, that is, enough properly-shaped arrays are passed to the function, the missing ones are built from this information, ensuring the correct number of rows and columns. Otherwise a ValueError is raised.
A, B, C, D : array_like, optional
State-space matrices. All of them are None (missing) by default. See ss2tf for format.
A, B, C, D : array
Properly shaped state-space matrices.
ValueError
If not enough information on the system was provided.
`ss2tf`(A, B, C, D, input=0)
rState-space to transfer function.
A, B, C, D defines a linear state-space system with p inputs, q outputs, and n state variables.
A : array_like
State (or system) matrix of shape `(n, n)`
B : array_like
Input matrix of shape `(n, p)`
C : array_like
Output matrix of shape `(q, n)`
D : array_like
Feedthrough (or feedforward) matrix of shape `(q, p)`
input : int, optional
For multiple-input systems, the index of the input to use.
num : 2-D ndarray
Numerator(s) of the resulting transfer function(s). num has one row for each of the system’s outputs. Each row is a sequence representation of the numerator polynomial.
den : 1-D ndarray
Denominator of the resulting transfer function(s). den is a sequence representation of the denominator polynomial.
Convert the state-space representation:
```>>> A = [[-2, -1], [1, 0]]
>>> B = [[1], [0]] # 2-dimensional column vector
>>> C = [[1, 2]] # 2-dimensional row vector
>>> D = 1
```
to the transfer function:
```>>> from scipy.signal import ss2tf
>>> ss2tf(A, B, C, D)
(array([[1, 3, 3]]), array([ 1., 2., 1.]))
```
`zpk2ss`(z, p, k)
Zero-pole-gain representation to state-space representation
z, p : sequence
Zeros and poles.
k : float
System gain.
A, B, C, D : ndarray
State space representation of the system, in controller canonical form.
`ss2zpk`(A, B, C, D, input=0)
State-space representation to zero-pole-gain representation.
A, B, C, D defines a linear state-space system with p inputs, q outputs, and n state variables.
A : array_like
State (or system) matrix of shape `(n, n)`
B : array_like
Input matrix of shape `(n, p)`
C : array_like
Output matrix of shape `(q, n)`
D : array_like
Feedthrough (or feedforward) matrix of shape `(q, p)`
input : int, optional
For multiple-input systems, the index of the input to use.
z, p : sequence
Zeros and poles.
k : float
System gain.
`cont2discrete`(system, dt, method="zoh", alpha=None)
Transform a continuous to a discrete state-space system.
system : a tuple describing the system or an instance of lti
The following gives the number of elements in the tuple and the interpretation:
• 1: (instance of lti)
• 2: (num, den)
• 3: (zeros, poles, gain)
• 4: (A, B, C, D)
dt : float
The discretization time step.
method : {“gbt”, “bilinear”, “euler”, “backward_diff”, “zoh”}, optional
Which method to use:
• gbt: generalized bilinear transformation
• bilinear: Tustin’s approximation (“gbt” with alpha=0.5)
• euler: Euler (or forward differencing) method (“gbt” with alpha=0)
• backward_diff: Backwards differencing (“gbt” with alpha=1.0)
• zoh: zero-order hold (default)
alpha : float within [0, 1], optional
The generalized bilinear transformation weighting parameter, which should only be specified with method=”gbt”, and is ignored otherwise
sysd : tuple containing the discrete system
Based on the input type, the output will be of the form
• (num, den, dt) for transfer function input
• (zeros, poles, gain, dt) for zeros-poles-gain input
• (A, B, C, D, dt) for state-space system input
By default, the routine uses a Zero-Order Hold (zoh) method to perform the transformation. Alternatively, a generalized bilinear transformation may be used, which includes the common Tustin’s bilinear approximation, an Euler’s method technique, or a backwards differencing technique.
The Zero-Order Hold (zoh) method is based on [1], the generalized bilinear approximation is based on [2] and [3].
[3] G. Zhang, X. Chen, and T. Chen, Digital redesign via the generalized bilinear transformation, Int. J. Control, vol. 82, no. 4, pp. 741-754, 2009. (http://www.mypolyuweb.hk/~magzhang/Research/ZCC09_IJC.pdf) | 1,629 | 5,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2019-22 | latest | en | 0.663361 |
https://www.aqua-calc.com/calculate/gravel-rectangular-solid/substance/caribsea-coma-and-blank-freshwater-coma-and-blank-super-blank-naturals-coma-and-blank-zen-blank-garden | 1,716,037,497,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057422.43/warc/CC-MAIN-20240518121005-20240518151005-00091.warc.gz | 579,980,624 | 8,743 | # Rectangular: CaribSea Freshwater Super Naturals Zen Garden
## rectangular aquarium (tank or pond) sand, gravel, substrate calcu...50
### The weight of CaribSea, Freshwater, Super Naturals, Zen Garden, 1 inch high, in a rectangular aquarium with the length and width of a base of 20 and 10 inches respectively
gram 4 829.92 ounce 170.37 kilogram 4.83 pound 10.65
### The volume of CaribSea, Freshwater, Super Naturals, Zen Garden, 1 inch high, in a rectangular aquarium with the length and width of a base of 20 and 10 inches respectively
centimeter³ 3 277.41 inch³ 200 foot³ 0.12 meter³ 0
• About CaribSea, Freshwater, Super Naturals, Zen Garden
• CaribSea, Freshwater, Super Naturals, Zen Garden weighs 1.47 gram per cubic centimeter or 1 473.7 kilogram per cubic meter, i.e. density of caribSea, Freshwater, Super Naturals, Zen Garden is equal to 1 473.7 kg/m³. In Imperial or US customary measurement system, the density is equal to 92 pound per cubic foot [lb/ft³], or 0.85 ounce per cubic inch [oz/inch³] .
• CaribSea, Freshwater, Super Naturals, Zen Garden weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
• A rectangular tank, aquarium or a pond has a shape of a box with six (6) sides, all of each are rectangles, and the lower rectangle is known as the base of the tank. The rectangular aquarium gravel calculator allows you to specify the length (l) and width (w) of the base rectangle, and desired height (h) of a selected gravel or substrate. The number in square brackets, to the right of the gravel name, is the gravel specific gravity or relative density, as compared to pure water.
The calculator computes the weight (1) and volume (2) of gravel using the following formulas:
(1) m = ρ × V and (2) V = l × w × h, where ρ is the density, V — volume, m — weight (or mass), and h is the height of gravel; l and w and are the length and width of a base of an aquarium respectively.
See also our gravel calculators for quarter cylinder and cylindrical shaped aquariums and ponds.
#### Foods, Nutrients and Calories
Beans, string, green, with chickpeas, cooked, fat added in cooking weigh(s) 147 grams per metric cup or 4.9 ounces per US cup, and contain(s) 93 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
Cryptoxanthin, alpha in Eggs, Grade A, Large, egg whole
#### Gravels, Substances and Oils
CaribSea, Freshwater, Flora Max, Midnight weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
2,4-Hexadienoic acid potassium salt [CH3CH=CHCH=CHCOOK, C6H7O2K or C6H7KO2] weighs 1 360 kg/m³ (84.90203 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-403B, liquid (R403B) with temperature in the range of -56.67°C (-70.006°F) to 48.89°C (120.002°F)
#### Weights and Measurements
A millijoule per kelvin is a SI-multiple (see prefix milli) of the entropy unit joule per kelvin and equal to one thousandth of an joule per kelvin (0.001 J/K)
Capacitance is a scalar quantity of a capacitor, that identifies the amount of charge, that can be stored by this capacitor for a given electric potential between its plates.
oz/ml to troy/metric tsp conversion table, oz/ml to troy/metric tsp unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a right circular cone and its surface area | 1,055 | 3,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-22 | latest | en | 0.728788 |
https://www.physicsforums.com/threads/does-this-series-converege.214466/ | 1,480,781,581,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540975.18/warc/CC-MAIN-20161202170900-00025-ip-10-31-129-80.ec2.internal.warc.gz | 996,079,259 | 15,451 | Does this series converege?
1. Feb 10, 2008
Mattofix
1. The problem statement, all variables and given/known data
The series is shown in the answer
3. The attempt at a solution
2. Feb 10, 2008
Mattofix
i feel like im going in the totally wrong direction - if so please can someone please point me the right way.
3. Feb 10, 2008
Dick
Expand the denominator first. The n^2 terms cancel so the denominator is basically proportional to n. The numerator, I think, tends to 1. Do you agree? So do you think it converges or diverges? It's good to have opinion to organize the strategy before you start trying to prove anything.
4. Feb 10, 2008
rootX
I am getting:
1+exp(-n)
----------
4n.e^n
Now, using ratio test I can prove that 1/(4n.e^n) is converging, and I also know that 1+exp(-n) is converging so that means there product is converging?
My Second approach
lim n--> inf (this series)/ [(1+exp(n))/exp(n)] is equal to 0, so this is equivalent to:
exp(n)+1
----------
exp(n)
now this seems wrong because my series does not converge. What have done wrong here?
5. Feb 11, 2008
Dick
The nth term in the series is (1+e^(-n))/((n+1)^2-(n-1)^2). The denominator is 4n. The numerator converges to one. It's begging for a comparison test.
6. Feb 11, 2008
Mattofix
thanks.
(1+exp(-n))/4n > 1/4n
and since 1/4n diverges so must the series. | 406 | 1,355 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-50 | longest | en | 0.918438 |
https://mypapersupport.com/tag/assignment-inferential-statistics/ | 1,685,839,091,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649348.41/warc/CC-MAIN-20230603233121-20230604023121-00072.warc.gz | 454,296,906 | 16,336 | ## Assignment: Inferential Statistics
Assignment: Inferential Statistics
Assignment: Inferential Statistics
Assignment: Inferential Statistics
Week 6 discussion Data Results and Analysis After the data are collected, it is time to analyze the results! Discuss one of the four basic rules for understanding results in a research study. Compare clinical significance and statistical significance. Which one is more meaningful when considering applying evidence to your practice? Compare descriptive statistics and inferential statistics in research. Please give an example of each type that could be collected in a study that would be done on your nursing clinical issue you identified in previous weeks.
Source: https://www.homeworkjoy.com/questions/health-care/577483-DeVry-nr439-full-course-latest-2017-october/
Statistical inference makes propositions about a population, using data drawn from the population with some form of sampling. Given a hypothesis about a population, for which we wish to draw inferences, statistical inference consists of (first) selecting a statistical model of the process that generates the data and (second) deducing propositions from the model.[citation needed]
Konishi & Kitagawa state, The majority of the problems in statistical inference can be considered to be problems related to statistical modeling.[2] Relatedly, Sir David Cox has said, How [the] translation from subject-matter problem to statistical model is done is often the most critical part of an analysis.[3]
The conclusion of a statistical inference is a statistical proposition.[4] Some common forms of statistical proposition are the following:
a point estimate, i.e. a particular value that best approximates some parameter of interest;
an interval estimate, e.g. a confidence interval (or set estimate), i.e. an interval constructed using a dataset drawn from a population so that, under repeated sampling of such datasets, such intervals would contain the true parameter value with the probability at the stated confidence level;
a credible interval, i.e. a set of values containing, for example, 95% of posterior belief;
rejection of a hypothesis;[note 1]
clustering or classification of data points into groups. | 437 | 2,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-23 | latest | en | 0.888491 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/hackem/word-break-2kj5 | 1,723,066,229,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713269.38/warc/CC-MAIN-20240807205613-20240807235613-00751.warc.gz | 359,525,341 | 20,563 | hackem
Posted on
# Word Break
## Problem Description
Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.
Note that the same word in the dictionary may be reused multiple times in the segmentation.
Example:
Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".
Example:
Input: s = "catsandog", wordDict =["cats","dog","sand","and","cat"]
Output: false
## Concept
It is rather straightforward or 'human-like' to brute force this problem. To segment s, we simply check if we can create a joint of words in wordDict to create s. A simple approach would be to check whether for each segment in s there is a match in wordDict. If the segment is a match, it also has to have a connected path of matched segments until the end of s . Example: It is not enough that 'cats' and 'and' (from wordDict) are matches in 'catsandog' if we do not find a match for segment 'og'.
A dynamic programming solution is applicable here as we are creating a new subproblem each time we find a segment matching in s, the subproblem is s minus the matched segment. Our base case being an empty s.
I did a recursive approach by starting from the beginning of s. The function then breaks off into all subproblems created by all matchings in s and returns True if one subproblem reaches an empty s, i.e. all parts of s matched. I tried this solution however I got a Time Limit Exceeded. I then chanced upon a clever trick in the leetcode discussions section to save computational time.
For this solution, we create a dp array for all indices of s plus one for the base case. Set all value of the array to False except our base case. Then we iterate from the end of s checking for each element whether it is a match and whether that match is connected to the end by checking dp[i + len(matched_word)]. The advantage of this method is that we save computational time by caching our results.
## Implementation
My initial solution, recursion
class Solution:
def wordBreak(self,s, wordDict, i=0):
res = []
for w in wordDict:
if (i+len(w))<=len(s) and (s[i:i+len(w)]==w):
if i+len(w) == len(s): return True
else:
res.append(self.wordBreak(s[i+len(w):],wordDict))
if True in res:
return True
return False
The bottom-up dp caching solution
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
dp = [False] * (len(s) + 1)
dp[len(s)] = True
for i in range(len(s)-1,-1,-1):
for w in wordDict:
if ((i+len(w))<=len(s)) and (s[i:i+len(w)]==w):
dp[i] = dp[i+len(w)]
if dp[i]:
break
return dp[0]
## Remarks
This problem appeared uncomplicated but had an interesting elusiveness that made it different from other dp problems. The solutions remind us to consider computational time beyond just time complexity. Both solutions were
$O(n^2*m)$
where n is the length of s and m the length of wordDict. Yet, only the caching solution was accepted. Remarkception, still trying to figure out dev.to's mathematical text support, currently using katex. | 760 | 3,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-33 | latest | en | 0.862398 |
https://www.hackmath.net/en/math-problem/362 | 1,642,646,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301670.75/warc/CC-MAIN-20220120005715-20220120035715-00640.warc.gz | 860,155,047 | 8,089 | # 4-gon
It is true that a 4-gon whose two sides are parallel and the other two has equal length, is a parallelogram?
Result
### Step-by-step explanation:
e.g. isosceles trapezoid is a parallelogram
Did you find an error or inaccuracy? Feel free to write us. Thank you!
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem: | 106 | 422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | latest | en | 0.881928 |
https://www.inchcalculator.com/how-to-use-tape-measure/ | 1,722,974,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00398.warc.gz | 639,553,485 | 21,346 | # How to Read a Tape Measure
A tape measure, also called a measuring tape, is a roll of metal (sometimes cloth) tape with evenly graduated markings used for measuring the length of objects or spaces. The tape is often yellow and rolled in a plastic case.
Tape measures are commonly used in construction, architecture, building, home projects, crafts, and woodworking. They usually come in lengths from 6 feet to 35 feet long.
Tape measures may have measurements in imperial and metric, imperial-only, or metric-only.
## How to Read an Imperial Tape Measure
On an imperial tape measure, the markings represent lengths in inches and fractions of an inch.
The largest ticks each represent one inch (1″) increments, and the smaller ticks between them represent the following fractions: 116“, 18“, 316“, 14“, 516“, 38“, 716“, 12“, 916“, 58“, 1116“, 34“, 1316“, 78“, and 1516“. Therefore, the large ticks denote inches, and the smaller ticks denote fractions of an inch.
To read a tape measure, find the number next to the large tick, and then find how many small ticks past it the measurement is. Add the number next to the large tick with the fractional tick to get the measurement. For instance, if the length of an object measures five small ticks past the number 4 tick, then the measurement is 4 516“.
### What Do All the Markings Mean?
To read a measuring tape, you need to understand what all the markings mean. The large ticks are spaced 1″ apart, and the small ticks are all spaced fractions of an inch apart. The numbers next to the large ticks indicate the number of inches from the end of the tape measure.
The second largest ticks halfway between the inch markings are the half-inch markings. There is 12” between each inch marking and half-inch marking.
The next largest ticks that are halfway between the inch markings and half-inch markings are the quarter-inch markings. There is 14” between the one-inch marking and the quarter-inch marking, as well as between each quarter-inch marking and the subsequent half-inch marking.
The second smallest ticks are the eighth-inch markings, and there is 18” between each eighth-inch marking and the adjacent quarter-inch, half-inch, and one-inch markings.
The smallest ticks on a tape measure are the sixteenth-inch markings. There is 116” between each adjacent marking on the tape measure.
### Inch Fractions for Each Mark
See the table below for the decimal equivalents of all the fractions on a tape measure. You might also like our inch fraction calculator for converting between decimal and inch fractions as well as getting decimal equivalents for fractions.
### Inch Fraction, Decimal, and Millimeter Equivalents
Chart showing equivalent fraction, decimal, and millimeter measurements
Fraction Decimal Millimeters
116 0.0625 1.5875
18 0.125 3.175
316 0.1875 4.7625
14 0.25 6.35
516 0.3125 7.9375
38 0.375 9.525
716 0.4375 11.1125
12 0.5 12.7
916 0.5625 14.2875
58 0.625 15.875
1116 0.6875 17.4625
34 0.75 19.05
1316 0.8125 20.6375
78 0.875 22.225
1516 0.9375 23.8125
1″ 1 25.4
## How to Read a Metric Tape Measure
Metric tape measures have similar markings to imperial models, but the markings represent centimeters and millimeters instead of inches and fractions of an inch. The larger markings on a metric tape that are labeled with numbers are the centimeters, and the smaller marks are millimeters.
Because there are 10 millimeters in a centimeter, there are 9 millimeter tick marks between each centimeter on the tape.
On a metric tape measure, there is 1 cm between each large numbered tick mark, and there is 1 mm between each smaller, non-numbered tick.
## How to Use a Tape Measure
Get the most out of your measuring tape with the following tips for using it.
### How to Use the Tape Lock
Almost all tape measures have a lock that will hold the tape measure in place and prevent it from recoiling. This is useful if you need to take tension off of the tape measure or if you need to set the tape measure aside while it is extended.
On this Stanley FatMax model, the slide lock is the large black button on the top. Sliding this down will lock the tape in place to prevent recoil.
As an Amazon Associate, we may earn commissions from qualifying purchases from Amazon. Learn more
### How to Use the Sliding Hook
A signature feature of a tape measure is the hook on the end of the tape. This serves a dual purpose to:
1. prevent the tape from rolling into the case
2. allow the tape to be hooked onto the end of items for convenient measuring
You might notice that the end hook slides or moves just a bit. This is by design to account for the thickness of the hook, so when hooking to an object to be measured, the hook slides outward so that the end of the tape can sit flush with the end of the object.
When butting up against a surface instead, the hook slides slightly inward, so, once again, the end of the tape can sit flush with the surface. Both setups allow for better accuracy when performing various measurements.
Be mindful of tape measures that do not have a sliding hook, as they will not be as accurate.
### How to Use the Framing Stud Hints
Most tape measures have red markings at certain intervals: 16″, 32″, 48″, 64″, and so on. These numbers are significant in that they indicate the placement of studs, which are usually spaced at 16 inches on center intervals, a property useful for framing.
Some tape measures also feature a black diamond symbol at intervals that are 19.2″ apart. These diamonds are also used to indicate joist spacing or framing intervals for a wider stud.
### How to Use the Nail Grab
The hook on a tape measure often has a small hole or groove in it. This is actually used to allow attaching the hook on a nail or screw so it doesn’t slide off when making longer measurements.
This is especially useful for longer measurements, such as measuring across the length of a room or patio.
On the hook in the image above, the oval cutout is used to hook onto a nail or screw.
### Use the Sides of the Hook
Some tape measures, especially framing tapes, have large hooks that can be used to grip surfaces on the side of the hook. Using these can improve the hook’s gripping ability, attach to a wider range of surfaces or edges, and improve the accuracy of measurements since the tape measure will not need to be twisted to read the markings.
## How to Choose the Right Tape Measure
There are many tape measures on the market, and many serve very different purposes. When choosing the one that’s right for you, consider what you’ll be using it for, how long you need it to be, and how much you’d prefer to spend.
When choosing a tape measure, consider the following features to find a tape measure that’s right for you and your needs.
• Size and ease of reading the markings
• Imperial or metric markings
• Length of the tape
• Physical size of the tape and housing
• Standout length for measuring longer lengths
• Locking features
• Durability
• Price
See our review of the best tape measures to find which tape measure we found to be the best and for reviews on several leading tape measures on the market. In a pinch, you can even print a tape measure to save a trip to the store.
### Is a tape measure in centimeters or inches?
An imperial tape measure has inch markings, while a metric tape measure has centimeter markings. However, you will sometimes encounter tape measures that have metric markings on one side and imperial markings on the other.
### How do you read a tape measure in inches and feet?
When measuring the length of an object, extend the tape measure from one end of the object to the other. Where the tape measure meets the end of the object is where you will read the measurement. To read the length of the object in feet and inches, first, find the total number of whole feet that you can read on the extended tape.
Then find the total number of whole inches after the last whole foot. Finally, find the fraction of an inch from the last whole inch, and add all three measurements together (you can use our feet and inches calculator if needed). That will be the length of your object in feet and inches.
### What can you use as a tape measure if you don’t have one?
If you don’t have a tape measure, you can instead use a ruler for shorter measurements, or if you know the length of a specific object, you may use that object. For example, if you have a standard piece of paper, you know it is 11 inches long.
Therefore, if the object you are trying to measure is equal to the length of two pieces of paper, then you know the object is 22 inches long. We cover this process in detail in our guide on measuring without a tape measure.
### Can you use your phone as a tape measure?
There are some apps out there that know the size of your phone screen and can thus display an image of a tape measure that maps to the correct tape measure sizing. There are also some devices that are equipped with better sensors to facilitate measuring the length of distance to objects.
However, be cautious with these as the measurements are not necessarily as accurate as a physical tape measure.
### How accurate is a tape measure?
Tape measures are accurate to their smallest tick. For example, if the smallest tick on your tape measure is 1/16th of an inch, then that tape measure is only accurate down to 1/16th of an inch. | 2,132 | 9,400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.94098 |
https://www.jiskha.com/display.cgi?id=1308071738 | 1,516,611,098,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891196.79/warc/CC-MAIN-20180122073932-20180122093932-00691.warc.gz | 954,692,617 | 4,928 | # PHYSICS
posted by .
A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet.
Compared to the acceleration due to gravity near Earth’s surface, the acceleration due to gravity near the
surface of the planet is approximately
(1) the same (3) one-half as great
(2) twice as great (4) four times as great
• PHYSICS -
PLEASE ANSWER. I HAVE PHYSIS REGENTS TOMMORW MUY IMPORTANTE
• PHYSICS -
Hmmmm. You have to be a highschool boy. You use the immature name PIMP, and you expect serious responses from adults. Maybe you are just joking about you need help, how are we to know?
• PHYSICS -
On earth, the distance an object would fall in 3.0 seconds is
(g/2)*t^2 = 4.9*9 = 44.1 m
On planet X, it falls 22 meters.
The distance it falls in a given time t is proportional to the acceleration of gravity.
Reach your own conclusion from that.
• PHYSICS -
7 On the surface of Earth, a spacecraft has a mass
of 2.00 × 104 kilograms. What is the mass of the
spacecraft at a distance of one Earth radius
above Earth’s surface?
(1) 5.00 × 103 kg (3) 4.90 × 104 kg
(2) 2.00 × 104 kg (4) 1.96 × 105 kg
• PHYSICS -
The mass does not change with position.
• PHYSICS -
bobpursley, instead of wasting time on speculating about the poster, just actually help out. Clearly this person needs help because he deliberately came on the site, asking for help. Immature name or not.
Anyway, the answer is one half as great. Here's how you find out:
First the givens are:
vi=0 (b/c it starts at rest)
m= 5kg
d= 22m
t= 3s
a on Earth always equals 9.81 m/s*s
Given these variable, I surmised that the best formula to use is d=vi(t)+(1/2)at^2.
When you plug in the variables given, you are left with:
22=(1/2)a(3)^(2) ,which gives you:
22=4.5(a)
Now its basic algebra. Divide both sides by 4.5 and you end up getting a=4.8 repeating. When you divide 9.81 by 2, you get 4.905, which is close enough, showing you that it one half as great.
Hope that helps! And good luck on the regents.
• PHYSICS -
>Uses the name PIMP
>Expects a serious homework question
## Similar Questions
1. ### Physics
An object is allowed to fall freely near the surface of an unknown planet. The object falls 80 meters from rest in 5.0 seconds. The acceleration due to gravity on that planet is?
2. ### phyics-universal gravity
A hypothetical planet has a radius 3.3 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?
3. ### physics
acceleration due to gravity questions when an object is dropped from a height of 10m above the surface of the planet z, it takes 1.2 seconds for the object to reach the surface. what is the acceleration of a falling object near the …
4. ### physics
A hypothetical planet has a radius 1.1 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?
5. ### physics
A hypothetical planet has a radius 2.1 times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?
6. ### physics
An object rests on the surface of planet y. Its mass is unknown. Find the acceleration due to gravity near the surface of planet y. Answer in m/s/s. G=6.673x10^-23 radius of planet y=2439700m mass of planet y=3.302x10^23kg radius of …
7. ### physics
A rock falls from rest a vertical distance of 0.63 meters to the surface of a planet in 0.48 seconds. What is the magnitude of the acceleration due to gravity on that planet?
8. ### physics
9. The surface of the earth is approximately 6,400 km from its center. If the mass of the earth is 6.0 x 1024 kg, what is the acceleration due to gravity near the surface?
9. ### earth
A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the acceleration due to gravity near Earth’s surface, the acceleration due to gravity near the surface of …
10. ### physic
Determine the time it takes for a satellite to orbit the Saturn in a circular "near-Saturn" orbit. A "near-Saturn" orbit is at a height above the surface of the Saturn that is very small compared to the radius of the Saturn. [Hint: …
More Similar Questions | 1,132 | 4,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-05 | latest | en | 0.916254 |
https://www.hackmath.net/en/problem/6237 | 1,561,057,645,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999273.24/warc/CC-MAIN-20190620190041-20190620212041-00134.warc.gz | 749,294,368 | 7,001 | # Soaps
Each box has the same number of soaps. A quarter of all boxes contain only white soaps, and in each of the remaining 120 boxes there are always half the white soaps and half the green. White soaps total 1200.
(a) the number of all soap boxes;
(b) the smallest number of boxes in which all white soaps should be inserted;
(c) the number of all soaps.
d) How many are green soaps?
Result
k = 160
b = 100
t = 1920
z = 720
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this verbal math problem are needed these knowledge from mathematics:
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation?
## Next similar examples:
1. Three figures - numbers
The sum of three numbers, if each is 10% larger than the previous one, is 662. Determine the figures.
2. Three days
During the three days sold in stationery 1490 workbooks. The first day sold about workbooks more than third day. The second day 190 workbooks sold less than third day. How many workbooks sold during each day?
3. Mushrooms
Eva and Jane collected 114 mushrooms together. Eve found twice as much as Jane. How many mushrooms found each of them?
4. Three friends
The three friends spent 600 KC in a teahouse. Thomas paid twice as much as Paul. Paul a half less than Zdeněk. How many each paid?
5. Family parcels
In father will he divided the land so that the older son had three bigger part than younger son. Later elder son gave 2.5 ha field to younger and they had both the same. Determine the area of family parcel.
6. The larger
The larger of two numbers is nine more than four times the smaller number. The sum of the two numbers is fifty-nine. Find the two numbers.
7. Football match 4
In a football match with the Italy lost 3 goals with Germans. Totally fell 5 goals in the match. Determine the number of goals of Italy and Germany.
8. Two numbers
We have two numbers. Their sum is 140. One-fifth of the first number is equal to half the second number. Determine those unknown numbers.
9. Belgium vs Italy
Belgium played a match with Italy and Belgium win by 2 goals. The match fell a total 6 goals. Determine the number of goals scored by Belgium and by Italy.
10. Fifth of the number
The fifth of the number is by 24 less than that number. What is the number?
11. Equations
Solve following system of equations: 6(x+7)+4(y-5)=12 2(x+y)-3(-2x+4y)=-44
12. Trees
Along the road were planted 250 trees of two types. Cherry for 60 CZK apiece and apple 50 CZK apiece. The entire plantation cost 12,800 CZK. How many was cherries and apples?
13. Mom and daughter
Mother is 39 years old. Her daughter is 15 years. For many years will mother be four times older than the daughter?
14. Substitution
solve equations by substitution: x+y= 11 y=5x-25
15. Viju
viju has 40 chickens and rabbits. If in all there are 90 legs. How many rabbits are there with viju??
16. Hotel rooms
In the 45 rooms, there were 169 guests, some rooms were three-bedrooms and some five-bedrooms. How many rooms were?
17. Linear system
Solve this linear system (two linear equations with two unknowns): x+y =36 19x+22y=720 | 831 | 3,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-26 | latest | en | 0.956032 |
http://gundemonline.org/free-online-games-no-downloads/rate-of-return-formula-excel-2010 | 1,490,793,188,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190295.65/warc/CC-MAIN-20170322212950-00325-ip-10-233-31-227.ec2.internal.warc.gz | 156,898,370 | 6,874 | # Rate of return formula excel 2010
1, Calculating IRR in Excel. 2. 3, This worksheet demonstrates examples of using an Excel function to find the internal rate of return (IRR). 4. 5. 6, IRR is the.
The Excel Irr Function - Returns the Internal Rate of Return for a Supplied Series of Periodic Cash Flows - Function Description, Examples, and Common Errors.
The Microsoft Excel IRR function returns the internal rate of return for a series of Excel Excel Excel 2011 for Mac, Excel 2010, Excel Excel.
### Rate of return formula excel 2010 - best
If the IRR function returns a NUM! Note that, in the above example, the initial investment is a negative value, as this is an outgoing payment, and the income payments are represented by positive values. The syntax for the IRR function in Microsoft Excel is:. The monetary value of all the finished goods and services produced within a country's borders in a specific time period,... The XIRR function is: Again, what is needed are the beginning and ending investing values, and date periods. Because of the time value of the dollar, the IRR calculation can be very simple or very complicated depending on the project in question. It is your guess at the internal rate of return. If this question or a similar one is answered twice in this section, please click here to let us know. Fortunately, financial calculators and Microsoft Excel make the process amazingly simple. Note: This is only a value for Excel to start off working with - Excel then uses an iterative procedure to converge to the IRR. By using our site you accept the terms of our Privacy Policy. | 350 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-13 | longest | en | 0.854953 |
https://web2.0calc.com/questions/parabola_75342 | 1,720,980,603,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00343.warc.gz | 541,436,406 | 5,591 | +0
# Parabola
-1
47
2
+1347
A parabola with equation \$y=x^2+bx+c\$ passes through the points \$(-1,-11)\$ and \$(3,-17)\$. What is \$c\$?
Aug 9, 2023
#1
+129659
+1
We have these equations
-11 = (-1)^2 + b(-1) + c
-17 = (3)^2 + b(3) + c simplify
-11 = 1 -b + c
-17= 9 + 3b + c
-12 = -b + c
-26 = 3b + c subtract these
14 = -4b
14/ -4 = b = -7/2
And we can find c as
-12 = - (-7/2) + c
-12 = 7/2 + c
-12 - 7/2 = c = -24/2 - 7/2 = -31/2
Aug 9, 2023 | 256 | 484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-30 | latest | en | 0.433373 |
https://www.teacherspayteachers.com/Store/Math-Adventures | 1,542,277,402,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742666.36/warc/CC-MAIN-20181115095814-20181115121814-00255.warc.gz | 1,006,705,324 | 27,524 | (70)
United States - Texas - kemp
4.0
"I think outside of the box cause its to crouded inside."
My Products
sort by:
Best Seller
view:
Give your student practice using Dot Plots and Frequency Tables by using this activity in your classroom. Collect the information in groups or ass a class and fill in the frequency table and dot plot. This activity covers the NEW 2014-2015 Math TEK
Subjects:
Math, Statistics
3rd, 4th, 5th, 6th
Types:
Activities, Math Centers, Cooperative Learning
CCSS:
6.SP.B.4
\$0.95
183 ratings
4.0
Assess your students understanding of strip diagrams with this worksheet.This worksheet goes with TEK 4.5A represent multi-step problems involving the four operations with whole numbers using strip diagrams and equations with a letter standing for
Subjects:
Math
3rd, 4th, 5th
Types:
Worksheets, Assessment
\$1.00
25 ratings
4.0
Assess your students knowledge of dot plots using this worksheet. Students are given data that they must fill in on the dot plot. There are also high level comprehension questions to go along with the dot plot to assess your students knowledge. This
Subjects:
Math, Statistics
3rd, 4th, 5th, 6th
Types:
Worksheets, Assessment
\$0.95
19 ratings
4.0
Students can practice decomposing fractions using these Matching cards. Each composed fraction card has 3 decomposed fraction cards that go with it. You can use these cards for Math Centers or for Math Stations. You can also play concentration with
Subjects:
Math, Fractions
3rd, 4th, 5th, 6th
Types:
Activities, Math Centers, Cooperative Learning
CCSS:
4.NF.B.3b
\$1.50
12 ratings
4.0
This activity covers the newly revised math TEK 4.2(A) interpret the value of each place-value position as 10 times the position to the right and as one-tenth of the value of the place to its left. It also covers Common Core State Standard 4.NBT.A.1
Subjects:
Math
4th
Types:
\$2.00
5 ratings
4.0
This google sheet lets you select the TEKS and Student Expectations for Math, Science, Reading, and Writing in 4th grade instead of typing the TEKS in. Once you have selected the Student Expectation for that day the Vertical Alignment for the Year
Subjects:
English Language Arts, Math, Science
4th
Types:
Lesson Plans (Individual), Lesson Plans (Bundled), Excel Spreadsheets
\$4.99
7 ratings
4.0
Assess your students knowledge of decimals and fractions on a number line.
Subjects:
Math
4th
Types:
Worksheets, Assessment
\$1.00
4 ratings
4.0
This is a weekly student agenda or planner for the 2014-2015 school year. It is great to use with the AVID program. There is room at the bottom of the pages for teacher notes, post-it notes, or Avery sticker that have reminders. I like to have
Subjects:
For All Subject Areas
3rd, 4th, 5th, 6th, 7th, 8th
Types:
Homework, Classroom Forms
\$1.99
not yet rated
Introduce Dot Plots by using this activity in your classroom. Collect the information with your class and fill in the dot plot. There are also high level comprehension questions to go along with the dot plot to assess your students knowledge. This
Subjects:
Math, Statistics
3rd, 4th, 5th, 6th
Types:
Worksheets, Activities
CCSS:
6.SP.B.4
FREE
16 ratings
4.0
Please Leave Feedback! Students can practice decomposing fractions using this set of Matching cards. You can use the cards for Math Centers or for Math Stations. You can also use them to play concentration. New TEKS 2014-2015 4.3(A) represent a
Subjects:
Math, Fractions
3rd, 4th, 5th, 6th
Types:
Activities, Math Centers, Cooperative Learning
CCSS:
4.NF.B.3
FREE
7 ratings
4.0
I made this chart for a classroom activity that involved recording my students height. You can hang it on the wall. This would also be a great way to observe how much your students have grown from the beginning to the end of the year. Please leave
Subjects:
Math, Measurement
Kindergarten, 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
Printables
CCSS:
K.MD.A.1, 1.MD.A.2, 2.MD.A.1, 3.MD.B.4, 4.MD.A.1, 5.MD.A.1
FREE
2 ratings
4.0
I made this chart for a classroom activity that involved recording my students height. You can hang it on the wall. This would also be a great way to observe how much your students have grown from the beginning to the end of the year. Please leave
Subjects:
Math, Measurement
Kindergarten, 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th
Types:
Printables
CCSS:
K.MD.A.1, 1.MD.A.2, 2.MD.A.1, 3.MD.B.4, 4.MD.A.1, 5.MD.A.1
FREE
not yet rated
This chart is great for interactive notebooks. You could also laminate it and use it in stations. I am going to have my student roll dice and fill in the top part of the chart with the numbers that they roll. Afterwards they will represent that
Subjects:
Math, Numbers
3rd, 4th, 5th
Types:
Math Centers
CCSS:
4.NBT.A.2
FREE
3 ratings
4.0
showing 1-13 of 13
### Ratings
Digital Items
4.0
Overall Quality:
4.0
Accuracy:
4.0
Practicality:
4.0
Thoroughness:
4.0
Creativity:
4.0
Clarity:
4.0
Total:
281 total vote(s)
TEACHING EXPERIENCE
MY TEACHING STYLE
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
I went to school at Texas A&M- Commerce and recieved a degree in Teaching Math Grades 4-8. | 1,500 | 5,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-47 | latest | en | 0.910117 |
http://www.coursehero.com/file/5863231/Econ1B03-qCh3/ | 1,411,308,459,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657135558.82/warc/CC-MAIN-20140914011215-00107-ip-10-234-18-248.ec2.internal.warc.gz | 455,650,016 | 16,128 | This preview has intentionally blurred parts. Sign up to view the full document
View Full Document
Unformatted Document Excerpt
1 Chapter 3 Supply and Demand: Theory 1. The law of demand states that price and quantity demanded are a. directly related, ceteris paribus . b. inversely related, ceteris paribus . c. independent. d. positively related, ceteris paribus . ANS: b 2. Which of the following illustrates the law of demand? a. Jeff buys more pens at \$1 per pen than at \$2 per pen, ceteris paribus . b. Sara buys more ice cream at \$2 per half-gallon than at \$1 per half gallon, ceteris paribus . c. Chris buys fewer trousers at \$30 per pair than at \$25 per pair, ceteris paribus . d. a, b, and c e. a and c ANS: e 3. At a price of \$15, Marta buys 3 CDs per month. When the price increases to \$20, Marta buys 2 CDs per month. Leah says that Martas demand for CDs has decreased. Is Leah correct? a. Yes, Leah is correct. b. No, Leah is incorrect. Martas demand has increased. c. No, Leah is incorrect. Martas quantity demanded has decreased, but her demand has stayed the same. d. No, Leah is incorrect. Martas quantity demanded has increased, but her demand has stayed the same. e. No, Leah is incorrect. Martas quantity demanded has decreased and her demand has increased. ANS: c 4. A major reason for the law of demand is that a. one price changing requires at least one other price to change in the opposite direction. b. people substitute lower-priced goods for higher-priced goods. c. a higher price never reduces quantity demanded by enough to lower total revenue. d. people are willing to produce more units at a higher price. ANS: b 5. The price of pizzas rises, and Mei increases her monthly purchases of pizza. Therefore, the law of demand a. doesnt hold for any good. b. doesnt hold during a month. c. doesnt hold for pizzas. d. doesnt hold for Mei. e. none of the above ANS: e 64 65 Chapter 3 6. A demand schedule is a listing of a. incomes and quantities demanded. b. costs and quantities demanded. c. prices and quantities demanded. d. assets and quantities demanded. ANS: c 7. If an increase in income results in an increase in the demand for chicken, then chicken is a. a neutral good. b. a luxury good. c. a normal good. d. an inferior good. ANS: c 8. If Maxs demand for hot dogs falls as his income rises, then hot dogs are a. a bad good. b. an inferior good. c. a preferential good. d. a normal good. e. a neutral good. ANS: b 9. A decrease in the number of buyers in an area will result in a a. movement up the demand curve. b. movement down the demand curve. c. leftward shift in the demand curve. d. rightward shift in the demand curve. ANS: c 10. If the demand curve for a good shifts leftward, a. quantity demanded is less at each price. b. quantity demanded remains constant at each price.... View Full Document
End of Preview | 732 | 2,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2014-41 | latest | en | 0.886552 |
deanyztmd.blog5.net | 1,544,443,199,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823322.49/warc/CC-MAIN-20181210101954-20181210123454-00013.warc.gz | 66,348,973 | 6,183 | # Facts About statistics homework help Revealed
Exposure to unsafe consuming h2o, insufficient sanitation and poor hygiene is a number one reason for cholera and various infectious and tropical ailments in the African Area. fifteen
"The connection in between the data and the things they explain just demonstrates the fact that particular varieties of statistical statements can have reality values which are not invariant less than some transformations. Whether a transformation is sensible to ponder is determined by the issue one particular is attempting to answer" (Hand, 2004, p. eighty two).[21]
As of March 1st, 2017, We are going to use demographic (population) based on facts published through the United Nations (UN), and contained of their outstanding website at: United Nations - Inhabitants Division, exactly where It's also possible to uncover specific demographic info for the whole world and for the many nations of the entire world.
This alphabetical country listing has been Increased with the image for each nation or location, The present range of World wide web consumers and penetration amount, the population as well as land dimensions.
How unfortunate certainly to put dying as “good friends” just bash on close by. I’ve heard about circumstances where by they partied for eight hours or even more Following their “mates” experienced by now died nearby. Include to that those who do get up from their unique exciting night time on prescription drugs to find a useless entire body beside them and people who tucked their buddies into mattress and remaining completely, assuming they might wake each morning……Nonetheless they didn’t.
The common this contact form solution[22] is to test a null hypothesis in opposition to an alternate hypothesis. A vital region could be the list of values of your estimator that contributes to refuting the null hypothesis. The probability of type I error is as a result the chance that the estimator belongs into the significant region given that null hypothesis is true (statistical importance) and the chance of style II mistake would be the probability the estimator does not belong towards the crucial area provided that the alternative speculation is real.
The final wave, which mainly observed the refinement and expansion of previously developments, emerged from your collaborative get the job done involving Egon Pearson and Jerzy Neyman within the 1930s.
Offering methamphetamine or cocaine or other stimulants to test to counteract the depressant results of a GHB coma will not be successful or sane.
Learners, Generally, explore challenging to discover their responses provided instances and special conditions and independent the solutions for his or her troubles. So we're right here to simplify your problems you simply have to go to our Web-site and can avail our solutions.
All over again, this checklist is maintained in order to maintain data index as it is available in with the hope that further more analysis and documentation might be finished as feasible.
Mitchell Trubisky has demonstrated he can seem as good off the field as on. With the fashion world turned east for the beginning of Big apple Style 7 days, take a look at our dapper quarterback.
Social and Local community company supervisors coordinate and supervise social assistance courses and Group companies. They regulate staff who deliver social services to the general public. see profile »
Job and complex education and learning teachers instruct college students in different technological and vocational topics, for example automobile fix, healthcare, and culinary arts. They educate educational and technological information to offer college students with the talents and understanding necessary to enter an occupation. view profile »
Medical experts carry out investigate geared toward strengthening All round human well being. They frequently use scientific trials and other investigative techniques to arrive at their findings. check out profile » | 740 | 4,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-51 | latest | en | 0.945378 |
https://vustudents.ning.com/group/sta301statisticsandprobability/forum/topics/question-about-gbd-why-an-average-computed-from-a-frequency?commentId=3783342%3AComment%3A5018386&groupId=3783342%3AGroup%3A59512 | 1,586,008,301,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370524043.56/warc/CC-MAIN-20200404134723-20200404164723-00181.warc.gz | 762,965,311 | 19,471 | We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
www.vustudents.ning.com
www.bit.ly/vucodes + Link For Assignments, GDBs & Online Quizzes Solution www.bit.ly/papersvu + Link For Past Papers, Solved MCQs, Short Notes & More
Dear Students! Share your Assignments / GDBs / Quizzes files as you receive in your LMS, So it can be discussed/solved timely. Add Discussion
# STA301 Statistics and Probability GDB No.1 Solution and Discussion Spring 2015 Due Date: May 15, 2015
STA301 Statistics and Probability GDB No.1 Solution
Dear Students,
Graded Discussion Board (GDB) for STA301 will be opened on May 14, 2015 and will remain open till May 15, 2015. The topic of this GDB is,
“Why an average computed from a frequency distribution is not exactly the same as computed from the raw data? Give the reason”.
NOTE: Your comments MUST be posted with in due date and it will NEVER accept through e-mail.
+ How to Join Subject Study Groups & Get Helping Material?
+ How to become Top Reputation, Angels, Intellectual, Featured Members & Moderators?
+ VU Students Reserves The Right to Delete Your Profile, If?
Views: 6835
.
+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)
+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)
### Replies to This Discussion
AoA! Rajput bhai... 1 post kafi si na....
hahahahah
Average frequency reports the average number of times.It is a basic measure of schedule intensity.average data easily distributed in various due to his grouping while the raw data can not be manipulated further.
Nice Knowledge Sharing...Thanks
JAZAK ALLAH
2 hogi tu kia howa yr
Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
Readers are cautioned that the mean calculated from the frequency distribution is not exactly the same as that calculated from the raw data. It approaches the mean calculated from the raw data as the number of intervals increase
The standard statistical notations (in relation to measures of central tendency) are mentioned in
Average frequency reports the average number of times.It is a basic measure of schedule intensity.average data easily distributed in various due to his grouping while the raw data can not be manipulated further.
thnx u Sir for sharing your idea :thank u very much
## Latest Activity
30 minutes ago
31 minutes ago
1 hour ago
Farrukh Merchant updated their profile
1 hour ago
1 hour ago
1 hour ago
1 hour ago
2 hours ago
2 hours ago
2 hours ago
2 hours ago
2 hours ago
1
2
3 | 656 | 2,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.850537 |
https://9dok.net/article/argument-of-latitude-auxiliary-computations-d.q05217vg | 1,723,254,672,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782288.54/warc/CC-MAIN-20240809235615-20240810025615-00642.warc.gz | 56,129,293 | 29,012 | • Keine Ergebnisse gefunden
# Argument of latitude
## Auxiliary computations D
### D.3 Argument of latitude
The argument of latitudeu describes the position of the satellite along its orbit. u is the angle between the ascending node and the satellite, measured in the satellite’s orbital plane, cf.
Figure D.4.
The argument of latitude is defined as
u=ω+v (D.8)
whereω andvare the Keplerian orbital elements: ω-argument of perigee,v-true anomaly. The values of the argument of latitude are withinu=h0; 360i. The argument of latitude can be derived from the satellite’s positionrsat = (x, y, z)T and velocity vsat= (vx, vy, vz)T vectors in inertial coordinates as
h= rsat×vsat
|rsat×vsat| =
hx hy hz
(D.9)
u= arctan z
−x·hy+y·hx
!
(D.10)
Ω ω ϒ
e
### i u
Figure D.4:Keplerian orbital elements: a- semi-major axis,e- numerical eccentricity,i- orbit inclination, Ω - right ascension of ascending node,ω- argument of perigee,v- true anomaly,u- argument of latitude
In this thesis, some of the parameters of interest are plotted as a function of time and argument of latitude. This allows to plot the values along the whole orbit for unlimited period of time, which is advantageous especially for observing the long-term systematic effects.
Figure D.5 shows the argument of latitude as a funtion of time for 1 day. The values of the parameter of interest are then expressed by color, cf. e.g. Figure 3.3.
90°
180°
270°
360°
equator North pole South pole
equator equator
Flight direction
Figure D.5:Argument of latitude
A, G., Wahr, J., and Zhong, S. (2013). Computations of the viscoelastic response of a 3-D compressible Earth to surface loading: an application to Glacial Isostatic Adjustment in Antarctica and Canada. Geophysical Journal International, 192, Pages 557–572, doi: 10.1093/gji/ggs030.
Anderson, D. (1984). Alignment of resonant optical cavities. Applied Optics, 23, Pages 2944–2949, doi:
10.1364/AO.23.002944.
Arbinger, C., D’Amico, S., Feucht, U., and Finzi, A. (2003). The GRACE formation: science mode pointing performance. 3rd International Workshop on Satellite Constellations and Formations Flying, Pisa, Italy, Feb 24-26, 2003.
Bandikova, T. and Flury, J. (2014). Improvement of the GRACE star camera data based on the revision of the combination method. Advances in Space Research, 54, Pages 1818–1827, doi: 10.1016/j.asr.2014.07.004.
Bandikova, T., Flury, J., and Ko, U.-D. (2010). Analysis of systematic effects in GRACE inter-satellite pointing.
GRACE Science Team Meeting, Nov 11-12, 2010, Potsdam. GSTM proceedings 2010.
Bandikova, T., Flury, J., and Ko, U.-D. (2012). Characteristics and accuracies of the GRACE inter-satellite pointing. Advances in Space Research, 50, Pages 123–135, doi: 10.1016/j.asr.2012.03.011.
Baur, O., Kuhn, M., and Featherstone, W. (2013). Continental mass change from GRACE over 2002–2011 and its impact on sea level. Journal of Geodesy, 87, Pages 117–125, doi: 10.1007/s00190-012-0583-2.
Bender, P., Wiese, D., and Nerem, R. (2008). A possible dual-GRACE mission with 90 degree and 63 degree inclination orbits. InProc. 3rd Int. Symp. on Formation Flying, Missions and Technologies, Noordwijk ,The Netherlands.
Bettadpur, S. (2012). Gravity Recovery and Climate Experiment: Product specification document. Technical Report GRACE 327-720, Center for Space Research, The University of Texas at Austin.
Bettadpur, S. and the CSR Level-2 Team (2012). Insights into the Earth System mass variability from CSR-RL05 GRACE gravity fields.EGU General Assembly, 2012.
Beutler, G., Jäggi, A., Mervart, L., and Meyer, U. (2010). The celestial mechanics approach: theoretical foundations. Journal of Geodesy, 84, Pages 605–624, doi: 10.1007/s00190-010-0401-7.
Blockley, R. and Shyy, W. (2010). Encyclopedia of Aerospace Engineering. Volume 5 Dynamics and Control.
Wiley, ISBN: 978-0-470-75440-5.
Bock, R. and Lühr, H. (2001). CHAMP attitude aberration correction. Technical Report CH-GFZ-TN-2702, GFZ Potsdam.
Bruinsma, S., Lemoine, J.-M., Biancale, R., and Vales, N. (2010). CNES/GRGS 10-day gravity field models (release 2) and their evaluation.Advances in Space Research, 45, Page 587–601, doi: 10.1016/j.asr.2009.10.012.
Case, K., Kruizinga, G., and Wu, S.-C. (2010). GRACE level-1B data product user handbook. Technical Report JPL D-22027, Jet Propulsion Laboratory.
Cazenave, A. and Llovel, W. (2010). Contemporary sea level rise. Annual Review of Marine Science, 2, Pages 145–173, doi: 10.1146/annurev-marine-120308-081105.
Crassidis, J., Lai, K.-L., and Harman, R. (2005). Real-time attitude-independent three-axis magnetometer calibration. Journal of Guidance, Control and Dynamics, 28, Pages 115–120, doi: 10.2514/1.6278.
Dahle, C., Flechtner, F., Gruber, C., König, D., König, R., Michalak, G., and Neumayer, K.-H. (2014).
Observation of the System Earth from Space - CHAMP, GRACE, GOCE and future missions, Chapter GFZ RL05: An Improved Time-Series of Monthly GRACE Gravity Field Solutions, Pages 29–39. Springer, doi:
10.1007/978-3-642-32135-1_4.
de Angelis, M., Bertoldi, A., Cacciapuoti, L., Giorgini, A., Lamporesi, G., Prevedelli, M., Saccorotti, G., Sorrentino, F., and Tino, G. (2009). Precision gravimetry with atomic sensors. Measurement Science and Technology, 20, Page 022001, doi: 10.1088/0957-0233/20/2/022001.
Doll, B. and Wolters, R. (1999). GRACE Coarse Earth- & Sun Sensor Specification 327-40230. Technical Report GR-DSS-SP-2230, DaimlerChrysler Aerospace.
Doornbos, E., Förster, M., Fritsche, B., van Helleputte, T., van den Ijssel, J., and et al. (2009). Air density models derived from multi-satellite drag observations. Technical Report ESTEC contract 210022/07/NL/HE, TU Delft.
Eisenman, A. R. and Liebe, C. C. (1998). The advancing state-of-the-art in second generation star trackers. In Aerospace Conference, 1998 IEEE, Volume 1, Pages 111–118. doi: 10.1109/AERO.1998.686810.
Elsaka, B. (2012).Simulated satellite formation flight for detecting the temporal variations of the Earth’s gravity field. PhD thesis, University of Bonn.
Elsaka, B., Raimondo, J.-C., Brieden, P., Reubelt, T., Kusche, J., Flechtner, F., Iran Pour, S., Sneeuw, N., and Müller, J. (2013). Comparing seven candidate mission configurations for temporal gravity field retrieval through full-scale numerical simulation.Journal of Geodesy, 88, Pages 31–43, doi: 10.1007/s00190-013-0665-9.
ESA (1999). Gravity Field and Steady-State Ocean Circulation Mission. Report for mission selection of the four candidate Earth Explorer missions ESA SP-1233(1), European Space Agency.
Famiglietti, J. S. and Rodell, M. (2013). Water in the balance. Science, 340, Pages 1300–1301, doi:
10.1126/science.1236460 .
Flechtner, F. (2004). GRACE Science Data System Monthly Report February 2004. Technical Report, JPL/GFZ.
Flechtner, F., Morton, P., Watkins, M., and Webb, F. (2014). Status of the GRACE Follow-On mission. In Gravity, Geoid and Height Systems. Springer, doi: 10.1007/978-3-319-10837-7_15.
Flury, J., Bettadpur, S., and Tapley, B. (2008). Precise accelerometry onboard the GRACE gravity field satellite mission.Advances in Space Research, 42, Pages 1414–1423, doi: 10.1016/j.asr.2008.05.004.
Förste, C., Bruinsma, S., Abrikosov, O., Lemoine, J.-M., Schaller, T., Götze, H.-J., and et al. (2014). The latest combined global gravity field model including GOCE data up to degree and order 2190 of GFZ Potsdam and GRGS Toulouse.5th GOCE User Workshop, Paris. http://icgem.gfz-potsdam.de/ICGEM/documents/Foerste-et-al-EIGEN-6C4.pdf.
Frappart, F., Papa, F., Santos da Silva, J., Ramillien, G., Prigent, C., Seyler, F., and Calmant, S. (2012). Surface freshwater storage and dynamics in the Amazon basin during the 2005 exceptional drought. Environmental Research Letters, 4, Page 044010, doi: 10.1088/1748-9326/7/4/044010.
Friis-Christensen, E., Lühr, H., Knudsen, D., and Haagmans, R. (2008). Swarm – an Earth observation mission investigating geospace. Advances in Space Research, 41, Pages 210–2016, doi: 10.1016/j.asr.2006.10.008.
Frommknecht, B. (2008).Integrated Sensor Analysis of the GRACE Mission. PhD thesis, Technische Universität München.
Goodkind, J. (1999). The superconductiong gravimeter. Review of Scientific Instruments, 70, Pages 4131–4152, doi: 10.1063/1.1150092.
Han, S.-C., Jekeli, C., and Shum, C. (2004). Time-variable aliasing effects of ocean tides, atmosphere, and continental water mass on monthly mean GRACE gravity field. Journal of Geophysical Research, 109, Page B04403, doi: 10.1029/2003JB002501.
Han, S.-C., Shum, C. K., Bevis, M., Ji, C., and Kuo, C.-Y. (2006). Crustal Dilatation Observed by GRACE After the 2004 Sumatra-Andaman Earthquake. Science, 313, Pages 658–662, doi: 10.1126/science.1128661.
Heinzel, G., Sheard, B., Brause, N., Danzmann, K., Dehne, M., Gerberding, O., Mahrdt, C., and et al. (2012).
Laser Ranging Interferometer for GRACE Follow-On. InProceedings of the International Conference on Space Optics (ICSO2012). ISCO proceedings.
Herman, J. and Steinhoff, M. (2012). Balancing, turning, Saving: Special AOCS operations to extend the GRACE mission.SpaceOps 2012, Stockholm, Sweden, June 11-15. doi: 10.2514/6.2012-1275114.
Herman, J., Presti, D., Codazzi, A., and Belle, C. (2004). Attitude control for GRACE: the first low-flying satellite formation. InESA SP-548: 18th International Symposium on Space Flight Dynamics.
Herman, J., Davis, A., Chin, K., Kinzler, M., Scholz, S., and Steinhoff, M. (2012). Life with a weak heart -prolonging the GRACE mission despite degraded batteries. InSpaceOps 2012, Stockholm, Sweden, June 11-15.
Hofmann-Wellenhof, B. and Moritz, H. (2005). Physical geodesy. SpringerWienNewYork, ISBN: 3-211-23584-1.
Hofmann-Wellenhof, B., Lichtenegger, H., and Wasle, E. (2008). GNSS - Global navigation satellite system.
GPS, GLONASS, Galileo & more. SpringerWienNewYork, ISBN: 978-3-211-73012-6.
Horwath, M., Lemoine, J.-M., Biancale, R., and Bourgogne, S. (2010). Improved GRACE science results after adjustment of geometric biases in the Level-1B K-band ranging data. Journal of Geodesy, 85, Pages 23–38, doi: 10.1007/s00190-010-0414-2.
Hudson, D. (2003). In-flight characterisation and calibration of the SuperSTAR accelerometer. Master thesis, CSR, The University of Texas at Austin.
Inacio, P., Ditmar, P., Klees, R., and Hashemi Farahani, H. (2015). Analysis of star camera errors in GRACE data and their impact on monthly gravity field models. Journal of Geodesy, 89, Pages 551–571, doi:
10.1007/s00190-015-0797-1.
Jäggi, A., Hugentobler, U., Bock, H., and Beutler, G. (2007). Precise orbit determination for GRACE using undifferenced or doubly differenced GPS data. Advances in Space Research, 39, Pages 1612–1619, doi:
10.1016/j.asr.2007.03.012.
Jäggi, A., Dach, R., Montenbruck, O., Hugentobler, U., Bock, H., and Beutler, G. (2009). Phase center modeling for LEO GPS receiver antennas and its impact on precise orbit determination.Journal of Geodesy, 83, Pages 1145–1162, doi: 10.1007/s00190-009-0333-2.
Jäggi, A., Beutler, G., Meyer, U., Prange, L., Dach, R., and Mervart, L. (2012). AIUB-GRACE02S: Status of GRACE Gravity Field Recovery Using the Celestial Mechanics Approach. In S. Kenyon, M. C. Pacino, and U. Marti (Ed.),Geodesy for Planet Earth, Volume 136,International Association of Geodesy Symposia, Pages 161–169. Springer Berlin Heidelberg, doi: 10.1007/978-3-642-20338-1_20.
Jekeli, C. (2001). Inertial navigation systems with geodetic applications. de Gruyter, ISBN: 3-11-015903-1.
Jørgensen, J. L. (2000). In-orbit performance of a fully autonomous star tracker. In Spacecraft Guidance, Navigation and Control Systems. 4th ESA International Conference, Oct 18-21, 199, ESTEC, Noordwijk.
Jørgensen, J. L. and Liebe, C. C. (1996). The advanced stellar compass, development and operations. Acta Astronautica, 39, Pages 775–783.
Jørgensen, J. L. and Pickles, A. (1998). Fast and robust pointing and tracking using a second generation star tracker. Proceedings of the International Society for Optical Engineering, 3351, Pages 51–61.
Jørgensen, P., Jørgensen, J., Denver, T., and Betto, M. (2004). In-flight quality and accuracy of attitude measurements from the CHAMP advanced stellar compass. Acta Astronautica, 56, Pages 181–186, doi:
10.1016/j.actaastro.2004.09.012.
Jørgensen, P. S., Jørgensen, J. L., and Denver, T. (2008). On-the-fly merging of attitude solutions. In Sandau et al. (Ed.),Small satellites for Earth observation. Springer, ISBN: 978-1-4020-6942-0.
Kang, Z., Tapley, B., Bettadpur, S., Ries, J., and Nagel, P. (2006). Precise orbit determination for GRACE using accelerometer data. Advances in Space Research, 38, Pages 2131–2136, doi: 10.1016/j.asr.2006.02.021.
Kim, J. (2000). Simulation study of a low-low satellite-to-Satellite tracking mission. PhD thesis, University of Texas at Austin.
Kim, J. and Tapley, B. (2002). Error Analysis of a Low–Low Satellite-to-Satellite Tracking Mission. Journal of Guidance, Control and Dynamics, 25, Page 6.
Klinger, B. and Mayer-Gürr, T. (2014). Combination of GRACE star camera and angular acceleration data.
Geophysical Research Abstracts, 16, Pages EGU2014–5340.
Ko, U.-D. (2008). Analysis of the characteristics of the GRACE dual one-way ranging system. PhD thesis, University of Texas at Austin.
Kruizinga, G., Bertiger, W., Byun, S., Finch, C., and et al. (2012). GRACE Level-1 processing status. GRACE Science Team Meeting, Potsdam, Sep 17-19, 2012. GSTM 2012 proceedings.
Kruizinga, G., Bertiger, W., Byun, S., Finch, C., and et al. (2013). GRACE Level-1 processing status. GRACE Science Team Meeting, Austin, TX, Oct 23-25, 2013. GSTM 2013 proceedings.
Kuipers, J. B. (1999). Quaternions and ratation sequences. A primer with applications to orbits, aerospace, and virtual reality. Princeton University Press, Princeton, New Jersey, ISBN: 0-691-05872-5.
Liebe, C. C. (2002). Accuracy performance of star trackers - a tutorial. IEEE Transactions on Aerospace Electronic Systems, 38, Pages 587–599, doi: 10.1109/TAES.2002.1008988.
Liu, X. (2008). Global gravity field recovery from satellite-to-satellite tracking data with acceleration approach.
PhD thesis, Technical University of Delft.
Lundquist, C. A. and Veis, G. (1966). Geodetic Parameters for a 1966 Smithsonian Institution Standard Earth.
Sao special report #200, Smithsonian Astrophysical Observatory.
Luthcke, S. B., Zwally, H., Abdalati, W., Rowlands, D. D., Ray, R. D., Nerem, R., Lemoine, F., McCarthy, J. J., and Chinn, D. (2006). Recent Greenland ice mass loss by drainage system from satellite gravity observations.
Science, 314, Pages 1286–1289, doi: 10.1126/science.1130776.
Luthcke, S. B., Arendt, A. A., Rowlands, D. D., McCarthy, J. J., and Larsen, C. F. (2008). Recent glacier mass changes in the Gulf of Alaska region from GRACE mascon solutions. Journal of Glaciology, 54, Pages 767–777, doi: 10.3189/002214308787779933.
Mahrdt, C.rdt, C. (2014). Laser link acquisition for the GRACE Follow-On laser ranging Interferometry. PhD thesis, Gottfried Wilhelm Leibniz Universität Hannover.
Mandea, M., Holschneider, M., Lesur, V., and Lühr, H. (2010). System Earth via geodetic-geophysical space techniques, Chapter The Earth’s magnetic field at the CHAMP satellite epoch. Springer, doi: 10.1007/978-3-642-10228-8_42.
Mayer-Gürr, T. (2006). Gravitationsfeldbestimmung aus der Analyse Kurzer Bahnbögen am Beispiel der Satellitenmissionen CHAMP und GRACE. PhD thesis, University of Bonn.
Mayer-Gürr, T., Rieser, D., Höck, E., Brockmann, J.-M., Schuh, W.-D., Krasbutter, I., and at al. (2012).
The new combined satellite only model GOCO03s. International Symposium on Gravity, Geoid and Height Systems GGHS 2012.
McCarthy, D. (1996). IERS Conventions (1996). Technical Report IERS Technical Note 21, U.S. Naval Observatory.
Meyer, U., Jäggi, A., and Beutler, G. (2012). Monthly gravity field solutions based on GRACE observations generated with the Celestial Mechanics Approach. Earth and Planetary Science Letters, 345–348, Pages 72 – 80, doi: 10.1016/j.epsl.2012.06.026.
Michaelis, I. (2015). Star tracker / VFM activities. Swarm - National CalVal-Meeting, 12.-13.5.2015, GFZ Potsdam.
Montenbruck, O. and Gill, E. (2000). Satellite orbits: models, mmethod, applications. Springer, ISBN:
3-540-67280-X.
Montenbruck, O., van Helleputte, T., Kroes, R., and Gill, E. (2005). Reduced dynamic orbit determination using GPS code and carrier measurements. Aerospace Science and Technology, 9, Pages 261–271, doi:
10.1016/j.ast.2005.01.003.
Montenbruck, O., Garcia-Fernandez, M., Yoon, Y., Schön, S., and Jäggi, A. (2008). Antenna phase center calibration for precise positioning of LEO satellites.GPS Solutions, 13, Pages 23–34, doi: 10.1007/s10291-008-0094-z.
Müller, J., Naeimi, M., Gitlein, O., Timmen, L., and Denker, H. (2012). A land uplift model in Fennoscandia combining GRACE and absolute gravimetry data. Physics and Chemistry of the Earth, 53-54, Pages 54–60, doi: 10.1016/j.pce.2010.12.006.
NGGM-D Team, with contributions by Baldesarra, M., Brieden, P., Danzmann, K., Daras, I., Doll, B., Feili, D., Flechtner, F., Flury, J., Gruber, T., Heinzel, G., Iran Pour, S., Kusche, J., Langemenn, M., Löcher, A., Müller, J., Müller, V., Murböck, M., Naeimi, M., Pail, R., Raimondo, J., Reiche, J., Reubelt, T., Sheard, B., Sneeuw, N., and Wang, X. (2014). e2.motion- Earth System Mass Transport Mission (Square) - Concept for a Next Generation Gravity Field Mission - Final Report of Project "Satellite Gravimetry of the Next Generation (NGGM-D)". Deutsche Geodätische Kommission, ISBN: 978-3-7696-8596-1.
Northrop Grumman (2013). LN-200s Inertial measurement unit.http://www.northropgrumman.com/Capabilities/
LN200FOG/Documents/ln200s.pdf.
Panet, I., Flury, J., Biancale, R., Gruber, T., Johannessen, J., van den Broeke, M., and van Dam, T. (2013).
Earth System Mass Transport Mission (e.motion): A Concept for Future Earth Gravity Field Measurements from Space. Surveys in Geophysics, 34, Pages 141–163, doi: 10.1007/s10712-012-9209-8.
Peterseim, N. (2014). Twangs - high-frequency disturbing signals in 10 Hz accelerometer data of the GRACE satellites. PhD thesis, Technische Universität München.
Presti, D., Herman, J., and Codazzi, A. (2004). Mission operations system design and adaptations for the twin-satellite mission GRACE. Space OPS 2004 Conference. doi: 10.2514/6.2004-382-219.
Psiaki, M., Martel, F., and Pal, P. (1990). Three-axis attitude determination via Kalman filtering of magnetometer data. Journal of Guidance, Control and Dynamics, 13, Pages 506–514, doi: 10.2514/3.25364.
Ray, R. D. and Luthcke, S. B. (2006). Tide model errors and GRACE gravimetry: towards a more realistic assess-ment. Geophysical Journal International, 167, Pages pp. 1055–1059, doi: 10.1111/j.1365-246X.2006.03229.x.
Reigber, C., Schwintzer, P., and Lühr, H. (1999). The CHAMP geopotential mission. Bollettino di Geofisica Teorica ed Applicata, 40, Pages 285–289.
Rodell, M., Famiglietti, J. S., Chen, J., Seneviratne, S. I., Viterbo, P., Holl, S., and Wilson, C. R. (2004). Basin scale estimates of evapotranspiration using GRACE and other observations. Geophysical Research Letters, 31, Page 20504, doi: 10.1029/2004GL020873.
Rodell, M., Velicogna, I., and Famiglietti, J. S. (2009). Satellite-based estimates of groundwater depletion in India. Nature, 460, Pages 999–1003, doi: 10.1038/nature08238.
Romans, L. (2003). Optimal combination of quaternions from multiple star cameras. Technical Report, JPL Internal Memorandum.
Sasgen, I., Konrad, H., Ivins, E., Van den Broeke, M., Bamber, J., Martinec, Z., and Kleemann, V. (2013).
Antarctic ice-mass balance 2003 to 2012: regional reanalysis of GRACE satellite gravimetry measurements with improved estimate of glacial-isostatic adjustment based on GPS uplift rates. The Cryosphere, 7, Pages 1499–1512, doi: 10.5194/tc-7-1499-2013.
Schelkle, M. (2000). The GRACE cold gas attitude and orbit control system. In3rd International conference on spacecraft propulsion, ESA SP-465.
Schütze, D. (2015). Intersatellite laser interferometry. Test environments for GRACE Follow-On. PhD thesis, Gottfried Wilhelm Leibniz Universität Hannover.
Schütze, D., Stede, G., Müller, V., Gerberding, O., Bandikova, T., Sheard, B., Heinzel, G., and Danzmann, K.
(2014a). Laser beam steering for GRACE Follow-On intersatellite Interferometry. Optics Express, 22, Pages 24117–24132, doi: 10.1364/OE.22.024117.
Schütze, D., Müller, V., Stede, G., Sheard, B., Heinzel, G., Danzmann, K., Sutton, A., and Shaddock, D.
(2014b). Retroreflector for GRACE Follow-On: Vertex vs. point of minimal coupling. Optics Express, 22, Pages 9324–9333, doi: 10.1364/OE.22.009324.
Sheard, B. S., Heinzel, G., Danzmann, K., Shaddock, D. A., and et al. (2012). Intersatellite laser ranging instrument for the GRACE Follow-On mission.Journal of Geodesy, 86, Pages 1083–1095, doi: 10.1007/s00190-012-0566-3.
Siemes, C. (2011). GOCE gradiometer calibration and Level 1b data processing. Technical Report EWP-2384, European Space Agency.
Stanton, R. (2000). Science & Mission requirements document. Technical Report JPL D-15928, Jet Propulsion Laboratory, University of Texas Center for Space Research, GeoForschungszentrum Potsdam.
Steffen, H., Denker, H., and Müller, J. (2008). Glacial isostatic adjustment in Fennoscandia from GRACE data and comparison with geodynamical models. Journal of Geodynamics, 46, Pages 155–164, doi:
10.1016/j.jog.2008.03.002.
Stummer, C., Fecher, T., and Pail, R. (2011). Alternative method for angular rate determination within the GOCE gradiometer processing. Journal of Geodesy, 85, Pages 585–596, doi: 10.1007/s00190-011-0461-3.
Svendsen, P., Andersen, O., and Nielsen, A. (2013). Acceleration of the Greenland ice sheet mass loss as observed by GRACE: Confidence and sensitivity. Earth and Planetary Science Letters, 364, Pages 24–29, doi:
10.1016/j.epsl.2012.12.010.
Tapley, B., Bettadpur, S., Watkins, M., and Reigber, C. (2004). The Gravity Recovery and Climate Experiment:
Mission overview and early results.Geophysical Research Letters, 31, doi: 10.1029/2004GL019920.
Tapley, B., Flechtner, F., Bettadpur, S., and Watkins, M. (2013). The status and future prospect for GRACE after the first decade.AGU, Fall Meeting 2013, abstract G32A-01.
Thomas, J. (1999). An analysis of gravity-field estimation based on intersatellite dual 1-way biased ranging.
Technical Report JPL 98-15, Jet Propulsion Laboratory.
Torge, W. (1989). Gravimetry. Walter de Gruyter, Berlin, New York, ISBN: 3-11-010702-3.
Torge, W. and Müller, J. (2012). Geodesy. De Gruyter, ISBN: 978-3-11-020718-7.
Touboul, P., Foulon, B., and Willemenot, E. (1999). Electrostatic space accelerometers for present and future missions. Acta Astronautica, 45, Pages 605–617, doi: 10.1016/S0094-5765(99)00132-0.
Usbeck, T., Wohlfart, J., and Schelkle, M. (2004). A flexible cold gas propulsion system concept for different space applications. In4th Int. Spacecraft Propulsion Conference, June 2-4, 2004.
van Helleputte, T. and Visser, P. (2008). GPS based orbit determination using accelerometer data.Aerospace Science and Technology, 12, Pages 478–484, doi: 10.1016/j.ast.2007.11.002.
Velicogna, I. and Wahr, J. (2006). Measurements of time-variable gravity show mass loss in Antarctica.Science, 311, Pages 1754–1756, doi: 10.1126/science.1123785.
Vey, S., Steffen, H., Müller, J., and Boike, J. (2013). Inter-annual water mass variations from GRACE in central Siberia. Journal of Geodesy, 87, Pages 287–299, doi: 10.1007/s00190-012-0597-9.
Wang, F. (2003). Study on Center of Mass Calibration and K-band Ranging System Calibration of the GRACE Mission. PhD thesis, The University of Texas at Austin.
Wang, L., Shum, C. K., Simons, F., Tassara, A., Erkan, K., Jekeli, C., and et al. (2012). Coseismic slip of the 2010 Mw 8.8 Great Maule, Chile, earthquake quantified by the inversion of GRACE observations. Earth and Planetary Science Letters, 335–336, Pages 167–179, doi: 10.1016/j.epsl.2012.04.044.
Wang, W., Shi, Y., Sun, W., and Zhang, J. (2011). Viscous lithospheric structure beneath Sumatra inferred from post-seismic gravity changes detected by GRACE. Science China Earth Sciences, 54, Pages 1257–1267, doi: 10.1007/s11430-011-4217-y.
Watkins, M., Gruber, T., and Bettadpur, S. (2000). Science data system development plan. Technical Report 327-710, JPL, GFZ, CSR.
Watkins, M., Flechtner, F., and Morton, P. (2013). Status of the GRACE Follow-on mission.GRACE Science Team Meeting. GSTM proceedings.
Wertz, J. R. (1978).Spacecraft Attitude Determination and Control. D. Reidel Publishing company, Dordrecht, Holland, ISBN: 90-277-0959-9.
Wertz, J. R. (2001). Orbit and Constellation Design and Management. Microcosm Press, Springer, ISBN:
1881883078.
Wertz, J. R. and Larson, W. (1999). Space mission analysis and design. Microcosm Press, Springer, ISBN:
978-1881883-10-4.
Wickert, J., Beyerle, G., Koenig, R., Heise, S., Grunwaldt, L., Michalak, G., Reigber, C., and Schmidt, T.
(2005). GPS radio occultation with CHAMP and GRACE: A first look at a new and promising satellite configuration for global atmospheric sounding. Annales Geophysicae, 23, Pages 653–658.
Wickert, J., Michalak, G., Schmidt, T., Beyerle, G., CChen, C.-Z., Healy, S., Heise, S., and et al. (2009). GPS Radio Occultation: Results from CHAMP, GRACE and FORMOSAT-3/COSMIC. Terrestrial Atmospheric And Oceanic Sciences, 20, Pages 35–50, doi: 10.3319/TAO.2007.12.26.01(F3C).
Wiese, D., Nerem, R., and Lemoine, F. (2011). Design considerations for a dedicated gravity recovery satellite mission consisting of two pairs of satellites. Journal of Geodesy, 86, Pages 81–98, doi: 10.1007/s00190-011-0493-8.
Witkowski, M. and Massmann, F.-H. (2011). Status GRACE mission operations.GRACE Science Team Meeting.
GSTM proceedings 2011.
Witkowski, M. and Massmann, F.-H. (2012). Status GRACE mission operations.GRACE Science Team Meeting.
GSTM proceedings 2012.
Witkowski, M. and Massmann, F.-H. (2014). Status GRACE mission operations.GRACE Science Team Meeting.
GSTM proceedings 2014.
Wolff, M. (1969). Direct measurements of the Earth’s gravitational potential using a satellite pair.Journal of Geophysical Research, 74, Pages 5295–5300.
Wu, S.-C., Kruizinga, G., and Bertiger, W. (2006). Algorithm theoretical basis document for GRACE Level-1B data processing v1.2. Technical Report GRACE 327-741, Jet Propulsion Laboratory.
Wuchenich, D., Mahrdt, C., Sheard, B., Francis, S., Spero, R., Miller, J., Mow-Lowry, C., Ward, R., Klipstein, W., Heinzel, G., Danzmann, K., McClelland, D., and Shaddock, D. (2014). Laser link acquisition demonstration for the GRACE Follow-On mission. Optics Express, 22, Pages 11351–11366, doi: 10.1364/OE.22.011351.
Yuan, D.-N., Byun, S., and Kuang, D., e. a. (2012). GRACE JPL L2 RL05 status. GRACE Science Team Meeting, Potsdam, Sep 17-19, 2012. GSTM 2012 proceedings.
ZARM Technik (2010). Magnetic torquers for spacecraft attitude control. http://www.zarm-technik.de/
Zenner, L. (2013). Atmospheric and oceanic mass variations and their role for gravity field determination. PhD thesis, Technische Universität München.
Zuber, M., Smith, D., Watkins, M., Asmar, S., Konopliv, A., Lemoine, F., and et al. (2013). Gravity field of the Moon from the Gravity Recovery and Interior Laboratory (GRAIL) mission. Science, 339 (6120), Pages 668–671, doi: 10.1126/science.1231507.
1.1 The role of attitude determination for inter-satellite ranging mission . . . . 11
2.1 The Gravity Recovery and Climate Experiment (GRACE). . . . 13
2.2 GRACE groundtrack . . . . 15
2.3 GRACE measurement principle . . . . 15
2.4 Orientation of the GRACE satellites relative to the orbit trajectory. . . . 16
2.5 The principle of dual one way ranging system . . . . 17
2.6 Modeled error of the KBR range measurement . . . . 18
2.7 Accelerometer error model . . . . 19
2.8 GRACE payload . . . . 21
2.9 Comparison of the GRACE baseline accuracy with the accuracy of the monthly gravity field models 23 2.10 Static gravity field model in terms of geoid . . . . 23
2.11 Monthly gravity field models in terms of geoid . . . . 24
3.1 The attitude determination and control loop. . . . 26
3.2 Coarse Earth/Sun sensor (CESS) . . . . 28
3.3 CESS temperature data in 2008. . . . 28
3.4 Vector components of the Earth’s magnetic flux density . . . . 30
3.5 Inertial measurement unit - sensor unit. . . . 31
3.6 Inertial measurement unit - angular rates time series . . . . 31
3.7 Star camera sensor . . . . 32
3.8 Star camera mounting onboard GRACE . . . . 33
3.9 Block diagram of a star camera . . . . 33
3.10 Block diagram of star camera attitude determination . . . . 34
3.11 Star camera data availability in 2008 . . . . 35
3.12 Demonstration of the anisotropic noise of the star camera measurement . . . . 37
3.13 Star camera measurement accuracy . . . . 39
3.14 Number of stars in the star camera field-of-view. . . . 40
3.15 Star camera inter-boresight angle . . . . 42
3.16 Sensitivity of the star cameras to temperature changes . . . . 42
3.17 Magnetic torquer and its principle of operation . . . . 43
3.18 Earth’s magnetic field direction in 2010 . . . . 44
3.19 Electric current flow through magnetic torquer rods . . . . 45
3.20 Frequency switch in magnetic torquer current flow . . . . 45
3.21 Cold gas thrusters on GRACE . . . . 46
3.22 The principle of attitude control using cold gas thrusters. . . . 47
3.23 Geographical location of thruster activations . . . . 47
3.24 Propellant consumption 2002-2014 . . . . 49
3.25 Solar activity during GRACE mission operation period. . . . 49
4.1 Sketch of the ideal and the real inter-satellite pointing . . . . 52
4.2 Mutual orientation of the K-Frame and the LOS-Frame . . . . 52
4.3 Flowchart of the in-flight determination of the inter-satellite pointing. . . . 54
4.4 Flowchart of the on-ground determination of the inter-satellite pointing . . . . 55
4.5 Inter-satellite pointing variations plotted as time series and along groundtrack. . . . 57
4.6 Inter-satellite pointing pitch angle variations in 2007 and 2008. . . . 58
4.7 Visibility of satellite operation events in inter-satellite pointing angles . . . . 58
4.8 Sketch of the antenna offset correction for range . . . . 59
4.9 Rotation from SCF into KF using QKS, QSA and VKB calibration parameters . . . . 60
4.10 Comparison of the in-flight and on-ground determined pointing angles . . . . 63
4.11 Bias of the inter-satellite pointing angles in 2007 and 2008 . . . . 64
4.12 Effect of the pointing bias on the antenna offset correction. . . . 65
4.13 Comparison of the AOC derived from the in-flight and on-ground computed pointing angles . . 66
5.1 Demonstration of noise propagation after SCA data rotation from SCF into SRF . . . . 69
5.2 Demonstration of the unexpected higher noise of the SCA1B RL02 data . . . . 69
5.3 Scheme of SCA1A to SCA1B data processing while implementing the method for SCA data combination based on a weight matrix . . . . 72
Outline
ÄHNLICHE DOKUMENTE | 9,794 | 31,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.595232 |
https://www.coursehero.com/file/p6hjdej5/a-sorting-technique-to-sort-an-array-or-we-can-say-to-sort-a-list-of-many/ | 1,579,604,027,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250603761.28/warc/CC-MAIN-20200121103642-20200121132642-00399.warc.gz | 824,506,867 | 58,287 | a sorting technique to sort an array or we can say to sort a list of many
# A sorting technique to sort an array or we can say to
• 7
This preview shows page 2 - 3 out of 7 pages.
a sorting technique to sort an array, or we can say to sort a list of many numbers and insertion sort is a sorting algorithm that places the input element at its suitable place in each pass while the merge sort technique is based on divide and conquer technique it will divide the data set into smaller parts and merge them into a larger piece in sorted order. Radix sort works on the integer keys by grouping digits which share the same position and value and lastly counting sort is used to sort objects according the keys that are small numbers. It counts the number of keys whose key values are same. For example 2. What makes binary search faster than sequential search? When can you only use binary search? - Binary search faster because is it has a time complexity of O(log n) and it works by finding the middle element of a sorted array and comparing it to your target element while the sequential search looks for an item in a list using a target value called the key and it only the search begins when the
#### You've reached the end of your free preview.
Want to read all 7 pages?
• Fall '16
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 474 | 2,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-05 | latest | en | 0.927922 |
http://learn2cube.com/cfop/f2l/index.php | 1,722,830,040,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00890.warc.gz | 19,640,399 | 4,759 | Fridrich (CFOP) Method
First 2 Layers (F2L)
Introduction
F2L (First 2 Layers) is a way of continuing after solving the cross. In this step, a first-layer corner (henceforth referred to as "the corner") and the middle-layer edge above it ("the edge") are inserted at the same time into their designated locations ("the slot"). F2L combines steps 2 and 3 of my beginner solution, but it requires approximately the same number of moves as step 3 alone. When done at full speed, F2L can be completed in less than 10 seconds!
Most F2L cases consist of two parts. One - the pairing of the corner and edge (aka "F2L pair") - and two - the insertion of the F2L pair. The moves are based solely on intuition, so algorithms are not required. However, many other sites do have so-called "algorithms" for this step. In most cases those algorithms are simply the intuitive pair-and-insert procedure optimized with moves such as x, y, d, etc. Don't start by learning those algorithms. Learn the intuitive steps first and fully understand how the moves work. It will greatly reduce the amount of memorization needed. After you get all 42 cases down (1 solved pair, 19 cases + their mirrors, and 3 symmetrical cases), then you can learn the "algorithms" if you want.
Note: In the diagrams of the cases (made with Lars Vandenbergh's amazing ImageCube script), only three sides can be shown, so sometimes, stickers from the edge cubie will be hidden from view. The red-blue F2L pair is used in all of these diagrams so the sticker that is hidden would be either red or blue, depending on which sticker is shown. For example, if the edge only shows a blue sticker, the unseen sticker would be red.
Another Note: I have not had time to make sure all these moves are correct. If you spot an error, please contact me.
The F2L Cases
In these cases, it is assumed that the corner and edge are either in their respective locations or in the top layer. If one or both pieces are stuck in another slot, you can simply move them out by inserting random pieces from the top layer into that slot. Or you can check the advanced F2L section.
For each case, there will be a "method" (how to convert it to one of the basic cases) and an "algorithm" in with a bullet in front of it. Ignore the algorithm if you are just learning F2L.
Basic Cases
Case 2
Corner and Edge in Top Layer (Type 1)
Case 3
Corner and Edge in Top Layer (Type 2)
Case 4
Corner on Bottom, Edge on Top
Case 5
Corner on Top, Edge in Middle
Case 6
Corner on Bottom, Edge in Middle
THIS SECTION IS UNDER CONSTRUCTION
These miscellaneous techniques can help you improve your speed in the step. They are for after you get the basics down so don't read these until after you do. Otherwise, they will only confuse you! | 661 | 2,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.927477 |
https://coin.fyi/news/bitcoin/there-may-be-between-400-000-and-800-000-people-who-hold-at-least-one-bitco-f3om1m | 1,591,095,590,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347424174.72/warc/CC-MAIN-20200602100039-20200602130039-00466.warc.gz | 299,641,552 | 81,676 | Bitcoin
\$10,125.70 6.12%
BTC · 15w
# There may be between 400,000 and 800,000 people who hold at least one Bitcoin. If they would live in a community, they’d make a city whose population would compare to that of Las Vegas, US (583,756)
Bitcoin has consolidated above \$10,000 this week. This is a pretty decent amount for one coin, so you might wonder how many people hold at least 1 BTC.
We’ll begin by checking the distribution of Bitcoin by addresses as presented by BitInfoCharts.
As you can see, there are 633,008 addresses that have at least 1 BTC (1.68 million coins in total), which is 2.18% of all addresses out there. However, one address doesn’t necessarily equal one person, so we’ll have to dig deeper in order to obtain a realistic range.
We already have the first scenario, in which one person has a single wallet with no less than one Bitcoin. That’s the best-case scenario for us – it’s the most convenient one for counting.
But the situation gets trickier in the second scenario, in which one person can possess a total amount of 1+ Bitcoin but in several different addresses. For example, one might hold 0.8 BTC in one wallet and 0.2 in another one, or keep 0.1 BTC each in 10 different wallets. In these cases, the addresses would have less than one coin, so they should be displayed in other rows of the distribution table – the ones that we initially ignored. Counting those who hold at least 1BTC in the second scenario would be much more difficult, as they possess any of the 98% of addresses that contain less than 1 BTC.
There is also a third scenario. For example, you would expect that we should count a wallet holding 10 BTC. However, it might belong to a company or association where each member is eligible to less than 1 Bitcoin from that wallet, so he/she shouldn’t make our list.
Anyway, based on the distribution table, we can be sure that between 500,000 and 1 million people have at least 1 BTC.
Right – we shouldn’t ignore this aspect. If I hold 1 BTC and it happens that I lose the private keys, do I still possess the funds? Not really. If we consider th...
Continue on bitcoinist.com
forbes.com
Bitcoin
\$10,125.70
28
coindesk.com
Bitcoin
\$10,125.70
14
zycrypto.com
#### “Bitcoin is the peaceful protest,” CZ says as U.S. riot heightens
Bitcoin
\$10,125.70
10
Recent news
BTC +6.12% · rusty.ozlabs.org · 2h
#### (Some) Exchanges Are Now The Enemy of Bitcoin
BTC +6.12% · nytimes.com · 6h
#### Hong Kong was once the only way to get money out of China, now Bitcoin is the only escape
Beijing is using fear and pressure to drum up support for its increasingly hard-line stance in the Asian financial capital, threatening its status as a global business center.
BTC +6.12% · anchor.fm · 15h
#### Bitcoin Rapid-Fire: Bheki Mahlobo: Bitcoin in South Africa
Bheki Mahlobo is a student at the University of South Africa, majoring in economics. He was recently featured in the documentary “Banking on Africa: The Bitcoin Revolution", on Amazon Prime Video. I’m...
BTC +6.12% · ccnews24.net · 16h
#### Bitcoin has maintained a good position and continues to grow •
As we wrote in the last analysis, overcoming \$ 9K really worked as a strong impetus for the recovery of the bulls and the market was more optimistic.
BTC +6.12% · bitboycrypto.com · 16h
#### Battle of Privacy: Bitcoin May be Superior to Ethereum
Research shows that Ethereum transactions are much easier to be de-anonymized, when compared to Bitcoin. This may be a problem.
BTC +6.12% · decrypt.co · 16h
#### Glassnode: Over 60% of Bitcoin has not moved in a year
Over 60 percent of Bitcoin supply has not moved in over a year, data from Glassnode shows. Not since 2016 has this much Bitcoin been HODLed.
BTC +6.12% · medium.com · 18h
#### In the Post-Dollar World That Is Shaping Up for the Future, Bitcoin Will Have a Major Role to Play
The beginning of a New Cold War between the U.S. and China is a trend that will lead to a post-dollar world in which Bitcoin will have a major role to play.
BTC +6.12% · twitter.com · 19h
#### Now more than ever, It’s Time for Plan ₿
Our tax dollars have bought us a system that has repeatedly #failed everyone: citizens are not protected and businesses are allowed to #burn. It should be #defunded. It is time for a #decentralized so...
BTC +6.12% · zycrypto.com · 19h
#### “Bitcoin is the peaceful protest,” CZ says as U.S. riot heightens
As the riot and protest in the United States continues, Binance CEO Changpang “CZ” Zhao says Bitcoin is the peaceful version of the protest
10
BTC +6.12% · forbes.com · 20h
#### Russia Sort Of Dropped The Hammer On Bitcoin
The new law is designed to stop tax fraud. But it makes it nearly impossible for exchanges to operate in Russia. The same may hold for mining operations.
28
BTC +6.12% · coindesk.com · 21h
#### Maker of Coldcard Bitcoin Wallet Rolls Out an Extra-Strength 'USB Condom'
CoinKite, maker of the Coldcard hardware wallet, has introduced two accessories that underscore the near-paranoia required to safely hold bitcoin.
14
BTC +6.12% · decrypt.co · 21h
#### He directed a Taylor Swift music video. Now he's trading Bitcoin.
Multi award-winning video director Joseph Kahn says “If anyone has any ideas on crypto I'm open.” Anyone care to oblige him?
BTC +6.12% · beincrypto.com · 21h
#### Bitcoin 401(K) Plans Introduced by Gemini-Partnered Payroll Service Provider
Crypto payroll company Bitwage has launched the world's first Bitcoin 401(k) plan and has partnered with Gemini on the project.
BTC +6.12% · cryptothenews.com · 23h
#### Kleiman Legal Team Presents New Evidence of Craig Wright’s “Fabrication
The Kleiman estate's legal team has entered into evidence last week's revelation that 145 addresses claimed by Craig Wright are not controlled by him.
BTC +6.12% · medium.com · 23h | 1,465 | 5,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-24 | latest | en | 0.958028 |
https://caracolche.com/south-australia/value-at-risk-calculation-example.php | 1,590,947,361,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00058.warc.gz | 281,074,697 | 6,675 | Value at risk calculation example
South Australia - 2019-09-25
# At calculation example risk value
## The estimation of Value at Risk and Expected Shortfall R. Value at Risk Calculation Portfolio Risk Analysis.
Lecture 7: Value At Risk (VAR) • Example of one-asset VaR to Calculate? 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40. Is it possible calculate Value at Risk on an asset without a time horizon? What kind of variables do you need? Variables that are on the table are value, standard.
The “Value at Risk” Concept for Insurance Companies Willi Ufer In order to run a Value-at-Risk calculation on a portfolio, all transaction data Value at Risk (VAR) is defined as An example of a parametric VAR calculation is as has been fully incorporated in the risk calculation without the forced
An Introduction to Value at Risk (VAR) VAR and the three basic methods of calculating it. in mind as we give some examples of variations of the question 5/10/2015 · Component versus Incremental value at risk Is this the correct method to delete the EUR VAR to calculate the CAD Incremental VAR? For example, if I
Anybody can do Value at Risk: A Nonparametric Teaching Study Abstract Value at Risk VaR calculation can. Value, optimise and manage Value-at-Risk for Energy Derivatives. Detailed example VaR calculation for an energy portfolio;.
“Value at Risk Historical Simulation”.
Here is my shot at doing Historical Simulation to find the Value at Risk of your portfolio.Due to the method it is not a great method for risk management - but can.
A risk management model that calculates the largest possible loss that an institution or other investor could incur on a portfolio. Value at risk calculation can. Value, optimise and manage Value-at-Risk for Energy Derivatives. Detailed example VaR calculation for an energy portfolio;. What is Value at Risk (VaR)?|How to Calculate Value at Risk to risk adjusted value and historical data above to calculate portfolio VaR. Example:.
Learn how MATLAB uses various mathematical techniques to calculate value-at-risk (VaR) to predict the potential loss in different types of risk exposure. VaR is used Example: Probability (\$1 Value at Risk is only about Market Risk under normal market conditions. - Calculate portfolio expected return and standard deviation;
Evaluation of Value-at-Risk Models Using Historical Data For example, in its recent risk-based a risk manager can use a single calculation of the portfolio Over the last two months, we have explained how variance-covariance methods of calculating Value at Risk work. However, we have also highlighted some of the problems
## An example of cultural diffusion
Cultural Diffusion Meaning Types and Examples. Get an answer for 'What are some examples of cultural diffusion in technology?' and find homework help for other History questions at eNotes …
## Long Double In Java Example
Java 2 Variables(int double String) - YouTube. For double conversion, for example, Java 6 has an interesting sequence of calls One thought on “ Using double/long vs BigDecimal for monetary calculations ” … | 673 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-24 | latest | en | 0.831079 |
https://jarviscodinghub.com/product/cpts355-lab-1-haskell-solution/ | 1,709,009,585,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474670.19/warc/CC-MAIN-20240227021813-20240227051813-00801.warc.gz | 320,142,633 | 24,100 | # CptS355 – Lab 1 (Haskell) solution
\$30.00
Original Work ?
Category:
## Description
Lab problems:
1.insert
Write a function insert that takes an integer “n”, a value “item”, and a list “iL” and inserts the
“item” at index “n” in the list “iL”. “n” is a 1-based index, i.e., “item” should be inserted after n
th
element in the list. The type of insert can be one of the following:
insert :: (Num t1) => t1 -> t2 -> [t2] -> [t2]
insert :: (Eq t1, Num t1) => t1 -> t2 -> [t2] -> [t2]
insert :: (Ord t1, Num t1) => t1 -> t2 -> [t2] -> [t2]
If “n” is greater than the length of the input list, the “item” will not be inserted. If “n” is 0, “item” will
be inserted to the beginning of the list. (You may assume that n >=0.)
Examples:
> insert 3 100 [1,2,3,4,5,6,7,8]
[1,2,3,100,4,5,6,7,8]
> insert 8 100 [1,2,3,4,5,6,7,8]
[1,2,3,4,5,6,7,8,100]
> insert 9 100 [1,2,3,4,5,6,7,8]
[1,2,3,4,5,6,7,8]
> insert 3 100 []
[]
2.insertEvery
Write a function insertEvery that takes an integer “n”, a value “item”, and a list “iL” and inserts
the “item” at every n
th index in “iL”. “n” is a 1-based index, i.e., “item” should be inserted after n
th
,
2n
th,3nth
, etc. elements in the list. The type of insertEvery can be one of the following:
insertEvery :: (Num t) => t -> a -> [a] -> [a]
insertEvery :: (Eq t, Num t) => t -> a -> [a] -> [a]
If “n” is greater than the length of the input list, the “item” will not be inserted. If “n” is 0, “item” will
be inserted to the beginning of the list. (You may assume that n >=0.)
Examples:
> insertEvery 3 100 [1,2,3,4,5,6,7,8,9,10]
[1,2,3,100,4,5,6,100,7,8,9,100,10]
> insertEvery 8 100 [1,2,3,4,5,6,7,8]
[1,2,3,4,5,6,7,8,100]
> insertEvery 9 100 [1,2,3,4,5,6,7,8]
[1,2,3,4,5,6,7,8]
> insertEvery 3 100 []
[]
3.getSales
Assume that you have an online sales business and you sell products on Amazon, Ebay, Etsy, etc. and
you keep track of your daily sales (in \$) for each online store. You maintain the log of your sales in a
First consider the sales log for a single store, for example:
storelog = [(“Mon”,50),(“Fri”,20), (“Tue”,20),(“Fri”,10),(“Wed”,25),(“Fri”,30)]
storelog is a list of (day, sale-amount) pairs. Note that the store may have multiple sales on the same
day of the week or may not have any sales on some days.
Write a function getSales that takes a “day” abbreviation (e.g. “Mon”, “Tue”, etc.) and a store’s sales
log as input and returns the total sales in that store for the given day.
The type of getSales should be getSales :: (Num p, Eq t) => t -> [(t, p)] -> p
Examples:
> getSales “Fri” storelog
60
> getSales “Mon” storelog
50
> getSales “Sat” storelog
0
4.sumSales
Now we combine the sales logs for all stores into one single list. An example list is given below:
sales = [(“Amazon”,[(“Mon”,30),(“Wed”,100),(“Sat”,200)]),
(“Etsy”,[(“Mon”,50),(“Tue”,20),(“Wed”,25),(“Fri”,30)]),
(“Ebay”,[(“Tue”,60),(“Wed”,100),(“Thu”,30)]),
(“Etsy”,[(“Tue”,100),(“Thu”,50),(“Sat”,20),(“Tue”,10)])]
The list includes tuples where the first value in the tuple is the store name and the second value is the
list of (day, sale amount) pairs. Note that the list may include multiple entries (tuples) for the same
store.
Write a function, sumSales, that takes a store name, a day-of-week, and a sales log list (similar to
“sales”) and returns the total sales of that store on that day-of-week.
(Hint: You can make use of getSales function you defined in part-a.)
The type of sumSales can be:
sumSales:: (Num p)=> String -> String -> [(String,[(String,p)])] -> p
> sumSales “Etsy” “Tue” mysales
130
> sumSales “Etsy” “Sun” mysales
0
> sumSales “Amazon” “Mon” mysales
30
5.split
You should implement your own split logic for this problem. You are not allowed to use any built-in or
aWrite a function split that takes a delimiter value “c” and a list “iL”, and it splits the input list with
respect to the delimiter “c”. The goal is to produce a result in which the elements of the original list have
been collected into ordered sub-lists each containing the elements between the occurrences of the
delimiter in the input list. The delimiter value should be excluded from the sub-lists.
The type of split can be one of the following:
split :: Eq a => a -> [a] -> [[a]]
split :: a -> [a] -> [[a]]
Examples:
> split ‘,’ “Courses:,CptS355,CptS322,CptS451,CptS321”
[“Courses:”,”CptS355″,”CptS322″,”CptS451″,”CptS321″]
> split 0 [1,2,3,0,4,0,5,0,0,6,7,8,9,10]
[[1,2,3],[4],[5],[],[6,7,8,9,10]]
6. nSplit
Write a function nSplit that takes a delimiter value “c”, an integer “n”, and a list “iL”, and it splits
the input list with respect to the delimiter “c” up to “n” times. Unlike split, it should not split the
input list at every delimiter occurrence, but only for the first “n” occurrences of it.
The type of nSplit can be one of the following:
nSplit :: (Ord a1, Num a1, Eq a2) => a2 -> a1 -> [a2] -> [[a2]]
nSplit :: (Num a1, Eq a2) => a2 -> a1 -> [a2] -> [[a2]]
nSplit :: (Num a1) => a2 -> a1 -> [a2] -> [[a2]]
Examples:
> nSplit ‘,’ 1 “Courses:,CptS355,CptS322,CptS451,CptS321”
[“Courses:”,”CptS355,CptS322,CptS451,CptS321″]
> nSplit ‘,’ 2 “Courses:,CptS355,CptS322,CptS451,CptS321”
[“Courses:”,”CptS355″,”CptS322,CptS451,CptS321″]
> nSplit ‘,’ 4 “Courses:,CptS355,CptS322,CptS451,CptS321”
[“Courses:”,”CptS355″,”CptS322″,”CptS451″,”CptS321″]
> nSplit ‘,’ 5 “Courses:,CptS355,CptS322,CptS451,CptS321”
[“Courses:”,”CptS355″,”CptS322″,”CptS451″,”CptS321″]
> nSplit 0 3 [1,2,3,0,4,0,5,0,0,6,7,8,9,10]
[[1,2,3],[4],[5],[0,6,7,89,10]]
The Lab1SampleTests.zip file includes 6 .hs files where each one includes the HUnit tests for a
different lab problem. The tests compare the actual output to the expected (correct) output and raise an
exception if they don’t match. The test files import the Lab1 module (Lab1.hs file) which will include
your implementations of the lab problems.
You will write your solutions to Lab1.hs file. To test your solution for the first lab problem run the
following commands on the command line window (i.e., terminal):
\$ ghci
\$ :l Q1_tests.hs
*Q1_tests> main
Repeat the above for other lab problems by changing the test file name, i.e. , Q2_tests.hs,
Q3_tests.hs, etc. | 2,161 | 6,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.797771 |
https://codingwithnick.in/dictionaries-and-maps-30-days-of-code-solution/ | 1,713,027,241,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00616.warc.gz | 153,522,510 | 50,597 | Home30 Days Of CodeHackerrank Day 8 : Dictionaries and Maps 30 days of code solution
# Hackerrank Day 8 : Dictionaries and Maps 30 days of code solution
In this Hackerrank Day 8 : Dictionaries and Maps 30 days of code solution we’re learning about Key-Value pair mappings using a Map or Dictionary data structure.
Given n names and phone numbers, assemble a phone book that maps friends’ names to their respective phone numbers. You will then be given an unknown number of names to query your phone book for. For each name queried, print the associated entry from your phone book on a new line in the form `name=phoneNumber`; if an entry for name is not found, print `Not found` instead.
Note: Your phone book should be a Dictionary/Map/HashMap data structure.
### Input Format
The first line contains an integer, n, denoting the number of entries in the phone book.
Each of the n subsequent lines describes an entry in the form of 2 space-separated values on a single line. The first value is a friend’s name, and the second value is an 8-digit phone number.
After the n lines of phone book entries, there are an unknown number of lines of queries. Each line (query) contains a name to look up, and you must continue reading lines until there is no more input.
Note: Names consist of lowercase English alphabetic letters and are first names only.
### Constraints
• 1 <= n <= 105
• 1 <= queries <= 105
### Output Format
On a new line for each query, print `Not found` if the name has no corresponding entry in the phone book; otherwise, print the full name and phoneNumber in the format `name=phoneNumber`.
### Sample Input
``````3
sam 99912222
tom 11122222
harry 12299933
sam
edward
harry``````
### Sample Output
``````sam=99912222
harry=12299933``````
### Explanation
We add the following n = 3 (Key,Value) pairs to our map so it looks like this:
phoneBook = {(sam, 99912222), (tom, 11122222), (harry, 12299933)}
We then process each query and print `key=value` if the queried key is found in the map; otherwise, we print `Not found`.
Query 0: `sam`
Sam is one of the keys in our dictionary, so we print `sam=99912222`.
Query 1: `edward`
Edward is not one of the keys in our dictionary, so we print `Not found`.
Query 2: `harry`
Harry is one of the keys in our dictionary, so we print `harry=12299933`.
## Dictionaries and Maps HackerRank Solution in C
```#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <strings.h>
typedef struct pair {
char* first;
char* second;
struct pair* next;
} pair;
typedef struct dict {
int size;
pair** table;
} dict;
unsigned int hash(char* s) {
unsigned int hashval = 1337;
for (int i=0; i<(int)strlen(s); i++) {
hashval = hashval * s[i] + 0xdeadbeef;
hashval %= 0x3f3f3f3f;
}
return hashval;
}
void setsize(dict* d, int s) {
d->table = malloc(s * sizeof(pair*));
d->size = s;
bzero(d->table, s * sizeof(pair*));
}
void insert(dict* d, char* k, char* v) {
pair* p = malloc(sizeof(pair));
char* s = malloc(strlen(k) * sizeof(char) + 1);
char* t = malloc(strlen(v) * sizeof(char) + 1);
strcpy(s, k);
strcpy(t, v);
p->first = s;
p->second = t;
p->next = NULL;
unsigned int hashval = hash(k);
if (d->table[hashval % d->size] == NULL)
d->table[hashval % d->size] = p;
else {
pair* q = d->table[hashval % d->size];
while (q->next)
q = q->next;
q->next = p;
}
}
char* retreive(dict* d, char* k) {
unsigned int hashval = hash(k);
pair* p = d->table[hashval % d->size];
if (!p) return NULL;
while (strcmp(p->first, k) != 0) {
if (p->next) p = p->next;
else return NULL;
}
return p->second;
}
int main(void) {
dict* d = malloc(sizeof(dict));
setsize(d, 10000007);
int T;
scanf("%d", &T);
char s[37];
char t[37];
while(T--) {
scanf("%s %s", s, t);
insert(d, s, t);
}
while(scanf("%s", s) != EOF) {
if (retreive(d, s) != NULL)
printf("%s=%s\n", s, retreive(d, s));
else
}
return 0;
}```
## Dictionaries and Maps HackerRank Solution in C ++
```#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
string temp;
getline(cin, temp);
int N = stoi(temp);
map<string, string> phoneList;
for(int n = 0; n < N; n++){
string name;
string number;
getline(cin, name);
getline(cin, number);
phoneList.insert(std::pair<string, string>(name, number));
}
string name;
while(getline(cin, name))
{
std::map<string, string>::iterator it;
it = phoneList.find(name);
if (it == phoneList.end()){
} else {
cout << name << "=" << it->second << endl;
}
}
return 0;
}```
## Dictionaries and Maps HackerRank Solution in Java
```//Complete this code or write your own from scratch
import java.util.*;
import java.io.*;
class Solution{
public static void main(String []argh){
Map<String,Integer> phoneBook = new HashMap<String,Integer>();
Scanner in = new Scanner(System.in);
int n = in.nextInt();
in.nextLine();
for(int i=0;i<n;i++){
String name = in.nextLine();
int phone = Integer.parseInt(in.nextLine());
phoneBook.put(name, phone);
}
while(in.hasNext()){
String s = in.nextLine();
Integer phoneNumber = phoneBook.get(s);
System.out.println(
(phoneNumber != null)
? s + "=" + phoneNumber
);
}
in.close();
}
}
```
## Dictionaries and Maps HackerRank Solution in Python 3
```# Enter your code here. Read input from STDIN. Print output to STDOUT
x = int(input())
dictt = {}
for i in range(x):
text = input().split()
dictt[text[0]] = text[1]
while True:
try:
inpt = input()
if inpt in dictt:
print(inpt+"="+dictt[inpt])
else:
except EOFError:
break```
## Dictionaries and Maps HackerRank Solution in JavaScript
```function processData(input) {
var input = input.split('\n');
var numLines = input[0];
var phoneBook = {};
for (var i = 1; i < numLines*2; i=i+2){
// 1,2; 3,4; 5,6
phoneBook[input[i]] = input[i+1];
}
for (var j = numLines*2 + 1; j < input.length; j++){
if (input[j] in phoneBook) console.log(input[j] + '=' + phoneBook[input[j]]);
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});```
30 Days of Code HackerRank Solutions List – Day 0 to Day 29
RELATED ARTICLES | 1,806 | 6,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-18 | longest | en | 0.595326 |
http://www.jiskha.com/display.cgi?id=1321367051 | 1,495,587,506,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607726.24/warc/CC-MAIN-20170524001106-20170524021106-00043.warc.gz | 560,195,609 | 3,838 | # chemistry
posted by on .
A rolaids tablet contains calcium carbonate which neutralizes stomach acid. If rolaids tablet neutralizes 42.15 mL of 0.320M hydrochloric acid, how many milligrams of calcium carbonate are in a Rolaids tablet
CaCO3 + 2 HCl ------> CaCl2 + H2O + CO2
Pls show steps
• chemistry - ,
mols HCl used = M x L = ?
Now use the coefficients in the balanced equation to convert moles HCl to moles CaCO3. moles HCl x (1 mole CaCO3/2 moles HCl) = ?moles HCl x 1/2 = xx moles CaCO3.
grams CaCO3 = moles x molar mass and convert to mg. | 174 | 551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-22 | latest | en | 0.700554 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-2-solving-equations-get-ready-page-77/12 | 1,726,886,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00689.warc.gz | 713,629,082 | 14,314 | Algebra 1: Common Core (15th Edition)
When we multiply fractions, we can simply multiply the numerators (tops of the fractions) by the denominators (bottoms of the fractions). Thus, we obtain: $\frac{7}{1}\times-\frac{6}{14}=-\frac{42}{14}=-3$ | 74 | 244 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.780874 |
butterflybloglife.blogspot.com | 1,503,453,854,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.82/warc/CC-MAIN-20170823020201-20170823040201-00636.warc.gz | 64,591,438 | 9,808 | ## Thursday, October 29, 2015
### don't give me that altitude (and median)
On these triangles I tried to find the medians and the altitudes for both acute triangles and obtuse triangles.
If a line segment is drawn from a vertex of a triangle and 1. splits the opposite side into two congruent segments 2. bisects the opposite side of the angle 3. goes to the midpoint of the opposite side of the triangle..., then it's a median.
If a line segment is drawn from a vertex and 1. is perpendicular to the opposite side 2. forms two right angles when it intersects the opposite side 3. makes two 90 degree angles when it intersects the opposite side..., then it is an altitude. | 157 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-34 | longest | en | 0.901899 |
https://codefying.com/2016/09/15/a-tutorial-on-hidden-markov-model-with-a-stock-price-example/?shared=email&msg=fail | 1,627,691,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154032.75/warc/CC-MAIN-20210730220317-20210731010317-00324.warc.gz | 178,619,178 | 27,878 | A Tutorial on Hidden Markov Model with a Stock Price Example – Part 1
This tutorial is on a Hidden Markov Model. What is a Hidden Markov Model and why is it hiding?
I have split the tutorial in two parts. Part 1 will provide the background to the discrete HMMs. I will motivate the three main algorithms with an example of modeling stock price time-series. In part 2 I will demonstrate one way to implement the HMM and we will test the model by using it to predict the Yahoo stock price!
A Hidden Markov Model (HMM) is a statistical signal model. This short sentence is actually loaded with insight! A statistical model estimates parameters like mean and variance and class probability ratios from the data and uses these parameters to mimic what is going on in the data. A signal model is a model that attempts to describe some process that emits signals. Putting these two together we get a model that mimics a process by cooking-up some parametric form. Then we add “Markov”, which pretty much tells us to forget the distant past. And finally we add ‘hidden’, meaning that the source of the signal is never revealed. BTW, the later applies to many parametric models.
Setting up the Scene
HMM is trained on data that contains an observed sequence of signals (and optionally the corresponding states the signal generator was in when the signals were emitted). Once the HMM is trained, we can give it an unobserved signal sequence and ask:
1. How probable is that this sequence was emitted by this HMM? This would be useful for a problem like credit card fraud detection.
2. What is the most probable set of states the model was in when generating the sequence? This would be useful for a problem of time-series categorization and clustering.
It is remarkable that the model that can do so much was originally designed in the 1960-ies! Here we will discuss the 1-st order HMM, where only the current and the previous model states matter. Compare this, for example, with the nth-order HMM where the current and the previous n states are used. Here, by “matter” or “used” we will mean used in conditioning of states’ probabilities. For example, we will be asking about the probability of the HMM being in some state $q_{t}=s_{i}$ given that the previous state was $q_{t-1}=s_{j}$. Such probabilities can be expressed in 2 dimensions as a state transition probability matrix. Let’s pick a concrete example. Let’s image that on the 4th of January 2016 we bought one share of Yahoo Inc. stock. Let’s say we paid $32.4 for the share. From then on we are monitoring the close-of-day price and calculating the profit and loss (PnL) that we could have realized if we sold the share on the day. The states of our PnL can be described qualitatively as being up, down or unchanged. Here being “up” means we would have generated a gain, while being down means losing money. The PnL states are observable and depend only on the stock price at the end of each new day. What generates this stock price? The stock price is generated by the market. The states of the market influence whether the price will go down or up. The states of the market can be inferred from the stock price, but are not directly observable. In other words they are hidden. So far we have described the observed states of the stock price and the hidden states of the market. Let’s imagine for now that we have an oracle that tells us the probabilities of market state transitions. Generally the market can be described as being in bull or bear state. We will call these “buy” and “sell” states respectively. Table 1 shows that if the market is selling Yahoo stock, then there is a 70% chance that the market will continue to sell in the next time frame. We also see that if the market is in the buy state for Yahoo, there is a 42% chance that it will transition to selling next. Table 1 Sell Buy Sell 0.70 0.30 Buy 0.42 0.58 The oracle has also provided us with the stock price changes probabilities per market state. Table 2 shows that if the market is selling Yahoo, there is an 80% chance that the stock price will drop below our purchase price of$32.4 and will result in negative PnL. We also see that if the market is buying Yahoo, then there is a 10% chance that the resulting stock price will not be different from our purchase price and the PnL is zero. As I said, let’s not worry about where these probabilities come from. It will become clear later on. Note that row probabilities add to 1.0
Table 2
Down Up Unchanged
Sell 0.80 0.15 0.05
It is February 10th 2016 and the Yahoo stock price closes at $27.1. If we were to sell the stock now we would have lost$5.3. Before becoming desperate we would like to know how probable it is that we are going to keep losing money for the next three days.
To put this in the HMM terminology, we would like to know the probability that the next three time-step sequence realised by the model will be {down, down, down} for t=1, 2, 3. This sequence of PnL states can be given a name $O=\{O_{1}, O_{2},O_{3}\}$. So far the HMM model includes the market states transition probability matrix (Table 1) and the PnL observations probability matrix for each state (Table 2). Let’s call the former A and the latter B. We need one more thing to complete our HMM specification – the probability of stock market starting in either sell or buy state. Intuitively, it should be clear that the initial market state probabilities can be inferred from what is happening in Yahoo stock market on the day. But, for the sake of keeping this example more general we are going to assign the initial state probabilities as $\pi=\{0.5, 0.5\}$. Thus we are treating each initial state as being equally likely. So, the market is selling and we are interested to find out $P(O|\lambda(A,B,\pi))$. This is the probability to observe sequence $O$ given the current HMM parameterization. It is clear that sequence $O$ can occur under 2^3=8 different market state sequences. To solve the posed problem we need to take into account each state and all combinations of state transitions. The probability of this sequence being emitted by our HMM model is the sum over all possible state transitions and observing sequence values in each state.
This gives rise to very long sum!
$P(O|\lambda)=P(\pi = sell)P(emit \, down \, from \, sell)P(move \, to \, sell \, from \, sell) \\ P(emit \, down \, from \, sell)P(move \, to \, sell \, from \, sell)P(emit \, down \, from \, sell) \\ + P(\pi = sell)P(emit \, down \, from \, sell)P(move \, to \, sell \, from \, sell)P(emit \, down \, from \, sell) \\ P(move \, to \, buy\, from \, sell)P(emit \, down \, from \, buy) \\ + ... \\ + P(\pi = buy)P(emit \, down \, from \, buy)P(move \, to \, sell \, from \, buy)P(emit \, down \, from \, sell) \\ P(move \, to \, sell \, from \, sell)P(emit \, down \, from \, sell) \\ + ... \\ + P(\pi = buy)P(emit \, down \, from \, buy)P(move \, to \, buy \, from \, buy)P(emit \, down \, from \, buy) \\ P(move \, to \, buy \, from \, buy)P(emit \, down \, from \, buy)$
In total we need to consider 2*3*8=48 multiplications (there are 6 in each sum component and there are 8 sums). That is a lot and it grows very quickly. Please note that emission probability is tied to a state and can be re-written as a conditional probability of emitting an observation while in the state. If we perform this long calculation we will get $P(O|\lambda=(A,B,\pi))=0.192$. There is an almost 20% chance that the next three observations will be a PnL loss for us!
Oracle is No More
In the previous section we have gained some intuition about HMM parameters and some of the things the model can do for us. However, the model is hidden, so there is no access to oracle! The state and emission transition matrices we used to make projections must be learned from the data. The hidden nature of the model is inevitable, since in life we do not have access to the oracle. In life we have access to historical data/observations and a magic methods of “maximum likelihood estimation” (MLE) and Bayesian inference. The MLE essentially produces distributional parameters that maximize the probability of observing the data at hand (i.e. it gives you the parameters of the model that is most likely have had generated the data). The HMM has three parameters $\lambda=(A,B,\pi)$. A fully specified HMM should be able to do the following:
1. Given a sequence of observed values, provide us with a probability that this sequence was generated by the specified HMM. This can be re-phrased as the probability of the sequence occurring given the model. This is what we have calculated in the previous section.
2. Given a sequence of observed values, provide us with the sequence of states the HMM most likely has been in to generate such values sequence.
3. Given a sequence of observed values we should be able to adjust/correct our model parameters $A$, $B$ and $\pi$.
When looking at the three ‘should’, we can see that there is a degree of circular dependency. That is, we need the model to do steps 1 and 2, and we need the parameters to form the model in step 3. Where do we begin? There are three main algorithms that are part of the HMM to perform the above tasks. These are: the forward&backward algorithm that helps with the 1st problem, the Viterbi algorithm that helps to solve the 2nd problem, and the Baum-Welch algorithm that puts it all together and helps to train the HMM model. Let’s discuss them next.
Going Back and Forth
That long sum we performed to calculate $P(O|\lambda=(A,B,\pi))$ grows exponentially in the number of states and observed values. The Forward and Backward algorithm is an optimization on the long sum. If you look back at the long sum, you should see that there are sum components that have the same sub-components in the product. For example,
$P(O|\lambda)=P(\pi = sell)P(emit \, down \, from \, sell)P(move \, to \, sell \, from \, sell)...$
appears twice. The HMM Forward and Backward (HMM FB) algorithm does not re-compute these, but stores the partial sums as a cache. There are $N$ states, $M$ discrete values that can be emitted from each of $s_{i,...,N}$ states and $i=1,...,N$. There are $T$ observations in the considered sequence $O$. The state transition matrix is $A$, where $a_{i,j}$ is an individual entry and $j=1,...,N$. The emission matrix is $B$, where $b_{j}(O_{t})$ is an individual entry $b_{j}(T_{t})=P(emitting \, O_{t}|q_{t}=j)$, and $t=1,...,T$, $q_{t}$ is state at time t. For initial states we have $\pi_{i}=P(q_{1}=s_{i})$. The problem of finding the probability of sequence $O$ given an HMM is $P(O|\lambda=(A,B,\pi))$. Formally this probability can be expressed as a sum:
$P(O|\lambda)=\sum_{q_{1},q_{2},...,q_{T}}\pi_{q_{1}}b_{q_{1}}(O_{1})a_{q_{1}q_{2}}b_{q_{2}}(O_{2})a_{q_{2}q_{3}}b_{q_{3}}(O_{3})...a_{q_{T-1}q_{T}}b_{q_{T}}(O_{T})$
HMM FB calculates this sum efficiently by storing the partial sum calculated up to time $t$. HMM FB is defined as follows:
1. Start by initializing the “cache” $\alpha_{1}(i)=\pi_{i}b_{i}(O_{1})$, where $1 \le i \le N$.
2. Calculate over all remaining observation sequences and states the partial sums:
$\alpha_{t+1}(j)=\big[\sum_{j=1}^{N} \alpha_{ij}\big] b_{j}(O_{t+1})$, where $1 \le t \le T-1$ and $1 \le j \le N$. Since at a single iteration we hold $i$ constant, this results in a trellis-like calculation along the columns of the transition matrix.
3. The probability is then given by $P(O|\lambda)=\sum_{i=1}^{N}\alpha_{T}(i)$. This is where we gather the individual states probabilities together.
The above is the Forward algorithm which requires only $TN^{2}$ calculations. The algorithm moves forward. We can imagine an algorithm that performs similar calculation, but backwards, starting from the last observation in $O$. Why do we need this? Is the Forward algorithm not enough? It is enough to solve the 1st poised problem. But it is not enough to solve the 3rd problem, as we will see later. So, let’s define the Backward algorithm now.
1. Start by initializing the end “cache” $\beta_{T}(i)=1$, for $1 \le i \le N$. This is an arbitrary assignment.
2. Calculate over all remaining observation sequences and states the partial sums (moving back to the start of the observation sequence):
$\beta_{t}(i)=\sum_{j=1}^{N} \alpha_{ij}b_{j}(O_{t+1})\beta_{t+1}(j)$, where $t=T-1,T-2,...,1$ and $1 \le j \le N$. Since at a single iteration we hold $i$ constant, this results in a trellis-like calculation along the columns of the transition matrix. However, it is performed backwards in time.
The $\beta$ matrix also contains probabilities of observing sequence $O$. However, the actual values in $\beta$ are different from those in $\alpha$ because of the arbitrary assignment of $\beta_{T}$ to 1.
The State of the Affairs
We are now ready to solve the 2nd problem of the HMM – given the model and a sequence of observations, provide the sequence of states the model likely was in to generate such a sequence. Strictly speaking, we are after the optimal state sequence for the given $O$. Optimal often means maximum of something. At each state and emission transitions there is a node that maximizes the probability of observing a value in a state. Let’s take a closer look at the $\alpha$ and $\beta$ matrices we calculated for the $\{down, \, down, \, down\}$ example.
I have circled the values that are maximum. All of these correspond to the Sell market state. Our HMM would have told us that the most likely market state sequence that produced $\{down, \, down, \, down\}$ was $\{Sell, \, Sell, \, Sell\}$. Which makes sense. The HMM algorithm that solves the 2nd problem is called Viterbi Algorithm, named after its inventor Andrew Viterbi. The essence of Viterbi algorithm is what we have just done – find the path in the trellis that maximizes each node probability. It is a little bit more complex than just looking for the max, since we have to ensure that the path is valid (i.e. it is reachable in the specified HMM).
We are after the best state sequence $Q=\{q_{1},q_{2},...,q_{T}\}$ for the given $O$. The best state sequence maximizes the probability of the path. We will be taking the maximum over probabilities and storing the indices of states that result in the max.
1. Start by initializing the end “cache” $\delta_{1}(j)=\pi_{i}b_{i}(O_{1})$, for $1 \le i \le N$. We also need to initialize to zero a variable used to track the index of the max node: $\phi_{1}(i)=0$.
2. Calculate over all remaining observation sequences and states the partial max and store away the index that delivers it:
$\delta_{t}(j)=\max_{1\le i \le N} \small[ \delta_{t-1}(i)a_{ij}\small]b_{j}(O_{t})$, where $2 \le t \le T$ and $1 \le j \le N$.
Update the sequence of indices of the max nodes as:
$\phi_{t}=argmax_{1 \le i \le N} \small[ \delta_{t-1}(i) a_{ij}\small]$ for j and t as above.
3. Termination (probability and state):
$P*=\max_{1 \le i \le N} \small[\delta_{T}(i)\small]$
$q_{T}*=argmax_{1 \le i \le N} \small[\delta_{T}(i)\small]$
4. The optimal state sequence $Q$ is given by the path:
$q_{t}^{*}=\phi_{t+1}(q_{t+1}^{*})$, for $t=T-1,T-2,...,1$
Expect the Unexpected
And now what is left is the most interesting part of the HMM – how do we estimate the model parameters from the data? We will now describe the Baum-Welch Algorithm to solve this 3rd poised problem.
The reason we introduced the Backward Algorithm is to be able to express a probability of being in some state i at time t and moving to a state j at time t+1. Imagine again the probabilities trellis. Pick a model state node at time t, use the partial sums for the probability of reaching this node, trace to some next node j at time t+1, and use all the possible state and observation paths after that until T. This gives the probability of being in state $s_{t}(i)$ and move to $s_{t+1}(j)$. The figure below graphically illustrates this point.
To make this transition into a proper probability, we need to scale it by all possible transitions in $\lambda$ and $O$. It can now be defined as follows:
$\gamma_{t}(i,j)=\frac{\alpha_{t}(i)a_{ij}b_{j}(O_{t+1})\beta_{t+1}(j)}{P(O|\lambda)}$
So, $\gamma_{t}(i,j)$ is a probability of being in state i at time t and moving to state j at time t+1. It makes perfect sense as long as we have true estimates for $\alpha$, $A$, $B$ and $\pi$. We will come to these in a moment…
Summing $\gamma_{t}(i,j)$ across $j$ gives the probability of being in state i at time t under $O$ and $\lambda$: $\gamma_{t}(i)=\sum_{j=1}^{N}\gamma_{t}(i,j)$. Let’s look at an example to make things clear. Let’s consider $O=\{0,1,1\}=\{down, up, up\}$. We will use the same $A$ and $B$ from Table 1 and Table 2. The $alpha$ learned for this observation sequence is shown below:
The (un-scaled) $\beta$ is shown below:
So, what is $\gamma_{1}(0,1)$? It is (0.7619*0.30*0.65*0.176)/0.05336=49%, where the denominator is calculated for $t=1$. The denominator is calculated across all i and j at $t=1$, thus it is a normalizing factor. If we calculate $\gamma_{2}(0,1)$ and sum the two estimates together, we will get the expected number of transitions from state $s(i)$ to $s(j)$. Note, we do transition between two time-steps, but not from the final time-step as it is absorbing. Similarly, the sum over $\gamma_{t}(i)$, where $t=1,2$ gives the expected number of transitions from $s(i)$. It should now be easy to recognize that $\frac{\sum_{t=1}^{T-1}\gamma_{t}(i,j)}{\sum_{t=1}^{T-1}\gamma_{t}(i)}$ is the transition probability $a_{ij}$, and this is how we estimate it:
$\hat{a_{ij}}=\frac{\sum_{t=1}^{T-1}\gamma_{t}(i,j)}{\sum_{t=1}^{T-1}\gamma_{t}(i)}$
We can derive the update to $b_{j}(O_{t})$ in a similar fashion. The matrix $B$ stores probabilities of observing a value from $O$ in some state. For $k=1,2,...,M$ $b_{j}(k)=\frac{\sum_{t\in\{1,...,T\}O_{t}=k}\gamma_{t}(j)}{\sum_{t=1}^{T}\gamma_{t}(j)}$ is the probability of observing symbol $k$ in state j. The update rule becomes:
$\hat{b_{j}}(k)=\frac{\sum_{t\in\{1,...,T\}O_{t}=k}\gamma_{t}(j)}{\sum_{t=1}^{T}\gamma_{t}(j)}$
$\pi$ stores the initial probabilities for each state. This parameter can be updated from the data as:
$\pi_{i}=\gamma_{1}(i)$
We now have the estimation/update rule for all parameters in $\lambda$. The Baum-Welch algorithm is the following:
1. Repeat until convergence:
2. Initialize $A, B, \pi$ to random values such that $\pi_{i}\approx \frac{1}{N}$, $a_{ij}\approx \frac{1}{N}$ and $b_{j}(k)\approx \frac{1}{M}$. It is important to ensure that row probabilities add up to 1 and probabilities are not uniform as this may result in convergence to a local maximum.
3. Compute $\alpha_{t}(i)$, $\beta_{t}(i)$, and $\gamma_{t}(j)$ and $\gamma_{t}(i,j)$.
4. Re-estimate $A$, $B$, and $\pi$.
The convergence can be assessed as the maximum change achieved in values of $A$ and $B$ between two iterations. The authors of this algorithm have proved that either the initial model defines the optimal point of the likelihood function and $\lambda = \hat{\lambda}$, or the converged solution provides model parameters that are more likely for a given sequence of observations $O$. The described algorithm is often called the expectation maximization algorithm and the observation sequence works like a pdf and is the “pooling” factor in the update of $\lambda$. I have described the discrete version of HMM, however there are continuous models that estimate a density from which the observation come from, rather than a discrete time-series.
I hope some of you may find this tutorial revealing and insightful. I will share the implementation of this HMM with you next time.
Reference: L.R.Rabiner. An introduction to Hidden Markov models and selected applications in speech recognition. Proceedings of the IEEE, 77(2):257-268, 1989. | 5,301 | 19,798 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 132, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-31 | latest | en | 0.947976 |
https://www.transtutors.com/homework-help/networks-systems/two-port-network/z-abcd-parameters/ | 1,516,303,344,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887600.12/warc/CC-MAIN-20180118190921-20180118210921-00469.warc.gz | 1,006,583,583 | 19,484 | ## What is Example On Z And Abcd Parameters?
Determine the z and ABCD parameters of the network shown in Pig. a
FIG.
SOLUTION:
Determination of z parameters
Figure b shows the currents in different branches of the network after the application of KCL at nodes a and b.
Application of KVL in the left-hand mesh gives
V1 – l × I1 – 1 × (2I1 + I2) = 0
Or V1 = 3I1 + I2
Also, V1 = z11I1 + z12I2
Comparison of above equations,it gives
z11 = 3Ω, z12 = 1Ω
Application of KVL in niesh abc a gives
1 × (I1 + I2) – Vbc + 1 × (2L1 + I2) = 0
But Vbc = V2
I1 + I2 – V2 + 2I1 + I1 = 0
Or V2 = 3I1 + 2I2
Also, V2 = z21I1 + z22I2
Comparison of these two equation,we get
z21 = 3Ω, z22 = 2Ω
Determination of ABCD parameters
In order to determine ABCD parameters we shall rearrange above network equations in the form of the defining equations of the ABCD parameters elimination of I1 gives
V1 = V2 − I2
Also, V1 = AV2 − BI2
Comparison of these two equations gives
A = 1, B = 1?
Now, we have
3I1 = V2 − 2I2
I1 = 1/3 V2 – 2/3 I2
Also I1 = CV2 – DI2
Therefore, as a result we get,
C = 1/3 S, D = 2/3
## Email Based, Online Homework Assignment Help in Z & ABCD Parameters
Transtutors comprises highly qualified and certified teachers, college professors, subject professionals in various subjects like network systems etc. All our tutors are highly experienced and can clear your doubt regarding Z & ABCD parameters and can explain the different concepts to you effectively.
## Related Topics
All Computer Science Topics
More Q&A | 493 | 1,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-05 | latest | en | 0.730496 |
https://www.jiskha.com/display.cgi?id=1353988955 | 1,516,724,673,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891980.75/warc/CC-MAIN-20180123151545-20180123171545-00663.warc.gz | 942,472,133 | 3,895 | # math
posted by .
A ship sail S 40 degrees W for 4 hours at 30 kph. make a scale drawing
• math -
We cannot make a drawing for you on our posts.
## Similar Questions
1. ### math
a ship sail S 40 degrees W four 4 hours how do i figure out this question
2. ### Math help,plz :(
So I'm drawing a scale drawing of A room (Yes, I know what a Scale drawing is) and I don't know how to do it adding proportions.... and what does it mean to Use dimensions to scale the room?
3. ### math
Brett made a scale drawing of a rectangular room in his house. The actual length of the room is 12 4/5 ft. The scale used to make the drawing was 1/4 in.=1 ft. What is the length, in inches, of the room on the drawing?
4. ### trig
Help! Ships A and B leave port at the same time and sail on straight paths making an angle of 60 degrees with each other. HOw far apart are the ships at the end of 1 hour if the speed of ship A is 25 km/h and that of ship B is 15 km/h?
5. ### Math
A ship leaves port and sails northwest for 2 hours, and then sails N35degreesE for three hours. If it does not change speed, what direction must the ship sail to return directly to port?
6. ### math, scale drawing
charlies swimming pool is rectangular. it has an area of 120ft2. charlie drew a scale drawing of the pool using the scale 3/2 centimeters= 2 feet. what was the area of the scale drawing?
7. ### Algebra
Suppose you wish to make a scale drawing of your living room, which measures 18 ft by 23 ft. If the shorter side of the drawing is 6 in., how long is the longer side of the scale drawing?
8. ### math
two cruise ships sail from NY to FL.ONE ship make the round trip in 8 days.the other ship make the round trip in 10 days.both ships sail from ny today.which is the fewest number of days before both ships will sail again from newyork …
9. ### trigonometry
A ship started sailing 42.58 degrees west of south at the rate of 15 kph. after 2 hours, ship B started at the same port going 46.33 degrees west of north at the rate of 7 kph. after how many hours will the second ship be exactly north …
10. ### trigonometry
A ship started sailing 42.58 degrees west of south at the rate of 5 kph. after 2 hours, ship B started at the same port going 46.33 degrees west of north at the rate of 7 kph. after how many hours will the second ship be exactly north …
More Similar Questions | 623 | 2,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-05 | latest | en | 0.937341 |
https://www.quesba.com/questions/write-quantificational-formulas-that-state-the-following-a-the-unary-functi-40644 | 1,606,542,525,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195069.35/warc/CC-MAIN-20201128040731-20201128070731-00633.warc.gz | 837,164,358 | 18,326 | # Write quantificational formulas that state the following. (a) The unary function f is bisection...
Write quantificational formulas that state the following.
(a) The unary function f is bisection between the universe and itself.
(b) The binary function g does not depend on the second argument.
Mar 24 2020| 06:03 AM
## Related Questions
### FUNCTION POINT ANALYSIS Function point analysis was developed by Allan Albrecht at IBM in 1979....
FUNCTION POINT ANALYSIS Function point analysis was developed by Allan Albrecht at IBM in 1979. Function point analysis divides code into categories:  ● Outputs ● Inquiries  ● Inputs ● Internal files ● External interfaces Once the code is categorized, it is assessed for complexity and assigned a number that is then further examined and given adjustments and...
Apr 07 2020
### 1. 2What is printed from fun(19) as coded below? Explain what the function does. 2. Write a tail...
1.       2What is printed from fun(19) as coded below? Explain what the function does. 2.       Write a tail version of the factorial function and add a no recursive function to call it. Test the function in a program.
Aug 26 2020
### 1. Write a function that converts all characters in a C++ string to uppercase. Test the function...
1.       Write a function that converts all characters in a C++ string to uppercase. Test the function in a program. 2.       Write a function that removes all occurrences of a given character from a C++ string. Test the function in a program.
Aug 26 2020
### 1. Write a function that removes duplicate characters from a string (keeping only one instance of...
1.       Write a function that removes duplicate characters from a string (keeping only one instance of each character). Test the function in a program. 2.       Write a function that, given two strings, creates another string that contains only the common characters of the two strings. One solution is to first remove the duplicate...
Aug 26 2020
### 1. Write the definition of a function that, given an array of integers named arr, deletes the...
1.       Write the definition of a function that, given an array of integers named arr, deletes the element at the given index. Note that the function should assert the validity of the given index. 2.       Write the definition of a function that, given an array of integers named arr, inserts a given value at the given index. Note that...
Aug 26 2020
## Recent Questions
### Imagine you have decided to adopt a vegetarian diet. a. List foods that you would like to eat to...
Imagine you have decided to adopt a vegetarian diet. a. List foods that you would like to eat to form a complete protein. b. While you are generally satisfied with your new eating plan, you have noticed that you have decreased energy during your workouts and have unintentionally lost weight. You worry that maybe you are not eating enough calories to fuel your exercise. i. Describe the concept of...
Nov 27 2020
### After this experience Brian shares that he is allergic to shellfish. He said that he never told...
After this experience Brian shares that he is allergic to shellfish. He said that he never told you because you never asked. On further questioning he shares that he had to be hospitalized when he was 12 years old with anaphylaxis to shrimp. You ask him if he carries an epinephrine pen with him. He replies “no.” Outline how you might proceed to discuss the management of food allergies...
Nov 27 2020
### You learn that Brian has been trying to eat healthier and ate a lunch which included a salad with...
You learn that Brian has been trying to eat healthier and ate a lunch which included a salad with tomatoes and celery. a. Given this history, what is a possible cause of Brian’s symptoms? b. How could Brian avoid this occurring in the future? c. What, if any, changes should Brian make in his everyday eating plan, given this event?
Nov 27 2020
### 1. How does a vegan diet affect athletic performance? What are some general considerations vegan...
1. How does a vegan diet affect athletic performance? What are some general considerations vegan athletes should consider to optimize performance while adhering to a vegan diet? 2. What does it mean to be a vegan? List specific types of foods that are and are not consumed in a vegan diet.
Nov 27 2020
### What vitamins and minerals do vegans face increased risk of deficiency for? What are some good...
What vitamins and minerals do vegans face increased risk of deficiency for? What are some good food sources of these vitamins and minerals that are allowable in the vegan diet? For which, if any, of the vitamins and minerals would a vegan need to take a supplement in order to avoid deficiency?
Nov 27 2020 | 1,165 | 5,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-50 | latest | en | 0.804417 |
https://what.thedailywtf.com/topic/20883/accalia-rolls-dice-for-starting-stats | 1,505,873,572,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686117.24/warc/CC-MAIN-20170920014637-20170920034637-00493.warc.gz | 730,694,189 | 26,179 | # Accalia Rolls Dice for starting stats
• the title says it all
@yamibot roll 6x4d6 for stating stats
4d6: 6 3 6 4 = 19
Total: 19
4d6: 2 4 1 2 = 9
Total: 9
4d6: 4 5 1 3 = 13
Total: 13
4d6: 5 3 3 2 = 13
Total: 13
4d6: 6 5 5 5 = 21
Total: 21
4d6: 1 3 4 2 = 10
Total: 10
Grand Total: 85
• 15+8+12+11+16+9 = 71
this could take a while
@yamibot roll 6x4d6 for stating stats
4d6: 6 2 6 2 = 16
Total: 16
4d6: 6 6 6 4 = 22
Total: 22
4d6: 4 1 6 1 = 12
Total: 12
4d6: 3 1 4 5 = 13
Total: 13
4d6: 3 1 4 5 = 13
Total: 13
4d6: 6 6 3 2 = 17
Total: 17
Grand Total: 93
• 14+18+11+12+12+15=82
damn.... i liked that roll
@yamibot roll 6x4d6
4d6: 3 3 6 2 = 14
Total: 14
4d6: 1 6 1 4 = 12
Total: 12
4d6: 3 2 1 4 = 10
Total: 10
4d6: 5 4 3 6 = 18
Total: 18
4d6: 5 6 3 4 = 18
Total: 18
4d6: 5 2 6 4 = 17
Total: 17
Grand Total: 89
• @yamibot 12+11+9+15+15+15=77
and again
@yamibot roll 6x4d6
4d6: 3 6 5 6 = 20
Total: 20
4d6: 5 1 6 5 = 17
Total: 17
4d6: 6 5 4 2 = 17
Total: 17
4d6: 5 3 1 3 = 12
Total: 12
4d6: 2 6 5 6 = 19
Total: 19
4d6: 6 3 3 1 = 13
Total: 13
Grand Total: 98
• If anything, what I'd say is, do a set of 6x4d6 rolls as the baseline, then do a follow-up 6x4d6 for replacements, swapping the lowest rolls of the previous set for the highest of the new set until you get over 85 (or have to roll again). That way you're (hopefully) not spamming up as many rerolls.
... or, you know, get over 85 on the last set you rolled. That works too.
• 17+16+15+11+17+12=88
well that took less time than i thought it would.
• Maybe we could open this up for anyone else that wants to roll for stats?
• sure i guess. go to town.
• @yamibot roll 2d6*10 for starting Gold
2d6: 3 2 = 5
Total: 50
• Rolling for age, height, and weight:
@yamibot roll 7d6
@yamibot roll 2d4
@yamibot roll 2d4*7
7d6: 6 1 2 2 5 5 5 = 26
Total: 26
You rolled 2d4:**Your rolls:** 2d4: 3 2 = 5 **Total**: 5
You rolled 2d4*7:**Your rolls:** 2d4: 3 4 = 7 **Total**: 49
• You rolled 7d6:**Your rolls:** 7d6: 6 1 2 2 5 5 5 = 26 **Total**: 26
You rolled 2d4:**Your rolls:** 2d4: 3 2 = 5 **Total**: 5
You rolled 2d4*7:**Your rolls:** 2d4: 3 4 = 7 **Total**: 49
@Yamikuronue bug with the formatting?
• @dangeRuss Kinda mixed with a bug with the Markdown parsing. The parser doesn't really handle Markdown after a <hr> very well.
**See**?
vs.
See?
``````<hr>
**See**?
vs.
---
**See**?
``````
• @dangeRuss Kinda mixed with a bug with the Markdown parsing. The parser doesn't really handle Markdown after a <hr> very well.
**See**?
vs.
See?
``````<hr>
**See**?
vs.
---
**See**?
``````
sounds like using <hr> is
• Rolling for Age
@yamibot roll 2d4+20
2d4: 2 1 = 3
Total: 23
• Derp, I'm a ranger, which means self-taught starting age (I guess)
@yamibot roll 3d6+20
3d6: 6 5 4 = 15
Total: 35
• Rolling for Height
@yamibot roll 2d4+32
in inches total
2d4: 1 4 = 5
Total: 37
• @yamibot roll 2d6*10
Starting gold for Cy
2d6: 2 6 = 8
Total: 80
• @Yamikuronue I blame Markdown.
• Gonna see about rapidly creating a more "neutral" healer for my group.
@yamibot roll 6x4d6
4d6: 3 4 1 6 = 14
Total: 14
4d6: 4 1 5 1 = 11
Total: 11
4d6: 4 1 6 6 = 17
Total: 17
4d6: 6 4 3 6 = 19
Total: 19
4d6: 3 6 5 6 = 20
Total: 20
4d6: 5 3 4 3 = 15
Total: 15
Grand Total: 96
• 1 point under, rerolling the 10 (and padding rolls, first roll after accepted will be starting gold):
@yamibot roll 6x4d6
4d6: 6 2 4 2 = 14
Total: 14
4d6: 2 6 6 6 = 20
Total: 20
4d6: 4 4 3 2 = 13
Total: 13
4d6: 1 3 1 1 = 6
Total: 6
4d6: 5 3 6 1 = 15
Total: 15
4d6: 5 3 6 4 = 18
Total: 18
Grand Total: 86
• Well, that was a waste of the "18" stat... instead I get 200 gold.
• @yamibot roll 43+2d4 for rasins
2d4: 4 2 = 6
Total: 49
• @yamibot roll 43+2d4 for rasins
Total: 49
That's a lot of raisins.
• @yamibot thanks, now ....
4d6: 6 2 5 6 = 19
Total: 59
• @yamibot
!roll 16x4d20*1.7
4d20: 6 16 4 1 = 27
Total: 45.9
4d20: 1 18 16 5 = 40
Total: 68
4d20: 8 6 12 12 = 38
Total: 64.6
4d20: 10 4 6 2 = 22
Total: 37.4
4d20: 8 2 12 20 = 42
Total: 71.39999999999999
4d20: 18 8 10 2 = 38
Total: 64.6
4d20: 1 7 3 15 = 26
Total: 44.199999999999996
4d20: 7 13 7 7 = 34
Total: 57.8
4d20: 11 1 20 3 = 35
Total: 59.5
4d20: 20 1 10 14 = 45
Total: 76.5
4d20: 14 17 10 5 = 46
Total: 78.2
4d20: 20 13 14 16 = 63
Total: 107.1
4d20: 10 8 12 7 = 37
Total: 62.9
4d20: 7 4 18 19 = 48
Total: 81.6
4d20: 20 16 13 20 = 69
Total: 117.3
4d20: 15 1 15 19 = 50
Total: 85
Grand Total: 1122
• @yamibot Kinda wish it would put the GT in the Summary section...
• I think adding syntax like "6x4C3d6" is useful for dicing starting stats.
Roll 6 groups of (topmost)"3 out of 4" on 6 face dice.
I think it's rarely need to roll for bottommost value, is it?
• @cheong I'd suggest "6x3+d6", with the '+' indicating 'roll one extra die, keep the highest values'. Might be trickier to parse though.
• This post is deleted!
• I think adding syntax like
The reason I've never added a command for it is because every option I've looked at for syntax is gross.
6x4C3d6
Who will ever remember that? `x` vs `*` already confuses people.
with the '+' indicating 'roll one extra die, keep the highest values'.
ew as well. Nobody will remember that and it'll get confused for addition.
If I do add it, it'd probably be something like `!roll stats` that does the 6x4d6-drop-low baked in. We also have things like
@yamibot roll 2dF
and
@yamibot rollww 3d10
• 6x4C3d6
Who will ever remember that? `x` vs `*` already confuses people.
Humm... I think it's well recognized that "nCr" is the mathematical notation for "number of combinations in r out of n". But whatever.
• mathematical notation
I found the problem!
• If I do add it, it'd probably be something like !roll stats that does the 6x4d6-drop-low baked in. We also have things like
eeh. i'd prefer the syntax `!roll stats 6x4d6` because that allows one to change how many stats they are rolling for (the `6x` part) and the base roll from which to discard the lowest roll (the `4d6` part) (you should probably default the `6x4d6` as the stats to roll so you don't need to write it when you are rolling standard stats)
otherwise that seems cromulent
• eeh
See? The syntax is hard >.>
Looks like your connection to What the Daily WTF? was lost, please wait while we try to reconnect. | 2,760 | 6,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-39 | longest | en | 0.787892 |
http://webhost.bridgew.edu/hburgiel/MATH113/quadrilateral2.html | 1,511,469,558,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806939.98/warc/CC-MAIN-20171123195711-20171123215711-00393.warc.gz | 333,075,567 | 1,217 | ### MATH 113: Area of a Quadrilateral
This worksheet combines a geometric exploration of irregular quadrilaterals with practice in estimating and calculating area.
1. Use a ruler to draw an irregular convex quadrilateral on a sheet of stiff paper.
3. Estimate the area of the quadrilateral. It may help to calculate the area of the sheet of paper you started with.
4. For each edge of the quadrilateral, use your ruler to mark the midpoint of the edge. Check your work by gently folding the corners of the quadrilateral together -- if you were to crease the paper along that fold, the crease should go through the mark.
5. Use your ruler to connect the marks on the edges into a convex polygon. This polygon should have some special properties; what are they? (Is it a rectangle? a trapezoid?)
6. Cut out the convex polygon you just drew. Compare its area to the area of the corner pieces from the original polygon. What do you find?
7. Using your ruler, find the area of the cut out polygon.
9. Are your estimates in 3 and 8 close? If not, what might have gone wrong? | 242 | 1,073 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2017-47 | longest | en | 0.905845 |
http://www.robives.com/blog/pendex | 1,369,352,737,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704075359/warc/CC-MAIN-20130516113435-00051-ip-10-60-113-184.ec2.internal.warc.gz | 696,678,310 | 8,538 | 1
Mon 2nd Apr 2012
See the Membership page for more details.
Which is best, a paper hinge or a hinge balancing on a knife edge. There's only one way to find out. Experiments!
I've put together three pendulum in box experiments. One with a paper hinge, one with a card hinge and another where the edge of a triangular tube is resting against the side of a box like a knife edge. I then set the pendulums a-swinging to see which would run for the longest. Bet you can't guess which worked best!
If you are a member you can download the parts and try the experiment for yourself. Print the parts out onto thin card (230 micron /67 lb) Score along the dotted lines then carefully cut out the parts then follow the instructions for construction below. Scroll to the end if you can't wait to find out which was best.
There are two slightly different layouts for the two designs. You'll need two of the first page and one of the second page to make both types.
The first instructions are for the paper hinge pendulum. There are red circles on the parts specific to this model.
Cut out the hole in he top of the box following the black line. You can see clearly on the picture below right where the hole is. Fold up the flaps on the sides of the box to make right angled triangle sections as shown above.
Make up the triangle stiffener piece with the red circle and glue it next to the hole so that it runs underneath where the blue lines are.
Glue the box ends into place.
The knife edge box is slightly different. Cut out the hole on the top to include the area surrounded by the blue line and cut out the two slots outlined in blue. Glue the two longer triangle pieces long ways along the underside of the box as shown.
Complete the box in the same way as the other one.
Assemble the hinge as shown above and add coins to the end. The weight on the coin is not critical so long as they are the same on both models. I used UK 2p coins, roughly 7 grams each.
Assemble the second pendulum as shown above, this time adding a hinge made of the material of your choice. I tested paper and card.
Fit the pendulums into place and start them swinging. Which will run the longest?
In mine, the paper hinge was slightly better that the knife edge. That really surprised me. It was close though with one at 75 seconds and the other at 68 seconds. My other test, with the card hinge ran for only 20 seconds.
Let me know what you find out!
Rate this post: | 549 | 2,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2013-20 | longest | en | 0.942506 |
https://www.nag.com/numeric/nl/nagdoc_latest/clhtml/f01/f01hbc.html | 1,656,880,930,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00161.warc.gz | 956,921,994 | 8,488 | NAG CL Interfacef01hbc (complex_gen_matrix_actexp_rcomm)
Settings help
CL Name Style:
1Purpose
f01hbc computes the action of the matrix exponential ${e}^{tA}$, on the matrix $B$, where $A$ is a complex $n×n$ matrix, $B$ is a complex $n×m$ matrix and $t$ is a complex scalar. It uses reverse communication for evaluating matrix products, so that the matrix $A$ is not accessed explicitly.
2Specification
#include
void f01hbc (Integer *irevcm, Integer n, Integer m, Complex b[], Integer pdb, Complex t, Complex tr, Complex b2[], Integer pdb2, Complex x[], Integer pdx, Complex y[], Integer pdy, Complex p[], Complex r[], Complex z[], Complex ccomm[], double comm[], Integer icomm[], NagError *fail)
The function may be called by the names: f01hbc or nag_matop_complex_gen_matrix_actexp_rcomm.
3Description
${e}^{tA}B$ is computed using the algorithm described in Al–Mohy and Higham (2011) which uses a truncated Taylor series to compute the ${e}^{tA}B$ without explicitly forming ${e}^{tA}$.
The algorithm does not explicity need to access the elements of $A$; it only requires the result of matrix multiplications of the form $AX$ or ${A}^{\mathrm{H}}Y$. A reverse communication interface is used, in which control is returned to the calling program whenever a matrix product is required.
4References
Al–Mohy A H and Higham N J (2011) Computing the action of the matrix exponential, with an application to exponential integrators SIAM J. Sci. Statist. Comput. 33(2) 488-511
Higham N J (2008) Functions of Matrices: Theory and Computation SIAM, Philadelphia, PA, USA
5Arguments
Note: this function uses reverse communication. Its use involves an initial entry, intermediate exits and re-entries, and a final exit, as indicated by the argument irevcm. Between intermediate exits and re-entries, all arguments other than b2, x, y, p and r must remain unchanged.
1: $\mathbf{irevcm}$Integer * Input/Output
On initial entry: must be set to $0$.
On intermediate exit: ${\mathbf{irevcm}}=1$, $2$, $3$, $4$ or $5$. The calling program must:
1. (a)if ${\mathbf{irevcm}}=1$: evaluate ${B}_{2}=AB$, where ${B}_{2}$ is an $n×m$ matrix, and store the result in b2;
if ${\mathbf{irevcm}}=2$: evaluate $Y=AX$, where $X$ and $Y$ are $n×2$ matrices, and store the result in y;
if ${\mathbf{irevcm}}=3$: evaluate $X={A}^{\mathrm{H}}Y$ and store the result in x;
if ${\mathbf{irevcm}}=4$: evaluate $p=Az$ and store the result in p;
if ${\mathbf{irevcm}}=5$: evaluate $r={A}^{\mathrm{H}}z$ and store the result in r.
2. (b)call f01hbc again with all other parameters unchanged.
On final exit: ${\mathbf{irevcm}}=0$.
Note: any values you return to f01hbc as part of the reverse communication procedure should not include floating-point NaN (Not a Number) or infinity values, since these are not handled by f01hbc. If your code inadvertently does return any NaNs or infinities, f01hbc is likely to produce unexpected results.
2: $\mathbf{n}$Integer Input
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
3: $\mathbf{m}$Integer Input
On entry: the number of columns of the matrix $B$.
Constraint: ${\mathbf{m}}\ge 0$.
4: $\mathbf{b}\left[\mathit{dim}\right]$Complex Input/Output
Note: the dimension, dim, of the array b must be at least ${\mathbf{pdb}}×{\mathbf{m}}$.
The $\left(i,j\right)$th element of the matrix $B$ is stored in ${\mathbf{b}}\left[\left(j-1\right)×{\mathbf{pdb}}+i-1\right]$.
On initial entry: the $n×m$ matrix $B$.
On intermediate exit: if ${\mathbf{irevcm}}=1$, contains the $n×m$ matrix $B$.
On intermediate re-entry: must not be changed.
On final exit: the $n×m$ matrix ${e}^{tA}B$.
5: $\mathbf{pdb}$Integer Input
On entry: the stride separating matrix row elements in the array b.
Constraint: ${\mathbf{pdb}}\ge {\mathbf{n}}$.
6: $\mathbf{t}$Complex Input
On entry: the scalar $t$.
7: $\mathbf{tr}$Complex Input
On entry: the trace of $A$. If this is not available then any number can be supplied ($0.0$ is a reasonable default); however, in the trivial case, $n=1$, the result ${e}^{{\mathbf{tr}}t}B$ is immediately returned in the first row of $B$. See Section 9.
8: $\mathbf{b2}\left[\mathit{dim}\right]$Complex Input/Output
Note: the dimension, dim, of the array b2 must be at least ${\mathbf{pdb2}}×{\mathbf{m}}$.
The $\left(i,j\right)$th element of the matrix is stored in ${\mathbf{b2}}\left[\left(j-1\right)×{\mathbf{pdb2}}+i-1\right]$.
On initial entry: need not be set.
On intermediate re-entry: if ${\mathbf{irevcm}}=1$, must contain $AB$.
On final exit: the array is undefined.
9: $\mathbf{pdb2}$Integer Input
On entry: the stride separating matrix row elements in the array b2.
Constraint: ${\mathbf{pdb2}}\ge {\mathbf{n}}$.
10: $\mathbf{x}\left[\mathit{dim}\right]$Complex Input/Output
Note: the dimension, dim, of the array x must be at least ${\mathbf{pdx}}×2$.
The $\left(i,j\right)$th element of the matrix $X$ is stored in ${\mathbf{x}}\left[\left(j-1\right)×{\mathbf{pdx}}+i-1\right]$.
On initial entry: need not be set.
On intermediate exit: if ${\mathbf{irevcm}}=2$, contains the current $n×2$ matrix $X$.
On intermediate re-entry: if ${\mathbf{irevcm}}=3$, must contain ${A}^{\mathrm{H}}Y$.
On final exit: the array is undefined.
11: $\mathbf{pdx}$Integer Input
On entry: the stride separating matrix row elements in the array x.
Constraint: ${\mathbf{pdx}}\ge {\mathbf{n}}$.
12: $\mathbf{y}\left[\mathit{dim}\right]$Complex Input/Output
Note: the dimension, dim, of the array y must be at least ${\mathbf{pdy}}×2$.
The $\left(i,j\right)$th element of the matrix $Y$ is stored in ${\mathbf{y}}\left[\left(j-1\right)×{\mathbf{pdy}}+i-1\right]$.
On initial entry: need not be set.
On intermediate exit: if ${\mathbf{irevcm}}=3$, contains the current $n×2$ matrix $Y$.
On intermediate re-entry: if ${\mathbf{irevcm}}=2$, must contain $AX$.
On final exit: the array is undefined.
13: $\mathbf{pdy}$Integer Input
On entry: the stride separating matrix row elements in the array y.
Constraint: ${\mathbf{pdy}}\ge {\mathbf{n}}$.
14: $\mathbf{p}\left[{\mathbf{n}}\right]$Complex Input/Output
On initial entry: need not be set.
On intermediate re-entry: if ${\mathbf{irevcm}}=4$, must contain $Az$.
On final exit: the array is undefined.
15: $\mathbf{r}\left[{\mathbf{n}}\right]$Complex Input/Output
On initial entry: need not be set.
On intermediate re-entry: if ${\mathbf{irevcm}}=5$, must contain ${A}^{\mathrm{H}}z$.
On final exit: the array is undefined.
16: $\mathbf{z}\left[{\mathbf{n}}\right]$Complex Input/Output
On initial entry: need not be set.
On intermediate exit: if ${\mathbf{irevcm}}=4$ or $5$, contains the vector $z$.
On intermediate re-entry: must not be changed.
On final exit: the array is undefined.
17: $\mathbf{ccomm}\left[{\mathbf{n}}×\left({\mathbf{m}}+2\right)\right]$Complex Communication Array
18: $\mathbf{comm}\left[3×{\mathbf{n}}+14\right]$double Communication Array
19: $\mathbf{icomm}\left[2×{\mathbf{n}}+40\right]$Integer Communication Array
20: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value.
NE_INT
On entry, ${\mathbf{m}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{m}}\ge 0$.
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 0$.
On initial entry, ${\mathbf{irevcm}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{irevcm}}=0$.
On intermediate re-entry, ${\mathbf{irevcm}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{irevcm}}=1$, $2$, $3$, $4$ or $5$.
NE_INT_2
On entry, ${\mathbf{pdb}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdb}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{pdb2}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdb2}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{pdx}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdx}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{pdy}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdy}}\ge {\mathbf{n}}$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NW_SOME_PRECISION_LOSS
${e}^{tA}B$ has been computed using an IEEE double precision Taylor series, although the arithmetic precision is higher than IEEE double precision.
7Accuracy
For an Hermitian matrix $A$ (for which ${A}^{\mathrm{H}}=A$) the computed matrix ${e}^{tA}B$ is guaranteed to be close to the exact matrix, that is, the method is forward stable. No such guarantee can be given for non-Hermitian matrices. See Section 4 of Al–Mohy and Higham (2011) for details and further discussion.
8Parallelism and Performance
f01hbc is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
9.1Use of $\mathbit{T}\mathbit{r}\left(\mathbit{A}\right)$
The elements of $A$ are not explicitly required by f01hbc. However, the trace of $A$ is used in the preprocessing phase of the algorithm. If $Tr\left(A\right)$ is not available to the calling function then any number can be supplied ($0$ is recommended). This will not affect the stability of the algorithm, but it may reduce its efficiency.
9.2When to use f01hbc
f01hbc is designed to be used when $A$ is large and sparse. Whenever a matrix multiplication is required, the function will return control to the calling program so that the multiplication can be done in the most efficient way possible. Note that ${e}^{tA}B$ will not, in general, be sparse even if $A$ is sparse.
If $A$ is small and dense then f01hac can be used to compute ${e}^{tA}B$ without the use of a reverse communication interface.
The real analog of f01hbc is f01gbc.
9.3Use in Conjunction with NAG Library Functions
To compute ${e}^{tA}B$, the following skeleton code can normally be used:
```do {
f01hbc(&irevcm,n,m,b,tdb,t,tr,b2,tdb2,x,tdx,y,tdy,p,r,z,ccomm,comm, &
icomm,&fail);
if (irevcm == 1) {
.. Code to compute B2=AB ..
}
else if (irevcm == 2){
.. Code to compute Y=AX ..
}
else if (irevcm == 3){
.. Code to compute X=A^H Y ..
}
else if (irevcm == 4){
.. Code to compute P=AZ ..
}
else if (irevcm == 5){
.. Code to compute R=A^H Z ..
}
} (while irevcm !=0)```
The code used to compute the matrix products will vary depending on the way $A$ is stored. If all the elements of $A$ are stored explicitly, then f16zac can be used. If $A$ is triangular then f16zfc should be used. If $A$ is Hermitian, then f16zcc should be used. If $A$ is symmetric, then f16ztc should be used. For sparse $A$ stored in coordinate storage format f11xnc and f11xsc can be used.
10Example
This example computes ${e}^{tA}B$ where
$A = ( 0.7+0.8i -0.2+0.0i 1.0+0.0i 0.6+0.5i 0.3+0.7i 0.7+0.0i 0.9+3.0i 1.0+0.8i 0.3+3.0i -0.7+0.0i 0.2+0.6i 0.7+0.5i 0.0+0.9i 4.0+0.0i 0.0+0.0i 0.2+0.0i ) ,$
$B = ( 0.1+0.0i 1.2+0.1i 1.3+0.9i -0.2+2.0i 4.0+0.6i -1.0+0.8i 0.4+0.0i -0.9+0.0i )$
and
$t=1.1+0.0i .$
$A$ is stored in compressed column storage format (CCS) and matrix multiplications are performed using the function matmul.
10.1Program Text
Program Text (f01hbce.c)
10.2Program Data
Program Data (f01hbce.d)
10.3Program Results
Program Results (f01hbce.r) | 3,922 | 12,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 163, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.724846 |
https://mittenmen.org/bennington/structural-analysis-with-the-finite-element-method-pdf.php | 1,606,527,710,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141194982.45/warc/CC-MAIN-20201128011115-20201128041115-00272.warc.gz | 415,242,380 | 9,208 | PDF Structural Analysis with the Finite Element Method The aim of this journal is to provide ideas and information involving the use of the finite element method and its variants, both in scientific inquiry and in professional practice. The scope is intentionally broad, encompassing use of the finite element method …
## 2.094— Finite Element Analysis of Solids and Fluids
Structural Analysis Finite Element Method BrainKart. · Finite element-Small elements used for subdividing the given domain tobe analysed are called finite elements. The seelements may be 1D, 2D or 3D elements depend in on the type of structure. The seelements may be 1D, 2D or 3D elements depend in on the type of structure., Finite Element Methods (in Solid and Structural Mechanics) An Introduction to the Finite Element Method 3nd Edition, McGraw Hill J. N. Reddy (2004) An Introduction to Nonlinear Finite Element Analysis Oxford University Publication . 13 Recommended Textbooks M. Asghar Bhatti (2005) Fundamental Finite Element Analysis and Applications: with Mathematica and Matlab ….
An Object-Oriented Smartphone Application for Structural Finite Element Analysis B.J. Mac Donald Faculty of Engineering and Computing Dublin City University, Dublin 9, Ireland Abstract—Smartphones are becoming increasingly ubiquitous both in general society and the workplace. Recent increases in mobile processing power have shown the current generation of smartphones has … The finite element method (FEM) is a numerical technique for finding approximate solutions to various computational domains. Numerical analysis done using FEM is commonly referred to as finite element analysis …
Thermal and Structural Analysis of a Gas Turbine Casing Using Finite Element Method 48 Structural Analysis: Symmetric boundary conditions are applied in the structural analysis. The finite element method is a powerful technique to solve the complex problems in structural engineering. The finite element models for the considered domes were
Structural Analysis with Finite Elements PDF-ebook in english (with Adobe DRM) This book provides a solid introduction to the foundation and the application of the finite element method in structural analysis. Finite Element Methods (in Solid and Structural Mechanics) An Introduction to the Finite Element Method 3nd Edition, McGraw Hill J. N. Reddy (2004) An Introduction to Nonlinear Finite Element Analysis Oxford University Publication . 13 Recommended Textbooks M. Asghar Bhatti (2005) Fundamental Finite Element Analysis and Applications: with Mathematica and Matlab …
A First Course in the Finite Element Analysis provides a simple, basic approach to the finite element method that can be understood by both undergraduate and graduate students. It does not have the usual prerequisites (such as structural analysis) require Finite element analysis (FEA) is a computerised method for predicting how a product reacts to real-world forces, vibration, heat, fluid flow and other physical effects.
The aim of this journal is to provide ideas and information involving the use of the finite element method and its variants, both in scientific inquiry and in professional practice. The scope is intentionally broad, encompassing use of the finite element method … An Introduction to Finite Element Analysis With Emphasis on Altair HyperWorks Suite Applications for FSAE December 3rd, Finite Element Analysis (FEA) is a numerical technique of obtaining solutions to the differential equations that describe or approximate a physical problem. • FEA uses the finite element method (FEM) to discretize a region (CAD model) into many smaller regions (elements
commonly used in the finite element method of structural analysis and specifically for the analysis of flat plates. They are written in 1900 Fortran The improved finite element model with splitting criterion was able to model the local pull-through failures in the less ductile G550 steel claddings th at are initiated by …
Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition Thermal and Structural Analysis of a Gas Turbine Casing Using Finite Element Method 48 Structural Analysis: Symmetric boundary conditions are applied in the structural analysis.
Mechanics of Structures, 2nd year, Mechanical Engineering, Cairo University Matrix Structural Analysis – the Stiffness Method Matrix structural analyses solve practical problems of … An Introduction to Finite Element Analysis With Emphasis on Altair HyperWorks Suite Applications for FSAE December 3rd, Finite Element Analysis (FEA) is a numerical technique of obtaining solutions to the differential equations that describe or approximate a physical problem. • FEA uses the finite element method (FEM) to discretize a region (CAD model) into many smaller regions (elements
The application of finite element method (FEM) to the analysis of discontinuous structural systems has received a significant interest in recent years. Examples of problems in which discontinuities play prominent role in the physical behavior of a system are numerous. From mathematical point of view, analytical solutions are possible only for a limited class of such problems. The complexities 12/05/2016 · Algebra - Solving Linear Equations by using the Gauss-Jordan Elimination Method 2/2
OГ±ate E. Structural Analysis with the Finite Element. The application of finite element method (FEM) to the analysis of discontinuous structural systems has received a significant interest in recent years. Examples of problems in which discontinuities play prominent role in the physical behavior of a system are numerous. From mathematical point of view, analytical solutions are possible only for a limited class of such problems. The complexities, In this study, a numerical analysis is presented to investigate the post buckling behavior of plane structural steel members subjected to static loads using finite element method. The material of.
### Structural Analysis Finite Element Method BrainKart
An Object-Oriented Smartphone Application for Structural. The improved finite element model with splitting criterion was able to model the local pull-through failures in the less ductile G550 steel claddings th at are initiated by …, 2.094 — Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 3 - Finite element formulation for solids and structures Prof. K.J. Bathe MIT OpenCourseWare.
How to learn finite element method for structural analysis. The aim of this journal is to provide ideas and information involving the use of the finite element method and its variants, both in scientific inquiry and in professional practice. The scope is intentionally broad, encompassing use of the finite element method …, 12/05/2016 · Algebra - Solving Linear Equations by using the Gauss-Jordan Elimination Method 2/2.
### structural analysis Buckling Finite Element Method
MODELLING AND STRUCTURAL ANALYSIS OF LEAF SPRING. The finite element method (FEM) is a numerical technique for finding approximate solutions to various computational domains. Numerical analysis done using FEM is commonly referred to as finite element analysis … Finite element analysis (FEA) is a computerised method for predicting how a product reacts to real-world forces, vibration, heat, fluid flow and other physical effects..
The finite element method (FEM) is a numerical technique for finding approximate solutions to various computational domains. Numerical analysis done using FEM is commonly referred to as finite element analysis … Finite element analysis has broad application and in different contexts terms may have different meanings, so this section defines terms as used in this course. The second portion of the course provides a number of FE analysis examples /
commonly used in the finite element method of structural analysis and specifically for the analysis of flat plates. They are written in 1900 Fortran Finite element analysis has broad application and in different contexts terms may have different meanings, so this section defines terms as used in this course. The second portion of the course provides a number of FE analysis examples /
The finite element method (FEM) is a numerical technique for finding approximate solutions to various computational domains. Numerical analysis done using FEM is commonly referred to as finite element analysis … Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition
The finite element method (FEM) is a numerical technique for finding approximate solutions to various computational domains. Numerical analysis done using FEM is commonly referred to as finite element analysis … A First Course in the Finite Element Analysis provides a simple, basic approach to the finite element method that can be understood by both undergraduate and graduate students. It does not have the usual prerequisites (such as structural analysis) require
Solving the Laplacian Equation in 3D using Finite Element Method in C# for Structural Analysis Article (PDF Available) · July 2011 with 570 Reads Export this citation The application of finite element method (FEM) to the analysis of discontinuous structural systems has received a significant interest in recent years. Examples of problems in which discontinuities play prominent role in the physical behavior of a system are numerous. From mathematical point of view, analytical solutions are possible only for a limited class of such problems. The complexities
Finite Element Methods (in Solid and Structural Mechanics) An Introduction to the Finite Element Method 3nd Edition, McGraw Hill J. N. Reddy (2004) An Introduction to Nonlinear Finite Element Analysis Oxford University Publication . 13 Recommended Textbooks M. Asghar Bhatti (2005) Fundamental Finite Element Analysis and Applications: with Mathematica and Matlab … Thermal and Structural Analysis of a Gas Turbine Casing Using Finite Element Method 48 Structural Analysis: Symmetric boundary conditions are applied in the structural analysis.
Finite element analysis (FEA) is a computerised method for predicting how a product reacts to real-world forces, vibration, heat, fluid flow and other physical effects. Structural Analysis with Finite Elements PDF-ebook in english (with Adobe DRM) This book provides a solid introduction to the foundation and the application of the finite element method in structural analysis.
Structural Analysis with Finite Elements PDF-ebook in english (with Adobe DRM) This book provides a solid introduction to the foundation and the application of the finite element method in structural analysis. The aim of this journal is to provide ideas and information involving the use of the finite element method and its variants, both in scientific inquiry and in professional practice. The scope is intentionally broad, encompassing use of the finite element method …
9/06/2016 · Algebra - Solving Linear Equations by using the Gauss-Jordan Elimination Method 2/2 · Finite element-Small elements used for subdividing the given domain tobe analysed are called finite elements. The seelements may be 1D, 2D or 3D elements depend in on the type of structure. The seelements may be 1D, 2D or 3D elements depend in on the type of structure.
## How to learn finite element method for structural analysis
How to learn finite element method for structural analysis. The application of finite element method (FEM) to the analysis of discontinuous structural systems has received a significant interest in recent years. Examples of problems in which discontinuities play prominent role in the physical behavior of a system are numerous. From mathematical point of view, analytical solutions are possible only for a limited class of such problems. The complexities, The finite element method is a powerful technique to solve the complex problems in structural engineering. The finite element models for the considered domes were.
### Thermal and Structural Analysis of a Gas Turbine Casing
*PDF* Finite Element Methods lovemombook.com. 9/06/2016 · Algebra - Solving Linear Equations by using the Gauss-Jordan Elimination Method 2/2, Finite element analysis (FEA) is a computerised method for predicting how a product reacts to real-world forces, vibration, heat, fluid flow and other physical effects..
The aim of this journal is to provide ideas and information involving the use of the finite element method and its variants, both in scientific inquiry and in professional practice. The scope is intentionally broad, encompassing use of the finite element method … The application of finite element method (FEM) to the analysis of discontinuous structural systems has received a significant interest in recent years. Examples of problems in which discontinuities play prominent role in the physical behavior of a system are numerous. From mathematical point of view, analytical solutions are possible only for a limited class of such problems. The complexities
The improved finite element model with splitting criterion was able to model the local pull-through failures in the less ductile G550 steel claddings th at are initiated by … Analysis into the Direct Stiffness Method from 1934 through 1970. This was the subject of a This was the subject of a separate essay [238], which is also given in Appendix H.
An Introduction to Finite Element Analysis With Emphasis on Altair HyperWorks Suite Applications for FSAE December 3rd, Finite Element Analysis (FEA) is a numerical technique of obtaining solutions to the differential equations that describe or approximate a physical problem. • FEA uses the finite element method (FEM) to discretize a region (CAD model) into many smaller regions (elements Finite element analysis has broad application and in different contexts terms may have different meanings, so this section defines terms as used in this course. The second portion of the course provides a number of FE analysis examples /
An Object-Oriented Smartphone Application for Structural Finite Element Analysis B.J. Mac Donald Faculty of Engineering and Computing Dublin City University, Dublin 9, Ireland Abstract—Smartphones are becoming increasingly ubiquitous both in general society and the workplace. Recent increases in mobile processing power have shown the current generation of smartphones has … MODELLING AND STRUCTURAL ANALYSIS OF LEAF SPRING USING FINITE ELEMENT METHOD M.Sai kumar1, Dr C.Bhaskara Reddy2 1PG Scholar, Department of Mechanical Engineering, Sri Kalahasteeswara Institute of Technology (SKIT), Srikalahasti, 517640 2
· Finite element-Small elements used for subdividing the given domain tobe analysed are called finite elements. The seelements may be 1D, 2D or 3D elements depend in on the type of structure. The seelements may be 1D, 2D or 3D elements depend in on the type of structure. Report No. 56-7, "A Finite-Element Analysis of Structural Frames" by T. Allan Haliburton and Hudson Matlock, describes a method of analysis for rectangular …
The aim of this journal is to provide ideas and information involving the use of the finite element method and its variants, both in scientific inquiry and in professional practice. The scope is intentionally broad, encompassing use of the finite element method … Finite element analysis has broad application and in different contexts terms may have different meanings, so this section defines terms as used in this course. The second portion of the course provides a number of FE analysis examples /
Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition 9/06/2016 · Algebra - Solving Linear Equations by using the Gauss-Jordan Elimination Method 2/2
The aim of the course is to provide the participants an overview on Finite Element Method, Material models, and Applications in Civil Engineering. At the end of the course, the participants are expected to have fair understanding of: Basics of Finite Element Analysis. Available material models for structural materials, soils and interfaces/joints. Modeling of engineering systems and Soil Solving the Laplacian Equation in 3D using Finite Element Method in C# for Structural Analysis Article (PDF Available) · July 2011 with 570 Reads Export this citation
PDF Structural Analysis with the Finite Element Method. Mechanics of Structures, 2nd year, Mechanical Engineering, Cairo University Matrix Structural Analysis – the Stiffness Method Matrix structural analyses solve practical problems of …, Report No. 56-7, "A Finite-Element Analysis of Structural Frames" by T. Allan Haliburton and Hudson Matlock, describes a method of analysis for rectangular ….
### IIT Gandhinagar Short Course on FEM
(PDF) APPLICATIONS OF FINITE ELEMENT METHOD IN STRUCTURAL. Finite Element Methods (in Solid and Structural Mechanics) An Introduction to the Finite Element Method 3nd Edition, McGraw Hill J. N. Reddy (2004) An Introduction to Nonlinear Finite Element Analysis Oxford University Publication . 13 Recommended Textbooks M. Asghar Bhatti (2005) Fundamental Finite Element Analysis and Applications: with Mathematica and Matlab …, Report No. 56-7, "A Finite-Element Analysis of Structural Frames" by T. Allan Haliburton and Hudson Matlock, describes a method of analysis for rectangular ….
### *PDF* Finite Element Methods lovemombook.com
An Object-Oriented Smartphone Application for Structural. Mechanics of Structures, 2nd year, Mechanical Engineering, Cairo University Matrix Structural Analysis – the Stiffness Method Matrix structural analyses solve practical problems of … commonly used in the finite element method of structural analysis and specifically for the analysis of flat plates. They are written in 1900 Fortran.
Finite element analysis (FEA) is a computerised method for predicting how a product reacts to real-world forces, vibration, heat, fluid flow and other physical effects. Structural Analysis with Finite Elements PDF-ebook in english (with Adobe DRM) This book provides a solid introduction to the foundation and the application of the finite element method in structural analysis.
Mechanics of Structures, 2nd year, Mechanical Engineering, Cairo University Matrix Structural Analysis – the Stiffness Method Matrix structural analyses solve practical problems of … 2.094 — Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 3 - Finite element formulation for solids and structures Prof. K.J. Bathe MIT OpenCourseWare
MODELLING AND STRUCTURAL ANALYSIS OF LEAF SPRING USING FINITE ELEMENT METHOD M.Sai kumar1, Dr C.Bhaskara Reddy2 1PG Scholar, Department of Mechanical Engineering, Sri Kalahasteeswara Institute of Technology (SKIT), Srikalahasti, 517640 2 · Finite element-Small elements used for subdividing the given domain tobe analysed are called finite elements. The seelements may be 1D, 2D or 3D elements depend in on the type of structure. The seelements may be 1D, 2D or 3D elements depend in on the type of structure.
In this study, a numerical analysis is presented to investigate the post buckling behavior of plane structural steel members subjected to static loads using finite element method. The material of Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition
Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition
Finite element analysis has broad application and in different contexts terms may have different meanings, so this section defines terms as used in this course. The second portion of the course provides a number of FE analysis examples / Finite element analysis (FEA) is a computerised method for predicting how a product reacts to real-world forces, vibration, heat, fluid flow and other physical effects.
2.094 — Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 3 - Finite element formulation for solids and structures Prof. K.J. Bathe MIT OpenCourseWare Finite element analysis (FEA) is a computerised method for predicting how a product reacts to real-world forces, vibration, heat, fluid flow and other physical effects.
Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition The finite element method is a powerful technique to solve the complex problems in structural engineering. The finite element models for the considered domes were
Structural Analysis with the Finite Element Method. Linear Statics: Volume 2: Beams, Plates and Shells (Lecture Notes on Numerical Methods in Engineering and Sciences) Softcover reprint of the original 1st ed. 2013 Edition Thermal and Structural Analysis of a Gas Turbine Casing Using Finite Element Method 48 Structural Analysis: Symmetric boundary conditions are applied in the structural analysis. | 4,373 | 21,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-50 | latest | en | 0.815185 |
https://courses.lumenlearning.com/suny-microeconomics/chapter/the-structure-of-costs-in-the-short-run/ | 1,718,790,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00004.warc.gz | 157,166,461 | 19,978 | ## The Structure of Costs in the Short Run
The cost of producing a firm’s output depends on how much labor and physical capital the firm uses. A list of the costs involved in producing cars will look very different from the costs involved in producing computer software or haircuts or fast-food meals. However, the cost structure of all firms can be broken down into some common underlying patterns. When a firm looks at its total costs of production in the short run, a useful starting point is to divide total costs into two categories: fixed costs that cannot be changed in the short run and variable costs that can be changed.
The breakdown of total costs into fixed and variable costs can provide a basis for other insights as well. The first five columns of Table 7.3 duplicate the previous table, but the last three columns show average total costs, average variable costs, and marginal costs. These new measures analyze costs on a per-unit (rather than a total) basis and are reflected in the curves shown in Figure 7.4.
Table 7.3. Output and Total Costs
Labor Quantity Fixed Cost Variable Cost Total Cost
1 16 $160$80 $240 2 40$160 $160$320
3 60 $160$240 $400 4 72$160 $320$480
5 80 $160$400 $560 6 84$160 $480$640
7 82 $160$560 $720 Figure 7.4. Cost Curves at the Clip Joint. The information on total costs, fixed cost, and variable cost can also be presented on a per-unit basis. Average total cost (ATC) is calculated by dividing total cost by the total quantity produced. The average total cost curve is typically U-shaped. Average variable cost (AVC) is calculated by dividing variable cost by the quantity produced. The average variable cost curve lies below the average total cost curve and is typically U-shaped or upward-sloping. Marginal cost (MC) is calculated by taking the change in total cost between two levels of output and dividing by the change in output. The marginal cost curve is upward-sloping. Average total cost is total cost divided by the quantity of output. Since the total cost of producing 40 haircuts is$320, the average total cost for producing each of 40 haircuts is $320/40, or$8 per haircut. Average cost curves are typically U-shaped, as Figure 7.4 shows. Average total cost starts off relatively high, because at low levels of output total costs are dominated by the fixed cost; mathematically, the denominator is so small that average total cost is large. Average total cost then declines, as the fixed costs are spread over an increasing quantity of output. In the average cost calculation, the rise in the numerator of total costs is relatively small compared to the rise in the denominator of quantity produced. But as output expands still further, the average cost begins to rise. At the right side of the average cost curve, total costs begin rising more rapidly as diminishing returns kick in.
Average variable cost obtained when variable cost is divided by quantity of output. For example, the variable cost of producing 80 haircuts is $400, so the average variable cost is$400/80, or $5 per haircut. Note that at any level of output, the average variable cost curve will always lie below the curve for average total cost, as shown in Figure 7.4. The reason is that average total cost includes average variable cost and average fixed cost. Thus, for Q = 80 haircuts, the average total cost is$8 per haircut, while the average variable cost is $5 per haircut. However, as output grows, fixed costs become relatively less important (since they do not rise with output), so average variable cost sneaks closer to average cost. Average total and variable costs measure the average costs of producing some quantity of output. Marginal cost is somewhat different. Marginal cost is the additional cost of producing one more unit of output. So it is not the cost per unit of all units being produced, but only the next one (or next few). Marginal cost can be calculated by taking the change in total cost and dividing it by the change in quantity. For example, as quantity produced increases from 40 to 60 haircuts, total costs rise by 400 – 320, or 80. Thus, the marginal cost for each of those marginal 20 units will be 80/20, or$4 per haircut. The marginal cost curve is generally upward-sloping, because diminishing marginal returns implies that additional units are more costly to produce. A small range of increasing marginal returns can be seen in the figure as a dip in the marginal cost curve before it starts rising. There is a point at which marginal and average costs meet, as explained below.
### Where do marginal and average costs meet?
The marginal cost line intersects the average cost line exactly at the bottom of the average cost curve—which occurs at a quantity of 72 and cost of \$6.60 in Figure 7.4. The reason why the intersection occurs at this point is built into the economic meaning of marginal and average costs. If the marginal cost of production is below the average cost for producing previous units, as it is for the points to the left of where MC crosses ATC, then producing one more additional unit will reduce average costs overall—and the ATC curve will be downward-sloping in this zone. Conversely, if the marginal cost of production for producing an additional unit is above the average cost for producing the earlier units, as it is for points to the right of where MC crosses ATC, then producing a marginal unit will increase average costs overall—and the ATC curve must be upward-sloping in this zone. The point of transition, between where MC is pulling ATC down and where it is pulling it up, must occur at the minimum point of the ATC curve.
This idea of the marginal cost “pulling down” the average cost or “pulling up” the average cost may sound abstract, but think about it in terms of your own grades. If the score on the most recent quiz you take is lower than your average score on previous quizzes, then the marginal quiz pulls down your average. If your score on the most recent quiz is higher than the average on previous quizzes, the marginal quiz pulls up your average. In this same way, low marginal costs of production first pull down average costs and then higher marginal costs pull them up.
The numerical calculations behind average cost, average variable cost, and marginal cost will change from firm to firm. However, the general patterns of these curves, and the relationships and economic intuition behind them, will not change.
## Self Check: Marginal, Average, and Total Cost
Answer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does not count toward your grade in the class, and you can retake it an unlimited number of times.
You’ll have more success on the Self Check if you’ve completed the two Readings in this section.
Use this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section. | 1,492 | 6,954 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.93422 |
https://physics.stackexchange.com/questions/734969/how-can-a-plane-wave-induce-perpendicular-motion-in-particles-when-the-plane-wa | 1,686,054,879,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00189.warc.gz | 496,659,048 | 35,756 | # How can a plane wave induce perpendicular motion in particles, when the plane wave only carries longitudinal momentum? [closed]
An electromagnetic plane wave moving in the $$z$$-direction
$$E(x,t) = E_x \cos(k z - \omega t), \hspace{1cm} B(x,t) = B_y \sin(kz-\omega t)$$
has field momentum in $$z$$-direction. But according to this paper, it induces motion in $$x$$ and $$y$$ direction in a charged particle. How is this possible? Does conservation of momentum not apply? If not, are field and particle momentum completely separated concepts?
Remark: My question is less focused on the precise movement described in the paper, but generally how it is possible for a wave with momentum in $$z$$ direction to induce motion in $$x$$ and $$y$$ direction.
• When the particle acquires x momentum, the field acquires negative x momentum. (Don’t forget that charged particles make their own fields.) Nov 2, 2022 at 20:48
• Can this be true? The particles momentum is dependent on the mass, but the charged particles field is indipendent of the mass. So how can they cancel out reliably? Or did I missunderstand you? Nov 2, 2022 at 22:07
• The field the charged particle makes depends on how fast it's moving, which in turn depends on its mass, because $F = ma$. Nov 2, 2022 at 22:10
• Nov 3, 2022 at 6:01
• The paper you link in your question is too involved for me to check, In the linked answer to a similar question it is seen that the oscilations happen in the direction of the propagation of the wave, Nov 3, 2022 at 6:43
It is easiest to think of this in terms of scattering. A photon, with momentum in $$z$$ collides elastically with a particle at rest. After the collision the particle has momentum along $$x$$ so the photon is scattered and has equal momentum along $$-x$$ in addition to its momentum along $$z$$
In the simple classical picture, the electromagnetic field does no work on the charged particle (because the E-field and particle velocity are $$\pi/2$$ out of phase) in a time-averaged sense. Similarly, there is no time-averaged momentum transferred to the particle because it oscillates up and down along the polarisation vector of the E-field. In a bit more detail we would include the Lorentz force due to the magnetic part of the field and find that there was some momentum transferred to the particle along the z-axis. | 569 | 2,347 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-23 | latest | en | 0.930606 |
http://math.stackexchange.com/questions/tagged/logic+higher-order-logic | 1,409,132,023,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500828050.28/warc/CC-MAIN-20140820021348-00265-ip-10-180-136-8.ec2.internal.warc.gz | 128,326,304 | 22,659 | # Tagged Questions
92 views
107 views
### There is a second-order sentence that is valid in standard semantics but not valid in Henkin semantics?
Let $\Sigma^\mathrm{ST}$ be a set of sentences that is valid in standard semantics and $\Sigma^\mathrm{Henk}$ be a set of sentences that is valid in Henkin semantics. Since ...
37 views
### How is it possible that the well-ordering theorem is strictly stronger than the axiom of choice in second-order logic? [duplicate]
If I am not wrong, the well-ordering theorem is strictly stronger than the axiom of choice in second-order logic. I am not sure to understand how this is possible. The reason is that second order ...
88 views
### Why is better to work with first-order Peano's axioms than with second-order Peano's axioms?
In the case of Peano's axioms the second-order version is categorical, but the first-order is not. Besides, with the second-order version the operations: addition, multiplication and exponentiation ...
94 views
### Why is better to work with first-order logic than with second-order logic? [duplicate]
Why is better to work with first-order logic than with second-order logic? In the case of Peano's axioms the second-order version is categorical, but the first-order is not. Besides, with the ...
135 views
### Any “natural” examples of true statements in number theory not provable in 2nd order systems?
I know that there are a few theorems in number theory that are somehow known to be true, but have been shown not to be provable in first-order Peano arithmetic (PA). Have any so-called "natural" ...
85 views
### What is the intuition behind $\Delta_1^0$ sets and $\Delta_1^1$ sets?
In the context of first-order arithmetic, if $\phi$ is a formula with only bounded quantifiers, then if you put existential quantifiers in front it becomes a $\Sigma_1^0$ formula according to the ...
43 views
### What subsystem of second-order arithmetic can interpret the theory of real closed fields?
Real numbers can be encoded as sets of natural numbers, because they can be encoded as Dedekind cuts or Cauchy sequences of rational numbers, and a rational number can be encoded by a natural number. ...
104 views
### Who first proved that the second-order theory of real numbers is categorical?
The second-order theory of real numbers is obtained by taking the axioms of ordered fields and adding a (Dedekind) completeness axiom, which states that every set which has an upper bound has a least ...
52 views
### What is the proof-theoretic strength of the predicative second-order theory of real numbers?
The first-order theory of real numbers, AKA the theory of real closed fields, is obtained by added to the axioms for ordered fields an axiom schema of completeness, which states that for each formula ...
77 views
### What is $M_x$ in Frege's Basic Law IIb?
Gottlob Frege's magnum opus, "The Basic Laws of Arithmetic" (Die Grundgesetze der Arithmetic in German) constitutes one of most impressive and meticulous attempts at developing a rigorous foundation ...
108 views
### Can equinumerosity by defined in monadic second-order logic?
Two properties (or concepts) $F$ and $G$ are said to be equinumerous if they have the same cardinality, i.e. if they can be put in one-to-one correspondence with each other. This can be very easily ...
125 views
### Peano arithmetic with the second-order induction axiom
I am in the middle of my PhD and I am trying to reinforce my knowledge of mathematics by studying the foundations of Analysis. The first task is to get the bases of the natural numbers. So for this I ...
196 views
### How can one quantify on a function in ZFC?
I have read that $ZFC$ and first-order logic could formalize all the mathematics, but I do not manage to conceive that. First, let me show what my understanding of $ZFC$ is. I have read that $ZFC$ was ...
628 views
### Relationship between propositional logic, first-order logic, second-order logic higher-order logic, and type theory
I understand there is propositional logic, first-order logic, second-order logic higher-order logic, and type theory, where the latter logics are extensions of the former logics. Can someone explain ...
190 views
### what are first and second order logics? [duplicate]
The only knowledge I have on logic is due to a book I read a couple of years ago called Introduction to logic: and to the methodology of deductive sciences by Alfred Tarski. And in it he talks about ...
230 views
### Do isomorphic structures always satisfy the same second-order sentences?
I know that if two mathematical structures are isomorphic, then they satisfy the same first-order sentences. The converse is false. This is probably a completely obvious question, but is it true that ...
138 views
### Henkin vs. “Full” Semantics for Second-order Logic and Multi-Sorted First Order Interpretations
In this paper by Jeff Ketland, he notes: With Henkin semantics, the Completeness, Compactness and Löwenheim-Skolem Theorems all hold, because Henkin structures can be re-interpreted as many-sorted ...
144 views
### Is higher order type theory the same as higher order logic?
The internal language of a topos is higher order intuitionistic type theory (or logic). Here the higher order simply refers to allowing function types. In mathematical logic we have higher-order ...
292 views
### Is second order logic even a logic?
Second order logic is a language, but, is it a logic? My understanding is that a logic (or "logical system") is an ordered pair; it is a formal system together with a semantics. However, the language ...
145 views
### Is the expressive power of infinitary logic language $L(\infty,\infty)$ larger, the same or smaller than that of ZFC+large cardinal axioms?
In a previous question I learned that the power of statements of the form $\Pi_m^n$ or $\Sigma_m^n$ for arbitrary positive $m$ and $n$, is smaller than that of ZFC. For instance, the GCH cannot be ...
220 views
### Monadic second order logic without constants, functions and equality
Leibniz's law of the identity of indiscernibles can be stated in monadic second order logic: $$\forall x\forall y (x=y \leftarrow \forall P (Px \leftrightarrow Py))$$ This law is true for standard ...
150 views
### Are there statements in set theory about arithmetic beyond the reach of the analytical hierarchy?
Even if the answer were negative for arithmetics(I have no idea), in the more general case: Can any mathematical statement be expressed as a $\Delta_m^n$ (with n, m belongs to N) statement in a chosen ...
188 views
### Can second order logic express each (computable) infinitary logic sentence?
In chapter 9 of Ebbinghaus et. al, the logical systems $\mathcal{L}_\text{II}$ ("full" second order logic with standard semantics) and $\mathcal{L}_{\omega_1\omega}$ (countable infinitary logic with ...
165 views
### Is every φ above the second level of the arithmetical hierarchy independent of PA?
If I am not wrong, every $\Sigma_n$ (or $\Pi_n$ ) statement $\phi$ is equivalent to a statement that says that a given Turing machine halts (or doesn't halt) on input $C$ using a ...
109 views
### Can decidability results for monadic second-order logic be extended to monadic higher-order logics?
Call a higher-order logic fully monadic if and only if all of its predicate constants (at any order) and higher-order variables (at any order) are monadic (and it has no function symbols). In Solvable ...
261 views
### How are the full semantics of SOL and HOL specified?
In relation to this question about the "fundamental" character of possible logical systems, I realized that I just had an intuitive (and so, inadequate) understanding of the way logics higher than FOL ...
506 views
### Is First Order Logic (FOL) the only fundamental logic?
I'm far from being an expert in the field of mathematical logic, but I've been reading about the academic work invested in the foundations of mathematics, both in a historical and objetive sense; and ...
417 views
### Second-order logic - monadic version and Henkin semantics
After looking at texts and Wikipedia, I am getting some confusion on difference between monadic second-order logic and full second-order logic and difference between Henkin semantic and full semantic. ...
423 views
### advantage of first-order logic over second-order logic
As I look over the post that has the similar question, I began to wonder: The only reason I found is that first-order logic can prove validity of some second-order logic formula/sentences, as some of ...
235 views
### What makes higher/second-order logic incompatible with completeness theorem?
Completeness theorem and compactness theorem do not hold in full second and higher-order logic. What makes them incompatible with the theorems? Is it somehow related to Russell's paradox or ...
368 views
### First-order logic advantage over second-order logic
What is the advantage of using first-order logic over second-order logic? Second-order logic is more expressive and there is also a way to overcome Russell's paradox... So what makes first-order ...
94 views
### Formal second-order statements of Archimedean and completeness properties
I am trying to translate the following statements from English to second-order logic, and I want to know if I got them right. I have a language for an ordered field $(F,+,\cdot,0,1,\leq)$, i.e., I ...
225 views
### Independence results in first-order PA and second-order PA
There are statements $\varphi$ that are independent of first-order Peano Axioms. Are these statements also independent of second-order Peano Axioms? I'm reading Wikipedia articles around ...
159 views
### Are $\{X \in 2^{A}:|X|=2\}$, $\{X \subset 2^{A}:|X|=2\}$, $\{X \subset A:|X|=2\}$ first order or second order?
Are those statements first order or second order: $$\{X\in 2^{A}:|X|=2\} \\ \{X \subset 2^{A}:|X|=2\} \\ \{X \subset A:|X|=2\}$$ why?
79 views
### Introductory text about different stratification methods in higher-order logic and set theory
Could someone recommend me a good overview text about stratification of predicates, comprehension axioms, and other methods of avoiding the paradoxes in untyped or only loosely/relatively typed ...
71 views
### Differential fields and rings
If one is to compute the derivative of $$y=3x+2$$ by $$\frac{\mathrm{d}(3x+2)}{\mathrm{d} x}$$ Would I be working with differential fields? Since differential fields is a first-order ...
155 views
### Is there such a thing as “second-order-undecidability”? And what about higher order Undecidability statements?
I know that there are statements that are neither provable nor disprovable within some set of axioms, and I also know that such statements are called undecidable. Please allow me to call these ... | 2,446 | 10,809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2014-35 | longest | en | 0.922456 |
https://oeis.org/A228825 | 1,502,964,821,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00459.warc.gz | 792,649,129 | 4,261 | This site is supported by donations to The OEIS Foundation.
"Email this user" was broken Aug 14 to 9am Aug 16. If you sent someone a message in this period, please send it again.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A228825 Delayed continued fraction of e. 3
2, 2, -1, -1, -1, -2, 2, -2, 1, 1, 1, 2, -2, 2, -2, 2, -1, -1, -1, -2, 2, -2, 2, -2, 2, -2, 1, 1, 1, 2, -2, 2, -2, 2, -2, 2, -2, 2, -1, -1, -1, -2, 2, -2, 2, -2, 2, -2, 2, -2, 2, -2, 1, 1, 1, 2, -2, 2, -2, 2, -2, 2, -2, 2, -2, 2, -2, 2, -1, -1, -1, -2, 2, -2 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS An algorithm for the (usual) continued fraction of r > 0 follows: x(0) = r, a(n) = floor(x(n)), x(n+1) = 1/(x(n) - a(n)). The accelerated continued fraction uses "round" instead of "floor" (cf. A133593, A133570, A228667), where round(x) is the integer nearest x. The delayed continued fraction (DCF) uses "second nearest integer", so that all the terms are in {-2, -1, 1, 2}. If s/t and u/v are consecutive convergents of a DCF, then |s*x-u*t| = 1. Regarding DCF(e), after the initial (2,2), the strings (-1,-1,-1) and (1,1,1) alternate with odd-length strings of the forms (-2,2,...,-2) and (2,-2,...,2). The string lengths form the sequence 2,3,3,3,5,3,7,3,9,3,11,3,13,3,... Comparison of convergence rates is indicated by the following approximate values of x-e, where x is the 20th convergent: for delayed CF, x-e = 5.4x10^-7; for classical CF, x-e = 6.1x10^-16; for accelerated CF, x-e = -6.6x10^-27. The convergents for accelerated CF are a proper subset of those for classical CF, which are a proper subset of those for delayed CF (which are sampled in Example). LINKS EXAMPLE Convergents: 2, 5/2, 3, 8/3, 11/4, 30/11, 49/18, 68/25, 19/7, 87/32, 106/39, 299/110, 492/181,... MATHEMATICA \$MaxExtraPrecision = Infinity; x[0] = E; s[x_] := s[x] = If[FractionalPart[x] < 1/2, Ceiling[x], Floor[x]]; a[n_] := a[n] = s[Abs[x[n]]]*Sign[x[n]]; x[n_] := 1/(x[n - 1] - a[n - 1]); t = Table[a[n], {n, 0, 100}] CROSSREFS Cf. A133570, A228826 Sequence in context: A025451 A184257 A275656 * A201208 A006513 A105224 Adjacent sequences: A228822 A228823 A228824 * A228826 A228827 A228828 KEYWORD cofr,sign,easy AUTHOR Clark Kimberling, Sep 04 2013 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent | More pages
The OEIS Community | Maintained by The OEIS Foundation Inc. | 979 | 2,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-34 | longest | en | 0.752468 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=10004&CurriculumID=24&NQ=2&Num=9.18 | 1,582,578,795,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145981.35/warc/CC-MAIN-20200224193815-20200224223815-00481.warc.gz | 83,881,067 | 3,731 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
Example: The sides of a quadrilateral ABCD are 5 cm, 6 cm, 9 cm and 8 cm. Find its perimeter. Solution: Perimeter of the Quadrilateral (P) = S1 + S2 + S3 + S4 Where P is the perimeter of the quadrilateral and S1, S2, S3 and S4 are the sides of the quadrilateral. Given that, S1 = 5 cm, S2 = 6 cm, S3 = 9 cm and S4 = 8 cm. Hence P = 5 + 6 + 9 + 8 P = 28 cm. Directions: Read the above example carefully and answer the following questions. Also draw 10 different quadrilaterals and find their perimeters.
Q 1: The sides of a quadrilateral ABCD are 2564 cm, 1235 cm, x cm and 1250 cm. Find x if its perimeter is given by 6618 cm.1569 cm1238 cm1258 cm1548 cm Question 2: This question is available to subscribers only! | 277 | 882 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-10 | latest | en | 0.833796 |
https://recalll.co/app/?q=algorithms | 1,568,664,644,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572934.73/warc/CC-MAIN-20190916200355-20190916222355-00225.warc.gz | 638,712,951 | 44,240 | ## Why we care for the asymptotic bound of an algorithm?
First let's understand what big O, big Theta and big Omega are. They are all sets of functions.
Big O is giving upper asymptotic bound, while big Omega is giving a lower bound. Big Theta gives both.
`T(n)`
`(f(n))`
`O(f(n))`
`Omega(f(n))`
For example, merge sort worst case is both O(n*log(n)) and Omega(n*log(n)) - and thus is also (n*log(n)), but it is also O(n^2), since n^2 is asymptotically "bigger" than it. However, it is not (n^2), Since the algorithm is not Omega(n^2).
O(n) is asymptotic upper bound. If T(n) is O(f(n)), it means that from a certain n0, there is a constant C such that T(n) <= C * f(n). On the other hand, big-Omega says there is a constant C2 such that T(n) >= C2 * f(n))).
Not to be confused with worst, best and average cases analysis: all three (Omega, O, Theta) notation are not related to the best, worst and average cases analysis of algorithms. Each one of these can be applied to each analysis.
We usually use it to analyze complexity of algorithms (like the merge sort example above). When we say "Algorithm A is O(f(n))", what we really mean is "The algorithms complexity under the worst1 case analysis is O(f(n))" - meaning - it scales "similar" (or formally, not worse than) the function f(n).
Well, there are many reasons for it, but I believe the most important of them are:
• It is much harder to determine the exact complexity function, thus we "compromise" on the big-O/big-Theta notations, which are informative enough theoretically.
• The exact number of ops is also platform dependent. For example, if we have a vector (list) of 16 numbers. How much ops will it take? The answer is: it depends. Some CPUs allow vector additions, while other don't, so the answer varies between different implementations and different machines, which is an undesired property. The big-O notation however is much more constant between machines and implementations.
To demonstrate this issue, have a look at the following graphs:
It is clear that f(n) = 2*n is "worse" than f(n) = n. But the difference is not quite as drastic as it is from the other function. We can see that f(n)=logn quickly getting much lower than the other functions, and f(n) = n^2 is quickly getting much higher than the others. So - because of the reasons above, we "ignore" the constant factors (2* in the graphs example), and take only the big-O notation.
In the above example, f(n)=n, f(n)=2*n will both be in O(n) and in Omega(n) - and thus will also be in Theta(n). On the other hand - f(n)=logn will be in O(n) (it is "better" than f(n)=n), but will NOT be in Omega(n) - and thus will also NOT be in Theta(n). Symetrically, f(n)=n^2 will be in Omega(n), but NOT in O(n), and thus - is also NOT Theta(n).
1Usually, though not always. when the analysis class (worst, average and best) is missing, we really mean the worst case.
+1 for the nice explanation, I have a confusion ,you have written in last line that ,, f(n)=n^2 will be in Omega(n). but not in O(n) .and thus is also not Theta(n) >how?
@krishnaChandra: f(n) = n^2 is asymptotically stronger then n, and thus is Omega(n). However it is not O(n) (because for large n values, it is bigger then c*n, for all n). Since we said Theta(n) is the intersection of O(n) and Omega(n), since it is not O(n), it cannot be Theta(n) as well.
It's great to see someone explain how big-O notation isn't related to the best/worst case running time of an algorithm. There are so many websites that come up when I google the topic that say O(T(n)) means the worse case running time.
@almel It's 2*n (2n, two times n) not 2^n
@VishalK 1. Big O is the upper bound as n tends to infinity. 2. Omega is the lower bound as n tends to infinity. 3. Theta is both the upper and lower bound as n tends to infinity. Note that all bounds are only valid "as n tends to infinity", because the bounds do not hold for low values of n (less than n0). The bounds hold for all n n0, but not below n0 where lower order terms become dominant.
## algorithm - What exactly does big Ө notation represent? - Stack Overfl...
algorithm computer-science big-o notation big-theta
This is one of the faster algorithms, up to 170!. It fails inexplicably beyond 170!, and it's relatively slow for small factorials, but for factorials between 80 and 170 it's blazingly fast compared to many algorithms.
`curl http://www.google.com/search?q=170!`
Let me know if you find a bug, or faster implementation for large factorials.
This algorithm is slightly slower, but gives results beyond 170:
`curl http://www58.wolframalpha.com/input/?i=171!`
Use MPFR (mpfr.org). It allows floats with exponents in the 2^(2^32) range, or so...
... and the formatting of the page is destroyed!
## Factorial Algorithms in different languages - Stack Overflow
algorithm language-agnostic
There is precisely one algorithm with runtime O(1/n), the "empty" algorithm.
For an algorithm to be O(1/n) means that it executes asymptotically in less steps than the algorithm consisting of a single instruction. If it executes in less steps than one step for all n > n0, it must consist of precisely no instruction at all for those n. Since checking 'if n > n0' costs at least 1 instruction, it must consist of no instruction for all n.
Summing up: The only algorithm which is O(1/n) is the empty algorithm, consisting of no instruction.
This is the only correct answer in this thread, and (despite my upvote) it is at zero votes. Such is StackOverflow, where "correct-looking" answers are voted higher than actually correct ones.
No, its rated 0 because it is incorrect. Expressing a big-Oh value in relation to N when it is independent of N is incorrect. Second, running any program, even one that just exists, takes at least a constant amount of time, O(1). Even if that wasn't the case, it'd be O(0), not O(1/n).
Any function that is O(0) is also O(1/n), and also O(n), also O(n^2), also O(2^n). Sigh, does no one understand simple definitions? O() is an upper bound.
@kenj0418 You managed to be wrong in every single sentence. "Expressing a big-Oh value in relation to N when it is independent of N is incorrect." A constant function is a perfectly goof function. "Second, running any program, even one that just exists, takes at least a constant amount of time, O(1)." The definition of complexity doesn't say anything about actually running any programs. "it'd be O(0), not O(1/n)". See @ShreevatsaR's comment.
it seems that you are not plumber. good answer.
## theory - Are there any O(1/n) algorithms? - Stack Overflow
theory complexity-theory big-o
To describe a permutation of n elements, you see that for the position that the first element ends up at, you have n possibilities, so you can describe this with a number between 0 and n-1. For the position that the next element ends up at, you have n-1 remaining possibilities, so you can describe this with a number between 0 and n-2. Et cetera until you have n numbers.
As an example for n = 5, consider the permutation that brings abcde to caebd.
• a, the first element, ends up at the second position, so we assign it index 1.
• b ends up at the fourth position, which would be index 3, but it's the third remaining one, so we assign it 2.
• d ends up at the last remaining position, which (out of only two remaining positions) is 1.
So we have the index sequence {1, 2, 0, 1, 0}.
Now you know that for instance in a binary number, 'xyz' means z + 2y + 4x. For a decimal number, it's z + 10y + 100x. Each digit is multiplied by some weight, and the results are summed. The obvious pattern in the weight is of course that the weight is w = b^k, with b the base of the number and k the index of the digit. (I will always count digits from the right and starting at index 0 for the rightmost digit. Likewise when I talk about the 'first' digit I mean the rightmost.)
The reason why the weights for digits follow this pattern is that the highest number that can be represented by the digits from 0 to k must be exactly 1 lower than the lowest number that can be represented by only using digit k+1. In binary, 0111 must be one lower than 1000. In decimal, 099999 must be one lower than 100000.
Encoding to variable-base The spacing between subsequent numbers being exactly 1 is the important rule. Realising this, we can represent our index sequence by a variable-base number. The base for each digit is the amount of different possibilities for that digit. For decimal each digit has 10 possibilities, for our system the rightmost digit would have 1 possibility and the leftmost will have n possibilities. But since the rightmost digit (the last number in our sequence) is always 0, we leave it out. That means we're left with bases 2 to n. In general, the k'th digit will have base b[k] = k + 2. The highest value allowed for digit k is h[k] = b[k] - 1 = k + 1.
Our rule about the weights w[k] of digits requires that the sum of h[i] * w[i], where i goes from i = 0 to i = k, is equal to 1 * w[k+1]. Stated recurrently, w[k+1] = w[k] + h[k] * w[k] = w[k]*(h[k] + 1). The first weight w[0] should always be 1. Starting from there, we have the following values:
```k h[k] w[k]
0 1 1
1 2 2
2 3 6
3 4 24
... ... ...
n-1 n n!```
(The general relation w[k-1] = k! is easily proved by induction.)
The number we get from converting our sequence will then be the sum of s[k] * w[k], with k running from 0 to n-1. Here s[k] is the k'th (rightmost, starting at 0) element of the sequence. As an example, take our {1, 2, 0, 1, 0}, with the rightmost element stripped off as mentioned before: {1, 2, 0, 1}. Our sum is 1 * 1 + 0 * 2 + 2 * 6 + 1 * 24 = 37.
Note that if we take the maximum position for every index, we'd have {4, 3, 2, 1, 0}, and that converts to 119. Since the weights in our number encoding were chosen so that we don't skip any numbers, all numbers 0 to 119 are valid. There are precisely 120 of these, which is n! for n = 5 in our example, precisely the number of different permutations. So you can see our encoded numbers completely specify all possible permutations.
Decoding from variable-base Decoding is similar to converting to binary or decimal. The common algorithm is this:
```int number = 42;
int base = 2;
int[] bits = new int[n];
for (int k = 0; k < bits.Length; k++)
{
bits[k] = number % base;
number = number / base;
}```
```int n = 5;
int number = 37;
int[] sequence = new int[n - 1];
int base = 2;
for (int k = 0; k < sequence.Length; k++)
{
sequence[k] = number % base;
number = number / base;
base++; // b[k+1] = b[k] + 1
}```
This correctly decodes our 37 back to {1, 2, 0, 1} (sequence would be {1, 0, 2, 1} in this code example, but whatever ... as long as you index appropriately). We just need to add 0 at the right end (remember the last element always has only one possibility for its new position) to get back our original sequence {1, 2, 0, 1, 0}.
Permuting a list using an index sequence You can use the below algorithm to permute a list according to a specific index sequence. It's an O(n) algorithm, unfortunately.
```int n = 5;
int[] sequence = new int[] { 1, 2, 0, 1, 0 };
char[] list = new char[] { 'a', 'b', 'c', 'd', 'e' };
char[] permuted = new char[n];
bool[] set = new bool[n];
for (int i = 0; i < n; i++)
{
int s = sequence[i];
int remainingPosition = 0;
int index;
// Find the s'th position in the permuted list that has not been set yet.
for (index = 0; index < n; index++)
{
if (!set[index])
{
if (remainingPosition == s)
break;
remainingPosition++;
}
}
permuted[index] = list[i];
set[index] = true;
}```
Common representation of permutations Normally you would not represent a permutation as unintuitively as we've done, but simply by the absolute position of each element after the permutation is applied. Our example {1, 2, 0, 1, 0} for abcde to caebd is normally represented by {1, 3, 0, 4, 2}. Each index from 0 to 4 (or in general, 0 to n-1) occurs exactly once in this representation.
Applying a permutation in this form is easy:
```int[] permutation = new int[] { 1, 3, 0, 4, 2 };
char[] list = new char[] { 'a', 'b', 'c', 'd', 'e' };
char[] permuted = new char[n];
for (int i = 0; i < n; i++)
{
permuted[permutation[i]] = list[i];
}```
```for (int i = 0; i < n; i++)
{
list[i] = permuted[permutation[i]];
}```
Converting from our representation to the common representation Note that if we take our algorithm to permute a list using our index sequence, and apply it to the identity permutation {0, 1, 2, ..., n-1}, we get the inverse permutation, represented in the common form. ({2, 0, 4, 1, 3} in our example).
To get the non-inverted premutation, we apply the permutation algorithm I just showed:
```int[] identity = new int[] { 0, 1, 2, 3, 4 };
int[] inverted = { 2, 0, 4, 1, 3 };
int[] normal = new int[n];
for (int i = 0; i < n; i++)
{
normal[identity[i]] = list[i];
}```
Or you can just apply the permutation directly, by using the inverse permutation algorithm:
```char[] list = new char[] { 'a', 'b', 'c', 'd', 'e' };
char[] permuted = new char[n];
int[] inverted = { 2, 0, 4, 1, 3 };
for (int i = 0; i < n; i++)
{
permuted[i] = list[inverted[i]];
}```
Note that all the algorithms for dealing with permutations in the common form are O(n), while applying a permutation in our form is O(n). If you need to apply a permutation several times, first convert it to the common representation.
In "Permuting a list using an index sequence", you mention a quadratic algorithm. This is certainly fine because n is probably going to be very small. This can "easily" be reduced to O(nlogn) though, through an order statistics tree (pine.cs.yale.edu/pinewiki/OrderStatisticsTree), i.e. a red-black tree which initially will contains the values 0, 1, 2, ..., n-1, and each node contains the number of descendants below it. With this, one can find/remove the kth element in O(logn) time.
This algorithm is awesome, but I just found several cases to be wrong. Take the string "123"; the 4th permutation should be 231, but according to this algorithm, it will be 312. say 1234, the 4th permutation should be 1342, but it will be mistaken to be "1423". Correct me if I observed wrong. Thanks.
@IsaacLi, if i am correct, f(4) = {2, 0, 0} = 231. And f'(312) = {1, 1, 0} = 3. For 1234, f(4) = {0, 2, 0, 0} = 1342. And f'(1423) = {0, 1 1, 0} = 3. This algorithm is really inspiring. I wonder it is the original work from the OP. i have studied and analysed it for a while. And i believe it is correct :)
How to convert from "our representation" to "common representation", {1, 2, 0, 1, 0} --> {1, 3, 0, 4, 2}? And vice versa? Is it possible? (by not converting between {1, 2, 0, 1, 0} <--> {C, A, E, B, D}, which needs O(n^2).) If "our style" and "common style" are not convertible, they are in fact two different separate things, isn't it? Thanks x
## math - Fast permutation -> number -> permutation mapping algorithms - ...
algorithm math permutation combinatorics
## Analysing a random algorithm by computing the distribution
My personal approach about correctness of probability-using algorithms: if you know how to prove it's correct, then it's probably correct; if you don't, it's certainly wrong.
Said differently, it's generally hopeless to try to analyse every algorithm you could come up with: you have to keep looking for an algorithm until you find one that you can prove correct.
I know of one way to "automatically" analyse a shuffle (or more generally a random-using algorithm) that is stronger than the simple "throw lots of tests and check for uniformity". You can mechanically compute the distribution associated to each input of your algorithm.
The general idea is that a random-using algorithm explores a part of a world of possibilities. Each time your algorithm asks for a random element in a set ({true, false} when flipping a coin), there are two possible outcomes for your algorithm, and one of them is chosen. You can change your algorithm so that, instead of returning one of the possible outcomes, it explores all solutions in parallel and returns all possible outcomes with the associated distributions.
In general, that would require rewriting your algorithm in depth. If your language supports delimited continuations, you don't have to; you can implement "exploration of all possible outcomes" inside the function asking for a random element (the idea is that the random generator, instead of returning a result, capture the continuation associated to your program and run it with all different results). For an example of this approach, see oleg's HANSEI.
An intermediary, and probably less arcane, solution is to represent this "world of possible outcomes" as a monad, and use a language such as Haskell with facilities for monadic programming. Here is an example implementation of a variant of your algorithm, in Haskell, using the probability monad of the probability package :
```import Numeric.Probability.Distribution
shuffleM :: (Num prob, Fractional prob) => [a] -> T prob [a]
shuffleM [] = return []
shuffleM [x] = return [x]
shuffleM (pivot:li) = do
(left, right) <- partition li
sleft <- shuffleM left
sright <- shuffleM right
return (sleft ++ [pivot] ++ sright)
where partition [] = return ([], [])
partition (x:xs) = do
(left, right) <- partition xs
uniform [(x:left, right), (left, x:right)]```
You can run it for a given input, and get the output distribution :
```*Main> shuffleM [1,2]
fromFreqs [([1,2],0.5),([2,1],0.5)]
*Main> shuffleM [1,2,3]
fromFreqs
[([2,1,3],0.25),([3,1,2],0.25),([1,2,3],0.125),
([1,3,2],0.125),([2,3,1],0.125),([3,2,1],0.125)]```
You can see that this algorithm is uniform with inputs of size 2, but non-uniform on inputs of size 3.
The difference with the test-based approach is that we can gain absolute certainty in a finite number of steps : it can be quite big, as it amounts to an exhaustive exploration of the world of possibles (but generally smaller than 2^N, as there are factorisations of similar outcomes), but if it returns a non-uniform distribution we know for sure that the algorithm is wrong. Of course, if it returns an uniform distribution for [1..N] and 1 <= N <= 100, you only know that your algorithm is uniform up to lists of size 100; it may still be wrong.
: this algorithm is a variant of your Erlang's implementation, because of the specific pivot handling. If I use no pivot, like in your case, the input size doesn't decrease at each step anymore : the algorithm also considers the case where all inputs are in the left list (or right list), and get lost in an infinite loop. This is a weakness of the probability monad implementation (if an algorithm has a probability 0 of non-termination, the distribution computation may still diverge), that I don't yet know how to fix.
Here is a simple algorithm that I feel confident I could prove correct:
• Pick a random key for each element in your collection.
• If the keys are not all distinct, restart from step 1.
You can omit step 2 if you know the probability of a collision (two random numbers picked are equal) is sufficiently low, but without it the shuffle is not perfectly uniform.
If you pick your keys in [1..N] where N is the length of your collection, you'll have lots of collisions (Birthday problem). If you pick your key as a 32-bit integer, the probability of conflict is low in practice, but still subject to the birthday problem.
If you use infinite (lazily evaluated) bitstrings as keys, rather than finite-length keys, the probability of a collision becomes 0, and checking for distinctness is no longer necessary.
Here is a shuffle implementation in OCaml, using lazy real numbers as infinite bitstrings:
```type 'a stream = Cons of 'a * 'a stream lazy_t
let rec real_number () =
Cons (Random.bool (), lazy (real_number ()))
let rec compare_real a b = match a, b with
| Cons (true, _), Cons (false, _) -> 1
| Cons (false, _), Cons (true, _) -> -1
| Cons (_, lazy a'), Cons (_, lazy b') ->
compare_real a' b'
let shuffle list =
List.map snd
(List.sort (fun (ra, _) (rb, _) -> compare_real ra rb)
(List.map (fun x -> real_number (), x) list))```
Algorithmic considerations: the complexity of the previous algorithm depends on the probability that all keys are distinct. If you pick them as 32-bit integers, you have a one in ~4 billion probability that a particular key collides with another key. Sorting by these keys is O(n log n), assuming picking a random number is O(1).
If you infinite bitstrings, you never have to restart picking, but the complexity is then related to "how many elements of the streams are evaluated on average". I conjecture it is O(log n) in average (hence still O(n log n) in total), but have no proof.
After more reflexion, I think (like douplep), that your implementation is correct. Here is an informal explanation.
Each element in your list is tested by several random:uniform() < 0.5 tests. To an element, you can associate the list of outcomes of those tests, as a list of booleans or {0, 1}. At the beginning of the algorithm, you don't know the list associated to any of those number. After the first partition call, you know the first element of each list, etc. When your algorithm returns, the list of tests are completely known and the elements are sorted according to those lists (sorted in lexicographic order, or considered as binary representations of real numbers).
So, your algorithm is equivalent to sorting by infinite bitstring keys. The action of partitioning the list, reminiscent of quicksort's partition over a pivot element, is actually a way of separating, for a given position in the bitstring, the elements with valuation 0 from the elements with valuation 1.
The sort is uniform because the bitstrings are all different. Indeed, two elements with real numbers equal up to the n-th bit are on the same side of a partition occurring during a recursive shuffle call of depth n. The algorithm only terminates when all the lists resulting from partitions are empty or singletons : all elements have been separated by at least one test, and therefore have one distinct binary decimal.
A subtle point about your algorithm (or my equivalent sort-based method) is that the termination condition is probabilistic. Fisher-Yates always terminates after a known number of steps (the number of elements in the array). With your algorithm, the termination depends on the output of the random number generator.
There are possible outputs that would make your algorithm diverge, not terminate. For example, if the random number generator always output 0, each partition call will return the input list unchanged, on which you recursively call the shuffle : you will loop indefinitely.
However, this is not an issue if you're confident that your random number generator is fair : it does not cheat and always return independent uniformly distributed results. In that case, the probability that the test random:uniform() < 0.5 always returns true (or false) is exactly 0 :
• the probability that the first N calls return true is 2^{-N}
• the probability that all calls return true is the probability of the infinite intersection, for all N, of the event that the first N calls return 0; it is the infimum limit of the 2^{-N}, which is 0
More generally, the algorithm does not terminate if and only if some of the elements get associated to the same boolean stream. This means that at least two elements have the same boolean stream. But the probability that two random boolean streams are equal is again 0 : the probability that the digits at position K are equal is 1/2, so the probability that the N first digits are equal is 2^{-N}, and the same analysis applies.
Therefore, you know that your algorithm terminates with probability 1. This is a slightly weaker guarantee that the Fisher-Yates algorithm, which always terminate. In particular, you're vulnerable to an attack of an evil adversary that would control your random number generator.
With more probability theory, you could also compute the distribution of running times of your algorithm for a given input length. This is beyond my technical abilities, but I assume it's good : I suppose that you only need to look at O(log N) first digits on average to check that all N lazy streams are different, and that the probability of much higher running times decrease exponentially.
My real question here, though, is what empirical tests can I throw at the output of my shuffler to see if it is plausibly shuffled? For example, that "pair a random weight with each element" approach tested badly even with my limited ability to test this stuff. (I tested the sequence [1,2] repeatedly and found a huge imbalance.)
[min_int..max_int] is not enough to bring the probability of conflict close to 0, because of the birthday problem you mentioned: with 32-bit ints, you already reach a 0.5 chance of conflict with a list of only ~77,000 items.
Also, note that in general, making any sort-based shuffle perfectly uniform/correct is probably much harder than it seems at first: for some of the problems, see Oleg's writeup, and my answer's comments. If a perfect shuffle is important at all, it is certainly much easier and simpler to just use the FisherYates algorithm.
I edited to mention your warning about [min_int..max_int] : you're right and it doesn't scale to big sequences. I also included an implementation of the real-number based sort. I agree that Fisher-Yates is simpler, but I'm not sure Oleg's proposal is.
@AJMansfield: Actually, with 64-bit keys, you only need ~5 billion selections to expect a collision with 50% probability. After 10 billion selections, the probability of a collision increases to ~93%. This counter-intuitive result is the Birthday Problem.
## functional programming - What, if anything, is wrong with this shuffli...
algorithm functional-programming shuffle
We begin by assembling the algorithmic building blocks from the Standard Library:
```#include <algorithm> // min_element, iter_swap,
// upper_bound, rotate,
// partition,
// inplace_merge,
// make_heap, sort_heap, push_heap, pop_heap,
// is_heap, is_sorted
#include <cassert> // assert
#include <functional> // less
#include <iterator> // distance, begin, end, next```
• the iterator tools such as non-member std::begin() / std::end() as well as with std::next() are only available as of C++11 and beyond. For C++98, one needs to write these himself. There are substitutes from Boost.Range in boost::begin() / boost::end(), and from Boost.Utility in boost::next().
• the std::is_sorted algorithm is only available for C++11 and beyond. For C++98, this can be implemented in terms of std::adjacent_find and a hand-written function object. Boost.Algorithm also provides a boost::algorithm::is_sorted as a substitute.
• the std::is_heap algorithm is only available for C++11 and beyond.
C++14 provides transparent comparators of the form std::less<> that act polymorphically on their arguments. This avoids having to provide an iterator's type. This can be used in combination with C++11's default function template arguments to create a single overload for sorting algorithms that take < as comparison and those that have a user-defined comparison function object.
In C++11, one can define a reusable template alias to extract an iterator's value type which adds minor clutter to the sort algorithms' signatures:
```template<class It>
using value_type_t = typename std::iterator_traits<It>::value_type;
template<class It, class Compare = std::less<value_type_t<It>>>
void xxx_sort(It first, It last, Compare cmp = Compare{});```
In C++98, one needs to write two overloads and use the verbose typename xxx<yyy>::type syntax
```template<class It, class Compare>
void xxx_sort(It first, It last, Compare cmp); // general implementation
template<class It>
void xxx_sort(It first, It last)
{
xxx_sort(first, last, std::less<typename std::iterator_traits<It>::value_type>());
}```
• Another syntactical nicety is that C++14 facilitates wrapping user-defined comparators through polymorphic lambdas (with auto parameters that are deduced like function template arguments).
• C++11 only has monomorphic lambdas, that require the use of the above template alias value_type_t.
• In C++98, one either needs to write a standalone function object or resort to the verbose std::bind1st / std::bind2nd / std::not1 type of syntax.
• Boost.Bind improves this with boost::bind and _1 / _2 placeholder syntax.
• C++11 and beyond also have std::find_if_not, whereas C++98 needs std::find_if with a std::not1 around a function object.
There is no generally acceptable C++14 style yet. For better of for worse, I closely follow Scott Meyers's draft Effective Modern C++ and Herb Sutter's revamped GotW. I use the following style recommendations:
• Herb Sutter's "Almost Always Auto" and Scott Meyers's "Prefer auto to specific type declarations" recommendation, for which the brevity is unsurpassed, although its clarity is sometimes disputed.
• Scott Meyers's "Distinguish () and {} when creating objects" and consistently choose braced-initialization {} instead of the good old parenthesized initialization () (in order to side-step all most-vexing-parse issues in generic code).
• Scott Meyers's "Prefer alias declarations to typedefs". For templates this is a must anyway, and using it everywhere instead of typedef saves time and adds consistency.
• I use a for (auto it = first; it != last; ++it) pattern in some places, in order to allow for loop invariant checking for already sorted sub-ranges. In production code, the use of while (first != last) and a ++first somewhere inside the loop might be slightly better.
Selection sort does not adapt to the data in any way, so its runtime is always O(N^2). However, selection sort has the property of minimizing the number of swaps. In applications where the cost of swapping items is high, selection sort very well may be the algorithm of choice.
To implement it using the Standard Library, repeatedly use std::min_element to find the remaining minimum element, and iter_swap to swap it into place:
Note that selection_sort has the already processed range [first, it) sorted as its loop invariant. The minimal requirements are forward iterators, compared to std::sort's random access iterators.
`if (std::distance(first, last) <= 1) return;`
`if (first == last || std::next(first) == last) return;`
Although it is one of the elementary sorting algorithms with O(N^2) worst-case time, insertion sort is the algorithm of choice either when the data is nearly sorted (because it is adaptive) or when the problem size is small (because it has low overhead). For these reasons, and because it is also stable, insertion sort is often used as the recursive base case (when the problem size is small) for higher overhead divide-and-conquer sorting algorithms, such as merge sort or quick sort.
To implement insertion_sort with the Standard Library, repeatedly use std::upper_bound to find the location where the current element needs to go, and use std::rotate to shift the remaining elements upward in the input range:
Note that insertion_sort has the already processed range [first, it) sorted as its loop invariant. Insertion sort also works with forward iterators.
• insertion sort can be optimized with an early test if (std::distance(first, last) <= 1) return; (or for forward / bidirectional iterators: if (first == last || std::next(first) == last) return;) and a loop over the interval [std::next(first), last), because the first element is guaranteed to be in place and doesn't require a rotate.
• for bidirectional iterators, the binary search to find the insertion point can be replaced with a reverse linear search using the Standard Library's std::find_if_not algorithm.
Four Live Examples (C++14, C++11, C++98 and Boost, C++98) for the fragment below:
```using RevIt = std::reverse_iterator<BiDirIt>;
auto const insertion = std::find_if_not(RevIt(it), RevIt(first),
[=](auto const& elem){ return cmp(*it, elem); }
).base();```
• For random inputs this gives O(N^2) comparisons, but this improves to O(N) comparisons for almost sorted inputs. The binary search always uses O(N log N) comparisons.
• For small input ranges, the better memory locality (cache, prefetching) of a linear search might also dominate a binary search (one should test this, of course).
When carefully implemented, quick sort is robust and has O(N log N) expected complexity, but with O(N^2) worst-case complexity that can be triggered with adversarially chosen input data. When a stable sort is not needed, quick sort is an excellent general-purpose sort.
Even for the simplest versions, quick sort is quite a bit more complicated to implement using the Standard Library than the other classic sorting algorithms. The approach below uses a few iterator utilities to locate the middle element of the input range [first, last) as the pivot, then use two calls to std::partition (which are O(N)) to three-way partition the input range into segments of elements that are smaller than, equal to, and larger than the selected pivot, respectively. Finally the two outer segments with elements smaller than and larger than the pivot are recursively sorted:
```template<class FwdIt, class Compare = std::less<>>
void quick_sort(FwdIt first, FwdIt last, Compare cmp = Compare{})
{
auto const N = std::distance(first, last);
if (N <= 1) return;
auto const pivot = *std::next(first, N / 2);
auto const middle1 = std::partition(first, last, [=](auto const& elem){
return cmp(elem, pivot);
});
auto const middle2 = std::partition(middle1, last, [=](auto const& elem){
return !cmp(pivot, elem);
});
quick_sort(first, middle1, cmp); // assert(std::is_sorted(first, middle1, cmp));
quick_sort(middle2, last, cmp); // assert(std::is_sorted(middle2, last, cmp));
}```
However, quick sort is rather tricky to get correct and efficient, as each of the above steps has to be carefully checked and optimized for production level code. In particular, for O(N log N) complexity, the pivot has to result into a balanced partition of the input data, which cannot be guaranteed in general for an O(1) pivot, but which can be guaranteed if one sets the pivot as the O(N) median of the input range.
• the above implementation is particularly vulnerable to special inputs, e.g. it has O(N^2) complexity for the "organ pipe" input 1, 2, 3, ..., N/2, ... 3, 2, 1 (because the middle is always larger than all other elements).
• median-of-3 pivot selection from randomly chosen elements from the input range guards against almost sorted inputs for which the complexity would otherwise deteriorate to O(N^2).
• 3-way partitioning (separating elements smaller than, equal to and larger than the pivot) as shown by the two calls to std::partition is not the most efficient O(N) algorithm to achieve this result.
`O(N log N)`
`std::nth_element(first, middle, last)`
`quick_sort(first, middle, cmp)`
`quick_sort(middle, last, cmp)`
• this guarantee comes at a cost, however, because the constant factor of the O(N) complexity of std::nth_element can be more expensive than that of the O(1) complexity of a median-of-3 pivot followed by an O(N) call to std::partition (which is a cache-friendly single forward pass over the data).
If using O(N) extra space is of no concern, then merge sort is an excellent choice: it is the only stable O(N log N) sorting algorithm.
It is simple to implement using Standard algorithms: use a few iterator utilities to locate the middle of the input range [first, last) and combine two recursively sorted segments with a std::inplace_merge:
```template<class BiDirIt, class Compare = std::less<>>
void merge_sort(BiDirIt first, BiDirIt last, Compare cmp = Compare{})
{
auto const N = std::distance(first, last);
if (N <= 1) return;
auto const middle = std::next(first, N / 2);
merge_sort(first, middle, cmp); // assert(std::is_sorted(first, middle, cmp));
merge_sort(middle, last, cmp); // assert(std::is_sorted(middle, last, cmp));
std::inplace_merge(first, middle, last, cmp); // assert(std::is_sorted(first, last, cmp));
}```
Merge sort requires bidirectional iterators, the bottleneck being the std::inplace_merge. Note that when sorting linked lists, merge sort requires only O(log N) extra space (for recursion). The latter algorithm is implemented by std::list<T>::sort in the Standard Library.
Heap sort is simple to implement, performs an O(N log N) in-place sort, but is not stable.
The first loop, O(N) "heapify" phase, puts the array into heap order. The second loop, the O(N log N) "sortdown" phase, repeatedly extracts the maximum and restores heap order. The Standard Library makes this extremely straightforward:
```template<class RandomIt, class Compare = std::less<>>
void heap_sort(RandomIt first, RandomIt last, Compare cmp = Compare{})
{
lib::make_heap(first, last, cmp); // assert(std::is_heap(first, last, cmp));
lib::sort_heap(first, last, cmp); // assert(std::is_sorted(first, last, cmp));
}```
In case you consider it "cheating" to use std::make_heap and std::sort_heap, you can go one level deeper and write those functions yourself in terms of std::push_heap and std::pop_heap, respectively:
```namespace lib {
// NOTE: is O(N log N), not O(N) as std::make_heap
template<class RandomIt, class Compare = std::less<>>
void make_heap(RandomIt first, RandomIt last, Compare cmp = Compare{})
{
for (auto it = first; it != last;) {
std::push_heap(first, ++it, cmp);
assert(std::is_heap(first, it, cmp));
}
}
template<class RandomIt, class Compare = std::less<>>
void sort_heap(RandomIt first, RandomIt last, Compare cmp = Compare{})
{
for (auto it = last; it != first;) {
std::pop_heap(first, it--, cmp);
assert(std::is_heap(first, it, cmp));
}
}
} // namespace lib```
The Standard Library specifies both push_heap and pop_heap as complexity O(log N). Note however that the outer loop over the range [first, last) results in O(N log N) complexity for make_heap, whereas std::make_heap has only O(N) complexity. For the overall O(N log N) complexity of heap_sort it doesn't matter.
Details omitted: O(N) implementation of make_heap
Here are four Live Examples (C++14, C++11, C++98 and Boost, C++98) testing all five algorithms on a variety of inputs (not meant to be exhaustive or rigorous). Just note the huge differences in the LOC: C++11/C++14 need around 130 LOC, C++98 and Boost 190 (+50%) and C++98 more than 270 (+100%).
While I disagree with your use of auto (and many people disagree with me), I enjoyed seeing the standard library algorithms being used well. I'd been wanting to see some examples of this kind of code after seeing Sean Parent's talk. Also, I had no idea std::iter_swap existed, although it seems strange to me that it's in <algorithm>.
@sbabbi The entire standard library is based on the principle that iterators are cheap to copy; it passes them by value, for example. If copying an iterator isn't cheap, then you're going to suffer performance problems everywhere.
@gnzlbg The asserts you can comment out, of course. The early test can be tag-dispatched per iterator category, with the current version for random access, and if (first == last || std::next(first) == last). I might update that later. Implementing the stuff in the "omitted details" sections is beyond the scope of the question, IMO, because they contain links to entire Q&As themselves. Implementing real-word sorting routines is hard!
Great post. Though, you've cheated with your quicksort by using nth_element in my opinion. nth_element does half a quicksort already (including the partitioning step and a recursion on the half that includes the n-th element you're interested in).
@DavidStone You are right of course. This Q&A is not meant to be the definite guide to write real life optimized sort routines, but rather to show how to combine basic building blocks. See e.g. the cpp-sort library on how much extra details require careful attention in real life :)
## How to implement classic sorting algorithms in modern C++? - Stack Ove...
c++ algorithm sorting c++14 c++-faq
We begin by assembling the algorithmic building blocks from the Standard Library:
```#include <algorithm> // min_element, iter_swap,
// upper_bound, rotate,
// partition,
// inplace_merge,
// make_heap, sort_heap, push_heap, pop_heap,
// is_heap, is_sorted
#include <cassert> // assert
#include <functional> // less
#include <iterator> // distance, begin, end, next```
• the iterator tools such as non-member std::begin() / std::end() as well as with std::next() are only available as of C++11 and beyond. For C++98, one needs to write these himself. There are substitutes from Boost.Range in boost::begin() / boost::end(), and from Boost.Utility in boost::next().
• the std::is_sorted algorithm is only available for C++11 and beyond. For C++98, this can be implemented in terms of std::adjacent_find and a hand-written function object. Boost.Algorithm also provides a boost::algorithm::is_sorted as a substitute.
• the std::is_heap algorithm is only available for C++11 and beyond.
C++14 provides transparent comparators of the form std::less<> that act polymorphically on their arguments. This avoids having to provide an iterator's type. This can be used in combination with C++11's default function template arguments to create a single overload for sorting algorithms that take < as comparison and those that have a user-defined comparison function object.
In C++11, one can define a reusable template alias to extract an iterator's value type which adds minor clutter to the sort algorithms' signatures:
```template<class It>
using value_type_t = typename std::iterator_traits<It>::value_type;
template<class It, class Compare = std::less<value_type_t<It>>>
void xxx_sort(It first, It last, Compare cmp = Compare{});```
In C++98, one needs to write two overloads and use the verbose typename xxx<yyy>::type syntax
```template<class It, class Compare>
void xxx_sort(It first, It last, Compare cmp); // general implementation
template<class It>
void xxx_sort(It first, It last)
{
xxx_sort(first, last, std::less<typename std::iterator_traits<It>::value_type>());
}```
• Another syntactical nicety is that C++14 facilitates wrapping user-defined comparators through polymorphic lambdas (with auto parameters that are deduced like function template arguments).
• C++11 only has monomorphic lambdas, that require the use of the above template alias value_type_t.
• In C++98, one either needs to write a standalone function object or resort to the verbose std::bind1st / std::bind2nd / std::not1 type of syntax.
• Boost.Bind improves this with boost::bind and _1 / _2 placeholder syntax.
• C++11 and beyond also have std::find_if_not, whereas C++98 needs std::find_if with a std::not1 around a function object.
There is no generally acceptable C++14 style yet. For better of for worse, I closely follow Scott Meyers's draft Effective Modern C++ and Herb Sutter's revamped GotW. I use the following style recommendations:
• Herb Sutter's "Almost Always Auto" and Scott Meyers's "Prefer auto to specific type declarations" recommendation, for which the brevity is unsurpassed, although its clarity is sometimes disputed.
• Scott Meyers's "Distinguish () and {} when creating objects" and consistently choose braced-initialization {} instead of the good old parenthesized initialization () (in order to side-step all most-vexing-parse issues in generic code).
• Scott Meyers's "Prefer alias declarations to typedefs". For templates this is a must anyway, and using it everywhere instead of typedef saves time and adds consistency.
• I use a for (auto it = first; it != last; ++it) pattern in some places, in order to allow for loop invariant checking for already sorted sub-ranges. In production code, the use of while (first != last) and a ++first somewhere inside the loop might be slightly better.
Selection sort does not adapt to the data in any way, so its runtime is always O(N^2). However, selection sort has the property of minimizing the number of swaps. In applications where the cost of swapping items is high, selection sort very well may be the algorithm of choice.
To implement it using the Standard Library, repeatedly use std::min_element to find the remaining minimum element, and iter_swap to swap it into place:
Note that selection_sort has the already processed range [first, it) sorted as its loop invariant. The minimal requirements are forward iterators, compared to std::sort's random access iterators.
`if (std::distance(first, last) <= 1) return;`
`if (first == last || std::next(first) == last) return;`
Although it is one of the elementary sorting algorithms with O(N^2) worst-case time, insertion sort is the algorithm of choice either when the data is nearly sorted (because it is adaptive) or when the problem size is small (because it has low overhead). For these reasons, and because it is also stable, insertion sort is often used as the recursive base case (when the problem size is small) for higher overhead divide-and-conquer sorting algorithms, such as merge sort or quick sort.
To implement insertion_sort with the Standard Library, repeatedly use std::upper_bound to find the location where the current element needs to go, and use std::rotate to shift the remaining elements upward in the input range:
Note that insertion_sort has the already processed range [first, it) sorted as its loop invariant. Insertion sort also works with forward iterators.
• insertion sort can be optimized with an early test if (std::distance(first, last) <= 1) return; (or for forward / bidirectional iterators: if (first == last || std::next(first) == last) return;) and a loop over the interval [std::next(first), last), because the first element is guaranteed to be in place and doesn't require a rotate.
• for bidirectional iterators, the binary search to find the insertion point can be replaced with a reverse linear search using the Standard Library's std::find_if_not algorithm.
Four Live Examples (C++14, C++11, C++98 and Boost, C++98) for the fragment below:
```using RevIt = std::reverse_iterator<BiDirIt>;
auto const insertion = std::find_if_not(RevIt(it), RevIt(first),
[=](auto const& elem){ return cmp(*it, elem); }
).base();```
• For random inputs this gives O(N^2) comparisons, but this improves to O(N) comparisons for almost sorted inputs. The binary search always uses O(N log N) comparisons.
• For small input ranges, the better memory locality (cache, prefetching) of a linear search might also dominate a binary search (one should test this, of course).
When carefully implemented, quick sort is robust and has O(N log N) expected complexity, but with O(N^2) worst-case complexity that can be triggered with adversarially chosen input data. When a stable sort is not needed, quick sort is an excellent general-purpose sort.
Even for the simplest versions, quick sort is quite a bit more complicated to implement using the Standard Library than the other classic sorting algorithms. The approach below uses a few iterator utilities to locate the middle element of the input range [first, last) as the pivot, then use two calls to std::partition (which are O(N)) to three-way partition the input range into segments of elements that are smaller than, equal to, and larger than the selected pivot, respectively. Finally the two outer segments with elements smaller than and larger than the pivot are recursively sorted:
```template<class FwdIt, class Compare = std::less<>>
void quick_sort(FwdIt first, FwdIt last, Compare cmp = Compare{})
{
auto const N = std::distance(first, last);
if (N <= 1) return;
auto const pivot = *std::next(first, N / 2);
auto const middle1 = std::partition(first, last, [=](auto const& elem){
return cmp(elem, pivot);
});
auto const middle2 = std::partition(middle1, last, [=](auto const& elem){
return !cmp(pivot, elem);
});
quick_sort(first, middle1, cmp); // assert(std::is_sorted(first, middle1, cmp));
quick_sort(middle2, last, cmp); // assert(std::is_sorted(middle2, last, cmp));
}```
However, quick sort is rather tricky to get correct and efficient, as each of the above steps has to be carefully checked and optimized for production level code. In particular, for O(N log N) complexity, the pivot has to result into a balanced partition of the input data, which cannot be guaranteed in general for an O(1) pivot, but which can be guaranteed if one sets the pivot as the O(N) median of the input range.
• the above implementation is particularly vulnerable to special inputs, e.g. it has O(N^2) complexity for the "organ pipe" input 1, 2, 3, ..., N/2, ... 3, 2, 1 (because the middle is always larger than all other elements).
• median-of-3 pivot selection from randomly chosen elements from the input range guards against almost sorted inputs for which the complexity would otherwise deteriorate to O(N^2).
• 3-way partitioning (separating elements smaller than, equal to and larger than the pivot) as shown by the two calls to std::partition is not the most efficient O(N) algorithm to achieve this result.
`O(N log N)`
`std::nth_element(first, middle, last)`
`quick_sort(first, middle, cmp)`
`quick_sort(middle, last, cmp)`
• this guarantee comes at a cost, however, because the constant factor of the O(N) complexity of std::nth_element can be more expensive than that of the O(1) complexity of a median-of-3 pivot followed by an O(N) call to std::partition (which is a cache-friendly single forward pass over the data).
If using O(N) extra space is of no concern, then merge sort is an excellent choice: it is the only stable O(N log N) sorting algorithm.
It is simple to implement using Standard algorithms: use a few iterator utilities to locate the middle of the input range [first, last) and combine two recursively sorted segments with a std::inplace_merge:
```template<class BiDirIt, class Compare = std::less<>>
void merge_sort(BiDirIt first, BiDirIt last, Compare cmp = Compare{})
{
auto const N = std::distance(first, last);
if (N <= 1) return;
auto const middle = std::next(first, N / 2);
merge_sort(first, middle, cmp); // assert(std::is_sorted(first, middle, cmp));
merge_sort(middle, last, cmp); // assert(std::is_sorted(middle, last, cmp));
std::inplace_merge(first, middle, last, cmp); // assert(std::is_sorted(first, last, cmp));
}```
Merge sort requires bidirectional iterators, the bottleneck being the std::inplace_merge. Note that when sorting linked lists, merge sort requires only O(log N) extra space (for recursion). The latter algorithm is implemented by std::list<T>::sort in the Standard Library.
Heap sort is simple to implement, performs an O(N log N) in-place sort, but is not stable.
The first loop, O(N) "heapify" phase, puts the array into heap order. The second loop, the O(N log N) "sortdown" phase, repeatedly extracts the maximum and restores heap order. The Standard Library makes this extremely straightforward:
```template<class RandomIt, class Compare = std::less<>>
void heap_sort(RandomIt first, RandomIt last, Compare cmp = Compare{})
{
lib::make_heap(first, last, cmp); // assert(std::is_heap(first, last, cmp));
lib::sort_heap(first, last, cmp); // assert(std::is_sorted(first, last, cmp));
}```
In case you consider it "cheating" to use std::make_heap and std::sort_heap, you can go one level deeper and write those functions yourself in terms of std::push_heap and std::pop_heap, respectively:
```namespace lib {
// NOTE: is O(N log N), not O(N) as std::make_heap
template<class RandomIt, class Compare = std::less<>>
void make_heap(RandomIt first, RandomIt last, Compare cmp = Compare{})
{
for (auto it = first; it != last;) {
std::push_heap(first, ++it, cmp);
assert(std::is_heap(first, it, cmp));
}
}
template<class RandomIt, class Compare = std::less<>>
void sort_heap(RandomIt first, RandomIt last, Compare cmp = Compare{})
{
for (auto it = last; it != first;) {
std::pop_heap(first, it--, cmp);
assert(std::is_heap(first, it, cmp));
}
}
} // namespace lib```
The Standard Library specifies both push_heap and pop_heap as complexity O(log N). Note however that the outer loop over the range [first, last) results in O(N log N) complexity for make_heap, whereas std::make_heap has only O(N) complexity. For the overall O(N log N) complexity of heap_sort it doesn't matter.
Details omitted: O(N) implementation of make_heap
Here are four Live Examples (C++14, C++11, C++98 and Boost, C++98) testing all five algorithms on a variety of inputs (not meant to be exhaustive or rigorous). Just note the huge differences in the LOC: C++11/C++14 need around 130 LOC, C++98 and Boost 190 (+50%) and C++98 more than 270 (+100%).
While I disagree with your use of auto (and many people disagree with me), I enjoyed seeing the standard library algorithms being used well. I'd been wanting to see some examples of this kind of code after seeing Sean Parent's talk. Also, I had no idea std::iter_swap existed, although it seems strange to me that it's in <algorithm>.
@sbabbi The entire standard library is based on the principle that iterators are cheap to copy; it passes them by value, for example. If copying an iterator isn't cheap, then you're going to suffer performance problems everywhere.
Great post. Regarding the cheating part of [std::]make_heap. If std::make_heap is considered cheating, so would std::push_heap. I.e. cheating = not implementing the actual behaviour defined for a heap structure. I would find it instructive have push_heap included as well.
@gnzlbg The asserts you can comment out, of course. The early test can be tag-dispatched per iterator category, with the current version for random access, and if (first == last || std::next(first) == last). I might update that later. Implementing the stuff in the "omitted details" sections is beyond the scope of the question, IMO, because they contain links to entire Q&As themselves. Implementing real-word sorting routines is hard!
Great post. Though, you've cheated with your quicksort by using nth_element in my opinion. nth_element does half a quicksort already (including the partitioning step and a recursion on the half that includes the n-th element you're interested in).
## How to implement classic sorting algorithms in modern C++? - Stack Ove...
c++ algorithm sorting c++14 c++-faq
Skewness and Kurtosis
For the on-line algorithms for Skewness and Kurtosis (along the lines of the variance), see in the same wiki page here the parallel algorithms for higher-moment statistics.
Median is tough without sorted data. If you know, how many data points you have, in theory you only have to partially sort, e.g. by using a selection algorithm. However, that doesn't help too much with billions of values. I would suggest using frequency counts, see the next section.
Median and Mode with Frequency Counts
If it is integers, I would count frequencies, probably cutting off the highest and lowest values beyond some value where I am sure that it is no longer relevant. For floats (or too many integers), I would probably create buckets / intervals, and then use the same approach as for integers. (Approximate) mode and median calculation than gets easy, based on the frequencies table.
If it is normally distributed, I would use the population sample mean, variance, skewness, and kurtosis as maximum likelihood estimators for a small subset. The (on-line) algorithms to calculate those, you already now. E.g. read in a couple of hundred thousand or million datapoints, until your estimation error gets small enough. Just make sure that you pick randomly from your set (e.g. that you don't introduce a bias by picking the first 100'000 values). The same approach can also be used for estimating mode and median for the normal case (for both the sample mean is an estimator).
All the algorithms above can be run in parallel (including many sorting and selection algorithm, e.g. QuickSort and QuickSelect), if this helps.
I have always assumed (with the exception of the section on the normal distribution) that we talk about sample moments, median, and mode, not estimators for theoretical moments given a known distribution.
In general, sampling the data (i.e. only looking at a sub-set) should be pretty successful given the amount of data, as long as all observations are realizations of the same random variable (have the same distributions) and the moments, mode and median actually exist for this distribution. The last caveat is not innocuous. For example, the mean (and all higher moments) for the Cauchy Distribution do not exist. In this case, the sample mean of a "small" sub-set might be massively off from the sample mean of the whole sample.
## statistics - "On-line" (iterator) algorithms for estimating statistica...
algorithm statistics iterator median
`algorithms`
`algorithm`
Thanks. This package seems to be very strong in more theoretical algirthm discussion, i know it from many math books. But i wont emphasize this so much (prerequsites, if, else), i would like to have a formatting like the one above.
I was only talking about the algorithm environment, not algorithmic. algorithm is a floating container, which looks pretty nice. You can put whatever you'd like inside, even the listing mentioned elsethread.
## LaTeX source code listing like in professional books - Stack Overflow
latex
What you're looking for are called String Metric algorithms. There a significant number of them, many with similar characteristics. Among the more popular:
• Levenshtein Distance : The minimum number of single-character edits required to change one word into the other. Strings do not have to be the same length
• Hamming Distance : The number of characters that are different in two equal length strings.
• SmithWaterman : A family of algorithms for computing variable sub-sequence similarities.
• SrensenDice Coefficient : A similarity algorithm that computes difference coefficients of adjacent character pairs.
Have a look at these as well as others on the wiki page on the topic.
## language agnostic - What are some algorithms for comparing how similar...
algorithm language-agnostic string-comparison stdstring heuristics
Statistically informed algorithms solve this problem using fewer passes than deterministic approaches.
If very large integers are allowed then one can generate a number that is likely to be unique in O(1) time. A pseudo-random 128-bit integer like a GUID will only collide with one of the existing four billion integers in the set in less than one out of every 64 billion billion billion cases.
If integers are limited to 32 bits then one can generate a number that is likely to be unique in a single pass using much less than 10MB. The odds that a pseudo-random 32-bit integer will collide with one of the 4 billion existing integers is about 93% (4e9 / 2^32). The odds that 1000 pseudo-random integers will all collide is less than one in 12,000 billion billion billion (odds-of-one-collision ^ 1000). So if a program maintains a data structure containing 1000 pseudo-random candidates and iterates through the known integers, eliminating matches from the candidates, it is all but certain to find at least one integer that is not in the file.
I'm pretty sure the integers are bounded. If they weren't, then even a beginner programmer would think of the algorithm "take one pass through the data to find the maximum number, and add 1 to it"
Literally guessing a random output probably won't get you many points on an interview
@Adrian, your solution seems obvious (and it was to me, I used it in my own answer) but it's not obvious to everybody. It's a good test to see if you can spot obvious solutions or if you're going to over-complicate everything you touch.
## algorithm - Find an integer not among four billion given ones - Stack ...
algorithm
Statistically informed algorithms solve this problem using fewer passes than deterministic approaches.
If very large integers are allowed then one can generate a number that is likely to be unique in O(1) time. A pseudo-random 128-bit integer like a GUID will only collide with one of the existing four billion integers in the set in less than one out of every 64 billion billion billion cases.
If integers are limited to 32 bits then one can generate a number that is likely to be unique in a single pass using much less than 10MB. The odds that a pseudo-random 32-bit integer will collide with one of the 4 billion existing integers is about 93% (4e9 / 2^32). The odds that 1000 pseudo-random integers will all collide is less than one in 12,000 billion billion billion (odds-of-one-collision ^ 1000). So if a program maintains a data structure containing 1000 pseudo-random candidates and iterates through the known integers, eliminating matches from the candidates, it is all but certain to find at least one integer that is not in the file.
I'm pretty sure the integers are bounded. If they weren't, then even a beginner programmer would think of the algorithm "take one pass through the data to find the maximum number, and add 1 to it"
Literally guessing a random output probably won't get you many points on an interview
@Adrian, your solution seems obvious (and it was to me, I used it in my own answer) but it's not obvious to everybody. It's a good test to see if you can spot obvious solutions or if you're going to over-complicate everything you touch.
## algorithm - Find an integer not among four billion given ones - Stack ...
algorithm
This has the correct and optimal solution. I wanted to expand it into a separate answer to help people stuck in the paywall but... the algorithm is pretty messy.
It's asymptotically optimal but not especially practical I expect. It's certainly messy enough not to fit in the confines of an interview.
@missingno: Not so optimal. I have a more efficient average running time solution to this problem at stackoverflow.com/questions/5940420/ and it isn't that messy of an algorithm.
-1 This answer is totally useless, the article is no longer available. The answer-er has not been online since answering this question.
## Selection algorithms on sorted matrix - Stack Overflow
algorithm
The author of Capo, a transcription program for the Mac, has a pretty in-depth blog. The entry "A Note on Auto Tabbing" has some good jumping off points:
I started researching different methods of automatic transcription in mid-2009, because I was curious about how far along this technology was, and if it could be integrated into a future version of Capo.
Each of these automatic transcription algorithms start out with some kind of intermediate represenation of the audio data, and then they transfer that into a symbolic form (i.e. note onsets, and durations).
This is where I encountered some computationally expensive spectral representations (The Continuous Wavelet Transform (CWT), Constant Q Transform (CQT), and others.) I implemented all of these spectral transforms so that I could also implement the algorithms presented by the papers I was reading. This would give me an idea of whether they would work in practice.
Capo has some impressive technology. The standout feature is that its main view is not a frequency spectrogram like most other audio programs. It presents the audio like a piano roll, with the notes visible to the naked eye.
(Note: The hard note bars were drawn by a user. The fuzzy spots underneath are what Capo displays.)
## audio - Chord detection algorithms? - Stack Overflow
algorithm audio music signal-processing
What you have coded in your example is very similar to a depth first search. So, that's one answer.
A depth first search algorithm without any special characteristics ( like re-convergent paths that can be optimized out ), should be n^n.
This is actually not a contrived example. Chess programs operate on the same algorithm. Each move there are n moves to consider ( i.e. branches ), and you search d moves deep. So that becomes O(n^d)
Ah, but OP asked about problems which cannot be solved in better than O(n^n). An O(n^n) algorithm for a problem doesn't demonstrate that the problem can't be solved by a more efficient algorithm.
@Ted: No more efficient algorithm has been found for solving chess so far. There are optimizations - like alpha-beta pruning - but that doesn't change the fundamental characteristic of chess solving algorithm being O(n^d).
I totally agree that there are O(n^n) problems. I was just making the point that proving that a problem is O(n^n) involves more than showing that there's an O(n^n) algorithm that solves it. Chess is a good example of this because it is a finite game (there are a finite number of board positions). Theoretically, then it has O(1) complexity. But all known (practical) algorithms are very inefficient.
@Ted: Ok fair point. Though the OP did ask the question in 2 ways ( your interpretation being the second ). The first way he/she was asking whether there any algorithm that was O(n^n) that is not a gimmick.
Thank you, I guess your example with Chess is the most realistic one, even though there are better Algorithms to play Chess. And DFS based Algorithms are afaik not uncommon.
## complexity theory - Are there any real O(n^n) algorithms? - Stack Over...
algorithm complexity-theory big-o
You can do it in O(n) (where n is the number of digits) like this:
Starting from the right, you find the first pair-of-digits such that the left-digit is smaller than the right-digit. Let's refer to the left-digit by "digit-x". Find the smallest number larger than digit-x to the right of digit-x, and place it immediately left of digit-x. Finally, sort the remaining digits in ascending order - since they were already in descending order, all you need to do is reverse them (save for digit-x, which can be placed in the correct place in O(n)).
Let's use capital letters to define digit-strings and lower-case for digits. The syntax AB means "the concatenation of strings A and B". < is lexicographical ordering, which is the same as integer ordering when the digit-strings are of equal length.
Our original number N is of the form AxB, where x is a single digit and B is sorted descending. The number found by our algorithm is AyC, where y B is the smallest digit > x (it must exist due to the way x was chosen, see above), and C is sorted ascending.
Assume there is some number (using the same digits) N' such that AxB < N' < AyC. N' must begin with A or else it could not fall between them, so we can write it in the form AzD. Now our inequality is AxB < AzD < AyC, which is equivalent to xB < zD < yC where all three digit-strings contain the same digits.
In order for that to be true, we must have x <= z <= y. Since y is the smallest digit > x, z cannot be between them, so either z = x or z = y. Say z = x. Then our inequality is xB < xD < yC, which means B < D where both B and D have the same digits. However, B is sorted descending, so there is no string with those digits larger than it. Thus we cannot have B < D. Following the same steps, we see that if z = y, we cannot have D < C.
Therefore N' cannot exist, which means our algorithm correctly finds the next largest number.
nice solution! have one question. say "the smallest digit larger than x" is y. can we just swap x and y, then reverse x.index+1 -> end?
@Sterex, it's not just 99999; any number whose digits are already fully sorted in descending order is the max (so 98765 also has no solution, for example). This is easy to detect programatically because step 1 of the algorithm will fail (there is no pair of consecutive digits such that "the left-digit is smaller than the right-digit").
@TMN: 9 is larger than 8, so you'd move 9 to the left of 8: 9 832 then sort everything to the right of 9: 9238
@Kent for your solution to work you will have to change find the smallest digit larger than 4 to the right to find the smallest digit larger than 4 from the right. Otherwise, for example, 1234567849876 55 4321 will result in 1234567851234 54 6789 (instead of 1234567851234 45 6789). A nitpick :-)
## algorithm - Given a number, find the next higher number which has the ...
algorithm
If the data you are sorting has a known distribution, I would use a Bucket Sort algorithm. You could add some extra logic to it so that you calculated the size and/or positions of the various buckets based upon properties of the distribution (ex: for Gaussian, you might have a bucket every (sigma/k) away from the mean, where sigma is the standard deviation of the distribution).
By having a known distribution and modifying the standard Bucket Sort algorithm in this way, you would probably get the Histogram Sort algorithm or something close to it. Of course, your algorithm would be computationally faster than the the Histogram Sort algorithm because there would probably not be a need to do the first pass (described in the link) since you already know the distribution.
Edit: given your new criteria of your question, (though my previous answer concerning Histogram Sort links to the respectable NIST and contains performance information), here is a peer review journal article from the International Conference on Parallel Processing:
The authors claim this algorithm has better performance (up to 30% better) than the popular Quick-Sort Algorithm.
Considering Quick-Sort as the reference sorting algorithm is quite skewed. IntroSort improves on it by special-casing the small arrays that occur in the recursion, TimSort (and a few other variations) also improve on it by detecting patterns (ascending / descending blocks) on the fly. Interesting paper still :)
## performance - Sorting algorithms for data of known statistical distrib...
algorithm performance sorting statistics complexity-theory
You can do this in Go using interfaces. A function that takes an interface type is generic in the sense that it doesn't care about the data representation of the underlying concrete type. It does everything through method calls.
To make a generic version of your algorithm then, you have to identify all of the capabilities that the algorithm requires of the data objects and you have to define methods that abstract these capabilities. The abstract method signatures become method sets of interfaces.
To make a type compatible with this kind of generic algorithm, you define methods on the type to satisfy the interface of the algorithm parameter.
I'll take your example code and show one way to do this. Most of the required capabilities happen to be covered by sort.Interface so I chose to embed it. Only one other capability is needed, one to make a copy of the data.
```type algoContainer interface {
sort.Interface
Copy() algoContainer
}```
```package main
import (
"fmt"
"sort"
)
func main() {
s1 := sortableString("abc")
c1 := Algo(s1)
fmt.Println(s1, <-c1)
s2 := sortable3Ints([3]int{1,2,3})
c2 := Algo(&s2)
fmt.Println(s2, <-c2)
}
type algoContainer interface {
sort.Interface
Copy() algoContainer
}
type sortableString []byte
func (s sortableString) Len() int { return len(s) }
func (s sortableString) Swap(i, j int) { s[i], s[j] = s[j], s[i] }
func (s sortableString) Less(i, j int) bool { return s[i] < s[j] }
func (s sortableString) Copy() algoContainer {
return append(sortableString{}, s...)
}
func (s sortableString) String() string { return string(s) }
type sortable3Ints [3]int
func (sortable3Ints) Len() int { return 3 }
func (s *sortable3Ints) Swap(i, j int) {
(*s)[i], (*s)[j] = (*s)[j], (*s)[i]
}
func (s sortable3Ints) Less(i, j int) bool { return s[i] < s[j] }
func (s sortable3Ints) Copy() algoContainer { c := s; return &c }
func Algo(list algoContainer) chan algoContainer {
n := list.Len()
out := make(chan algoContainer)
go func () {
for i := 0; i < n; i++ {
result := list.Copy()
// actually useful:
if result.Less(n-1, 0) {
result.Swap(n-1, 0)
}
out <- result
}
close(out)
}()
return out
}```
## making generic algorithms in go - Stack Overflow
algorithm go
A detailed discussion on this problem has been discussed in Jon Bentley "Column 1. Cracking the Oyster" Programming Pearls Addison-Wesley pp.3-10
Bentley discusses several approaches, including external sort, Merge Sort using several external files etc., But the best method Bentley suggests is a single pass algorithm using bit fields, which he humorously calls "Wonder Sort" :) Coming to the problem, 4 billion numbers can be represented in :
`4 billion bits = (4000000000 / 8) bytes = about 0.466 GB`
The code to implement the bitset is simple: (taken from solutions page )
```#define BITSPERWORD 32
#define SHIFT 5
#define N 10000000
int a[1 + N/BITSPERWORD];
void set(int i) { a[i>>SHIFT] |= (1<<(i & MASK)); }
void clr(int i) { a[i>>SHIFT] &= ~(1<<(i & MASK)); }
int test(int i){ return a[i>>SHIFT] & (1<<(i & MASK)); }```
Bentley's algorithm makes a single pass over the file, setting the appropriate bit in the array and then examines this array using test macro above to find the missing number.
If the available memory is less than 0.466 GB, Bentley suggests a k-pass algorithm, which divides the input into ranges depending on available memory. To take a very simple example, if only 1 byte (i.e memory to handle 8 numbers ) was available and the range was from 0 to 31, we divide this into ranges of 0 to 7, 8-15, 16-22 and so on and handle this range in each of 32/8 = 4 passes.
I dont know the book, but no reason to call it "Wonder Sort", as it is just a bucketsort, with a 1-bit counter.
In the book, Jon Bentley is just being humorous...and calls it "Wonder Sort"
Although more portable, this code will be annihilated by code written to utilize hardware-supported vector instructions. I think gcc can automatically convert code to using vector operations in some cases though.
@brian I don't think Jon Bentley was allowing such things into his book on algorithms.
@BrianGordon, the time spent in ram will be negligible compared to the time spent reading the file. Forget about optimizing it.
## algorithm - Find an integer not among four billion given ones - Stack ...
algorithm
A detailed discussion on this problem has been discussed in Jon Bentley "Column 1. Cracking the Oyster" Programming Pearls Addison-Wesley pp.3-10
Bentley discusses several approaches, including external sort, Merge Sort using several external files etc., But the best method Bentley suggests is a single pass algorithm using bit fields, which he humorously calls "Wonder Sort" :) Coming to the problem, 4 billion numbers can be represented in :
`4 billion bits = (4000000000 / 8) bytes = about 0.466 GB`
The code to implement the bitset is simple: (taken from solutions page )
```#define BITSPERWORD 32
#define SHIFT 5
#define N 10000000
int a[1 + N/BITSPERWORD];
void set(int i) { a[i>>SHIFT] |= (1<<(i & MASK)); }
void clr(int i) { a[i>>SHIFT] &= ~(1<<(i & MASK)); }
int test(int i){ return a[i>>SHIFT] & (1<<(i & MASK)); }```
Bentley's algorithm makes a single pass over the file, setting the appropriate bit in the array and then examines this array using test macro above to find the missing number.
If the available memory is less than 0.466 GB, Bentley suggests a k-pass algorithm, which divides the input into ranges depending on available memory. To take a very simple example, if only 1 byte (i.e memory to handle 8 numbers ) was available and the range was from 0 to 31, we divide this into ranges of 0 to 7, 8-15, 16-22 and so on and handle this range in each of 32/8 = 4 passes.
I dont know the book, but no reason to call it "Wonder Sort", as it is just a bucketsort, with a 1-bit counter.
In the book, Jon Bentley is just being humorous...and calls it "Wonder Sort"
Although more portable, this code will be annihilated by code written to utilize hardware-supported vector instructions. I think gcc can automatically convert code to using vector operations in some cases though.
@brian I don't think Jon Bentley was allowing such things into his book on algorithms.
@BrianGordon, the time spent in ram will be negligible compared to the time spent reading the file. Forget about optimizing it.
algorithm | 18,191 | 76,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-39 | latest | en | 0.941434 |
http://math.stackexchange.com/questions/318814/how-to-use-the-power-series-of-ex-to-find-the-derivative-of-ex | 1,469,606,468,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826736.89/warc/CC-MAIN-20160723071026-00315-ip-10-185-27-174.ec2.internal.warc.gz | 158,898,101 | 22,784 | # How to use the power series of $e^x$ to find the derivative of $e^x$?
I am currently reading Roger Penrose's The Road to Reality and in the book, the author poses various problems with three different levels of difficultly easy, hard and really hard, according to the author this is easy. The problem I am looking at is as follows:
Using the power series of $e^x$ show that $de^x = e^x \, dx$
I have no idea as to how to tackle this problem.
If someone could provide some key points to solving the problem that would be great. Please do not provide the full steps, just key ideas or things to note.
Thanks!
EDIT:
I believe I understand this now because when you take the derivative of a power series you can do it term by term. The power series for $e^x$ is:
$$e^x = \sum_{i=0}^\infty \frac{x^n}{n!}$$
But more expanded it looks like this:
$$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} +\ldots$$
If the derivative of each term then I get: $0 + 1 + x + \frac{x^2}{2} + \ldots$ So in essence, I'm coming back to the original series. Therefore, the derivatives are the same.
-
That's about right. But no one really explained why the fact that it is a power series implies that you can differentiate term by term. Do you know why? – Git Gud Mar 2 '13 at 19:27
@GitGud No I don't, can you just think of it as a polynomial that is begin added? Like one that looks like this: $a_nx^n + a_{n-1}x^n-1 \ldots$ but more compact? And that is why you can do it term for term. If it it was $\prod$ instead of $\sum$ we would have to use the product rule instead of the power rule right? If my initial assertion is correct. – Jeel Shah Mar 2 '13 at 19:29
Do you know anything about function series? – Git Gud Mar 2 '13 at 19:32
@GitGud Not really. I understand the basic idea behind $\sum$ and how to do some basic things with it but I don't know much beyond that but is what I said correct? Like the basic idea of it? – Jeel Shah Mar 2 '13 at 19:33
Then I think you're not supposed to think about why it is legal to differentiate term by term. It's Physics anyway, so just do it. – Git Gud Mar 2 '13 at 19:35
The power series representation of $e^x$ is $$e^x = \sum_{i=0}^{\infty} \frac{x^n}{n!}$$ When you take the derivative of a power series you can do it term by term: \begin{align} \frac{d}{dx}e^x &= \frac{d}{dx}\sum_{i=0}^{\infty} \frac{x^n}{n!} \\ &= \sum_{i=0}^{\infty} \frac{d}{dx}\frac{x^n}{n!} \\ &= \sum_{i=1}^{\infty}\frac{x^{n-1}}{(n-1)!} \end{align} This is the same as $e^x$ (think about it).
To add a bit more detail. Remember that there is a rule that says that $$\frac{d}{dx} x^n = nx^{n-1}.$$ You for example have that $$\frac{d}{dx} x^2 = 2x \quad\quad\frac{d}{dx} x^3 = 3x^2.$$ So in the above we have used that $$\frac{d}{dx}\frac{x^n}{n!} =\frac{1}{n!} \frac{d}{dx} x^n = \frac{1}{n!}nx^{n-1} = \frac{n}{n\cdot(n-1)!}x^{n-1} = \frac{x^{n-1}}{(n-1)!}.$$
-
I have edited the post. Is my understanding correct? The fact that when you find each derivative, you will get the answer as $n-1$ term of the original series? – Jeel Shah Mar 2 '13 at 19:27
@gekkostate: What you say in your edited question is correct. So the derivative of each term gives the term just before it. So in all you end up with the same series. – Thomas Mar 2 '13 at 19:35
Why do you change the index from starting at 0 to starting at 1 when you take the derivative? – Jeff Jul 5 at 0:05
@Jeff: Because the first term is a constant. – Thomas Jul 5 at 12:13
Thanks. But you need a reason to change the starting index. If you don't change it, you have a term $\frac{x^{n-1}}{(n-1)!}$ where $n=0$ that leads to $\frac{1}{x(-1)!}$. Where's that term go? And what is -1 factorial? – Jeff Jul 5 at 20:21
Just write it out,
$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\;,$$
and differentiate it term by term. What’s the derivative of $\dfrac{x^n}{n!}$?
-
I have not yet encountered how to find the derivative of $\frac{x^n}{n!}$, we haven't been taught this in school. What's the idea behind find that derivative? – Jeel Shah Mar 2 '13 at 19:09
@gekkostate: Do you know how to find the derivative of $x^n$? That’s pretty much a prerequisite for answering the question. – Brian M. Scott Mar 2 '13 at 19:11
@gekkostate I think you know how to differentiate that. You're probably confusing something. Can you find the derivatives of $\displaystyle x, \frac{x^2}{2}$ and $\displaystyle \frac{x^3}{6}$? – Git Gud Mar 2 '13 at 19:11
@gekkostate: $n!$ is just a number. When you differentiate $\dfrac{x^3}{3!}$, you’re just differentiating $\dfrac{x^3}6$, so you get $\dfrac{x^2}2$. But it’s more useful here to think of that as differentiating $\dfrac{x^3}{3\cdot2\cdot1}$ and getting $\dfrac{3x^2}{3\cdot2\cdot1}=\dfrac{x^2}{2\cdot1}$. – Brian M. Scott Mar 2 '13 at 19:15
@gekkostate note that $n\in \Bbb N$ is already a fixed constant. Therefore $\displaystyle d\frac{x^n}{n!}=\frac{1}{n!}dx^n$. And to answer your question: yes, it is the same principle. – Git Gud Mar 2 '13 at 19:17
We denote $\displaystyle f_n(x)=\frac{x^n}{n!}$, so we have $\displaystyle e^x=\sum_{n=0}^{\infty}f_n(x).$
Let $[a,b]$ an arbitrary interval.
To justify the term by term differntiation of the series on $[a,b]$ we verify this points:
• the series $\displaystyle \sum_{n=0}^{\infty}f_n(x_0)$ at some point $x_0\in[a,b]$ and
• the series of derivatives $\displaystyle \sum_{n=0}^{\infty}f'_n(x)$ converges uniformly on $[a,b]$ to, say, $g$,
in this case, the series $\displaystyle \sum_{n=0}^{\infty}f_n(x)$ converges at every $x\in [a,b]$ and $$\displaystyle \left(\sum_{n=0}^{\infty}f_n(x)\right)'=g(x).$$
In our question, the two points are straightforward, for instance, we verify the second point: we have $$|f'_n(x)|=\frac{|x|^{n-1}}{(n-1)!}\leq \frac{\max(|a|,|b|)^{n-1}}{(n-1)!}=c_n,$$ so the series $\displaystyle\sum_nf'_n(x)$ converges uniformly on $[a,b]$ by Weierstrass M-test since the series $\displaystyle\sum_n c_n$ is convergent.
-
If you are having trouble to visualize this series with sum notation, just write down term by term and derive them. Doing this you will get the idea.
$\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots$
$\displaystyle \frac{d}{dx}e^x=\frac{d}{dx}\big(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\big)=$
$\displaystyle =\frac{d}{dx}\big(1\big)+\frac{d}{dx}\big(\frac{x}{1!}\big)+\frac{d}{dx}\big(\frac{x^2}{2!}\big)+\frac{d}{dx}\big(\frac{x^3}{3!}\big)+\frac{d}{dx}\big(\frac{x^4}{4!}\big)+\ldots=$
$\displaystyle =0+\frac{1}{1!}\frac{d}{dx}(x)+\frac{1}{2!}\frac{d}{dx}(x^2)+\frac{1}{3!}\frac{d}{dx}(x^3)+\frac{1}{4!}\frac{d}{dx}(x^4)+\ldots=$
$\displaystyle =\frac{1}{1!}+\frac{1}{2!}2x+\frac{1}{3!}3x^2+\frac{1}{4!}4x^3+\ldots=$
$\displaystyle =\sum_{n=1}^\infty \frac{nx^{n-1}}{n!}=\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$
I hope this clarify all the process done.
-
We can differentiate each term separately because without the $\sum$, the function would be like any other polynomial of the form $a_nx^n + a_{n-1}x^{n-1} \ldots a_1 + a_0$ and that is why we can use the power rule on each term right? Is that the logical behind being able to differentiate each term separately? – Jeel Shah Mar 2 '13 at 19:45
You can think that way because your teacher probably will make you solve only "good" series. But the reason is a little more deep. – Integral Mar 2 '13 at 19:51
hmm, okay. Thanks for your answer! – Jeel Shah Mar 2 '13 at 19:54
If you want to understand all the details, you'll need to study real analysis. – Integral Mar 2 '13 at 19:54 | 2,619 | 7,584 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-30 | latest | en | 0.957042 |
https://learnfrenzy.com/puzzles/puzzles/number-puzzles/ | 1,632,288,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057329.74/warc/CC-MAIN-20210922041825-20210922071825-00006.warc.gz | 399,807,394 | 7,931 | # Number Puzzles
What is Puzzle
Puzzle is a problem that test imagination, cleverness and skill of the solver. Puzzles are often contrived as a form of Entertainment, But they can also be considered as mathematical or logical problems in most of cases.
People with a high inductive reasoning aptitude may be better in solving those puzzles. puzzles may be solved by good deduction skills.
Number Puzzles
A number puzzle in which the numbers 1 through 9 must be placed into a grid of cells so that each row or column contains only one of each number. acrostic, word square. a puzzle where you fill a square grid with words reading the same down as across.
Number puzzles have specific set of rules, you first has to figure out the pattern being followed and then answer the puzzle according to the pattern.
Number puzzles are a part of many competitive examinations, these also helps you improve your logical thinking and above all these are fun to crack !!!
HOW TO SOLVE PUZZLES
In general to solve puzzles logical reasoning is necessary. In logic, there are three kinds of logical reasoning. they are
1.deduction
2.induction
3.abduction
precondition and rule implies a conclusion. so based on precondition, induction, abduction we can define above three as below.
DEDUCTION means determining the conclusion. It is using precondition and rule to make conclusion. Mathematicians are commonly used this type of logical reasoning.
INDUCTION means determining the rule. It is learning the rule after numerous examples of conclusion following the precondition. scientists are commonly used this type of logical reasoning.
ABDUCTION means determining the precondition.
It is using conclusion and rule to support that precondition could explain the conclusion. diagnosticians and detectives commonly used this type of logical reasoning.
MIND IT !
To solve puzzles logical reasoning is necessary. With out reasoning we can’t solve any type of logical problems. solving logical puzzles can also improves IQ level and maturity level.
In this section you can practice Number puzzles and improve your skills in order to face the Interview, Competitive examination and various entrance test with full confidence. | 432 | 2,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-39 | longest | en | 0.94867 |
https://ask.csdn.net/questions/1099979 | 1,638,123,702,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00057.warc.gz | 193,476,459 | 21,247 | 2020-10-18 16:01
# matlab实现Bp神经网络拟合函数100*(x2-x1^2)^2+（1-x1)^2=f误差太大
clc,clear
Data = -5:1:5;
Value = zeros(11,11);
for i = 1:11
for j = 1:11
x1 = Data(i);
x2 = Data(j);
Value(i,j) = 100*(x2-x1^2)^2+(1-x1)^2;
end
end
W1 = rand(20,1);%输入层x1与隐含层之间的权重
W12 = rand(20,1);%输入层x2与隐含层之间的权重
B1 = rand(20,1);%隐含层神经元的阈值
W2 = rand(1,20);%隐含层与输出层之间的权重
B2 = rand();%输出层神经元的阈值
yita = 0.0005;
loop = 5000;
E = zeros(1,loop);%误差随迭代次数的变化
for loopi = 1:loop
tempsume = 0;
Y = zeros(11,11);%模型输出的结果
for i = 1:11
for k = 1:11
x1= Data(i);%输入层输入数据
x2= Data(k);
hidein = x1*W1+x2*W12-B1;%隐含层的输入数据
hideout = zeros(20,1);%隐含层的输出数据
for j = 1:20
hideout(j) = sigmod(hidein(j));
end
y = W2*hideout-B2;%输出
Y(i,k) = y;
e = y-Value(i,k);%误差
%反馈，修改参数
dB2 = -1*yita*e;
dW2 = e*yita*hideout';
dB1 = zeros(20,1);
for j = 1:20
dB1(j) = W2(j)*sigmod(hidein(j))*(1-sigmod(hidein(j)))*(-1)*e*yita;
end
dW1 = zeros(20,1);
dW12 =zeros(20,1);
for j = 1:20
dW1(j) = W2(j)*sigmod(hidein(j))*(1-sigmod(hidein(j)))*x1*e*yita;
dW12(j) = W2(j)*sigmod(hidein(j))*(1-sigmod(hidein(j)))*x2*e*yita;
end
W1 = W1-dW1;
W12 = W12-dW12;
B1 = B1-dB1;
W2 = W2-dW2;
B2 = B2-dB2;
tempsume = tempsume + abs(e);
end
end
E(loopi) = tempsume;
end
figure(1)
plot(E,'-r')
hold on
grid on
Data1=-5:0.5:5;
Value1 = zeros(21,21);
for i = 1:21
for j = 1:21
x1 = Data1(i);
x2 = Data1(j);
Value1(i,j) = 100*(x2-x1^2)^2+(1-x1)^2;
end
end
Y1 = zeros(21,21);
for i = 1:21
for k = 1:21
x1= Data1(i);%输入层输入数据
x2= Data1(k);
hidein = x1*W1+x2*W12-B1;%隐含层的输入数据
hideout = zeros(20,1);%隐含层的输出数据
for j = 1:20
hideout(j) = sigmod(hidein(j));
end
y = W2*hideout-B2;%输出
Y1(i,k) = y;
e(i,k) = y-Value1(i,k);%误差
end
end
figure(2)
plot(e,'-g')
hold on
grid on
• 写回答
• 好问题 提建议
• 关注问题
• 收藏
• 邀请回答 | 942 | 1,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-49 | latest | en | 0.369511 |
http://technodocbox.com/Computer_Networking/122321912-High-level-view-ee-122-ethernet-and-random-access-protocols-medium-access-protocols.html | 1,603,622,183,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107888931.67/warc/CC-MAIN-20201025100059-20201025130059-00435.warc.gz | 102,120,556 | 23,845 | # High Level View. EE 122: Ethernet and Random Access protocols. Medium Access Protocols
Size: px
Start display at page:
Download "High Level View. EE 122: Ethernet and Random Access protocols. Medium Access Protocols"
Transcription
1 High Level View EE 122: Ethernet and Ion Stoica September 18, 2002 Goal: share a communication medium among multiple hosts connected to it Problem: arbitrate between connected hosts Solution goals: - High resource utilization - Avoid starvation - Simplicity (non-decentralized algorithms) (* this talk is based in part on the on-line slides of J. Kurose & K. Rose) 2 Medium Access Protocols Random Access protocols Channel partitioning - Divide channel into smaller pieces (e.g., time slots, frequency) - Allocate a piece to node for exclusive use Random access - Allow collisions - recover from collisions Taking-turns - Tightly coordinate shared access to avoid collisions When node has packet to send - Transmit at full channel data rate R. - No a priori coordination among nodes Two or more transmitting nodes -> collision, Random access MAC protocol specifies: - How to detect collisions - How to recover from collisions Examples of random access MAC protocols: - Slotted ALOHA - CSMA and CSMA/CD 3 4 1
2 Slotted Aloha Slotted Aloha Efficiency Time is divided into equal size slots (= packet transmission time) Node with new arriving pkt: transmit at beginning of next slot If collision: retransmit pkt in future slots with probability p, until successful. What is the maximum fraction of successful transmissions? Suppose N stations have packets to send - Each transmits in slot with probability p - Prob. successful transmission S is (very approximated analysis!): by a particular node: S = p (1-p) (N-1) Success (S), Collision (C), Empty (E) slots by any of N nodes S = Prob (only one transmits) = N p (1-p)(N-1) <= 1/e = CSMA: Carrier Sense Multiple Access CS (Carrier Sense) means that each node can distinguish between an idle and a busy link Sender operations: - If channel sensed idle: transmit entire packet - If channel sensed busy, defer transmission Persistent CSMA: retry immediately with probability p when channel becomes idle Non-persistent CSMA: retry after a random time interval 7 Collisions can occur: propagation delay means two nodes may not hear each other s transmission Collision: entire packet transmission time wasted Note: role of distance and propagation delay in determining collision prob. CSMA collisions 8 2
3 CSMA/CD (Collision Detection) CSMA/CD collision detection Collisions detected within short time Colliding transmissions aborted, reducing channel wastage Easy in wired LANs: measure signal strengths, compare transmitted, received signals Difficult in wireless LANs 9 10 Ethernet Wireless (802.11) Overview Ethernet Dominant LAN technology CSMA/CD protocol Cheap \$20 for 100Mbs!
5 Minimum Packet Size Why put a minimum packet size? Give a host enough time to detect collisions In Ethernet, minimum packet size = 64 bytes (two 6-byte addresses, 2-byte type, 4-byte CRC, and 46 bytes of data) If host has less than 46 bytes to send, the adaptor pads (adds) bytes to make it 46 bytes What is the relationship between minimum packet size and the length of the LAN? a) Time = t; Host 1 starts to send frame b) Time = t + d; Host 2 starts to send a frame just before it hears from host 1 s frame c) Time = t + 2*d; Host 1 hears Host 2 s frame detects collision Minimum Packet Size (more) Host 1 Host 2 propagation delay (d) Host 1 Host 2 propagation delay (d) Host 1 Host 2 propagation delay (d) 17 LAN length = (min_frame_size)*(light_speed)/(2*bandwidth) = = (8*64b)*(2.5*10 8 mps)/(2*10 7 bps) = 6400m approx 18 Ethernet Technologies: 10Base2 10BaseT and 100BaseT 10: 10Mbps; 2: under 200 meters max cable length Thin coaxial cable in a bus topology 10/100 Mbps rate; latter called fast ethernet T stands for Twisted Pair Hub to which nodes are connected by twisted pair, thus star topology CSMA/CD implemented at hub Repeaters used to connect up to multiple segments Repeater repeats bits it hears on one interface to its other interfaces: physical layer device only!
6 10BaseT and 100BaseT (more) Gbit Ethernet Max distance from node to Hub is 100 meters Hub can disconnect jabbering adapter Hub can gather monitoring information, statistics for display to LAN administrators Use standard Ethernet frame format Allows for point-to-point links and shared broadcast channels In shared mode, CSMA/CD is used; short distances between nodes to be efficient Uses hubs, called here Buffered Distributors Full-Duplex at 1 Gbps for point-to-point links Interconnecting LANs Why not just one big LAN? - Limited amount of supportable traffic: on single LAN, all stations must share bandwidth - Limited length - Large collision domain (can collide with many stations) Ethernet Wireless (802.11) Overview
7 Wireless (802.11) Physical Link Designed for use in limited geographical area (i.e., couple of hundreds of meters) Designed for three physical media (run at either 1Mbps or 2 Mbps) - Two based on spread spectrum radio - One based on diffused infrared Frequency hoping - Transmit the signal over multiple frequencies - The sequence of frequencies is pseudo-random, i.e., both sender and receiver use the same algorithm to generate their sequences Direct sequence - Represent each bit by multiple (e.g., n) bits in a frame; XOR signal with a pseudo-random generated sequence with a frequency n times higher Infrared signal - Sender and receiver do not need a clear line of sight - Limited range; order of meters Collision Avoidance: The Problems Multiple Access with Collision Avoidance (MACA) Reachability is not transitive: if A can reach C, and C can reach D, it doesn t necessary mean that A can reach D sender RTS CTS data receiver other node in sender s range A B C D ACK Hidden nodes: A and C send a packet to B; neither A nor C will detect the collision! Exposed node: B sends a packet to A; C hears this and decides not to send a packet to D (despite the fact that this will not cause interference)! 27 Before every data transmission - Sender sends a Request to Send (RTS) frame containing the length of the transmission - Receiver respond with a Clear to Send (CTS) frame - Sender sends data - Receiver sends an ACK; now another sender can send data When sender doesn t get a CTS back, it assumes collision 28 7
8 Summary Arbitrate between multiple hosts sharing a common communication media Wired solution: Ethernet (use CSMA/CD protocol) - Detect collisions - Backoff exponentially on collision Wireless solution: Use MACA protocol - Cannot detect collisions; try to avoid them 29 8
### EE 122: Ethernet and
EE 122: Ethernet and 802.11 Ion Stoica September 18, 2002 (* this talk is based in part on the on-line slides of J. Kurose & K. Rose) High Level View Goal: share a communication medium among multiple hosts
### Summary of MAC protocols
Summary of MAC protocols What do you do with a shared media? Channel Partitioning, by time, frequency or code Time Division, Code Division, Frequency Division Random partitioning (dynamic) ALOHA, S-ALOHA,
### COMP/ELEC 429/556 Introduction to Computer Networks
COMP/ELEC 429/556 Introduction to Computer Networks Broadcast network access control Some slides used with permissions from Edward W. Knightly, T. S. Eugene Ng, Ion Stoica, Hui Zhang T. S. Eugene Ng eugeneng
### Medium Access Protocols
Medium Access Protocols Summary of MAC protocols What do you do with a shared media? Channel Partitioning, by time, frequency or code Time Division,Code Division, Frequency Division Random partitioning
### Lecture 5 The Data Link Layer. Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it
Lecture 5 The Data Link Layer Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it Link Layer: setting the context two physically connected devices: host-router, router-router, host-host,
### Lecture 6 The Data Link Layer. Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it
Lecture 6 The Data Link Layer Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it Link Layer: setting the context two physically connected devices: host-router, router-router, host-host,
### Review. Error Detection: CRC Multiple access protocols. LAN addresses and ARP Ethernet. Slotted ALOHA CSMA/CD
Review Error Detection: CRC Multiple access protocols Slotted ALOHA CSMA/CD LAN addresses and ARP Ethernet Some slides are in courtesy of J. Kurose and K. Ross Overview Ethernet Hubs, bridges, and switches
### CS 43: Computer Networks. 27: Media Access Contd. December 3, 2018
CS 43: Computer Networks 27: Media Access Contd. December 3, 2018 Last Class The link layer provides lots of functionality: addressing, framing, media access, error checking could be used independently
### Topics. Link Layer Services (more) Link Layer Services LECTURE 5 MULTIPLE ACCESS AND LOCAL AREA NETWORKS. flow control: error detection:
1 Topics 2 LECTURE 5 MULTIPLE ACCESS AND LOCAL AREA NETWORKS Multiple access: CSMA/CD, CSMA/CA, token passing, channelization LAN: characteristics, i basic principles i Protocol architecture Topologies
### Getting Connected (Chapter 2 Part 4) Networking CS 3470, Section 1 Sarah Diesburg
Getting Connected (Chapter 2 Part 4) Networking CS 3470, Section 1 Sarah Diesburg Five Problems Encoding/decoding Framing Error Detection Error Correction Media Access Five Problems Encoding/decoding Framing
Link Layer and Ethernet 14-740: Fundamentals of Computer Networks Bill Nace Material from Computer Networking: A Top Down Approach, 6 th edition. J.F. Kurose and K.W. Ross traceroute Data Link Layer Multiple
### Computer Networks Medium Access Control. Mostafa Salehi Fall 2008
Computer Networks Medium Access Control Mostafa Salehi Fall 2008 2008 1 Outline Issues ALOHA Network Ethernet Token Ring Wireless 2 Main Issues Local Area Network (LAN) : Three or more machines are physically
### CSMA/CD (Collision Detection)
CSMA/CD (Collision Detection) CD (collision detection): easy in wired LANs: measure signal strengths, compare transmitted, received signals difficult in wireless LANs: received signal strength overwhelmed
Link Layer and Ethernet 14-740: Fundamentals of Computer Networks Bill Nace Material from Computer Networking: A Top Down Approach, 6 th edition. J.F. Kurose and K.W. Ross traceroute Data Link Layer Multiple
### CS 43: Computer Networks Media Access. Kevin Webb Swarthmore College November 30, 2017
CS 43: Computer Networks Media Access Kevin Webb Swarthmore College November 30, 2017 Multiple Access Links & Protocols Two classes of links : point-to-point dial-up access link between Ethernet switch,
### The Link Layer II: Ethernet
Monday Recap The Link Layer II: Ethernet q Link layer services q Principles for multiple access protocols q Categories of multiple access protocols CSC 249 March 24, 2017 1 2 Recap: Random Access Protocols
### Links Reading: Chapter 2. Goals of Todayʼs Lecture. Message, Segment, Packet, and Frame
Links Reading: Chapter 2 CS 375: Computer Networks Thomas Bressoud 1 Goals of Todayʼs Lecture Link-layer services Encoding, framing, and error detection Error correction and flow control Sharing a shared
### Lecture 8 The Data Link Layer part I. Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it
Lecture 8 The Data Link Layer part I Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it Link Layer: setting the context two physically connected devices: host-router, router-router,
### CSE 461: Multiple Access Networks. This Lecture
CSE 461: Multiple Access Networks This Lecture Key Focus: How do multiple parties share a wire? This is the Medium Access Control (MAC) portion of the Link Layer Randomized access protocols: 1. Aloha 2.
### Chapter 5 Link Layer and LANs
Chapter 5 Link Layer and LANs A note on the use of these ppt slides: All material copyright 1996-2007 J.F Kurose and K.W. Ross, All Rights Reserved Computer Networking: A Top Down Approach 4 th edition.
### Computer Network Fundamentals Spring Week 3 MAC Layer Andreas Terzis
Computer Network Fundamentals Spring 2008 Week 3 MAC Layer Andreas Terzis Outline MAC Protocols MAC Protocol Examples Channel Partitioning TDMA/FDMA Token Ring Random Access Protocols Aloha and Slotted
### Ethernet. Typical Setup. Names. Operations. Operations Switch: No Collisions EECS 122. Hub: Single Collision Domain
Overview Physical Layer MAC Bridged VLAN Link Aggregation XON/XOFF 802.11 Summary Overview Typical Setup Names Operations Perspective TOC TOC Overview Typical Setup Names Structure [rate][modulation][media
### Computer Networks. Today. Principles of datalink layer services Multiple access links Adresavimas, ARP LANs Wireless LANs VU MIF CS 1/48 2/48
Computer Networks VU MIF CS 1/48 Today Principles of datalink layer services Multiple access links Adresavimas, ARP LANs Wireless LANs 2/48 1 Link layer: introduction terminology: hosts and routers: nodes
### ECE 4450:427/527 - Computer Networks Spring 2017
ECE 4450:427/527 - Computer Networks Spring 2017 Dr. Nghi Tran Department of Electrical & Computer Engineering Lecture 5.4: Multiple Access Protocols Dr. Nghi Tran (ECE-University of Akron) ECE 4450:427/527
### 1-1. Switching Networks (Fall 2010) EE 586 Communication and. November 8, Lecture 30
EE 586 Communication and Switching Networks (Fall 2010) Lecture 30 November 8, 2010 1-1 Announcements Quiz on Wednesday Next Monday hands-on training on Contiki OS Bring your laptop 4-2 Multiple Access
### Outline. Introduction to Networked Embedded Systems - Embedded systems Networked embedded systems Embedded Internet - Network properties
Outline Introduction to Networked Embedded Systems - Embedded systems Networked embedded systems Embedded Internet - Network properties Layered Network Architectures - OSI framework descriptions of layers
### CMPE 257: Wireless and Mobile Networking
CMPE 257: Wireless and Mobile Networking Katia Obraczka Computer Engineering UCSC Baskin Engineering Lecture 3 CMPE 257 Winter'11 1 Announcements Accessing secure part of the class Web page: User id: cmpe257.
### Data Link Layer, Part 5. Medium Access Control
CS 455 Medium Access Control, Page 1 Data Link Layer, Part 5 Medium Access Control These slides are created by Dr. Yih Huang of George Mason University. Students registered in Dr. Huang s courses at GMU
### The Link Layer and LANs. Chapter 6: Link layer and LANs
The Link Layer and LANs EECS3214 2018-03-14 4-1 Chapter 6: Link layer and LANs our goals: understand principles behind link layer services: error detection, correction sharing a broadcast channel: multiple
### Multiple Access Links and Protocols
Multiple Access Links and Protocols Two types of links : point-to-point PPP for dial-up access point-to-point link between Ethernet switch and host broadcast (shared wire or medium) old-fashioned Ethernet
### CMPE 150/L : Introduction to Computer Networks. Chen Qian Computer Engineering UCSC Baskin Engineering Lecture 16
CMPE 150/L : Introduction to Computer Networks Chen Qian Computer Engineering UCSC Baskin Engineering Lecture 16 1 Final project demo Please do the demo next week to the TAs. So basically you may need
### LAN PROTOCOLS. Beulah A AP/CSE
LAN PROTOCOLS Beulah A AP/CSE IEEE STANDARDS In 1985, the Computer Society of the IEEE started a project, called Project 802, to set standards to enable intercommunication among equipment from a variety
### Link Layer and LANs 안상현서울시립대학교컴퓨터 통계학과.
Link Layer and LANs 안상현서울시립대학교컴퓨터 통계학과 ahn@venus.uos.ac.kr Data Link Layer Goals: understand principles behind data link layer services: error detection, correction sharing a broadcast channel: multiple
### Computer Networking Lecture 5 Data link Layer Access Control. Based on slides by Peter Steenkiste Copyright, Carnegie Mellon
15-441 Computer Networking Lecture 5 Data link Layer Access Control Based on slides by Peter Steenkiste Copyright, Carnegie Mellon 2007-12 1 Datalink Functions Framing: encapsulating a network layer datagram
### Data Link Layer: Multi Access Protocols
Digital Communication in the Modern World Data Link Layer: Multi Access Protocols http://www.cs.huji.ac.il/~com1 com1@cs.huji.ac.il Some of the slides have been borrowed from: Computer Networking: A Top
Chapter 5: Link layer our goals: v understand principles behind link layer services: error detection, correction sharing a broadcast channel: multiple access link layer addressing local area networks:
### Chapter 5 Link Layer and LANs
Chapter 5 Link Layer and LANs A note on the use of these ppt slides: We re making these slides freely available to all (faculty, students, readers). They re in PowerPoint form so you can add, modify, and
Reminder: Datalink Functions 15-441 15 441 15-641 Computer Networking Lecture 5 Media Access Control Peter Steenkiste Fall 2015 www.cs.cmu.edu/~prs/15-441-f15 Framing: encapsulating a network layer datagram
### CMPE 257: Wireless and Mobile Networking
CMPE 257: Wireless and Mobile Networking Katia Obraczka Computer Engineering UCSC Baskin Engineering Lecture 3 CMPE 257 Spring'15 1 Next week Announcements April 14: ICN (Spencer Sevilla) April 16: DTN
Link Layer and LANs Instructor: Anirban Mahanti Office: ICT 745 Email: mahanti@cpsc.ucalgary.ca Class Location: ICT 121 Lectures: MWF 12:00 12:50 hours Notes derived from Computer Networking: A Top Down
### Data Link Layer, Part 3 Medium Access Control. Preface
Data Link Layer, Part 3 Medium Access Control These slides are created by Dr. Yih Huang of George Mason University. Students registered in Dr. Huang's courses at GMU can make a single machine-readable
### Physical Layer. Medium Access Links and Protocols. Point-to-Point protocols. Modems: Signaling. Modems Signaling. Srinidhi Varadarajan
P Physical Layer Srinidhi Varadarajan Medium Access Links and Protocols Three types of links : point-to-point (single wire, e.g. PPP, SLIP) broadcast (shared wire or medium; e.g, Ethernet, Wavelan, etc.)
### Lectures Ethernet
Lectures 22-23 Ethernet EECS 122 University of California Berkeley Overview Robert Metcalfe at Xerox in Palo Alto developed the original Ethernet in 1973. Digital Equipment Corporation, Intel, and Xerox
### Chapter 5 Link Layer and LANs
Chapter 5 Link Layer and LANs Computer Networking: A Top Down Approach 4 th edition. Jim Kurose, Keith Ross Addison-Wesley, July 2007. All material copyright 1996-2007 J.F Kurose and K.W. Ross, All Rights
### CSE 461: Multiple Access. Homework: Chapter 2, problems 1, 8, 12, 18, 23, 24, 35, 43, 46, and 58
CSE 461: Multiple Access Homework: Chapter 2, problems 1, 8, 12, 18, 23, 24, 35, 43, 46, and 58 Next Topic Key Focus: How do multiple parties share a wire? This is the Medium Access Control (MAC) portion
### Data and Computer Communications. Chapter 11 Local Area Network
Data and Computer Communications Chapter 11 Local Area Network LAN Topologies Refers to the way in which the stations attached to the network are interconnected Bus Topology Used with multipoint medium
### Contention Protocols and Networks
4/13/2005 314 Lecture Contention Protocols and Networks 1 Contention Protocols and Networks Contention Protocols CSMA/CD Network Topologies Ethernet 4/13/2005 314 Lecture Contention Protocols and Networks
### CS 455/555 Intro to Networks and Communications. Link Layer
CS 455/555 Intro to Networks and Communications Link Layer Dr. Michele Weigle Department of Computer Science Old Dominion University mweigle@cs.odu.edu http://www.cs.odu.edu/~mweigle/cs455-s13 1 Link Layer
### COS 140: Foundations of Computer Science
COS 140: Foundations of Computer Science ALOHA Network Protocol Family Fall 2017 Homework 2 Introduction 3 Network Protocols.......................................................... 3 Problem.................................................................
### Ethernet. Lecture 6. Outline. Ethernet - Physical Properties. Ethernet - Physical Properties. Ethernet
Lecture 6 Ethernet Reminder: Homework 2, Programming Project 2 due on 9/20/12. Thick-net Thin-net Twisted Pair Thursday, September 13 CS 475 Networks - Lecture 6 1 Thursday, September 13 CS 475 Networks
Chapter 5: DataLink Layer Course on Computer Communication and Networks, CTH/GU The slides are adaptation of the slides made available by the authors of the course s main textbook Slides with darker background
### CSC 4900 Computer Networks: The Link Layer
CSC 4900 Computer Networks: The Link Layer Professor Henry Carter Fall 2017 Last Time We talked about intra-as routing protocols: Which routing algorithm is used in RIP? OSPF? What techniques allow OSPF
### Random Access. 1. Aloha. 2. Slotted Aloha 3. CSMA 4. CSMA/CD
Random Access 1. Aloha 2. Slotted Aloha 3. CSMA 4. CSMA/CD Background Communication medium B No Collision collision A C Modern Local Area Networks (LANs) operate as follows Users are connected to communication
### Lecture 6. Reminder: Homework 2, Programming Project 2 due on Thursday. Questions? Tuesday, September 13 CS 475 Networks - Lecture 6 1
Lecture 6 Reminder: Homework 2, Programming Project 2 due on Thursday. Questions? Tuesday, September 13 CS 475 Networks - Lecture 6 1 Outline Chapter 2 - Getting Connected 2.1 Perspectives on Connecting
### Master Course Computer Networks IN2097
Chair for Network Architectures and Services Prof. Carle Department for Computer Science TU München Master Course Computer Networks IN2097 Prof. Dr.-Ing. Georg Carle Christian Grothoff, Ph.D. Dr. Nils
Link Layer: Retransmissions Context on Reliability Where in the stack should we place reliability functions? Application Transport Network Link Physical CSE 461 University of Washington 2 Context on Reliability
Chapter V: Link Layer UG3 Computer Communications & Networks (COMN) Myungjin Lee myungjin.lee@ed.ac.uk Slides copyright of Kurose and Ross Link layer services framing, link access: encapsulate datagram
### MULTIPLE ACCESS PROTOCOLS 2. 1
MULTIPLE ACCESS PROTOCOLS AND WIFI 1 MULTIPLE ACCESS PROTOCOLS 2. 1 MULTIPLE ACCESS LINKS, PROTOCOLS Two types of links : point-to-point broadcast (shared wire or medium) POINT-TO-POINT PPP for dial-up
### Lecture 9: Bridging. CSE 123: Computer Networks Alex C. Snoeren
Lecture 9: Bridging CSE 123: Computer Networks Alex C. Snoeren Lecture 9 Overview Finishing up media access Ethernet Contention-free methods (rings) Moving beyond one wire Link technologies have limits
### Computer Networks Principles LAN - Ethernet
Computer Networks Principles LAN - Ethernet Prof. Andrzej Duda duda@imag.fr http://duda.imag.fr 1 Interconnection structure - layer 3 interconnection layer 3 router subnetwork 1 interconnection layer 2
Link Layer: Introduction daptors ommunicating hosts and routers are nodes links connect nodes wired links wireless links layer-2 packet is a frame, encapsulates datagram datagram controller sending host
### Redes de Computadores. Medium Access Control
Redes de Computadores Medium Access Control Manuel P. Ricardo Faculdade de Engenharia da Universidade do Porto 1 » How to control the access of computers to a communication medium?» What is the ideal Medium
### Medium Access Control. CSCI370 Lecture 5 Michael Hutt New York Institute of Technology
Medium Access Control CSCI370 Lecture 5 Michael Hutt New York Institute of Technology The Data Link Layer Logical Link Control (LLC) IEEE 802.2 Standard RFC 1042 Provides three service options Unreliable
### Module 10 Data Link Layer CS655! 10-1!
Module 10 Data Link Layer CS655! 10-1! Please note: Most of these slides come from this book. Note their copyright notice below! A note on the use of these ppt slides: We re making these slides freely
### CS 716: Introduction to communication networks. - 9 th class; 19 th Aug Instructor: Sridhar Iyer IIT Bombay
CS 716: Introduction to communication networks - 9 th class; 19 th Aug 2011 Instructor: Sridhar Iyer IIT Bombay Contention-based MAC: ALOHA Users transmit whenever they have data to send Collisions occur,
### CSC 4900 Computer Networks: Link Layer (2)
CSC 4900 Computer Networks: Link Layer (2) Professor Henry Carter Fall 2017 Link Layer 6.1 Introduction and services 6.2 Error detection and correction 6.3 Multiple access protocols 6.4 LANs addressing,
### Protocols for Multiaccess Networks
Protocols for Multiaccess Networks Hosts broadcast packets When a collision occurs, all transmitted packets are lost Lost packets have to be retransmitted => Need Multiaccess Protocol Model - Slotted Aloha
### Aloha and slotted aloha
CSMA 2/13/06 Aloha and slotted aloha Slotted aloha: transmissions are synchronized and only start at the beginning of a time slot. Aloha sender A sender B collision sender C t Slotted Aloha collision sender
### Lecture 9 The Data Link Layer part II. Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it
Lecture 9 The Data Link Layer part II Antonio Cianfrani DIET Department Networking Group netlab.uniroma1.it Physical Addresses Physical (or LAN or MAC) address: 48 bit string Hexadecimal representation
### Principles behind data link layer services
Data link layer Goals: Principles behind data link layer services Error detection, correction Sharing a broadcast channel: Multiple access Link layer addressing Reliable data transfer, flow control: Done!
### COS 140: Foundations of Computer Science
COS 140: Foundations of C S Network Protocol Family Fall 2017 Copyright c 2002 2017 UMaine School of Computing and Information S 1 / 25 Homework Homework Slides, book Chapter 24 on line Homework: All exercises
### Ethernet. Introduction. CSE 3213 Fall 2011
Ethernet CSE 3213 Fall 2011 19 October 2011 1 Introduction Rapid changes in technology designs Broader use of LANs New schemes for high-speed LANs High-speed LAN technologies: Fast and gigabit Ethernet
### Master Course Computer Networks IN2097
Chair for Network Architectures and Services Prof. Carle Department for Computer Science TU München Master Course Computer Networks IN2097 Prof. Dr.-Ing. Georg Carle Christian Grothoff, Ph.D. Chair for
### Goals of Today s Lecture. Adaptors Communicating
Goals of Today s Lecture EE 122: Link Layer Ion Stoica TAs: Junda Liu, DK Moon, David Zats http://inst.eecs.berkeley.edu/~ee122/ (Materials with thanks to Vern Paxson, Jennifer Rexford, and colleagues
### Chapter 5 Link Layer. Computer Networking: A Top Down Approach. 6 th edition Jim Kurose, Keith Ross Addison-Wesley March 2012
Chapter 5 Link Layer A note on the use of these ppt slides: We re making these slides freely available to all (faculty, students, readers). They re in PowerPoint form so you see the animations; and can
### CCM 4300 Lecture 6 Computer Networks, Wireless and Mobile Communications. Dr Shahedur Rahman. Room: T115
CCM 4300 Lecture 6 Computer Networks, Wireless and Mobile Communications Dr Shahedur Rahman s.rahman@mdx.ac.uk Room: T115 1 Recap of Last Session Access Control ALOHA FDMA, TDMA, CDMA OSI Model Internetworking
Links 1 Goals of Today s Lecture Link-layer services Encoding, framing, and error detection Error correction and flow control Sharing a shared media Channel partitioning Taking turns Random access Shared
### Chapter 4. The Medium Access Control Sublayer. Points and Questions to Consider. Multiple Access Protocols. The Channel Allocation Problem.
Dynamic Channel Allocation in LANs and MANs Chapter 4 The Medium Access Control Sublayer 1. Station Model. 2. Single Channel Assumption. 3. Collision Assumption. 4. (a) Continuous Time. (b) Slotted Time.
Link Layer (continued) Topics 1. Framing Delimiting start/end of frames 2. Error detection and correction Handling errors 3. Retransmissions Handling loss 4. Multiple Access 802.11, classic Ethernet 5.
### Chapter 6 Medium Access Control Protocols and Local Area Networks
Chapter 6 Medium Access Control Protocols and Local Area Networks Part I: Medium Access Control Part II: Local Area Networks CSE 3213, Winter 2010 Instructor: Foroohar Foroozan Chapter Overview Broadcast
### ECEN 5032 Data Networks Medium Access Control Sublayer
ECEN 5032 Data Networks Medium Access Control Sublayer Peter Mathys mathys@colorado.edu University of Colorado, Boulder c 1996 2005, P. Mathys p.1/35 Overview (Sub)networks can be divided into two categories:
### Wireless MACs: MACAW/802.11
Wireless MACs: MACAW/802.11 Mark Handley UCL Computer Science CS 3035/GZ01 Fundamentals: Spectrum and Capacity A particular radio transmits over some range of frequencies; its bandwidth, in the physical
### Data Link Layer. Our goals: understand principles behind data link layer services: instantiation and implementation of various link layer technologies
Data Link Layer Our goals: understand principles behind data link layer services: link layer addressing instantiation and implementation of various link layer technologies 1 Outline Introduction and services
### RMIT University. Data Communication and Net-Centric Computing COSC 1111/2061/1110. Lecture 8. Medium Access Control Methods & LAN
RMIT University Data Communication and Net-Centric Computing COSC 1111/2061/1110 Medium Access Control Methods & LAN Technology Slide 1 Lecture Overview During this lecture, we will Look at several Multiple
### CS 3640: Introduction to Networks and Their Applications
CS 3640: Introduction to Networks and Their Applications Fall 2018, Lecture 7: The Link Layer II Medium Access Control Protocols Instructor: Rishab Nithyanand Teaching Assistant: Md. Kowsar Hossain 1 You
Chapter 5 Link Layer Computer Networking: A Top Down Approach 6 th edition Jim Kurose, Keith Ross Addison-Wesley March 2012 Link layer, LANs: outline 5.1 introduction, services 5.2 error detection, correction
### CS 123: Lecture 12, LANs, and Ethernet. George Varghese. October 24, 2006
CS 123: Lecture 12, LANs, and Ethernet George Varghese October 24, 2006 Selective Reject Modulus failure Example w = 2, Max = 3 0 0 1 3 0 A(1) A(2) 1 0 retransmit A(1) A(2) buffer Case 1 Case 2 reject
Chapter V: Link Layer UG3 Computer Communications & Networks (COMN) Myungjin Lee myungjin.lee@ed.ac.uk Slides copyright of Kurose and Ross Link layer: introduction terminology: hosts and routers: nodes
### CSCD 433 Network Programming Fall Lecture 7 Ethernet and Wireless
CSCD 433 Network Programming Fall 2016 Lecture 7 Ethernet and Wireless 802.11 1 Topics 802 Standard MAC and LLC Sublayers Review of MAC in Ethernet MAC in 802.11 Wireless 2 IEEE Standards In 1985, Computer
### Lecture 8 Link Layer: functionality, error detection, media access algorithm
Lecture 8 Link Layer: functionality, error detection, media access algorithm From Kurose & Ross Book slightly modified by Romaric Duvignau duvignau@chalmers.se Thanks and enjoy! JFK/KWR All material copyright
### Data Link Layer: Overview, operations
Data Link Layer: Overview, operations Chapter 3 1 Outlines 1. Data Link Layer Functions. Data Link Services 3. Framing 4. Error Detection/Correction. Flow Control 6. Medium Access 1 1. Data Link Layer
### COMPUTER NETWORKS - Local area networks
Local area networks Telecommunication Networks Group firstname.lastname@polito.it http://www.telematica.polito.it/ COMPUTER NETWORKS LANs - 1 Copyright Quest opera è protetta dalla licenza Creative Commons
Local area networks Telecommunication Networks Group firstname.lastname@polito.it http://www.telematica.polito.it/ COMPUTER NETWORKS LANs - 1 Copyright Quest opera è protetta dalla licenza Creative Commons
### Lecture 19. Principles behind data link layer services Framing Multiple access protocols
Link Layer Lecture 19 Principles behind data link layer services Framing Multiple access protocols ALOHA *The slides are adapted from ppt slides (in substantially unaltered form) available from Computer | 7,470 | 31,856 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-45 | latest | en | 0.84339 |
http://www.slideserve.com/thi/chem10-topic-01-thermochemistry | 1,508,236,004,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187821017.9/warc/CC-MAIN-20171017091309-20171017111309-00168.warc.gz | 597,218,140 | 13,929 | Chem10 Topic 01 - Thermochemistry
1 / 16
# Chem10 Topic 01 - Thermochemistry - PowerPoint PPT Presentation
Chem10 Topic 01 - Thermochemistry. Science 10 CT01D04. Stoichi what?. Stoikheion ( greek )= element Metron ( greek )= to measure Stoichiometry: the calculation of amounts of substances involved in chemical reactions This will be Topic 01 in IB Chemistry. Moles.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## PowerPoint Slideshow about ' Chem10 Topic 01 - Thermochemistry' - thi
An Image/Link below is provided (as is) to download presentation
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
### Chem10 Topic 01 - Thermochemistry
Science 10
CT01D04
Stoichi what?
• Stoikheion (greek)= element
• Metron (greek)= to measure
• Stoichiometry: the calculation of amounts of substances involved in chemical reactions
• This will be Topic 01 in IB Chemistry
Moles
• Moles (latin): heap or pile
• 1 Mole (mol) = 6.02 x 1023 representative particles (atoms or molecules)
• This is called Avogadro’s number
Molar Mass
• The atomic mass of one mole of an element or molecule expressed in grams
• Round to 2 decimal places, your periodic table should already have this!
Calculating Molar Mass
• Step One: Find the molar mass (atomic mass) of each element on the PT
• round to 2 decimal places
• Unit is grams (g) per mole (mol)
• Step Two: Multiply the number of atoms of the element by the molar mass
• Remember – the subscript tells you how many atoms you have of each element
• Step Three: Add up the masses of all elements in the compound
Units for Molar Mass
• H2O is 18.02 g/mol
• Stated as “grams per mol”
• Meaning 18.02 grams = 1 mole
• OR 1 mole = 18.02 grams
• Use like common conversions
• How many meters in 1.000 km?
• Use the conversion of 1000 m = 1.000 km
Moles to Volume
(Molar Volume of a gas 22.4)
Moles to Molecules
Moles to Mass
(use Molar Mass)
Simple Conversions:
- Mole / Mass Conversions -
Use the Molar Mass of a substance to convert from Moles to Mass and Mass to Moles
80. g CuSO4
1 mol CuSO4
Mass to Moles
= 0.50 mol CuSO4
159.5 g CuSO4
0.50 mol CuSO4
159.5 g CuSO4
Moles to Mass
= 80. g CuSO4
1 mol CuSO4
Simple Conversions:
- Mole / Molecule Conversions -
Use Avogadro’s Number : 6.022 x 1023 molecules (mc) in one mole of the substance
2 mol CuSO4
6.022x1023 (mc) CuSO4
Moles to (mc)
= 1.2x1024 (mc) CuSO4
1 mol CuSO4
1.2x1024 (mc) CuSO4
1 mol CuSO4
(mc) to Moles
= 2 mol CuSO4
6.022x1023 (mc) CuSO4
Let’s use it, but first…
• You needed to calculate q (heat change) for NaOHand Na2S2O3. Lets use NaOH as our example:
• qNaOH cannot be found directly, but qH2O can be
• qNaOH = - (qH2O) (this is the conservation of energy)
• qH2O = msΔT
• m = 50.00 g
• s = 4.184 J/goc
• ΔT =20.00 oc
• qH2O = (50.00 g)(4.184 J/goc)(20.00oc)
• qH2O = 4,184 J
• qNaOH = -4,184 J
• Now, how many grams of NaOH were used to produce this amount of heat? (for my data it was 1.24 g NaOH)
First Law of Thermodynamics!
Energy lost = Energy gained
q = -q
Calculate Molar Heat of Solution
• For a 7 on the Heat Change Lab you were asked to calculate the molar heat change
• In my reaction example, with 1.24 g NaOHI produced a value of qNaOH= -4,184 J
• So, lets write this in units of Joules per gram
• qNaOH =
• You need to find Joules per mole (molar heats of solution) so that you can relate to the heat production of the other salt.
• NaOH(s) Na+(aq) + OH-(aq) qNaOH = -134 kJ/mol
• Na2S2O3(s) 2 Na+(aq) + S2O32-(aq) qNa2S2O3 = + or - ??
Enthalpy: Thermochemical Equations
NaOH(s)
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
Na+(aq) + OH-(aq)
134 kJ are released for every 1 mole of sodium hydroxide that is dissolved into water.
• NaOH(s) Na+(aq) + OH-(aq) qNaOH= -134 kJ/mol
• ΔH = -134 kJ/mol | 1,376 | 4,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-43 | latest | en | 0.810599 |
https://www.mathalino.com/tag/forum/geometric-progression | 1,709,450,855,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476205.65/warc/CC-MAIN-20240303043351-20240303073351-00880.warc.gz | 853,220,706 | 11,286 | # Geometric Progression
## Calculate $0.5 + 0.5^2 + 0.5^3 + \ldots + 0.5^{20}$
Using your calculator, find the value of
$0.5 + 0.5^2 + 0.5^3 + \, ... \, + 0.5^{20}$ | 81 | 166 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-10 | latest | en | 0.363932 |
http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/Digi_Img_Pro/chapter_1/1_1b.html | 1,386,220,129,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163040022/warc/CC-MAIN-20131204131720-00078-ip-10-33-133-15.ec2.internal.warc.gz | 126,768,950 | 3,333 | Module 1: Concept of Visual Information
Common values
There are standard values for the various parameters encountered in digital image processing. These values can be caused by video standards, by algorithmic requirements, or by the desire to keep digital circuitry simple. Table 1 gives some comm
Parameter Symbol Typical values Rows N 256,512,525,625,1024,1035 Columns M 256,512,768,1024,1320 Gray Levels L 2,64,256,1024,4096,16384
Table 1: Common values of digital image parameters
Quite frequently we see cases of M=N=2k where .This can be motivated by digital circuitry or by the use of certain algorithms such as the (fast) Fourier transform.
The number of distinct gray levels is usually a power of 2, that is, where B is the number of bits in the binary representation of the brightness levels. When we speak of a gray-level image; when we speak of a binary image. In a binary image there are just two gray levels which can be referred to, for example, as "black" and "white" or "0" and "1".
Suppose that a continuous image is approximated by equally spaced samples arranged in the form of an array as:
(1)
Each element of the array refered to as "pixel" is a discrete quantity. The array represents a digital image.
The above digitization requires a decision to be made on a value for N a well as on the number of discrete gray levels allowed for each pixel.
It is common practice in digital image processing to let N=2n and G = number of gray levels = . It is assumed that discrete levels are equally spaced between 0 to L in the gray scale.
Therefore the number of bits required to store a digitized image of size is In other words a image with 256 gray levels (ie 8 bits/pixel) required a storage of bytes.
The representation given by equ (1) is an approximation to a continuous image.
Reasonable question to ask at this point is how many samples and gray levels are required for a good approximation? This brings up the question of resolution. The resolution (ie the degree of discernble detail) of an image is strangely dependent on both N and m. The more these parameters are increased, the closer the digitized array will approximate the original image.
Unfortunately this leads to large storage and consequently processing requirements increase rapidly as a function of N and large m. | 514 | 2,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2013-48 | latest | en | 0.915136 |
http://www.delorie.com/gnu/docs/calc/calc_59.html | 1,532,180,541,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00012.warc.gz | 453,800,930 | 2,248 | www.delorie.com/gnu/docs/calc/calc_59.html search
GNU Emacs Calc 2.02 Manual
[ < ] [ > ] [ << ] [ Up ] [ >> ] [Top] [Contents] [Index] [ ? ]
### 3.7.25 List Tutorial Exercise 7
Here's one solution. First, compute the triangular list from the previous exercise and type 1 - to subtract one from all the elements.
```1: [ [0], [0, 1], [0, 1, 2], ... 1 - ```
The numbers down the lefthand edge of the list we desire are called the "triangular numbers" (now you know why!). The nth triangular number is the sum of the integers from 1 to n, and can be computed directly by the formula n * (n+1) / 2.
```2: [ [0], [0, 1], ... ] 2: [ [0], [0, 1], ... ] 1: [0, 1, 2, 3, 4, 5] 1: [0, 1, 3, 6, 10, 15] . . v x 6 RET 1 - V M ' \$ (\$+1)/2 RET ```
Adding this list to the above list of lists produces the desired result:
```1: [ [0], [1, 2], [3, 4, 5], [6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19, 20] ] . V M + ```
If we did not know the formula for triangular numbers, we could have computed them using a V U + command. We could also have gotten them the hard way by mapping a reduction across the original triangular list.
```2: [ [0], [0, 1], ... ] 2: [ [0], [0, 1], ... ] 1: [ [0], [0, 1], ... ] 1: [0, 1, 3, 6, 10, 15] . . RET V M V R + ```
(This means "map a V R + command across the vector," and since each element of the main vector is itself a small vector, V R + computes the sum of its elements.)
webmaster donations bookstore delorie software privacy Copyright © 2003 by The Free Software Foundation Updated Jun 2003 | 550 | 1,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-30 | longest | en | 0.851108 |
https://bioedonline.org/lessons-and-more/lessons-by-topic/forces-and-motion/ | 1,720,807,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00855.warc.gz | 104,460,206 | 16,947 | Forces and Motion
Potential energy builds in rollercoaster cars as they move up, and is transferred to kinetic energy as they race down.
Mass, acceleration, change, momentum, Newton's laws, push, pull, friction, kinetic, potential
• 3...2...1...Puff! (rocket stability)
Lesson
Students learn about rocket stability by constructing and flying small paper rockets.
• Arm Model
Lesson
Students construct a model arm and learn how muscles and bones work together to achieve efficient movement.
• Balloon Blast
Lesson
Students devise a plan to measure the distance of a balloon’s flight, predict the direction a balloon will travel as it deflates, and learn about Newton's Laws of Motion.
• Center of Gravity
Lesson
Students learn about center of gravity and how the body adjusts to the force of gravity to remain balanced.
• Heavy Lifting (launching with payloads)
Lesson
Students design and build balloon-powered rockets with which to launch a payload to the classroom ceiling.
• Hero Engine (third law of motion)
Lesson
Students use the thrust produced by falling water to investigate Newton’s Third Law of Motion.
• Launch Altitude Tracker
Lesson
Students construct a launch altitude tracker and use it, along with basic mathematics, to indirectly measure the altitude achieved by their rockets.
• Newton Car (second law of motion)
Lesson
Students investigate the relationship among mass, acceleration and force, as described in Newton's Second Law of Motion.
Lesson
Students design and construct advanced, high-powered paper rockets for specific flight missions.
• Paper Rockets - Basic
Lesson
Students design and construct simple, three-sided rockets.
• Paper Rockets - Intermediate
Lesson
Students construct paper rockets, fly them using the launchers built earlier, evaluate their rockets' flight, and modify their designs to improve flight performance.
• Rocket Cars: Rocket Races
Lesson
Students learn about Newton’s Three Laws of Motion as they construct and test a lightweight “rocket” car propelled by the action/reaction force of air escaping from an inflated balloon.
• Rocket Launcher (for all paper rockets)
Lesson
Students create an air pressure launcher for use with their paper rockets.
• Water Rocket Launcher (teacher only)
Article
Instructions for building and operating a platform from which to launch water rockets used in student experiments.
• Water Rockets
Lesson
Student teams construct water rockets from empty two-liter soft drink bottles, and qualify their rockets for flight by conducting string tests.
• Wind Tunnel (performance tester)
Lesson
Students construct a wind tunnel to test the performance of their paper rockets, and use data generated from the test to help them design better rockets.
• X-51 Project
Lesson
Students apply rocket science principles to design, construct, test, and launch rockets using a real-world problem-solving simulation.
• Force, Motion, Friction and Energy
Student teams use the 2016 Hess Toy Truck and Dragster in a series of hands-on classroom activities that stimulate exploration of STEM concepts and engage students in a powerful STEM learning experience that is enriching and fun! (7 activities)
• Introduction to Force and Motion
Discover a way to introduce force and motion to your students by examining the science of rocketry. (3 contact hours)
Log-in Enter Module >>
• Science Concepts Explained
Fundamental science concepts explained quickly in a fun and entertaining way.
• Simple and Compound Machines
Student teams use the 2017 Hess Dump Truck and Loader to explore science, technology, engineering and mathematics concepts. Hands-on activities engage students in powerful STEM learning experiences that are captivating and exciting! (8 activities) | 726 | 3,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-30 | latest | en | 0.934441 |
https://www.enotes.com/homework-help/x-2-12x-12-x-3-4x-write-partial-fraction-608832?en_action=hh-question_click&en_label=hh-sidebar&en_category=internal_campaign | 1,603,319,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878633.8/warc/CC-MAIN-20201021205955-20201021235955-00579.warc.gz | 719,283,299 | 10,609 | # `(x^2 + 12x + 12)/(x^3 - 4x)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.
mathace | Certified Educator
calendarEducator since 2015
starTop subject is Math
`[x^2+12x+12]/[x^3-4x]=[x^2+12x+12]/[x(x+2)(x-2)]`
`[x^2+12x+12]/(x^3-4x)=A/x+B/(x+2)+C/(x-2)`
Multiply through by the LCD `x^3+4x.`
`x^2+12x+12=A(x^2-4)+Bx(x-2)+Cx(x+2)`
`x^2+12x+12=Ax^2-4A+Bx^2-2Bx+Cx^2+2Cx`
`x^2+12x+12=(A+B+C)x^2+(-2B+2C)x+(-4A)`
Equate coefficient of like terms. Then solve for A, B, and C.
`1=A+B+C`
`12=-2B+2C`
`12=-4A`
`A=-3`
`1=A+B+C`
`1=-3+B+C`
`4=B+C`
Solve for B and C using the elimination method.
`4=B+C`
`12=-2B+2C`
Multiply the first equation by 2. Then solve using the elimination method.
`8=2B+2C`
`12=-2B+2C`
_________________
20=4C
`C=5`
`1=A+B+C`
`1=-3+B+5`
`B=-1`
`A=-3, B=-1, C=5`
`(x^2+12x+12)/(x^3-4x)=-3/x-1/(x+2)+5/(x-2)`
check Approved by eNotes Editorial | 439 | 946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-45 | latest | en | 0.566146 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.