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Asked by Prashant DIGHE 7th December 2019, 10:22 PM
Let the wire bent 90° be on the y and x-axis as shown in figure.
Let P be the point on z-axis at a distance 35 cm from bend point of wire.
At point P, we have magnteic flux density Bx due to wire lying on y-axis .
Similarly we have magnetic flux density By due to wire lying on x-axis.
| Bx | = | By | = μo i / (4π × 0.35 ) = (1/7) × 10-6 T
Resutant magnetic flux density = { Bx2 + By2 }1/2 = ( √2/7 )× 10-6 T ≈ 2 μT
Answered by Expert 8th December 2019, 4:05 PM
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Forum Index > Test Paper Related > Primary 5 Matters p5 maths problem sum
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By: lisatang (offline) Tuesday, April 24 2012 @ 10:52 AM CDT (Read 1662 times)
lisatang
Dear all
May I check is there a better or shorter way to solve this question?
Junior
Registered: 01/26/12
Posts: 19
By: MathIzzzFun (offline) Tuesday, April 24 2012 @ 05:26 PM CDT
MathIzzzFun
Quote by: lisatang
Dear all
May I check is there a better or shorter way to solve this question?
Hi
this is a triangular pattern.
Pattern 1 = 1
Pattern 2 = 1 + 2
Pattern 3 = 1+2+3
Pattern N = 1+ 2+3+4+ ...+N = N (N+1)/2
Pattern 48 = 48 x 49 /2
cheers.
Chatty
Registered: 04/02/10
Posts: 36
By: lisatang (offline) Thursday, April 26 2012 @ 11:24 PM CDT
lisatang
Hi
Thanks for your help but does primary school student know this?
regards
lisa
Junior
Registered: 01/26/12
Posts: 19
By: MathIzzzFun (offline) Sunday, April 29 2012 @ 11:06 PM CDT
MathIzzzFun
Quote by: lisatang
Hi
Thanks for your help but does primary school student know this?
regards
lisa
Hi
the students are taught how to add 1+2+3+4.... N, but may not be taught the general formula ie N(N+1)/2 - students learned this in maths olympiad training though.
However, G&C are taught so students can use the "look for pattern" heuristics see that the sum is N(N+1)/2 eg. Pattern 1 = 1
Pattern 2 = 1+2 = 3 = 2x3/2, Pattern 3 = 1+2+3 = 6 = 3x4/2.. then using G&C, they can then get the final answer ie 48
cheers.
Chatty
Registered: 04/02/10
Posts: 36
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View Anonymous Posts Able to Post HTML Allowed Censored Content | 669 | 2,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-34 | longest | en | 0.803585 |
http://mathhelpforum.com/math-topics/44049-2-part-question-i-ve-done-first-bit-but-i-don-t-understand-second-bit.html | 1,527,205,193,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866894.26/warc/CC-MAIN-20180524224941-20180525004941-00050.warc.gz | 187,777,518 | 11,133 | # Thread: 2 Part Question: I've done the first bit but I don't understand the second bit
1. ## 2 Part Question: I've done the first bit but I don't understand the second bit
a) Find the equation of the normal to y = x^2 / 3 at the point where x = 1
b) This normal cuts the coordinate axis at P and Q. Given that P and Q are adjacent vertices of a rhombus, and that the diagonals of this rhombus intersect at the origin, determine the area of this rhombus
[I don't understand this part and I'm not too sure on what to do can someone please help]
My working out for part a [Please tell me if i made a mistake]
y = x^2 / 3
dy/dx = (x^2 x 0) - (3 x 2x)
_________________
9
= x^2 - 6x
________
9
when x = 1
m(tangent) = -5/9
So m(normal) = 9/5
Therefore Equation of normal = y = 9/5x - 22/15
[Sorry I missed a few lines and skipped straight to the final answer]
2. Yeah, that's not right. Sorry.
Looks like you're trying to use the quotient rule, which is making things difficult. Three problems with what you've done:
• The top should be $\displaystyle (3 \cdot 2x) - (x^2 \cdot 0)$. You've got it backwards.
• $\displaystyle x^2 \cdot 0 = 0$, not $\displaystyle x^2$.
• You've forgotten to square the bottom.
An easier way to do this might be to look at the 'dividing by three' as a coefficient:
$\displaystyle \frac{1}{3}x^2$
3. Hello, sweetG!
a) Find the equation of the normal to $\displaystyle y \:= \:\frac{1}{3}x^2$ where $\displaystyle x = 1$
When $\displaystyle x = 1,\;y = \frac{1}{3}$
We have: .$\displaystyle y' \:=\:\frac{2}{3}x$
When $\displaystyle x = 1$, the slope of the tangent is: .$\displaystyle m \:=\:\frac{2}{3}$
. . Hence, the slope of the normal is: $\displaystyle -\frac{3}{2}$
The equation of the normal is: .$\displaystyle y - \frac{1}{3}\:=\:-\frac{3}{2}(x-1) \quad\Rightarrow\quad \boxed{y \:=\:-\frac{3}{2}x + \frac{11}{6}}$
b) This normal cuts the coordinate axis at P and Q.
Given that P and Q are adjacent vertices of a rhombus,
and that the diagonals of this rhombus intersect at the origin,
determine the area of this rhombus.
The intercept of the normal are: .$\displaystyle P\left(\frac{11}{9},\:0\right)\,\text{ and }\,Q\left(0,\:\frac{11}{6}\right)$
If the diagonals intersect at the origin, the rhombus looks like this:
Code:
|
Q o (0,11/6)
* | *
* | *
R * | * P
- - - o - - - - - + - - - - - o - - -
(-11/9,0) * | * (11/9,0)
* | *
* | *
S o (0,-11/6)
|
The area of a rhomus is one-half the product of its diagonals.
The horizontal diagonal has length: .$\displaystyle 2 \times \frac{11}{9} \:=\:\frac{22}{9}$
The vertical diagonal has length: .$\displaystyle 2 \times \frac{11}{6} \:=\:\frac{11}{3}$
The area is: .$\displaystyle \frac{1}{2} \times \frac{22}{9} \times \frac{11}{3} \;=\;\boxed{\frac{121}{27}\text{ units}^2}$
*
4. Thanks so much guys !
=)
that really helped and wow i made a lot of silly mistakes there
and I just realized how easy this question really was I just wasn't thinking hard enough Thanks once again ! | 993 | 3,067 | {"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} | 4.34375 | 4 | CC-MAIN-2018-22 | latest | en | 0.849299 |
http://www.math-only-math.com/division-of-algebraic-fractions.html | 1,500,778,503,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424239.83/warc/CC-MAIN-20170723022719-20170723042719-00617.warc.gz | 509,393,291 | 8,569 | Division of Algebraic Fractions
To solve the problems on division of algebraic fractions we will follow the same rules that we already learnt in dividing fractions in arithmetic.
From division of fractions we know,
First fraction ÷ Second fraction = First fraction × $$\frac{1}{Second fraction}$$
In algebraic fractions, the quotient can be determined in the same way i.e.
First algebraic fraction ÷ Second algebraic fraction
= First algebraic fraction × $$\frac{1}{Second algebraic fraction}$$
1. Determine the quotient of the algebraic fractions: $$\frac{p^{2}r^{2}}{q^{2}s^{2}} \div \frac{qr}{ps}$$
Solution:
$$\frac{p^{2}r^{2}}{q^{2}s^{2}} \div \frac{qr}{ps}$$
= $$\frac{p^{2}r^{2}}{q^{2}s^{2}} \times \frac{ps}{qr}$$
= $$\frac{p^{2}r^{2} \cdot ps}{q^{2}s^{2} \cdot qr}$$
= $$\frac{p^{3}r^{2}s}{q^{3}rs^{2}}$$
In the numerator and denominator of the quotient, the common factor is ‘rs’ by which if the numerator and denominator are divided, its lowest form will be = $$\frac{p^{3}r}{q^{3}s}$$
2. Find the quotient of the algebraic fractions: $$\frac{x(y + z)}{y^{2} - z^{2}} \div \frac{y + z}{y - z}$$
Solution:
$$\frac{x(y + z)}{y^{2} - z^{2}} \div \frac{y + z}{y - z}$$
= $$\frac{x(y + z)}{y^{2} - z^{2}} \times \frac{y - z}{y + z}$$
= $$\frac{x(y + z)}{(y + z)(y - z)} \times \frac{y - z}{y + z}$$
= $$\frac{x(y + z) \cdot (y - z)}{(y + z)(y - z) \cdot (y + z)}$$
= $$\frac{x(y + z)(y - z)}{(y + z)(y - z)(y + z)}$$
We observe that the common factor in the numerator and denominator of the quotient is (y + z) (y – z) by which, if the numerator and the denominator are divided, its lowest form will be $$\frac{x}{y + z}$$.
3. Divide the algebraic fractions and express in the lowest form:
$$\frac{m^{2} - m - 6}{m^{2} + 4m - 5} \div \frac{m^{2} - 4m + 3}{m^{2} + 6m + 5}$$
Solution:
$$\frac{m^{2} - m - 6}{m^{2} + 4m - 5} \div \frac{m^{2} - 4m + 3}{m^{2} + 6m + 5}$$
= $$\frac{m^{2} - m - 6}{m^{2} + 4m - 5} \times \frac{m^{2} + 6m + 5}{m^{2} - 4m + 3}$$
= $$\frac{m^{2} - 3m + 2m - 6}{m^{2} + 5m - m - 5} \times \frac{m^{2} + 5m + m + 5}{m^{2} - 3m - m + 3}$$
= $$\frac{m(m - 3) + 2(m - 3)}{m(m + 5) - 1(m + 5)} \times \frac{m(m + 5) + 1(m + 5)}{m(m - 3) - 1(m - 3)}$$
= $$\frac{(m - 3)(m + 2)}{(m + 5) (m - 1)} \times \frac{(m + 5) (m + 1)}{(m - 3) (m - 1)}$$
= $$\frac{(m - 3)(m + 2) \cdot (m + 5) (m + 1)}{(m + 5) (m - 1) \cdot (m - 3) (m - 1)}$$
= $$\frac{(m - 3)(m + 2)(m + 5) (m + 1)}{(m + 5) (m - 1)(m - 3) (m - 1)}$$
We observe that the common factor in the numerator and denominator of the quotient is (m - 3) (m + 5), by which if the numerator and the denominator of the quotient is divided, $$\frac{(m + 2) (m + 1)}{(m - 1) (m - 1)}$$ i.e. $$\frac{(m + 2) (m + 1)}{(m - 1)^{2}}$$ will be its reduced lowest form. | 1,177 | 2,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2017-30 | longest | en | 0.607195 |
https://www.hollywoodreporter.com/lists/oscars-math-points-a-spider-man-animated-feature-victory-1188600/item/best-original-song-oscars-math-points-a-spider-man-animated-feature-victory-1188611 | 1,563,214,413,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715200238-00027.warc.gz | 717,422,535 | 14,804 | # Oscars: The Math Points to a 'Spider-Man' Animated Feature Victory
6:15 AM 2/21/2019
by Ben Zauzmer
## Lady Gaga's "Shallow" has the best chance of collecting the best original song Academy Award.
So which movie has the best chance of winning the best animated feature Oscar, and does any film have a lock on the documentary feature trophy at Sunday night's 91st Oscars?
In yesterday's article, I announced percentages for each nominee in the eight most prominent Oscar categories: best picture, best director, the four acting awards and the two screenplay awards. Those are the categories that people pay the most attention to.
In the final two portions of my eighth annual Oscar predictions, we arrive at the categories that draw less attention, which has a number of implications. For one, there are fewer precursors, making some of them harder to predict. Two, math is arguably more important in these categories since there are fewer indicators pointing the way, so we need to handle each one correctly. Three, and perhaps of most interest, these are the categories where Oscar pools are won and lost.
I make no promises that these percentages will help win your Oscar pools; upsets happen every year. But if you'd like to know who the mathematical favorites are before making your picks, this is the article for you.
Tomorrow: Film Editing, Visual Effects, Costume Design, Makeup and Hairstyling, Sound Mixing and Sound Editing.
Ben Zauzmer (@BensOscarMath) uses math to predict and write about the Oscars for The Hollywood Reporter. A Harvard graduate with a degree in applied math, he works as a baseball analyst for the Los Angeles Dodgers.
• # Best Animated Feature
Spider-Man: Into the Spider-Verse flew through awards season, picking up wins from the Golden Globes, Producers Guild, and BAFTAs, among many others. Both math and common sense point to it as the film to beat.
But this win isn’t as automatic as some would suggest. Wes Anderson’s Isle of Dogs quietly scooped up a number of honors that come with less fanfare but are still somewhat predictive of this category, including quite a few critics’ circle nods. You should still probably pick Spider-Man in your Oscar pool, but the math says this one remains a race.
• # Best Documentary Feature
Last year, when 20 of the 21 films my model identified as the favorites won their Oscars, the lone category to witness an upset was best documentary feature. What happened? Jane won most of the major precursors before surprisingly being left off the Academy’s nominee list, leaving the model relatively little data to work with.
This year, we may have the same situation. Won’t You Be My Neighbor? took home many prizes throughout awards season yet wasn’t invited to the Oscars. That left RBG and Free Solo fighting for any unclaimed trophies, and statistics is basically shrugging on this one. National Board of Review winner RBG emerges just 1.7 percentage points ahead in a thrilling battle.
• # Best Foreign Language Film
Why isn’t Roma at 100 percent? How could a film be the only foreign language entrant among the best eight movies of the year and yet not be the year’s best foreign-language film?
The Academy may have to answer precisely that question if Cold War, whose director Pawel Pawlikowski is also nominated for best director, pulls off the upset. It could come down to different groups of people voting for nominees versus winners. Or perhaps voters will anticipate Roma wins in other categories and look to spread the wealth. But in all likelihood, the logical conclusion that Roma is the best foreign language film will hold true Sunday night.
• # Best Production Design
The Art Directors Guild doesn’t make this easy, as they split their top live-action film award in three by genre. Crazy Rich Asians won the contemporary category but wasn’t nominated at the Oscars. Black Panther received the fantasy honor. The Favourite earned its trophy in the period film category.
So why does the math prefer The Favourite over Black Panther? Part of it has to do with the categories themselves, as historically more period film winners than fantasy film winners go on to repeat at the Oscars. But the bigger part is the other awards shows, including the BAFTAs, which tended to favor The Favourite, though not overwhelmingly so.
• # Best Cinematography
Alfonso Cuarón already made history by being nominated for directing and cinematography in the same year, and now he is the odds-on favorite to win both. It’s an intriguing group with more nominees who filmed foreign-language nominees than English-language ones, and the American Society of Cinematographers chose another one of those foreign-language contenders, Cold War.
But the math is more impressed by Roma’s BAFTA win and its sheer number of nominations. Not since Legends of the Fall (1994) has a film won best cinematography with fewer than five total nominations.
• # Best Original Score
When all five percentages are in double-digits, that’s the math’s way of saying that it has relatively little confidence on this one. First Man’s score won the Golden Globe and Critics Choice Awards, while A Star Is Born earned best original music at the BAFTAs, and neither was nominated for the Oscar.
Among the nation’s critic circles, however, a shaky consensus emerged in favor of Nicholas Britell’s score for If Beale Street Could Talk. That said, if you’re trying to win an Oscar pool and you anticipate a number of your friends are skimming this article and picking all the favorites, this isn’t a bad category to try and distinguish your picks, in hopes of getting one right that others may miss.
• # Best Original Song
Though Lady Gaga’s hit “Shallow” lost the Grammys for song and record of the year — although it did win best song written for visual media and she and Bradley Cooper won for best pop duo/group performance — here it faces no competition from Childish Gambino’s Grammy-winning “This Is America.” This is one of the rare categories in which one nominee has led wire-to-wire: from the moment the film premiered, “Shallow” has been dubbed the song to beat, and no awards since then have changed anyone’s mind.
If another song does manage to take down “Shallow,” look to fellow Grammy nominee “All the Stars” by Kendrick Lamar and SZA for Black Panther as the most likely victor. | 1,324 | 6,361 | {"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} | 2.671875 | 3 | CC-MAIN-2019-30 | latest | en | 0.94721 |
http://math.stackexchange.com/questions/97147/open-sets-and-poincares-inequality | 1,394,327,404,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999669442/warc/CC-MAIN-20140305060749-00074-ip-10-183-142-35.ec2.internal.warc.gz | 117,830,608 | 15,112 | # Open sets and Poincaré's inequality
In many references, Poincaré inequality is presented in the following way :
Let $\Omega\subset \mathbb R^d$ an open bounded set. We can find a constant $C$ which depend of $\Omega$ such that for all $u\in H^1_0(\Omega)$, we have $$\lVert u\rVert_{L^2}\leq C\lVert \nabla u\rVert_{(L^2(\Omega))^d}.$$
In fact it works if $\Omega$ is bounded in one direction. An other sufficient condition is that we can find $v\neq 0$ such that Lebesgue measure of $\{\lambda\in\mathbb R,\lambda v\in \Omega\}$ is finite).
My question, maybe a little vague, is the following: is there a "nice" necessary and sufficient condition on $\Omega$ to have Poincaré's inequality?
-
You know Ziemers Book 'Weakly Differentiable Functions'? Chapter 4 is dedicated to Poincaré type inequalities. – user20266 Jan 7 '12 at 11:19
Yes, but when I looked at it I didn't think about this question. And some pages are missing in Gooble book (which is normal). Anyway, this book is at the library of my university, so I will have a look at it Monday. – Davide Giraudo Jan 7 '12 at 11:29
One generalization I know from one of my teachers can be found here: mathproblems123.wordpress.com/2011/10/05/… This needs $\Omega$ to have Lipschitz boundary, and increases the space of admissible functions $H_0^1(\Omega)$ to a closed subspace of $H^1(\Omega)$ which does not contain the non-zero constant functions. – Beni Bogosel Jan 7 '12 at 17:38
Beni Bogosel: Thanks, I didn't know this result. @Thomas I look at this book, but I didn't find the answer. Maybe should I ask it at MathOverfow. – Davide Giraudo Jan 11 '12 at 10:40
@DavideGiraudo: I suppose such condition can be that $\Omega$ is regular enough such that the Rellich Kondrachov theorem holds. en.wikipedia.org/wiki/Rellich%E2%80%93Kondrachov_theorem – Beni Bogosel Apr 30 '12 at 8:02
show 1 more comment | 561 | 1,874 | {"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": 1, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2014-10 | latest | en | 0.903849 |
https://gretlml.univpm.it/hyperkitty/list/gretl-users@gretlml.univpm.it/message/VQOSDO5LXMXOGHPOJRCFAG4W3KRLAK5H/attachment/3/attachment.html | 1,723,024,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00505.warc.gz | 229,107,956 | 2,931 | Hi there,
I'd like to ask what the interaction operator 'makes'?
I was working on a function that generates new interaction series for each step combination of factors.
This is working for a list of factors and generates bivariate combinations as in A*B
I'm stuck with making it possible to generate A*B*C and so on.
This should be possible with a recursive function!
I think with a new gretl operato '*' the whole procedure would get much easier to implement.
I also encountered some bugs (at least on W8):
- having 'opened' a bundle and then using the '= null' command gretl crashes
- 'smpl full' does 'sometimes' not work in functions - I think if it's outside a loop.
- 'delete --type==' does not update the symbols window automatically
Here is the code for the function as well (gonna be implemented in my new multivariate statistic package).
Perhaps someone has an idea how to solve the sorting issue for more than A*B.
Cheers
Leon
04.01.2013 11:52, schrieb Pindar:
Hi,
I'm of the opinion that a "*" operator for interaction variables is a very useful means!
A few weeks ago we discussed such possibilities in terms of MANOVA.
This would be the solution to such problems of descriptive statistics and further analysis.
What should be achieved? Well, I think list creation and usage in a model command like "dummify" now.
Would be great if this functionality would be implemented!
Have nice weekend
Leon
Am 30.12.2012 13:15, schrieb Riccardo (Jack) Lucchetti:
On Sun, 30 Dec 2012, Allin Cottrell wrote:
With regard to dummify, I'm not sure that accepting a list argument
is a great idea, since a (single) series argument already produces a
list result. Producing a "list of lists" seems to me on the verge of
being out of control.
Partially OT: I'm asking myself if we should go back to an idea that we briefly considered when we introduced operators for lists, and ended up being put on hold.
The idea is to use the "*" operator for a specialised version of the Cartesian product, commony known among applied economists as "interaction" variables. To be more explicit: suppose you have two lists, X and Y, with the only requirement that X should contain discrete variables only. Then, the construct "X*Y" would perform what is accomplished through the following script.
<hansl>
open keane.gdt
list X = choice status
list Y = wage educ
loop foreach j X --quiet
matrix v = values(\$j)
scalar n = rows(v)
loop foreach i Y --quiet
loop for k=1..n --quiet
xi = v[k]
sprintf vname "\$i_\$j_%d", xi
series @vname = (\$j == xi)*\$i
end loop
end loop
end loop
</hansl>
Note that the dummify command/function would be a special case of this for "Y=const". I have the feeling that people working with micro data would love this.
Thoughts?
-------------------------------------------------------
Riccardo (Jack) Lucchetti
Dipartimento di Scienze Economiche e Sociali (DiSES)
Università Politecnica delle Marche
(formerly known as Università di Ancona)
r.lucchetti@univpm.it
http://www2.econ.univpm.it/servizi/hpp/lucchetti
-------------------------------------------------------
```_______________________________________________
Gretl-users mailing list
Gretl-users@lists.wfu.edu
http://lists.wfu.edu/mailman/listinfo/gretl-users``` | 792 | 3,248 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-33 | latest | en | 0.936199 |
https://www.c-sharpcorner.com/blogs/accessing-a-tuple-value-to-another-tuple | 1,521,823,338,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648404.94/warc/CC-MAIN-20180323161421-20180323181421-00532.warc.gz | 746,963,165 | 11,942 | # Accessing A Tuple Value To Another Tuple
Introduction
In this blog, I will explain about accessing a Tuple value to another Tuple. I will try to access the value of the Tuple to another Tuple.
Software Requirement
Python 3.5.2
Simple Programming
1. print("Welcome To C#Corner...........")
2. # Accessing a Tuple Through Another Tuple
3.
4. a=("Zero","First","Second","Third")
5. b=(a,"B's 1st Element")
6. print("%s" % b[1])
7. print("%s" % b[0][0])
8. print("%s" % b[0][1])
9. print("%s" % b[0][2])
10. print("%s" % b[0][3])
Explanation
In this blog, I will explain about accessing the value through another tuple. In the first code, I will assign the "a" value and in the second code, I will display the "a" value by the code in the "a" value.
Output | 238 | 762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | longest | en | 0.574834 |
https://www.jiskha.com/display.cgi?id=1367660281 | 1,503,407,669,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110774.86/warc/CC-MAIN-20170822123737-20170822143737-00357.warc.gz | 907,391,857 | 3,883 | # science - physics
posted by .
A potential charge of magnitude 2.0uC is moved between two points X and Y Point X is at a potential of 6.0V and Point Y is at a potential of 9.0 V Find the gain in potential energy of Point charge
## Similar Questions
1. ### Physics
An electric force moves a charge of +1.40 x 10^-4 C from point A to point B and performs 4.70 x 10^-3 J of work on the charge. (a) What is the difference (EPEA - EPEB) between the electric potential energies of the charge at the two …
2. ### physics electricity
An electric force moves a charge of +1.65 multiplied by 10-4 C from point A to point B and performs 6.40 multiplied by 10-3 J of work on the charge. (a) What is the difference (EPEA - EPEB) between the electric potential energies of …
3. ### Physics
A)Two point charges, +3.90 ìC and -7.50 ìC, are separated by 3.30 m. What is the electric potential midway between them?
4. ### physics
the elecric potential at a certain point is space is 12 V. What is the electric potential energy of a -3.0uc charge placed at that point?
5. ### physics
Point X and Y are 30.0 mm and 58.0 mm away from a charge of +8.0 C. a. How much work must be done in moving a +2.0 C charge from point Y to point X?
6. ### physics, help! urgent!
A positive point charge q = +2.50 nC is located at x = 1.20 m and a negative charge of ƒ{2q = ƒ{5.00 nC is located at the origin. (a) Sketch the electric potential verses x for points on the x-axis in the range -1.50 m < x < …
7. ### Physics
Electric charge charge is distributed uniformly along a thin rod of length "a", with total charge Q. Take the potential to be zero at infinity find the potential at P. If z>>a, what would be the potential at p?
8. ### physics
The potential energy of a +2 x 10-6 C charge decreases from 0.15 J to 0.05 J when it is moved from point A to point B. What is the change in electric potential between these two points? | 549 | 1,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.934225 |
http://fmwww.bc.edu/repec/bocode/_/_ghvar.ado | 1,702,145,028,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00559.warc.gz | 17,170,215 | 1,714 | * This is an example of a program that can be used to estimate other RIFs that are not yet available in rifvar * but the user may still want to use it with the oaxaca_rif and rifhdreg command. * This program is meant to estimate the RIF of what i call Half variance: * half variance positive hvp(x)=sum((X-E(X))^2*(X>E(X))) * half variance negative hvn(x)=sum((X-E(X))^2*(X(`mn'[_N]/`tt'[_N]))) } if `option'==2 { by `touse' `by':egen double `hvar'=sum((`exp'-`mn'[_N]/`tt'[_N])^2*`weight'/`tt'[_N]*(`exp'<(`mn'[_N]/`tt'[_N]))) } ** if Function tempvar N n jkvar jfrif gen double `jfrif'=. by `touse' `by':gen double `N'=_N by `touse' `by':gen double `n'=_n qui:sum `N' local nmax=r(max) ** JK estimates the statistic for ALL statistics excluding observation i forvalues i=1/`nmax' { capture drop `mn' capture drop `t' capture drop `jkvar' by `touse' `by':egen double `mn'=sum(`exp'*`weight'*(`n'!=`i')) by `touse' `by':egen double `t'=sum(`weight'*(`n'!=`i')) if `option'==1 { by `touse' `by':egen double `jkvar'=sum((`exp'-`mn'/`t')^2*`weight'/`t'*(`exp'>(`mn'/`t'))*(`n'!=`i')) } if `option'==2 { by `touse' `by':egen double `jkvar'=sum((`exp'-`mn'/`t')^2*`weight'/`t'*(`exp'<(`mn'/`t'))*(`n'!=`i')) } replace `jfrif'= `hvar'+(`hvar'-`jkvar')*`tt'[_N]/`weight' if `n'==`i' } } ** saving results qui: by `touse' `by': gen `typlist' `varlist' = `jfrif' if `touse' if `option'==1 { label var `varlist' "RIF for Positive Half variance of `exp'" } if `option'==2 { label var `varlist' "RIF for Negative Half variance of `exp'" } end | 576 | 1,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-50 | latest | en | 0.578455 |
fkdservices.com | 1,721,086,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00546.warc.gz | 227,847,355 | 34,023 | # How to Use Regular Expression in Eloqua to Format Phone numbers
As an Eloqua user, you will occasionally have a need to cleanse or format your data to make it usable for campaigning. In this article, we’ll cover how to use regular expression in Eloqua to format phone numbers for an SMS campaign.
This comes in handy when outputting mobile numbers and prefixing it with the country code.
To be able to do this, you require the Contact Washing Machine cloud app installed on your instance. The app is a must for anyone using Eloqua and greatly extends Eloqua’s functionality when it comes to manipulating data. You can find out about all the app action here
We’ll be using Regular Expression (Regex) within the Contact Washing Machine App to format our data.
For those who are new to Regex, I would advise you to have a quick look at https://regex101.com/ for more information and also to give you a platform to test your own scripts.
## Challenge
Format 07999999999 into +447999999999
Oracle provides some guidelines on regex but you will need to get your head around it first before you can make it work for you.
The example below picks up the number in the source field and breaks it into different components that allow it to be manipulated
• Source field`+1111234567891`
• Regular expression to find`^(\\+[\\d]{1,3}|0)?(\\d{3})(\\d{3})(\\d{4})\$`
• Regular expression replace`(\$1) \$2 \$3-\$4`
• Destination field result`(+111) 123 456-7891`
## Using Regular Expression in Eloqua Contact Washing Machine
The key is to break the number in the source field into bite-size chunks then write an expression to alter each of the sections.
• Source field07999999999
• Regular expression to find([\d]{1})(\d{10})\$
• Regular expression replace+44\$2
• Destination field result+447999999999
The source text has 11 characters and the goal is to split it into two sections. ([\d]{1}) isolates the first character (0) and the (\d{10}) isolates the rest of the numbers.
• ([\d]{1}) = 0
• ([\d]{10}) = 7999999999
• \$1\$2 = 07999999999
## Final Output
Final step is to replace the \$1 part with the +44
+44\$2 = +447999999999 | 535 | 2,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-30 | latest | en | 0.870989 |
https://www.thestudentroom.co.uk/showthread.php?t=9387&page=3 | 1,531,774,498,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589455.35/warc/CC-MAIN-20180716193516-20180716213516-00532.warc.gz | 972,679,399 | 43,193 | You are Here: Home >< Maths
# What's the integral of sin(x)sinh(x)? watch
1. i did hyperbolics in P4 (AQA syllabus A)
2. How is sinh pronounced?
3. (Original post by Linda)
Nevermind, mis read your question. What do you mean exactly? sin^2(x)or what?
Er....no.
4. (Original post by ZJuwelH)
How is sinh pronounced?
Shine or I've heard it pronounced Sinch (by my lecturers)
5. (Original post by Bigcnee)
Shine or I've heard it pronounced Sinch (by my lecturers)
Shine? OK...
What about cosh (I imagine that's just "cosh") and tanh?
6. (Original post by ZJuwelH)
Shine? OK...
What about cosh (I imagine that's just "cosh") and tanh?
shine cosh tanch shec coshec coth
7. (Original post by elpaw)
shine cosh tanch shec coshec coth
Ive heard it pronounced Than.
8. (Original post by Bigcnee)
Ive heard it pronounced Than.
Those were the ways i was indoctrinated to pronounce them. i still like to use my own pronounciation, sine haych, cos haitch, tan haych, dot dot dot
9. (Original post by elpaw)
shine cosh tanch shec coshec coth
My teacher says "setch" rather than "shec"
10. (Original post by Ben.S.)
I don't have the answer to this particular, indefinite integral, and I need to know if I've got it right. Also, I can't find it anywhere on the internet!
WPGG;EBEULGJH
1/2 [sin(x)cosh(x) - cos(x)sinh(x)] + C
Thanks,
Ben
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Top tips from students who have already aced their exams | 557 | 1,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-30 | longest | en | 0.935278 |
https://www.jiskha.com/display.cgi?id=1313087954 | 1,501,132,052,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427429.9/warc/CC-MAIN-20170727042127-20170727062127-00676.warc.gz | 816,582,064 | 3,535 | # calculus
posted by .
x=2cosa-cos2a,y=2sina-sin2a
## Respond to this Question
First Name School Subject Your Answer
## Similar Questions
1. ### trig maths
rewrite 7 sin2A – 5 sin A + cos2A = o to solve for A
2. ### trig
CosA x cos2A + ((sin2A)^2)/2cosA
3. ### math
how do you solve this.prove: cos2A plus cosA divided by sin2A minus sinA equal to cos2A plus 1 divided by sinA
4. ### trig
Prove: 1/cos2A+sin2A/cos2A=sinA+cosA/cosA-sinA
5. ### trig
Prove: 1/cos2A+sin2A/cos2A=sinA+cosA/cosA-sinA
6. ### further Mathematics
Let A=(cosA -sinA) (sinA cosA) show that A^2=(cos2A sin2A) (sin2A cos2A)
7. ### MathS triG
1 .If tanA=1/3 and tanB=1/7 (both A and B are acute),calculate 50sin(2A +B) 2.1Prove that sin2A+2cosA-2cos^3A/1+sinA =sin2A 2.2 for which values of A in the interval[-360;360] is the identity in 2.1 undefined?
8. ### maths Pls help trig
1 .If tanA=1/3 and tanB=1/7 (both A and B are acute),calculate 50sin(2A +B) 2.1Prove that sin2A+2cosA-2cos^3A/1+sinA =sin2A 2.2 for which values of A in the interval[-360;360] is the identity in 2.1 undefined?
9. ### Math
If sinA=0.5 , what is the value of cos2A and sin2A?
10. ### math
prove that 2sinA + 1 / cosA+sin2A=secA
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Post a New Question | 501 | 1,237 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.666681 |
https://www.edx.org/learn/data-analysis/the-georgia-institute-of-technology-probability-and-statistics-iii-a-gentle-introduction-to-statistics?webview=false&campaign=Probability+and+Statistics+III%3A++A+Gentle+Introduction+to+Statistics&source=edx&product_category=course | 1,695,649,472,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508977.50/warc/CC-MAIN-20230925115505-20230925145505-00230.warc.gz | 818,367,203 | 163,241 | # The Georgia Institute of Technology: Probability and Statistics III: A Gentle Introduction to Statistics
This course provides an introduction to basic statistical concepts. We begin by walking through a library of probability distributions – including the normal distribution, which in turn leads to the Central Limit Theorem. We then discuss elementary descriptive statistics and estimation methods.
4 weeks
6–10 hours per week
Self-paced
Free
This course provides an introduction to basic statistical concepts.
We begin by walking through a library of probability distributions, where we motivate their uses and go over their fundamental properties.
These distributions include such important folks as the Bernoulli, binomial, geometric, Poisson, uniform, exponential, and normal distributions, just to name a few. Particular attention is paid to the normal distribution, because it leads to the Central Limit Theorem (the most-important mathematical result in the universe, actually), which enables us to make probability calculations for arbitrary averages and sums of random variables.
We then discuss elementary descriptive statistics and estimation methods, including unbiased estimation, maximum likelihood estimation, and the method of moments – you gotta love your MoM! Finally, we describe the t, X2, and F sampling distributions, which will prove to be useful in upcoming statistical applications.
### At a glance
• Institution: GTx
• Subject: Data Analysis & Statistics
• Level: Intermediate
• Prerequisites:
Learners will be expected to come in knowing a bit of set theory and basic calculus, as well as the material from the first two courses in this series (the Gentle Introduction to Probability and Random Variables courses). The prerequisite material is all available for you to access; and in any event, we will try to make the current course as self-contained as possible. In addition, this course will involve a bit of computer programming, so it would be nice to have at least a little experience in something like Excel and/or the R freeware statistical package.
• Language: English
• Video Transcript: English
• Associated programs:
• Associated skills: Sampling (Statistics), Statistics, Random Variables, Normal Distribution, Probability, Probability And Statistics, Maximum Likelihood, Probability Distribution, Basic Math
# What you'll learn
Skip What you'll learn
Upon completion of this course, learners will be able to:
• Review a library of discrete and continuous probability distributions
• Recognize the normal distribution and the Central Limit Theorem, and how they are applied in practice
• Recognize elementary methods of descriptive statistics
• Describe methods that can be used to estimate the unknown parameters of a distribution
• Identify statistical sampling distributions
# Syllabus
Skip Syllabus
“FCPS” refers to the free text, A First Course in Probability and Statistics: free access is provided via a PDF file or as a book
Module 1: Distributions
• Lesson 1: Bernoulli and Binomial Distributions (FCPS §4.1.1)
• Lesson 2: Hypergeometric Distribution (FCPS §4.1.2)
• Lesson 3: Geometric and Negative Binomial Distributions (FCPS §4.1.3)
• Lesson 4: Poisson Distribution (FCPS §4.1.4)
• Lesson 5: Uniform, Exponential, and Friends (FCPS §4.2.1–4.2.2)
• Lesson 6: Other Continuous Distributions (FCPS §4.2.3)
• Lesson 7: Normal Distribution: Basics (FCPS §4.3.1)
• Lesson 8: Standard Normal Distribution (FCPS §4.3.2)
• Lesson 9: Sample Mean of Normals (FCPS §4.3.3)
• Lesson 10: The Central Limit Theorem + OPTIONAL Proof (FCPS §4.3.4)
• Lesson 11: Central Limit Theorem Examples (FCPS §4.3.5)
• Lesson 12 [OPTIONAL]: Extensions – Multivariate Normal Distribution (FCPS §4.4.1)
• Lesson 13 [OPTIONAL]: Extensions – Lognormal Distribution (FCPS §4.4.2)
• Lesson 14: Computer Stuff, including OPTIONAL Box-Muller Proof (FCPS §4.5)
Module 2: Getting Started with Statistics
• Lesson 1: Introduction to Descriptive Statistics (FCPS §5.1.1)
• Lesson 2: Summarizing Data (FCPS §5.1.2)
• Lesson 3: Candidate Distributions (FCPS §5.1.3)
• Lesson 4: Introduction to Estimation (FCPS §5.2.1)
• Lesson 5: Unbiased Estimation (FCPS §5.2.2)
• Lesson 6: Mean Squared Error (FCPS §5.2.3)
• Lesson 7: Maximum Likelihood Estimation (FCPS §5.2.4)
• Lesson 8: Trickier MLE Examples (FCPS §5.2.4)
• Lesson 9: Invariance Property of MLEs (FCPS §5.2.4)
• Lesson 10: Method of Moments Estimation (FCPS §5.2.5)
• Lesson 11: Sampling Distributions (FCPS §5.3) | 1,095 | 4,504 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.913486 |
https://stat.ethz.ch/R-manual/R-devel/RHOME/library/base/html/numeric.html | 1,695,475,265,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00517.warc.gz | 609,518,091 | 3,193 | numeric {base} R Documentation
## Numeric Vectors
### Description
Creates or coerces objects of type "numeric". is.numeric is a more general test of an object being interpretable as numbers.
### Usage
numeric(length = 0)
as.numeric(x, ...)
is.numeric(x)
### Arguments
length a non-negative integer specifying the desired length. Double values will be coerced to integer: supplying an argument of length other than one is an error. x object to be coerced or tested. ... further arguments passed to or from other methods.
### Details
numeric is identical to double. It creates a double-precision vector of the specified length with each element equal to 0.
as.numeric is a generic function, but S3 methods must be written for as.double. It is identical to as.double.
is.numeric is an internal generic primitive function: you can write methods to handle specific classes of objects, see InternalMethods. It is not the same as is.double. Factors are handled by the default method, and there are methods for classes "Date", "POSIXt" and "difftime" (all of which return false). Methods for is.numeric should only return true if the base type of the class is double or integer and values can reasonably be regarded as numeric (e.g., arithmetic on them makes sense, and comparison should be done via the base type).
### Value
for numeric and as.numeric see double.
The default method for is.numeric returns TRUE if its argument is of mode "numeric" (type "double" or type "integer") and not a factor, and FALSE otherwise. That is, is.integer(x) || is.double(x), or (mode(x) == "numeric") && !is.factor(x).
### Warning
If x is a factor, as.numeric will return the underlying numeric (integer) representation, which is often meaningless as it may not correspond to the factor levels, see the ‘Warning’ section in factor (and the 2nd example below).
### S4 methods
as.numeric and is.numeric are internally S4 generic and so methods can be set for them via setMethod.
To ensure that as.numeric and as.double remain identical, S4 methods can only be set for as.numeric.
### Note on names
It is a historical anomaly that R has two names for its floating-point vectors, double and numeric (and formerly had real).
double is the name of the type. numeric is the name of the mode and also of the implicit class. As an S4 formal class, use "numeric".
The potential confusion is that R has used mode "numeric" to mean ‘double or integer’, which conflicts with the S4 usage. Thus is.numeric tests the mode, not the class, but as.numeric (which is identical to as.double) coerces to the class.
### References
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.
double, integer, storage.mode.
### Examples
## Conversion does trim whitespace; non-numeric strings give NA + warning
as.numeric(c("-.1"," 2.7 ","B"))
## Numeric values are sometimes accidentally converted to factors.
## Converting them back to numeric is trickier than you'd expect.
f <- factor(5:10)
as.numeric(f) # not what you might expect, probably not what you want
## what you typically meant and want:
as.numeric(as.character(f))
## the same, considerably more efficient (for long vectors):
as.numeric(levels(f))[f]
[Package base version 4.4.0 Index] | 767 | 3,279 | {"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": 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} | 2.65625 | 3 | CC-MAIN-2023-40 | latest | en | 0.788561 |
http://write-up.semantic-db.org/2-defining-some-terms.html | 1,656,606,694,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103850139.45/warc/CC-MAIN-20220630153307-20220630183307-00313.warc.gz | 65,739,045 | 2,212 | # defining some terms
So, the notation I use is motivated by the bra-ket notation as used in quantum mechanics, and invented by the famous physicist Paul Dirac. Note though that the mathematics of my scheme and that of QM are vastly different.
Let's define the terms:
<x| is called a bra
|x> is called a ket
Essentially any text (currently ASCII only) can be inside bra/kets except <, |, > and \r and \n. Though we do have some conventions, more on that later.
Next, we have operators, that are again text.
The python defining valid operators is:
```def valid_op(op):
if not op[0].isalpha() and not op[0] == '!':
return False
return all(c in ascii_letters + '0123456789-+!?' for c in op) ```
Next, we have what I call "superpositions" (again borrowed from QM). A superposition is just the sum of one or more kets.
eg:
|a> + |b> + |c>
But a full superposition can also have coeffs (I almost always write coefficients as coeffs).
3|a> + 7.25|b> + 21|e> + 9|d>
The name superpositions is partly motivated by Schrodinger's poor cat:
is-alive |cat> => 0.5 |yes> + 0.5 |no>
This BTW, is what we call a "learn rule" (though there are a couple of other variants).
They have the general form:
OP KET => SUPERPOSITION
Next, we have some math rules for all this, though for now it will suffice to mention only these:
```1) <x||y> == 0 if x != y.
2) <x||y> == 1 if x == y.
7) applying bra's is linear. <x|(|a> + |b> + |c>) == <x||a> + <x||b> + <x||c>
8) if a coeff is not given, then it is 1. eg, <x| == <x|1 and 1|x> == |x>
9) bra's and ket's commute with the coefficients. eg, <x|7 == 7 <x| and 13|x> == |x>13
13) kets in superpositions commute. |a> + |b> == |b> + |a>
18) |> is the identity element for superpositions. sp + |> == |> + sp == sp.
|a> + |a> + |a> = 3|a>
|a> + |b> + |c> + 6|b> = |a> + 7|b> + |c>```
And that is it for now. Heaps more to come!
Update: I guess you could call superpositions labelled, sparse vectors. By giving each element in a vector a name, it gives us the freedom to drop elements with coeff = 0. For large sparse vectors, this is a big win. For large dense vectors, there is of course a price to pay for all those labels. And since we have labels we can change the order of the elements without changing the meaning of the superposition. Say if we want to sort by coeff size. This is harder if you use standard unlabeled vectors. I guess the other thing to note about superpositions is that they allow you to define operators with respect to vector labels, and not vector positions.
Home
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next: context learn and recall
updated: 19/12/2016
by Garry Morrison
email: garry -at- semantic-db.org | 791 | 2,669 | {"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} | 3.53125 | 4 | CC-MAIN-2022-27 | latest | en | 0.884333 |
https://sanfrancisco.cbslocal.com/2015/06/10/the-tour-de-france-in-terms-of-jelly-donuts/ | 1,639,010,156,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363641.20/warc/CC-MAIN-20211209000407-20211209030407-00597.warc.gz | 559,701,264 | 21,195 | Elite cyclists burn up to 32 per day in energy
By Ben P. Stein, Director with Inside Science
(Inside Science Currents Blog) — With today being National Donut Day, and this year’s Tour de France coming up in the next month, I recalled how a late MIT physicist known for his work in the Manhattan Project once combined donuts and the famous bicycle race to teach a lesson about energy. Count on science to mash together two different things in unexpectedly fun ways. So in honor of National Donut Day, we present to you a slightly updated version of one of the very first Inside Science News Service articles I wrote with Rory McGee Richards all the way back in July 22 1999:
Like jelly donuts? How about 30 a day? Well, that’s the number of donuts you would have to eat to get enough calories to compete for just “one day” during the Tour de France.
Traveling approximately 2000 miles in 23 days, a Tour de France cyclist eats the equivalent of 30-32 jelly donuts per day in calories, according to an estimate once performed by the late MIT physicist Philip Morrison. Morrison conducted an investigation of the Tour de France for his 1987 PBS documentary series “The Ring of Truth.” (In comparison, a normal adult male eats the equivalent of 12 jelly donuts a day, assuming that each jelly donut is about 250 calories.)
So where do all those donuts go? During a typical day’s race, Morrison said, a biker burns off about 1 or 2 jelly donuts worth of energy in overcoming friction and manipulating his bike. Propelling the bike forward in a typical day’s race requires only about 6 jelly donuts of energy.
Where do the remaining 25 donuts of energy get spent? You guessed right–it’s wasted–in the form of heat. Now, the human body is a relatively efficient machine. But like all machines, it unavoidably generates heat which cannot be employed to perform useful work. This heat–20-25 donuts’ worth, according to Morrison–is mainly carried away in the form of sweat created by exertion.
Competitors drink about 4 gallons of water a day, and this water evaporates from their bodies to carry away the heat. What keeps them cool enough to endure the race, says Morrison, is the streaming wind that hits their face.
In the years since we wrote that Inside Science piece, I have enjoyed finding other examples of how athletes can have fun thinking in terms of donuts. If you’re a runner, this infographic approximates how many more donuts you will be able to eat without gaining weight after running different distances, from a 5K to a full marathon. And it’s worth mentioning another unrelated combination of bikes and donuts, known as the Tour de Donut, in which riders’ times are reduced by a few minutes with every donut they eat at a rest stop. But even if you’re not a donut-lover or bicyclist, it’s not hard to find many ways of how one form of energy transforms into another.
Reprinted with permission from Inside Science, an editorially independent news product of the American Institute of Physics, a nonprofit organization dedicated to advancing, promoting and serving the physical sciences. | 675 | 3,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-49 | latest | en | 0.936442 |
https://learn.saylor.org/mod/book/view.php?id=37119 | 1,718,321,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00741.warc.gz | 313,730,328 | 29,275 | ## The Limit of a Function
Read this section for an introduction to connecting derivatives to quantities we can see in the real world. Work through practice problems 1-4.
### The Idea, Informally
Calculus has been called the study of continuous change, and the limit is the basic concept which allows us to describe and analyze such change. An understanding of limits is necessary to understand derivatives, integrals, and other fundamental topics of calculus.
The limit of a function describes the behavior of the function when the variable is near, but does not equal, a specified number (Fig. 1). If the values of $\mathrm{f}(\mathrm{x})$ get closer and closer, as close as we want, to one number $L$ as we take values of $x$ very close to (but not equal to) a number $\mathrm{c}$, then we
say "the limit of $\mathbf{f}(\mathbf{x})$, as x approaches $\mathbf{c}$, is $\mathbf{L}$ " $\quad$ and we
write " $\lim\limits_{x \rightarrow c}\mathbf{f}(\mathbf{x})=\mathbf{L. "} \quad$ (The symbol " $\rightarrow$ " means "approaches" or "gets very close to").
$f(c)$ is a single number that describes the behavior (value) of f AT the point $x=c$.
$\lim\limits_{x \rightarrow c} \mathbf{f}(\mathbf{x}) \quad$ is a single number that describes the behavior of $f$ NEAR, BUT NOT AT, the point $\mathrm{x}=\mathrm{c}$.
If we have a graph of the function near $x=c$, then it is usually easy to determine $\lim\limits_{x \rightarrow c} f(x)$.
Example 1: Use the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ in Fig. 2 to determine the following limits:
(a) $\lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x})$
(b) $\lim\limits_{x \rightarrow 2} \mathrm{f}(\mathrm{x})$
(c) $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})$
(d) $\lim\limits_{x \rightarrow 4} \mathrm{f}(\mathrm{x})$
Solution: (a) $\quad \lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x})=2$. When $\mathrm{x}$ is very close to 1 , the values of $\mathrm{f}(\mathrm{x})$ are very close to $\mathrm{y}=2 .$ In this example, it happens that $\mathrm{f}(1)=2$, but that is irrelevant for the limit. The only thing that matters is what happens for $x$ close to 1 but $x \neq 1$.
(b) $\mathrm{f}(2)$ is undefined, but we only care about the behavior of $\mathrm{f}(\mathrm{x})$ for $\mathrm{x}$ close to 2 and not equal to 2. When $\mathrm{x}$ is close to 2 , the values of $\mathrm{f}(\mathrm{x})$ are close to 3 . If we restrict $\mathrm{x}$ close enough to 2 , the values of $y$ will be as close to 3 as we want, so $\lim\limits_{x \rightarrow 2} f(x)=3$.
(c) When $\mathrm{x}$ is close to 3 (or as $\mathrm{x}$ approaches the value 3 ), the values of $\mathrm{f}(\mathrm{x})$ are close to 1 (or approach the value 1), so $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})=1$. For this limit it is completely irrelevant that $\mathrm{f}(3)=2$, We only care about what happens to $\mathrm{f}(\mathrm{x})$ for $\mathrm{x}$ close to and not equal to 3.
(d) This one is harder and we need to be careful. When $x$ is close to 4 and slightly less than 4 ($x$ is just to the left of 4 on the $x$-axis), then the values of $f(x)$ are close to $2.$ But if $x$ is close to 4 and slightly larger than 4 then the values of $f(x)$ are close to $3.$ If we only know that $x$ is very close to 4, then we cannot say whether $y=f(x)$ will be close to 2 or close to 3 - it depends on whether $x$ is on the right or the left side of $4.$ In this situation, the $f(x)$ values are not close to a single number so we say $\lim\limits_{x \rightarrow 4} \mathrm{f}(\mathrm{x})$ does not exist. It is irrelevant that $\mathrm{f}(4)=1.$ The limit, as $\mathrm{x}$ approaches 4, would still be undefined if $\mathrm{f}(4)$ was 3 or 2 or anything else.
Practice 1: Use the graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ in Fig. 3 to determine the following limits:
(a) $\lim\limits_{x \rightarrow 1} \mathrm{f}(\mathrm{x})$
(b) $\lim\limits_{t \rightarrow 2} \mathrm{f}(\mathrm{t})$
(c) $\lim\limits_{x \rightarrow 3} \mathrm{f}(\mathrm{x})$
(d) $\lim\limits_{w \rightarrow 4} \mathrm{f}(\mathrm{w})$
Example 2: Determine the value of $\lim\limits_{x \rightarrow 3} \frac{2 x^{2}-x-1}{x-1}$.
Solution: We need to investigate the values of $f(x)$ when $x$ is close to 3. If the $f(x)$ values get arbitrarily close to or even equal some number $\mathrm{L}$, then $\mathrm{L}$ will be the limit. One way to keep track of both the $x$ and the $f(x)$ values is to set up a table and to pick several $x$ values which are closer and closer (but not equal) to $3.$ We can pick some values of $x$ which approach 3 from the left, say $x=2.91,2.9997$ $2.999993$, and $2.9999999$, and some values of $x$ which approach 3 from the right, say $x=3.1,3.004$ $3.0001$, and $3.000002$. The only thing important about these particular values for $x$ is that they get closer and closer to 3 without equaling $3.$ You should try some other values "close to 3" to see what happens.
\begin{aligned} &\begin{array} {l|l} {\mathrm{x}} & \mathrm{f}(\mathrm{x}) \\ \hline 2.9 & 6.82 \\ 2.9997 & 6.9994 \\ 2.999993 & 6.999986 \\ 2.9999999 & 6.9999998 \end{array} \\ &\begin{array}{ll} & \quad \downarrow & \qquad \qquad \quad \downarrow \\ & \quad 3 & \qquad \qquad \quad 7 \end{array}\end{aligned} \begin{aligned} &\begin{array}{l|l} {\mathrm{x}} & {\mathrm{f}(\mathrm{x})} \\ \\ \hline 3.1 & 7.2 \\ 3.004 & 7.008 \\ 3.0001 & 7.0002 \\ 3.000002 & 7.000004 \end{array}\\ &\begin{array}{ll}\qquad \downarrow & \qquad \qquad \downarrow \\ \qquad 3 & \qquad \qquad 7 \end{array} \end{aligned}
As the $x$ values get closer and closer to 3, the $f(x)$ values are getting closer and closer to $7$. In fact, we can get $\mathrm{f}(\mathrm{x})$ as close to 7 as we want ("arbitrarily close") by taking the values of $\mathrm{x}$ very close ("sufficiently close") to 3. $\lim\limits_{x \rightarrow 3} \frac{2 x^{2}-x-1}{x-1}=7$.
Instead of using a table of values, we could have graphed $y=f(x)$ for $x$ close to 3, Fig. 4, and used the graph to answer the limit question. This graphic approach is easier, particularly if you have a calculator or computer do the graphing work for you, but it is really very similar to the "table of values" method: in each case you need to evaluate $\mathrm{y}=\mathrm{f}(\mathrm{x})$ at many values of $\mathrm{x}$ near 3.
You might have noticed that if we just evaluate $f(3)$, then we get the correct answer $7.$ That works for this particular problem, but it often fails. The next example illustrates the difficulty.
Example 3: Find $\lim\limits_{x \rightarrow 1} \frac{2 x^{2}-x-1}{x-1}$. (Same as Example 2 but with $\mathrm{x} \rightarrow 1$).
Solution: You might try to evaluate $f(x)=\frac{2 x^{2}-x-1}{x-1}$ at $x=1$, but $\mathrm{f}$ is not defined at $\mathrm{x}=1$. It is tempting, but wrong, to conclude that this function does not have a limit as $\mathrm{x}$ approaches 1.
Table Method: Trying some "test" values for $\mathrm{x}$ which get closer and closer to 1 from both the left and the right, we get
\begin{aligned} &\begin{array}{l|l} {\mathrm{x}} & \mathrm{f}(\mathrm{x}) \\ \hline 0.9 & 2.82 \\ 0.9998 & 2.9996 \\ 0.999994 & 2.999988 \\ 0.9999999 & 2.9999998 \end{array} \\ &\begin{array}{ll}\qquad \downarrow & \qquad \qquad \downarrow \\ \qquad 1 & \qquad \qquad 3 \end{array} \end{aligned} \begin{align} \begin{array}{l|l} \mathrm{x} & \mathrm{f}(\mathrm{x}) \\ \hline 1.1 & 3.2 \\ 1.003 & 3.006 \\ 1.0001 & 3.0002 \\ 1.000007 & 3.000014 \\ \quad \downarrow & \quad \downarrow \\ \quad \mathbf{1} & \quad \mathbf{3} \end{array} \end{align}
The function $\mathrm{f}$ is not defined at $\mathrm{x}=1$, but when $\mathrm{x}$ is close to 1, the values of $\mathrm{f}(\mathrm{x})$ are getting very close to 3. We can get $\mathrm{f}(\mathrm{x})$ as close to 3 as we want by taking $\mathrm{x}$ very close to 1 so $\lim\limits_{x \rightarrow 1} \frac{2 x^{2}-x-1}{x-1}=3.$
Graph Method: We can graph $y=f(x)=\frac{2 x^{2}-x-1}{x-1}$ for $x$ close to 1 , Fig. 5, and notice that whenever $\mathrm{x}$ is close to 1 , the values of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ are close to $3 . \mathrm{f}$ is not defined at $\mathrm{x}=1$, so the graph has a hole above $x=1$, but we only care about what $f(x)$ is doing for $x$ close to but not equal to 1.
Algebra Method: We could have found the same result by noting that $f(x)=\frac{2 x^{2}-x-1}{x-1}=\frac{(2 x+1)(x-1)}{x-1}=2 x+1$ as long as $\mathrm{x} \neq 1$. (If $\mathrm{x} \neq 1$, then $\mathrm{x}-1 \neq 0$ so it is valid to divide the numerator and denominator by the factor $x-1.$) The "$x \rightarrow 1$" part of the limit means that $x$ is close to 1 but not equal to $1$, so our division step is valid and
$\lim\limits_{x \rightarrow 1} \frac{2 x^{2}-x-1}{x-1}=\lim _{x \rightarrow 1} 2 x+1=3$, the correct answer. | 2,893 | 8,765 | {"found_math": true, "script_math_tex": 135, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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} | 4.71875 | 5 | CC-MAIN-2024-26 | latest | en | 0.694851 |
https://getrevising.co.uk/revision-tests/nnumerical_concepts_3 | 1,527,400,138,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868003.97/warc/CC-MAIN-20180527044401-20180527064401-00030.warc.gz | 545,362,420 | 12,969 | # Nnumerical Concepts
"^" means "to the power of"
HideShow resource information
• Created by: Tish
• Created on: 04-06-14 06:22
## 1. 2/3 is a/an..
• rational number, and real number
• positive integer
• integer
• real number, natural number
1 of 11
## 2. irrational numbers belong to the set of..
• real numbers
• positive integers
• natural numbers
• integers
• 125.6
• 124.6
• 125
• 124.55
• 124.5547
• 124.6
• 125
• 124.555
• 0
• 0.00
• 0.006
• 0.005 | 175 | 463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-22 | latest | en | 0.418719 |
https://volumeconverter.intemodino.com/conversion/m3-to-gi-and-gi-to-m3.html | 1,721,468,826,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00213.warc.gz | 530,188,775 | 5,855 | # Cubic Meters and Gills Converter, Conversion Formula and Conversion Table
Length Conversion Settings
Number of decimal places
Round fraction to the nearest
## M³ to gi conversion factor and conversion formula
The to gi conversion factor is 8453.50567903
1 = 8453.50567903 gi
To convert X cubic meters to gills you should use the following conversion formula:
X cubic meters × 8453.50567903 = Result in gills
## Cubic Meters to Gills Conversion Table
1 m³ = 8453.50567903 gi
2 m³ = 16907.01135807 gi
3 m³ = 25360.5170371 gi
4 m³ = 33814.02271614 gi
5 m³ = 42267.52839517 gi
6 m³ = 50721.0340742 gi
7 m³ = 59174.53975324 gi
8 m³ = 67628.04543227 gi
9 m³ = 76081.5511113 gi
10 m³ = 84535.05679034 gi
11 m³ = 92988.56246937 gi
12 m³ = 101442.06814841 gi
13 m³ = 109895.57382744 gi
14 m³ = 118349.07950647 gi
15 m³ = 126802.58518551 gi
16 m³ = 135256.09086454 gi
17 m³ = 143709.59654358 gi
18 m³ = 152163.10222261 gi
19 m³ = 160616.60790164 gi
20 m³ = 169070.11358068 gi
21 m³ = 177523.61925971 gi
22 m³ = 185977.12493874 gi
23 m³ = 194430.63061778 gi
24 m³ = 202884.13629681 gi
25 m³ = 211337.64197585 gi
26 m³ = 219791.14765488 gi
27 m³ = 228244.65333391 gi
28 m³ = 236698.15901295 gi
29 m³ = 245151.66469198 gi
30 m³ = 253605.17037102 gi
31 m³ = 262058.67605005 gi
32 m³ = 270512.18172908 gi
33 m³ = 278965.68740812 gi
34 m³ = 287419.19308715 gi
35 m³ = 295872.69876618 gi
36 m³ = 304326.20444522 gi
37 m³ = 312779.71012425 gi
38 m³ = 321233.21580329 gi
39 m³ = 329686.72148232 gi
40 m³ = 338140.22716135 gi
41 m³ = 346593.73284039 gi
42 m³ = 355047.23851942 gi
43 m³ = 363500.74419845 gi
44 m³ = 371954.24987749 gi
45 m³ = 380407.75555652 gi
46 m³ = 388861.26123556 gi
47 m³ = 397314.76691459 gi
48 m³ = 405768.27259362 gi
49 m³ = 414221.77827266 gi
50 m³ = 422675.28395169 gi
51 m³ = 431128.78963073 gi
52 m³ = 439582.29530976 gi
53 m³ = 448035.80098879 gi
54 m³ = 456489.30666783 gi
55 m³ = 464942.81234686 gi
56 m³ = 473396.31802589 gi
57 m³ = 481849.82370493 gi
58 m³ = 490303.32938396 gi
59 m³ = 498756.835063 gi
60 m³ = 507210.34074203 gi
61 m³ = 515663.84642106 gi
62 m³ = 524117.3521001 gi
63 m³ = 532570.85777913 gi
64 m³ = 541024.36345817 gi
65 m³ = 549477.8691372 gi
66 m³ = 557931.37481623 gi
67 m³ = 566384.88049527 gi
68 m³ = 574838.3861743 gi
69 m³ = 583291.89185333 gi
70 m³ = 591745.39753237 gi
71 m³ = 600198.9032114 gi
72 m³ = 608652.40889044 gi
73 m³ = 617105.91456947 gi
74 m³ = 625559.4202485 gi
75 m³ = 634012.92592754 gi
76 m³ = 642466.43160657 gi
77 m³ = 650919.93728561 gi
78 m³ = 659373.44296464 gi
79 m³ = 667826.94864367 gi
80 m³ = 676280.45432271 gi
81 m³ = 684733.96000174 gi
82 m³ = 693187.46568077 gi
83 m³ = 701640.97135981 gi
84 m³ = 710094.47703884 gi
85 m³ = 718547.98271788 gi
86 m³ = 727001.48839691 gi
87 m³ = 735454.99407594 gi
88 m³ = 743908.49975498 gi
89 m³ = 752362.00543401 gi
90 m³ = 760815.51111305 gi
91 m³ = 769269.01679208 gi
92 m³ = 777722.52247111 gi
93 m³ = 786176.02815015 gi
94 m³ = 794629.53382918 gi
95 m³ = 803083.03950821 gi
96 m³ = 811536.54518725 gi
97 m³ = 819990.05086628 gi
98 m³ = 828443.55654532 gi
99 m³ = 836897.06222435 gi
100 m³ = 845350.56790338 gi
## Gi to m³ conversion formula and conversion factor
The gi to conversion factor is 0.00011829
1 gi = 0.00011829
To convert X gills to cubic meters you should use the following conversion formula:
X gills × 0.00011829 = Result in cubic meters
## Gills to Cubic Meters Conversion Chart
1 gi = 0.00011829 m³
2 gi = 0.00023659 m³
3 gi = 0.00035488 m³
4 gi = 0.00047318 m³
5 gi = 0.00059147 m³
6 gi = 0.00070976 m³
7 gi = 0.00082806 m³
8 gi = 0.00094635 m³
9 gi = 0.00106465 m³
10 gi = 0.00118294 m³
11 gi = 0.00130124 m³
12 gi = 0.00141953 m³
13 gi = 0.00153782 m³
14 gi = 0.00165612 m³
15 gi = 0.00177441 m³
16 gi = 0.00189271 m³
17 gi = 0.002011 m³
18 gi = 0.00212929 m³
19 gi = 0.00224759 m³
20 gi = 0.00236588 m³
21 gi = 0.00248418 m³
22 gi = 0.00260247 m³
23 gi = 0.00272076 m³
24 gi = 0.00283906 m³
25 gi = 0.00295735 m³
26 gi = 0.00307565 m³
27 gi = 0.00319394 m³
28 gi = 0.00331224 m³
29 gi = 0.00343053 m³
30 gi = 0.00354882 m³
31 gi = 0.00366712 m³
32 gi = 0.00378541 m³
33 gi = 0.00390371 m³
34 gi = 0.004022 m³
35 gi = 0.00414029 m³
36 gi = 0.00425859 m³
37 gi = 0.00437688 m³
38 gi = 0.00449518 m³
39 gi = 0.00461347 m³
40 gi = 0.00473176 m³
41 gi = 0.00485006 m³
42 gi = 0.00496835 m³
43 gi = 0.00508665 m³
44 gi = 0.00520494 m³
45 gi = 0.00532324 m³
46 gi = 0.00544153 m³
47 gi = 0.00555982 m³
48 gi = 0.00567812 m³
49 gi = 0.00579641 m³
50 gi = 0.00591471 m³
51 gi = 0.006033 m³
52 gi = 0.00615129 m³
53 gi = 0.00626959 m³
54 gi = 0.00638788 m³
55 gi = 0.00650618 m³
56 gi = 0.00662447 m³
57 gi = 0.00674276 m³
58 gi = 0.00686106 m³
59 gi = 0.00697935 m³
60 gi = 0.00709765 m³
61 gi = 0.00721594 m³
62 gi = 0.00733424 m³
63 gi = 0.00745253 m³
64 gi = 0.00757082 m³
65 gi = 0.00768912 m³
66 gi = 0.00780741 m³
67 gi = 0.00792571 m³
68 gi = 0.008044 m³
69 gi = 0.00816229 m³
70 gi = 0.00828059 m³
71 gi = 0.00839888 m³
72 gi = 0.00851718 m³
73 gi = 0.00863547 m³
74 gi = 0.00875376 m³
75 gi = 0.00887206 m³
76 gi = 0.00899035 m³
77 gi = 0.00910865 m³
78 gi = 0.00922694 m³
79 gi = 0.00934524 m³
80 gi = 0.00946353 m³
81 gi = 0.00958182 m³
82 gi = 0.00970012 m³
83 gi = 0.00981841 m³
84 gi = 0.00993671 m³
85 gi = 0.010055 m³
86 gi = 0.01017329 m³
87 gi = 0.01029159 m³
88 gi = 0.01040988 m³
89 gi = 0.01052818 m³
90 gi = 0.01064647 m³
91 gi = 0.01076476 m³
92 gi = 0.01088306 m³
93 gi = 0.01100135 m³
94 gi = 0.01111965 m³
95 gi = 0.01123794 m³
96 gi = 0.01135624 m³
97 gi = 0.01147453 m³
98 gi = 0.01159282 m³
99 gi = 0.01171112 m³
100 gi = 0.01182941 m³
Wednesday, December 16, 2020 | 2,692 | 5,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-30 | latest | en | 0.160544 |
https://bhavinionline.com/2014/09/page/5/ | 1,632,028,280,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00482.warc.gz | 190,792,796 | 9,825 | ## Whatsapp Puzzles: Guess Madhuri Dixit Songs From Emoticons and Smileys
Create puzzles and Riddles with Whatsapp emoticons and smileys…and send it to friends and let them crack the puzzles. Here’s one more that’s doing rounds now. This one is a real tough one. The difficulties are getting crazier with these emoji puzzles. For other puzzles you would usually find a solution on Google but finding
## Fill the Boxes: [] + [] + [] + [] + [] = 30
Here is a tough mathematical puzzle only for the genius. This question came in UPSC Final Exams and only 1 person was able to solve it. So the difficulty on this riddle is “Very Hard”. [] + [] + [] + [] + [] = 30 Fill the boxes using one of these numbers 1, | 173 | 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} | 2.5625 | 3 | CC-MAIN-2021-39 | latest | en | 0.936377 |
http://orac.amt.edu.au/cgi-bin/train/problem.pl?problemid=521 | 1,638,320,196,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00377.warc.gz | 58,642,765 | 3,763 | # Problem: The Virus
Want to try solving this problem? You can submit your code online if you log in or register.
## The Virus
Input File: standard input
Output File: standard output
Time Limit: 1 second
Memory Limit: 32 MB
No one knows how it all started, just that it happened so quickly. Two weeks ago you were watching the news about political problems in far away nations, today you are on the run from your own neighbours who have fallen victim to The Virus. There is no known cure for The Virus which is plaguing the nation - once a person has contracted it, they cannot be cured. Thankfully though, Dr. Yes has come to you with a curious biological weapon, which is used to deploy a vaccine against the virus. Using his weapon, the vaccine can be administered to everybody within a city instantly, effectively stopping the spread of The Virus within that city.
The eccentric Dr. Yes hastily describes to you in a raspy voice his plan to save the nation. Together you will travel to every city in the country and use the weapon on each city, saving those you can by vaccinating the healthy individuals against The Virus.
Each city's mayor has determined the rate of spread of The Virus in their city, measured in victims per hour. You need to perform this rescue option to minimise the number of victims. Some roads take longer to travel on than others, and each wasted hour means more people lost to The Virus. Given a map of all the roads in the country and the number of citizens in each city falling victim to The Virus every hour, determine a route passing through every city such that the total number of people infected by The Virus is as small as possible. Furthermore, in these dangerous times with thieves and murderers abound, you must not travel on any road more than twice.
There exist N cities in the country, and N-1 bi-directional roads which connect the cities such that there is a path from every city to every other city. You know from your days as an informatics student that the graph formed is called a "tree". You also may assume that city populations are sufficiently large that they will never lose all their citizens to The Virus.
You are about to start coding, when Dr. Yes interrupts you with a final remark.
"Remember, lives are in your hands. Code fast and we will be the heroes of tomorrow."
### Input
The first line of input will contain a single integer N, the number of cities (1 ≤ N ≤ 100 000). The cities are numbered from 1 to N. You must commence your rescue operation in city 1.
The second line of input will contain N non-negative integers. The ith integer represents the number of citizens succumbing to The Virus each hour in city i. This value will be at most 100.
This will be followed by N-1 lines of input. Each line will contain three space-separated integers a, b and h, representing a road from city a to city b that takes h hours to travel (1 ≤ h ≤ 100).
For 60% of the available marks, N ≤ 1 000.
For 20% of the available marks, N ≤ 10.
### Output
Output should consist of one line with a single integer: the minimum number of victims of The Virus before every city is saved.
```5
9 10 2 5 1
1 2 2
1 4 4
3 4 3
4 5 5
```
```101
```
### Explanation
The optimal path is:
1 --> 2 2 hours 2 --> 1 2 hours 1 --> 4 4 hours 4 --> 3 3 hours 3 --> 4 3 hours 4 --> 5 5 hours
City Hours until saved Infected each hour Citizens lost to The Virus 1 0 9 0 2 2 10 20 3 11 2 22 4 8 5 40 5 19 1 19 Total: 101
### Scoring
The score for each input scenario will be 100% if the correct answer is written to the output file, and 0% otherwise.
Privacy statement
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https://www.coursehero.com/file/5921212/ma5c-hw1-soln/ | 1,519,516,509,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816068.93/warc/CC-MAIN-20180224231522-20180225011522-00653.warc.gz | 814,759,520 | 64,684 | {[ promptMessage ]}
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ma5c-hw1-soln
# ma5c-hw1-soln - M a 5c HOMEWORK 1 SOLUTION SPRING 09 The...
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Ma 5 c HOMEWORK 1 SOLUTION SPRING 09 The exercises are taken from the text, Abstract Algebra (third edi- tion) by Dummit and Foote, unless stated otherwise. Page 519, 4 . Define φ : Q [ 2] Q [ 2] via a + b 2 7→ a - b 2, where a,b Q . Then φ ( a + b 2) + φ ( c + d 2) = a - b 2 + c - d 2 = a + c - ( b + d ) 2 = φ (( a + c ) + ( b + d ) 2) = φ ( a + b 2 + c + d 2). φ ( a + b 2) φ ( c + d 2) = ( a - b 2)( c - d 2) = ( ac +2 bd ) - ( ad + bc ) 2 = φ ( ac + 2 bd + ( ad + bc ) 2) = φ (( a + b 2)( c + d 2)). Therefore φ is a homomorhpism. It’s obvious that φ is both injective and surjective. Page 530, 12 . We know that [ E : F ] divides [ K : F ] = p . Therefore [ E : F ] = 1, in which case E = F since E contains F , or [ E : F ] = p = [ K : F ], in which case E = K , since E is a subfield of K . Page 530, 13 . Note that [ Q ( α 1 ) : Q ] | 2 since α 1 is a root of x 2 - α 2 1 Q [ x ]. Similarly, for any
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Ask a homework question - tutors are online | 558 | 1,525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.765147 |
https://www.developerfaqs.com/1988/check-for-identical-rows-in-different-numpy-arrays | 1,571,790,401,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00540.warc.gz | 851,980,342 | 8,679 | # check for identical rows in different numpy arrays
• A+
Category：Languages
how do I get a row-wise comparison between two arrays, in the result of a row-wise true/false array?
Given datas:
``a = np.array([[1,0],[2,0],[3,1],[4,2]]) b = np.array([[1,0],[2,0],[4,2]]) ``
Result step 1:
``c = np.array([True, True,False,True]) ``
Result final:
``a = a[c] ``
So how do I get the array `c` ????
P.S.: In this example the arrays `a` and `b` are sorted, please give also information if in your solution it is important that the arrays are sorted
Here's a vectorised solution:
``res = (a[:, None] == b).all(-1).any(-1) print(res) array([ True, True, False, True]) ``
Note that `a[:, None] == b` compares each row of `a` with `b` element-wise. We then use `all` + `any` to deduce if there are any rows which are all `True` for each sub-array:
``print(a[:, None] == b) [[[ True True] [False True] [False False]] [[False True] [ True True] [False False]] [[False False] [False False] [False False]] [[False False] [False False] [ True True]]] `` | 336 | 1,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.694649 |
https://www.emathinstruction.com/emath-february-2020-newsletter/ | 1,719,057,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862396.85/warc/CC-MAIN-20240622112740-20240622142740-00308.warc.gz | 678,554,134 | 29,433 | Posted on
Happy February my math teacher friends. Hopefully you all had a long Presidents’ Day weekend. I know many in my part of the state have this entire week off from classes. We’ve been working diligently at eMATHinstruction on our new middle school courses and on the add-ons for the high school courses. We’ll talk more about middle school in a bit. For now, let’s get into those add-ons.
In our Common Core Algebra I Add-Ons this month we have three new resources to share. To begin, we bring you the Form D assessment for Unit 7 (Polynomials). We also have created the Unit 8 (Quadratic Functions) Exit Tickets. Finally, we bring you a new lesson in Unit 8 on the factored form of a polynomial. We plan to eventually add this lesson to the Next Generation version of Algebra I, which we will begin to finalize next year.
For Common Core Geometry Add-Ons this month we bring you two resources for Unit 7 and one for Unit 8. In Unit 7 we created an activity where students explore the centroid of a triangle. Strangely enough, the topic of the centroid is not addressed anywhere in the Common Core Standards (easy enough to search the term centroid). We discuss the centroid in Lesson 10 of this unit when we look at the medians of a triangle. But, in this activity students find the centroid, cut out the triangles, and see how the centroid becomes its balancing point (center of mass). We then bring you the Unit 7 (Similarity) Form C Assessment. Finally, we have the Unit 8 (Right Triangle Trigonometry) Exit Tickets.
In Common Core Algebra II Add-Ons this month we first bring you a quiz for Unit 7 (Transformations of Functions). We are trying to round out unit quizzes in all of our courses this year (no matter how short the unit). We also bring you the Unit 7 Form D assessment. Finally we also offer the Unit 8 (Radicals and the Quadratic Formula) Exit Tickets.
Our Algebra 2 with Trigonometry Add-Ons for the month begin with the Unit 7 (Trigonometric Functions) Form C Assessment. We also bring you the Unit 8 (Trigonometric Algebra) Exit Tickets. Finally, we also have a Unit 8 Quiz for after students have completed Lesson #3 of the unit.
Besides working on the add-ons, we’ve been very busy with our three new middle school courses, which are aligned to the Next Generation Mathematics Learning Standards. We’ve been getting fantastic feedback from teachers about the courses and we’ve integrated some of that feedback into the lessons for each course. We are now very close to having both the Math 6 and Math 7 books done in their first edition versions. Unlike our high school courses, we will only be offering our middle school workbooks in a single binding that is both spiral bound and three hole punched. The sheets also have a fine perforation so that students can rip pages out and then place them in a 3-ring binder. Here are a couple of pictures of our N-Gen Math™ 6 book:
If you look very hard, you can even see the fine perforation in that second picture. We’ve designed the books to be as light as possible with lower weight paper. We are always trying to balance that weight issue with the quality of the paper.
Work continues on the supplemental materials for these courses, like the videos, unit reviews, answer keys, and assessments. We plan to have the Teacher Plus Answer Keys and workbooks available for use by July 1, 2020. Until then, we can certainly provide quotes for school districts interested in these for next year.
Well, that’s about it for now. Time to get back to work. Have a great rest of February. May it end soon and may spring be upon us again! | 797 | 3,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-26 | latest | en | 0.935513 |
https://www.shaalaa.com/textbook-solutions/c/rd-sharma-solutions-class-12-maths-chapter-20-definite-integrals_645 | 1,695,456,450,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00887.warc.gz | 1,102,384,372 | 34,283 | # RD Sharma solutions for Class 12 Maths chapter 20 - Definite Integrals [Latest edition]
## Solutions for Chapter 20: Definite Integrals
Below listed, you can find solutions for Chapter 20 of CBSE, Karnataka Board PUC RD Sharma for Class 12 Maths.
Exercise 20.1Exercise 20.2Exercise 20.3Exercise 20.4Exercise 20.5Exercise 20.6Very Short AnswersMCQRevision Exercise
Exercise 20.1 [Pages 16 - 18]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Exercise 20.1 [Pages 16 - 18]
Exercise 20.1 | Q 1 | Page 16
$\int\limits_4^9 \frac{1}{\sqrt{x}} dx$
Exercise 20.1 | Q 2 | Page 16
$\int\limits_{- 2}^3 \frac{1}{x + 7} dx$
Exercise 20.1 | Q 3 | Page 16
$\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx$
Exercise 20.1 | Q 4 | Page 16
$\int\limits_0^1 \frac{1}{1 + x^2} dx$
Exercise 20.1 | Q 5 | Page 16
$\int\limits_2^3 \frac{x}{x^2 + 1} dx$
Exercise 20.1 | Q 6 | Page 16
$\int\limits_0^\infty \frac{1}{a^2 + b^2 x^2} dx$
Exercise 20.1 | Q 7 | Page 16
$\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx$
Exercise 20.1 | Q 8 | Page 16
$\int\limits_0^\infty e^{- x} dx$
Exercise 20.1 | Q 9 | Page 16
$\int\limits_0^1 \frac{x}{x + 1} dx$
Exercise 20.1 | Q 10 | Page 16
$\int\limits_0^{\pi/2} \left( \sin x + \cos x \right) dx$
Exercise 20.1 | Q 11 | Page 16
$\int\limits_{\pi/4}^{\pi/2} \cot x\ dx$
Exercise 20.1 | Q 12 | Page 16
$\int\limits_0^{\pi/4} \sec x dx$
Exercise 20.1 | Q 13 | Page 16
$\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx$
Exercise 20.1 | Q 14 | Page 16
$\int\limits_0^1 \frac{1 - x}{1 + x} dx$
Exercise 20.1 | Q 15 | Page 16
$\int\limits_0^\pi \frac{1}{1 + \sin x} dx$
Exercise 20.1 | Q 16 | Page 16
$\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx$
Exercise 20.1 | Q 17 | Page 16
$\int\limits_0^{\pi/2} \cos^2 x\ dx$
Exercise 20.1 | Q 18 | Page 16
$\int\limits_0^{\pi/2} \cos^3 x\ dx$
Exercise 20.1 | Q 19 | Page 16
$\int\limits_0^{\pi/6} \cos x \cos 2x\ dx$
Exercise 20.1 | Q 20 | Page 16
$\int\limits_0^{\pi/2} \sin x \sin 2x\ dx$
Exercise 20.1 | Q 21 | Page 16
$\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx$
Exercise 20.1 | Q 22 | Page 16
$\int\limits_0^{\pi/2} \cos^4\ x\ dx$
Exercise 20.1 | Q 23 | Page 16
$\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx$
Exercise 20.1 | Q 24 | Page 16
$\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx$
Exercise 20.1 | Q 25 | Page 16
$\int\limits_0^{\pi/2} \sqrt{1 + \cos x}\ dx$
Exercise 20.1 | Q 26 | Page 16
Evaluate the following definite integrals:
$\int_0^\frac{\pi}{2} x^2 \sin\ x\ dx$
Exercise 20.1 | Q 27 | Page 17
$\int\limits_0^{\pi/2} x \cos\ x\ dx$
Exercise 20.1 | Q 28 | Page 17
$\int\limits_0^{\pi/2} x^2 \cos\ x\ dx$
Exercise 20.1 | Q 29 | Page 17
$\int\limits_0^{\pi/4} x^2 \sin\ x\ dx$
Exercise 20.1 | Q 30 | Page 17
$\int\limits_0^{\pi/2} x^2 \cos\ 2x\ dx$
Exercise 20.1 | Q 31 | Page 17
$\int\limits_0^{\pi/2} x^2 \cos^2 x\ dx$
Exercise 20.1 | Q 32 | Page 17
$\int\limits_1^2 \log\ x\ dx$
Exercise 20.1 | Q 33 | Page 17
$\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx$
Exercise 20.1 | Q 34 | Page 17
$\int\limits_1^e \frac{e^x}{x} \left( 1 + x \log x \right) dx$
Exercise 20.1 | Q 35 | Page 17
$\int\limits_1^e \frac{\log x}{x} dx$
Exercise 20.1 | Q 36 | Page 17
$\int\limits_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} dx$
Exercise 20.1 | Q 37 | Page 17
$\int\limits_1^2 \frac{x + 3}{x \left( x + 2 \right)} dx$
Exercise 20.1 | Q 38 | Page 17
$\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx$
Exercise 20.1 | Q 39 | Page 17
$\int\limits_0^2 \frac{1}{4 + x - x^2} dx$
Exercise 20.1 | Q 40 | Page 17
$\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx$
Exercise 20.1 | Q 41 | Page 17
$\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx$
Exercise 20.1 | Q 42 | Page 17
$\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx$
Exercise 20.1 | Q 43 | Page 17
$\int\limits_0^4 \frac{1}{\sqrt{4x - x^2}} dx$
Exercise 20.1 | Q 44 | Page 17
$\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx$
Exercise 20.1 | Q 45 | Page 17
$\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx$
Exercise 20.1 | Q 46 | Page 17
$\int\limits_0^1 x \left( 1 - x \right)^5 dx$
Exercise 20.1 | Q 47 | Page 17
$\int\limits_1^2 \left( \frac{x - 1}{x^2} \right) e^x dx$
Exercise 20.1 | Q 48 | Page 17
$\int\limits_0^1 \left( x e^{2x} + \sin\frac{\ pix}{2} \right) dx$
Exercise 20.1 | Q 49 | Page 17
$\int\limits_0^1 \left( x e^x + \cos\frac{\pi x}{4} \right) dx$
Exercise 20.1 | Q 50 | Page 17
$\int\limits_{\pi/2}^\pi e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$
Exercise 20.1 | Q 51 | Page 17
$\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx$
Exercise 20.1 | Q 52 | Page 17
$\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx$
Exercise 20.1 | Q 53 | Page 17
$\int_0^\pi e^{2x} \cdot \sin\left( \frac{\pi}{4} + x \right) dx$
Exercise 20.1 | Q 54 | Page 17
$\int\limits_0^1 \frac{1}{\sqrt{1 + x} - \sqrt{x}} dx$
Exercise 20.1 | Q 55 | Page 17
$\int\limits_1^2 \frac{x}{\left( x + 1 \right) \left( x + 2 \right)} dx$
Exercise 20.1 | Q 56 | Page 17
$\int\limits_0^{\pi/2} \sin^3 x\ dx$
Exercise 20.1 | Q 57 | Page 17
$\int\limits_0^\pi \left( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} \right) dx$
Exercise 20.1 | Q 58 | Page 17
$\int\limits_1^2 e^{2x} \left( \frac{1}{x} - \frac{1}{2 x^2} \right) dx$
Exercise 20.1 | Q 59 | Page 17
Evaluate the following definite integral:
$\int_0^1 \frac{1}{\sqrt{\left( x - 1 \right)\left( 2 - x \right)}}dx$
Exercise 20.1 | Q 60 | Page 17
$\int\limits_0^k \frac{1}{2 + 8 x^2} dx = \frac{\pi}{16},$ find the value of k.
Exercise 20.1 | Q 61 | Page 18
$\int\limits_0^a 3 x^2 dx = 8,$ find the value of a.
Exercise 20.1 | Q 62 | Page 18
$\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx$
Exercise 20.1 | Q 63 | Page 18
$\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx$
Exercise 20.1 | Q 64 | Page 18
$\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx$
Exercise 20.1 | Q 65 | Page 18
$\int_0^1 x\log\left( 1 + 2x \right)dx$
Exercise 20.1 | Q 66 | Page 18
$\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx$
Exercise 20.1 | Q 67 | Page 18
$\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx$
Exercise 20.1 | Q 68 | Page 18
$\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx$
Exercise 20.2 [Pages 38 - 40]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Exercise 20.2 [Pages 38 - 40]
Exercise 20.2 | Q 1 | Page 38
$\int\limits_2^4 \frac{x}{x^2 + 1} dx$
Exercise 20.2 | Q 2 | Page 38
$\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx$
Exercise 20.2 | Q 3 | Page 38
$\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx$
Exercise 20.2 | Q 4 | Page 38
$\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx$
Exercise 20.2 | Q 5 | Page 38
$\int\limits_0^a \frac{x}{\sqrt{a^2 + x^2}} dx$
Exercise 20.2 | Q 6 | Page 38
$\int\limits_0^1 \frac{e^x}{1 + e^{2x}} dx$
Exercise 20.2 | Q 7 | Page 38
$\int\limits_0^1 x e^{x^2} dx$
Exercise 20.2 | Q 8 | Page 38
$\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx$
Exercise 20.2 | Q 9 | Page 38
$\int\limits_0^1 \frac{2x}{1 + x^4} dx$
Exercise 20.2 | Q 10 | Page 38
$\int\limits_0^a \sqrt{a^2 - x^2} dx$
Exercise 20.2 | Q 11 | Page 39
$\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi$
Exercise 20.2 | Q 12 | Page 39
$\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx$
Exercise 20.2 | Q 13 | Page 39
$\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta$
Exercise 20.2 | Q 14 | Page 39
$\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx$
Exercise 20.2 | Q 15 | Page 39
$\int\limits_0^1 \frac{\sqrt{\tan^{- 1} x}}{1 + x^2} dx$
Exercise 20.2 | Q 16 | Page 39
$\int\limits_0^2 x\sqrt{x + 2}\ dx$
Exercise 20.2 | Q 17 | Page 39
$\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx$
Exercise 20.2 | Q 18 | Page 39
$\int\limits_0^{\pi/2} \frac{\sin x \cos x}{1 + \sin^4 x} dx$
Exercise 20.2 | Q 19 | Page 39
$\int\limits_0^{\pi/2} \frac{dx}{a \cos x + b \sin x}a, b > 0$
Exercise 20.2 | Q 20 | Page 39
$\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx$
Exercise 20.2 | Q 21 | Page 39
$\int\limits_0^\pi \frac{\sin x}{\sin x + \cos x} dx$
Exercise 20.2 | Q 22 | Page 39
$\int\limits_0^\pi \frac{1}{3 + 2 \sin x + \cos x} dx$
Exercise 20.2 | Q 23 | Page 39
$\int\limits_0^1 \tan^{- 1} x\ dx$
Exercise 20.2 | Q 24 | Page 39
$\int_0^\frac{1}{2} \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}dx$
Exercise 20.2 | Q 25 | Page 39
$\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx$
Exercise 20.2 | Q 26 | Page 39
$\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx$
Exercise 20.2 | Q 27 | Page 39
$\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx$
Exercise 20.2 | Q 28 | Page 39
$\int\limits_0^{\pi/2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx$
Exercise 20.2 | Q 29 | Page 39
$\int\limits_0^{\pi/2} \frac{x + \sin x}{1 + \cos x} dx$
Exercise 20.2 | Q 30 | Page 39
$\int\limits_0^1 \frac{\tan^{- 1} x}{1 + x^2} dx$
Exercise 20.2 | Q 31 | Page 39
$\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx$
Exercise 20.2 | Q 32 | Page 39
$\int\limits_0^1 x \tan^{- 1} x\ dx$
Exercise 20.2 | Q 33 | Page 39
$\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx$
Exercise 20.2 | Q 34 | Page 39
$\int\limits_0^1 \frac{24 x^3}{\left( 1 + x^2 \right)^4} dx$
Exercise 20.2 | Q 35 | Page 39
$\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx$
Exercise 20.2 | Q 36 | Page 39
$\int\limits_0^{\pi/2} x^2 \sin\ x\ dx$
Exercise 20.2 | Q 37 | Page 39
$\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx$
Exercise 20.2 | Q 38 | Page 39
$\int\limits_0^1 \frac{1 - x^2}{\left( 1 + x^2 \right)^2} dx$
Exercise 20.2 | Q 39 | Page 39
$\int\limits_{- 1}^1 5 x^4 \sqrt{x^5 + 1} dx$
Exercise 20.2 | Q 40 | Page 39
$\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx$
Exercise 20.2 | Q 41 | Page 39
$\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt$
Exercise 20.2 | Q 42 | Page 39
$\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta$
Exercise 20.2 | Q 43 | Page 39
$\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta$
Exercise 20.2 | Q 44 | Page 39
$\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx$
Exercise 20.2 | Q 45 | Page 40
$\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx$
Exercise 20.2 | Q 46 | Page 40
$\int\limits_0^{\pi/2} \cos^5 x\ dx$
Exercise 20.2 | Q 47 | Page 40
$\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx$
Exercise 20.2 | Q 48 | Page 40
$\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx$
Exercise 20.2 | Q 49 | Page 40
$\int\limits_0^{\pi/2} 2 \sin x \cos x \tan^{- 1} \left( \sin x \right) dx$
Exercise 20.2 | Q 50 | Page 39
$\int\limits_0^{\pi/2} \sin 2x \tan^{- 1} \left( \sin x \right) dx$
Exercise 20.2 | Q 51 | Page 40
$\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx$
Exercise 20.2 | Q 52 | Page 40
$\int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx$
Exercise 20.2 | Q 53 | Page 40
$\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx$
Exercise 20.2 | Q 54 | Page 40
$\int\limits_0^a x \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} dx$
Exercise 20.2 | Q 55 | Page 40
$\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx$
Exercise 20.2 | Q 56 | Page 40
$\int\limits_0^{\pi/2} \frac{\sin x \cos x}{\cos^2 x + 3 \cos x + 2} dx$
Exercise 20.2 | Q 57 | Page 40
$\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx$
Exercise 20.2 | Q 58 | Page 40
$\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx$
Exercise 20.2 | Q 59 | Page 40
$\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx$
Exercise 20.2 | Q 60 | Page 40
$\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx$
Exercise 20.2 | Q 61 | Page 40
$\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx$
Exercise 20.2 | Q 62 | Page 40
$\int_0^\frac{\pi}{2} \frac{\cos x}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^n}dx$
Exercise 20.3 [Pages 55 - 56]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Exercise 20.3 [Pages 55 - 56]
Exercise 20.3 | Q 1.1 | Page 55
$\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}$
Exercise 20.3 | Q 1.2 | Page 55
$\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}$
Exercise 20.3 | Q 1.3 | Page 55
$\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}$
Exercise 20.3 | Q 2 | Page 56
Evaluate the following integral:
$\int\limits_{- 4}^4 \left| x + 2 \right| dx$
Exercise 20.3 | Q 3 | Page 56
Evaluate the following integral:
$\int\limits_{- 3}^3 \left| x + 1 \right| dx$
Exercise 20.3 | Q 4 | Page 56
Evaluate the following integral:
$\int\limits_{- 1}^1 \left| 2x + 1 \right| dx$
Exercise 20.3 | Q 5 | Page 56
Evaluate the following integral:
$\int\limits_{- 2}^2 \left| 2x + 3 \right| dx$
Exercise 20.3 | Q 6 | Page 56
Evaluate the following integral:
$\int\limits_0^2 \left| x^2 - 3x + 2 \right| dx$
Exercise 20.3 | Q 7 | Page 56
Evaluate the following integral:
$\int\limits_0^3 \left| 3x - 1 \right| dx$
Exercise 20.3 | Q 8 | Page 56
Evaluate the following integral:
$\int\limits_{- 6}^6 \left| x + 2 \right| dx$
Exercise 20.3 | Q 9 | Page 56
Evaluate the following integral:
$\int\limits_{- 2}^2 \left| x + 1 \right| dx$
Exercise 20.3 | Q 10 | Page 56
Evaluate the following integral:
$\int\limits_1^2 \left| x - 3 \right| dx$
Exercise 20.3 | Q 11 | Page 56
Evaluate the following integral:
$\int\limits_0^{\pi/2} \left| \cos 2x \right| dx$
Exercise 20.3 | Q 12 | Page 56
Evaluate the following integral:
$\int\limits_0^{2\pi} \left| \sin x \right| dx$
Exercise 20.3 | Q 13 | Page 56
Evaluate the following integral:
$\int\limits_{- \pi/4}^{\pi/4} \left| \sin x \right| dx$
Exercise 20.3 | Q 14 | Page 56
Evaluate the following integral:
$\int\limits_2^8 \left| x - 5 \right| dx$
Exercise 20.3 | Q 15 | Page 56
Evaluate the following integral:
$\int\limits_{- \pi/2}^{\pi/2} \left\{ \sin \left| x \right| + \cos \left| x \right| \right\} dx$
Exercise 20.3 | Q 16 | Page 56
Evaluate the following integral:
$\int\limits_0^4 \left| x - 1 \right| dx$
Exercise 20.3 | Q 17 | Page 56
Evaluate the following integral:
$\int\limits_1^4 \left\{ \left| x - 1 \right| + \left| x - 2 \right| + \left| x - 4 \right| \right\} dx$
Exercise 20.3 | Q 18 | Page 56
Evaluate the following integral:
$\int\limits_{- 5}^0 f\left( x \right) dx, where\ f\left( x \right) = \left| x \right| + \left| x + 2 \right| + \left| x + 5 \right|$
Exercise 20.3 | Q 19 | Page 56
Evaluate the following integral:
$\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx$
Exercise 20.3 | Q 20 | Page 56
$\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx$
Exercise 20.3 | Q 21 | Page 56
$\int_{- 2}^2 x e^\left| x \right| dx$
Exercise 20.3 | Q 22 | Page 56
$\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx$
Exercise 20.3 | Q 23 | Page 56
$\int_0^\pi \cos x\left| \cos x \right|dx$
Exercise 20.3 | Q 24 | Page 56
$\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx$
Exercise 20.3 | Q 25 | Page 56
$\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx$
Exercise 20.3 | Q 26 | Page 56
$\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx$
Exercise 20.3 | Q 27 | Page 56
$\int_0^2 2x\left[ x \right]dx$
Exercise 20.3 | Q 28 | Page 56
$\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx$
Exercise 20.4 [Page 61]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Exercise 20.4 [Page 61]
Exercise 20.4 | Q 1 | Page 61
Evaluate each of the following integral:
$\int_0^{2\pi} \frac{e^\ sin x}{e^\ sin x + e^{- \ sin x}}dx$
Exercise 20.4 | Q 2 | Page 61
Evaluate each of the following integral:
$\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx$
Exercise 20.4 | Q 3 | Page 61
Evaluate each of the following integral:
$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + \sqrt{\cot x}}dx$
Exercise 20.4 | Q 4 | Page 61
Evaluate each of the following integral:
$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$
Exercise 20.4 | Q 5 | Page 61
Evaluate each of the following integral:
$\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{\tan^2 x}{1 + e^x}dx$
Exercise 20.4 | Q 6 | Page 61
Evaluate each of the following integral:
$\int_{- a}^a \frac{1}{1 + a^x}dx$, a > 0
Exercise 20.4 | Q 7 | Page 61
Evaluate each of the following integral:
$\int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^\ tan\ x}dx$
Exercise 20.4 | Q 8 | Page 61
Evaluate each of the following integral:
$\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{\cos^2 x}{1 + e^x}dx$
Exercise 20.4 | Q 9 | Page 61
Evaluate each of the following integral:
$\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{x^{11} - 3 x^9 + 5 x^7 - x^5 + 1}{\cos^2 x}dx$
Exercise 20.4 | Q 10 | Page 61
Evaluate each of the following integral:
$\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2$
Exercise 20.4 | Q 11 | Page 61
$\int\limits_0^{\pi/2} \left( 2 \log \cos x - \log \sin 2x \right) dx$
Exercise 20.4 | Q 12 | Page 61
$\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx$
Exercise 20.4 | Q 13 | Page 61
$\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx$
Exercise 20.4 | Q 14 | Page 61
$\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx$
Exercise 20.4 | Q 15 | Page 61
$\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx$
Exercise 20.4 | Q 16 | Page 61
If $f\left( a + b - x \right) = f\left( x \right)$ , then prove that $\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx$
Exercise 20.5 [Pages 94 - 96]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Exercise 20.5 [Pages 94 - 96]
Exercise 20.5 | Q 1 | Page 94
$\int\limits_0^{\pi/2} \frac{1}{1 + \tan x}$
Exercise 20.5 | Q 2 | Page 94
$\int\limits_0^{\pi/2} \frac{1}{1 + \cot x} dx$
Exercise 20.5 | Q 3 | Page 94
$\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx$
Exercise 20.5 | Q 4 | Page 94
$\int\limits_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx$
Exercise 20.5 | Q 5 | Page 94
$\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx$
Exercise 20.5 | Q 6 | Page 94
$\int\limits_0^{\pi/2} \frac{1}{1 + \sqrt{\tan x}} dx$
Exercise 20.5 | Q 7 | Page 95
$\int\limits_0^a \frac{1}{x + \sqrt{a^2 - x^2}} dx$
Exercise 20.5 | Q 8 | Page 95
$\int\limits_0^\infty \frac{\log x}{1 + x^2} dx$
Exercise 20.5 | Q 9 | Page 95
$\int\limits_0^1 \frac{\log\left( 1 + x \right)}{1 + x^2} dx$
Exercise 20.5 | Q 10 | Page 95
$\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx$
Exercise 20.5 | Q 11 | Page 95
$\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx$
Exercise 20.5 | Q 12 | Page 95
$\int\limits_0^\pi x \sin x \cos^4 x\ dx$
Exercise 20.5 | Q 13 | Page 95
$\int\limits_0^\pi x \sin^3 x\ dx$
Exercise 20.5 | Q 14 | Page 95
$\int\limits_0^\pi x \log \sin x\ dx$
Exercise 20.5 | Q 15 | Page 95
$\int\limits_0^\pi \frac{x \sin x}{1 + \sin x} dx$
Exercise 20.5 | Q 16 | Page 95
$\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi$
Exercise 20.5 | Q 17 | Page 95
$\int\limits_0^\pi x \cos^2 x\ dx$
Exercise 20.5 | Q 18 | Page 95
Evaluate the following integral:
$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \cot^\frac{3}{2} x}dx$
Exercise 20.5 | Q 19 | Page 95
Evaluate the following integral:
$\int_0^\frac{\pi}{2} \frac{\tan^7 x}{\tan^7 x + \cot^7 x}dx$
Exercise 20.5 | Q 20 | Page 95
Evaluate the following integral:
$\int_2^8 \frac{\sqrt{10 - x}}{\sqrt{x} + \sqrt{10 - x}}dx$
Exercise 20.5 | Q 21 | Page 95
Evaluate the following integral:
$\int_0^\pi x\sin x \cos^2 xdx$
Exercise 20.5 | Q 22 | Page 95
$\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx$
Exercise 20.5 | Q 23 | Page 95
$\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx$
Exercise 20.5 | Q 24 | Page 95
$\int\limits_{- \pi/2}^{\pi/2} \sin^4 x\ dx$
Exercise 20.5 | Q 25 | Page 95
$\int\limits_{- 1}^1 \log\left( \frac{2 - x}{2 + x} \right) dx$
Exercise 20.5 | Q 26 | Page 95
$\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx$
Exercise 20.5 | Q 27 | Page 95
$\int\limits_0^\pi \log\left( 1 - \cos x \right) dx$
Exercise 20.5 | Q 28 | Page 95
$\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx$
Exercise 20.5 | Q 29 | Page 95
Evaluate the following integral:
$\int_{- \pi}^\pi \frac{2x\left( 1 + \sin x \right)}{1 + \cos^2 x}dx$
Exercise 20.5 | Q 30 | Page 95
Evaluate the following integral:
$\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta$
Exercise 20.5 | Q 31 | Page 95
Evaluate the following integral:
$\int_{- 2}^2 \frac{3 x^3 + 2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx$
Exercise 20.5 | Q 32 | Page 95
Evaluate the following integral:
$\int_{- \frac{3\pi}{2}}^{- \frac{\pi}{2}} \left\{ \sin^2 \left( 3\pi + x \right) + \left( \pi + x \right)^3 \right\}dx$
Exercise 20.5 | Q 33 | Page 95
$\int\limits_0^2 x\sqrt{2 - x} dx$
Exercise 20.5 | Q 34 | Page 95
$\int\limits_0^1 \log\left( \frac{1}{x} - 1 \right) dx$
Exercise 20.5 | Q 35 | Page 95
Evaluate the following integral:
$\int_{- 1}^1 \left| xcos\pi x \right|dx$
Exercise 20.5 | Q 36 | Page 95
Evaluate the following integral:
$\int_0^\pi \left( \frac{x}{1 + \sin^2 x} + \cos^7 x \right)dx$
Exercise 20.5 | Q 37 | Page 95
Evaluate
$\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx$
Exercise 20.5 | Q 38 | Page 95
Evaluate the following integral:
$\int_0^{2\pi} \sin^{100} x \cos^{101} xdx$
Exercise 20.5 | Q 39 | Page 95
Evaluate the following integral:
$\int_0^\frac{\pi}{2} \frac{a\sin x + b\sin x}{\sin x + \cos x}dx$
Exercise 20.5 | Q 40 | Page 95
int_0^(2a)f(x)dx
Exercise 20.5 | Q 41 | Page 95
$\int_0^1 | x\sin \pi x | dx$
Exercise 20.5 | Q 42 | Page 95
Evaluate :
$\int\limits_0^{3/2} \left| x \sin \pi x \right|dx$
Exercise 20.5 | Q 43 | Page 96
If f is an integrable function such that f(2a − x) = f(x), then prove that
$\int\limits_0^{2a} f\left( x \right) dx = 2 \int\limits_0^a f\left( x \right) dx$
Exercise 20.5 | Q 44 | Page 96
If f(2a − x) = −f(x), prove that
$\int\limits_0^{2a} f\left( x \right) dx = 0 .$
Exercise 20.5 | Q 45.1 | Page 96
If f is an integrable function, show that
$\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx$
Exercise 20.5 | Q 45.2 | Page 96
If f is an integrable function, show that
$\int\limits_{- a}^a x f\left( x^2 \right) dx = 0$
Exercise 20.5 | Q 46 | Page 96
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
$\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx$
Exercise 20.5 | Q 47 | Page 96
If $f\left( a + b - x \right) = f\left( x \right)$ , then prove that
$\int_a^b xf\left( x \right)dx = \left( \frac{a + b}{2} \right) \int_a^b f\left( x \right)dx$
Exercise 20.5 | Q 48 | Page 96
If f(x) is a continuous function defined on [−aa], then prove that
$\int\limits_{- a}^a f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( - x \right) \right\} dx$
Exercise 20.5 | Q 49 | Page 96
Prove that:
$\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx$
Exercise 20.6 [Pages 110 - 111]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Exercise 20.6 [Pages 110 - 111]
Exercise 20.6 | Q 1 | Page 110
$\int\limits_0^3 \left( x + 4 \right) dx$
Exercise 20.6 | Q 2 | Page 110
$\int\limits_0^2 \left( x + 3 \right) dx$
Exercise 20.6 | Q 3 | Page 110
$\int\limits_1^3 \left( 3x - 2 \right) dx$
Exercise 20.6 | Q 4 | Page 110
$\int\limits_{- 1}^1 \left( x + 3 \right) dx$
Exercise 20.6 | Q 5 | Page 110
$\int\limits_0^5 \left( x + 1 \right) dx$
Exercise 20.6 | Q 6 | Page 110
$\int\limits_1^3 \left( 2x + 3 \right) dx$
Exercise 20.6 | Q 7 | Page 110
$\int\limits_3^5 \left( 2 - x \right) dx$
Exercise 20.6 | Q 8 | Page 110
$\int\limits_0^2 \left( x^2 + 1 \right) dx$
Exercise 20.6 | Q 9 | Page 110
$\int\limits_1^2 x^2 dx$
Exercise 20.6 | Q 10 | Page 110
$\int\limits_2^3 \left( 2 x^2 + 1 \right) dx$
Exercise 20.6 | Q 11 | Page 110
$\int\limits_1^2 \left( x^2 - 1 \right) dx$
Exercise 20.6 | Q 12 | Page 110
$\int\limits_0^2 \left( x^2 + 4 \right) dx$
Exercise 20.6 | Q 13 | Page 111
$\int\limits_1^4 \left( x^2 - x \right) dx$
Exercise 20.6 | Q 14 | Page 111
$\int\limits_0^1 \left( 3 x^2 + 5x \right) dx$
Exercise 20.6 | Q 15 | Page 111
$\int\limits_0^2 e^x dx$
Exercise 20.6 | Q 16 | Page 111
$\int\limits_a^b e^x dx$
Exercise 20.6 | Q 17 | Page 111
$\int\limits_a^b \cos\ x\ dx$
Exercise 20.6 | Q 18 | Page 111
$\int\limits_0^{\pi/2} \sin x\ dx$
Exercise 20.6 | Q 19 | Page 111
$\int\limits_0^{\pi/2} \cos x\ dx$
Exercise 20.6 | Q 20 | Page 111
$\int\limits_1^4 \left( 3 x^2 + 2x \right) dx$
Exercise 20.6 | Q 21 | Page 111
$\int\limits_0^2 \left( 3 x^2 - 2 \right) dx$
Exercise 20.6 | Q 22 | Page 111
$\int\limits_0^2 \left( x^2 + 2 \right) dx$
Exercise 20.6 | Q 23 | Page 111
$\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx$
Exercise 20.6 | Q 23 | Page 111
$\int\limits_0^4 \left( x + e^{2x} \right) dx$
Exercise 20.6 | Q 24 | Page 111
$\int\limits_0^2 \left( x^2 + x \right) dx$
Exercise 20.6 | Q 25 | Page 111
$\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx$
Exercise 20.6 | Q 26 | Page 111
$\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx$
Exercise 20.6 | Q 27 | Page 111
$\int\limits_a^b x\ dx$
Exercise 20.6 | Q 28 | Page 111
$\int\limits_0^5 \left( x + 1 \right) dx$
Exercise 20.6 | Q 29 | Page 111
$\int\limits_2^3 x^2 dx$
Exercise 20.6 | Q 30 | Page 111
$\int\limits_1^4 \left( x^2 - x \right) dx$
Exercise 20.6 | Q 31 | Page 111
$\int\limits_0^2 \left( x^2 - x \right) dx$
Exercise 20.6 | Q 32 | Page 111
$\int\limits_1^3 \left( 2 x^2 + 5x \right) dx$
Exercise 20.6 | Q 33 | Page 111
Evaluate the following integrals as limit of sums:
$\int_1^3 \left( 3 x^2 + 1 \right)dx$
Very Short Answers [Pages 115 - 116]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Very Short Answers [Pages 115 - 116]
Very Short Answers | Q 1 | Page 115
$\int\limits_0^{\pi/2} \sin^2 x\ dx .$
Very Short Answers | Q 2 | Page 115
$\int\limits_0^{\pi/2} \cos^2 x\ dx .$
Very Short Answers | Q 3 | Page 115
$\int\limits_{- \pi/2}^{\pi/2} \sin^2 x\ dx .$
Very Short Answers | Q 4 | Page 115
$\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .$
Very Short Answers | Q 5 | Page 111
$\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx .$
Very Short Answers | Q 6 | Page 115
$\int\limits_{- \pi/2}^{\pi/2} x \cos^2 x\ dx .$
Very Short Answers | Q 7 | Page 115
$\int\limits_0^{\pi/4} \tan^2 x\ dx .$
Very Short Answers | Q 8 | Page 115
$\int\limits_0^1 \frac{1}{x^2 + 1} dx$
Very Short Answers | Q 9 | Page 115
$\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .$
Very Short Answers | Q 10 | Page 115
$\int\limits_0^\infty e^{- x} dx .$
Very Short Answers | Q 11 | Page 115
$\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .$
Very Short Answers | Q 12 | Page 115
$\int\limits_0^3 \frac{1}{x^2 + 9} dx .$
Very Short Answers | Q 13 | Page 115
$\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .$
Very Short Answers | Q 14 | Page 115
$\int\limits_0^{\pi/2} \log \tan x\ dx .$
Very Short Answers | Q 15 | Page 115
$\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .$
Very Short Answers | Q 16 | Page 115
$\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .$
Very Short Answers | Q 17 | Page 115
$\int\limits_0^\pi \cos^5 x\ dx .$
Very Short Answers | Q 18 | Page 115
$\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta$
Very Short Answers | Q 19 | Page 115
$\int\limits_{- 1}^1 x\left| x \right| dx .$
Very Short Answers | Q 20 | Page 115
$\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .$
Very Short Answers | Q 21 | Page 115
$\int\limits_0^1 \frac{1}{1 + x^2} dx$
Very Short Answers | Q 22 | Page 115
Evaluate each of the following integral:
$\int_0^\frac{\pi}{4} \tan\ xdx$
Very Short Answers | Q 23 | Page 115
$\int\limits_2^3 \frac{1}{x}dx$
Very Short Answers | Q 24 | Page 115
$\int\limits_0^2 \sqrt{4 - x^2} dx$
Very Short Answers | Q 25 | Page 115
$\int\limits_0^1 \frac{2x}{1 + x^2} dx$
Very Short Answers | Q 26 | Page 115
Evaluate each of the following integral:
$\int_0^1 x e^{x^2} dx$
Very Short Answers | Q 27 | Page 115
Evaluate each of the following integral:
$\int_0^\frac{\pi}{4} \sin2xdx$
Very Short Answers | Q 28 | Page 115
Evaluate each of the following integral:
$\int_e^{e^2} \frac{1}{x\log x}dx$
Very Short Answers | Q 29 | Page 115
Evaluate each of the following integral:
$\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx$
Very Short Answers | Q 30 | Page 115
Solve each of the following integral:
$\int_2^4 \frac{x}{x^2 + 1}dx$
Very Short Answers | Q 31 | Page 116
If $\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,$ find the value of k.
Very Short Answers | Q 32 | Page 116
If $\int\limits_0^a 3 x^2 dx = 8,$ write the value of a.
Very Short Answers | Q 33 | Page 116
If $f\left( x \right) = \int_0^x t\sin tdt$, the write the value of $f'\left( x \right)$
Very Short Answers | Q 34 | Page 116
If $\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}$ , find the value of a.
Very Short Answers | Q 35 | Page 116
Write the coefficient abc of which the value of the integral
$\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx$ is independent.
Very Short Answers | Q 36 | Page 116
Evaluate :
$\int\limits_2^3 3^x dx .$
Very Short Answers | Q 37 | Page 116
$\int\limits_0^2 \left[ x \right] dx .$
Very Short Answers | Q 38 | Page 116
$\int\limits_0^{15} \left[ x \right] dx .$
Very Short Answers | Q 39 | Page 116
$\int\limits_0^1 \left\{ x \right\} dx,$ where {x} denotes the fractional part of x.
Very Short Answers | Q 40 | Page 116
$\int\limits_0^1 e^\left\{ x \right\} dx .$
Very Short Answers | Q 41 | Page 116
$\int\limits_0^2 x\left[ x \right] dx .$
Very Short Answers | Q 42 | Page 116
$\int\limits_0^1 2^{x - \left[ x \right]} dx$
Very Short Answers | Q 43 | Page 116
$\int\limits_1^2 \log_e \left[ x \right] dx .$
Very Short Answers | Q 44 | Page 116
$\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .$
Very Short Answers | Q 45 | Page 116
If $\left[ \cdot \right] and \left\{ \cdot \right\}$ denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
$\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx$
MCQ [Pages 117 - 120]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals MCQ [Pages 117 - 120]
MCQ | Q 1 | Page 117
$\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx$ equals
• π/2
• π/4
• π/6
• π/8
MCQ | Q 2 | Page 117
$\int\limits_0^\pi \frac{1}{1 + \sin x} dx$ equals
• 0
• 1/2
• 2
• 3/2
MCQ | Q 3 | Page 117
The value of $\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx$ is __________ .
• $\frac{\pi^2}{4}$
• $\frac{\pi^2}{2}$
• $\frac{3 \pi^2}{2}$
• $\frac{\pi^2}{3}$
MCQ | Q 4 | Page 117
The value of $\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx$ is
• 0
• 2
• 8
• 4
MCQ | Q 5 | Page 117
The value of the integral $\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ is
• 0
• π/2
• π/4
• none of these
MCQ | Q 6 | Page 117
$\int\limits_0^\infty \frac{1}{1 + e^x} dx$ equals
• log 2 − 1
• log 2
• log 4 − 1
• − log 2
MCQ | Q 7 | Page 117
$\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx$ equals
• 2
• 1
• π/4
• π2/8
MCQ | Q 8 | Page 117
$\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx$ equals
• $\log\left( \frac{2}{3} \right)$
• $\log\left( \frac{3}{2} \right)$
• $\log\left( \frac{3}{4} \right)$
• $\log\left( \frac{4}{3} \right)$
MCQ | Q 9 | Page 117
$\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx$ equals
• $\frac{1}{3} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)$
• $\frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right)$
• $\sqrt{3} \tan^{- 1} \left( \sqrt{3} \right)$
• $2\sqrt{3} \tan^{- 1} \sqrt{3}$
MCQ | Q 10 | Page 117
int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"
• $\frac{\pi}{2}$
• $\frac{\pi}{2} - 1$
• $\frac{\pi}{2} + 1$
• π + 1
• None of these
MCQ | Q 11 | Page 117
$\int\limits_0^\pi \frac{1}{a + b \cos x} dx =$
• $\frac{\pi}{\sqrt{a^2 - b^2}}$
• $\frac{\pi}{ab}$
• $\frac{\pi}{a^2 + b^2}$
• (a + b) π
MCQ | Q 12 | Page 118
$\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\cot}x} dx$ is
• π/3
• π/6
• π/12
• π/2
MCQ | Q 13 | Page 118
Given that $\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},$ the value of $\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},$
• $\frac{\pi}{60}$
• $\frac{\pi}{20}$
• $\frac{\pi}{40}$
• $\frac{\pi}{80}$
MCQ | Q 14 | Page 118
$\int\limits_1^e \log x\ dx =$
• 1
• e − 1
• e + 1
• 0
MCQ | Q 15 | Page 118
$\int\limits_1^\sqrt{3} \frac{1}{1 + x^2} dx$ is equal to
• $\frac{\pi}{12}$
• $\frac{\pi}{6}$
• $\frac{\pi}{4}$
• $\frac{\pi}{3}$
MCQ | Q 16 | Page 118
$\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =$
• $\frac{\pi}{12} + \log\left( 2\sqrt{2} \right)$
• $\frac{\pi}{2} + \log\left( 2\sqrt{2} \right)$
• $\frac{\pi}{6} + \log\left( 2\sqrt{2} \right)$
• $\frac{\pi}{3} + \log\left( 2\sqrt{2} \right)$
MCQ | Q 17 | Page 118
The value of the integral $\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx$
• $\frac{\pi}{2}$
• $\frac{\pi}{4}$
• $\frac{\pi}{6}$
• $\frac{\pi}{3}$
MCQ | Q 18 | Page 118
$\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx$ is equal to
• 1
• 2
• − 1
• − 2
MCQ | Q 19 | Page 118
$\int\limits_0^{\pi/2} \frac{1}{1 + \tan x} dx$ is equal to
• $\frac{ \pi}{4}$
• $\frac{\pi}{3}$
• $\frac{\pi}{2}$
• π
MCQ | Q 20 | Page 118
The value of $\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx$ is
• 1
• e − 1
• 0
• − 1
MCQ | Q 21 | Page 118
If $\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},$ then a equals
• $\frac{\pi}{2}$
• $\frac{1}{2}$
• $\frac{\pi}{4}$
• 1
MCQ | Q 22 | Page 118
If $\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx$ equals
• 4a2
• 0
• 2a2
• none of these
MCQ | Q 23 | Page 119
The value of $\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx$ is
• $\frac{\pi^4}{2}$
• $\frac{\pi^4}{4}$
• 0
• none of these
MCQ | Q 24 | Page 119
$\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx$ is equal to
• loge 3
• $\log_e \sqrt{3}$
• $\frac{1}{2}\log\left( - 1 \right)$
• log (−1)
MCQ | Q 25 | Page 119
$\int\limits_{- 1}^1 \left| 1 - x \right| dx$ is equal to
• −2
• 2
• 0
• 4
MCQ | Q 26 | Page 119
The derivative of $f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),$ is
• $\frac{1}{3 \ln x}$
• $\frac{1}{3 \ln x} - \frac{1}{2 \ln x}$
• (ln x)−1 x (x − 1)
• $\frac{3 x^2}{\ln x}$
MCQ | Q 27 | Page 119
If $I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,$ then the value of I10 + 90I8 is
• $9 \left( \frac{\pi}{2} \right)^9$
• $10 \left( \frac{\pi}{2} \right)^9$
• $\left( \frac{\pi}{2} \right)^9$
• $9 \left( \frac{\pi}{2} \right)^8$
MCQ | Q 28 | Page 119
$\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =$
• 15/16
• 3/16
• -3/16
• -16/3
MCQ | Q 29 | Page 119
$\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}$ is equal to
• $\ln\left( \frac{1}{3} \right)$
• $\ln\left( \frac{2}{3} \right)$
• $\ln\left( \frac{3}{2} \right)$
• $\ln\left( \frac{4}{3} \right)$
MCQ | Q 30 | Page 118
The value of the integral $\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx$ is ________ .
• 4
• 2
• −2
• 0
MCQ | Q 31 | Page 119
$\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx$ is equal to
• 0
• 1
• π/2
• π/4
MCQ | Q 32 | Page 119
$\int\limits_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx$ equals to
MCQ | Q 33 | Page 120
$\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx$ is equal to
• 0
• π
• π/2
• π/4
MCQ | Q 34 | Page 120
$\int\limits_0^{\pi/2} x \sin x\ dx$ is equal to
• π/4
• π/2
• π
• 1
MCQ | Q 35 | Page 120
$\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx$ is equal to
• π
• π/2
• 0
MCQ | Q 36 | Page 120
The value of $\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx$ is
• π/4
• π/8
• π/2
• 0
MCQ | Q 37 | Page 120
$\int\limits_0^\infty \log\left( x + \frac{1}{x} \right) \frac{1}{1 + x^2} dx =$
• π ln 2
• −π ln 2
• 0
• $- \frac{\pi}{2}\ln 2$
MCQ | Q 38 | Page 120
$\int\limits_0^{2a} f\left( x \right) dx$ is equal to
• $2 \int\limits_0^a f\left( x \right) dx$
• 0
• $\int\limits_0^a f\left( x \right) dx + \int\limits_0^a f\left( 2a - x \right) dx$
• $\int\limits_0^a f\left( x \right) dx + \int\limits_0^{2a} f\left( 2a - x \right) dx$
MCQ | Q 39 | Page 120
If f (a + b − x) = f (x), then $\int\limits_a^b$ x f (x) dx is equal to
• $\frac{a + b}{2} \int\limits_a^b f\left( b - x \right) dx$
• $\frac{a + b}{2} \int\limits_a^b f\left( b + x \right) dx$
• $\frac{b - a}{2} \int\limits_a^b f\left( x \right) dx$
• $\frac{b + a}{2} \int\limits_a^b f\left( x \right) dx$
MCQ | Q 40 | Page 120
The value of $\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,$ is
• 1
• 0
• −1
• π/4
MCQ | Q 41 | Page 120
The value of $\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx$ is
• 2
• $\frac{3}{4}$
• 0
• −2
MCQ | Q 42 | Page 120
The value of $\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx,$ is
• 0
• 2
• π
• 1
Revision Exercise [Pages 121 - 123]
### RD Sharma solutions for Class 12 Maths Chapter 20 Definite Integrals Revision Exercise [Pages 121 - 123]
Revision Exercise | Q 1 | Page 121
$\int\limits_0^4 x\sqrt{4 - x} dx$
Revision Exercise | Q 2 | Page 121
$\int\limits_1^2 x\sqrt{3x - 2} dx$
Revision Exercise | Q 3 | Page 121
$\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx$
Revision Exercise | Q 4 | Page 121
$\int\limits_0^1 \cos^{- 1} x dx$
Revision Exercise | Q 5 | Page 121
$\int\limits_0^1 \tan^{- 1} x dx$
Revision Exercise | Q 6 | Page 121
$\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx$
Revision Exercise | Q 7 | Page 121
$\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx$
Revision Exercise | Q 8 | Page 121
$\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx$
Revision Exercise | Q 9 | Page 121
$\int\limits_0^1 \frac{1 - x}{1 + x} dx$
Revision Exercise | Q 10 | Page 121
$\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx$
Revision Exercise | Q 11 | Page 121
$\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx$
Revision Exercise | Q 12 | Page 121
$\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx$
Revision Exercise | Q 13 | Page 121
$\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx$
Revision Exercise | Q 14 | Page 121
$\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx$
Revision Exercise | Q 15 | Page 121
$\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx$
Revision Exercise | Q 16 | Page 121
$\int\limits_0^{\pi/4} \sin 2x \sin 3x dx$
Revision Exercise | Q 17 | Page 121
$\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx$
Revision Exercise | Q 18 | Page 121
$\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx$
Revision Exercise | Q 19 | Page 121
$\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx$
Revision Exercise | Q 20 | Page 121
$\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx$
Revision Exercise | Q 21 | Page 121
$\int\limits_0^{\pi/2} x^2 \cos 2x dx$
Revision Exercise | Q 22 | Page 121
$\int\limits_0^1 \log\left( 1 + x \right) dx$
Revision Exercise | Q 23 | Page 121
Evaluate the following integrals :-
$\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx$
Revision Exercise | Q 24 | Page 121
$\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx$
Revision Exercise | Q 25 | Page 121
$\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx$
Revision Exercise | Q 26 | Page 121
$\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx$
Revision Exercise | Q 27 | Page 122
$\int\limits_0^{\pi/4} e^x \sin x dx$
Revision Exercise | Q 28 | Page 122
$\int\limits_0^{\pi/4} \tan^4 x dx$
Revision Exercise | Q 29 | Page 122
$\int\limits_0^1 \left| 2x - 1 \right| dx$
Revision Exercise | Q 30 | Page 122
$\int\limits_1^3 \left| x^2 - 2x \right| dx$
Revision Exercise | Q 31 | Page 122
$\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx$
Revision Exercise | Q 32 | Page 122
$\int\limits_0^1 \left| \sin 2\pi x \right| dx$
Revision Exercise | Q 33 | Page 122
$\int\limits_1^3 \left| x^2 - 4 \right| dx$
Revision Exercise | Q 34 | Page 122
$\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx$
Revision Exercise | Q 35 | Page 122
$\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx$
Revision Exercise | Q 36 | Page 122
$\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx$
Revision Exercise | Q 37 | Page 122
$\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx$
Revision Exercise | Q 38 | Page 122
$\int\limits_0^{2\pi} \cos^7 x dx$
Revision Exercise | Q 39 | Page 122
$\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx$
Revision Exercise | Q 40 | Page 122
$\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx$
Revision Exercise | Q 41 | Page 122
$\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx$
Revision Exercise | Q 42 | Page 122
$\int\limits_0^\pi x \sin x \cos^4 x dx$
Revision Exercise | Q 43 | Page 122
$\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx$
Revision Exercise | Q 44 | Page 122
$\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx$
Revision Exercise | Q 45 | Page 122
$\int\limits_0^{15} \left[ x^2 \right] dx$
Revision Exercise | Q 46 | Page 122
$\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx$
Revision Exercise | Q 47 | Page 12
$\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx$
Revision Exercise | Q 48 | Page 122
$\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx$
Revision Exercise | Q 49 | Page 122
$\int\limits_0^\pi \cos 2x \log \sin x dx$
Revision Exercise | Q 50 | Page 122
$\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1$
Revision Exercise | Q 51 | Page 122
$\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx$
Revision Exercise | Q 52 | Page 122
$\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx$
Revision Exercise | Q 53 | Page 122
$\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx$
Revision Exercise | Q 54 | Page 122
$\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx$
Revision Exercise | Q 55 | Page 122
$\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx$
Revision Exercise | Q 56 | Page 122
$\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx$
Revision Exercise | Q 57 | Page 122
$\int\limits_0^\pi \frac{dx}{6 - \cos x}dx$
Revision Exercise | Q 58 | Page 122
$\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx$
Revision Exercise | Q 59 | Page 123
$\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx$
Revision Exercise | Q 60 | Page 123
$\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx$
Revision Exercise | Q 61 | Page 123
$\int\limits_0^4 x dx$
Revision Exercise | Q 62 | Page 123
$\int\limits_0^2 \left( 2 x^2 + 3 \right) dx$
Revision Exercise | Q 63 | Page 123
$\int\limits_1^4 \left( x^2 + x \right) dx$
Revision Exercise | Q 64 | Page 123
$\int\limits_{- 1}^1 e^{2x} dx$
Revision Exercise | Q 65 | Page 123
$\int\limits_2^3 e^{- x} dx$
Revision Exercise | Q 66 | Page 123
$\int\limits_1^3 \left( 2 x^2 + 5x \right) dx$
Revision Exercise | Q 67 | Page 123
$\int\limits_1^3 \left( x^2 + 3x \right) dx$
Revision Exercise | Q 68 | Page 123
$\int\limits_0^2 \left( x^2 + 2 \right) dx$
Revision Exercise | Q 69 | Page 123
$\int\limits_0^3 \left( x^2 + 1 \right) dx$
## Solutions for Chapter 20: Definite Integrals
Exercise 20.1Exercise 20.2Exercise 20.3Exercise 20.4Exercise 20.5Exercise 20.6Very Short AnswersMCQRevision Exercise
## RD Sharma solutions for Class 12 Maths chapter 20 - Definite Integrals
Shaalaa.com has the CBSE, Karnataka Board PUC Mathematics Class 12 Maths CBSE, Karnataka Board PUC solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. RD Sharma solutions for Mathematics Class 12 Maths CBSE, Karnataka Board PUC 20 (Definite Integrals) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.
Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.
Concepts covered in Class 12 Maths chapter 20 Definite Integrals are Definite Integrals Problems, Indefinite Integral Problems, Comparison Between Differentiation and Integration, Integrals of Some Particular Functions, Indefinite Integral by Inspection, Some Properties of Indefinite Integral, Integration Using Trigonometric Identities, Introduction of Integrals, Evaluation of Definite Integrals by Substitution, Properties of Definite Integrals, Fundamental Theorem of Calculus, Definite Integral as the Limit of a Sum, Evaluation of Simple Integrals of the Following Types and Problems, Methods of Integration: Integration by Parts, Methods of Integration: Integration Using Partial Fractions, Methods of Integration: Integration by Substitution, Integration as an Inverse Process of Differentiation, Geometrical Interpretation of Indefinite Integrals.
Using RD Sharma Class 12 Maths solutions Definite Integrals exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in RD Sharma Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board PUC Class 12 Maths students prefer RD Sharma Textbook Solutions to score more in exams.
Get the free view of Chapter 20, Definite Integrals Class 12 Maths additional questions for Mathematics Class 12 Maths CBSE, Karnataka Board PUC, and you can use Shaalaa.com to keep it handy for your exam preparation. | 20,519 | 47,068 | {"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} | 3.9375 | 4 | CC-MAIN-2023-40 | latest | en | 0.482166 |
https://numberworld.info/458443762 | 1,670,153,180,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00443.warc.gz | 465,710,388 | 3,888 | # Number 458443762
### Properties of number 458443762
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
1b534bf2
Base 32:
dl6ivi
sin(458443762)
-0.80177577568453
cos(458443762)
0.59762497063415
tan(458443762)
-1.3416035391455
ln(458443762)
19.943348185649
lg(458443762)
8.6612860677027
sqrt(458443762)
21411.299867126
Square(458443762)
### Number Look Up
Look Up
458443762 (four hundred fifty-eight million four hundred forty-three thousand seven hundred sixty-two) is a very impressive number. The cross sum of 458443762 is 43. If you factorisate the number 458443762 you will get these result 2 * 7 * 32745983. The number 458443762 has 8 divisors ( 1, 2, 7, 14, 32745983, 65491966, 229221881, 458443762 ) whith a sum of 785903616. The number 458443762 is not a prime number. The number 458443762 is not a fibonacci number. 458443762 is not a Bell Number. The number 458443762 is not a Catalan Number. The convertion of 458443762 to base 2 (Binary) is 11011010100110100101111110010. The convertion of 458443762 to base 3 (Ternary) is 1011221122100121121. The convertion of 458443762 to base 4 (Quaternary) is 123110310233302. The convertion of 458443762 to base 5 (Quintal) is 1414330200022. The convertion of 458443762 to base 8 (Octal) is 3324645762. The convertion of 458443762 to base 16 (Hexadecimal) is 1b534bf2. The convertion of 458443762 to base 32 is dl6ivi. The sine of the number 458443762 is -0.80177577568453. The cosine of 458443762 is 0.59762497063415. The tangent of the figure 458443762 is -1.3416035391455. The square root of 458443762 is 21411.299867126.
If you square 458443762 you will get the following result 210170682916712644. The natural logarithm of 458443762 is 19.943348185649 and the decimal logarithm is 8.6612860677027. that 458443762 is impressive number! | 670 | 1,963 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-49 | latest | en | 0.721057 |
https://quiltnparty.com/2017/01/02/2017-stash-bee-for-january/ | 1,500,749,066,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424090.53/warc/CC-MAIN-20170722182647-20170722202647-00514.warc.gz | 713,113,693 | 39,834 | # 2017 Stash Bee For January…
Welcome to 2017! I hope you had a nice New Year celebration, however you rang in the New Year. I’m very excited to start my second year of the Stash Bee as a Hive Mama for Hive 4. That means that I get to share my tutorial in January. Let’s get to my block. I’ve seen this block done as a 12″ or 14″ block but the math was a hassle. To simplify things I upsized the block to 16″ finished.
Fabric Requirements – This is a chance to use up your scrappy scraps. The only thing I ask is that you not use civil war fabrics as they run more brown than I want. For the low volume pieces it’s ok if the low volume pieces have color in them as long as the read dominantly white, gray or off white.
Cutting Directions
Fabric A 40 – 2.5″ x 2.5″ scrappy squares
Fabric 2 – 4.5″ x 4.5″ low volume squares
Fabric C 2 – 4.5″ x 8.5″ low volume rectangles
Piecing Directions: use a 1/4″ seam allowance
1. Using Fabric A squares, layout 4 – 2.5″ squares, to create a 4 patch square. Sew rows together. Repeat so you have 2 – 4 patch squares.
2. Sew 1 Fabric B square to a 4 patch from step 1. Repeat so you have 2.
3. Sew 1 Fabric C rectangle to the unit from step 2. This unit should be 8 1/2″ x 8 1/2″. Repeat so you have 2.
4. Using the rest of the Fabric A squares, layout 16 – 2.5″ squares, to create a 16 patch block. Sew rows together. This unit should be 8 1/2″ x 8 1/2″. Repeat so you have 2 -16 patch squares.
5. Layout the units as shown in the photo. Sew rows together.
6. Sew the last seam to finish the block. The block should measure 16 1/2″ x 16 1/2″.
## One thought on “2017 Stash Bee For January…”
1. Pingback: Stash Bee 2017 Finish… | 514 | 1,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-30 | longest | en | 0.841569 |
http://encyclopedia2.thefreedictionary.com/discriminants | 1,477,636,190,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721558.87/warc/CC-MAIN-20161020183841-00412-ip-10-171-6-4.ec2.internal.warc.gz | 84,180,326 | 11,414 | # Discriminant
(redirected from discriminants)
Also found in: Dictionary, Wikipedia.
## discriminant
[di′skrim·ə·nənt]
(mathematics)
The quantity b 2- 4 ac, where a,b,c are coefficients of a given quadratic polynomial: ax 2+ bx + c.
More generally, for the polynomial equation a0 xn + a1 xn -1+···+ anx0= 0, a02 n-2times the product of the squares of all the differences of the roots of the equation, taken in pairs.
## Discriminant
The discriminant of a polynomial
P(x) = a0xn + a1xn−1 + … + an
is the expression
in which the product is distributed over all possible differences of the roots α1, β2, … , αn of the equation P (x) = 0. The discriminant vanishes if and only if there are equal roots among the roots of the polynomial. The discriminant can be expressed through the coefficients of the polynomial P(x) by representing it in the form of a determinant consisting of these coefficients. Thus, for the second-degree polynomial ax2 + bx + c, the discriminant is b2 − 4ac. For x3 + px + q, the discriminant is −4p3 −27q2. The discriminant differs only by a factor a0 from the resultant R(P, P′) of the polynomial P(x) and its derivative P′(x).
References in periodicals archive ?
with the module [DELTA] = [alpha][delta] - [beta][gamma], we get another quadratic form f', with its discriminant being D'.
When patterns are numeric, discriminants are based on one of the following concepts:
Solutions found by means of the use of neural networks allow to obtain discriminants of the first type.
The usual practice is to look for a pattern recognition system which is able to learn in an adaptive way from the experiences of several discriminants, each of them corresponding to a specific purpose.
Within the pattern recognition area, one of the most important concepts is that of discriminant.
In general, a discriminant is a function or operator that, when applied to a pattern, allows to obtain an output corresponding to an estimation of the class to which it belongs, or an estimation of one or more of the attributes of the pattern.
The discriminant is a surface dividing the space of input data from where patterns are classified according to the sector to which they belong.
The discriminant is a distance measurement, and patterns are classified according to the class to which their closest neighbor [Simp 92], [Simp 93], or the closest prototype [ Torb 98], or the closest class center [Mega 98], [Shie 95] belong.
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Open / Close | 594 | 2,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2016-44 | longest | en | 0.88416 |
http://mathhelpforum.com/advanced-math-topics/index7.html | 1,503,203,524,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105970.61/warc/CC-MAIN-20170820034343-20170820054343-00552.warc.gz | 280,732,679 | 15,391 | Page 7 of 76 First ... 345678910111656 ... Last
University Math Topics Help Forum: For all other university math help topics including chemistry and physics
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Allows you to choose the data by which the thread list will be sorted. | 781 | 2,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-34 | longest | en | 0.716646 |
https://assignments.lrde.epita.fr/compiler_stages/tc_6/samples/canonicalization_samples.html | 1,718,905,535,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00307.warc.gz | 90,126,574 | 5,528 | TC-6 Canonicalization Samples
The first task in TC-6 is getting rid of all the `eseq`. To do this, you have to move the statement part of an `eseq` at the end of the current sequence point, and keep the expression part in place.
For instance, compare the HIR to the LIR in the following case:
preincr-1.tig
```let
function print_ints(a: int, b: int) =
(print_int(a); print(", "); print_int(b); print("\n"))
var a := 0
in
print_ints(1, (a := a + 1; a))
end
```
One possible HIR translation is:
tc -eH preincr-1.tig
```\$ tc -eH preincr-1.tig
/* == High Level Intermediate representation. == */
label l1
", "
label l2
"\n"
# Routine: print_ints
label l0
# Prologue
move
temp t2
temp fp
move
temp fp
temp sp
move
temp sp
binop sub
temp sp
const 4
move
mem
temp fp
temp i0
move
temp t0
temp i1
move
temp t1
temp i2
# Body
seq
sxp
call
name print_int
temp t0
call end
sxp
call
name print
name l1
call end
sxp
call
name print_int
temp t1
call end
sxp
call
name print
name l2
call end
seq end
# Epilogue
move
temp sp
temp fp
move
temp fp
temp t2
label end
# Routine: _main
label main
# Prologue
# Body
seq
seq
move
temp t3
const 0
sxp
call
name l0
temp fp
const 1
eseq
move
temp t3
temp t3
const 1
temp t3
call end
seq end
sxp
const 0
seq end
# Epilogue
label end
\$ echo \$?
0
```
A possible canonicalization is then:
tc -eL preincr-1.tig
```\$ tc -eL preincr-1.tig
/* == Low Level Intermediate representation. == */
label l1
", "
label l2
"\n"
# Routine: print_ints
label l0
# Prologue
move
temp t2
temp fp
move
temp fp
temp sp
move
temp sp
binop sub
temp sp
const 4
move
mem
temp fp
temp i0
move
temp t0
temp i1
move
temp t1
temp i2
# Body
seq
label l3
sxp
call
name print_int
temp t0
call end
sxp
call
name print
name l1
call end
sxp
call
name print_int
temp t1
call end
sxp
call
name print
name l2
call end
label l4
seq end
# Epilogue
move
temp sp
temp fp
move
temp fp
temp t2
label end
# Routine: _main
label main
# Prologue
# Body
seq
label l5
move
temp t3
const 0
move
temp t5
temp fp
move
temp t3
temp t3
const 1
sxp
call
name l0
temp t5
const 1
temp t3
call end
label l6
seq end
# Epilogue
label end
\$ echo \$?
0
```
The example above is simple because `1` commutes with `(a := a + 1; a)`: the order does not matter. But if you change the `1` into `a`, then you cannot exchange `a` and `(a := a + 1; a)`, so the translation is different. Compare the previous LIR with the following:
preincr-2.tig
```let
function print_ints(a: int, b: int) =
(print_int(a); print(", "); print_int(b); print("\n"))
var a := 0
in
print_ints(a, (a := a + 1; a))
end
```
tc -eL preincr-2.tig
```\$ tc -eL preincr-2.tig
/* == Low Level Intermediate representation. == */
label l1
", "
label l2
"\n"
# Routine: print_ints
label l0
# Prologue
move
temp t2
temp fp
move
temp fp
temp sp
move
temp sp
binop sub
temp sp
const 4
move
mem
temp fp
temp i0
move
temp t0
temp i1
move
temp t1
temp i2
# Body
seq
label l3
sxp
call
name print_int
temp t0
call end
sxp
call
name print
name l1
call end
sxp
call
name print_int
temp t1
call end
sxp
call
name print
name l2
call end
label l4
seq end
# Epilogue
move
temp sp
temp fp
move
temp fp
temp t2
label end
# Routine: _main
label main
# Prologue
# Body
seq
label l5
move
temp t3
const 0
move
temp t6
temp fp
move
temp t5
temp t3
move
temp t3
temp t3
const 1
sxp
call
name l0
temp t6
temp t5
temp t3
call end
label l6
seq end
# Epilogue
label end
\$ echo \$?
0
```
As you can see, the output is the same for the HIR and the LIR:
tc -eH preincr-2.tig > preincr-2.hir
```\$ tc -eH preincr-2.tig > preincr-2.hir
\$ echo \$?
0
```
havm preincr-2.hir
```\$ havm preincr-2.hir
0, 1
\$ echo \$?
0
```
tc -eL preincr-2.tig > preincr-2.lir
```\$ tc -eL preincr-2.tig > preincr-2.lir
\$ echo \$?
0
```
havm preincr-2.lir
```\$ havm preincr-2.lir
0, 1
\$ echo \$?
0
```
Be very careful when dealing with `mem`. For instance, rewriting something like:
```call(foo, eseq(move(temp t, const 51), temp t))
```
into
```move temp t1, temp t
move temp t, const 51
call(foo, temp t)
```
is wrong: `temp t` is not a subexpression, rather it is being defined here. You should produce:
```move temp t, const 51
call(foo, temp t)
```
Another danger is the handling of `move(mem, )`. For instance:
```move(mem foo, x)
```
must be rewritten into:
```move(temp t, foo)
move(mem(temp t), x)
```
not as:
```move(temp t, mem(foo))
move(temp t, x)
```
In other words, the first subexpression of `move(mem(foo), )` is `foo`, not `mem(foo)`. The following example is a good crash test against this problem:
move-mem.tig
```let type int_array = array of int
var tab := int_array [2] of 51
in
tab[0] := 100;
tab[1] := 200;
print_int(tab[0]); print("\n");
print_int(tab[1]); print("\n")
end
```
tc -eL move-mem.tig > move-mem.lir
```\$ tc -eL move-mem.tig > move-mem.lir
\$ echo \$?
0
```
havm move-mem.lir
```\$ havm move-mem.lir
100
200
\$ echo \$?
0
```
You also ought to get rid of nested calls.
nested-calls.tig
```print(chr(ord("\n")))
```
tc -L nested-calls.tig
```\$ tc -L nested-calls.tig
/* == Low Level Intermediate representation. == */
label l0
"\n"
# Routine: _main
label main
# Prologue
# Body
seq
label l1
move
temp t1
call
name ord
name l0
call end
move
temp t2
call
name chr
temp t1
call end
sxp
call
name print
temp t2
call end
label l2
seq end
# Epilogue
label end
\$ echo \$?
0
```
There are only two valid call forms: `sxp(call(...))`, and `move(temp(...), call(...))`.
Contrary to C, the HIR and LIR always denote the same value. For instance the following Tiger code:
seq-point.tig
```let
var a := 1
function a(t: int) : int =
(a := a + 1;
print_int(t); print(" -> "); print_int(a); print("\n");
a)
var b := a(1) + a(2) * a(3)
in
print_int(b); print("\n")
end
```
should always produce:
tc -L seq-point.tig > seq-point.lir
```\$ tc -L seq-point.tig > seq-point.lir
\$ echo \$?
0
```
havm seq-point.lir
```\$ havm seq-point.lir
1 -> 2
2 -> 3
3 -> 4
14
\$ echo \$?
0
```
independently of which IR you ran. It has nothing to do with operator precedence!
In C, you have no such guarantee: the following program can give different results with different compilers and/or on different architectures.
```#include <stdio.h>
int a_ = 1;
int
a(int t)
{
++a_;
printf("%d -> %d\n", t, a_);
return a_;
}
int
main(void)
{
int b = a(1) + a(2) * a(3);
printf("%d\n", b);
return 0;
}
``` | 2,304 | 6,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.427232 |
https://www.teachervision.com/measurement/pro-dev/57076.html | 1,462,168,504,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125175.9/warc/CC-MAIN-20160428161525-00166-ip-10-239-7-51.ec2.internal.warc.gz | 1,039,297,051 | 19,498 | |
Discover everything you need to successfully teach measurement to elementary students, from references and lesson plans to printable worksheets and mathematical graphic organizers. These resources will improve your teaching skills by helping you understand how the topic of measurement can be approached and taught for your students' maximum absorption and retention.
## Powerful Ideas Related to Measurement
From linear measurement to volume and weight, you'll find the references you need to improve your teaching skills and help your students understand the tremendously practical and wide-ranging topic of measurement.
## Developing Measurement Concepts
Take a student-centered approach to teaching elementary mathematics. These printables contain great strategies for developing children's measurement concepts.
## Instruction for Students with Learning Problems
Ensure that all of your students with learning disabilities and exceptionalities are prepared to study measurement in an in-depth manner.
## Teaching Measurement: Videos
Discover a wealth of knowledge about elementary children's concepts of measurement. These videos, featuring renowned math instructor John Van de Walle, include teacher workshops, demonstrations of classroom activities, and interviews with math educators.
## Measurement Worksheets & Graphic Organizers
Explore units of measurement and geometric shapes, with these mathematics printables. Use these resources to practice measuring space and time with your students.
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Earth Day
Kids love hearing about the Earth and ways we can be better to our environment! We've gathered some great resources to help you celebrate Earth Day (April 22) with your class. Some of our most popular activities include this Pollution Matching Worksheet, Recycling Videos and Activities, and Renewable and Non-renewable Energy Worksheet, Recycled Art Lesson Plan, and a Trash & Climate Change Activities Packet!
Videos
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April is full of events that you can incorporate into your standard curriculum. Our Educators' Calendar outlines activities for each event, including: Encourage a Young Writer Day (4/10), Library Week (4/10-16), Passover Begins (4/22), Earth Day (4/22), Tell a Story Day (4/27), and International Dance Day (4/20). Plus, celebrate Math Education Month, Poetry Month, Garden Month, School Library Month all April long!
Coding & Computer Science
Introduce your students to basic coding and computer science! Our Top 5 Free Coding Tools for Kids, Top 5 Free Coding Tools for Teens, and Hour of Code resources make a great introduction to the computer science skills all students will benefit from. | 600 | 3,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.919042 |
https://en.xen.wiki/w/1559edo | 1,720,802,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00735.warc.gz | 185,206,784 | 9,459 | 1559edo
← 1558edo 1559edo 1560edo →
Prime factorization 1559 (prime)
Step size 0.769724¢
Fifth 912\1559 (701.988¢)
Semitones (A1:m2) 148:117 (113.9¢ : 90.06¢)
Consistency limit 9
Distinct consistency limit 9
1559 equal divisions of the octave (abbreviated 1559edo or 1559ed2), also called 1559-tone equal temperament (1559tet) or 1559 equal temperament (1559et) when viewed under a regular temperament perspective, is the tuning system that divides the octave into 1559 equal parts of about 0.77 ¢ each. Each step represents a frequency ratio of 21/1559, or the 1559th root of 2.
Theory
1559edo is consistent to the 9-odd-limit. Using the patent val, it tempers out [38 -2 -15 (luna comma) and [71 -99 37 (raider comma) in the 5-limit; 43046721/43025920, 78125000/78121827 and [32 1 -6 -7 in the 7-limit, 3025/3024, 180224/180075, 50014503/50000000 and 56953125/56942116 in the 11-limit.
Prime harmonics
Approximation of prime harmonics in 1559edo
Harmonic 2 3 5 7 11 13 17 19 23 29 31
Error Absolute (¢) +0.000 +0.033 +0.088 +0.257 -0.195 +0.011 -0.273 +0.370 -0.179 +0.314 +0.314
Relative (%) +0.0 +4.3 +11.4 +33.4 -25.4 +1.4 -35.5 +48.1 -23.3 +40.8 +40.8
Steps
(reduced)
1559
(0)
2471
(912)
3620
(502)
4377
(1259)
5393
(716)
5769
(1092)
6372
(136)
6623
(387)
7052
(816)
7574
(1338)
7724
(1488)
Subsets and supersets
1559edo is the 246th prime edo.
Regular temperament properties
Subgroup Comma List Mapping Optimal
8ve Stretch (¢)
Tuning Error
Absolute (¢) Relative (%)
2.3 [2471 -1559 [1559 2471]] -0.0106 0.0106 1.38
2.3.5 [38 -2 -15, [71 -99 37 [1559 2471 3620]] -0.0196 0.0155 2.01
2.3.5.7 43046721/43025920, 78125000/78121827, 12884901888/12867859375 [1559 2471 3620 4377]] -0.0376 0.0339 4.40
2.3.5.7.11 3025/3024, 180224/180075, 50014503/50000000, 56953125/56942116 [1559 2471 3620 4377 5393]] -0.0188 0.0483 6.27
2.3.5.7.11.13 1716/1715, 3025/3024, 4096/4095, 492128/492075, 5282739/5281250 [1559 2471 3620 4377 5393 5769]] -0.0162 0.0445 5.78
Rank-2 temperaments
Table of rank-2 temperaments by generator
Periods
per 8ve
Generator* Cents* Associated
Ratio*
Temperaments
1 251\1559 193.201 262144/234375 Luna
1 446\1559 343.297 8000/6561 Raider
* octave-reduced form, reduced to the first half-octave, and minimal form in parentheses if it is distinct
Music
Francium
• "confluenceprettype" from albumwithoutspaces (2024) – Spotify | Bandcamp | YouTube – luna[25] in 1559edo tuning | 999 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.578016 |
https://nrich.maths.org/public/leg.php?code=230&cl=3&cldcmpid=912 | 1,508,713,225,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825473.61/warc/CC-MAIN-20171022222745-20171023002745-00306.warc.gz | 816,886,493 | 5,001 | Search by Topic
Resources tagged with Frequency distribution similar to Master Minding:
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Broad Topics > Handling, Processing and Representing Data > Frequency distribution
Master Minding
Stage: 3 Challenge Level:
Your partner chooses two beads and places them side by side behind a screen. What is the minimum number of guesses you would need to be sure of guessing the two beads and their positions?
Dicing with Numbers
Stage: 3 Challenge Level:
In how many ways can you arrange three dice side by side on a surface so that the sum of the numbers on each of the four faces (top, bottom, front and back) is equal?
Nines and Tens
Stage: 3 Challenge Level:
Explain why it is that when you throw two dice you are more likely to get a score of 9 than of 10. What about the case of 3 dice? Is a score of 9 more likely then a score of 10 with 3 dice?
Oware, Naturally
Stage: 3 Challenge Level:
Could games evolve by natural selection? Take part in this web experiment to find out! | 244 | 1,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2017-43 | latest | en | 0.935201 |
http://betterlesson.com/lesson/resource/2038871/1-13-whiteboard-practice-docx | 1,477,352,682,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719815.3/warc/CC-MAIN-20161020183839-00445-ip-10-171-6-4.ec2.internal.warc.gz | 29,625,714 | 24,442 | 1.13 Whiteboard Practice.docx - Section 4: Whiteboard Practice
1.13 Whiteboard Practice.docx
Equivalent Numerical Expressions, Day 1 of 2
Unit 1: Intro to 6th Grade Math & Number Characteristics
Lesson 13 of 16
Big Idea: How can you represent the area of a diagram using numerical expressions? Students apply their knowledge of area and order of operations to match area diagrams with numerical expressions.
Print Lesson
23 teachers like this lesson
Standards:
Subject(s):
Math, order of operations, Number Sense and Operations, 6th grade, master teacher project, equivalent expressions, area model
61 minutes
Andrea Palmer
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Environment: Urban | 277 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-44 | longest | en | 0.805366 |
https://www.physicsforums.com/threads/find-distance-from-com-using-torque.868020/ | 1,723,460,509,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00831.warc.gz | 700,676,979 | 20,918 | # Find distance from COM using torque
• Notre Dame
In summary, the problem involves finding the distance from a woman's feet to her center of mass while lying on a reaction board. This can be done using equations of static equilibrium and balancing torque. It is important to have a good understanding of static equilibrium in order to solve this problem accurately.
Notre Dame
## Homework Statement
Word for word, from the problem:
"A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal 2.5-m-long, 6.1 kg reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A 60 kg woman lies on the reaction board with her feet over the pivot. The scale reads 25 kg. What is the distance from the woman’s feet to her center of mass?"
## Homework Equations
T=Iα (torque = moment of inertia times angular acceleration)
T=rFsin∅ (torque = radius times the force times sine of the angle between the radius and the force)
T=rFt (torque = radius times component of force perpendicular to the moment arm)
l=rmvsin∅ (angular momentum = radius times mass times velocity times the sine of the angle between the radius and the force)
?
## The Attempt at a Solution
So the scale reads 25kg, and I thought maybe that force could be found using F = ma. So I did that and got a force of 245N. I figure that force is perpendicular to the moment arm, so I multiplied it by 2.5m and got a torque of 612.5N⋅m. From there I'm kind of stuck though. I think it maybe also has something to do with angular momentum?
Notre Dame said:
## Homework Statement
Word for word, from the problem:
"A person’s center of mass is easily found by having the person lie on a reaction board. A horizontal 2.5-m-long, 6.1 kg reaction board is supported only at the ends, with one end resting on a scale and the other on a pivot. A 60 kg woman lies on the reaction board with her feet over the pivot. The scale reads 25 kg. What is the distance from the woman’s feet to her center of mass?"
## Homework Equations
T=Iα (torque = moment of inertia times angular acceleration)
T=rFsin∅ (torque = radius times the force times sine of the angle between the radius and the force)
T=rFt (torque = radius times component of force perpendicular to the moment arm)
l=rmvsin∅ (angular momentum = radius times mass times velocity times the sine of the angle between the radius and the force)
?
## The Attempt at a Solution
So the scale reads 25kg, and I thought maybe that force could be found using F = ma. So I did that and got a force of 245N. I figure that force is perpendicular to the moment arm, so I multiplied it by 2.5m and got a torque of 612.5N⋅m. From there I'm kind of stuck though. I think it maybe also has something to do with angular momentum?
You're flailing at this. The woman lies on a board which is supported at two ends. A scale is fitted at one end to record the support reaction there.
Knowing the mass of the board and its length, you know where the center of mass of the board is located from one end. You know the mass of the woman, but you don't know where her center of mass is located.
This is a simple problem in static equilibrium. You don't need angular momentum or the special theory of relativity to find the answer.
Draw a free body diagram of the reaction board first to help with your analysis.
[\CENTER]
Okay, so I tried the free-body diagram. It's on a fixed pivot point, and supported on both ends, so the acceleration would be 0? I just thought it would be torque because the problem was posed in the notes for the chapter about torque.
Also, my professor emailed back...the answer is .91m. Which is great, but I would really like to understand how to do the problem.
Notre Dame said:
Okay, so I tried the free-body diagram. It's on a fixed pivot point, and supported on both ends, so the acceleration would be 0? I just thought it would be torque because the problem was posed in the notes for the chapter about torque.
Also, my professor emailed back...the answer is .91m. Which is great, but I would really like to understand how to do the problem.
Do you know what the equations of static equilibrium are?
You just need to balance the torque and you'll be good, SteamKing gives a good visualisation of how a reaction board works. Drawing the FBD youll only find one unknown force and balancing torque about that point will eliminate the need of that force. You can concentrate the mass of the woman at her COM and let it be at a distance x from that point. Solving the equation youll get x
Last edited:
I looked up static equilibrium in my textbook. We skipped that chapter. I'm actually really, really stupid when it comes to physics. Biology, chemistry, any other science, I'm good. Even calculus, I'm okay. Physics, not so much.
Notre Dame said:
I looked up static equilibrium in my textbook. We skipped that chapter. I'm actually really, really stupid when it comes to physics. Biology, chemistry, any other science, I'm good. Even calculus, I'm okay. Physics, not so much.
It's hardly a physics course if you skipped static equilibrium.
Regardless of whether or not it's a physics course, it has professors and a test, the nearest of which is in twelve hours.
Tom.G
Notre Dame said:
Regardless of whether or not it's a physics course, it has professors and a test, the nearest of which is in twelve hours.
Yes, but there are good professors and bad professors, and it appears that the one you drew is toward the bad end of the spectrum.
BTW, make sure that "spectrum" is one of the topics your physics course covers when it comes to studying light.
## 1. How is torque used to find the distance from the center of mass (COM)?
Torque is a measure of the rotational force applied to an object. By using the equation Torque = Force x Distance, we can find the distance from the COM by rearranging the equation to Distance = Torque / Force. This means that if we know the torque and the force acting on an object, we can determine the distance from the COM.
## 2. What is the center of mass (COM) and why is it important in finding distance using torque?
The center of mass is the point in an object where all of its mass can be considered to be concentrated. It is important in finding distance using torque because it is the point around which an object rotates. By calculating the distance from the COM, we can determine how an object will rotate under the influence of a torque.
## 3. Can torque be used to find the distance from the COM for any object?
Yes, torque can be used to find the distance from the COM for any object. However, it is important to note that the object must be in equilibrium, meaning that all external forces acting on the object must be balanced. This allows us to use the torque equation to determine the distance from the COM.
## 4. How does the direction of the force affect the distance from the COM?
The direction of the force has a significant impact on the distance from the COM. If the force is applied perpendicular to the axis of rotation, it will result in a larger distance from the COM. However, if the force is applied parallel to the axis of rotation, it will not contribute to the torque and therefore will not affect the distance from the COM.
## 5. What are some real-world applications of finding distance from the COM using torque?
One important application is in the design of structures, such as bridges, buildings, and cranes. By calculating the torque and distance from the COM, engineers can ensure that these structures are stable and can withstand external forces without collapsing. Another application is in sports, such as gymnastics and diving, where athletes must be aware of their center of mass in order to perform complex maneuvers and maintain balance.
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ALL-IN-ONE 2nd GRADE PLACE VALUE MATH UNIT! 332 PAGES of differentiated, fun, and engaging 2nd Grade Common Core-Aligned Place Value Strategy WORKSHEETS, CENTERS, ACTIVITIES & CRAFTS designed specifically to meet the rigorous Common Core Math Standards. Example problems with explanations an
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Fun in the Sun Place Value - Hundreds Charts & Fact Family Practice The activities in this file are perfect for Common Core, Numbers and Operations, Place Value, Fluency, and Subitizing. Students will use the hundreds chart to quickly identify numbers (subitizing), skip count by 2s, 5s, and 10
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This set of 360 9-step cards helps to teach place value by breaking down the learning process needed to transition into reading and understanding quantities of numbers up to 4-digits. These cards can be used to give children practice when learning to read and say numbers properly. They can also be
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After you’ve taught place value, these activities are perfect for keeping the concept fresh all year long! These activities are also great for nudging along those students who don’t quite get it yet. Use one or two of these warm-ups every day as you start your math class. Once they get the hang o
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These task cards were designed to review many skills with decimals and fractions, with some place value of whole numbers mixed in, too. Cards can be used in groups for review or as a SCOOT game, whichever you choose. There are two sets included so that you may use one set for review and the other
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Did you just finish teaching your unit on 2.NBT.1 for place value? Are you a paperless classroom and need a Place Value Digital Assessment using Google Forms? This is a download for a completely editable digital assessment on understanding and using an thousands, hundreds, and tens base ten number
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PDF (17.24 MB)
A variety of independent Place Value activities including a partner game and a reference sheet that can be laminated, used in math centers or in math notebooks.This is an overall place value bundle for 1st grade students that include. Included With This Purchase: -Reference Sheet- Understanding Pl
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\$4.25
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3.9
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This 12 page set of fun hundreds chart activities varies in degree of difficulty so every student can be successful! these pictures are great for morning work, math centers, homework, inside recess, plus much more! This set contains: *3 hidden picture answer sheets *3 easy worksheets where students
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3.9
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Students often need extra practice with the important skill of adding and subtracting 2, 3, and/or 4 digit numbers with and without regrouping. Look no further than this activity for a fun, hands-on activity that can be easily differentiated to meet the diverse needs of all of your students. NOTE:
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Updated Jan. 2015! This Place Value Set covers a variety of skills typically for 4th grade students or can be used to challenge students at the 3rd grade ADV level. Place Value Quiz, Rounding Game and Concentration Game Included!You May Also Like These Place Value Task Cards CLICK HERE Included wit
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3.8
PDF (1.45 MB)
This bundle contains all of my Christmas themed Hundreds Chart Hidden Pictures sets. You'll get 4 answer keys, 4 hundreds charts, 4 numbers in order clue sheets, and 4 numbers out of order clue sheets for each hidden picture. Students will enjoy coloring to find the hidden pictures of Santa, a Chri
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\$4.95
6 Ratings
4.0
ZIP (2.25 MB)
Place Value Heroes- Place Value Worksheets- Practice counting base ten blocks with a place value chart. Counting base ten to the hundreds and thousands. Question sheet included so that students can practice explaining their thinking and math reasoning.2 LEVELS INCLUDED! 2 levels make it easy to diff
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\$4.10
\$2.20
6 Ratings
4.0
PDF (8.73 MB)
Flash Card Pack Includes: - Numbers 1-100 - 100 - 900 (by hundreds) - Base 10 Fluency Drill Cards (ones, tens, hundreds, thousands) - Place Value Drill Cards Using Numerics in Word Form (ie: 1 Ten and 2 ones) - Addition, Subtraction, and Equal Signs
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Addition With Regrouping into the Hundreds Place This FALL Addition with Regrouping into the Hundreds Place is a great way to practice regrouping the ones and the tens into the hundreds place. Print the pages out and use as classwork or homework. Each page has a fall themed word problem as well.
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\$2.75
3 Ratings
4.0
PDF (8.3 MB)
60 task cards to use to reinforce or teach comparing numbers up to the hundreds place These could be used as reinforcement or review of comparing numbers concepts. They are perfect to use as a math center, scoot activity, guided math group, partner work, scavenger hunt, early finishers or RTI.
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5 Ratings
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This bundle consists of 3 of my hidden picture sets. You will get 9 pictures for students to create, along with 3 levels of difficulty, so everyone in your classroom can be successful AND your can have your students practice different skills. Great to use for math centers, morning work, homework
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\$8.00
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ZIP (4.33 MB)
Math T Charts are a resource to help students with practicing addition and subtractions for places ones, tens, and/or hundreds. During the first few uses, it is suggested to explain why it is a T chart. Later, students may be encouraged to draw their own T charts if they still need them.
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\$5.75
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I had the students keep this place value bookmark in their math book and use it to help read large whole numbers and decimals. It spans from hundred thousands to hundred thousandths. Laminating will make it last much longer. Students can then also use their pencil to "write in" the numbers to help
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\$1.66
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A variety of independent Place Value activities including a partner game and a reference sheet that can be laminated, used in math centers or in math notebooks.This is an overall place value bundle for 2nd grade students that include. Included With This Purchase: -Reference Sheet- Understanding Pl
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\$4.00
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My students have a hard time with place value, they often struggle with anything past the millions or into the decimals. I created this product to display around the room to help remind them of Place Values. This product can be used in multiple ways, as an intervention tool, whole group, math work
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Are you looking for something to go in depth with place value?Here are some number talks to build on place value understanding!Have fun and watch your kids become more confident in their abilities as a mathematician!! When you do number talks daily, their confidence and skill grows exponentially! CG
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Also included in: CGI Math Number Talks
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PPT (8 MB)
This useful hundred, tens, and ones graphic organizer will help your student add or subtract! I have created a chart with the place value disk visual and one without. There are also place value disks that you can print as well. There are 3 pages! 2 charts and one page of place value disks! Enjoy
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Ten More Ten Less One More One Less on a Hundreds Chart Use place value understanding and properties of operations to add and subtract. I can find one more, one less, ten more, and ten less. CCSS.Math.1.NBT.C.4 CCSS.Math.1.NBT.C.5 CCSS.Math.1.NBT.C.6
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Place Value Math Game / Activity to Improve Thinking Skills! Roll the dice to create the problems! The activity is different every time you use it!Skills Include:Whole Number Place Value to Ten Thousands PlaceWhole Number Place Value to Hundred Thousands PlaceDecimal Place Value to The Ten Thousandt
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http://mathhelpforum.com/discrete-math/150650-why-permutation.html | 1,481,290,492,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542712.12/warc/CC-MAIN-20161202170902-00248-ip-10-31-129-80.ec2.internal.warc.gz | 169,900,805 | 10,039 | 1. ## Why permutation ?
we have 4 distinct ice cream flavors to choose from
How many different combinations of flavors of three scoops of ice cream are possible if it is permissible to make all three scoops the same flavor ?
all diff + all similar + 2 similar and 3rd diff
4 choose 3 + 4 choose 1 + 4P2 (the last one is the permutation)
Can any one justify 4P2 ?
2. Originally Posted by Hitman6267
we have 4 distinct ice cream flavors to choose from
How many different combinations of flavors of three scoops of ice cream are possible if it is permissible to make all three scoops the same flavor ?
Yeah 4P2 works, although I just thought of it as $\displaystyle 2\cdot\binom{4}{2}$ since once you choose the two flavours, you have to choose which one is the double scoop, meaning multiply by two. | 199 | 804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2016-50 | longest | en | 0.949166 |
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All Right Reserved. Copyright 2005-2024 | 1,231 | 5,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-38 | latest | en | 0.900012 |
http://math.stackexchange.com/questions/192856/why-is-this-proof-on-rational-set-valid | 1,469,337,181,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823947.97/warc/CC-MAIN-20160723071023-00105-ip-10-185-27-174.ec2.internal.warc.gz | 158,993,808 | 17,258 | # Why is this proof on rational set valid?
Prove that if $y$ is rational and $x$ is irrational, then $y + x$ and $yx$ (assume y $\neq 0$) is irrational.
I kinda guessed the proof and it turned out to be right with the key, but it doesn't answer my question. Here is my quick and dirty proof
Assume $x + y$ is rational such that $x + y = r$ ($r$ is rational), then $x + y = r \iff x = r - y \iff x = r + (-y)$ But this is a contradiction since $\mathbb{Q}$ is closed under addition.
Q1. Okay so I showed that by contradiction that my claim is wrong, but showing that something isn't doesn't tell us what the original thing was now does it? Or does this work because rational and irrational are polar opposites and I don't need to fill in?
The proof for multiplication is similar:
Assume $xy$ is rational such that $xy = r$ ($r$ is rational), then $xy = r \iff x = r/y$ But this is a contradiction since $\mathbb{Q}$ is closed under division
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## Linear Algebra and Its Applications, Exercise 2.3.23
Exercise 2.3.23. Let through be vectors in . Answer the following questions: a) Are the nine vectors linearly independent? Not linearly independent? Might be linearly independent? b) Do the nine vectors span ? Not span ? Might span ? c) … Continue reading
Posted in linear algebra | 1 Comment
## Linear Algebra and Its Applications, Exercise 2.3.22
Exercise 2.3.22. Given a vector space of dimension 7 and a subspace of of dimension 4, state whether the following are true or false: 1) You can create a basis for by adding three vectors to any set of vectors … Continue reading
## Linear Algebra and Its Applications, Exercise 2.3.21
Exercise 2.3.21. Suppose is a 64 by 17 matrix and has rank 11. How many independent vectors are solutions to the system ? What about the system ? Answer: If the rank of is 11 then performing elimination on produces … Continue reading
## Linear Algebra and Its Applications, Exercise 2.3.20
Exercise 2.3.20.Consider the set of all 2 by 2 matrices that have the sum of their rows equal to the sum of their columns. What is a basis for this subspace? Consider the analogous set of 3 by 3 matrices … Continue reading
## Linear Algebra and Its Applications, Exercise 2.3.19
Exercise 2.3.19. Suppose is an by matrix, with columns taken from . What is the rank of if its column vectors are linearly independent? What is the rank of if its column vectors span ? What is the rank of … Continue reading
## Linear Algebra and Its Applications, Exercise 2.3.18
Exercise 2.3.18. Indicate whether the following statements are true or false: a) given a matrix whose columns are linearly independent, the system has one and only solution for any right-hand side b) if is a 5 by 7 matrix then … Continue reading
## Linear Algebra and Its Applications, Exercise 2.3.17
Exercise 2.3.17. Suppose that and are subspaces of , each with dimension 3. Show that and must have at least one vector in common other than the zero vector. Answer: Since and each have dimension 3 their respective bases each … Continue reading
## Linear Algebra and Its Applications, Exercise 2.3.16
Exercise 2.3.16. What is the dimension of the vector space consisting of all 3 by 3 symmetric matrices? What is a basis for it? Answer: There are nine possible entries that can be set in a 3 b 3 matrix, … Continue reading
Posted in linear algebra | 2 Comments
## Linear Algebra and Its Applications, Exercise 2.3.15
Exercise 2.3.15. If the vector space has dimension show that a) if a set of vectors in is linearly independent then that set forms a basis b) if a set of vectors in spans then that set forms a basis … Continue reading | 672 | 2,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-14 | latest | en | 0.876036 |
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Thirty two of this amount make up a Quart?
Updated: 9/26/2023
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32 fluid ounces
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Q: Thirty two of this amount make up a Quart?
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Related questions
Thirty-two of this amount makeup a quart?
1 US quart = 32 US fl oz.
How many quarts are in thirty two ounces?
Thirty-two ounces equal one quart.
How many ounces are in a qurt?
Thirty-two (32) ounces equals 1 quart, so there is 1/32 of a quart in an ounce.
How much pints is to make a quart?
There are two pints in every quart.
3 pints and 3 quart equals?
Two pints make a quart. Do the math yourself.
How many cups are in a 1 US liquid quarts?
one cup is equal to eight ounces. one quart is thirty two ounces. four cups equal one quart.
How many pints equal one quart of thirty two ounces?
There are 2 pints (16 fluid oz each) in 1 quart. 1 quart is 32 fluid ounces, 4 cups, 2 pints, or 1/4 gallon.
How many quarts make one pint?
There are two pints in a quart therefore half a quart in a pint.
Do two thirty-two ounce make a gallon?
No. Two thirty-two ounce containers equal 64 ounces, which is a half-gallon. It takes 128 ounces to make a gallon.
Which two capacity units are about the same amount?
If you are thinking of obsolete measures, then a quart is similar to a litre.
How many ounces in 6 quarts?
In one quart, there are thirty two US ounces. This means that there are 192 US ounces in six quarts.
Is 1 quart bigger than 3 pints?
No, it's actually one half of a quart. One cup equals 8 ounces; 16 ounces equals one pint, and 32 ounces equals one quart. So two cups make a pint and two pints make a quart. | 446 | 1,667 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.927111 |
https://www.unitconverters.net/length/megaparsec-to-angstrom.htm | 1,726,489,688,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00602.warc.gz | 965,268,409 | 4,266 | Home / Length Conversion / Convert Megaparsec to Angstrom
# Convert Megaparsec to Angstrom
Please provide values below to convert megaparsec [Mpc] to angstrom [A], or vice versa.
From: megaparsec To: angstrom
### Megaparsec to Angstrom Conversion Table
Megaparsec [Mpc]Angstrom [A]
0.01 Mpc3.08567758128E+30 A
0.1 Mpc3.08567758128E+31 A
1 Mpc3.08567758128E+32 A
2 Mpc6.17135516256E+32 A
3 Mpc9.25703274384E+32 A
5 Mpc1.54283879064E+33 A
10 Mpc3.08567758128E+33 A
20 Mpc6.17135516256E+33 A
50 Mpc1.54283879064E+34 A
100 Mpc3.08567758128E+34 A
1000 Mpc3.08567758128E+35 A
### How to Convert Megaparsec to Angstrom
1 Mpc = 3.08567758128E+32 A
1 A = 3.2407792896664E-33 Mpc
Example: convert 15 Mpc to A:
15 Mpc = 15 × 3.08567758128E+32 A = 4.62851637192E+33 A | 328 | 764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-38 | latest | en | 0.405961 |
http://www.accountingtutorials.net/2013/02/p12-4b-isaacs-auto-repair-is.html | 1,555,805,486,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530100.28/warc/CC-MAIN-20190421000555-20190421022555-00551.warc.gz | 193,243,267 | 13,504 | ## Pages
This Website Has Been Moved to a New Link
### P12-4B Isaac’s Auto Repair is considering the purchase
Price: \$2.50
Isaac’s Auto Repair is considering the purchase of a new tow truck. The garage
doesn’t currently have a tow truck, and the \$65,000 price tag for a new truck would represent
a major expenditure for the garage. Isaac Mayer, owner of the garage, has compiled
the following estimates in trying to determine whether to purchase the truck.
Initial cost \$65,000 Estimated useful life 8 years Net annual cash inflows from towing \$9,600 Overhaul costs (end of year 4) \$7,000 Salvage value \$16,000
Isaac’s good friend, Brad Jolie, stopped by. He is trying to convince Isaac that the tow truck
will have other benefits that Isaac hasn’t even considered. First, he says, cars that need
towing need to be fixed. Thus, when Isaac tows them to his facility his repair revenues will
increase. Second, he notes that the tow truck could have a plow mounted on it, thus saving
Isaac will plow Brad’s driveway.) Third, he notes that the truck will generate goodwill; that
is, people who are rescued by Isaac and his tow truck will feel grateful and might be more
inclined to use his service station in the future or buy gas there. Fourth, the tow truck will
have “Isaac’s Auto Repair” on its doors, hood, and back tailgate—a form of free advertising
wherever the tow truck goes.
Brad estimates that, at a minimum, these benefits would be worth the following.
Additional annual net cash flows from repair work \$2,600 Annual savings from plowing 600 Additional annual net cash flows from customer "goodwill" 1,200 Additional annual net cash flows resulting from free advertising 500
The company’s cost of capital is 10%.
Instructions
(a) Calculate the net present value, ignoring the additional benefits described by Brad.
Should the tow truck be purchased?
(b) Calculate the net present value, incorporating the additional benefits suggested by
Brad. Should the tow truck be purchased?
(c) Suppose Brad has been overly optimistic in his assessment of the value of the additional
benefits. At a minimum, how much would the additional benefits have to be
worth in order for the project to be accepted? | 502 | 2,216 | {"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} | 2.796875 | 3 | CC-MAIN-2019-18 | latest | en | 0.959079 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Book%3A_Physical_Methods_in_Chemistry_and_Nano_Science_(Barron)/07%3A_Molecular_and_Solid_State_Structure/7.01%3A_Crystal_Structure | 1,606,538,371,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195069.35/warc/CC-MAIN-20201128040731-20201128070731-00707.warc.gz | 235,757,941 | 35,993 | # 7.1: Crystal Structure
In any sort of discussion of crystalline materials, it is useful to begin with a discussion of crystallography: the study of the formation, structure, and properties of crystals. A crystal structure is defined as the particular repeating arrangement of atoms (molecules or ions) throughout a crystal. Structure refers to the internal arrangement of particles and not the external appearance of the crystal. However, these are not entirely independent since the external appearance of a crystal is often related to the internal arrangement. For example, crystals of cubic rock salt (NaCl) are physically cubic in appearance. Only a few of the possible crystal structures are of concern with respect to simple inorganic salts and these will be discussed in detail, however, it is important to understand the nomenclature of crystallography.
## Crystallography
### Bravais Lattice
The Bravais lattice is the basic building block from which all crystals can be constructed. The concept originated as a topological problem of finding the number of different ways to arrange points in space where each point would have an identical “atmosphere”. That is each point would be surrounded by an identical set of points as any other point, so that all points would be indistinguishable from each other. Mathematician Auguste Bravais discovered that there were 14 different collections of the groups of points, which are known as Bravais lattices. These lattices fall into seven different "crystal systems”, as differentiated by the relationship between the angles between sides of the “unit cell” and the distance between points in the unit cell. The unit cell is the smallest group of atoms, ions or molecules that, when repeated at regular intervals in three dimensions, will produce the lattice of a crystal system. The “lattice parameter” is the length between two points on the corners of a unit cell. Each of the various lattice parameters are designated by the letters a, b, and c. If two sides are equal, such as in a tetragonal lattice, then the lengths of the two lattice parameters are designated a and c, with b omitted. The angles are designated by the Greek letters α, β, and γsize 12{γ} {}, such that an angle with a specific Greek letter is not subtended by the axis with its Roman equivalent. For example, α is the included angle between the b and c axis.
Table $$\PageIndex{1}$$ shows the various crystal systems, while Figure $$\PageIndex{1}$$ shows the 14 Bravais lattices. It is important to distinguish the characteristics of each of the individual systems. An example of a material that takes on each of the Bravais lattices is shown in Table $$\PageIndex{2}$$.
System Axial Lengths and Angles Unit Cell Geometry cubic a=b=c, α = β = γ = 90° tetragonal a = b ≠ c, α = β = γ= 90° orthorhombic a ≠ b ≠ c, α = β = γ= 90° rhombohedral a = b = c, α = β = γ ≠ 90° hexagonal a = b ≠ c, α = β = 90°, γ = 120° monoclinic a ≠ b ≠ c, α = γ = 90°, β ≠ 90° triclinic a ≠ b ≠ c, α ≠ β ≠ γ
Crystal System Example triclinic K2S2O8 monoclinic As4S4, KNO2 rhombohedral Hg, Sb hexagonal Zn, Co, NiAs orthorhombic Ga, Fe3C tetragonal In, TiO2 cubic Au, Si, NaCl
The cubic lattice is the most symmetrical of the systems. All the angles are equal to 90°, and all the sides are of the same length (a = b = c). Only the length of one of the sides (a) is required to describe this system completely. In addition to simple cubic, the cubic lattice also includes body-centered cubic and face-centered cubic (Figure $$\PageIndex{1}$$. Body-centered cubic results from the presence of an atom (or ion) in the center of a cube, in addition to the atoms (ions) positioned at the vertices of the cube. In a similar manner, a face-centered cubic requires, in addition to the atoms (ions) positioned at the vertices of the cube, the presence of atoms (ions) in the center of each of the cubes face.
The tetragonal lattice has all of its angles equal to 90°, and has two out of the three sides of equal length (a = b). The system also includes body-centered tetragonal (Figure $$\PageIndex{1}$$.
In an orthorhombic lattice all of the angles are equal to 90°, while all of its sides are of unequal length. The system needs only to be described by three lattice parameters. This system also includes body-centered orthorhombic, base-centered orthorhombic, and face-centered orthorhombic (Figure $$\PageIndex{1}$$.
A base-centered lattice has, in addition to the atoms (ions) positioned at the vertices of the orthorhombic lattice, atoms (ions) positioned on just two opposing faces.
The rhombohedral lattice is also known as trigonal, and has no angles equal to 90°, but all sides are of equal length (a = b = c), thus requiring only by one lattice parameter, and all three angles are equal (α = β = γ).
A hexagonal crystal structure has two angles equal to 90°, with the other angle ( γsize 12{γ} {}) equal to 120°. For this to happen, the two sides surrounding the 120° angle must be equal (a = b), while the third side (c) is at 90° to the other sides and can be of any length.
The monoclinic lattice has no sides of equal length, but two of the angles are equal to 90°, with the other angle (usually defined as β) being something other than 90°. It is a tilted parallelogram prism with rectangular bases. This system also includes base-centered monoclinic (Figure $$\PageIndex{2}$$).
In the triclinic lattice none of the sides of the unit cell are equal, and none of the angles within the unit cell are equal to 90°. The triclinic lattice is chosen such that all the internal angles are either acute or obtuse. This crystal system has the lowest symmetry and must be described by 3 lattice parameters (a, b, and c) and the 3 angles (α, β, and γ).
## Atom Positions, Crystal Directions and Miller Indices
### Atom Positions and Crystal Axes
The structure of a crystal is defined with respect to a unit cell. As the entire crystal consists of repeating unit cells, this definition is sufficient to represent the entire crystal. Within the unit cell, the atomic arrangement is expressed using coordinates. There are two systems of coordinates commonly in use, which can cause some confusion. Both use a corner of the unit cell as their origin. The first, less-commonly seen system is that of Cartesian or orthogonal coordinates (X, Y, Z). These usually have the units of Angstroms and relate to the distance in each direction between the origin of the cell and the atom. These coordinates may be manipulated in the same fashion are used with two- or three-dimensional graphs. It is very simple, therefore, to calculate inter-atomic distances and angles given the Cartesian coordinates of the atoms. Unfortunately, the repeating nature of a crystal cannot be expressed easily using such coordinates. For example, consider a cubic cell of dimension 3.52 Å. Pretend that this cell contains an atom that has the coordinates (1.5, 2.1, 2.4). That is, the atom is 1.5 Å away from the origin in the x direction (which coincides with the a cell axis), 2.1 Å in the y (which coincides with the b cell axis) and 2.4 Å in the z (which coincides with the c cell axis). There will be an equivalent atom in the next unit cell along the x-direction, which will have the coordinates (1.5 + 3.52, 2.1, 2.4) or (5.02, 2.1, 2.4). This was a rather simple calculation, as the cell has very high symmetry and so the cell axes, a, b and c, coincide with the Cartesian axes, X, Y and Z. However, consider lower symmetry cells such as triclinic or monoclinic in which the cell axes are not mutually orthogonal. In such cases, expressing the repeating nature of the crystal is much more difficult to accomplish.
Accordingly, atomic coordinates are usually expressed in terms of fractional coordinates, (x, y, z). This coordinate system is coincident with the cell axes (a, b, c) and relates to the position of the atom in terms of the fraction along each axis. Consider the atom in the cubic cell discussion above. The atom was 1.5 Å in the a direction away from the origin. As the a axis is 3.52 Å long, the atom is (1.5/3.52) or 0.43 of the axis away from the origin. Similarly, it is (2.1/3.52) or 0.60 of the b axis and (2.4/3.5) or 0.68 of the c axis. The fractional coordinates of this atom are, therefore, (0.43, 0.60, 0.68). The coordinates of the equivalent atom in the next cell over in the a direction, however, are easily calculated as this atom is simply 1 unit cell away in a. Thus, all one has to do is add 1 to the x coordinate: (1.43, 0.60, 0.68). Such transformations can be performed regardless of the shape of the unit cell. Fractional coordinates, therefore, are used to retain and manipulate crystal information.
### Crystal Directions
The designation of the individual vectors within any given crystal lattice is accomplished by the use of whole number multipliers of the lattice parameter of the point at which the vector exits the unit cell. The vector is indicated by the notation [hkl], where h, k, and l are reciprocals of the point at which the vector exits the unit cell. The origination of all vectors is assumed defined as [000]. For example, the direction along the a-axis according to this scheme would be [100] because this has a component only in the a-direction and no component along either the b or c axial direction. A vector diagonally along the face defined by the a and baxis would be [110], while going from one corner of the unit cell to the opposite corner would be in the [111] direction. Figure $$\PageIndex{2}$$ shows some examples of the various directions in the unit cell. The crystal direction notation is made up of the lowest combination of integers and represents unit distances rather than actual distances. A [222] direction is identical to a [111], so [111] is used. Fractions are not used. For example, a vector that intercepts the center of the top face of the unit cell has the coordinates x = 1/2, y = 1/2, z = 1. All have to be inversed to convert to the lowest combination of integers (whole numbers); i.e., [221] in Figure $$\PageIndex{2}$$. Finally, all parallel vectors have the same crystal direction, e.g., the four vertical edges of the cell shown in Figure $$\PageIndex{2}$$ all have the crystal direction [hkl] = [001].
Crystal directions may be grouped in families. To avoid confusion there exists a convention in the choice of brackets surrounding the three numbers to differentiate a crystal direction from a family of direction. For a direction, square brackets [hkl] are used to indicate an individual direction. Angle brackets <hkl> indicate a family of directions. A family of directions includes any directions that are equivalent in length and types of atoms encountered. For example, in a cubic lattice, the [100], [010], and [001] directions all belong to the <100> family of planes because they are equivalent. If the cubic lattice were rotated 90°, the a, b, and cdirections would remain indistinguishable, and there would be no way of telling on which crystallographic positions the atoms are situated, so the family of directions is the same. In a hexagonal crystal, however, this is not the case, so the [100] and [010] would both be <100> directions, but the [001] direction would be distinct. Finally, negative directions are identified with a bar over the negative number instead of a minus sign.
### Crystal Planes
Planes in a crystal can be specified using a notation called Miller indices. The Miller index is indicated by the notation [hkl] where h, k, and l are reciprocals of the plane with the x, y, and z axes. To obtain the Miller indices of a given plane requires the following steps:
1. The plane in question is placed on a unit cell.
2. Its intercepts with each of the crystal axes are then found.
3. The reciprocal of the intercepts are taken.
4. These are multiplied by a scalar to insure that is in the simple ratio of whole numbers.
For example, the face of a lattice that does not intersect the y or z axis would be (100), while a plane along the body diagonal would be the (111) plane. An illustration of this along with the (111) and (110) planes is given in Figure $$\PageIndex{3}$$.
As with crystal directions, Miller indices directions may be grouped in families. Individual Miller indices are given in parentheses (hkl), while braces {hkl} are placed around the indices of a family of planes. For example, (001), (100), and (010) are all in the {100} family of planes, for a cubic lattice.
## Description of Crystal Structures
Crystal structures may be described in a number of ways. The most common manner is to refer to the size and shape of the unit cell and the positions of the atoms (or ions) within the cell. However, this information is sometimes insufficient to allow for an understanding of the true structure in three dimensions. Consideration of several unit cells, the arrangement of the atoms with respect to each other, the number of other atoms they in contact with, and the distances to neighboring atoms, often will provide a better understanding. A number of methods are available to describe extended solid-state structures. The most applicable with regard to elemental and compound semiconductor, metals and the majority of insulators is the close packing approach.
### Close Packed Structures: Hexagonal Close Packing and Cubic Close Packing
Many crystal structures can be described using the concept of close packing. This concept requires that the atoms (ions) are arranged so as to have the maximum density. In order to understand close packing in three dimensions, the most efficient way for equal sized spheres to be packed in two dimensions must be considered.
The most efficient way for equal sized spheres to be packed in two dimensions is shown in Figure $$\PageIndex{4}$$, in which it can be seen that each sphere (the dark gray shaded sphere) is surrounded by, and is in contact with, six other spheres (the light gray spheres in Figure $$\PageIndex{4}$$. It should be noted that contact with six other spheres the maximum possible is the spheres are the same size, although lower density packing is possible. Close packed layers are formed by repetition to an infinite sheet. Within these close packed layers, three close packed rows are present, shown by the dashed lines in Figure $$\PageIndex{4}$$.
The most efficient way for equal sized spheres to be packed in three dimensions is to stack close packed layers on top of each other to give a close packed structure. There are two simple ways in which this can be done, resulting in either a hexagonal or cubic close packed structures.
### Hexagonal Close Packed
If two close packed layers A and B are placed in contact with each other so as to maximize the density, then the spheres of layer B will rest in the hollow (vacancy) between three of the spheres in layer A. This is demonstrated in Figure $$\PageIndex{5}$$. Atoms in the second layer, B (shaded light gray), may occupy one of two possible positions (Figure $$\PageIndex{5}$$ a or b) but not both together or a mixture of each. If a third layer is placed on top of layer B such that it exactly covers layer A, subsequent placement of layers will result in the following sequence ...ABABAB.... This is known as hexagonal close packing or hcp.
The hexagonal close packed cell is a derivative of the hexagonal Bravais lattice system (Figure $$\PageIndex{6}$$ with the addition of an atom inside the unit cell at the coordinates (1/3,2/3,1/2). The basal plane of the unit cell coincides with the close packed layers (Figure $$\PageIndex{6}$$. In other words the close packed layer makes-up the {001} family of crystal planes.
The “packing fraction” in a hexagonal close packed cell is 74.05%; that is 74.05% of the total volume is occupied. The packing fraction or density is derived by assuming that each atom is a hard sphere in contact with its nearest neighbors. Determination of the packing fraction is accomplished by calculating the number of whole spheres per unit cell (2 in hcp), the volume occupied by these spheres, and a comparison with the total volume of a unit cell. The number gives an idea of how “open” or filled a structure is. By comparison, the packing fraction for body-centered cubic (Figure $$\PageIndex{5}$$) is 68% and for diamond cubic (an important semiconductor structure to be described later) is it 34%.
### Cubic Close Packed: Face-centered Cubic
In a similar manner to the generation of the hexagonal close packed structure, two close packed layers are stacked (Figure $$\PageIndex{7}$$ however, the third layer (C) is placed such that it does not exactly cover layer A, while sitting in a set of troughs in layer B (Figure $$\PageIndex{7}$$), then upon repetition the packing sequence will be ...ABCABCABC.... This is known as cubic close packing or ccp.
The unit cell of cubic close packed structure is actually that of a face-centered cubic (fcc) Bravais lattice. In the fcc lattice the close packed layers constitute the {111} planes. As with the hcp lattice packing fraction in a cubic close packed (fcc) cell is 74.05%. Since face centered cubic or fcc is more commonly used in preference to cubic close packed (ccp) in describing the structures, the former will be used throughout this text.
## Coordination Number
The coordination number of an atom or ion within an extended structure is defined as the number of nearest neighbor atoms (ions of opposite charge) that are in contact with it. A slightly different definition is often used for atoms within individual molecules: the number of donor atoms associated with the central atom or ion. However, this distinction is rather artificial, and both can be employed.
The coordination numbers for metal atoms in a molecule or complex are commonly 4, 5, and 6, but all values from 2 to 9 are known and a few examples of higher coordination numbers have been reported. In contrast, common coordination numbers in the solid state are 3, 4, 6, 8, and 12. For example, the atom in the center of body-centered cubic lattice has a coordination number of 8, because it touches the eight atoms at the corners of the unit cell, while an atom in a simple cubic structure would have a coordination number of 6. In both fcc and hcp lattices each of the atoms have a coordination number of 12.
## Octahedral and Tetrahedral Vacancies
As was mentioned above, the packing fraction in both fcc and hcp cells is 74.05%, leaving 25.95% of the volume unfilled. The unfilled lattice sites (interstices) between the atoms in a cell are called interstitial sites or vacancies. The shape and relative size of these sites is important in controlling the position of additional atoms. In both fcc and hcp cells most of the space within these atoms lies within two different sites known as octahedral sites and tetrahedral sites. The difference between the two lies in their “coordination number”, or the number of atoms surrounding each site. Tetrahedral sites (vacancies) are surrounded by four atoms arranged at the corners of a tetrahedron. Similarly, octahedral sites are surrounded by six atoms which make-up the apices of an octahedron. For a given close packed lattice an octahedral vacancy will be larger than a tetrahedral vacancy.
Within a face centered cubic lattice, the eight tetrahedral sites are positioned within the cell, at the general fractional coordinate of (n/4,n/4,n/4) where n = 1 or 3, e.g., (1/4,1/4,1/4), (1/4,1/4,3/4), etc. The octahedral sites are located at the center of the unit cell (1/2,1/2,1/2), as well as at each of the edges of the cell, e.g., (1/2,0,0). In the hexagonal close packed system, the tetrahedral sites are at (0,0,3/8) and (1/3,2/3,7/8), and the octahedral sites are at (1/3,1/3,1/4) and all symmetry equivalent positions.
## Important Structure Types
The majority of crystalline materials do not have a structure that fits into the one atom per site simple Bravais lattice. A number of other important crystal structures are found, however, only a few of these crystal structures are those of which occur for the elemental and compound semiconductors and the majority of these are derived from fcc or hcp lattices. Each structural type is generally defined by an archetype, a material (often a naturally occurring mineral) which has the structure in question and to which all the similar materials are related. With regard to commonly used elemental and compound semiconductors the important structures are diamond, zinc blende, Wurtzite, and to a lesser extent chalcopyrite. However, rock salt, β-tin, cinnabar and cesium chloride are observed as high pressure or high temperature phases and are therefore also discussed. The following provides a summary of these structures. Details of the full range of solid-state structures are given elsewhere.
### Diamond Cubic
The diamond cubic structure consists of two interpenetrating face-centered cubic lattices, with one offset 1/4 of a cube along the cube diagonal. It may also be described as face centered cubic lattice in which half of the tetrahedral sites are filled while all the octahedral sites remain vacant. The diamond cubic unit cell is shown in Figure $$\PageIndex{8}$$. Each of the atoms (e.g., C) is four coordinate, and the shortest interatomic distance (C-C) may be determined from the unit cell parameter (a).
$C-C\ =\ a \frac{\sqrt{3} }{4} \approx \ 0.422 a \label{1}$
### Zinc Blende
This is a binary phase (ME) and is named after its archetype, a common mineral form of zinc sulfide (ZnS). As with the diamond lattice, zinc blende consists of the two interpenetrating fcc lattices. However, in zinc blende one lattice consists of one of the types of atoms (Zn in ZnS), and the other lattice is of the second type of atom (S in ZnS). It may also be described as face centered cubic lattice of S atoms in which half of the tetrahedral sites are filled with Zn atoms. All the atoms in a zinc blende structure are 4-coordinate. The zinc blende unit cell is shown in Figure $$\PageIndex{9}$$. A number of inter-atomic distances may be calculated for any material with a zinc blende unit cell using the lattice parameter (a).
$Zn-S\ =\ a \frac{\sqrt{3} }{4} \approx \ 0.422 a \label{2}$
$Zn-Zn \ =\ S-S\ = \frac{a}{\sqrt{2}} \approx 0.707\ a \label{3}$
### Chalcopyrite
The mineral chalcopyrite CuFeS2 is the archetype of this structure. The structure is tetragonal (a = b ≠ c, α = β = γ = 90°, and is essentially a superlattice on that of zinc blende. Thus, is easiest to imagine that the chalcopyrite lattice is made-up of a lattice of sulfur atoms in which the tetrahedral sites are filled in layers, ...FeCuCuFe..., etc. (Figure $$\PageIndex{10}$$. In such an idealized structure c = 2a, however, this is not true of all materials with chalcopyrite structures.
### Rock Salt
As its name implies the archetypal rock salt structure is NaCl (table salt). In common with the zinc blende structure, rock salt consists of two interpenetrating face-centered cubic lattices. However, the second lattice is offset 1/2a along the unit cell axis. It may also be described as face centered cubic lattice in which all of the octahedral sites are filled, while all the tetrahedral sites remain vacant, and thus each of the atoms in the rock salt structure are 6-coordinate. The rock salt unit cell is shown in Figure $$\PageIndex{11}$$. A number of inter-atomic distances may be calculated for any material with a rock salt structure using the lattice parameter (a).
$Na-Cl\ =\ \frac{a}{2} \approx 0.5 a \label{4}$
$Na-Na \ =\ Cl-Cl \ =\ \frac{a}{\sqrt{2}} \approx 0.707\ a \label{5}$
### Cinnabar
Cinnabar, named after the archetype mercury sulfide, HgS, is a distorted rock salt structure in which the resulting cell is rhombohedral (trigonal) with each atom having a coordination number of six.
### Wurtzite
This is a hexagonal form of the zinc sulfide. It is identical in the number of and types of atoms, but it is built from two interpenetrating hcp lattices as opposed to the fcc lattices in zinc blende. As with zinc blende all the atoms in a wurtzite structure are 4-coordinate. The wurtzite unit cell is shown in Figure $$\PageIndex{12}$$. A number of inter atomic distances may be calculated for any material with a wurtzite cell using the lattice parameter (a).
$Zn-S\ =\ a \sqrt{3/8 } \ =\ 0.612\ a\ = \frac{3 c}{8} \ =\ 0.375\ c \label{6}$
$Zn- Zn \ =\ S-S\ =\ a\ =\ 1.632\ c \label{7}$
However, it should be noted that these formulae do not necessarily apply when the ratio a/c is different from the ideal value of 1.632.
### Cesium Chloride
The cesium chloride structure is found in materials with large cations and relatively small anions. It has a simple (primitive) cubic cell (Figure $$\PageIndex{13}$$) with a chloride ion at the corners of the cube and the cesium ion at the body center. The coordination numbers of both Cs+ and Cl-, with the inner atomic distances determined from the cell lattice constant (a).
$Cs-Cl\ =\ \frac{a \sqrt{3} }{2} \approx 0.866a \label{8}$
$Cs-Cs \ =\ Cl-Cl\ = a \label{9}$
### β-Tin
The room temperature allotrope of tin is β-tin or white tin. It has a tetragonal structure, in which each tin atom has four nearest neighbors (Sn-Sn = 3.016 Å) arranged in a very flattened tetrahedron, and two next nearest neighbors (Sn-Sn = 3.175 Å). The overall structure of β-tin consists of fused hexagons, each being linked to its neighbor via a four-membered Sn4 ring.
## Defects in Crystalline Solids
Up to this point we have only been concerned with ideal structures for crystalline solids in which each atom occupies a designated point in the crystal lattice. Unfortunately, defects ordinarily exist in equilibrium between the crystal lattice and its environment. These defects are of two general types: point defects and extended defects. As their names imply, point defects are associated with a single crystal lattice site, while extended defects occur over a greater range.
### Point Defects: "Too Many or Too Few" or "Just Plain Wrong"
Point defects have a significant effect on the properties of a semiconductor, so it is important to understand the classes of point defects and the characteristics of each type. Figure $$\PageIndex{13}$$ summarizes various classes of native point defects, however, they may be divided into two general classes; defects with the wrong number of atoms (deficiency or surplus) and defects where the identity of the atoms is incorrect.
### Interstitial Impurity
An interstitial impurity occurs when an extra atom is positioned in a lattice site that should be vacant in an ideal structure (Figure $$\PageIndex{13}$$ b).Since all the adjacent lattice sites are filled the additional atom will have to squeeze itself into the interstitial site, resulting in distortion of the lattice and alteration in the local electronic behavior of the structure. Small atoms, such as carbon, will prefer to occupy these interstitial sites. Interstitial impurities readily diffuse through the lattice via interstitial diffusion, which can result in a change of the properties of a material as a function of time. Oxygen impurities in silicon generally are located as interstitials.
### Vacancies
The converse of an interstitial impurity is when there are not enough atoms in a particular area of the lattice. These are called vacancies. Vacancies exist in any material above absolute zero and increase in concentration with temperature. In the case of compound semiconductors, vacancies can be either cation vacancies (Figure $$\PageIndex{13}$$ c) or anion vacancies (Figure $$\PageIndex{13}$$ d), depending on what type of atom are “missing”.
### Substitution
Substitution of various atoms into the normal lattice structure is common, and used to change the electronic properties of both compound and elemental semiconductors. Any impurity element that is incorporated during crystal growth can occupy a lattice site. Depending on the impurity, substitution defects can greatly distort the lattice and/or alter the electronic structure. In general, cations will try to occupy cation lattice sites (Figure $$\PageIndex{13}$$ e), and anion will occupy the anion site (Figure $$\PageIndex{13}$$ f). For example, a zinc impurity in GaAs will occupy a gallium site, if possible, while a sulfur, selenium and tellurium atoms would all try to substitute for an arsenic. Some impurities will occupy either site indiscriminately, e.g., Si and Sn occupy both Ga and As sites in GaAs.
### Antisite Defects
Antisite defects are a particular form of substitution defect, and are unique to compound semiconductors. An antisite defect occurs when a cation is misplaced on an anion lattice site or vice versa ( Figure $$\PageIndex{13}$$ g and h).Dependant on the arrangement these are designated as either AB antisite defects or BA antisite defects. For example, if an arsenic atom is on a gallium lattice site the defect would be an AsGa defect. Antisite defects involve fitting into a lattice site atoms of a different size than the rest of the lattice, and therefore this often results in a localized distortion of the lattice. In addition, cations and anions will have a different number of electrons in their valence shells, so this substitution will alter the local electron concentration and the electronic properties of this area of the semiconductor.
### Extended Defects: Dislocations in a Crystal Lattice
Extended defects may be created either during crystal growth or as a consequence of stress in the crystal lattice. The plastic deformation of crystalline solids does not occur such that all bonds along a plane are broken and reformed simultaneously. Instead, the deformation occurs through a dislocation in the crystal lattice. Figure shows a schematic representation of a dislocation in a crystal lattice. Two features of this type of dislocation are the presence of an extra crystal plane, and a large void at the dislocation core. Impurities tend to segregate to the dislocation core in order to relieve strain from their presence.
## Epitaxy
Epitaxy, is a transliteration of two Greek words epi, meaning "upon", and taxis, meaning "ordered". With respect to crystal growth it applies to the process of growing thin crystalline layers on a crystal substrate. In epitaxial growth, there is a precise crystal orientation of the film in relation to the substrate. The growth of epitaxial films can be done by a number of methods including molecular beam epitaxy, atomic layer epitaxy, and chemical vapor deposition, all of which will be described later.
Epitaxy of the same material, such as a gallium arsenide film on a gallium arsenide substrate, is called homoepitaxy, while epitaxy where the film and substrate material are different is called heteroepitaxy. Clearly, in homoepitaxy, the substrate and film will have the identical structure, however, in heteroepitaxy, it is important to employ where possible a substrate with the same structure and similar lattice parameters. For example, zinc selenide (zinc blende, a = 5.668 Å) is readily grown on gallium arsenide (zinc blende, a = 5.653 Å). Alternatively, epitaxial crystal growth can occur where there exists a simple relationship between the structures of the substrate and crystal layer, such as is observed between Al2O3 (100) on Si (100). Whichever route is chosen a close match in the lattice parameters is required, otherwise, the strains induced by the lattice mismatch results in distortion of the film and formation of dislocations. If the mismatch is significant epitaxial growth is not energetically favorable, causing a textured film or polycrystalline untextured film to be grown. As a general rule of thumb, epitaxy can be achieved if the lattice parameters of the two materials are within about 5% of each other. For good quality epitaxy, this should be less than 1%. The larger the mismatch, the larger the strain in the film. As the film gets thicker and thicker, it will try to relieve the strain in the film, which could include the loss of epitaxy of the growth of dislocations. It is important to note that the <100> directions of a film must be parallel to the <100> direction of the substrate. In some cases, such as Fe on MgO, the [111] direction is parallel to the substrate [100]. The epitaxial relationship is specified by giving first the plane in the film that is parallel to the substrate [100]. | 7,593 | 32,451 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-50 | latest | en | 0.945862 |
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## Problem 51E Chapter 1
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Problem 51E
The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants” (?Lubric. Engr., ?1984: 75–83) reported the following data on oxidation-induction time (min) for various commercial oils: a?? alculate the sample variance and standard deviation. b. ?If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression.
Step-by-Step Solution:
Answer : Step 1 : Given The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants”. Consider the data 87 145 103 153 130 152 160 138 180 87 195 99 132 93 145 119 211 129 105 Now we have to calculate the sample variance and standard deviation. 2 x xi 87 7569 103 10609 130 16900 160 25600 180 32400 195 38025 132 17424 145 21025 211 44521 105 11025 145 21025 153 23409 152 23104 138 19044 87 7569 99 9801 93 8649 119 14161 129 16641 x =2563 x =368501 i i So we have to find mean. The formula of the mean is xi x = n Substitute the value x = 20,179 and n=27. i 2,563 x = 19 x = 134.8947 The formula of the sample variance is. 2 ( x ) 2 (xi) n s = n1 Substitute the value i , n and x. ( 2563 ) 2 (368501) 19 s = 191 ( 658969) 2 (368501) 19 s = 18 s = (368501) 345735.2105 18 2 (22765.78947 s = 18 s = 1,264.7660 Therefore sample variance is 1,264.7660 Then we have to find standard deviation. Standard deviation is square root of the sample variance. We know that sample variance is 1,264.7660. s = 1,264.7660 s = 35.5635 Therefore the standard deviation is 35.5635.
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https://ask4knowledgebase.com/questions/3099219/ggplot-with-2-y-axes-on-each-side-and-different-scales | 1,624,190,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00601.warc.gz | 118,693,534 | 11,429 | ## ggplot with 2 y axes on each side and different scales
### Question
I need to plot a bar chart showing counts and a line chart showing rate all in one chart, I can do both of them separately, but when I put them together, I scale of the first layer (i.e. the `geom_bar`) is overlapped by the second layer (i.e. the `geom_line`).
Can I move the axis of the `geom_line` to the right?
2019/05/27
1
237
5/27/2019 7:58:38 PM
Sometimes a client wants two y scales. Giving them the "flawed" speech is often pointless. But I do like the ggplot2 insistence on doing things the right way. I am sure that ggplot is in fact educating the average user about proper visualization techniques.
Maybe you can use faceting and scale free to compare the two data series? - e.g. look here: https://github.com/hadley/ggplot2/wiki/Align-two-plots-on-a-page
2017/11/28
110
11/28/2017 9:11:04 PM
Starting with ggplot2 2.2.0 you can add a secondary axis like this (taken from the ggplot2 2.2.0 announcement):
``````ggplot(mpg, aes(displ, hwy)) +
geom_point() +
scale_y_continuous(
"mpg (US)",
sec.axis = sec_axis(~ . * 1.20, name = "mpg (UK)")
)``````
2018/07/21
Taking above answers and some fine-tuning (and for whatever it's worth), here is a way of achieving two scales via `sec_axis`:
Assume a simple (and purely fictional) data set `dt`: for five days, it tracks the number of interruptions VS productivity:
`````` when numinter prod
1 2018-03-20 1 0.95
2 2018-03-21 5 0.50
3 2018-03-23 4 0.70
4 2018-03-24 3 0.75
5 2018-03-25 4 0.60
``````
(the ranges of both columns differ by about factor 5).
The following code will draw both series that they use up the whole y axis:
``````ggplot() +
geom_bar(mapping = aes(x = dt\$when, y = dt\$numinter), stat = "identity", fill = "grey") +
geom_line(mapping = aes(x = dt\$when, y = dt\$prod*5), size = 2, color = "blue") +
scale_x_date(name = "Day", labels = NULL) +
scale_y_continuous(name = "Interruptions/day",
sec.axis = sec_axis(~./5, name = "Productivity % of best",
labels = function(b) { paste0(round(b * 100, 0), "%")})) +
theme(
axis.title.y = element_text(color = "grey"),
axis.title.y.right = element_text(color = "blue"))
``````
Here's the result (above code + some color tweaking):
The point (aside from using `sec_axis` when specifying the y_scale is to multiply each value the 2nd data series with 5 when specifying the series. In order to get the labels right in the sec_axis definition, it then needs dividing by 5 (and formatting). So a crucial part in above code is really `*5` in the geom_line and `~./5` in sec_axis (a formula dividing the current value `.` by 5).
In comparison (I don't want to judge the approaches here), this is how two charts on top of one another look like:
You can judge for yourself which one better transports the message (“Don’t disrupt people at work!”). Guess that's a fair way to decide.
The full code for both images (it's not really more than what's above, just complete and ready to run) is here: https://gist.github.com/sebastianrothbucher/de847063f32fdff02c83b75f59c36a7d a more detailed explanation here: https://sebastianrothbucher.github.io/datascience/r/visualization/ggplot/2018/03/24/two-scales-ggplot-r.html
2019/10/31
There are common use-cases duel y axes, e.g., the climatograph showing monthly temperature and precipitation. Here is a simple solution, generalized from Megatron's solution by allowing you to set the lower limit of the variables to something else than zero:
Example data:
``````climate <- tibble(
Month = 1:12,
Temp = c(-4,-4,0,5,11,15,16,15,11,6,1,-3),
Precip = c(49,36,47,41,53,65,81,89,90,84,73,55)
)
``````
Set the following two values to values close to the limits of the data (you can play around with these to adjust the positions of the graphs; the axes will still be correct):
``````ylim.prim <- c(0, 180) # in this example, precipitation
ylim.sec <- c(-4, 18) # in this example, temperature
``````
The following makes the necessary calculations based on these limits, and makes the plot itself:
``````b <- diff(ylim.prim)/diff(ylim.sec)
a <- b*(ylim.prim[1] - ylim.sec[1])
ggplot(climate, aes(Month, Precip)) +
geom_col() +
geom_line(aes(y = a + Temp*b), color = "red") +
scale_y_continuous("Precipitation", sec.axis = sec_axis(~ (. - a)/b, name = "Temperature")) +
scale_x_continuous("Month", breaks = 1:12) +
ggtitle("Climatogram for Oslo (1961-1990)")
``````
If you want to make sure that the red line corresponds to the right-hand y axis, you can add a `theme` sentence to the code:
``````ggplot(climate, aes(Month, Precip)) +
geom_col() +
geom_line(aes(y = a + Temp*b), color = "red") +
scale_y_continuous("Precipitation", sec.axis = sec_axis(~ (. - a)/b, name = "Temperature")) +
scale_x_continuous("Month", breaks = 1:12) +
theme(axis.line.y.right = element_line(color = "red"),
axis.ticks.y.right = element_line(color = "red"),
axis.text.y.right = element_text(color = "red"),
axis.title.y.right = element_text(color = "red")
) +
ggtitle("Climatogram for Oslo (1961-1990)")
``````
which colors the right-hand axis:
2019/11/19
You can create a scaling factor which is applied to the second geom and right y-axis. This is derived from Sebastian's solution.
``````library(ggplot2)
scaleFactor <- max(mtcars\$cyl) / max(mtcars\$hp)
ggplot(mtcars, aes(x=disp)) +
geom_smooth(aes(y=cyl), method="loess", col="blue") +
geom_smooth(aes(y=hp * scaleFactor), method="loess", col="red") +
scale_y_continuous(name="cyl", sec.axis=sec_axis(~./scaleFactor, name="hp")) +
theme(
axis.title.y.left=element_text(color="blue"),
axis.text.y.left=element_text(color="blue"),
axis.title.y.right=element_text(color="red"),
axis.text.y.right=element_text(color="red")
)
``````
Note: using `ggplot2` v3.0.0
2018/08/16
The technical backbone to the solution of this challenge has been provided by Kohske some 3 years ago [KOHSKE]. The topic and the technicalities around its solution have been discussed on several instances here on Stackoverflow [IDs: 18989001, 29235405, 21026598]. So i shall only provide a specific variation and some explanatory walkthrough, using above solutions.
Let us assume we do have some data y1 in group G1 to which some data y2 in group G2 is related in some way, e.g. range/scale transformed or with some noise added. So one wants to plot the data together on one plot with the scale of y1 on the left and y2 on the right.
`````` df <- data.frame(item=LETTERS[1:n], y1=c(-0.8684, 4.2242, -0.3181, 0.5797, -0.4875), y2=c(-5.719, 205.184, 4.781, 41.952, 9.911 )) # made up!
> df
item y1 y2
1 A -0.8684 -19.154567
2 B 4.2242 219.092499
3 C -0.3181 18.849686
4 D 0.5797 46.945161
5 E -0.4875 -4.721973
``````
If we now plot our data together with something like
``````ggplot(data=df, aes(label=item)) +
theme_bw() +
geom_segment(aes(x='G1', xend='G2', y=y1, yend=y2), color='grey')+
geom_text(aes(x='G1', y=y1), color='blue') +
geom_text(aes(x='G2', y=y2), color='red') +
theme(legend.position='none', panel.grid=element_blank())
``````
it doesnt align nicely as the smaller scale y1 obviosuly gets collapsed by larger scale y2.
The trick here to meet the challenge is to techncially plot both data sets against the first scale y1 but report the second against a secondary axis with labels showing the original scale y2.
So we build a first helper function CalcFudgeAxis which calculates and collects features of the new axis to be shown. The function can be amended to ayones liking (this one just maps y2 onto the range of y1).
``````CalcFudgeAxis = function( y1, y2=y1) {
Cast2To1 = function(x) ((ylim1[2]-ylim1[1])/(ylim2[2]-ylim2[1])*x) # x gets mapped to range of ylim2
ylim1 <- c(min(y1),max(y1))
ylim2 <- c(min(y2),max(y2))
yf <- Cast2To1(y2)
labelsyf <- pretty(y2)
return(list(
yf=yf,
labels=labelsyf,
breaks=Cast2To1(labelsyf)
))
}
``````
what yields some:
``````> FudgeAxis <- CalcFudgeAxis( df\$y1, df\$y2 )
> FudgeAxis
\$yf
[1] -0.4094344 4.6831656 0.4029175 1.0034664 -0.1009335
\$labels
[1] -50 0 50 100 150 200 250
\$breaks
[1] -1.068764 0.000000 1.068764 2.137529 3.206293 4.275058 5.343822
> cbind(df, FudgeAxis\$yf)
item y1 y2 FudgeAxis\$yf
1 A -0.8684 -19.154567 -0.4094344
2 B 4.2242 219.092499 4.6831656
3 C -0.3181 18.849686 0.4029175
4 D 0.5797 46.945161 1.0034664
5 E -0.4875 -4.721973 -0.1009335
``````
Now I wraped Kohske's solution in the second helper function PlotWithFudgeAxis (into which we throw the ggplot object and helper object of the new axis):
``````library(gtable)
library(grid)
PlotWithFudgeAxis = function( plot1, FudgeAxis) {
# based on: https://rpubs.com/kohske/dual_axis_in_ggplot2
plot2 <- plot1 + with(FudgeAxis, scale_y_continuous( breaks=breaks, labels=labels))
#extract gtable
g1<-ggplot_gtable(ggplot_build(plot1))
g2<-ggplot_gtable(ggplot_build(plot2))
#overlap the panel of the 2nd plot on that of the 1st plot
pp<-c(subset(g1\$layout, name=="panel", se=t:r))
ia <- which(g2\$layout\$name == "axis-l")
ga <- g2\$grobs[[ia]]
ax <- ga\$children[[2]]
ax\$widths <- rev(ax\$widths)
ax\$grobs <- rev(ax\$grobs)
ax\$grobs[[1]]\$x <- ax\$grobs[[1]]\$x - unit(1, "npc") + unit(0.15, "cm")
g <- gtable_add_cols(g, g2\$widths[g2\$layout[ia, ]\$l], length(g\$widths) - 1)
g <- gtable_add_grob(g, ax, pp\$t, length(g\$widths) - 1, pp\$b)
grid.draw(g)
}
``````
Now all can be put together: Below code shows, how the proposed solution could be used in a day-to-day environment. The plot call now doesnt plot the original data y2 anymore but a cloned version yf (held inside the pre-calculated helper object FudgeAxis), which runs of the scale of y1. The original ggplot objet is then manipulated with Kohske's helper function PlotWithFudgeAxis to add a second axis preserving the scales of y2. It plots as well the manipulated plot.
``````FudgeAxis <- CalcFudgeAxis( df\$y1, df\$y2 )
tmpPlot <- ggplot(data=df, aes(label=item)) +
theme_bw() +
geom_segment(aes(x='G1', xend='G2', y=y1, yend=FudgeAxis\$yf), color='grey')+
geom_text(aes(x='G1', y=y1), color='blue') +
geom_text(aes(x='G2', y=FudgeAxis\$yf), color='red') +
theme(legend.position='none', panel.grid=element_blank())
PlotWithFudgeAxis(tmpPlot, FudgeAxis)
``````
This now plots as desired with two axis, y1 on the left and y2 on the right
Above solution is, to put it straight, a limited shaky hack. As it plays with the ggplot kernel it will throw some warnings that we exchange post-the-fact scales, etc. It has to be handled with care and may produce some undesired behaviour in another setting. As well one may need to fiddle around with the helper functions to get the layout as desired. The placement of the legend is such an issue (it would be placed between the panel and the new axis; this is why I droped it). The scaling / alignment of the 2 axis is as well a bit challenging: The code above works nicely when both scales contain the "0", else one axis gets shifted. So definetly with some opportunities to improve...
In case on wants to save the pic one has to wrap the call into device open / close:
``````png(...)
PlotWithFudgeAxis(tmpPlot, FudgeAxis)
dev.off()
``````
2016/05/14 | 3,468 | 11,232 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.816009 |
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Feedback control theory makes it possible to control well even if • time domain: the settling time: r c(s)=r c/s system response is ()() ()1 2 ssk kur s rgs. Plotting system responses frequency-domain analysis is key to understanding right-click on the plot to display the peak response and settling time. Generate and visualize time-response data such as time-domain responses when you perform time-domain control system toolbox™ time-domain analysis. Design of pi and pid controllers with transient performance new tuning rules based on time-domain step response is to the actual system response. 256 chapter 5 time-domain analysis of control systems rotation of the shaft the ramp function has the ability to test how the system would respond to a signal that.
Develop an understanding of dynamic behavior and take a look at the concept of transfer function both are very useful for control system time-domain response. 612 relationship between closed-loop frequency response and the time-domain response section 610 has illustrated how the closed-loop frequency-domain response may. Modeling and experimental validation of a second order plant: response of the system, in laplace domain, the time domain response of the output takes the. Time domain and frequency domain measurement and the modulation-domain system time domain response,. Time domain analysis of control system steady state response of control system is a function of input transient state and steady state response of first.
Outline course info time-delay systems in control applications system-theoretic preliminaries basic properties delay systems in the frequency domain. Motor modeling and position control lab where ˝is the mechanical time constant of the system, it is possible to compute the system’s response to any input. 1 relationship between transfer function pole locations and time-domain step time-invariant (lti) system and the response of the system in the time domain to a.
Second-order systems unit 3: time response, part 2: second-order responses engineering 5821: control systems i faculty of engineering & applied science. Chapter eight time domain controller in section 83 we present common controllers used in linear system control improve both the system transient response.
Time-domain analysis of continuous-time systems systems are lti from now on unless otherwise stated recall course objectives. Industrial control systems are often designed using frequency correlation between time and frequency response frequency domain analysis of control system. Pid overview in this tutorial, we will consider the following unity-feedback system: the output of a pid controller, which is equal to the control input to the plant. 22451 dynamic systems – system response system response peter avitabile mechanical engineering department university of massachusetts lowell time domain frf y.
Control system response in time domain
Rated 3/5 based on 15 review
2018. | 799 | 4,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-43 | latest | en | 0.85727 |
https://brilliant.org/discussions/thread/coloring-a-cube/ | 1,537,506,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156780.2/warc/CC-MAIN-20180921033529-20180921053929-00207.warc.gz | 477,932,837 | 13,850 | # Coloring a cube
In how many ways a cube can be colored with n different colors such that no 2 neighboring or adjacent faces are colored with same color?
Note by Abdullah Ahmed
1 year, 8 months ago
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I have solved it .
- 1 year, 7 months ago
Great! How would you explain the approach?
Staff - 1 year, 7 months ago
Isn't the formula 1/24(n^6+12n^3+3 n^4+8n^2)? I know it from Burnside's lemma. but in the question two adjacent sides are not colored with same color. so what is the general formula for it? @Calvin Lin
- 1 year, 8 months ago | 477 | 1,419 | {"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} | 3 | 3 | CC-MAIN-2018-39 | latest | en | 0.816747 |
http://www.reddit.com/user/ihendley/comments/ | 1,408,531,635,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500804220.17/warc/CC-MAIN-20140820021324-00278-ip-10-180-136-8.ec2.internal.warc.gz | 554,874,037 | 16,987 | # reddit is a website about everything
[–] 0 points1 point
Good job! That's been my goal for a while but it still feels out of reach :(
[–] 0 points1 point
[pgn]
[Event "Live Chess"] [Site "Chess.com"] [Date "2014.06.28"] [White "oGsTheSTD"] [Black "clownopening"] [Result "0-1"] [WhiteElo "1165"] [BlackElo "1818"] [TimeControl "15|10"] [Termination "clownopening won by resignation"]
1.d4 Nf6 2.Nf3 g6 3.e3 Bg7 4.Nc3 O-O 5.b3 c5 6.Bc4 d5 7.Bd3 Bg4 8.Ba3 Qa5 9.Bb2 Nc6 10.O-O cxd4 11.exd4 Nxd4 12.Qd2 Bxf3 13.g3 Rac8 14.Rfe1 Bg4 15.Kf1 Bh3+ 16.Kg1 Nf3+ 0-1
[/pgn]
[–] 0 points1 point
Yes, sorry, different problem.
Well to add something useful to the discussion for this problem, if you want to do it with integers, then think about what the prime factorization of a number that ends in a zero looks like, and figure out how this helps you factor one million into two numbers that do not end in zeroes.
[–] -3 points-2 points
Your example violates two stipulations of the problem: It uses the digit 3 more than once, and it concatenates digits!
Edit: I was thinking about the 1,2,3 problem. In fact your answer is even worse than I thought because 3*333333.333... does not equal 19! :)
[–] 1 point2 points
So long as both don't contain zeros
This is the hard part. There are actually only two numbers that work, each giving you the other one when you divide it into one million.
[–] 1 point2 points
``````floor(3*tan(tan(tan(2))))*1 ?
``````
[–] 0 points1 point
The deadline to complete negotiations for your match is Wednesday, 18th of June at 23:59 GMT
I messaged my opponent on chess.com Monday night asking him when he would like to play, but he did get back to me.
[–] 0 points1 point
The deadline to complete negotiations for your match is Wednesday, 18th of June at 23:59 GMT.
When is the deadline to play the match?
[–] 0 points1 point
clownopening
[–] 0 points1 point
8.Bg5 looked more natural and this did turn out to be the mainline when I looked it up. It seems to put a stop to any Black Ne4 ideas immediately or later. And similarly I don't like 9.Nf3 as much as 9.e3, Because if Black plays 9...Ne4 then your Queen has to go to an awkward square because if 10.Qc2 then ...Bf5 is annoying. The difference after 9.e3 is that after ...Ne4 Qc2 Bf5 you have 11.Bd3.
[–] -2 points-1 points
What did you like about Buried? Every character in the movie was an idiot. I found it impossible to enjoy.
[–] 0 points1 point
I would strongly recommend at least triple checking the address. Not joking.
[–] 0 points1 point
I recommend the Vienna. By keeping your g1 Knight home you can get in an early f4 in a lot of lines, which often leads to a quick attack. In fact, I just played a Vienna in my last game a few minutes ago!
And the best Vienna game I ever played went like this:
[pgn] 1.e4 e5 2.Nc3 Nc6 3.f4 exf4 4.d4 Qh4+ 5.Ke2 d5 6.exd5 Bg4+ 7.Nf3 O-O-O 8.dxc6 Re8+ 9.Kd3 Bf5+ 10.Kc4 Qf6 11.Kb3 Qxc6 12.a4 Qb6+ 13.Ka2 Be6+ 14.Kb1 {Black resigns.} [/pgn]
[–] 0 points1 point
Nd5 is good and still winning, but I still think d4 is best :) When I analyzed this line a while ago I let my computer run a very long time on each move. Houdini should switch to d4 if you give it long enough.
[–] 0 points1 point
This opening is losing for Black but of course it's impossible to play accurately as White in blitz, especially against a strong opponent. The main ideas are setting up to trap the Black Queen while also playing against c7. As played d4 is actually the best move despite what the engine says at low depths. And after d6 in the game 9. Be3 is actually crushing. There are a lot of possible continuations but if for example Bh3 then Rg1 Qh2 Qh5 and Black is dropping the Bishop.
[–] 0 points1 point
writing debug messages to System.err (or out)
What are the best practices for writing debug messages?
[–] 1 point2 points
I can't breathe
[–] 0 points1 point
I got excited for a moment when I saw that there are stability improvements since Armory crashes almost every time I use it. Then I realized I am already running the latest release.
Can someone recommend something a little more lightweight? Waiting for Bitcoin-QT to sync and then waiting another 15 minutes for Armory to prepare databases and scan transaction history is also a huge hassle. Maybe running a full node is not for me.
[–] 0 points1 point
[–] 0 points1 point
Would an efficient indistinguishability obfuscation algorithm (problem 10) have any implications for useful computation in proof of work (problem 3)?
If I am understanding correctly, a useful computational problem could be obfuscated so that the output would be signed with some password if and only if the computation was run in full, right?
[–] 0 points1 point
I remember one summer in the mid 90's when my friend and I used to type a search into Webcrawler, go out and play a game of basketball to 5 points, and then come back in to see if the search had returned any results yet. I think this was on a 9600 modem.
[–] 0 points1 point
I love you too
[–] 2 points3 points
[–] 0 points1 point
Is it possible to set up 2FA on gmail? If not should I be using another email service?
[–] 2 points3 points
200k was exactly the amount they were shown to still be in possession of by analyzing the blockchain, was it not? | 1,541 | 5,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | latest | en | 0.925609 |
https://forum.dynamobim.com/t/group-similar-list-items-to-sub-lists/48919 | 1,606,798,702,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00024.warc.gz | 282,914,368 | 5,648 | # Group similar list items to sub-lists
Hi,
Who can help me to find a way how to group my list values by A001, B001, C001 in this case and create sublists? The string structure XX-XXX-XXXX would be the same at all times.
On the right is the list I’m looking for.
Thanks.
an issue with Python
``````lst_header = IN[0]
lst_b = IN[1]
fun = lambda x, lsty : [ x ]+[y for y in lsty if y.endswith(x)]
OUT = [fun(x,lst_b) for x in lst_header]
``````
or only with the full list XX-XXX-XXXX
``````full_list = IN[0]
set_header = set([(x.split('-'))[-1] for x in full_list if (x.split('-'))[-1]])
fun = lambda x, lsty : [x]+[y for y in lsty if y.endswith(x)]
OUT = [fun(x,full_list ) for x in set_header]``````
3 Likes
you can do something like this
1 Like
Look good. Thanks! But quite hard to read this syntax.
here is a version without lambda function and list comprehension
``````full_list = ['11-001-A001','03-003-B001', '05-002-C001', '02-023-B001', '06-013-A001','-']
lst_suff = []
for item in full_list:
#get suffix
suffix = item.split('-').pop(-1)
#check if value
if suffix :
lst_suff.append(suffix)
#remove duplicate | 345 | 1,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.684053 |
https://ell.stackexchange.com/questions/116460/is-the-phrase-the-long-way-of-the-oval-the-object-of-the-clause-and-should-the | 1,638,754,945,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363229.84/warc/CC-MAIN-20211206012231-20211206042231-00015.warc.gz | 300,959,590 | 29,081 | # Is the phrase "the long way of the oval" the object of the clause and should there be an "into" before it?
I encountered this sentence when reading "A Morbid Taste for Bones" by Ellis Peters:
Cadfael turned towards the cry, thrusting through thorn-branches, and came out in a narrow oval of grass surrounded every way with thick bushes, through which a used track no wider than a man's shoulders clove, the long way of the oval.
I under stand that the clause "through which a used track no wider than a man's shoulders clove" refers to "thick bushes", but what is the role of "the long way of the oval"? Is it the indirect object of "a used track"? So "a used track cleaves through the bushes into the oval"? But it does not have a preposition before it.
Edit:
This is what I imagine based on @TRomano's comments:
The black curves are the bushes, and the blue path is the track.
• The track goes through the thorn bushes and extends the length of the oval, not across its width. The long way of the oval is an absolute adjunct or an absolute appositive. You can understand it like this: a track ... (it being) the long way of the oval. Jan 21 '17 at 11:37
• @TRomano: Thank you, I see. It now sounds grammatically correct, but not easy to imagine. So the track has two parts, one part is through the bush, then it continues along the length of the oval, is that right? Jan 21 '17 at 11:41
• Imagine this as oval rather than round, and the perimeter of the circle (oval) a hedge of thorn bushes: Ø The narrow track cuts through the thorn bushes and runs the length of the oval. Jan 21 '17 at 11:42
• I don't imaging the narrow track ending at the far edge, but continuing through it, exiting on the far side (or entering there, if you're coming from that direction). The track cleaves through the thorn hedge. Also, I imagine the track extending along the long axis of the oval, not cleaving to the perimeter. It cleaves through, not to the thorn hedge. Cleave meaning "to cut through", not cleave meaning "to cling to". Jan 21 '17 at 12:03
• Yes, the long axis. (typo in my last comment, "imagine" not "imaging"). Jan 21 '17 at 12:34 | 543 | 2,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.941367 |
http://denethor.wlu.ca/pc320/scope/scope_lab.shtml | 1,723,294,072,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00641.warc.gz | 7,585,050 | 6,074 | # PC/CP320 Physical Computing
## Function Generator and Oscilloscope
### Objectives
A function generator or signal generator produces a voltage signal with a specified shape, frequency, amplitude, duty cycle, and DC offset. It's primary purpose is to produce a known accurate input signal to test a component's or circuit's response. One lead of the signal generator is attached to ground and the other(s) provide input to the device under test.
An oscilloscope, sometimes simply called a scope, is a device that measures the voltage difference between the positive and negative probes and displays the voltage difference as a function of voltage difference against time. The primary functions of an oscilloscope are to test circuit performance over time (i.e. meters are good for an instantaneous measurement) and to compare two signals (e.g. the input to the circuit against the output from a circuit).
Always make sure that your function generator, your oscilloscope, and your circuit are properly grounded. If ground is improperly setup, any results you may see are meaningless.
1. To gain familiarity with a function generator.
• how to properly connect a function generator to a circuit
• how to select the signal shape, frequency, amplitude, duty cycle and DC offset
• understand how the above signal characteristics are set on the function generator and how they impact the signal shape
2. To gain familiarity with an oscilloscope.
• how to properly connect an oscilloscope to a circuit
• how to determine the signal's shape, frequency, amplitude, duty cycle and DC offset
### Equipment
• bench power
• speaker - two connections: one to ground and one to the input signal
• signal/function generator - instek GFG-8217A
• oscilloscope - one of TDS 210, TDS 1002 (old and new numbers for the same oscilloscope)
### Procedure
#### Function Generator and Oscilloscope Introduction
1. Connect the signal generator to an oscilloscope as shown below. (Use a scope probe connected to the scope, and a normal BNC connector for the function generator.)
Use the signal generator to produce a sine wave at 1 kHz and peak amplitude of 1 Volt (2 volts peak-to-peak). Verify the amplitude and frequency are correct using the scope.
2. Play with the trigger settings on the scope to see what happens. For instance:
• Set the trigger to the channel that's not connected.
• Set the trigger to the channel that is connected, and then move the level around within the voltage range of the waveform, and then outside it.
• Try any other trigger settings you want to see what they do.
After playing around with the settings, use a setting that displays a stable waveform on the screen.
3. With your signal on the screen, see what happens when you change the "Probe" setting for Channel 1 on the scope to 10X. Does the signal change? Does the voltage scale for the channel change?
Now, with the "Probe" setting for Channel 1 still set to 10X, set the switch on the probe itself to 10X. What happened?
Make a note of how the probe switch and the probe setting on the scope need to go together.
Switch both the probe switch and the scope setting back to 1X.
4. Now that you understand how the oscilloscope works, sketch a signal in your lab notebook. Include the scope settings and show how to calculate amplitude, DC offset and period from the trace.
5. Notice how the rotary buttons along the bottom of the signal generator have two different labels; one of the labels applies when the button is pushed in, and the other applies when the button is pulled out. See what happens with the waveform as you test each button so that you understand what happens each way. After this go back to the required waveform.
6. Wire up a speaker (two connections: one to ground and one to the input signal) so that you will be able to both hear the sound of the waveform produced by the signal generator and see the waveform produced by the signal generator. (The signal generator is the input to both the speaker and the scope.)
• Did the voltage drop when you connected the speaker?
• By how much?
• What does this tell you?
7. Make sure the signal generator is set to a sine wave output for this part.
Infants can typically hear sinusoidal signals that range in frequency from 20 Hz to 20 kHz. Things go down hill as you age. The fine hairs in your ears that sense sound become brittle and break off. Usually the smallest hairs break first, making you lose your ability to hear high frequency sounds. You can speed this process up, significantly, by listening to loud noises like gunfire, jet aircraft, and music. Determine the highest and lowest frequency audio signal you can hear. Sometimes it is hard to tell if you can really hear a signal or are just imagining it. Have someone else turn the signal on/off while you are looking away. See if you can reliably tell if the signal is on or off. Do this for each person in the group.
Please keep the volume to a reasonable level! How did you adjust the volume?
8. Repeat the experiment above, but using square waves and triangle waves. Compare the sound of sine waves to square and triangular waves. Do they sound the same or different? Does your answer depend on the frequency of the signal? Can you hear the same range of frequencies for the different wave shapes?
Demonstrate your procedure to the lab supervisor. Be prepared to summarize your findings.
9. Select a square wave on the function generator. What happens when you vary the duty cycle? Does it affect the sound? How would you determine the duty cycle from the scope?
10. What happens when you vary the DC offset? Does it affect the sound?; How do you determine the DC offset from the scope?
Demonstrate your procedure to the lab supervisor. Be prepared to summarize your findings.
Before you leave, push in all of the buttons on the bottom row of the function generator; i.e. duty cycle, offset, etc.
## Resources
If you need to update a browser, you might try Firefox which is
• free
• open source
• available for several platforms
Since this page uses cascading style sheets for its layout, it will look best with a browser which supports the specifications as fully as possible.
If you are looking for an office package, with a word processor, spreadsheet, etc., you might try LibreOffice which is
• free
• open source
• available for several platforms
Go to the main page for the Department of Physics and Computer Science.
Wilfrid Laurier University | 1,368 | 6,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-33 | latest | en | 0.874621 |
https://studylib.net/doc/11168905/solution-of-quiz-4-march-14--2013 | 1,623,586,005,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608702.10/warc/CC-MAIN-20210613100830-20210613130830-00355.warc.gz | 479,591,676 | 11,142 | # SOLUTION OF QUIZ 4 March 14, 2013
```SOLUTION OF QUIZ 4
MINGFENG ZHAO
March 14, 2013
1. [5 Points] Let f : R2 → R be differentiable. Prove that
f (x, y) − f (a, b) = (x − a)fx (c, y) + (y − b)fy (a, d)
for some numbers c between a and x and d between b and y.
Proof. For any x, y, a, b ∈ R and fix them, by the mean value theorem, there exists some c ∈ (a, x)
and d ∈ (b, y) such that
f (x, y) − f (a, y) = fx (c, y)(x − a),
and f (a, y) − f (a, b) = fy (a, d)(y − b).
Hence we get
f (x, y) − f (a, b)
=
f (x, y) − f (a, y) + f (a, y) − f (a, b)
=
(x − a)fx (c, y) + (y − b)fy (a, d).
2. [5 Points] Let u(x1 , · · · , xn , t) : Rn+1 → R be C 2 . We say that u satisfies the n-dimensional wave
equation of speed c if
∂2u
= c2
∂t2
∂2u
∂2u
+ ··· +
2
∂x1
∂x2n
,
where t is the time variable, x1 , · · · , xn are the spatial variables, and the constant c is the propagation
speed of the wave. This equation appears naturally in physics. For example, it describes water waves,
sound waves, and light waves.
1
2
MINGFENG ZHAO
While it is not easy to find solutions to the wave equation in general, the solution to the onedimensional wave equation has a simpler form. For any C 2 functions f, g : R → R, we define
u(x, t) = f (x − ct) + g(x + ct).
Verify that u satisfies the one-dimensional wave equation of speed c, i.e.,
∂2u
∂2u
= c2 2 .
2
∂t
∂x
Proof. By the chain rule, we know that
∂u
(x, t)
∂t
= f 0 (x − ct)(−c) + g 0 (x + ct)c
= −cf 0 (x − ct) + cg 0 (x + ct)
∂2u
(x, t)
∂t2
= −cf 00 (x − ct)(−c) + cg 00 (x + ct)c
= c2 [f 00 (x − ct) + g 00 (x + tc)]
∂u
(x, t)
∂x
∂2u
(x, t)
∂x2
= f 0 (x − ct) + g 0 (x + ct)
= f 00 (x − ct) + g 00 (x − ct).
Hence we know that
2
∂2u
2∂ u
(x,
t)
−
c
(x, t)
∂t2
∂x2
=
c2 [f 00 (x − ct) + g 00 (x + tc)] − c2 [f 00 (x − ct) + g 00 (x + tc)]
=
0.
Therefore, u satisfies the one-dimensional wave equation of speed c.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009 | 875 | 2,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-25 | latest | en | 0.763662 |
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# CHAPTER 15 - PowerPoint PPT Presentation
CHAPTER 15. The Term Structure of Interest Rates. The yield curve is a graph that displays the relationship between yield and maturity. Information on expected future short term rates can be implied from the yield curve. Overview of Term Structure. Figure 15.1 Treasury Yield Curves.
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### CHAPTER 15
The Term Structure of Interest Rates
The yield curve is a graph that displays the relationship between yield and maturity.
Information on expected future short term rates can be implied from the yield curve.
Overview of Term Structure
Bond Pricing
• Yields on different maturity bonds are not all equal.
• We need to consider each bond cash flow as a stand-alone zero-coupon bond.
• Bond stripping and bond reconstitution offer opportunities for arbitrage.
• The value of the bond should be the sum of the values of its parts.
Example 15.1 Valuing Coupon Bonds
• Value a 3 year, 10% coupon bond using discount rates from Table 15.1:
• Price = \$1082.17 and YTM = 6.88%
• 6.88% is less than the 3-year rate of 7%.
Two Types of Yield Curves
Pure Yield Curve
The pure yield curve uses stripped or zero coupon Treasuries.
The pure yield curve may differ significantly from the on-the-run yield curve.
On-the-run Yield Curve
The on-the-run yield curve uses recently issued coupon bonds selling at or near par.
The financial press typically publishes on-the-run yield curves.
Yield Curve Under Certainty
• Suppose you want to invest for 2 years.
• Buy and hold a 2-year zero
-or-
• Rollover a series of 1-year bonds
• Equilibrium requires that both strategies provide the same return.
Yield Curve Under Certainty
Next year’s 1-year rate (r2) is just enough to make rolling over a series of 1-year bonds equal to investing in the 2-year bond.
Spot Rates vs. Short Rates
Spot rate – the rate that prevails today for a given maturity
Short rate – the rate for a given maturity (e.g. one year) at different points in time.
A spot rate is the geometric average of its component short rates.
Short Rates and Yield Curve Slope
When next year’s short rate, r2 , is greater than this year’s short rate, r1, the yield curve slopes up.
May indicate rates are expected to rise.
When next year’s short rate, r2 , is less than this year’s short rate, r1, the yield curve slopes down.
May indicate rates are expected to fall.
Forward Rates from Observed Rates
fn = one-year forward rate for period n
yn = yield for a security with a maturity of n
Example 15.4 Forward Rates
The forward interest rate is a forecast of a future short rate.
Rate for 4-year maturity = 8%, rate for 3-year maturity = 7%.
Interest Rate Uncertainty
Suppose that today’s rate is 5% and the expected short rate for the following year is E(r2) = 6%. The value of a 2-year zero is:
The value of a 1-year zero is:
Interest Rate Uncertainty
• The investor wants to invest for 1 year.
• Buy the 2-year bond today and plan to sell it at the end of the first year for \$1000/1.06 =\$943.40.
• 0r-
• Buy the 1-year bond today and hold to maturity.
Interest Rate Uncertainty
• What if next year’s interest rate is more (or less) than 6%?
• The actual return on the 2-year bond is uncertain!
Interest Rate Uncertainty
Investors require a risk premium to hold a longer-term bond.
This liquidity premium compensates short-term investors for the uncertainty about future prices.
Expectations
Liquidity Preference
Upward bias over expectations
Theories of Term Structure
Expectations Theory
Observed long-term rate is a function of today’s short-term rate and expected future short-term rates.
fn = E(rn) and liquidity premiums are zero.
The excess of fn over E(rn) is the liquidity premium.
The yield curve has an upward bias built into the long-term rates because of the liquidity premium.
Interpreting the Term Structure
• The yield curve reflects expectations of future interest rates.
• The forecasts of future rates are clouded by other factors, such as liquidity premiums.
• An upward sloping curve could indicate:
• Rates are expected to rise
• And/or
• Investors require large liquidity premiums to hold long term bonds.
Interpreting the Term Structure
• The yield curve is a good predictor of the business cycle.
• Long term rates tend to rise in anticipation of economic expansion.
• Inverted yield curve may indicate that interest rates are expected to fall and signal a recession.
Forward Rates as Forward Contracts
• In general, forward rates will not equal the eventually realized short rate
• Still an important consideration when trying to make decisions :
• Locking in loan rates | 1,226 | 5,281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-05 | latest | en | 0.872819 |
https://www.doubtnut.com/qna/96605962 | 1,721,213,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00266.warc.gz | 645,317,296 | 34,996 | # In terms of the ratio r of the amplitudes of two coherent waves producing an interference pattern, the ratio of the intensity maximum to intensity minimum in the pattern is
A
r+1r1
B
r1r+1
C
(r+1r1)2
D
(r1r+1)2
Text Solution
Verified by Experts
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Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## If the ratio of the intensities of two waves producing interference is 9:4, then the ratio of the resultant maximum and minimum intensities will be
A3:2
B4:9
C25:1
D5:1
• Question 2 - Select One
## Light waves producing interference have their amplitudes in the ratio 3:2. The intensity ratio of maximum and minimum interference fringes is
A36:1
B9:4
C25:1
D6:4
• Question 3 - Select One
## If the ratio of the amplitudes of two interfering wave is 4:3, then the ratio of the maximumum and minimum intensities in the interference pattern is
A9:16
B16:9
C49:1
D1:49
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1. Help, urgently needed in physics
Can someone teach how to solve formula manipulation/transformation in Physics??
2. Originally Posted by divinewisdom0218
Can someone teach how to solve formula manipulation/transformation in Physics??
You are going to have to be more specific. What kind of problem(s) are you thinking about?
-Dan
3. E.G. S=Vot + 1/2gt^2 = Vo=S - 1/2gt^2 / T
Problem no.1: from the equation, A=at+v, determine the formula for a
Problem no.2: S=Vo t+ 1/2 at^, for Vo,a
PROblem no.3: v= Vot + at^2, for Vo,a
4. Originally Posted by divinewisdom0218
Problem no.1: from the equation, A=at+v, determine the formula for a
You solve these the same way you would solve them if the A, t, and v are just numbers (which in reality they are.)
Step by step:
$\displaystyle A = at + v$
$\displaystyle A - v = at + v - v$
$\displaystyle A - v = at$
$\displaystyle \frac{A - v}{t} = \frac{at}{t}$
$\displaystyle a = \frac{A - v}{t}$
Originally Posted by divinewisdom0218
Problem no.2: S=Vo t+ 1/2 at^2, for Vo,a
For $\displaystyle V_0$:
$\displaystyle S = V_0t + \frac{1}{2}at^2$
$\displaystyle S - \frac{1}{2}at^2 = V_0t + \frac{1}{2}at^2 - \frac{1}{2}at^2$
$\displaystyle V_0t = S - \frac{1}{2}at^2$
$\displaystyle \frac{V_0t}{t} = \frac{S - \frac{1}{2}at^2}{t}$
$\displaystyle V_0 = \frac{S}{t} - \frac{1}{2}at$
You solve for a.
-Dan | 494 | 1,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.792229 |
https://www.studypool.com/discuss/13801692/end-of-chapter-questions-1 | 1,680,186,521,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949331.26/warc/CC-MAIN-20230330132508-20230330162508-00759.warc.gz | 1,116,100,969 | 80,628 | # Grand Canyon Chapter 6 & 7 Correlation Coefficient Statistics Questions
User Generated
FlzcngulCflpurq
Mathematics
Grand Canyon University
## Description
Complete the following exercises located at the end of each chapter and put them into a Word document to be submitted as directed by the instructor.
Show all relevant work; use the equation editor in Microsoft Word when necessary.
1. Chapter 6, numbers 6.7, 6.10, and 6.11
2. Chapter 7, numbers 7.8, 7.10, and 7.13
### Unformatted Attachment Preview
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Hello,Attached find the completed work together with the plagiarism report. Go through it and in case you need any corrections kindly let me know.All the best.
Running head: END OF CCHAPTER QUESTIONS
End of Chapter questions
Name
Institutional Affiliation
Date Submitted
1
2
END OF CHAPTER QUESTIONS
Question 6.7
Solution:
-
An almost linear trend can be observed from the scatterplot above.
The correlation coefficient
x
y
xy
x2
y2
64
66
4224
4096
4356
3
END OF CHAPTER QUESTIONS
Total
-
40
79
3160
1600
6241
30
98
2940
900
9604
71
65
4615
5041
4225
55
76
4180
3025
5776
31
83
2573
961
6889
61
68
4148
3721
4624
42
80
3360
1764
6400
57
72
4104
3249
5184
451
687
33304
24357
53299
The correlation coefficient is given by:
𝑛(∑ 𝑥𝑦)−(∑ 𝑥)(∑ 𝑦)
r=
√𝑛{(∑ 𝑥 2 )−(𝑥)2 }{√𝑛(∑ 𝑦 2 )−(∑ 𝑦)2
rx y...
### Review
Anonymous
Really great stuff, couldn't ask for more.
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Originally Posted by Bjoern
What has "voltage" to do with a current loop?
Originally Posted by ngeo
Well, in order to maintain a current you would need a potential.
Originally Posted by Bjoern
Wrong. For current loops, no potential is needed, and can't even be defined in a sensible way. Please read up on the basics of electrodynamics.
Perhaps you can explain your understanding of a current loop, and if you are interested in addressing "the rest", I will attempt to explain it. Otherwise I will take it your interest is merely technical.
[Moderator Note]
I have edited this post to reflect what I think is the original sequence of posts.
[/Moderator Note]
Last edited by Nereid; 2007-Jan-24 at 04:07 PM. Reason: attempt to put quoted text into quote boxes
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Could you have a go at editing this please ngeo? I can't quite work out who said what, and in what order, nor what the point of this thread is.
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Nereid I will have to figure out how quotes end up in the blue boxes, I haven't done it before. The thread is a continuation from the Q & A section, and the point (for me) is to find out whether the relation I outlined in that thread (I don't believe it is so ATM) is "numerology" as Bjoern has called it, or physical. I believe it is a physical relationship regardless of my "skill level". Sorry I will have to leave it there for a few hours.
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5. Originally Posted by ngeo
Perhaps you can explain your understanding of a current loop...
A current loop means that charges are flowing in a closed path; usually one talks about circular loops, but the following argument applies also to other forms.
For a circular current loop, you need a circular electric field. But for a circular electric field, the rotation (I mean the mathematical "rot" here - look it up if you don't know what that means) does not vanish. And it is a well-known result from vector calculus that for fields with non-vanishing rotation, no (well-defined) potential exists.
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I guess by “current loop” I am using terminology that already exists and which does not appear to apply here. In the system I have been describing, the flow of current is a flow of space, and this flow creates the charge. So it is not a system of flowing charges. A tension is created by the opposition of internal and external fields which is released by rotation (flow). So the particle is a surface or shell between the two fields.
7. the flow of current is a flow of space
so current is space ?
Guess I will have to read the other thread, because these messages together make as much sense as "fish with a bicycle".
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[Moderator Note]
ngeo's questions on space etc is the (at present) Q&A thread which lead to this ATM thread.
The relevant exchange/post immediately preceeding the OP (of this thread) is this one. Here is my reconstruction of it:
Originally Posted by ngeo
My last response to Bjoern here:
Bjoern: “What has "voltage" to do with a current loop?
Well, in order to maintain a current you would need a potential.
(Re potential is in internal and external fields)
Bjoern: “And what is that supposed to mean?”
“What does it mean to say that a current flow is "
from fields"?
“What does it mean to say that a current flow "
maintains a loop"?”
I don’t believe that you really do not understand. However, that is possible. I have been thinking on these lines for a long time, and I guess you haven’t.
Bjoern: “Well, then what does it mean to say that "a potential produces a frequency"?”
See “Josephson constant”, “Josephson effect“, etc.
(Re potential and frequency being same thing)
Bjoern: “Then why are they measured in different units?”
That is the problem of trying to understand a system. You have to break it down into parts. But you have to keep in mind that it works as a whole.
Bjoern: “But ignoring the units is an absolutely basic mathematical error.”
I don’t ignore them, I try to figure out what they mean. Now once you end up with a relationship that can be expressed in terms of alpha and electron-proton ratio, giving numbers that match the “unit” numbers, exactly what units would you want?
Bjoern: “h is probably Planck's constant, but what is q? And why on earth should J/C be the same as h/q in this case??? J (Joule) is the unit of energy, C (Coulomb) the unit of charge. So the denominator would match (charge). But the numerator does not match - h is not an energy! h has the unit Joule-seconds, not simply Joule!”
See below.
Bjoern: “Charge is current times time, not current per time! Come on, this is yet again absolutely basic!”
I did not say charge is measured in current per time, I said it is measured in current per second.
From Wikipedia:
1 coulomb is the amount of electric charge carried by a current of 1 ampere flowing for 1 second.
The ampere, in practice often shortened to amp, (symbol: A) is a unit of electric current, or amount of electric charge per second.
---------------
Bjoern: ““But h has units of Joules times seconds. Not Joules per second.”
Again from Wikipedia:
Planck's constant is stated in SI units of measurement, joules per hertz, or joules per (cycle per second) . . . .
In SI units Planck's constant is expressed in Joule-seconds.
It seems the wikipedia author is not that worried about how h is expressed or stated, and neither am I, because I am not thinking about how to express or state h, I am thinking about what it represents. Sometimes it’s hard to see the forest for the trees.
prep.
To, for, or by each; for every: Gasoline once cost 40 cents per gallon.
According to; by: Changes were made to the manuscript per the author's instructions.
By means of; through.
For each one; apiece: sold the cookies for one dollar per.
Per hour: was driving at 60 miles per.
ngeo, re editing, colours, quotes etc: if you know '(HTML) tags', it's easy - [ color ] and [ quote ] (without the spaces, and with the matching [/ ] tag).
If not, it's still easy: enclose the text you wish to be coloured with a pair of tags ([ color = blue ] and [/ color ], for example, without the spaces: [ color=blue ]text you wish to be coloured[ /color ] -> text you wish to be coloured).
Similarly, to make the enclosed text a quote, use [ quote ] and [/ quote ]; to add the attribution, add "={name}" after "quote"; for example: [ quote = Nereid ]To make the enclosed text a quote[ /quote ] ->
Originally Posted by Nereid
To make the enclosed text a quote
[/Moderator Note]
9. Originally Posted by ngeo
I guess by “current loop” I am using terminology that already exists and which does not appear to apply here.
Indeed. And judging from what you write below, you do that also for "field" and several other terms.
Originally Posted by ngeo
In the system I have been describing, the flow of current is a flow of space, and this flow creates the charge. So it is not a system of flowing charges. A tension is created by the opposition of internal and external fields which is released by rotation (flow). So the particle is a surface or shell between the two fields.
So few sentences, so much word salad...
What does "a flow of space" mean? How could such a flow create charge? (what about charge conservation?) What "internal and external fields" do you mean? What does it mean to say that there is opposition between fields? How does this opposition create tension? What is tensed here? What is "released by rotation" - the opposition??? What does "rotation (flow)" mean - that "rotation" and "flow" mean the same thing here? What does it mean to say that something is "between two fields"? How does it follow from the things you said in the previous sentences that a particle is a surface or shell?
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Originally Posted by Bjoern
Indeed. And judging from what you write below, you do that also for "field" and several other terms.
So few sentences, so much word salad...
What does "a flow of space" mean? How could such a flow create charge? (what about charge conservation?) What "internal and external fields" do you mean? What does it mean to say that there is opposition between fields? How does this opposition create tension? What is tensed here? What is "released by rotation" - the opposition??? What does "rotation (flow)" mean - that "rotation" and "flow" mean the same thing here? What does it mean to say that something is "between two fields"? How does it follow from the things you said in the previous sentences that a particle is a surface or shell?
In this scenario, a particle represents an internal field attempting to expand into an external field, and an external field attempting to expand into an internal field. This occurs because the fundamental ability of the universe is to expand, which it does at a constant rate (the limit of its ability) at its edge. However within the universe this ability is limited by its own existence everywhere within the universe. This limitation creates pressure. In response to this pressure, certain regions begin to spin. Inside these spinning regions is a small area seeking to expand. Outside these spinning regions is a larger area seeking to expand into these spinning regions. The energy of the two fields, internal and external, is absorbed in the constant acceleration represented by rotation of the surface that separates them. This is not a smooth surface but a vibrating surface. Each field is alternately compressed. It may be that therein lies the “tension” (a word I first hit upon yesterday as a salad ingredient).
That is the basic idea. So in this scenario there is no “nugget” of “matter”, no “kilogram” of “mass”. E=hf turns out to be a handy way to describe the electron and proton in this scenario. It gives a frequency for each particle, h as the work done each cycle, a current per cycle for each particle which must end up as q (elementary charge) in one second, and as a result a voltage or potential. But this potential is not the potential of the particle; it is the potential of the field to produce the rotation, “embodied” in the particle.
Now, digging around for patterns in the set of equations i.e. v = hf/q, vq = hf, etc. I find that the frequency of the electron divided by the square of the voltage of the proton = approximately 140. In addition the frequency of either particle is approximately ½ the product of the square of the voltage of that particle and the voltage of the other particle. When 140 is replaced by the inverse of alpha the fine structure constant (a), then the frequencies are exactly equal to twice the product of the square etc. The voltages and frequencies of each particle are slightly altered also. And a ratio (r) of approximately 1860.308707 appears between proton and electron voltages/frequencies. It is then possible to derive the (new) voltage of the electron as 2 x r / a and of the proton as 2 x r^2 / a, and also the frequencies, using r and a.
It also turns out that the product of electron and proton voltages in this method is exactly equal to the Josephson constant, 2 x q / h, the inverse of the magnetic flux quantum. And this allows the possibility of a system of superconducting current loops between which a quantized magnetic current flows.
The electron (potential in volts = 2r/a) emits half a magnetic flux quantum a^2/2r^3, at a frequency 4r^4/a^3 producing a potential in the field of 2r/a. The proton (potential = 2r^2/a) emits half a magnetic flux quantum a^2/2r^3 at a frequency of 4r^5/a^3, producing a potential in the field of 2r^2/a. The total potential of the field inside the system is then 4r^3/a^2 which is the Josephson constant. Half this potential, 2r^3/a^2, is transferred to each particle and combined with the potential of the particle produces the frequency in the proton of 4r^5/a^3 and in the electron of 4r^4/a^3. The potential of each particle itself, in the absence of this system, is maintained by the external field and the internal field of each particle.
It may be seen that the frequency of each particle is the potential of the particle multiplied by half the Josephson constant. The Josephson constant itself is the product of the potentials of each particle.
11. Originally Posted by ngeo
In this scenario, a particle represents an internal field attempting to expand into an external field, and an external field attempting to expand into an internal field.
Please explain what "field" means to you. As I already pointed out, you seem to use that term with a non-standard meaning, and as long I don't understand what you mean by that, I can't follow the rest of what you write.
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Originally Posted by Bjoern
Please explain what "field" means to you. As I already pointed out, you seem to use that term with a non-standard meaning, and as long I don't understand what you mean by that, I can't follow the rest of what you write.
I am not sure, but I think an internal field is a subfield with differencial movement of a larger field (external field) of which it is embedded. The problem I have is that the "pressure" differencials of the two field would not allow a mutual push hydrodynamically. How do you remove and ignore the nature of pressure dynamics and attraction and give preferrence to electrodynamics when the idea of intenal/external fields is based on pressure differencial created by the movement of space?
This is still the question of the day and you haven't answered it yet.
13. ## Not coincidental ratios for mass?
Originally Posted by ngeo
In this scenario, a particle represents an internal field attempting to expand into an external field, and an external field attempting to expand into an internal field. This occurs because the fundamental ability of the universe is to expand, which it does at a constant rate (the limit of its ability) at its edge. However within the universe this ability is limited by its own existence everywhere within the universe. This limitation creates pressure. In response to this pressure, certain regions begin to spin. Inside these spinning regions is a small area seeking to expand. Outside these spinning regions is a larger area seeking to expand into these spinning regions. The energy of the two fields, internal and external, is absorbed in the constant acceleration represented by rotation of the surface that separates them. This is not a smooth surface but a vibrating surface. Each field is alternately compressed. It may be that therein lies the “tension” (a word I first hit upon yesterday as a salad ingredient).
That is the basic idea. So in this scenario there is no “nugget” of “matter”, no “kilogram” of “mass”. E=hf turns out to be a handy way to describe the electron and proton in this scenario. It gives a frequency for each particle, h as the work done each cycle, a current per cycle for each particle which must end up as q (elementary charge) in one second, and as a result a voltage or potential. But this potential is not the potential of the particle; it is the potential of the field to produce the rotation, “embodied” in the particle.
Now, digging around for patterns in the set of equations i.e. v = hf/q, vq = hf, etc. I find that the frequency of the electron divided by the square of the voltage of the proton = approximately 140. In addition the frequency of either particle is approximately ½ the product of the square of the voltage of that particle and the voltage of the other particle. When 140 is replaced by the inverse of alpha the fine structure constant (a), then the frequencies are exactly equal to twice the product of the square etc. The voltages and frequencies of each particle are slightly altered also. And a ratio (r) of approximately 1860.308707 appears between proton and electron voltages/frequencies.. It is then possible to derive the (new) voltage of the electron as 2 x r / a and of the proton as 2 x r^2 / a, and also the frequencies, using r and a.
It also turns out that the product of electron and proton voltages in this method is exactly equal to the Josephson constant, 2 x q / h, the inverse of the magnetic flux quantum. And this allows the possibility of a system of superconducting current loops between which a quantized magnetic current flows.
The electron (potential in volts = 2r/a) emits half a magnetic flux quantum a^2/2r^3, at a frequency 4r^4/a^3 producing a potential in the field of 2r/a. The proton (potential = 2r^2/a) emits half a magnetic flux quantum a^2/2r^3 at a frequency of 4r^5/a^3, producing a potential in the field of 2r^2/a. The total potential of the field inside the system is then 4r^3/a^2 which is the Josephson constant. Half this potential, 2r^3/a^2, is transferred to each particle and combined with the potential of the particle produces the frequency in the proton of 4r^5/a^3 and in the electron of 4r^4/a^3. The potential of each particle itself, in the absence of this system, is maintained by the external field and the internal field of each particle.
It may be seen that the frequency of each particle is the potential of the particle multiplied by half the Josephson constant. The Josephson constant itself is the product of the potentials of each particle.
(bold mine)
Ngeo, did you just define the 'fine structure constant', viz. 1/137? I thought it had remained a mystery.
Also, isn't the ratio of proton mass (1.672E-27 kg) to electron mass (9.109E-31 kg) or about 1835? Are these alternative 'electron volt' definitions of mass kg?
I can't imagine these ratio constants are merely coincidental. Can mass, kilogram, be redefined as a function of electromagnetic energy about a nuclear point, electron-volts, to give us ordinary 'atomic matter' with mass? Interesting idea, if I understand correctly what you are saying. How would that fit in with ZPE, for example, as the energy locked inside ordinary space?
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Originally Posted by Bjoern
Please explain what "field" means to you. As I already pointed out, you seem to use that term with a non-standard meaning, and as long I don't understand what you mean by that, I can't follow the rest of what you write.
I am not a scientist or mathematician so I can only quote from Wikipedia:
“In physics, a field is an assignment of a physical quantity to every point in space (or, more generally, spacetime). A field is thus viewed as extending throughout a large region of space so that its influence is all-pervading. The strength of a field usually varies over a region.”
In this idea, the universe would be a single field. I don’t know what physical quantity would be assigned to every point in the field, but it would have to represent an initial spherical expansion “energy”, and this energy would have to be in opposition to the energy of every other point in the field. However when a direction of flow would be established, namely by rotation, the spherical expansion energy would become a directional movement.
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Originally Posted by bigsplit
I am not sure, but I think an internal field is a subfield with differencial movement of a larger field (external field) of which it is embedded. The problem I have is that the "pressure" differencials of the two field would not allow a mutual push hydrodynamically. How do you remove and ignore the nature of pressure dynamics and attraction and give preferrence to electrodynamics when the idea of intenal/external fields is based on pressure differencial created by the movement of space?
This is still the question of the day and you haven't answered it yet.
I am sorry I don't understand fully, I will study what you say, but as far as I understand what you are saying, in this idea there would not be a pressure differential once a complete rotation occurs. The internal field pressure would balance the external field pressure. Added external field "pressure" would produce a higher frequency rotation (a smaller radius), etc., so the rotation is determined by the energy in the field.
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Originally Posted by nutant gene 71
(bold mine)
Ngeo, did you just define the 'fine structure constant', viz. 1/137? I thought it had remained a mystery.
Also, isn't the ratio of proton mass (1.672E-27 kg) to electron mass (9.109E-31 kg) or about 1835? Are these alternative 'electron volt' definitions of mass kg?
I can't imagine these ratio constants are merely coincidental. Can mass, kilogram, be redefined as a function of electromagnetic energy about a nuclear point, electron-volts, to give us ordinary 'atomic matter' with mass? Interesting idea, if I understand correctly what you are saying. How would that fit in with ZPE, for example, as the energy locked inside ordinary space?
I don't think I defined anything except by accident. I just inserted alpha where the number was close to the value of alpha. I think there are a lot of ways alpha is used and there is equation I think. It's still a mystery to me anyway!
I also don't think the ratios are merely coincidental, they are a clue to a physical relationship.
There has been a move toward redefining the kilogram to take it away from an "artifact" (which has lost mass in the last 100 years) in Paris. Wikipedia and other places have article on it and you probably will understand them more than me, I found one quote which seems to have disappeared and replaced by a new set of proposals under "kilogram" in Wikipedia.
I think somebody with way more knowledge than me would have to figure out the deal with ZPE, but I think this idea does need a different way of thinking than the "vacuum".
17. Originally Posted by ngeo
I am not a scientist or mathematician so I can only quote from Wikipedia:
“In physics, a field is an assignment of a physical quantity to every point in space (or, more generally, spacetime). A field is thus viewed as extending throughout a large region of space so that its influence is all-pervading. The strength of a field usually varies over a region.”
Please not that according to this definition, a field gives a value to every point in space. In other words, a fields extends over the whole space. Hence it makes to sense to speak of an "internal" and an "external" field!
Originally Posted by ngeo
In this idea, the universe would be a single field. I don’t know what physical quantity would be assigned to every point in the field,
"every point in the field" makes no sense. Please read again what you quoted above, and try to understand it this time better, please.
And if you don't know what physical quantity would be assigned to every point, then it makes no sense at all even to talk about a field, since, according to what you yourself quoted above, a field is the assigment of a physical quantity (to every point of space)! If you don't know what is assigned to every point in space, you have no field!
Originally Posted by ngeo
but it would have to represent an initial spherical expansion “energy”,
Fields can't in any way "represent" energy. They can contain energy, or you can have an "energy field" (i. e. to every point in space, a value for its energy is assigned), but this is not "representing" energy.
But you put "energy" in quote marks. May I assume therefore that you don't mean the usual "energy" of physics, but something different? What does "expansion energy" mean? And to what does the word "spherical" refer here? To "expansion", or to "energy"?
Originally Posted by ngeo
and this energy would have to be in opposition to the energy of every other point in the field.
What on earth does it mean to say that energy is in opposition to another energy???
Originally Posted by ngeo
However when a direction of flow would be established, namely by rotation, the spherical expansion energy would become a directional movement.
Word salad. Sorry, I don't understand at all what you try to tell me. And the reason for that is very simple: you use words which have well-defined meanings in physics with your own, private meanings, which I don't know.
Perhaps you should write a thread which consists only of defining all these terms which you use with non-standard meanings, before going on? Or, even better: perhaps you should learn what all these words actually mean in physics and then use them properly? Otherwise, a discussion with you is pointless - we are simply talking past each other.
18. Originally Posted by ngeo
I also don't think the ratios are merely coincidental, they are a clue to a physical relationship.
The temperature of the CMBR is 2.73 K. A (tropical) month has a length of 27.3 days. Do you also suspect that the agreement between these two numbers is not coincidental, but a clue to a physical relationship?
19. Originally Posted by Bjoern
...
Word salad. Sorry, I don't understand at all what you try to tell me. And the reason for that is very simple: you use words which have well-defined meanings in physics with your own, private meanings, which I don't know.
Perhaps you should write a thread which consists only of defining all these terms which you use with non-standard meanings, before going on? Or, even better: perhaps you should learn what all these words actually mean in physics and then use them properly? Otherwise, a discussion with you is pointless - we are simply talking past each other.
Using the word salad argument is not productive. I seems to mean you are rejecting the ideas because they sound nice but until they are mainstream they aren't worth discussion. This is an ATM thread.
Using the "define all terms" argument is a strawman; saying that you can't really make a logical word argument that falsifies the posters position, but you can refuse to try to understand the word usage.
20. Originally Posted by Bogie
Using the word salad argument is not productive. I seems to mean you are rejecting the ideas because they sound nice but until they are mainstream they aren't worth discussion. This is an ATM thread.
No, it means simply, as I already wrote in other places: "I don't understand a word of what you mean; please clarify and tell me what on earth you mean when you say XYZ".
Originally Posted by Bogie
Using the "define all terms" argument is a strawman; saying that you can't really make a logical word argument that falsifies the posters position, but you can refuse to try to understand the word usage.
No again. It's the most natural thing to ask for when I can't make out what my discussion partner tries to tell me, and when he clearly misuses terms.
I can't help you if don't understand (or don't care about?) a very basic principle of discussion: talk/write in a way so that your discussion partner will understand what you mean.
21. Originally Posted by Bjoern
No, it means simply, as I already wrote in other places: "I don't understand a word of what you mean; please clarify and tell me what on earth you mean when you say XYZ".
No again. It's the most natural thing to ask for when I can't make out what my discussion partner tries to tell me, and when he clearly misuses terms.
I can't help you if don't understand (or don't care about?) a very basic principle of discussion: talk/write in a way so that your discussion partner will understand what you mean.
I'll grant you the fact that communication has to be the responsibility of both parties. However, when one uses the "define your term" approach that person should offer a definition of the word that they would understand.
In other words, say what you understand the word to mean, and then using that meaning show how it doesn't make sense to you in the way the poster used it.
That approach may take more time than anyone really has to devote to the discussion of course, and if that is the case, the termination of the discussion could be based on specific examples where the meanings that you attribute to particular words are offered so that the poster can at least learn from the exchange.
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Originally Posted by Bogie
I'll grant you the fact that communication has to be the responsibility of both parties. However, when one uses the "define your term" approach that person should offer a definition of the word that they would understand.
In other words, say what you understand the word to mean, and then using that meaning show how it doesn't make sense to you in the way the poster used it.
That approach may take more time than anyone really has to devote to the discussion of course, and if that is the case, the termination of the discussion could be based on specific examples where the meanings that you attribute to particular words are offered so that the poster can at least learn from the exchange.
Which may (or may not) be reasonable if BAUT's ATM section were explicitly devoted to assisting those with ATM ideas - giving them the bare-bones basics of the relevant aspects of the relevant modern physics theories, or the essentials of the math needed to address some foundation, or suggesting ways to fix up a glaring hole, or ...*
It may (or may not) also be reasonable if BAUT's ATM section were explicitly devoted to discussion of ATM ideas.*
However, IMHO, it most clearly is neither of these things - it is, IMHO, an opportunity for those with ATM ideas to present them, and expect they (the ideas) will be challenged and attacked (with glee and fervour), in order that the presenter may improve their work, that readers may see how weak the ATM idea is, that ...
So, if a BAUT member presents an idea full of words which appear to have idiosyncratic meanings (or simply meanings that are non-standard wrt the accepted usage in the field(s) of astronomy/physics/etc relevant to the scope of the ATM idea), the burden of clarifying those meanings falls on the one making the ATM claims.
Further, in my own experience, in a great many ATM threads, too many pages are wasted because demands for clarity, wrt the ATM claims being made, were not made firmly enough right at the start of the threads.
Anyway, those are just some views of Nereid, the ordinary BAUT member (not Nereid, the BAUT moderator).
*Of course, any BAUT member may choose to do just this - provide assistance, support, guidance, to the BAUT member proposing the ATM idea; or discuss the ATM idea without challenging or attacking it, or ... and, in the life of many an ATM thread, just such assistance, support, guidance, discussion, ... has been offered - read some of Grey's posts, to take just one example.
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Originally Posted by Bjoern
The temperature of the CMBR is 2.73 K. A (tropical) month has a length of 27.3 days. Do you also suspect that the agreement between these two numbers is not coincidental, but a clue to a physical relationship?
Why not? Very little in the ATM threads (and particularly this one) makes even this much sense.
Seriously, the general characteristic of the ATM threads seems to be wild flights of fancy based on fundamentally incorrect assumptions, a la von Daniken. There is a dandy thread elsewhere where the originator feels there are point mass(es) inside the electron. There is zero evidence to support this; BAUT member trinitree88 shoots this idea down without mercy. Yet the thread goes on for many pages following. I guess that’s why it’s called ‘Against the Mainstream’.
Generally, ATM posters, it would be nice if you had the math and physics straight first. There are good self-teaching courses available free on the net. Wikipedia has to be the greatest reference source ever, try checking there. I think Bjoern is right; you can’t make up the math and physics as you go along. Do like Newton and stand on the shoulders of giants; you can see a lot further.
24. Originally Posted by Bogie
I'll grant you the fact that communication has to be the responsibility of both parties. However, when one uses the "define your term" approach that person should offer a definition of the word that they would understand.
In other words, say what you understand the word to mean, and then using that meaning show how it doesn't make sense to you in the way the poster used it.
That approach may take more time than anyone really has to devote to the discussion of course, and if that is the case, the termination of the discussion could be based on specific examples where the meanings that you attribute to particular words are offered so that the poster can at least learn from the exchange.
In general, you are right. But here, ngeo himself provided already the standard definition from Wikipedia, and most what I wrote consisted in pointing out that his use of the word "field" does not match what he himself quoted. I then pointed out his misuse of "energy". That's such a totally basic term in physics that I simply think anyone who doesn't know its meaning should look it up for himself!!! (otherwise it's just lazyness) And the last sentence was simply totally incomprehensible to me - word salad, as I said.
25. Originally Posted by Nereid
Originally Posted by Nereid
However, IMHO, it most clearly is neither of these things - it is, IMHO, an opportunity for those with ATM ideas to present them, and expect they (the ideas) will be challenged and attacked (with glee and fervour), in order that the presenter may improve their work, that readers may see how weak the ATM idea is, that ...
I agree. An opportunity for real science should supersede the airing of pseudoscience
So, if a BAUT member presents an idea full of words which appear to have idiosyncratic meanings (or simply meanings that are non-standard wrt the accepted usage in the field(s) of astronomy/physics/etc relevant to the scope of the ATM idea), the burden of clarifying those meanings falls on the one making the ATM claims.
I agree. It really is hard to be the one who can’t put the right words together to convey an idea, and not know they can’t do that. It reminds me of Clinton saying, “It depends on what you mean by “is”. Thinking that words in English generally are commonly understood, it is a shock to be confronted with the request for simple meanings. Are you sure it is a failure on the part of the originator to use words properly, or can it be a failure on the part of the responder to apply the “commonly understood” meaning to words and then address the idea behind the supposed improper word usage instead of the improper usage itself?
Further, in my own experience, in a great many ATM threads, too many pages are wasted because demands for clarity, wrt the ATM claims being made, were not made firmly enough right at the start of the threads.
Anyway, those are just some views of Nereid, the ordinary BAUT member (not Nereid, the BAUT moderator).
Ordinarily when an OP hits the ATM board, if I am interested in it I will read it and think about the idea being conveyed (or attempting to be conveyed). I can usually determine in the first post if this is likely to be meaningful in either advancing science or at least will result in a discussion that I can learn from.
I think that other readers will also have criteria by which they judge the OP that will determine if they are interested enough to participate or that will determine if the thread is worth following even if they can’t make a meaningful contribution.
To then see a thread with an idea fall victim to the dreaded “word usage” argument instead of calm logical rejection of the idea just makes the ATM forum seem to fall short of the high standards that BAUT clearly has in the Internet community.
*Of course, any BAUT member may choose to do just this - provide assistance, support, guidance, to the BAUT member proposing the ATM idea; or discuss the ATM idea without challenging or attacking it, or ... and, in the life of many an ATM thread, just such assistance, support, guidance, discussion, ... has been offered - read some of Grey's posts, to take just one example.
It is true that there is assistance, support and guidance offered by many members along the way. Too often though, IMHO, the real sources of informed discussion on many of the threads are too quick on the draw, thus eliminating ATM as being a place for beginners to come, learn, and maybe even present an idea of their own to test the waters.
It can be both a venue for real science and a place for science minded to come and learn. Is it better to be the former and not the latter, or is it better to be both?
26. Originally Posted by Bjoern
In general, you are right. But here, ngeo himself provided already the standard definition from Wikipedia, and most what I wrote consisted in pointing out that his use of the word "field" does not match what he himself quoted. I then pointed out his misuse of "energy". That's such a totally basic term in physics that I simply think anyone who doesn't know its meaning should look it up for himself!!! (otherwise it's just lazyness) And the last sentence was simply totally incomprehensible to me - word salad, as I said.
Well said, thanks.
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## Second the Motion
Originally Posted by Bogie
Well said, thanks.
Well said, indeed!
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Originally Posted by Bjoern
Please not that according to this definition, a field gives a value to every point in space. In other words, a fields extends over the whole space. Hence it makes to sense to speak of an "internal" and an "external" field!
"every point in the field" makes no sense. Please read again what you quoted above, and try to understand it this time better, please.
And if you don't know what physical quantity would be assigned to every point, then it makes no sense at all even to talk about a field, since, according to what you yourself quoted above, a field is the assigment of a physical quantity (to every point of space)! If you don't know what is assigned to every point in space, you have no field!
Fields can't in any way "represent" energy. They can contain energy, or you can have an "energy field" (i. e. to every point in space, a value for its energy is assigned), but this is not "representing" energy.
But you put "energy" in quote marks. May I assume therefore that you don't mean the usual "energy" of physics, but something different? What does "expansion energy" mean? And to what does the word "spherical" refer here? To "expansion", or to "energy"?
What on earth does it mean to say that energy is in opposition to another energy???
Word salad. Sorry, I don't understand at all what you try to tell me. And the reason for that is very simple: you use words which have well-defined meanings in physics with your own, private meanings, which I don't know.
Perhaps you should write a thread which consists only of defining all these terms which you use with non-standard meanings, before going on? Or, even better: perhaps you should learn what all these words actually mean in physics and then use them properly? Otherwise, a discussion with you is pointless - we are simply talking past each other.
Perhaps when you comment, you should comment on the entire statement. Breaking a statement into parts may work for a mathematician, but used on words it can reduce sense to nonsense. Which it seems you do, since you seem unable to make sense - say, in the form of a mental picture - out of the words. I guess chopping up statements and sentences does make a word salad.
Again from Wikipedia, the full statement:
“In physics, a field is an assignment of a physical quantity to every point in space (or, more generally, spacetime). A field is thus viewed as extending throughout a large region of space so that its influence is all-pervading. The strength of a field usually varies over a region.”
When I say that the universe is a single field, I am saying that the region of space which is influenced by whatever form of energy is represented by a physical quantity, is the entire universe, as opposed to “a large region of space”. The form of energy is the form that produces an expansion of space. The energy that produces the expansion of space also expands within the already existing space, which then produces pressure. I don’t believe that form of energy is known to physics, which is why I put it in quotes. I do not mean Dark Energy or cosmological constant.
I will try to draw a word picture, which probably will not be satisfying in the absence of a mathematical formulation. Imagine a skinless balloon being blown up so that its radius expands at a constant rate. The interior of this skinless balloon is filled with smaller skinless balloons each expanding at a constant rate. The balloons cannot all expand into each other. So they become distorted. This distortion takes the form of rotation, or of a rotating surface. I believe rotation is acceleration. This acceleration absorbs the energy of the field so that in the region of the rotating surface the field does not expand. It also separates one part of the field from the remainder: the part of the field inside the rotating surface is separated by the surface from the part of the field outside the surface. Maybe this is not known to physics either.
The rotating surface is what I earlier called a current loop (possibly an electron) or rotating charge (possibly a proton). This is where the idea may come into contact with known physics.
If you can’t make sense of that, I can’t help you. If you want to chop it up into word salad, I can’t stop you.
As far as introducing the CMBR and a tropical month into this discussion, why on earth would someone bring an irrelevant argument to argue against something which he apparently has difficulty understanding to begin with? Usually when someone changes the subject like this, it indicates he or she does not want to deal with the subject at hand, which is the exact equality of the product of hypothetical electron and proton voltage with the Josephson constant, a known physical constant.
29. Originally Posted by ngeo
Perhaps when you comment, you should comment on the entire statement. Breaking a statement into parts may work for a mathematician, but used on words it can reduce sense to nonsense. Which it seems you do, since you seem unable to make sense - say, in the form of a mental picture - out of the words. I guess chopping up statements and sentences does make a word salad.
I did break up the statements into smaller parts because taken at a whole, they were merely word salad to me. Breaking them up into smaller parts gave me the opportunity to point out where the problems with your usage of terms lie - which are very probably at the root of the fact that your statements look like word salad to me!
May I remind you that a few posts ago, I didn't break up your statement into smaller parts - and that had then the consequence that I had to follow your single statement with about 10 questions about the meanings of terms etc.?
Originally Posted by ngeo
Again from Wikipedia, the full statement:
“In physics, a field is an assignment of a physical quantity to every point in space (or, more generally, spacetime). A field is thus viewed as extending throughout a large region of space so that its influence is all-pervading. The strength of a field usually varies over a region.”
When I say that the universe is a single field, I am saying that the region of space which is influenced by whatever form of energy is represented by a physical quantity, is the entire universe, as opposed to “a large region of space”. The form of energy is the form that produces an expansion of space. The energy that produces the expansion of space also expands within the already existing space, which then produces pressure. I don’t believe that form of energy is known to physics, which is why I put it in quotes. I do not mean Dark Energy or cosmological constant.
So, let's address this statement in full, once again (or do you wish that I address your whole post in full, instead of breaking it up into paragraphs?).
Saying that "the universe is a single field" has, if one takes the Wikipedia quote you provide into account, nothing to do with what you write in the rest of the paragraph (for starters, you still do not tell us what physical quantity is assigned to every point in space). What does it mean to say that a region of space (or the entire universe, take your pick) is "influenced" by energy? What does it mean to say that a "form of energy is represented by a physical quantity"? Why do you think that energy is needed in order to "produce an expansion of space"? What does it mean to say that energy "expands"? (that is the statement which I could make the most sense of among all the above - but let's see if you meant the same as I understood). Why should the "expanding" of energy produce pressure? Why is the assumption that this form of energy is not known to physics a reason to put "energy" in quote marks?
There you have it. The result of not splitting up your statement into smaller parts is simply a whole paragraph consisting only of questions... what's the big difference to breaking it up? Simple: breaking it up makes the discussion much more clean and neat.
Originally Posted by ngeo
I will try to draw a word picture, which probably will not be satisfying in the absence of a mathematical formulation. Imagine a skinless balloon being blown up so that its radius expands at a constant rate. The interior of this skinless balloon is filled with smaller skinless balloons each expanding at a constant rate. The balloons cannot all expand into each other. So they become distorted. This distortion takes the form of rotation, or of a rotating surface. I believe rotation is acceleration. This acceleration absorbs the energy of the field so that in the region of the rotating surface the field does not expand. It also separates one part of the field from the remainder: the part of the field inside the rotating surface is separated by the surface from the part of the field outside the surface. Maybe this is not known to physics either.
I don't know what you mean with a "skinless" balloon. How could a distortion "take the form of rotation", or "of a rotating surface"? No, rotation is not acceleration, rotation merely involves a special kind of acceleration (centripetal acceleration). What does it mean to say that "acceleration absorbs energy"? What "field", once again? In your balloon analogy, what represents there the "field"? How can a rotation "absorbing" something lead to something not expanding? Have you ever heard of the conservations laws for momentum and angular momentum? How can two parts of a field be "separated" from each other, if we take the definition for "field" you yourself quoted?
What you write is indeed not "known" to physics. In fact, the very few parts which are understandable at all contradict heavily known physics - as I said, look up the conservations laws for momentum and angular momentum.
Originally Posted by ngeo
The rotating surface is what I earlier called a current loop (possibly an electron) or rotating charge (possibly a proton). This is where the idea may come into contact with known physics.
Why do you call a rotating surface a "loop"? This goes even against the common-sense meaning of the word "loop".
Originally Posted by ngeo
If you can’t make sense of that, I can’t help you. If you want to chop it up into word salad, I can’t stop you.
As I said above already: not chopping it up makes it word salad. And as I also pointed out already, you can help me: by either clearly defining what your terms mean, or even better, using the terms with their standard meanings. Since you apparently refuse to do either, I can only conclude that you are not really interested in other people understanding you.
Originally Posted by ngeo
As far as introducing the CMBR and a tropical month into this discussion, why on earth would someone bring an irrelevant argument to argue against something which he apparently has difficulty understanding to begin with?
It was an analogy for searching for meaning in a coincidental agreement between two numbers. Try to understand the point.
Originally Posted by ngeo
Usually when someone changes the subject like this, it indicates he or she does not want to deal with the subject at hand, which is the exact equality of the product of hypothetical electron and proton voltage with the Josephson constant, a known physical constant.
Err, but providing an analogy is not changing the subject. And one point of my analogy, which you obviously completely missed, is that totally ignoring the units of physical quantities is a bad idea (both for K and days on the one hand side, and for the product of "electron and proton voltage" and Hz/V, which is the unit of the Josephson constant).
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Re the post above, I have done my best to explain in simple language, I don’t know whether I can make the “creation” part of this idea any clearer and it may involve rotation on the equivalent of three axes, which is a big headache even for me. I doubt I could answer any of the questions in a way that would satisfy you. So I am not going to try again.
Apart from this, the particles themselves are taken to be massless, so any physical description or interpretation which invokes bodies of matter (i.e. centripetal force) will not work here. The gravitational force (or spacetime geometry) is an effect of this hypothetical system.
The exact equivalence of the product of two hypothetical voltages with a known constant giving a frequency product of voltage according to 2q/h is in no way similar to the (non-)coincidence given in the analogy. Nor is it the invocation of some supernatural influence. The relationship between voltage and frequency is same as the relationship between q and h. In other words, in this hypothetical system, f/v=q/h. In this system which comes first, f or v? They seem to be in effect functions of each other. The fact that the product of the electron and proton voltages in this hypothetical system is exactly equal to a known constant does not mean that the electron and proton somehow combine their potentials produce this constant. It means, as far as I can figure out, that the constant is a common multiplier which produces the frequency of each particle. Given the ratio of the particle voltages and frequencies determined by inserting alpha into this set of equations, the constant can be described without units, simply in terms of alpha and “r”. It seems to me that the ability to describe a cause-and-effect relation in the electron-proton system in terms of alpha and a ratio derived from the particles themselves is a step forward to a fundamental explanation for these particles, especially since the known physical constant and its inverse, the magnetic flux quantum, are in the middle of this relationship and apply to superconducting currents.
I admit I don’t understand much of the formal “science” of this, maybe only that you can move alpha, “r“, f, v, q and h around in interesting ways. Putting units against the letters and numbers, you come up with “work done per cycle per unit current per second” and suchlike. That’s fine (and doesn‘t make a lot of sense to me in my ignorance), but these units point to some effect we can see in the end, however they are juggled. Maybe there is more than one way to describe the system that creates the effect, and the result I am giving may be a clue to a different - maybe even an easier - way of describing it. It is beyond me how a scientist with a straight face could call the relation a coincidence, or numerology, with the implication “not worth considering”.
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• | 11,514 | 52,184 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.960307 |
http://clay6.com/qa/36843/if-z-is-a-compressibility-factor-van-der-waals-equa-tion-at-low-pressure-ca | 1,527,313,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867311.83/warc/CC-MAIN-20180526053929-20180526073929-00543.warc.gz | 56,210,524 | 25,924 | # If Z is a compressibility factor, van der Waals equa- tion at low pressure can be written as:
$\begin{array}{1 1}(A)\;Z=1-\large\frac{Pb}{RT}\\(B)\;Z=1+\large\frac{Pb}{RT}\\(C)\;Z=1+\large\frac{RT}{Pb}\\(D)\;Z=1-\large\frac{a}{VRT}\end{array}$
The van der Waals equation of state
$(P+\large\frac{a}{V_m^2})$$(V_m-b)=RT At low pressure, V_m -b \approx V_m , the equation reduces to (P+\large\frac{a}{V_m^2})$$(V_m)=RT$
$\Rightarrow PV_m+\large\frac{a}{V_m}$$=RT Z=\large\frac{PV_m}{RT}$$=1-\large\frac{a}{V_mRT}$
Hence (D) is the correct answer. | 227 | 548 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-22 | latest | en | 0.441383 |
https://it.mathworks.com/matlabcentral/answers/120016-can-i-sort-find-function-base-on-rows?s_tid=prof_contriblnk | 1,726,123,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00162.warc.gz | 298,486,873 | 29,864 | # Can I sort Find function base on rows?
2 visualizzazioni (ultimi 30 giorni)
bkshn il 3 Mar 2014
Risposto: Julian il 3 Ago 2023
Hello
I use find function to find location of a vector on my Image, It works, but I have location Order by column.I mean at firs I have location of pixels on column1 then column2,..... Can I change find function to have location of pixels on row1,row2,.....?
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Image Analyst il 4 Mar 2014
Modificato: Image Analyst il 4 Mar 2014
I agree - totally confusing. Posting the image would probably help, as well as reading this. And say what you want to do once you have this. Because I suspect you don't even need it. I think you might be able to do what you want to do just with a binary image - no need to get (x,y) coordinates of every single non-zero pixel. I mean, why? Why do you think you need that? There are some situations (e.g. edge linking) but I'd like to know your reason.
bkshn il 5 Mar 2014
Modificato: Walter Roberson il 15 Mar 2014
Hello Image Analyst
I'm working on seam carving(one of content aware image resizing methods).Then I need to khnow location of seam to cut it from my Image.I try to attach my image or insert it, But I can't do it compeletly. as you see I have a vector on the left of my image.
Accedi per commentare.
### Risposte (2)
Julian il 3 Ago 2023
I couldn't find a nice solution in the documentation, so I assume the best way is instead of:
A = [1 0;
0 1;
1 0];
[row, col] = find(A);
% Results:
% row = [1; 3; 2]
% col = [1; 1; 2]
Transpose the matrix A and exchange the row and column:
[col, row] = find(A.');
% Results:
% row = [1; 2; 3]
% col = [1; 2; 1]
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
Accedi per commentare.
Walter Roberson il 4 Mar 2014
[row, col] = find(.....);
cr = sortrows([col(:), row(:)];
r = cr(:,2);
c = cr(:,1);
now r(K), c(K) is a (row, column) pair and the pairs are ordered so that column varies more slowly. But it normally would anyhow. So perhaps what you want is
[row, col = find(.....);
rc = sortrows([row(:), col(:)]);
r = rc(:,1);
c = rc(:,2);
and that should be ordered with row varying more slowly, "going along rows", the opposite of what would normally happen.
##### 6 CommentiMostra 4 commenti meno recentiNascondi 4 commenti meno recenti
bkshn il 16 Mar 2014
I have a 5*2 matrix like A=[3,4;2,5;1,6;4,7;5,9;]
your answer sort this like A=[1,4;2,5;3,6;4,7;5,9;] , as you see it sort rows and column
but I want to sort just rows like
A=[1,6;2,5;3,4;4,7;5,9;]
Walter Roberson il 16 Mar 2014
sort(A,2)
but you are now not working with vectors returned from find(), which are indices with row(K) corresponding to col(K) after the find and row(K), col(K) giving the location of what was found; you are now sorting by array content based upon the full array, which is a different task.
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Translated by | 939 | 3,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-38 | latest | en | 0.847811 |
https://clomid.website/inner/piston/6584-writing.aspx | 1,669,793,214,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710733.87/warc/CC-MAIN-20221130060525-20221130090525-00001.warc.gz | 207,979,189 | 10,895 | # Sampling Distribution Of A Statistic
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## In all possible populations that statistic of sampling distribution of probability
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Power Tools | 1,814 | 10,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-49 | latest | en | 0.924743 |
https://math.stackexchange.com/questions/2966925/probability-of-a-4th-die-roll-being-higher-than-one-of-the-first-3-rolls | 1,566,409,703,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00007.warc.gz | 531,843,106 | 32,117 | # Probability of a 4th die roll being higher than one of the first 3 rolls?
If I roll 3 dice with n sides, and then roll a 4th die of the same size, what are the odds of it being higher than at least one of the previous rolls?
I'm thinking it should be something like the odds of rolling a 1 in the first 3 dice times the odds of rolling a 2 or higher on the last die, giving me this equation.
$$(1 - ((n-1)/n)^3)*((n-1)/n)$$
Is this equation correct? I feel like I'm missing something.
• I guess I misread the title at first glance, thinking you wanted the fourth (and last) roll to be higher than all the first three rolls. But as I reread everything, it seems more likely you ask only that the last roll exceeds at least one of the first three rolls. In that case you want the distribution of the smallest of three rolls, from which you can derive the chance that the fourth roll is above that. – hardmath Oct 23 '18 at 1:25
• @hardmath exactly, I'm unsure of how to turn that into an equation though. – william porter Oct 23 '18 at 1:29
There are $$n^3$$ sequences for the first three rolls, of which
• 1 sequence has minimum roll $$n$$
• $$2^3-1^3=7$$ sequences have minimum roll $$n-1$$. The expression as a difference of cubes can be seen by drawing a table of the minimum function
• $$3^3-2^3=19$$ sequences have minimum roll $$n-2$$, etc.
Thus the probability that the fourth roll is less than or equal to the minimum of the first three rolls is $$\frac1{n^3}\sum_{k=1}^n((n-k+1)^3-(n-k)^3)\frac kn=\frac1{n^4}\sum_{k=1}^nk^3=\frac1{n^4}\left(\frac{n(n+1)}2\right)^2=\frac{(n+1)^2}{4n^2}$$ where the simplification of the sum comes first from the difference of cubes telescoping and then the formula for the sum of consecutive cubes. Therefore the probability that the fourth roll is higher than at least one of the previous three rolls is the complement of this, or $$1-\frac{(n+1)^2}{4n^2}=\frac{(n-1)^2}{4n^2}$$.
What is the probability that the 4th roll is not the smallest?
One of the 4 dice must be the smallest. It is just as likely to be the 4th die rolled as one of the previous 3.
But there is a possibility that there is a tie.
As n gets to be very large, I would expect our probability to approach $$\frac 14$$
There are $$n^4$$ possibilities for the 4 dice to come up.
if you roll a $$k$$ on your last die there are $$(n-k)$$ ways to roll better than that, and $$(n-k)^3$$ to roll 3 dice better than that.
$$\sum_\limits{k=1}^n (n-k)^3 = \sum_\limits{k=1}^{n-1} k^3 = \frac {n^2(n-1)^2}{4}$$
$$\frac {(n-1)^2}{4n^2}$$ | 764 | 2,555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2019-35 | latest | en | 0.910855 |
https://casper.astro.berkeley.edu/astrobaki/index.php?title=Stokes_parameters&oldid=5215 | 1,669,526,416,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00106.warc.gz | 189,867,777 | 12,877 | # Stokes parameters
STOKES PARAMETERS
FIXME: seems to be different from wikipedia. check on that. (i.e. right vs left polarization, a vs b).
## 1 A Note on Notation
In the following, remember that ${\displaystyle E_{x}}$ and ${\displaystyle E_{y}}$ are COMPLEX numbers!!! We will use the “physicist notation" when talking about complex numbers. Recall that for some complex number ${\displaystyle z=x+iy}$, where ${\displaystyle x,y\in {\mathcal {R}}}$, the complex conjugate is defined as ${\displaystyle z^{*}=x-iy}$ and the squared magnitude is given by ${\displaystyle z^{2}=zz^{*}=x^{2}+y^{2}}$.
## 2 Definition
Stokes parameters: used to describe polarization state of EM radiation.
{\displaystyle {\begin{aligned}P_{I}&=\langle E_{x}^{2}\rangle +\langle E_{y}^{2}\rangle \\P_{Q}&=\langle E_{x}^{2}\rangle -\langle E_{y}^{2}\rangle \\P_{U}&=\langle E_{x}E_{y}^{*}\rangle +\langle E_{y}E_{x}^{*}\rangle \\P_{V}&=i(\langle E_{x}E_{y}^{*}\rangle -\langle E_{y}E_{x}^{*}\rangle )\end{aligned}}\,\!}
These 4 ${\displaystyle P}$s are the Stokes parameters. ${\displaystyle P_{I}}$ is the total intensity, ${\displaystyle P_{Q}}$ is the polarization along the coordinate axes, ${\displaystyle P_{U}}$ is the polarization along the ${\displaystyle 45^{\circ }}$ line between the coordinate axes, and ${\displaystyle P_{V}}$ is circular polarization.
## 3 Changing Bases
We could just leave it at that, and take these definitions as given. Every definition has a story, so let’s understand it. However, a physical picture is not really obvious to me from these definitions. This is not very obvious (in my opinion) just from inspection of these formulas. Let’s break this down further, and switch bases to make this a little more immediate. Denote the Cartesian basis ${\displaystyle ({\hat {x}},{\hat {y}})}$, the ${\displaystyle 45^{\circ }}$ rotated Cartesian basis ${\displaystyle ({\hat {a}},{\hat {b}})}$, and the circular basis ${\displaystyle ({\hat {r}},{\hat {l}})}$, which are defined as follows:
{\displaystyle {\begin{aligned}({\hat {a}},{\hat {b}})&={\Big (}{\frac {{\hat {x}}+{\hat {y}}}{\sqrt {2}}},{\frac {{\hat {x}}-{\hat {y}}}{\sqrt {2}}}{\Big )}\\({\hat {r}},{\hat {l}})&={\Big (}{\frac {{\hat {x}}-i{\hat {y}}}{\sqrt {2}}},{\frac {{\hat {x}}+i{\hat {y}}}{\sqrt {2}}}{\Big )}\end{aligned}}\,\!}
and
{\displaystyle {\begin{aligned}E&=E_{x}{\hat {x}}+E_{y}{\hat {y}}\\&=E_{a}{\hat {a}}+E_{b}{\hat {b}}\\&=E_{l}{\hat {l}}+E_{r}{\hat {r}}\end{aligned}}\,\!}
Note that ${\displaystyle {\hat {x}}}$, ${\displaystyle {\hat {y}}}$, ${\displaystyle {\hat {a}}}$, ${\displaystyle {\hat {b}}}$, ${\displaystyle {\hat {l}}}$, and ${\displaystyle {\hat {r}}}$ are all unit vectors. From these definitions, we can solve for ${\displaystyle E_{a}}$, ${\displaystyle E_{b}}$, ${\displaystyle E_{l}}$, and ${\displaystyle E_{r}}$ in terms of ${\displaystyle E_{x}}$ and ${\displaystyle E_{y}}$:
{\displaystyle {\begin{aligned}E_{a}&={\frac {E_{x}+E_{y}}{\sqrt {2}}}\\E_{b}&={\frac {E_{x}-E_{y}}{\sqrt {2}}}\\E_{l}&={\frac {E_{x}-iE_{y}}{\sqrt {2}}}\\E_{r}&={\frac {E_{x}+iE_{y}}{\sqrt {2}}}\end{aligned}}\,\!}
It can now be shown that
{\displaystyle {\begin{aligned}P_{I}&=\langle E_{x}^{2}\rangle +\langle E_{y}^{2}\rangle \\&=\langle E_{a}^{2}\rangle +\langle E_{b}^{2}\rangle \\&=\langle E_{l}^{2}\rangle +\langle E_{r}^{2}\rangle \\P_{Q}&=\langle E_{x}^{2}\rangle -\langle E_{y}^{2}\rangle \\P_{U}&=\langle E_{a}^{2}\rangle -\langle E_{b}^{2}\rangle \\P_{V}&=\langle E_{l}^{2}\rangle -\langle E_{r}^{2}\rangle \end{aligned}}\,\!}
Things to notice:
• All the Stokes parameters are real numbers. It might not be obvious because of all the complex conjugation, but it is easy to prove it to yourself. It is left as a simple exercise for the reader. For example, to do this, define ${\displaystyle E_{x}=a+ib}$ and ${\displaystyle E_{y}=c+id}$, where ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle c}$, ${\displaystyle d}$ ${\displaystyle \in {\mathcal {R}}}$. Using these definitions, calculate the Stokes parameters- you’ll see that you only end up with real values.
## 4 Actually doing the math...
${\displaystyle E_{a}={\frac {E_{x}+E_{y}}{\sqrt {2}}},\,E_{b}={\frac {E_{x}-E_{y}}{\sqrt {2}}}\,\!}$
Calculating ${\displaystyle P_{U}}$ by substitution:
{\displaystyle {\begin{aligned}P_{U}&=\langle E_{a}^{2}\rangle -\langle E_{b}^{2}\rangle \\&={\Big \langle }{\Big (}{\frac {E_{x}+E_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}+E_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }-{\Big \langle }{\Big (}{\frac {E_{x}-E_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}-E_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }\\&={\frac {1}{2}}\langle E_{x}^{2}+E_{x}E_{y}^{*}+E_{x}^{*}E_{y}+E_{y}^{2}\rangle -{\frac {1}{2}}\langle E_{x}^{2}-E_{x}E_{y}^{*}-E_{x}^{*}E_{y}+E_{y}^{2}\rangle \\&=\langle E_{x}E_{y}^{*}\rangle +\langle E_{x}^{*}E_{y}\rangle \\\end{aligned}}\,\!}
where we use the linearity of expectation in the last line. And lo and behold- it matches the definition from earlier!!!
${\displaystyle E_{r}={\frac {E_{x}+iE_{y}}{\sqrt {2}}},\,E_{l}={\frac {E_{x}-iE_{y}}{\sqrt {2}}}\,\!}$
Calculating ${\displaystyle P_{V}}$ by substitution:
{\displaystyle {\begin{aligned}P_{V}&=\langle E_{r}^{2}\rangle -\langle E_{l}^{2}\rangle \\&={\Big \langle }{\Big (}{\frac {E_{x}-iE_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}+iE_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }-{\Big \langle }{\Big (}{\frac {E_{x}+iE_{y}}{\sqrt {2}}}{\Big )}{\Big (}{\frac {E_{x}^{*}-iE_{y}^{*}}{\sqrt {2}}}{\Big )}{\Big \rangle }\\&={\frac {1}{2}}\langle E_{x}^{2}-iE_{x}^{*}E_{y}+iE_{x}E_{y}^{*}+E_{y}^{2}\rangle -{\frac {1}{2}}\langle E_{x}^{2}+iE_{x}^{*}E_{y}-iE_{x}E_{y}^{*}+E_{y}^{2}\rangle \\&=i(\langle E_{x}E_{y}^{*}\rangle -\langle E_{x}^{*}E_{y}\rangle )\end{aligned}}\,\!} | 2,192 | 5,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 46, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-49 | latest | en | 0.819305 |
https://mtp.tools/converters/data-storage/kibibit-to-gigabit-calculator | 1,590,521,359,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00119.warc.gz | 465,110,785 | 5,586 | # Kibibit to Gigabit calculator (Kib to Gb)
Convert kibibits to gigabits (Kib to Gb) by typing the amount of kibibits in the input field below and then clicking in the "Convert" button. If you want to convert from gigabits to kibibits, you can use our gigabit to kibibit converter.
## Formula
Formula used to convert Kib to Gb:
F(x) = x / 976562.5
For example, if you want to convert 1 Kib to Gb, just replace x by 1 [Kib]:
1 Kib = 1 / 976562.5 = 0.000001024 Gb
## Steps
1. Divide the amount of kibibits by 976562.5.
2. The result will be expressed in gigabits.
## Kibibit to Gigabit Conversion Table
The following table will show the most common conversions for Kibibits (Kib) to Gigabits (Gb):
Kibibits (Kib) Gigabits (Gb)
0.001 Kib 0.000000001024 Gb
0.01 Kib 0.000000010240000000000001 Gb
0.1 Kib 0.0000001024 Gb
1 Kib 0.000001024 Gb
2 Kib 0.000002048 Gb
3 Kib 0.000003072 Gb
4 Kib 0.000004096 Gb
5 Kib 0.00000512 Gb
6 Kib 0.000006144 Gb
7 Kib 0.000007168 Gb
8 Kib 0.000008192 Gb
9 Kib 0.000009216 Gb
10 Kib 0.00001024 Gb
20 Kib 0.00002048 Gb
30 Kib 0.00003072 Gb
40 Kib 0.00004096 Gb
50 Kib 0.0000512 Gb
60 Kib 0.00006144 Gb
70 Kib 0.00007168 Gb
80 Kib 0.00008192 Gb
90 Kib 0.00009216 Gb
100 Kib 0.0001024 Gb
A kibibit is a unit of measurement for digital information and computer storage. The binary prefix kibi (which is expressed with the letters Ki) is defined in the International System of Quantities (ISQ) as a multiplier of 2^10. Therefore, 1 kibibit is equal to 1,024 bits. The symbol commonly used to represent a kibibit is Kib (sometimes as Kibit).
A gigabit is a unit of measurement for digital information and computer storage. The prefix giga (which is expressed with the letter G) is defined in the International System of Units (SI) as a multiplier of 10^9 (1 billion). Therefore, 1 gigabit is equal to 1,000,000,000 bits and equal to 1,000 megabits. The symbol commonly used to represent a gigabit is Gb (sometimes as Gbit).
## FAQs for Kibibit to Gigabit converter calculator
### What is Kibibit to Gigabit converter calculator?
Kibibit to Gigabit converter is a free and online calculator that converts Kibibits to Gigabits.
### How do I use Kibibit to Gigabit converter?
You just have to insert the amount of Kibibits you want to convert and press the "Convert" button. The amount of Gigabits will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Kibibit to Gigabit converter work on?
Kibibit to Gigabit converter calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 895 | 2,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-24 | latest | en | 0.765942 |
http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=P4099&l=en | 1,477,649,904,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721606.94/warc/CC-MAIN-20161020183841-00456-ip-10-171-6-4.ec2.internal.warc.gz | 534,249,096 | 4,884 | Order KöMaL! Competitions Portal
P. 4099. A sample of water of mass m1 and of temperature t1 is in a thermos flask, and warmer water of mass m2 and of temperature t2 is poured to it. After the themal equilibrium is reached the temperature is measured. Next time, because we are curious, we swop the two samples of water, so the colder water of mass m1 and of temperature t1 is poured into the thermos in which there is the other sample of water of mass m2 and of temperature t2. Surprisingly it is found that the equilibrium temperature in the second case is different from the one measured in the first time.
a) How can you explain this difference?
b) The difference between the two equilibrium temperatures is 1.9 oC. What can you deduce from this?
Data: m1=300 g, t1=20 oC, m2=600 g, t2=80 oC.
(4 points)
Deadline expired on 11 November 2008.
Statistics on problem P. 4099.
118 students sent a solution. 4 points: 80 students. 3 points: 15 students. 2 points: 15 students. 1 point: 4 students. 0 point: 2 students. Unfair, not evaluated: 2 solutions.
• Problems in Physics of KöMaL, October 2008
• Our web pages are supported by: ELTE Morgan Stanley | 320 | 1,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2016-44 | longest | en | 0.930067 |
https://www.physicsforums.com/threads/is-mathematical-induction-a-deductive-or-inductive-argument.358120/ | 1,527,486,341,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794871918.99/warc/CC-MAIN-20180528044215-20180528064215-00105.warc.gz | 818,485,960 | 14,334 | Homework Help: Is mathematical induction a deductive or inductive argument?
1. Nov 26, 2009
jeremy22511
1. The problem statement, all variables and given/known data
Is mathematical induction a deductive or inductive argument?
Would appreciate the help. Thanks.
Jeremy
2. Relevant equations
3. The attempt at a solution
Its name suggests that the process is inductive, yet I know all of mathematics depends on deductive reasoning...
2. Nov 26, 2009
tjackson3
Short answer: Your hunch was dead on. It's deductive reasoning (inductive is not accepted as a valid type of reasoning in most disciplines).
Long answer: The first step in mathematical induction is inductive. I'll use as an example the formula for summing all consecutive integers from 1 to n:
$$1 + 2 + ... + n = \frac{n(n+1)}{2}$$
The first step in proving this is to prove that it's true for n = 1. That is:
$$1 = \frac{2}{2} = 1$$
To stop there would be to use inductive reasoning - i.e., since it's true for n = 1, it must be true for all n. This is obviously not necessarily correct, and that's where the deductive part comes in. The purpose of a deductive argument is to prove that, given a hypothesis, its conclusion must be valid and follow directly from the hypothesis. That is, now that we know that the above is true for n = 1, we assume that it's true for some n (that's the hypothesis), and show that it must then be true for n + 1. Now that it's in general form like that, you've completed the deduction, and shown that it's true for all n in the domain of the problem (in this case, natural numbers). | 402 | 1,589 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.948355 |
https://2021.help.altair.com/2021.0.1/activate/business/en_us/topics/reference/oml_language/CoreMinimalInterpreter/ndgrid.htm | 1,685,759,658,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00227.warc.gz | 90,074,050 | 16,730 | ndgrid
Construct a grid of points from coordinate vectors.
Syntax
[y1, y2, ...] = ndgrid(x1)
[y1, y2, ...] = ndgrid(x1, x2, ...)
Inputs
x1, x2, ...
Coordinate vectors from which to construct a grid.
Type: double | integer
Dimension: scalar | vector | matrix
Outputs
y1
The matrix of the x1 coordinates.
y2
The matrix of the x2 coordinates.
Examples
One input 2D example.
x1 = [1:3];
[y1 y2] = ndgrid(x1)
y1 = [Matrix] 3 x 3
1 1 1
2 2 2
3 3 3
y2 = [Matrix] 3 x 3
1 2 3
1 2 3
1 2 3
Two input 2D example:
x1 = [1:3];
x2 = [5:8];
[y1 y2] = ndgrid(x1, x2)
y1 = [Matrix] 3 x 4
1 1 1 1
2 2 2 2
3 3 3 3
y2 = [Matrix] 3 x 4
5 6 7 8
5 6 7 8
5 6 7 8
5 6 7 8 | 308 | 659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-23 | latest | en | 0.637136 |
https://www.airmilescalculator.com/distance/gbb-to-ztu/ | 1,656,542,688,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103645173.39/warc/CC-MAIN-20220629211420-20220630001420-00581.warc.gz | 700,080,324 | 15,197 | # Distance between Gabala (GBB) and Zaqatala (ZTU)
Flight distance from Gabala to Zaqatala (Qabala International Airport – Zaqatala International Airport) is 74 miles / 120 kilometers / 65 nautical miles. Estimated flight time is 38 minutes.
Driving distance from Gabala (GBB) to Zaqatala (ZTU) is 101 miles / 163 kilometers and travel time by car is about 2 hours 4 minutes.
74
Miles
120
Kilometers
65
Nautical miles
## How far is Zaqatala from Gabala?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 74.465 miles
• 119.839 kilometers
• 64.708 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 74.405 miles
• 119.744 kilometers
• 64.656 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Gabala to Zaqatala?
Estimated flight time from Qabala International Airport to Zaqatala International Airport is 38 minutes.
## What is the time difference between Gabala and Zaqatala?
There is no time difference between Gabala and Zaqatala.
## Flight carbon footprint between Qabala International Airport (GBB) and Zaqatala International Airport (ZTU)
On average flying from Gabala to Zaqatala generates about 36 kg of CO2 per passenger, 36 kilograms is equal to 79 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Gabala to Zaqatala
Shortest flight path between Qabala International Airport (GBB) and Zaqatala International Airport (ZTU).
## Airport information
Origin Qabala International Airport
City: Gabala
Country: Azerbaijan
IATA Code: GBB
ICAO Code: UBBQ
Coordinates: 40°49′36″N, 47°42′45″E
Destination Zaqatala International Airport
City: Zaqatala
Country: Azerbaijan
IATA Code: ZTU
ICAO Code: UBBY
Coordinates: 41°33′43″N, 46°40′1″E | 561 | 2,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | latest | en | 0.843218 |
https://www.ah-studio.com/228721-intercept-form-of-a-parabola-five-important-life-lessons-intercept-form-of-a-parabola-taught-us | 1,582,242,123,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145316.8/warc/CC-MAIN-20200220224059-20200221014059-00101.warc.gz | 582,425,329 | 5,193 | # Intercept Form Of A Parabola Five Important Life Lessons Intercept Form Of A Parabola Taught Us
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Expanded Form Kindergarten Ten Gigantic Influences Of Expanded Form Kindergarten W13 Form Ucsd Seven Things You Need To Know About W13 Form Ucsd Today Pennywise The Clown True Form The Reason Why Everyone Love Pennywise The Clown True Form Form 8 Penalties 8 Easy Ways To Facilitate Form 8 Penalties W8 Form Arizona 8 Things You Most Likely Didn’t Know About W8 Form Arizona W10 Form Online 100110 Five Awesome Things You Can Learn From W10 Form Online 100110 Form 11 Template Five Precautions You Must Take Before Attending Form 11 Template Form 8 Transition Tax 8 Advantages Of Form 8 Transition Tax And How You Can Make Full Use Of It Form I 12 Additional Evidence You Will Never Believe These Bizarre Truth Behind Form I 12 Additional Evidence | 576 | 2,522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-10 | latest | en | 0.759371 |
https://www.kodytools.com/units/linearcurrent/from/abamppcm/to/amppmm | 1,726,645,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00345.warc.gz | 775,696,461 | 15,327 | # Abampere/Centimeter to Ampere/Millimeter Converter
1 Abampere/Centimeter = 1 Ampere/Millimeter
## One Abampere/Centimeter is Equal to How Many Ampere/Millimeter?
The answer is one Abampere/Centimeter is equal to 1 Ampere/Millimeter and that means we can also write it as 1 Abampere/Centimeter = 1 Ampere/Millimeter. Feel free to use our online unit conversion calculator to convert the unit from Abampere/Centimeter to Ampere/Millimeter. Just simply enter value 1 in Abampere/Centimeter and see the result in Ampere/Millimeter.
Manually converting Abampere/Centimeter to Ampere/Millimeter can be time-consuming,especially when you don’t have enough knowledge about Linear Current Density units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Abampere/Centimeter to Ampere/Millimeter converter tool to get the job done as soon as possible.
We have so many online tools available to convert Abampere/Centimeter to Ampere/Millimeter, but not every online tool gives an accurate result and that is why we have created this online Abampere/Centimeter to Ampere/Millimeter converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Abampere/Centimeter to Ampere/Millimeter (abA/cm to A/mm)
By using our Abampere/Centimeter to Ampere/Millimeter conversion tool, you know that one Abampere/Centimeter is equivalent to 1 Ampere/Millimeter. Hence, to convert Abampere/Centimeter to Ampere/Millimeter, we just need to multiply the number by 1. We are going to use very simple Abampere/Centimeter to Ampere/Millimeter conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Abampere/Centimeter} = 1 \times 1 = \text{1 Ampere/Millimeter}$$
## What Unit of Measure is Abampere/Centimeter?
Abampere per centimeter is a unit of measurement for linear current density. It is defined as flow of current in abampere through a conductor which is one centimeter wide.
## What is the Symbol of Abampere/Centimeter?
The symbol of Abampere/Centimeter is abA/cm. This means you can also write one Abampere/Centimeter as 1 abA/cm.
## What Unit of Measure is Ampere/Millimeter?
Ampere per millimeter is a unit of measurement for linear current density. It is defined as flow of current in ampere through a conductor which is one millimeter wide.
## What is the Symbol of Ampere/Millimeter?
The symbol of Ampere/Millimeter is A/mm. This means you can also write one Ampere/Millimeter as 1 A/mm.
## How to Use Abampere/Centimeter to Ampere/Millimeter Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Abampere/Centimeter and in the first input field, enter a value.
• From the second dropdown, select Ampere/Millimeter.
• Instantly, the tool will convert the value from Abampere/Centimeter to Ampere/Millimeter and display the result in the second input field.
## Example of Abampere/Centimeter to Ampere/Millimeter Converter Tool
Abampere/Centimeter
1
Ampere/Millimeter
1
# Abampere/Centimeter to Ampere/Millimeter Conversion Table
Abampere/Centimeter [abA/cm]Ampere/Millimeter [A/mm]Description
1 Abampere/Centimeter1 Ampere/Millimeter1 Abampere/Centimeter = 1 Ampere/Millimeter
2 Abampere/Centimeter2 Ampere/Millimeter2 Abampere/Centimeter = 2 Ampere/Millimeter
3 Abampere/Centimeter3 Ampere/Millimeter3 Abampere/Centimeter = 3 Ampere/Millimeter
4 Abampere/Centimeter4 Ampere/Millimeter4 Abampere/Centimeter = 4 Ampere/Millimeter
5 Abampere/Centimeter5 Ampere/Millimeter5 Abampere/Centimeter = 5 Ampere/Millimeter
6 Abampere/Centimeter6 Ampere/Millimeter6 Abampere/Centimeter = 6 Ampere/Millimeter
7 Abampere/Centimeter7 Ampere/Millimeter7 Abampere/Centimeter = 7 Ampere/Millimeter
8 Abampere/Centimeter8 Ampere/Millimeter8 Abampere/Centimeter = 8 Ampere/Millimeter
9 Abampere/Centimeter9 Ampere/Millimeter9 Abampere/Centimeter = 9 Ampere/Millimeter
10 Abampere/Centimeter10 Ampere/Millimeter10 Abampere/Centimeter = 10 Ampere/Millimeter
100 Abampere/Centimeter100 Ampere/Millimeter100 Abampere/Centimeter = 100 Ampere/Millimeter
1000 Abampere/Centimeter1000 Ampere/Millimeter1000 Abampere/Centimeter = 1000 Ampere/Millimeter
# Abampere/Centimeter to Other Units Conversion Table
ConversionDescription
1 Abampere/Centimeter = 1000 Ampere/Meter1 Abampere/Centimeter in Ampere/Meter is equal to 1000
1 Abampere/Centimeter = 10 Ampere/Centimeter1 Abampere/Centimeter in Ampere/Centimeter is equal to 10
1 Abampere/Centimeter = 1 Ampere/Millimeter1 Abampere/Centimeter in Ampere/Millimeter is equal to 1
1 Abampere/Centimeter = 0.001 Ampere/Micrometer1 Abampere/Centimeter in Ampere/Micrometer is equal to 0.001
1 Abampere/Centimeter = 0.000001 Ampere/Nanometer1 Abampere/Centimeter in Ampere/Nanometer is equal to 0.000001
1 Abampere/Centimeter = 914.4 Ampere/Yard1 Abampere/Centimeter in Ampere/Yard is equal to 914.4
1 Abampere/Centimeter = 304.8 Ampere/Foot1 Abampere/Centimeter in Ampere/Foot is equal to 304.8
1 Abampere/Centimeter = 25.4 Ampere/Inch1 Abampere/Centimeter in Ampere/Inch is equal to 25.4
1 Abampere/Centimeter = 1 Kiloampere/Meter1 Abampere/Centimeter in Kiloampere/Meter is equal to 1
1 Abampere/Centimeter = 0.01 Kiloampere/Centimeter1 Abampere/Centimeter in Kiloampere/Centimeter is equal to 0.01
1 Abampere/Centimeter = 0.001 Kiloampere/Millimeter1 Abampere/Centimeter in Kiloampere/Millimeter is equal to 0.001
1 Abampere/Centimeter = 0.000001 Kiloampere/Micrometer1 Abampere/Centimeter in Kiloampere/Micrometer is equal to 0.000001
1 Abampere/Centimeter = 1e-9 Kiloampere/Nanometer1 Abampere/Centimeter in Kiloampere/Nanometer is equal to 1e-9
1 Abampere/Centimeter = 0.9144 Kiloampere/Yard1 Abampere/Centimeter in Kiloampere/Yard is equal to 0.9144
1 Abampere/Centimeter = 0.3048 Kiloampere/Foot1 Abampere/Centimeter in Kiloampere/Foot is equal to 0.3048
1 Abampere/Centimeter = 0.0254 Kiloampere/Inch1 Abampere/Centimeter in Kiloampere/Inch is equal to 0.0254
1 Abampere/Centimeter = 1000000 Milliampere/Meter1 Abampere/Centimeter in Milliampere/Meter is equal to 1000000
1 Abampere/Centimeter = 10000 Milliampere/Centimeter1 Abampere/Centimeter in Milliampere/Centimeter is equal to 10000
1 Abampere/Centimeter = 1000 Milliampere/Millimeter1 Abampere/Centimeter in Milliampere/Millimeter is equal to 1000
1 Abampere/Centimeter = 1 Milliampere/Micrometer1 Abampere/Centimeter in Milliampere/Micrometer is equal to 1
1 Abampere/Centimeter = 0.001 Milliampere/Nanometer1 Abampere/Centimeter in Milliampere/Nanometer is equal to 0.001
1 Abampere/Centimeter = 914400 Milliampere/Yard1 Abampere/Centimeter in Milliampere/Yard is equal to 914400
1 Abampere/Centimeter = 304800 Milliampere/Foot1 Abampere/Centimeter in Milliampere/Foot is equal to 304800
1 Abampere/Centimeter = 25400 Milliampere/Inch1 Abampere/Centimeter in Milliampere/Inch is equal to 25400
1 Abampere/Centimeter = 1000000000 Microampere/Meter1 Abampere/Centimeter in Microampere/Meter is equal to 1000000000
1 Abampere/Centimeter = 10000000 Microampere/Centimeter1 Abampere/Centimeter in Microampere/Centimeter is equal to 10000000
1 Abampere/Centimeter = 1000000 Microampere/Millimeter1 Abampere/Centimeter in Microampere/Millimeter is equal to 1000000
1 Abampere/Centimeter = 1000 Microampere/Micrometer1 Abampere/Centimeter in Microampere/Micrometer is equal to 1000
1 Abampere/Centimeter = 1 Microampere/Nanometer1 Abampere/Centimeter in Microampere/Nanometer is equal to 1
1 Abampere/Centimeter = 914400000 Microampere/Yard1 Abampere/Centimeter in Microampere/Yard is equal to 914400000
1 Abampere/Centimeter = 304800000 Microampere/Foot1 Abampere/Centimeter in Microampere/Foot is equal to 304800000
1 Abampere/Centimeter = 25400000 Microampere/Inch1 Abampere/Centimeter in Microampere/Inch is equal to 25400000
1 Abampere/Centimeter = 0.001 Megaampere/Meter1 Abampere/Centimeter in Megaampere/Meter is equal to 0.001
1 Abampere/Centimeter = 0.00001 Megaampere/Centimeter1 Abampere/Centimeter in Megaampere/Centimeter is equal to 0.00001
1 Abampere/Centimeter = 0.000001 Megaampere/Millimeter1 Abampere/Centimeter in Megaampere/Millimeter is equal to 0.000001
1 Abampere/Centimeter = 1e-9 Megaampere/Micrometer1 Abampere/Centimeter in Megaampere/Micrometer is equal to 1e-9
1 Abampere/Centimeter = 1e-12 Megaampere/Nanometer1 Abampere/Centimeter in Megaampere/Nanometer is equal to 1e-12
1 Abampere/Centimeter = 0.0009144 Megaampere/Yard1 Abampere/Centimeter in Megaampere/Yard is equal to 0.0009144
1 Abampere/Centimeter = 0.0003048 Megaampere/Foot1 Abampere/Centimeter in Megaampere/Foot is equal to 0.0003048
1 Abampere/Centimeter = 0.0000254 Megaampere/Inch1 Abampere/Centimeter in Megaampere/Inch is equal to 0.0000254
1 Abampere/Centimeter = 100 Biot/Meter1 Abampere/Centimeter in Biot/Meter is equal to 100
1 Abampere/Centimeter = 1 Biot/Centimeter1 Abampere/Centimeter in Biot/Centimeter is equal to 1
1 Abampere/Centimeter = 0.1 Biot/Millimeter1 Abampere/Centimeter in Biot/Millimeter is equal to 0.1
1 Abampere/Centimeter = 0.0001 Biot/Micrometer1 Abampere/Centimeter in Biot/Micrometer is equal to 0.0001
1 Abampere/Centimeter = 1e-7 Biot/Nanometer1 Abampere/Centimeter in Biot/Nanometer is equal to 1e-7
1 Abampere/Centimeter = 91.44 Biot/Yard1 Abampere/Centimeter in Biot/Yard is equal to 91.44
1 Abampere/Centimeter = 30.48 Biot/Foot1 Abampere/Centimeter in Biot/Foot is equal to 30.48
1 Abampere/Centimeter = 2.54 Biot/Inch1 Abampere/Centimeter in Biot/Inch is equal to 2.54
1 Abampere/Centimeter = 100 Abampere/Meter1 Abampere/Centimeter in Abampere/Meter is equal to 100
1 Abampere/Centimeter = 0.1 Abampere/Millimeter1 Abampere/Centimeter in Abampere/Millimeter is equal to 0.1
1 Abampere/Centimeter = 0.0001 Abampere/Micrometer1 Abampere/Centimeter in Abampere/Micrometer is equal to 0.0001
1 Abampere/Centimeter = 1e-7 Abampere/Nanometer1 Abampere/Centimeter in Abampere/Nanometer is equal to 1e-7
1 Abampere/Centimeter = 91.44 Abampere/Yard1 Abampere/Centimeter in Abampere/Yard is equal to 91.44
1 Abampere/Centimeter = 30.48 Abampere/Foot1 Abampere/Centimeter in Abampere/Foot is equal to 30.48
1 Abampere/Centimeter = 2.54 Abampere/Inch1 Abampere/Centimeter in Abampere/Inch is equal to 2.54
1 Abampere/Centimeter = 2997924536843.1 Statampere/Meter1 Abampere/Centimeter in Statampere/Meter is equal to 2997924536843.1
1 Abampere/Centimeter = 29979245368.43 Statampere/Centimeter1 Abampere/Centimeter in Statampere/Centimeter is equal to 29979245368.43
1 Abampere/Centimeter = 2997924536.84 Statampere/Millimeter1 Abampere/Centimeter in Statampere/Millimeter is equal to 2997924536.84
1 Abampere/Centimeter = 2997924.54 Statampere/Micrometer1 Abampere/Centimeter in Statampere/Micrometer is equal to 2997924.54
1 Abampere/Centimeter = 2997.92 Statampere/Nanometer1 Abampere/Centimeter in Statampere/Nanometer is equal to 2997.92
1 Abampere/Centimeter = 2741302196489.4 Statampere/Yard1 Abampere/Centimeter in Statampere/Yard is equal to 2741302196489.4
1 Abampere/Centimeter = 913767398829.79 Statampere/Foot1 Abampere/Centimeter in Statampere/Foot is equal to 913767398829.79
1 Abampere/Centimeter = 76147283235.82 Statampere/Inch1 Abampere/Centimeter in Statampere/Inch is equal to 76147283235.82 | 3,698 | 11,263 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.894767 |
https://mathleaks.com/study/writing_linear_equations_in_slope-intercept_form | 1,660,151,027,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00280.warc.gz | 371,852,860 | 36,045 | {{ option.icon }} {{ option.label }} arrow_right
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# Writing Linear Equations in Slope-Intercept Form
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### Direct messages
Linear function rules (linear equations) can be written in different ways to highlight different characteristics. In this section, writing these rules in slope-intercept form will be explored.
## Writing the Equation of a Line in Slope-Intercept Form Using Two Points
To write linear equations in slope-intercept form, the slope m and the y-intercept b of the line must be known.
When only two points on the line are known, the following four-step method can be used. For example, the equation of the line that passes through the points (-4,1) and (8,4) will be written.
1
Find the Slope
expand_more
Given two points on a line, the slope of the line can be found by using the Slope Formula. In this case, the coordinates (-4,1) and (8,4) can be substituted in place of (x1,y1) and (x2,y2), respectively.
m=0.25
The slope m of the line passing through the two points is 0.25.
2
Replace m With the Slope
expand_more
Now that the value of the slope is known, it can be substituted for m in the slope-intercept form of an equation.
3
Find b Using a Point
expand_more
Next, the y-intercept can be found by substituting either of the given points into the equation and solving for b. In the considered example, (8,4) can be used. Substitute its coordinates into the equation from Step 2 and solve for b.
y=0.25x+b
4=0.25(8)+b
4=2+b
2=b
b=2
Therefore, the y-intercept is 2.
4
Write the Equation
expand_more
Lastly, the complete equation in slope-intercept form can be written by substituting the value of the y-intercept found above into the equation from Step 2.
The equation is now complete.
## Write the equation of the line that passes through the given points
fullscreen
Write the equation of the line that passes through the point (3,1) and has the same y-intercept as the line y=9x+4.
Show Solution expand_more
A line in slope-intercept form is given by the equation
y=mx+b,
where m is the slope of the line and b is the y-intercept. The line y=9x+4 has a y-intercept of (0,4). We want our line to have the same y-intercept. Therefore, the equation of the new line must also have the value b=4. This gives
y=mx+4.
Our line must also pass through the point (3,1). We can solve for m in the equation above by substituting this point for x and y.
y=mx+4
1=m3+4
1=3m+4
-3=3m
-1=m
m=-1
The slope of the new line is m=-1. Thus, we can write the complete equation as
y=-x+4.
## Writing the Equation of a Line in Slope-Intercept Form From a Graph
To write the equation in slope-intercept form of the graph of a line, the y-intercept b and the slope m of the line must be found.
The following method can be used. As an example, consider the line shown.
1
Find the y-Intercept
expand_more
The y-intercept is the point where the graph intersects the y-axis. From the diagram, it can be seen that the y-intercept is (0,-4).
2
Replace b With the y-Intercept
expand_more
The y-coordinate of the y-intercept can be substituted into the slope-intercept form equation for b. In the considered example, substituting -4 for b results in the following equation.
3
Find the Slope
expand_more
Next, the slope of the line must be determined. By the definition, the slope of a line is expressed as the quotient of the rise and run of the line.
Here, rise is the vertical distance between two points, and run is the horizontal distance. To find the slope, use any two points. On the diagram, the previously-found y-intercept and the point (2,2) were chosen.
As seen above, the rise equals 6 and the run equals 2. By substituting these values into the formula, the slope can be calculated.
4
Replace m With the Slope
expand_more
The complete equation of the line can now be written by substituting the value of m into the equation from Step 2. In this case, substitute m=3.
The equation is now complete.
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Community | 1,224 | 4,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-33 | latest | en | 0.793667 |
https://www.nickzom.org/blog/tag/mechanics/ | 1,620,770,870,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00407.warc.gz | 935,190,886 | 31,816 | ## How to Calculate and Solve for Mass, Angular Velocity, Radius and Centrifugal Force of a Body | The Calculator Encyclopedia
The image above represents the centrifugal force.
To compute for the centrifugal force, three essential parameters are needed and these parameters are Mass of the body (m), Angular Velocity of the body (w) and Radius (r).
The formula for calculating the centrifugal force:
F = mω²r
Where:
F = Centrifugal Force
m = mass of the body
ω = angular velocity
Let’s solve an example;
Find the centrifugal force with mass of the body as 12, angular velocity as 32 and a radius of 8.
This implies that;
m = mass of the body = 12
ω = angular velocity = 32
r = radius = 8
F = mω²r
F = 12 x 32² x 8
F = 12 x 1024 x 8
F = 98304
Therefore, the centrifugal force is 98304 N.
Calculating the Mass of the body (m) when the Centrifugal Force, Angular Velocity and Radius is Given.
m = F / w2r
Where;
m = mass of the body
F = Centrifugal Force
ω = angular velocity
Let’s solve an example;
Find the mass of a body when centrifugal force is 140 with an angular velocity of 24 and a radius of 10.
This implies that;
F = Centrifugal Force = 140
ω = angular velocity = 24
r = radius = 10
m = F / w2r
m = 140 / 24210
m = 140 / 576 x 10
m = 140 / 5760
m = 0.024
Therefore, the mass of the body is 0.024 kg.
## How to Calculate and Solve the Centre of Gravity of a Segment of a Sphere
The image above represents a segment of a sphere.
To compute the centre of gravity of a segment of a sphere requires two essential parameters. These parameters are the radius of the sphere and height of the segment of the sphere.
The formula for computing the centre of gravity of a sphere is:
C.G. = 3(2r – h)² / 4(3r – h)
Where:
C.G. = Centre of Gravity
r = Radius of the Sphere
h = Height of the Segment of the Sphere
Let’s solve an example
Find the centre of gravity of the segment of the sphere where the radius of the sphere is 10 m and the height of the segment of the sphere is 4 m.
This implies that:
r = Radius of the Sphere = 10
h = Height of the Segment of the Sphere = 4
C.G. = 3(2(10) – 4)² / 4(3(10) – 4)
C.G. = 3(20 – 4)² / 4(30 – 4)
C.G. = 3(16)² / 4(26)
C.G. = 3(256) / 104
C.G. = 768 / 104
C.G. = 7.38
Therefore, the centre of gravity of the segment of the sphere is 7.38.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a segment of a sphere at a height, h at a distance from the centre of the sphere measured along the height.
## How to Calculate and Solve for the Centre of Gravity of a Cube
The image above is a cube with a length of 5.2 cm.
To compute the centre of gravity of a cube, one essential parameter is needed and this parameter is the length of the cube (l).
The formula for calculating the centre of gravity of a cube is:
C.G. = 0.5(l)
Where:
l = Length of the Cube
C.G. = Centre of Gravity
Let’s solve an example:
Find the centre of gravity of a cube where the length of a side of the cube is 5.2 cm.
This implies that:
l = Length of the Cube = 5.2
## How to Calculate and Solve for the Centroid or Centre of Gravity of a Hemisphere
The image above is a hemisphere with a radius of 5.
To compute the centroid or centre of gravity of a hemisphere. You need one essential parameter and this parameter is the radius of the hemisphere (r).
The formula for calculating the centroid or centre of gravity of a hemisphere is:
C.G. = 3r / 8
Where
r = Radius of the hemisphere
As always let us try and solve an example:
Find the centroid or centre of gravity of a hemisphere where the radius is 5 cm.
From the formula this implies that:
r = Radius of the hemisphere = 5
C.G. = 3(5) / 8
C.G. = 15 / 8
C.G. = 1.875
Therefore, the centroid or centre of gravity of the hemisphere is 1.875.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a hemisphere at a distance from its base measured along the vertical radius.
To get the answer and workings of the center of gravity or centroid of a hemisphere using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.
You can get this app via any of these means:
## Nickzom Calculator Calculates the Centroid or Centre of Gravity of a Semicircle | Statics (Mechanics)
The image above is a semicircle with a radius of 7 cm.
To compute the centroid or centre of gravity of a semicircle, you need one essential parameter and this parameter is the radius of the semicircle.
The formula for calculating the centroid or centre of gravity of a semicircle is:
C.G.= 4r /
Where:
r = Radius of the Semicircle
π = Mathematical Constant = 3.142 (approximately)
Let’s solve an example:
Find the centroid or centre of gravity of a semicircle where the radius is 7 cm.
From the example above,
r = radius of the semicircle = 7
C.G. = 4(7) /
C.G. = 28 / 9.4247
C.G. = 2.97
Therefore, the centroid or centre of gravity of the semicircle is 2.97.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a semicircle at a distance from its base measured along the vertical radius.
To get the answer and workings of the center of gravity or centroid of a semicircle using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.
## How to Calculate and Solve for the Centroid or Centre of Gravity of a Sphere
The image above is a sphere and 18 m is the diameter of the sphere. In the computing of the centroid or centre of gravity of a sphere there is only one essential parameter which is the diameter of the sphere.
The formula for calculating the centroid or centre of gravity of a sphere is:
C.G. = d / 2
Where:
d is the diameter of the sphere
As always, let’s take an example:
Let’s solve an example
Find the centroid or centre of gravity of a sphere where the diameter is 18 m.
C.G. = 18 / 2
C.G. = 9
Therefore, the centroid or centre of gravity of the sphere is 9.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a sphere at a distance from every point.
To get the answer and workings of the center of gravity or centroid of a sphere using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.
You can get this app via any of these means:
## How to Calculate and Solve the Centre of Gravity of a Right Circular Cone
This image above is a display of what a right circular cone looks like. There is only one essential parameter for calculating the centroid or centre of gravity of a right circular cone. This parameter is the height of the cone (h).
The formula for calculating the the centroid or centre of gravity of a right circular cone is:
C.G. = h / 4
As always let’s solve an example.
Find the centroid or centre of gravity of a right circular cone where the height of the cone is 12cm.
This implies that:
h = height of the cone = 12
C.G. = 12 / 4
C.G. = 3
Therefore, the centroid or centre of gravity of the right circular cone is 3.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a right circular cone at a distance from its base measured along the vertical axis.
To get the answer and workings to center of gravity or centroid of a right circular cone using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.
You can get this app via any of these means:
## How to Calculate the Centre of Gravity of a Circular Sector in Statics | Mechanics
It is very possible to compute the centroid or centre of gravity of a circular sector. There are two highly important parameters one needs to know to compute the centre of gravity of a circular sector. These parameters are:
• Radius of the Sector (r)
• Semi Vertical Angle (α)
The formula for computing the centre of gravity of a circular sector is:
C.G. = 2rsinα /
Now, let’s take an example.
Let’s find the centroid or the centre of gravity of a circular sector that has a radius of 4m and a semi vertical angle of 30°.
This implies that:
r = Radius of the Sector = 4
α = Semi Vertical Angle = 30
Entering this values into the formula we have:
C.G. = 2(4)sin30° / 3(30)
C.G. = 8 . sin30° / 90
C.G. = 8 . (0.5) / 90
C.G. = 4 / 90
C.G. = 0.0444
Therefore, the centroid or centre of gravity of the circular sector is 0.0444.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the centre of gravity of a circular sector making a semi vertical angle α at a distance from the centre of the sector measured along the central axis.
To get the answer and workings to center of gravity or centroid of a circular sector. First, you need to obtain the Nickzom Calculator – The Calculator Encyclopedia app.
## Nickzom Calculator Calculates the Center of Gravity of a Trapezium in Statics Mechanics
According to Quora,
The center of gravity of a trapezoid can be estimated by dividing the trapezoid in two triangles.
Nickzom Calculator calculates the centre of gravity of a trapezium with parallel sides a and b at a distance measured from side b.
The formula for calculating this center of gravity is:
C.G. = h (b + 2a) / 3 (b + a)
Where, a and b are the length of the parallel sides of the trapezium (b being the base length and a being the top length) whereas h is the height of the trapezium.
Let’s take for Example: Find the center of gravity or centroid of a trapezium where a is 4, b is 8 and h is 2.
This implies:
a = 4
b = 8
h = 2
C.G. = 2 (8 + 2(4)) / 3 (8 + 4)
C.G. = 2 (8 + 8) / 3 (12)
C.G. = 2 (16) / 36
C.G. = 32 / 36
C.G. = 0.89
Therefore, the center of gravity or centroid of the trapezium is 0.89.
To get the answer and workings to center of gravity or centroid of a trapezium. First, you need to obtain the Nickzom Calculator – The Calculator Encyclopedia app. | 2,632 | 9,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-21 | latest | en | 0.826212 |
https://boards.straightdope.com/t/the-proper-way-to-add-odds/338617 | 1,722,930,486,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00552.warc.gz | 103,121,970 | 5,508 | # The proper way to add odds
What is the proper way to add odds given a know likelyhood of occurance?
For instance: a coin flip having 50/50 odds of getting heads each time you flip it, what are the odds that you would get heads A LEAST ONCE (not ONLY once) in x number of flips.
Or if there is a 20% chance of rain for each day in a four day stretch, what are the odds that it will rain at least once in that four day stretch?
Here is a DIY source of information on the very subject!
Calculate Odds
There’s no general method that will cover all cases, but there are large classes of problems that can be handled similarly.
Your first problem is like that–since each flip of the coin is completely independent of all the other flips before it, the number of heads follows what’s known as a binomial distribution. In this case, the probability of at least one head in n flips is 1 - 1/2[sup]n[/sup].
The second problem is stated so that the weather on any given day doesn’t depend on the weather from any previous days. Under that assumption, the probability of rain at least once is 1 - (4/5)[sup]4[/sup], or 369/625. However, the assumption of independence is very likely not valid, and more complicated models would be needed to address a situation like that.
The easiest way to work this out is to note that if the probability of an event occuring is equal to one minus the probability of the event not occuring. So, the probability of getting no rain for four straight days is 4/5 * 4/5 * 4/5 * 4/5, which is 256/625. The probabilty of getting at least one day of rain is the probability of not getting 4 straight days of no rain, which is 1 - 256/625, which is 369/625, or about 59%.
I think the most important step is distinguishing odds from probability, odds being the one that sounds better.
For example, if you draw a card at random from a standard deck, the odds are 1:3 that it will be a heart, because for every heart in the deck there are three non-hearts. You’re three times as likely to get something else as you are to get a heart.
The probability is 1/4, which means if you do this little experiment a zillion times, it should come up as a heart about 1/4 of the time.
AFAIK, the only way odds are useful in calculation is if they can be converted to probabilities first. | 557 | 2,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.95592 |
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[–] 1 point2 points (3 children)
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I always assumed that a stone was 8 pounds or 12 pounds or something else I'd be used to. Nope. 14 pounds. What the fuck, British people? How the hell are people supposed to calculate it? I mean 8 and 12 would also be annoying, sure, but at least we're used to those.
[–] 2 points3 points (2 children)
sorry, this has been archived and can no longer be voted on
14 pounds in a stone, 12 16 ounces in a pound. As a Brit, even I think it's time to go full metric instead of straddling the 2 systems as we do now. In fairness, I think schools teach metric but colloquially people retain metric.
In temperature, we use Fahrenheit for hot days and Celsius for cold! Ha!
Edit: I can't remember the stupid imperial system!
[–] 1 point2 points (1 child)
sorry, this has been archived and can no longer be voted on
16 ounces = 1 pound 14 pounds=1 stone 8 stones=1 hundredweight 20 hundredweight = 1ton
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
My god, I'm an idiot!
Yes, I'll just hand my passport in | 322 | 1,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-18 | latest | en | 0.961295 |
https://stonoga-klubmalucha.pl/crusher/2020_10_31_4924/ | 1,638,343,167,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00365.warc.gz | 612,449,338 | 7,767 | # Calculation Of Roller Mill Charge Volume
## free calculation for ball mill charge volume
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## calculation of charge in ball mill - gujaratgenomics
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## how to calculate alumina ball charge for dry mill
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## determine charge volume in a ball mill
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## calculations mill charge - namesprojectsouthflorida.org
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## Mill Charge Load Calculation - Stone Crushing Machine
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## how to calculate charge in ball mill - verhuur-doms.be
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## Charge Volume Crusher - hsw-consulting
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## grinding volume calculation in a ball mill - rrcser
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## Calculation Of Cement Mill Media Charging
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## how to calculate mill charge level - dzerves
How to handle the charge volume of a ball mill or rod mill. 201215-Gauss's Law may be used to calculate the charge distribution of the,As a result, mills are seldom run with charge levels greater than 45%...calculation of grinding media charging ball mill of cement,calculation for cement mill pdf celak.org. Whic. cement ball mill grinding media calculation The ball charge mill consists of Do you want to show 100tph cement grinding ball mill or other products ofMedia Charge_Optimal Ball Size - Ball Mill | Mill,,Scope : The Media Charge_Optimal Ball Size spreadsheet was designed to estimate the ideal make-up ball size to any given grinding application, on the basis of the empirical correlations independently proposed by Ettore Azzaroni and Allis Chalmers.
## Ball Charge Design_v | Mill (Grinding) | Density
Ball Charge Design_v - Download as Excel Spreadsheet (.xls), PDF File (.pdf), Text File (.txt) or read online. Scribd is the world's largestcement mill charge calculation - karo-carhire,calculation of ball mill charge volume MTM Crusher . Ball charges calculators The Cement Grinding Office. Ball charges This calculator gives the surface and the . Check price. calculate ball mill grinding media in cement YouTube.calculation of filling degree of cement mill,Cement mill - Wikipedia. A cement mill (or finish mill in North American usage) is the equipment used to grind the hard, nodular clinker from the cement kiln into the fine grey powder that is cement.Most cement is currently ground in ball mills and also vertical roller mills which are more effective than ball mills.
## cement mill charging formula - dmcmachining
alculation formula cement mill media overlandconnection The ball mill ball calculation formula The mill Read more Download Cement Cement The mill tube grinding media Grinding Media Charging In Cement Mill Cement grinding Vertical roller mills versus ball millsCement grinding Vertical roller mills versus ball which involved a tube mill with a,,, | 2,196 | 10,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2021-49 | latest | en | 0.814025 |
https://scicomp.stackexchange.com/tags/quantum-mechanics/hot?filter=all | 1,627,666,804,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00031.warc.gz | 516,322,308 | 31,056 | # Tag Info
25
The Schroedinger equation is effectively a reaction-diffusion equation $$i\frac{\partial\psi}{\partial t}=-\nabla^2\psi+V\psi\tag{1}$$ (all constants are 1). When it comes to any partial differential equation, there's two ways to solve it: Implicit method (adv: large time steps & unconditionally stable, disadv: requires matrix solver that can give bad ...
18
Yes, it is much more difficult to do so. For the $N$ body problem, all you need to compute are the trajectories $\mathbf x_i(t), i=1\ldots N$ which are just $N$ functions of a single variable. On the other hand, even for a single electron, the solution of the Schroedinger equation is a function $\Psi(x,y,z,t)$, i.e., a function of four variables. For two ...
13
Finding the eigenvalues for the Schrödinger equation is really similar to finding the eigenvalues for the wave equation. You start with your differential equation $$\left[-\frac{1}{2}\nabla'^2 + V(r)\right]\psi(\mathbf{r}) = E' \psi(\mathbf{r})$$ where we did the change of variable $(x,y,z) \rightarrow (a_0 x, a_0 y, a_0 z)$, with $a_0 \equiv 1$ Bohr, $E'= ... 13 The best approach is to use an ODE solver that is guaranteed to conserve the norm of the initial condition, i.e., for which$\|y_n\| = \|y_0\|$for all$n\in\mathbb{N}$. Such solvers exist, and are called geometric integrators, since they preserve geometric properties of the exact solution (in this case, that energy is conserved, i.e.,$\frac{d}{dt}\|y(t)\| =...
10
Kyle Kanos's answer looks to be very full, but I thought I'd add my own experience. The split-step Fourier method (SSFM) is extremely easy to get running and fiddle with; you can prototype it in a few lines of Mathematica and it is, extremely stable numerically. It involves imparting only unitary operators on your dataset, so it automatically conserves ...
9
256 equations is a relatively small number. All of the usual integrators, such as those included in Matlab, Maple or Mathematica should have no real problem with equations of this size and should be able to return answers in a fraction of the time it would take an algorithm you would implement yourself, because they use sophisticated explicit/implicit and ...
9
If your discrete equations represent your problem properly you should have the right eigenvalues, although numerically you can get a small imaginary part when solving the eigenvalue problem. As an example, let us consider the axisymmetric case of a cylindrical well. The Schrödinger equation would be written as $$\frac{\partial^2 \psi}{\partial r^2} + \frac{... 8 Kirill's answer is a good method but it is hampered that the fact that you need to tune it to a specific range in energy, so it is not appropriate if you have a broad-band wavepacket you need to absorb. An alternative approach is exterior complex scaling, which is reviewed well in Infinite-range exterior complex scaling as a perfect absorber in time-... 6 The solutions for the equation are in$$\psi \in \mathbb{C}^{3M}\times\mathbb{R}^+ \enspace .$$If the number of electrons is small enough you can just use any traditional method. Like a domain discretization method (Finite Difference, Finite Element, Boundary Element), or a pseudospectral method. Since solving this equation is not more difficult than ... 6 I didn't follow exactly your notation so I can't say for sure what this would look like for your example, but two suggestions: if you really want your life made simple, check out qutip, the Quantum Toolbox in Python. It has classes specifically for quantum operators and state vectors, supports tensor product of operators in different spaces and keeps track ... 5 I can recommend using the finite-difference time-domain (FDTD) method. I even wrote a tutorial some time back that should answer most of your questions: J. R. Nagel, "A review and application of the finite-difference time-domain algorithm applied to the Schrödinger equation," ACES Journal, Vol. 24, No. 1, February 2009 I have some Matlab codes that run ... 5 In the literature these boundary conditions go by the name of absorbing boundary conditions (or nonreflecting, open, radiation, invisible, far-field), and this is a well-known topic. One clear description I think is Absorbing Boundary Conditions for the Schrödinger Equation by Fevens and Jiang. Here is one approach (described in the above paper; I haven't ... 5 Caveat: I did not read beyond the statement So, what I did was to define functions for the potential and the second derivative, and use Euler's method. So there may be other issues with your code, but this one is already fundamental. You are trying to solve a wave equation with Euler's method. This is numerically unstable. To see why, note that the ... 5 As always with numerical problems, it's a good idea to simplify things as much as possible to isolate what's really going wrong. You can simplify the problem by considering the evolution of a pure state \psi instead of an entire density matrix, so solving$$i\dot\psi = H\psi$$for the vector \psi instead of the matrix \rho. Additionally, you can ... 5 I think that the main problem might be with the solver you are using. The Hamiltonian (matrix) in this case is Hermitian, it is even symmetric since it is purely real. You could use eigh instead of eig to take advantage of this. Furthermore, you are not removing only the first and last points but intervals of size 1 at each end. Following, I show you a ... 4 I would disagree with your statement that Euler and Runge-Kutta have to be sequential. I know that you are saying this because you cannot parallelize across time steps (or across inner time steps in the case of something like RK4), however both can be parallel as long as you do so within a single time step. You would simply evaluate the derivatives of your ... 4 Since you didn't post your MATLAB code, I'm not sure how you're calling ode45. I'm guessing you are changing the tspan vector (second argument) on each call to ode45. The first thing to understand is that the tspan vector has (almost) no effect on the time step used by ode45. The tspan vector simply allows you to pass to ode45 the time span of the ... 4 You should rewrite the equation dimensionless with dimensional analysis. Proper step size should be evident then and can be compared more easily. 4 From the comments, it sounds like your problem is stiff, and using an implicit integrator will help you a lot more than trying to parallelize. But in case someone comes here looking for information on parallel time integration, you can find a discussion of some simple parallel extrapolation and deferred correction methods in this preprint of mine. It's not ... 4 Numerically the sum isn't terribly easy: it behaves as$$ \sum_{n\geq 2, k\geq 1} \frac{1}{n(n^2-1)} \frac{e^{-4/k}}{(k^2-n^{-2})^2}, $$so it converges linearly. Most numerical methods that can be expected to do well only really work for either alternating, or quickly converging sums. With a sum that converges linearly, you can only really get anywhere if ... 4 I know this problem well from my own research: it is given by the fact that the equation is very stiff. Thus, it is likely that you're doing nothing wrong (--although I haven't inspected your code). So where does the stiffness come from? The problem In order to solve the equation numerically, it is common to pick a grid and use finite-differences to ... 4 The SciPy tutorial explicitly states Note that ARPACK is generally better at finding extremal eigenvalues: that is, eigenvalues with large magnitudes. In particular, using which = 'SM' may lead to slow execution time and/or anomalous results. A better approach is to use shift-invert mode. and goes on to describe that. Basically, if \lambda is the ... 4 I don't have enough reputation to comment, but I wanted to give a basic complement of gIS's answer. Simply put, if we have the decomposition over V=V_1\otimes V_2, which we represent with, say, the reshaping [2,4,2,4] why do we einsum over the first and third indices for trace over V_1 and second and fourth for trace over V_2? Maybe this is obvious ... 4 I think that your potential for the numerical case is wrong. The potential should be a big positive number, so the solution tends to zero outside the well when the value of the potential increases. If you check the solutions in Wikipedia for the first three (bounded) states they should be$$E_n = \frac{2\hbar^2 v_n^2}{m L^2}\, ,$$with v_1 = 1.28, v_2 =... 4 Regarding the boundary conditions: Don't be fooled by Wikipedia. Yes, the scenario in the picture suggests an absorption at the boundaries, and yes, one could use absorbing boundary conditions in order to reproduce that numerically. In the simple case of a wavepacket these are readily available, because in the end, there exists an analytical solution for the ... 4 Your problem was the lower limit of integration. It should have been -x_e instead of 0, since x_e is the equilibrium point for the potential and not the minimum distance. After correcting that, you get the following #%% Solution xe, lam = 1.0, 6.0 # parameters for potential xmax = 10 # Bval, Bval2 = wavefunction values at x = bound1, bound2 bound1, ... 3 A few answers and comments here conflate confusingly the TDSE with a wave equation; perhaps a semantics issue, to some extent. The TDSE is the quantized version of the classical non-relativistic hamiltonian$$H=\frac{p^2}{2m} + V(x)= E.$$With the rules$$p \rightarrow i\hbar\partial_x,\ \ E\rightarrow i\hbar\partial_t, \ \ x\rightarrow x, (as discussed in ...
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Fin Heat Transfer
Fin Heat Transfer
(OP)
Hello everyone,
I have I project that I am working on which involves heat transfer. I am quite conflicted and would greatly appreciate your input.
The setup is the following: There is hot water being pumped through a rectangular aluminum tube. The tube is surrounded by constant temperature cold water. Would welding aluminum rectangular thin plates (fins) inside the tube be an effective form of increasing heat transfer efficiency?
In all class exercises I ever solved, the fin was attached to a hot surface and convected heat to the atmosphere. In this case, the convection happens with the hot surrounding water and the heat is escaping through conduction.
Is this exactly the same as the usual fin problems and my brain is just fried?
Thank you!
RE: Fin Heat Transfer
Fluid is as fluid does. Heat transfer is about the movement of heat through fluid transfer, be they gases or liquids. The major difference is the convection coefficient magnitudes and whether there's any boiling involved.
TTFN (ta ta for now)
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RE: Fin Heat Transfer
It would be better to understand basic concept behind using "fins".
"Fins" actually accelerate heat removal from high temperature heat source to cold temperature source in a way to attain thermal equilibrium condition - both sources at same temperature. How is this achieved? By conduction heat transfer through fin (not much) + convection heat transfer from fin to outside fluid (by increasing the surface area, the convective heat transfer (Q=h*A*ΔT) is increased from the fin surface (which gats heated due to attached to high temperature source) to low temperature source.)
Now, the amount of heat transferred from fins to low temperature source is depend on the convection coefficient of fin surrounding fluid (h) and heat transfer area (A) of fin. If the fin is attached to inside of rectangular tube, it may not be efficient way to remove heat it since fin is exposed to high temperature fluid the heat transfer coefficient might be less although the area is increased. If the fins are attached to outside surface, fin will be exposed to cold temperature and heat transfer coefficient might be high along with the increased surface area.
You can compare heat transfer rate from both cases and decide whether using fins inside or outside is effective.
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Register now while it's still free! | 945 | 4,610 | {"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} | 2.703125 | 3 | CC-MAIN-2022-49 | longest | en | 0.913658 |
http://forum.allaboutcircuits.com/threads/transistor-info.37381/ | 1,454,749,735,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701146241.46/warc/CC-MAIN-20160205193906-00098-ip-10-236-182-209.ec2.internal.warc.gz | 82,589,399 | 14,940 | # Transistor info
Discussion in 'Homework Help' started by TsAmE, Apr 21, 2010.
1. ### TsAmE Thread Starter Member
Joined:
Apr 19, 2010
72
0
I am having some trouble understanding the transistor and have a couple of things that I would like to know about transistors.
1. Why is the output of a transistor at the collecter?
2. How does a transistor invert a signal?
3. When will a darlington resistor go on?
2. ### kingdano Member
Joined:
Apr 14, 2010
377
19
1
Functional model of an NPN transistor
The operation of a transistor is difficult to explain and understand in terms of its internal structure. It is more helpful to use this functional model:
• The base-emitter junction behaves like a diode.
• A base current IB flows only when the voltage VBE across the base-emitter junction is 0.7V or more.
• The small base current IB controls the large collector current Ic.
• Ic = hFE × IB (unless the transistor is full on and saturated)
hFE is the current gain (strictly the DC current gain), a typical value for hFE is 100 (it has no units because it is a ratio)
• The collector-emitter resistance RCE is controlled by the base current IB:
IB = 0 RCE = infinity transistor off
IB small RCE reduced transistor partly on
IB increased RCE = 0 transistor full on ('saturated')
3. ### kingdano Member
Joined:
Apr 14, 2010
377
19
2.
A transistor inverter (NOT gate)
Inverters (NOT gates) are available on logic ICs but if you only require one inverter it is usually better to use this circuit. The output signal (voltage) is the inverse of the input signal:
When the input is high (+Vs) the output is low (0V).
When the input is low (0V) the output is high (+Vs).
Any general purpose low power NPN transistor can be used. For general use RB = 10k and RC = 1k, then the inverter output can be connected to a device with an input impedance (resistance) of at least 10k such as a logic IC or a 555 timer (trigger and reset inputs). If you are connecting the inverter to a CMOS logic IC input (very high impedance) you can increase RB to 100k and RC to 10k, this will reduce the current used by the inverter.
4. ### kingdano Member
Joined:
Apr 14, 2010
377
19
Darlington pair
This is two transistors connected together so that the current amplified by the first is amplified further by the second transistor. The overall current gain is equal to the two individual gains multiplied together: Darlington pair current gain, hFE = hFE1 × hFE2
(hFE1 and hFE2 are the gains of the individual transistors)
This gives the Darlington pair a very high current gain, such as 10000, so that only a tiny base current is required to make the pair switch on.
A Darlington pair behaves like a single transistor with a very high current gain. It has three leads (B, C and E) which are equivalent to the leads of a standard individual transistor. To turn on there must be 0.7V across both the base-emitter junctions which are connected in series inside the Darlington pair, therefore it requires 1.4V to turn on.
Joined:
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377
19 | 774 | 3,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | longest | en | 0.887827 |
https://mattermodeling.stackexchange.com/questions/10170/maximum-bond-length-in-vesta | 1,685,791,147,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00558.warc.gz | 429,558,124 | 26,516 | # Maximum bond length in Vesta
I need some cutoff radii to count bonds between different atoms in my system. When a .cif file is opened in Vesta, there are some default values of min and max bond lengths between two atomic species. While the minimum is always 0, the maximum values are different. How is this maximum value determined? Is it good enough cutoff to describe actual bonds in the system?
Minimum and maximum atomic distances can be selected by users. From the crystallographic point of view the nearest neighbour distances depend on the atomic sizes of the involved elements (C-C distance is shorter than U-U distance, for example).
In any case, the best thing is to plot the structure; then, you can measure the distance between selected atoms using the "Distance" tool in the vertical panel on the left (see the image). At this point you have an idea of the distange range and you can select the minimum and maximum atomic distances to be considered in the "bonds" window.
• thanks for your response. What I was particularly interested to know was how do VESTA set the default maximum bond lengths? For example, in case of Ba-O bonds, the default maximum is always 3.14795 A. How do they arrive at such a particular value?
– rik
Jan 4 at 18:42
• What do you mean with " in case of Ba-O bonds, the default maximum is always 3.14795 A"? Is this true also for other structures? I don't know how Vesta select the default values. I guess it could use 2 different approaches. 1) takes into account the atomic sizes of the involved elements; 2) determines the distances between the atomic species using the crystallographic data. Jan 5 at 11:28
• It is true for all structures. The bond length min and max are fixed for any two pair in VESTA. I think your approach is fundamentally what needs to be done. But then we have to decide whether to use ionic radius/covalent radius, and decide the coordination number(because ionic radius depends on it). But to find coordination number from a structure, we need to know the radius to consider. So it sort of becomes a chicken-and-egg problem?
– rik
Jan 6 at 14:28
• @rik I cannot agree. Bond length is calculated by using atomic positions and lattice parameters. Ionic or covalent size do not matter, the distance will be always the same because it is between the centers of the atoms, not between their surfaces. To determine the coordination number, you must look at your structure and play with it. In any case ionic radii as those given by Shannon hold also for covalent structures. Jan 8 at 19:49
• BTW, ionic radii (for example those determined by Shannon) hold also for covalent structure. In the end, no purely ionic bond is known, a signifant percentage of covalent bonding is always present. Jan 8 at 19:54 | 640 | 2,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | longest | en | 0.890223 |
https://www.omnimaga.org/ti-calculator-programming-and-support/help~an-program-for-ti84ti83-to-get-it's-id-to-str0-(but-not-work)/ | 1,679,305,426,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00344.warc.gz | 1,008,980,926 | 8,259 | ### Author Topic: help~an program for ti84/ti83 to get it's ID to Str0 (but not work) (Read 2728 times)
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#### jknear
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##### help~an program for ti84/ti83 to get it's ID to Str0 (but not work)
« on: September 19, 2019, 10:41:20 pm »
HOW TO USE IT:'( HOW TO USE IT
The function of the program is to get the machine ID and store it in STR0.
Preliminary feeling is that L6 calculates problematic ID and calculates 0 10 (A) 2D16C 7EA5. These bits are correct, but why does 10 not show A?
Attached to the program is the hope of expert answers, QQ contact me: 844288567
before use it :2nd+LINK -> sendID then run GETID
log(ID)→Str8
DelVar XDelVar L₆"→L₆
1ᴇ12(fPart(Str8)1→L₁
1ᴇ12(fPart(Str8)/1→L₂
Str8
Ans+1-1=Ans
(1)Ans(1)
If Ans or sum(iPart(10Str8)≠4 or not(fPart(1ᴇ9Str8
“ID→Str8
Repeat not(L₁(1
not(X→X
If sum(fPart(L₁)=fPart(L₂
Then
10sum(fPart(L₁→L₆(1+dim(L₆
.1iPart(L₁→L₁
Else
10sum((1+fPart(L₁→L₆(1+dim(L₆
If X
L₂+1→L₂
.1iPart(L₁)-.1→L₁6
End
.1iPart(L₂→L₂
End
"0
For(X,dim(L₆),1,⁻1
Ans+sub("0123456789ABCDEF",L₆(X)+1,1
End
DelVar L₁DelVar L₂DelVar L₆DelVar XDelVar DDelVar IDelVar Str8sub(Ans,1,length(Ans)-4)+sub(Ans,length(Ans)-1,2)+sub(Ans,length(Ans)-3,2→Str0 | 550 | 1,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.656452 |
https://matheducators.stackexchange.com/questions/10583/how-can-we-neatly-explain-chain-rule-of-differentiation/10584 | 1,606,356,587,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141185851.16/warc/CC-MAIN-20201126001926-20201126031926-00537.warc.gz | 392,711,496 | 36,250 | # How can we neatly explain chain rule of differentiation
My students are often getting confused while using chain rule for complicated functions. For example
$$f(x)=\tan^3\left(\sqrt{x^2+x+1}\right)$$ Most of the students wrote $f'(x)$ wrongly as
$$f'(x)=3 \tan^2\left(\sqrt{x^2+x+1}\right) \times \frac{1}{2\sqrt{x^2+x+1}} \times (2x+1)$$
Can I have any comments on how better I can teach chain rule?
• This is interesting. I have not seen this kind of error before. It seems like they just missed out one step, but it is unusual that most of the students left out the same step. – Steven Gubkin Feb 18 '16 at 6:34
• I wonder if they would have made the same mistake if the function had been written $\left(\tan\left(\sqrt{x^2+x+1}\right)\right)^3$. – Adam Feb 18 '16 at 13:44
• Is this really a problem of misunderstanding the chain rule or a problem of being fooled by the notation convention used for powers of trig functions? – Todd Wilcox Feb 18 '16 at 15:02
• This may be in the same category as the kind of mistake where a student is trying to differentiate $\sin\cos x$ and applies the product rule. For students operating at that level, it can be difficult to get them to articulate why they do what they do. They may not be thinking of a specific and incorrect interpretation of the notation; they may simply not be thinking at all. – Ben Crowell Feb 20 '16 at 20:50
Here is one way to write the solution which cuts down on errors:
$$\frac{d}{dx} \tan^3(\sqrt{x^2+x+1}) = \frac{d(\tan^3(\sqrt{x^2+x+1}))}{d(\tan(\sqrt{x^2+x+1}))} \frac{d(\tan(\sqrt{x^2+x+1}))}{ d(\sqrt{x^2+x+1})} \frac{d\sqrt{x^2+x+1}}{d (x^2+x+1)} \frac{d(x^2+x+1)}{dx}$$
Forcing the student to set up all of the computations first, before performing them, seems to cut down on the number of errors. I think this helps to solve the problem of remembering which step you are on while you compute a given derivative.
• A few things: (1) Excellent suggestion (2) It might be useful to teach the students to use variable substitution to make it clear what to differentiate in each case. For instance, the first factor looks pretty confusing as written, but if you make the substitution $u = \tan\sqrt{x^2 + x + 1}$ it becomes quite obvious. (Of course you have to remember to reverse the substitution after taking the derivative.) (3) Mathematicians always complain about this, so it's probably worth saying that this is a memory aid, not a rigorous derivation of the chain rule. – David Z Feb 18 '16 at 8:15
• @David Z My personal objection to the $u$-formulas you find in certain elementary textbooks is that they seem to reinforce the student's idea that we need to memorize different chain rules for all the different elementary functions. Instead, I want them to think about one chain-rule that applies to all functions because I want them to understand more and memorize less. The notation of $u$ itself is not the objection (I think). – James S. Cook Feb 18 '16 at 10:53
• @JamesS.Cook I'm not sure if I'm familiar with these $u$-formulas. Can you give an example? The only chain rule I know of is the standard one, $dy/dx = dy/du\times du/dx$ (for single-variable functions under appropriate conditions). – David Z Feb 18 '16 at 13:18
• @DavidZ Some calculus books will incorporate the chain rule into the statement of every formal rule of differentiation, for example writing $\frac{d}{dx} u^n = nu^{n-1} \frac{d u }{d x}$. – Steven Gubkin Feb 18 '16 at 16:40
• Or, $\frac{d}{dx} \cos( u) = -\sin(u) \frac{du}{dx}$ or $\frac{d}{dx} \ln (u) = \frac{1}{u} \frac{du}{dx}$ and... – James S. Cook Feb 18 '16 at 16:56
This is a slight variation on Steven Gubkin's suggestion, incorporating David Z's suggestion in his comment on that answer.
Suggest to the students that whenever they encounter a complicated nexted function, they explicitly write it as a chain of several simpler functions, e.g.: $$y=u^3$$ $$u=\tan(v)$$ $$v=\sqrt{w}$$ $$w=x^2+x+1$$
Then the chain rule says simply that the derivative of the first variable $y$ with respect to the last variable $x$ is the product of the derivaives of all of the individual links in the chain:
$$\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dv}\cdot \frac{dv}{dw}\cdot \frac{dw}{dx}$$
So $$\frac{dy}{dx} = 3u^2 \cdot \sec^2(v) \cdot \frac{1}{2\sqrt{w}} \cdot (2x+1)$$ and the problem can be completed by back-substituting in the expressions for each of the variables in terms of the simpler ones.
• It would also be good if this kind of skill was taught in precalculus, when students are learning function transformations. – Steven Gubkin Feb 18 '16 at 22:43 | 1,309 | 4,603 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-50 | latest | en | 0.920236 |
https://gmatclub.com/forum/best-preparation-97160.html?fl=similar | 1,508,608,681,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824824.76/warc/CC-MAIN-20171021171712-20171021191712-00451.warc.gz | 694,475,901 | 45,314 | It is currently 21 Oct 2017, 10:58
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# Best preparation
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13 Jul 2010, 01:06
Hi everybody,
I passed 3 times the gmat which i prepare by myself to finish with a score of 540...
I am a french engineer with a Master's degree and I live in a Reunion Island. It is not easy for me to prepare the GMAT : no prep courses, bad broadband and nothing to improve my english...
I really want to improve my score ! but i don't know how ? What is the best thing to do ? Go in a preparation course online or in a real class ? ( the nearest one is in Singapore...) Buy all the Manhattan books...
Well, any advice will be a good things...
Thanks
Fred
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13 Jul 2010, 01:20
Welcome to GMAT Club Fred!
What were your split scores? Q and V?
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13 Jul 2010, 01:45
Hi,
Q : 43 and V: 23...
Fred
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13 Jul 2010, 02:00
Thanks - it looks like there is some potential improvement with Q and V but mostly V is what needs work.
1. Get english at the point where it is not a limitation. Unfrotunately GMAT is in English and you have no way but to get over it. I would not attempt much verbal practice in the MGMAT guides or a prep class until you are comfortable with most of English grammar and have a decent vocabulary.
2. MGMAT Guides are indeed very good if you can get them - would highly recommend (many others do as well). If you are looking for a prep class, see here for reviews and good deals: gmat-prep-course-class-reviews-ratings-discount-codes-78451.html?highlight=ml601
3. I would recommend GMAT Fiction (reading a lot for a few months that you are prepping for the GMAT) or even longer if you can/willing to do it.
Also, take a peek here: gmat-study-plan-for-gmat-novices-start-your-gmat-journey-80727.html?highlight=ml301 and here everything-you-need-to-prepare-for-the-gmat-revised-77983.html?highlight=ml303
As usual error log is recommended.
P.S. What score are you shooting for and what have you used to prepare in the past?
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13 Jul 2010, 07:20
I need to have at least 600...
Fred
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Re: Best preparation [#permalink] 13 Jul 2010, 07:20
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# Best preparation
Moderator: HiLine
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,091 | 3,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-43 | latest | en | 0.867722 |
https://www.thefreelibrary.com/Numerical+Reconstruction+of+the+Covariance+Matrix+of+a+Spherically...-a0542803142 | 1,627,642,025,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.52/warc/CC-MAIN-20210730091645-20210730121645-00448.warc.gz | 1,064,911,126 | 42,946 | # Numerical Reconstruction of the Covariance Matrix of a Spherically Truncated Multinormal Distribution.
1. Introduction
It has been more than forty years since Tallis [1] worked out the moment-generating function of a normal multivariate X [equivalent to] [{[X.sub.k]}.sup.v.sub.k=1] ~ [N.sub.v](0, [SIGMA]), subject to the conditional event
X [member of] [E.sub.v] ([rho]; [SIGMA]), [E.sub.v] ([rho]; [SIGMA]) [equivalent to] {x [member of] [R.sup.v] : [x.sup.T][[SIGMA].sup.-1] x [less than or equal to] [rho]}. (1)
The perfect match between the symmetries of the ellipsoid [E.sub.v]([rho]; [SIGMA]) and those of [N.sub.v](0, [SIGMA]) allows for an exact analytic result, from which the complete set of multivariate truncated moments can be extracted upon differentiation. Consider, for instance, the matrix [G.sub.E]([rho]; [SIGMA]) of the second truncated moments, expressing the covariances among the components of X within [E.sub.v]([rho]; [SIGMA]). From Tallis' paper it turns out that
[G.sub.E] ([rho]; [SIGMA]) = [c.sub.T] ([rho]) [SIGMA], [c.sub.T] ([rho]) [equivalent to] [[F.sub.v+2] ([rho])]/[F.sub.v] ([rho]), (2)
with [F.sub.v] denoting the cumulative distribution function of a [chi square]- variable with v degrees of freedom. Inverting (2)--so as to express [SIGMA] as a function of [S.sub.E]--is trivial, since [c.sub.T]([rho]) is a scalar damping factor independent of [SIGMA]. In this paper, we shall refer to such inverse relation as the reconstruction of [SIGMA] from [G.sub.E]. Unfortunately, life is not always so easy. In general, the effects produced on the expectation of functions of X by cutting off the probability density outside a generic domain D [subset] [R.sup.v] can be hardly calculated in closed form, especially if the boundary of D is shaped in a way that breaks the ellipsoidal symmetry of [N.sub.v](0, [SIGMA]). Thus, for instance, unlike (2), the matrix of the second truncated moments is expected in general to display a nonlinear/nontrivial dependence upon [SIGMA].
In the present paper, we consider the case where D is a v-dimensional Euclidean ball with center in the origin and square radius [rho]. Specifically, we discuss the reconstruction of [SIGMA] from the matrix [G.sub.B] of the spherically truncated second moments. To this aim, we need to mimic Tallis' calculation, with (1) being replaced by the conditional event
X [member of] [B.sub.v] ([rho]), [B.sub.v]([rho]) [equivalent to] {x [member of] [R.sub.v] : [x.sup.T]x [less than or equal to] [rho]}. (3)
This is precisely an example of the situation described in the previous paragraph: although [B.sub.v]([rho]) has a higher degree of symmetry than [E.sub.v]([rho]; [SIGMA]), still there is no possibility of obtaining a closed-form relation between [SIGMA] and [G.sub.B], since [B.sub.v]([rho]) breaks the ellipsoidal symmetry of [N.sub.v](0, [SIGMA]): technically speaking, in this case we cannot perform any change of variable under the defining integral of the moment-generating function, which may reduce the dimensionality of the problem, as in Tallis' paper.
In spite of that, the residual symmetries characterizing the truncated distribution help simplify the problem in the following respects: (i) the reflection invariance of the whole setup still yields E[[X.sub.k] | X [member of] [B.sub.v]([rho])] = 0 [for all]k and (ii) the rotational invariance of [B.sub.v]([rho]) preserves the possibility of defining the principal components of the distribution just like in the unconstrained case. In particular, the latter property means that [G.sub.B] and [SIGMA] share the same orthonormal eigenvectors. In fact, the reconstruction of [SIGMA] from [G.sub.B] amounts to solving a system of nonlinear integral equations, having the eigenvalues [lambda] [equivalent to] [{[[lambda].sub.k]}.sup.v.sub.k=1] of [SIGMA] as unknown variables and the eigenvalues [mu] [equivalent to] [{[[mu].sub.k]}.sup.v.sub.k=1] of [G.sub.B] as input parameters. In a lack of analytic techniques to evaluate exactly the integrals involved, we resort to a numerical algorithm, of which we investigate feasibility, performance, and optimization.
The paper is organized as follows. In Section 2, we describe a few examples illustrating the occurrence of spherical truncations in practical situations. In Section 3, we show that the aforementioned integral equations have the analytic structure of a fixed point vector equation; that is to say, [lambda] = T([lambda]). This suggests achieving the reconstruction of [lambda] numerically via suitably chosen iterative schemes. In Section 4, we prove the convergence of the simplest of them by inductive arguments, the validity of which relies upon the monotonicity properties of ratios of Gaussian integrals over [B.sub.v]([rho]). In Section 5, we review some numerical techniques for the computation of Gaussian integrals over [B.sub.v]([rho]) with controlled systematic error. These are based on and extend a classic work by Ruben [2] on the distribution of quadratic forms of normal variables. For the sake of readability, we defer proofs of statements made in this section to the appendix. In Section 6, we report on our numerical experiences: since the simplest iterative scheme, namely, the Gauss-Jacobi iteration, is too slow for practical purposes, we investigate the performance of its improved version based on overrelaxation; as expected, we find that the latter has a higher convergence rate; yet it still slows down polynomially in 1/[rho] as [rho] [right arrow] 0 and exponentially in v as b [right arrow] [infinity]; in order to reduce the slowing down, we propose an acceleration technique, which boosts the higher components of the eigenvalue spectrum. A series of Monte Carlo simulations enables us to quantify the speedup. In Section 7 we discuss the problems arising when [mu] is affected by statistical uncertainty and propose a regularization technique based on perturbation theory. To conclude, we summarize our findings in Section 8.
2. Motivating Examples
Spherical truncations of multinormal distributions may characterize different kinds of experimental devices and may occur in various problems of statistical and convex analysis. In this section, we discuss two motivating examples.
2.1. A Two-Dimensional Gedanken Experiment in Classical Particle Physics. Consider the following ideal situation. An accelerator physicist prepares an elliptical beam of classical particles with Gaussian transversal profile. The experimenter knows a priori the spatial distribution of the beam, that is, the covariance matrix [SIGMA] of the two-dimensional coordinates of the particles on a plane orthogonal to their flight direction. We can assume with no loss of generality that the transversal coordinate system has origin at the maximum of the beam intensity and axes along the principal components of the beam; thus it holds [SIGMA] = diag([[lambda].sub.1], [[lambda].sub.2]). The beam travels straightforward until it enters a linear coaxial pipeline with circular profile, schematically depicted in Figure 1, where the beam is longitudinally accelerated. While the outer part of the beam is stopped by an absorbing wall, the inner part propagates within the pipeline. At the end of the beam flight the physicist wants to know if the transversal distribution of the particles is changed, due to unknown disturbance factors arisen within the pipeline. Accordingly, he measures again the spatial covariance matrix of the beam. Unfortunately, the absorbing wall has cut off the Gaussian tail, thus damping the covariance matrix and making it no more comparable to the original one. To perform such a comparison in the general case [[lambda].sub.1] [not equal to] [[lambda].sub.2], the covariance matrix of the truncated circular beam has to go through the reconstruction procedure described in next sections.
2.2. A Multivariate Example: Connections to Compositional Data Analysis. Compositional Data Analysis (CoDA) has been the subject of a number of papers, pioneered by Aitchison [3] over the past forty years. As a methodology of statistical investigation, it finds application in all cases where the main object of interest is a multivariate with strictly positive continuous components to be regarded as portions of a total amount [kappa] (the normalization [kappa] = 1 is conventionally adopted in the mathematical literature). In other words, compositional variates belong to the [kappa]-simplex:
[S.sub.v] = {z [member of] [R.sup.v.sub.+] : [[absolute value of z].sub.1] = [kappa]}, v [greater than or equal to] 2, (4)
with [[absolute value of z].sub.1] = [[summation].sup.v.sub.k=1] [z.sub.k] the taxi-cab norm of z, while compositions with different norms can be always projected onto [S.sub.v] via the closure operator C x x [equivalent to] {[kappa][x.sub.1]/[[absolute value of x].sub.1], ..., [kappa][x.sub.v]/[[absolute value of x].sub.1]}. There are countless types of compositional data, whose analysis raises problems of interest for statistics [4], for example, geochemical data, balance sheet data, and election data. The simplex constraint induces a kind of dependency among the parts of a composition that goes beyond the standard concept of covariance. This invalidates many ordinary techniques of statistical analysis.
In order to measure distances on [S.sub.v], Aitchison introduced a positive symmetric function [d.sub.A] : [S.sub.v] x [S.sub.v] [right arrow] [R.sup.+], explicitly defined by
[d.sub.A] (x, y) = [square root of (1/2v [v.summation over (i,k=1)] [[log ([x.sub.i]/[x.sub.k]) - log ([y.sub.i]/[y.sub.k])].sub.2])]. (5)
The Aitchison distance is a key tool in CoDA. It is scale invariant in both its first and second arguments; that is, it is left invariant by redefinitions z [right arrow] {[alpha][z.sub.1], ..., [alpha][z.sub.v]} with [alpha] [member of] [R.sub.+]. Accordingly, its support can be extended to [R.sup.v.sub.+] x [R.sup.v.sub.+] by imposing
[d.sub.A] (x, y) [equivalent to] [d.sub.A] (C x x, C x y), x, y [member of] [R.sup.v.sub.+]. (6)
It was proved in [5] that [d.sub.A] is a norm-induced metric on [S.sub.v], provided the latter is given an appropriate normed vector space structure. Owing to the compositional constraint [[absolute value of (x)].sub.1] = [kappa], it holds dim [S.sub.v] = v - 1. Accordingly, the description of [S.sub.v] in terms of v components is redundant: an essential representation requires compositions to be properly mapped onto (v - 1)-tuples. Among various possibilities, the Isometric Log-Ratio (ILR) transform introduced in [5] is the only known map of this kind leaving [d.sub.A] invariant. More precisely, the ILR fulfills
[d.sub.A] (x, y) = [d.sub.E] (ilr(x), ilr(y)), [d.sub.E] (u, v) [equivalent to] [square root of ([v-1.summation over (k=1)] [([u.sub.k] - [v.sub.k]).sup.2])]. (7)
It is known from [6] that if X ~ log [N.sub.v]([mu], [SIGMA]) is a log-normal v-variate, then C x X ~ [L.sub.v]([mu]', [SIGMA]') is a logistic-normal v-variate (the reader should notice that in [6] the simplex is defined by [S.sub.v] = {z [member of] [R.sup.v.sub.+] : [[absolute value of (z)].sub.1] < 1}; thus property 2.2 of [6] is here reformulated so as to take into account such difference), with a known relation between ([mu], [SIGMA]) and ([mu]', [SIGMA]'). Analogously, it is not difficult to show that ilr(C x X) ~ [N.sub.v-1]([mu]", [SIGMA]") is a normal (v - 1)-variate, with ([mu]", [SIGMA]") related to ([mu]', [SIGMA]') via the change of basis matrices derived in [5]. Just to sum up, it holds
X ~ log [N.sub.v] ([mu], [SIGMA]) [??] C x X ~ [L.sub.v] ([mu]', [SIGMA]') [??] ilr(C x X) ~ [N.sub.v-1] ([mu]", [SIGMA]"). (8)
Now, suppose that (i) X fulfills (8) and has a natural interpretation as a composition, (ii) a representative set [D.sub.X] of observations of X is given, and (iii) we wish to select from [D.sub.X] those units which are compositionally closer to the center of the distribution, according to the Aitchison distance. To see that the problem is well posed, we first turn [D.sub.X] into a set [D.sub.CxX] [equivalent to] {y : y = C x x, x [member of] [D.sub.X]} of compositional observations of Y = C x X. Then, we consider the special point cen[Y] = C x exp{E[ln Y]}, representing the center of the distribution of Y in a compositional sense: cen[Y] minimizes the expression E[[d.sup.2.sub.A](Y, cen[Y])] over [S.sub.v] and fulfills cen[Y] = [ilr.sup.-1](E[ilr(Y)]), see [7]. By virtue of (8) this entails ilr(cen[Y]) = E[ilr(Y)] = [mu]". In order to select the observations which are closer to cen[Y], we set a threshold [delta] > 0 and consider only those elements y [member of] [D.sub.CxX] fulfilling [d.sup.2.sub.A](y, cen[y]) < [delta], with cen[y] being a sample estimator of cen[Y] on [D.sub.CxX]. Such selection rule operates a well-defined truncation of the distribution of Y. Moreover, in view of (7) and (8), we have
P[[d.sup.2.sub.A] (Y, cen [Y]) < [delta] | Y ~ [L.sub.v] ([mu]', [SIGMA]')] = P[[d.sup.2.sub.E] (Z, [mu]") < [delta] | Z ~ [N.sub.v-1] ([mu]", [SIGMA]")], (9)
with Z = ilr(C x X). As a consequence, we see that a compositional selection rule based on the Aitchison distance and (8) is equivalent to a spherical truncation of a multinormal distribution. Obviously, once Z has been spherically truncated, the covariance matrix of the remaining data is damped; thus an estimate of the full covariance matrix requires the reconstruction procedure described in next sections.
2.3. Compositional Outlier Detection via Forward Search Techniques. Outlier detection in CoDA is a practical problem where the considerations made in Section 2.2 find concrete application. Most of the established methods for outlier detection make use of the Mahalanobis metric. This is however unfit to describe the distance between compositions. Log-ratio transforms allow to get rid of the compositional constraint and make it possible to apply standard statistical methods [8]. Here we consider an innovative approach, namely, the Forward Search Algorithm (FSA), introduced in [9] and thoroughly discussed in [10]. The FSA admits an elegant extension to CoDA, of which the covariance reconstruction is a key step. In the next few lines we sketch the original algorithm and outline its extension to CoDA.
2.3.1. Construction of the Signal. In its standard formulation the FSA applies to normal data. It assumes a dataset [D.sub.X] = [{[x.sup.(k)]}.sup.N.sub.k=1] with [x.sup.(k)] [member of] [R.sup.(v).sub.+] for k = 1, ..., N. The null hypothesis is that all the elements of [D.sub.X] are independent realizations of a stochastic variable X ~ N([[mu].sub.0], [[SIGMA].sub.0]). The FSA consists of a sequence of steps where data are recursively sorted. Along the recursion a signal is built and tested.
As a preliminary step, [m.sub.0] observations are randomly chosen from the bulk of [D.sub.X]. Let S([m.sub.0]) be the set of these observations. S([m.sub.0]) is used to provide initial estimates [mu]([m.sub.0]), [SIGMA]([m.sub.0]) of the true mean [[mu].sub.0] and the true covariance matrix [[SIGMA].sub.0], respectively. For m = [m.sub.0] + 1, [m.sub.0] + 2, ..., the (m - [m.sub.0])th step of the algorithm goes as follows:
(i) sort the elements of [D.sub.X] according to the increasing values of the square Mahalanobis distance:
[d.sub.m] [(x).sup.2] = [[x - [mu] (m - 1)].sup.T] [SIGMA] [(m - 1).sup.-1] [x - [mu] (m - 1)]; (10)
(ii) take the mth element [x.sup.(m)] of the newly sorted dataset and regard [s.sub.m] = [d.sub.m][([x.sup.(m)]).sup.2] as the (m - [m.sub.0])th component of the signal;
(iii) let S(m) be the set of the first m observations of the newly sorted dataset;
(iv) use S(m) to provide new estimates [mu](m), [SIGMA](m) of the true mean and the true covariance matrix, respectively.
Notice that S(m) is a truncated dataset. Therefore, [SIGMA](m) must include the Tallis' correction factor, (2) with [rho] = [s.sub.m]. While the recursion proceeds, the inliers of [D.sub.X] populate progressively S(m). The recursion stops at the [bar.mth] step, when the first outlier [x.sup.([bar.m])] of [D.sub.X] produces a statistically relevant discontinuity in the signal.
2.3.2. Statistical Test of the Signal. Statistical tests are needed to assess the relevance of discontinuities observed in the signal. At each step of the algorithm a new test is actually performed. Specifically, at the mth step, [s.sub.m] is computed together with the lowest and highest values, respectively, [delta][s.sub.m,[alpha]] and [delta][s.sub.m,1-[alpha]], which are theoretically admissible for [s.sub.m] under the null hypothesis at (1 - [alpha]) significance level for some [alpha]. These values are nothing but the [alpha]- and (1 - [alpha])- percentage points of [s.sub.m]. Their envelopes for m - [m.sub.0] > 0 generate two curves that surround the signal when plotted versus m. More precisely, one curve lies below it and the other lies above, provided the null hypothesis is not broken. The violation of one of the curves by the signal is interpreted as the result of the entrance of an outlier into S(m). Although the distribution of [s.sub.m] cannot be calculated in closed form, its percentage points are obtained from a general formula, first derived in [11], yielding
[delta][s.sub.m,[alpha]] = [([[chi].sup.2.sub.v]).sup.-1] (m/[m + (N - m - 1) [f.sub.2(N-m-1),2m;1-[alpha]]]), (11)
with [f.sub.a,b;[alpha]] the [alpha]-percentage point of the Fisher distribution with parameters (a, b). Equation (11) holds with decent approximation, as confirmed by numerical simulations.
2.3.3. Extension of the Forward Search to CoDA. When [D.sub.X] is a compositional dataset, it is unnatural to assume that its elements are realizations of a multivariate X ~ N([[mu].sub.0], [[SIGMA].sub.0]).
In this case the use of the FSE as outlined above does not make sense at all. Sometimes it is reasonable to assume X ~ [L.sub.v]([[mu].sub.0], [[SIGMA].sub.0]), as first argued in [6]. In this case we can use the FSA to find outliers, provided that we first modify the algorithm in two respects:
(i) we replace the Mahalanobis distance by the Aitchison one;
(ii) we perform statistical tests consistently with the change of null hypothesis.
Specifically, at the mth step of the algorithm, we sort [D.sub.X] according to the increasing values of the square Aitchison distance:
[d.sub.A] [(x).sup.2] = 1/2v [v.summation over (i,k=1)] [[log ([x.sub.i]/[x.sub.k]) - log ([([c.sub.m-1]).sub.i]/[([c.sub.m-1]).sub.k])].sup.2], (12)
where [c.sub.m-1] = cen[y | y [member of] S(m - 1)] is the center of S(m - 1). Analogously, given the mth element [x.sup.(m)] of the newly sorted dataset, we regard [s.sub.m] = [d.sub.A][([x.sup.(m)]).sup.2] as the mth component of the signal. The percentage points of sm are obtained from ilr(S(m)) by using the probability correspondence established in (9). Since ilr(S(m)) is a spherically truncated dataset, the estimate of the covariance matrix [SIGMA](m) derived from it must undergo the reconstruction procedure described in next sections.
2.4. General Covariance Reconstruction Problem. The examples discussed in the previous subsections are special cases of a more general inverse problem, namely, the reconstruction of the covariance matrix [SIGMA] of a normal multivariate X on the basis of the covariance matrix [G.sub.D] of X truncated to some (convex) region D. This is the simplest yet nontrivial inverse problem, which can be naturally associated with the normal distribution. The case D = [B.sub.v]([rho]) corresponds to a setup where theoretical and practical aspects of the problem can be investigated with relatively modest mathematical effort.
It is certainly a well-defined framework where to study regularization techniques for nonlinear inverse problems in statistics, for which there is still much room for interesting work [12, 13].
3. Definitions and Setup
Let X [member of] [R.sup.v] be a random vector with jointly normal distribution [N.sub.v](0, [SIGMA]) in v [greater than or equal to] 1 dimensions. The probability that X falls within [B.sub.v]([rho]) is measured by the Gaussian integral
[mathematical expression not reproducible]. (13)
Since [SIGMA] is symmetric positive definite, it has orthonormal eigenvectors [SIGMA][v.sup.(i)] = [[lambda].sub.i][v.sup.(i)]. Let us denote by R [equivalent to] [{[v.sup.(j).sub.i]}.sup.v.sub.i,j=1] the orthogonal matrix having these vectors as columns and by [LAMBDA] [equivalent to] diag([lambda]) = [R.sup.T][SIGMA]R the diagonal counterpart of [SIGMA]. From the invariance of [B.sub.v]([rho]) under rotations, it follows that [alpha] depends upon [SIGMA] just by way of [lambda]. Accordingly, we rename the Gaussian probability content of [B.sub.v]([rho]) as
[mathematical expression not reproducible]. (14)
Note that (14) is not sufficient to fully characterize the random vector X under the spherical constraint, for which we need to calculate the distribution law P[X [member of] A | X [member of] [B.sub.v]([rho])] as a function of A [subset] [R.sup.v]. Alternatively, we can describe X in terms of the complete set of its truncated moments
[mathematical expression not reproducible]. (15)
As usual, these can be all obtained from the moment-generating function
[mathematical expression not reproducible], (16)
by differentiating the latter an arbitrary number of times with respect to the components of t; namely,
[mathematical expression not reproducible]. (17)
It will be observed that m(t) is in general not invariant under rotations of t. Therefore, unlike [alpha], the moments [mathematical expression not reproducible] depend effectively on both [lambda] and R. For instance, for the matrix of the second moments [G.sub.B] [equivalent to] [{[[partial derivative].sup.2]m/[partial derivative][t.sub.i] [partial derivative][t.sub.j][|.sub.t=0]}.sup.v.sub.i,j=1] such dependence amounts to
[mathematical expression not reproducible]. (18)
By parity, the only nonvanishing terms in the above sum are those with k = l. Hence, it follows that [SIGMA] and [G.sub.B] share R as a common diagonalizing matrix. In other words, if M [equivalent to] diag([mu]) is the diagonal matrix of the eigenvalues of [G.sub.B], then M = [R.sup.T][G.sub.B]R. Moreover, [[mu].sub.k] is related to [[lambda].sub.k] by
[mathematical expression not reproducible]. (19)
The ratios [[alpha].sub.k]/[alpha] are naturally interpreted as adimensional correction factors to the eigenvalues of [SIGMA], so they play the same role as [c.sub.T]([rho]) in (2). However, [[alpha].sub.k]/[alpha] depends explicitly on the subscript k; thus each eigenvalue is damped differently from the others as a consequence of the condition X [member of] [B.sub.v]([rho]).
Remark 1. In practical terms, (18)-(19) tell us that estimating the sample covariance matrix of X ~ [N.sub.v](0, [SIGMA]) from a spherically truncated population [{[x.sup.(m)]}.sup.M.sub.m=1], made of M independent units, via the classical estimator [Q.sub.ij] = [(M - 1).sup.-1] [[summation].sup.M.sub.m=1]([x.sup.(m).sub.i] - [[??].sup.(m).sub.i])([x.sup.(m).sub.j - [[??].sup.(m).sub.j), being [??] = [M.sup.-1] [[summation].sup.M.sub.m=1] [x.sup.(m)] the sample mean, yields a damped result. Nonetheless, the damping affects only the eigenvalues of the estimator, whereas its eigenvectors are left invariant.
3.1. Monotonicity Properties of Ratios of Gaussian Integrals. Equations (14) and (19) suggest introducing a general notation for the Gaussian integrals over [B.sub.v]([rho]), under the assumption [SIGMA] = [LAMBDA]. So, we define
[mathematical expression not reproducible], (20)
with each subscript q on the left-hand side addressing an additional factor of [x.sup.2.sub.q]/[[lambda].sub.q] under the integral sign on the right-hand side (no subscripts means [alpha]). Several analytic properties of such integrals are discussed in [14]. In the following proposition, we lay emphasis on some issues concerning the monotonicity trends of the ratios [[alpha].sub.k]/[alpha].
Proposition 2 (monotonicities). Let [[lambda].sub.(k)] [equivalent to] [{[[lambda].sub.i]}.sup.i[not equal to]k.sub.i=1, ..., v] denote the set of the full eigenvalues without [[lambda].sub.k]. The ratios [[alpha].sub.k]/[alpha] fulfill the following properties:
([p.sub.1]) [[lambda].sub.k]([[alpha].sub.k]/[alpha])([rho]; [lambda]) is a monotonic increasing function of [[lambda].sub.k] at fixed [rho] and [[lambda].sub.(k)];
([p.sub.2]) ([[alpha].sub.k]/[alpha])([rho]; [lambda]) is a monotonic decreasing function of [[lambda].sub.k] at fixed [rho] and [[lambda].sub.(k)];
([p.sub.3]) ([[alpha].sub.k]/[alpha])([rho]; [lambda]) is a monotonic decreasing function of [[lambda].sub.i] (i [not equal to] k) at fixed [rho] and [[lambda].sub.(i)],
where an innocuous abuse of notation has been made on writing ([[alpha].sub.k]/[alpha])([rho]; [lambda]) in place of [[alpha].sub.k]([rho]; [lambda])/[alpha]([rho]; [lambda]).
Proof. Let the symbol [[partial derivative].sub.k] [equivalent to] [partial derivative]/[partial derivative][[lambda].sub.k] denote a derivative with respect to [[lambda].sub.k]. In order to prove property ([p.sub.1]), we apply the chain rule of differentiation to [[lambda].sub.k][[alpha].sub.k]/[alpha] and then we pass [[partial derivative].sub.k] under the integral sign in [[partial derivative].sub.k][alpha] and [[partial derivative].sub.k][[alpha].sub.k]. In this way, we obtain
[mathematical expression not reproducible]. (21)
Moreover, since the truncated marginal density of [X.sup.2.sub.k] is positive within a set of nonzero measure in R, the monotonic trend of [[lambda].sub.k][[alpha].sub.k]/[alpha] in [[lambda].sub.k] is strict.
Properties ([p.sub.2]) and ([p.sub.3]) are less trivial than ([p.sub.1]). Indeed, the same reasoning as above now yields on the one hand
[mathematical expression not reproducible], (22)
and on the other
[mathematical expression not reproducible]. (23)
Despite being not a priori evident, the right-hand side of both (22) and (23) is negative (and vanishes in the limit [rho] [right arrow] [infinity]). The inequalities var([X.sup.2.sub.k]) [less than or equal to] 2[[lambda].sub.k]E[[X.sup.2.sub.k]] within Euclidean balls have been first discussed in [14], while the inequalities cov([X.sup.2.sub.j], [X.sup.2.sub.k]) within generalized Orlicz balls have been discussed in [15, 16] for the case where the probability distribution of X is flat instead of being normal. More recently, a complete proof of both inequalities has been given in [17]. Despite the technical difficulties in proving them, their meaning should be intuitively clear. The variance inequality quantifies the squeezing affecting [X.sup.2.sub.k] as a consequence of the truncation (in the unconstrained case it would be var([X.sup.2.sub.k]) = 2[[lambda].sup.2.sub.k]). The covariance inequality follows from the opposition arising among the square components in proximity of the boundary of [B.sub.v]([rho]). Indeed, if [X.sub.2.sub.j] [??] [rho], then [X.sup.2.sub.k] [??] 0 [for all]k [not equal to] j in order for X to stay within [B.sub.v]([rho]).
3.2. Definition Domain of the Reconstruction Problem. A consequence of Proposition 2 is represented by the following.
Corollary 3. Given v, [rho], and [lambda], [[mu].sub.k] is bounded by
[rho]/[r (v, [rho]/2[[lambda].sub.k])] [less than or equal to] [[mu].sub.k] [less than or equal to] [rho]/3, r (v, z) [equivalent to] (2v + 1) [M (v, v + 1/2, z)]/[M (v, v + 3/2, z)], (24)
with M denoting the Kummer function; namely,
M (a, b, z) = [[infinity].summation over (n=0)] 1/n! [(a).sub.n]/[(b).sub.n] [z.sup.n], [(x).sub.n] [equivalent to] [[GAMMA] (x + n)]/[[GAMMA] (x)]. (25)
Proof. The upper bound of (24) corresponds to the value of [[mu].sub.k] in the v-tuple limit [[lambda].sub.k] [right arrow] [infinity], [[lambda].sub.(k)] [right arrow] {0, ..., 0}. This is indeed the maximum possible value allowed for [[mu].sub.k] according to properties ([p.sub.1]) and ([p.sub.3]) of Proposition 2. In order to perform this limit, we observe that
[mathematical expression not reproducible], (26)
with the [delta] symbol on the right-hand side representing the Dirac delta function (the reader who is not familiar with the theory of distributions may refer, for instance, to [18] for an introduction). Accordingly,
[mathematical expression not reproducible]. (27)
The lower bound corresponds instead to the value taken by [[mu].sub.k] as [[lambda].sub.(k)] [right arrow] {[infinity], ...,[infinity]} and [[lambda].sub.k] is kept fixed. In this limit, all the Gaussian factors in the probability density function except the kth one flatten to one. Hence,
[mathematical expression not reproducible]. (28)
Numerator and denominator of the rightmost ratio are easily recognized to be integral representations of Kummer functions (see, e.g., [19, ch. 13]).
The upper bound of (24) can be sharpened, as clarified by the following.
Proposition 4 (bounds on the truncated moments). Let v, [rho], and [lambda] be given. If {[i.sub.1], ..., [i.sub.v]} is a permutation of {1, ..., v} such that [mathematical expression not reproducible], then the following upper bounds hold:
[mathematical expression not reproducible]. (29)
Proof. The overall upper bound on the sum of truncated moments follows from
[mathematical expression not reproducible]. (30)
At the same time, the sum can be split and bounded from below by
[mathematical expression not reproducible]. (31)
The single upper bounds on the [[mu].sub.k]'s are then obtained from (30)-(31). It will be noted that (29) (ii) is sharper than the upper bound of (24) only for v > 3 and k < v - 2.
From now on, we shall assume, with no loss of generality, that the eigenvalues of [SIGMA] are increasingly ordered, namely, 0 < [[lambda].sub.1] [less than or equal to] ... [less than or equal to] [[lambda].sub.v] (we can always permute the labels of the coordinate axes, so as to let this be the case). An important aspect related to the eigenvalue ordering is provided by the following.
Proposition 5 (eigenvalue ordering). Let v, [rho], and [lambda] be given. If [[lambda].sub.1] [less than or equal to] [[lambda].sub.2] [less than or equal to] ... [less than or equal to] [[lambda].sub.v], then [[mu].sub.1] [less than or equal to] [[mu].sub.2] [less than or equal to] ... [less than or equal to] [[mu].sub.v] holds as well.
Proof. In order to show that the spherical truncation does not violate the eigenvalue ordering, we make repeated use of the monotonicity properties of Proposition 2. Specifically, if i < j, then
[mathematical expression not reproducible], (32)
where the symbol "[??]" is used to explain where the inequality sign preceding it comes from and the "exchange symmetry" refers to the formal property of the one-index Gaussian integrals over [B.sub.v]([rho]) to fulfill [[alpha].sub.i]([rho]; {[[lambda].sub.1], ..., [[lambda].sub.i], ..., [[lambda].sub.j], ..., [[lambda].sub.v]}) = [[alpha].sub.j]([rho]; {[[lambda].sub.1], ..., [[lambda].sub.j], ..., [[lambda].sub.i], ..., [[lambda].sub.v]}).
Let us now focus on (19). They have to be solved in order to reconstruct [lambda] from [mu]. Formally, if we introduce a family of truncation operators [[tau].sub.[rho]] : [R.sup.v.sub.+ [right arrow] [R.sup.v.sub.+ (parametrically depending on [rho]), such that
[([[tau].sub.[rho]] x [lambda]).sub.k] [equivalent to] [[lambda].sub.k] [[alpha].sub.k]/[alpha] ([rho]; [lambda]), k = 1, ..., v, (33)
then the reconstruction of [lambda] from [mu] amounts to calculating [lambda] = [[tau].sup.-1.sub.[rho]] x [mu]. One should be aware that [[tau].sub.[rho]] is not a surjective operator in view of Corollary 3 and Proposition 4. Therefore, [[tau].sup.-1.sub.[rho]] is only defined within a bounded domain D([[tau].sup.- 1.sub.[rho]]). If we define
[D.sub.0] = {[mu] [member of] [R.sup.v.sub.+] : [[mu].sub.1] [less than or equal to] ... [less than or equal to] [[mu].sub.v], [[mu].sub.k] = [[lambda].sub.k] [[[alpha].sub.k]/[alpha]] for k = 1, ..., v, for some [lambda] [member of] [R.sup.v.sub.+]}, (34)
then we have D([[tau].sup.-1.sub.[rho]]) = {[mu] : [mu] = [sigma] x [[mu].sub.0] for some [[mu].sub.0] [member of] [D.sub.0], [sigma] [member of] [S.sub.v]}, where [S.sub.v] is the set of permutations of v elements. From Proposition 4 we conclude that [D.sub.0] [subset or equal to] [H.sub.v]([rho]), being
[H.sub.v] ([rho]) [equivalent to] {x [member of] [R.sup.v.sub.+] : [x.sub.k] [less than or equal to] min {[rho]/3, [rho]/[v - k + 1]}, [v.summation over (k=1)] [x.sub.k] [less than or equal to] [rho], [for all]k}. (35)
In fact, there are vectors [mu] [member of] [R.sup.v.sub.+] fulfilling [mu] [member of] [H.sub.v]([rho]) and [mu] [not member of] [D.sub.0]; thus we conclude that [D.sub.0] is a proper subset of [H.sub.v]([rho]). Numerical experiences based on the techniques discussed in the next sections show indeed that
[mathematical expression not reproducible]. (36)
A graphical representation of (36) in v = 2 and v = 3 dimensions is depicted in Figure 2. The reader should note that until Section 7 we shall always assume that [mu] comes from the application of [[tau].sub.[rho]] to some [lambda]; thus [mu] [member of] D([[tau].sup.-1.sub.[rho]]) by construction.
Now, we observe that (19) can be written in the equivalent form
[lambda] = T ([lambda]; [mu]; [rho]), (37)
T : [R.sup.v.sub.+] x [R.sup.v.sub.+] x [R.sub.+] [right arrow] [R.sup.v.sub.+]; [T.sub.k] ([lambda]; [mu]; [rho]) = [[mu].sub.k] [[alpha]/[[alpha].sub.k]] ([rho]; [lambda]), k = 1, ..., v. (38)
Since [rho] and [mu] are (nonindependent) input parameters for the covariance reconstruction problem (and in order to keep the notation light), in the sequel we shall leave the dependence of T upon [rho] and [mu] implicitly understood; that is, we shall write (37) as [lambda] = T([lambda]). Hence, we see that the full eigenvalue spectrum [lambda] is a fixed point of the operator T. This suggests obtaining it as the limit of a sequence
[[lambda].sup.(0)] = [mu],
[[lambda].sup.(n+1)] = T ([[lambda].sup.(n)]), n = 0, 1, ..., (39)
[mathematical expression not reproducible], (40)
provided that this can be shown to converge. Note that since [[alpha].sub.k] < [alpha], it follows that [T.sub.k]([[lambda].sup.(n)]) > [[mu].sub.k] [for all]n, so the sequence is bounded from below by [mu]. In particular, this holds for n = 0. Therefore, the sequence moves to the right direction at least at the beginning. A formal proof of convergence, based on the monotonicity properties stated by Proposition 2, is given in the next section.
4. Convergence of the Fixed Point Equation
We split our argument into three propositions, describing different properties of the sequence [[lambda].sup.(n)]. They assert, respectively, that (i) the sequence is componentwise monotically increasing; (ii) the sequence is componentwise bounded from above by any fixed point of T; and (iii) if T has a fixed point, this must be unique. Statements (i) and (ii) are sufficient to guarantee the convergence of the sequence to a finite limit (the unconstrained spectrum is a fixed point of T). In addition, the limit is easily recognized to be a fixed point of T. Hence, statement (iii) guarantees that the sequence converges to the unconstrained eigenvalue spectrum. We remark that all the monotonicities discussed in Proposition 2 are strict; that is, the ratios [[alpha].sub.k]/[alpha] have no stationary points at finite [rho] and [lambda], which is crucial for the proof.
Proposition 6 (increasing monotonicity). Given v, [rho], and [mu] [member of] D([[tau].sup.-1.sub.[rho]]), the sequence [[lambda].sup.(0)] = [mu], [[lambda].sup.(n+1)] = T([[lambda].sup.(n)]), n = 0, 1, ..., is monotically increasing; namely, [[lambda].sup.(n+1).sub.k] > [[lambda].sup.(n).sub.k] [for all]k = 1, ..., v.
Proof. The proof is by induction. We first notice that
[[lambda].sup.(1).sub.k] = [T.sub.k] ([[lambda].sup.(0)]) = [T.sub.k] ([mu]) = [[mu].sub.k] [alpha]/[[alpha].sub.k] ([rho]; [mu]) > [[mu].sub.k] = [[lambda].sup.(0).sub.k], k = 1, ..., v; (41)
the inequality follows from [[alpha].sub.k]([rho]; [mu]) < [alpha]([rho]; [mu]). Suppose now that the property of increasing monotonicity has been checked off up to the nth element of the sequence. Then,
[[lambda].sup.(n+1).sub.k] = [[mu].sub.k] [[alpha]/[[alpha].sub.k]] ([rho]; [[lambda].sup.(n)]) > [[mu].sub.k] [[alpha]/[[alpha].sub.k]] ([rho]; [[lambda].sup.(n-1)]) = [[lambda].sup.(n).sub.k]; (42)
the inequality follows this time from the inductive hypothesis and from properties ([p.sub.2]) and ([p.sub.3]) of Proposition 2.
Proposition 7 (boundedness). Given v, [rho], and [mu] [member of] D([[tau].sup.- 1.sub.[rho]]), the sequence [[lambda].sup.(0)] = [mu], [[lambda].sup.(n+1)] = T([[lambda].sup.(n)]), n = 0, 1, ..., is bounded from above; namely, [[lambda].sup.(n).sub.k] < [[lambda].sup.*.sub.k] [for all]k = 1, ..., v, [[lambda].sup.*] being a fixed point of T.
Proof. We proceed again by induction. We first notice that
[[lambda].sup.(0).sub.k] = [[mu].sub.k] < [[mu].sub.k] [alpha]/[[alpha].sub.k] ([rho]; [[lambda].sup.*]) = [[lambda].sup.*.sub.k], k = 1, ..., v; (43)
the inequality follows as previously from [[alpha].sub.k]([rho]; [[lambda].sup.*]) < [alpha]([rho]; [[lambda].sup.*]). Suppose now that the property of boundedness has been checked off up to the nth element of the sequence. Then,
[mathematical expression not reproducible]; (44)
the inequality follows for the last time from the inductive hypothesis and from properties ([p.sub.2]) and ([p.sub.3]) of Proposition 2.
According to Propositions 6 and 7, the sequence converges. Now, let [??] = [lim.sub.n[right arrow][infinity]][[lambda].sup.(n)] be the limit of the sequence. Effortlessly, we prove that [??] is a fixed point of T. Indeed,
[mathematical expression not reproducible]. (45)
Note that passing the limit over n under the integral sign is certainly allowed for Gaussian integrals.
Proposition 8 (uniqueness of the fixed point). Let [lambda]' = T([lambda]') and [lambda]" = T([lambda]") be two fixed points of T, corresponding to the same choice of v, [rho], and [mu] [member of] D([[tau].sup.-1.sub.[rho]]). Then, it must be that [lambda]' = [lambda]".
Proof. According to the hypothesis, [lambda]' and [lambda]" fulfill the equations
[mathematical expression not reproducible]. (46)
Hence,
[mathematical expression not reproducible], (47)
where J denotes the Jacobian matrix of [[tau].sub.[rho]] and is given by
[mathematical expression not reproducible], (48)
having set [[OMEGA].sub.kl] [equivalent to] (1/2)([[alpha].sub.kl]/[alpha] - [[alpha].sub.k][[alpha].sub.l]/[[alpha].sup.2]). It will be noted that [OMEGA] = [{[[OMEGA].sub.kl]}.sup.v.sub.k,l=1] is essentially the covariance matrix of the square components of X under spherical truncation (we have come across its matrix elements in (21)-(23)). As such, [OMEGA] is symmetric positive definite. Indeed,
[mathematical expression not reproducible]. (49)
On setting [Z.sub.k] = ([X.sup.2.sub.k] - E[[X.sup.2.sub.k] | X [member of] [B.sub.v]([rho])])/[square root of (2)][[lambda].sub.k], we can represent [OMEGA] as [OMEGA] = E[Z[Z.sup.T] | X [member of] [B.sub.v]([rho])]. If x [member of] [R.sup.v] is not the null vector, then [x.sup.T][OMEGA]x = E[[x.sup.T]Z[Z.sup.T]x | X [member of] [B.sub.v]([rho])] = E[[([x.sup.T]Z).sup.2] | X [member of] [B.sub.v]([rho])] > 0. Moreover, the eigenvalues of [OMEGA] fulfill the secular equation
0 = det([OMEGA] - [phi][I.sub.v]) = det [[[LAMBDA].sup.-1] ([OMEGA] - [phi][I.sub.v]) [LAMBDA]] = det ([[LAMBDA].sup.-1] [OMEGA][LAMBDA] - [phi][I.sub.v]) = det(J - [phi][I.sub.v]), (50)
whence it follows that J is positive definite as well (though it is not symmetric). Since the sum of positive definite matrices is positive definite, we conclude that [[integral].sup.1.sub.0 dt J([rho]; [lambda]" + t([lambda]' - [lambda]")) is positive definite too. As such, it is nonsingular. Therefore, from (47), we conclude that [lambda]' = [lambda]".
5. Numerical Computation of Gaussian Integrals over [B.sub.v]([rho])
Let us now see how to compute [[alpha].sub.klm...] with controlled precision. Most of the relevant work has been originally done by Ruben in [2], where the case of [alpha] is discussed. We extend Ruben's technique to Gaussian integrals containing powers of the integration variable. Specifically, it is shown in [2] that [alpha]([rho]; [lambda]) can be represented as a series of chi-square cumulative distribution functions:
[alpha] ([rho]; [lambda]) = [[infinity].summation over (m=0)] [c.sub.m] (s; [lambda]) [F.sub.v+2m] ([rho]/s). (51)
The scale factor s has the same physical dimension as [rho] and [lambda]. It is introduced in order to factorize the dependence of [alpha] upon [rho] and [lambda] at each order of the expansion. The series on the right-hand side of (51) converges uniformly on every finite interval of [rho]. The coefficients [c.sub.m] are given by
[c.sub.m] (s; [lambda]) = [1/m!] [[s.sup.v/2+m]/[[absolute value of ([LAMBDA])].sup.1/2]] [[GAMMA] (v/2 + m)]/[[GAMMA] (v/2)] M [[(-Q).sup.m]], m = 0, 1, ..., (52)
having defined Q(x) [equivalent to] [x.sup.T][[[LAMBDA].sup.-1] - [s.sup.- 1][I.sub.v]]x for x [member of] [R.sup.v] and M as the uniform average operator on the (v - 1)-sphere [partial derivative][B.sub.v](1) [equivalent to] {u [member of] [R.sup.v] : [u.sup.T]u = 1}; namely,
[mathematical expression not reproducible]. (53)
Unfortunately, (52) is not particularly convenient for numerical computations, since M[[(-Q).sup.m]] is only given in integral form. However, it is also shown in [2] that the coefficients [c.sub.m] can be extracted from the Taylor expansion (at [z.sub.0] = 0) of the generating function
[mathematical expression not reproducible]. (54)
This series converges uniformly for [absolute value of (z)] < [min.sub.i][[absolute value of (1 - s/[[lambda].sub.i])].sup.-1]. On evaluating the derivatives of [psi](z), it is then shown that [c.sub.m]'s fulfill the recursion:
[mathematical expression not reproducible]. (55)
Finally, the systematic error produced on considering only the lowest k terms of the chi-square series of (51) is estimated by
[mathematical expression not reproducible], (56)
with [eta] = [max.sub.i][absolute value of (1 - s/[[lambda].sub.i])].
Now, as mentioned, it is possible to extend the above expansion to all Gaussian integrals [[alpha].sub.klm...]. Here, we are interested only in [[alpha].sub.k] and [[alpha].sub.jk], since these are needed in order to implement the fixed point iteration and to compute the Jacobian matrix of [[tau].sub.[rho]]. The extension is provided by the following.
Theorem 9 (Ruben's expansions). The integrals [[alpha].sub.k] and [[alpha].sub.jk] admit the series representations:
[[alpha].sub.k] ([rho]; [lambda]) = [[infinity].summation over (m=0)] [c.sub.k;m] (s; [lambda]) [F.sub.v+2(m+1)] ([rho]/s), (57)
[[alpha].sub.jk] ([rho]; [lambda]) = [[infinity].summation over (m=0)] [c.sub.jk;m] (s; [lambda]) [F.sub.v+2(m+2)] ([rho]/s), (58)
with s being an arbitrary positive constant. The series coefficients are given, respectively, by
[mathematical expression not reproducible], (59)
[mathematical expression not reproducible], (60)
with [[delta].sub.jk] denoting the Kronecker symbol. The series on the right-hand side of (57)-(58) converge uniformly on every finite interval of [rho]. The functions
[mathematical expression not reproducible], (61)
[mathematical expression not reproducible], (62)
[mathematical expression not reproducible] (63)
are generating functions, respectively, for the coefficients [c.sub.k;m], [c.sub.kk;m] and [c.sub.jk;m] (j [not equal to] k); that is, they fulfill
[[psi].sub.k] (z) = [[infinity].summation over (m=0)][c.sub.k;m] (s; [lambda]) [z.sup.m], [[psi].sub.jk] (z) = [[infinity].summation over (m=0)][c.sub.jk;m] (s; [lambda]) [z.sub.m], (64)
for [absolute value of (z)] < [min.sub.i] [[absolute value of (1 - s/[[lambda].sub.i])].sup.-1]. Finally, the coefficients [c.sub.k;m], [c.sub.kk;m], and [c.sub.jk;m] (j [not equal to] k) can be obtained iteratively from the recursions
[mathematical expression not reproducible]; (65)
[mathematical expression not reproducible], (66)
where the auxiliary coefficients [e.sub.k;i] and [e.sub.jk;i] are defined by
[mathematical expression not reproducible], (67)
[mathematical expression not reproducible]. (68)
It is not difficult to further generalize this theorem, so as to provide a chi-square expansion for any Gaussian integral [[alpha].sub.klm...]. The proof follows closely the original one given by Ruben. We reproduce it in the appendix for [[alpha].sub.k], just to highlight the differences arising when the Gaussian integral contains powers of the integration variable.
Analogously to (56), it is possible to estimate the systematic error produced when considering only the lowest k terms of the chi-square series of [[alpha].sub.k] and [[alpha].sub.jk]. Specifically, we find that
[mathematical expression not reproducible]. (69)
In order to evaluate all Ruben series with controlled uncertainty, we first set (see once more [2] for an exhaustive discussion on how to choose s) s = 2[[lambda].sub.1][[lambda].sub.v]/([[lambda].sub.1] + [[lambda].sub.v]); then we choose a unique threshold [epsilon] representing the maximum tolerable systematic error; for example, [[epsilon].sub.dp] = 1.0 x [10.sup.-14] (roughly corresponding to double floating-point precision), for all [alpha], [[alpha].sub.k], and [[alpha].sub.jk], and finally for each [[alpha].sub.X] we compute the integer
and finally for each [[alpha].sub.X] we compute the integer
[mathematical expression not reproducible], (70)
providing the minimum number of chi-square terms, for which the upper bound [[Real part].sub.X;n] to the residual sum [[Real part].sub.X;n] lies below [epsilon]. Of course, this procedure overshoots the minimum number of terms really required for the R's to lie below [epsilon], since we actually operate on the R's instead of the R's. Nevertheless, the computational overhead is acceptable, as it will be shown in the next section. For the sake of completeness, it must be said that typically the values of kth for [alpha], [[alpha].sub.k], and [[alpha].sub.jk] with the same [epsilon] (and [rho], [lambda]) are not much different from each other.
To conclude, we notice that [k.sub.th] depends nontrivially upon [lambda]. By contrast, since [F.sub.v](x) is monotically increasing in x, we clearly see that kth is monotically increasing in [rho]. Now, should one evaluate [alpha] and the like for a given [lambda] at several values of [rho], say [[rho].sub.1] [less than or equal to] [[rho].sub.2] [less than or equal to] ... [less than or equal to] [[rho].sub.max], it is advisable to save computing resources and work out Ruben coefficients just once, up to the order [k.sub.th] corresponding to [[rho].sub.max], since [k.sub.th]([[rho].sub.1]) [less than or equal to] ... [less than or equal to] [k.sub.th] ([[rho].sub.max]). We made use of this trick throughout our numerical experiences, as reported in the sequel.
6. Numerical Analysis of the Reconstruction Process
The fixed point (39) represents the simplest iterative scheme that can be used in order to reconstruct the solution [lambda] = [[tau].sup.-1.sub.[rho]] x [mu]. In the literature of numerical methods, this scheme is known as a nonlinear Gauss-Jacobi (GJ) iteration (see, e.g., [20]). Accordingly, we shall rewrite it as [[lambda].sup.(n+1).sub.GJ,k] = [T.sub.k]([[lambda].sup.(n).sub.GJ]). As we have seen, the sequence [[lambda].sup.(n).sub.GJ] converges with no exception as n [right arrow] [infinity], provided [mu] [member of] D([[tau].sup.- 1.sub.[rho]]). Given [[epsilon].sub.T] > 0, the number of steps [n.sub.it] needed for an approximate convergence with relative precision [[epsilon].sub.T], that is,
[mathematical expression not reproducible], (71)
depends not only upon [epsilon]T, but also on [rho] and [mu] (note that the stopping rule is well conditioned, since [[parallel][[lambda].sup.(n)][parallel].sub.[infinity]] > 0 [for all]n and also [lim.sub.n[right arrow][infinity]] [[parallel][[lambda].sup.(n)][parallel].sub.[infinity]] > 0). In order to characterize statistically the convergence rate of the reconstruction process, we must integrate out the fluctuations of [n.sub.it] due to changes of [mu]; that is, we must average [n.sub.it] by letting [mu] fluctuate across its own probability space. In this way, we obtain the quantity [[bar.n].sub.it] [equivalent to] [E.sub.[mu]][[n.sub.it] | [[epsilon].sub.T], [rho]], which better synthesizes the cost of the reconstruction for given [[epsilon].sub.T] and [rho]. It should be evident that carrying out this idea analytically is hard, for on the one hand [n.sub.it] depends upon [mu] nonlinearly and on the other hand [mu] has a complicated distribution, as we briefly explain below.
6.1. Choice of the Eigenvalue Ensemble. Since [lambda] is the eigenvalue spectrum of a full covariance matrix, it is reasonable to assume its distribution to be a Wishart [W.sub.v](p, [[SIGMA].sub.0]) for some scale matrix [[SIGMA].sub.0] and for some number of degrees of freedom p [greater than or equal to] v. In the sequel, we shall make the ideal assumption [[SIGMA].sub.0] = [p.sup.-1] x [I.sub.v], so that the probability measure of [lambda] is (see, e.g., [21])
[mathematical expression not reproducible]. (72)
Under this assumption, the probability measure of [mu] is obtained by performing the change of variable [lambda] = [[tau].sup.- 1.sub.[rho]] x [mu] in (72). Unfortunately, we have no analytic representation of [[tau].sup.-1.sub.[rho]]. Thus, we have neither an expression for the distribution of [mu]. However, [mu] can be extracted numerically as follows:
(i) generate randomly [SIGMA] ~ [W.sub.v](p, [p.sup.-1] x [I.sub.v]) by means of the Bartlett decomposition [22];
(ii) take the ordered eigenvalue spectrum [lambda] of [SIGMA];
(iii) obtain [mu] by applying the truncation operator [[tau].sub.[rho]] to [lambda].
Note that since [W.sub.v](p, [p.sup.-1] x [I.sub.v]) is only defined for p [greater than or equal to] v, we need to rescale p as v increases. The simplest choice is to keep the ratio p/v fixed. The larger this ratio, the closer [SIGMA] fluctuates around [I.sub.v] (recall that if [SIGMA] ~ [W.sub.v](p, [p.sup.-1] x [I.sub.v]), then E[[[SIGMA].sub.ij]] = [[delta].sub.ij] and var([[SIGMA].sub.ij]) = [p.sup.- 1] [1 + [[delta].sub.ij]]). In view of this, large values of p/v are to be avoided, since they reduce the probability of testing the fixed point iteration on eigenvalue spectra characterized by large condition numbers [n.sub.cond] [equivalent to] [[lambda].sub.v]/[[lambda].sub.1]. For this reason, we have set p = 2v in our numerical study.
Having specified an ensemble of matrices from which the eigenvalue spectra are extracted, we are now ready to perform numerical simulations. To begin with, we report in Figure 3 the marginal probability density function of the ordered eigenvalues [{[[lambda].sub.k]}.sup.v.sub.k=1] and their truncated counterparts [{[[mu].sub.k]}.sup.v.sub.k=1] for the Wishart ensemble [W.sub.10](20, [20.sup.-1] x [I.sub.10]) at [rho] = 1, as obtained numerically from a rather large sample of matrices ([equivalent][10.sup.6] units). It will be noted that (i) the effect of the truncation is severe on the largest eigenvalues, as a consequence of the analytic bounds of Corollary 2.1 and Proposition 2.2; (ii) while the skewness of the lowest truncated eigenvalues is negative, it becomes positive for the largest ones. This is due to a change of relative effectiveness of (29) (i) with respect to (29) (ii).
6.2. Choice of the Simulation Parameters. In order to explore the dependence of [[bar.n].sub.it] upon [rho], we need to choose one or more simulation points for the latter. Ideally, it is possible to identify three different regimes in our problem: [rho] [??] [[lambda].sub.1] (strong truncation regime), [[lambda].sub.1] [??] [rho] [??] [[lambda].sub.v] (crossover), and [rho] [??] [[lambda].sub.v] (weak truncation regime). We cover all of them with the following set of points:
[rho] [member of] {Mo ([[lambda].sub.1]), ..., Mo ([[lambda].sub.v])} [union] {[1/2] Mo ([[lambda].sub.1]), 2Mo ([[lambda].sub.v])}, (73)
where Mo(x) stands for the mode. In principle, it is possible to determine Mo([[lambda].sub.k]) with high accuracy by using analytic representations of the marginal probability densities of the ordered eigenvalues [24]. In practice, the latter become computationally demanding at increasingly large values of v: for instance, the determination of the probability density of [[lambda].sub.2] requires [(v!).sup.2] sums, which is unfeasible even at v ~ 10. Moreover, to our aims, it is sufficient to choose approximate values, provided that these lie not far from the exact ones. Accordingly, we have determined the eigenvalue modes numerically from samples of N [equivalent] [10.sup.6] Wishart matrices. Our estimates are reported in Table 1 for v = 3, ..., 10. They have been obtained from Grenander's estimator [23]:
[mathematical expression not reproducible], (74)
with properly chosen parameters r, s.
We are now in the position to investigate numerically how many terms in Ruben's expansions must be considered as [epsilon] is set to [[epsilon].sub.dp] = 1.0 x [10.sup.-14], for our choice of the eigenvalue ensemble [lambda] ~ [W.sub.v](2v, [(2v).sup.-1] x [I.sub.v]) and with [rho] set as in Table 1. As an example, we report in Figure 4 the discrete distributions of [k.sub.th] for the basic Gaussian integral [alpha] at v = 10, the largest dimension that we have simulated. As expected, we observe an increase of [k.sub.th] with [rho]. Nevertheless, we see that the number of Ruben's components to be taken into account for a double precision result keeps altogether modest even in the weak truncation regime, which proves the practical usefulness of the chi-square expansions.
6.3. Fixed Point Iteration at Work. The GJ iteration is too slow to be of practical interest. For instance, at v = 10, [rho] [equivalent] Mo([[lambda].sub.1]) and [[epsilon].sub.T] = 1.0 x [10.sup.-7] (corresponding to a reconstruction of [lambda] with single floating-point precision), it is rather easy to extract realizations of [mu] which require [n.sub.it] [equivalent] 15,000 to converge. An improvement of the GJ scheme is achieved via overrelaxation (GJOR); that is,
[[lambda].sup.(0).sub.GJOR,k] = [[mu].sub.k], [[lambda].sup.(n+1).sub.GJOR,k] = [[lambda].sup.(n).sub.GJOR,k] + [omega] [[T.sub.k] ([[lambda].sup.(n).sub.GJOR]) - [[lambda].sup.(n).sub.GJOR,k]], k = 1, ..., v. (75)
Evidently, at [omega] = 1, the GJOR scheme coincides with the standard GJ one. The optimal value [[omega].sub.opt] of the relaxation factor [omega] is not obvious even in the linear Jacobi scheme, where [[omega].sub.opt] depends upon the properties of the coefficient matrix of the system. For instance, if the latter is symmetric positive definite, it is demonstrated that the best choice is provided by [[omega].sub.opt] [equivalent to] 2[(1 + [square root of (1 - [[sigma].sup.2])]).sup.-1], [sigma] being the spectral radius of the Jacobi iteration matrix [25]. In our numerical tests with the GJOR scheme, we found empirically that the optimal value of [omega] at [rho] [less than or equal tol] [[lambda].sub.v] is close to the linear prediction, provided that [sigma] is replaced by [[parallel]J[parallel].sub.[infinity]], J being defined as in Section 3 (note that [[parallel]J[parallel].sub.[infinity]] < 1).
By contrast, the iteration diverges after few steps with increasing probability as [rho]/[[lambda].sub.v] [right arrow] [infinity] if [omega] is kept fixed at [omega] = [[omega].sub.opt]; in order to restore the convergence, [omega] must be lowered towards [omega] = 1 as such limit is taken.
To give an idea of the convergence rate of the GJOR scheme, we show in Figure 5(a) a joint box-plot of the distributions of [n.sub.it] at v = 10 and [[epsilon].sub.T] = 1.0 x [10.sup.-7]. From the plot we observe that the distribution of [n.sub.it] shifts rightwards as [rho] decreases: clearly, the reconstruction is faster if [rho] is in the weak truncation regime (where [mu] is closer to [lambda]), whereas it takes more iterations in the strong truncation regime. The dependence of [[bar.n].sub.it] upon [rho], systematically displayed in Figure 6, is compatible with a scaling law
log [[bar.n].sub.it] ([rho], v, [[epsilon].sub.T])= a(v, [[epsilon].sub.T]) - b(v, [[epsilon].sub.T])log [rho], (76)
apart from small corrections occurring at large [rho]. Equation (76) tells us that [[bar.n].sub.it] increases polynomially in 1/[rho] at fixed v. In order to estimate the parameters a and b in the strong truncation regime (where the algorithm becomes challenging), we performed jackknife fits to (76) of data points with [rho] [??] 1. Results are collected in Figure 5(b), showing that b is roughly constant, while a increases almost linearly in v. Thus, while the cost of the eigenvalue reconstruction is only polynomial in 1/[rho] at fixed v, it is exponential in v at fixed [rho]. The scaling law of the GJOR scheme is therefore better represented by [[bar.n].sub.it] = [Ce.sup.[kappa]v]/[[rho].sup.b], with C being a normalization constant independent of [rho] and v and [kappa] representing approximately the slope of a as a function of v. Although the GJOR scheme improves the GJ one, the iteration reveals to be still inefficient in a parameter subspace, which is critical for the applications.
6.4. Boosting the GJOR Scheme. A further improvement can be obtained by letting [omega] depend on the eigenvalue index in the GJOR scheme. Let us discuss how to work out such an adjustment. On commenting on Figure 3, we have already noticed that the largest eigenvalues are affected by the truncation to a larger extent than the smallest ones. Therefore, they must perform a longer run through the fixed point iteration, in order to converge to the untruncated values. This is a possible qualitative explanation for the slowing down of the algorithm as [rho] [right arrow] 0. In view of it, we expect to observe some acceleration of the convergence rate, if [omega] is replaced, for instance, by
[omega] [right arrow] [[omega].sub.k] [equivalent to] (1 + [beta] x k) [[omega].sub.opt], [beta] [greater than or equal to] 0, k = 1, ..., v. (77)
The choice [beta] = 0 corresponds obviously to the standard GJOR scheme. Any other choice yields [[omega].sub.k] > [[omega].sub.opt]. Therefore, the new scheme is also expected to display a higher rate of failures than the GJOR one at [rho] [much greater than] [[lambda].sub.v], for the reason explained in Section 5.3. The componentwise overrelaxation proposed in (77) is only meant to enhance the convergence speed in the strong truncation regime and in the crossover, where the improvement is actually needed.
In order to confirm this picture, we have explored systematically the effect of [beta] on [[bar.n].sub.it] by simulating the reconstruction process at v = 3, ..., 10, with [beta] varying from 0 to 2 in steps of 1/5. First of all, we have observed that the rate of failures at large [rho] is fairly reduced if the first 30 / 50 iterations are run with [[omega].sub.k] = [[omega].sub.opt], and only afterwards [beta] is switched on. Having minimized the failures, we have checked that for each value of [beta], the scaling law assumed in (76) is effectively fulfilled. Then, we have computed jackknife estimates of the scaling parameters a and b. These are plotted in Figure 7 as functions of v. Each trajectory (represented by a dashed curve) corresponds to a given value of [beta]. Those with darker markers refer to smaller values of [beta] and the other way round. From the plots we notice that
(i) all the trajectories with [beta] > 0 lie below the one with [beta] = 0;
(ii) the trajectories of a display a clear increasing trend with v; yet their slope lessens as [beta] increases. By contrast, the trajectories of b develop a mild increasing trend with v as [beta] increases, though this is not strictly monotonic;
(iii) the trajectories of both a and b seem to converge to a limit trajectory as [beta] increases; we observe a saturation phenomenon, which thwarts the benefit of increasing [beta] beyond a certain threshold close to [[beta].sub.max] [equivalent] 2.
We add that pushing [beta] beyond [[beta].sub.max] is counterproductive, as the rate of failures becomes increasingly relevant in the crossover and eventually also in the strong truncation regime. By contrast, if [beta] [less than or equal tol] [[beta].sub.max] the rate of failures keeps very low for essentially all simulated values of [rho].
Our numerical results signal a strong reduction of the slowing down of the convergence rate. Indeed, (i) means qualitatively that C and b are reduced as [beta] increases. (ii) means that [kappa] is reduced as [beta] increases (this is the most important effect, as [kappa] is mainly responsible for the exponential slowing down with v). The appearance of a slope in the trajectories of b as [beta] increases indicates that a mild exponential slowing down is also developed at denominator of the scaling law [[bar.n].sub.it] = [Ce.sup.[kappa]v]/[[rho].sup.b], but the value of b is anyway smaller than at [beta]=0. Finally, (iii) means that choosing [beta] > [[beta].sub.max] has a minor impact on the performance of the algorithm. In Figure 8, we report a plot of the parameter [kappa] (obtained from least-squares fits of data to a linear model a = [a.sub.0] + [kappa] x v) as a function of [beta]. We see that [kappa]([beta] = 0)/[kappa]([beta] = 2) [equivalent] 4. This quantifies the maximum exponential speedup of the convergence rate, which can be achieved by our proposal. When [beta] is close to [[beta].sub.max], [[bar.n].sub.it] amounts to few hundreds at v = 10 and [rho] [equivalent] [[lambda].sub.1]/2.
7. On the Ill-Posedness of the Reconstruction in Sample Space
So far we have discussed the covariance reconstruction under the assumption that [mu]=[[tau].sub.[rho]] x [lambda] represents the exact truncated counterpart of some [lambda] [member of] [R.sup.v] and we have looked at the algorithmic properties of the iteration schemes which operatively define [[tau].sup.-1.sub.[rho]]. Such analysis is essential in order to characterize [[tau].sup.-1.sub.[rho]] mathematically; yet it is not sufficient in real situations, specifically when [mu] is perturbed by statistical noise.
In this section, we examine the difficulties arising when performing the covariance reconstruction in sample space. We first recall that, according to Hadamard [26], a mathematical problem is well posed provided that the following conditions are fulfilled:
([H.sub.1]) there exists always a solution to the problem;
([H.sub.2]) the solution is unique;
([H.sub.3]) the solution depends smoothly on the input data.
Inverse problems are often characterized by violation of one or more of them; see, for instance, [13]. In such cases, the standard practice consists in regularizing the inverse operator, that is, in replacing it by a stable approximation. With regard to our problem, the reader will recognize that ([H.sub.1]) is violated (and the problem becomes ill-posed) as soon as the space of the input data is allowed to be a superset of D([[tau].sup.-1.sub.[rho]]): once clarifying how [mu] is concretely estimated in the applications (Sections 7.1 and 7.2), we propose a perturbative regularization of [[tau].sup.-1.sub.[rho]], which improves effectively the fulfillment of ([H.sub.1]) (Section 7.3). By contrast, Proposition 8 guarantees that whenever a solution exists, it is also unique; thus ([H.sub.2]) is never of concern. Finally, the fulfillment of ([H.sub.3]) depends on how the statistical noise on [mu] is nonlinearly inflated by the action of [[tau].sup.-1.sub.[rho]]. For the sake of conciseness, in the present paper, we just sketch the main ideas underlying the perturbative regularization of [[tau].sup.-1.sub.[rho]], whereas a technical implementation of it and a discussion of (H3) are deferred to a separate paper [27].
7.1. Definition of the Sample Truncated Covariance Matrix. The examples of Section 2 assume that (i) spherical truncations are operated on a representative sample [P.sub.N] = [{[x.sup.(k)]}.sup.N.sub.k=1] of X ~ [N.sub.v](0, [SIGMA]) with finite size N, (ii) [rho] is known exactly, and (iii) the input budget for the covariance reconstruction is given by the subset
[Q.sub.M] = {x [member of] [P.sub.N] : [[parallel]x[parallel].sup.2] < [rho]}, with [absolute value of ([Q.sub.M])] = M [less than or equal to] N. (78)
As usual in the analysis of stochastic variables in sample space, we assume that the observations [x.sup.(k)] are realizations of i.i.d. stochastic variables [X.sup.(k)] ~ [N.sub.v](0, [SIGMA]), k = 1, ..., N. Thus, M is itself a stochastic variable in sample space, where it reads
[mathematical expression not reproducible], (79)
with [mathematical expression not reproducible] denoting the characteristic function of [B.sub.v]([rho]) and [mathematical expression not reproducible] being just a shortcut for its extended counterpart. It is easily recognized that M ~ B(N, [alpha]) is a binomial variate. If we indeed denote by E the sample expectation operator (i.e., the integral with respect to the product measure of the joint variables [{X(k)}.sup.N.sub.k=1]), then a standard calculation yields
[mathematical expression not reproducible]. (80)
Hence, we see that the relative dispersion of M is O([N.sup.-1/2]). Now, the simplest way to measure [SIGMA] and [G.sub.B], respectively, from the sets [P.sub.N] and [Q.sub.M] is via the classical estimators
[mathematical expression not reproducible], (81)
[mathematical expression not reproducible]. (82)
We define the sample estimates [??] and [??], respectively, of [lambda] and [mu] as the eigenvalue spectra of [??] and [[??].sub.B]. By symmetry arguments we see that [[??].sub.i] is unbiased. Indeed, it holds
E [[[??].sub.i]] = [N.summation over (k=1)]E [[X.sup.(k).sub.i] [I.sub.k]/[[summation].sup.N.sub.s=1] [I.sub.s]]. (83)
The right-hand side of (83) makes only sense if we conventionally define the integrand to be zero in the integration subdomain {[X.sup.(k)] [not member of] [B.sub.v]([rho]), [for all]k} or equivalently if we interpret E[[[??].sub.i]] as the conditional one E[[[??].sub.i] | M > 0] (the event M > 0 occurs a.s. only as N [right arrow] [infinity]). Since the sample measure is even under [X.sup.(k)] [right arrow] - [X.sup.(k)] while the integrand is odd, we immediately conclude that bias[[[??].sub.i]] = 0.
7.2. Bias of the Sample Truncated Covariance Matrix. The situation gets somewhat less trivial with [[??].sub.B]: the normalization factor [(M - 1).sup.-1], which has been chosen in analogy with (81), is not sufficient to remove completely the bias of [[??].sub.B] at finite N, though we aim at showing here that the residual bias is exponentially small and asymptotically vanishing. In order to see this, we observe that
[mathematical expression not reproducible], (84)
with Ediag denoting the sample expectation corresponding to a multinormal measure with diagonal covariance matrix [LAMBDA] = diag([lambda]) = [R.sup.T][SIGMA]R, conditioned to M > 1. Having diagonalized the product measure, we observe that the integrand on the right-hand side is odd for l [not equal to] r and even for l = r under the joint change of variables [X.sup.(k).sub.l] [right arrow] - [X.sup.(k).sub.l] for k = 1, ..., N, similarly to what we did in Section 2. As a consequence, it holds
[mathematical expression not reproducible], (85)
whence we infer that the matrix E[[??]B]is diagonalized by the same matrix R as [SIGMA]. From (85) we also conclude that
bias [[[??].sub.B]] = R diag (w) [R.sup.T], [w.sub.i] [equivalent to] [N.summation over (k=1)][E.sub.diag] [[([X.sup.(k).sub.i - [[??].sub.i]).sup.2] [I.sub.k]/[[summation].sup.N.sub.s=1 [I.sub.s] - 1] - [[mu].sub.i], (86)
i = 1, ..., v. It should be observed that in general [w.sub.i] [not equal to] bias[[[??].sub.i]]since the computation of [[??].sub.i] requires the diagonalization of [??]B, which is in general performed by a diagonalizing matrix [??] = R. Nevertheless, if w vanishes then bias[[[??].sub.B]] vanishes too. Now, we observe that [w.sub.i] splits into three contributions:
[mathematical expression not reproducible], (87)
which can be exactly calculated and expressed in terms of [[mu].sub.i], [alpha], and N. For instance,
[mathematical expression not reproducible]. (88)
Analogously, we have
[mathematical expression not reproducible]. (89)
Hence, it follows that
[w.sub.i] = -[[mu].sub.i] [1 + [alpha] (N - 1)][(1 - [alpha]).sup.N-1]. (90)
Since [alpha] > 0, we see that [lim.sub.N[right arrow][infinity]][w.sub.i] = 0. Thus, we conclude that [[??].sub.B] is asymptotically unbiased.
A discussion of the variance of the sample truncated covariance matrix is beyond the scope of the present paper.
We just observe that, apart from the above calculation, studying the sample properties of the truncated spectrum is made hard by the fact that eigenvalues and eigenvectors of a diagonalizable matrix are intimately related from their very definition; thus such study would require a careful consideration of the distribution of the sample diagonalizing matrix [mathematical expression not reproducible].
7.3. Perturbative Regularization of [[tau].sup.-1.sub.[rho]]. When [mu] is critically close to the internal boundary of D([[tau].sup.-1.sub.[rho]]), a sample estimate [??] may fall outside of it due to statistical fluctuations. In that case the iterative procedure described in the previous sections diverges. On the quantitative side, the ill-posedness of the reconstruction problem is measured by the failure probability
[p.sub.fail] ([rho], [SIGMA], N) = P[[??] [not member of] D ([[tau].sup.-1.sub.[rho]]) | [X.sup.(k)] ~ [N.sub.v] (0; [SIGMA]), k = 1, ..., N], (91)
which is a highly nontrivial function of [rho], [SIGMA], and N. An illustrative example of it is reported in Figure 9(a), which refers to a specific case with v = 4 and [SIGMA] = diag(0.1, 0.3, 0.8, 2.2). The plot suggests that the iterative procedure becomes severely illposed in the regime of strong truncation.
In order to regularize the problem, we propose to go back to (19) and consider it from a different perspective. Specifically, we move from the observation that a simplified framework occurs in the special circumstance when the eigenvalue spectra are fully degenerate, which is essentially equivalent to the setup of [1]. If [[mu].sub.1] = ... = [[mu].sub.v] [equivalent to] [??], by symmetry arguments it follows that [[lambda].sub.1] = ... =[[lambda].sub.v] [equivalent to] [??] and the other way round. Equation (19) reduces in this limit to
[mathematical expression not reproducible]. (92)
It can be easily checked that the function [T.sub.[rho]([??])] is monotonically increasing in [??]. In addition, we have
(i) [mathematical expression not reproducible],
(ii) [mathematical expression not reproducible]; (93)
thus (92) can be surely (numerically) inverted provided that 0 < [??] < [rho]/(v + 2). We can regard (92) as an approximation to the original problem (19). When [mu] is not degenerate, we must define [??] in terms of the components of [mu]. One possibility is to average them, that is, to choose
[??] = [1/v] [v.summation over (i=1)] [[mu].sub.i]. (94)
Subject to this, we expect [??] to lie somewhere between [[lambda].sub.1] and [[lambda].sub.v]. Equation (92) can be thought of as the lowest order approximation of a perturbative expansion of (19) around the point [[lambda].sub.T] = {[??], ..., [??]}. If the condition number of [SIGMA] is not extremely large, such an expansion is expected to quickly converge, so that a few perturbative corrections to [[lambda].sub.T] should be sufficient to guarantee a good level of approximation.
As mentioned above, a technical implementation of the perturbative approach and a thorough discussion of its properties are deferred to a separate paper [27]. Here, we limit ourselves to observing that the definition domain of perturbation theory is ultimately set by its lowest order approximation, since corrections to (92) are all algebraically built in terms of it, with no additional constraints. Following (94), the domain of [T.sup.-1.sub.[rho]] comes to be defined as
D ([T.sup.-1.sub.[rho]]) = {[mu] [member of] [R.sup.v.sub.+] : [N.summation over (i=1)] [[mu].sub.i] [less than or equal to] [rho]v/[v + 2]}, (95)
and it is clear that D([[tau].sup.-1.sub.[rho]]) [subset] D([T.sup.- 1.sub.[rho]]) (it is sufficient to sum term by term all the inequalities contributing to (36)). In Figure 10, we show the set difference D([T.sup.-1.sub.[rho]]) \ D([[tau].sup.- 1.sub.[rho]]) in v = 2 and v = 3 dimensions. When [mu] [member of] D([[tau].sup.-1.sub.[rho]]) but its estimate [??] [not member of] D([[tau].sup.-1.sub.[rho]]), it may well occur [??] [member of] D([T.sup.-1.sub.[rho]]); that is, the set difference acts as an absorbing shield of the statistical noise. Therefore, if we define the failure probability of the perturbative reconstruction as
[q.sub.fail] ([rho], [SIGMA], N) = P[[??] [not member of] D ([T.sup.- 1.sub.[rho]]) | [X.sup.(k)] ~ [N.sub.v] (0; [SIGMA]), k = 1, ..., N], (96)
we expect the inequality [q.sub.fail]([rho], [SIGMA], N) [much less than] [p.sub.fail] ([rho], [SIGMA], N) to generously hold. An example is given in Figure 9(b): we see that [q.sub.fail] becomes lower than [p.sub.fail] by orders of magnitude as soon as [rho] and N are not exceedingly small. In this sense, the operator [T.sup.-1.sub.[rho]] can be regarded as the lowest order approximation of a regularizing operator for [[tau].sup.-1.sub.[rho]].
8. Conclusions
In this paper we have studied how to reconstruct the covariance matrix [SIGMA] of a normal multivariate X ~ [N.sub.v](0, [SIGMA]) from the matrix [G.sub.B] of the spherically truncated second moments, describing the covariances among the components of X when the probability density is cut off outside a centered Euclidean ball. We have shown that [SIGMA] and [G.sub.B] share the same eigenvectors. Therefore, the problem amounts to relating the eigenvalues of [SIGMA] to those of [G.sub.B]. Such relation entails the inversion of a system of nonlinear integral equations, which admits unfortunately no closed-form solution. Having found a necessary condition for the invertibility of the system, we have shown that the eigenvalue reconstruction can be achieved numerically via a converging fixed point iteration. In order to prove the convergence, we rely ultimately upon some probability inequalities, known in the literature as square correlation inequalities, which have been recently proved in [17].
In order to explore the convergence rate of the fixed point iteration, we have implemented some variations of the nonlinear Gauss-Jacobi scheme. Specifically, we have found that overrelaxing the basic iteration enhances the convergence rate by a moderate factor. However, the overrelaxed algorithm still slows down exponentially in the number of eigenvalues and polynomially in the truncation radius of the Euclidean ball. We have shown that a significant reduction of the slowing down can be achieved in the regime of strong truncation by adapting the relaxation parameter to the eigenvalue that is naturally associated with, so as to boost the higher components of the spectrum.
We have also discussed how the iterative procedure works when the eigenvalue reconstruction is performed on sample estimates of the truncated covariance spectrum. Specifically, we have shown that the statistical fluctuations make the problem ill-posed. We have sketched a possible way out based on perturbation theory, which is thoroughly discussed in a separate paper [27].
A concrete implementation of the proposed approach requires the computation of a set of multivariate Gaussian integrals over the Euclidean ball. For this, we have extended to the case of interest a technique, originally proposed by Ruben for representing the probability content of quadratic forms of normal variables as a series of chi-square distributions. In the paper, we have shown the practical feasibility of the series expansion for the integrals involved in our computations.
http://dx.doi.org/10.1155/2017/6579537
Appendix
Proof of Theorem 9. As already mentioned in Section 4, the proof follows in the tracks of the original one of [2]. We detail the relevant steps for [[alpha].sub.k], while for [[alpha].sub.jk] we only explain why it is necessary to distinguish between equal or different indices and the consequences for either case.
In order to prove (57), we first express [[alpha].sub.k] in spherical coordinates; that is, we perform the change of variable x = ru, being r = [parallel]x[parallel] and u [member of] [partial derivative][B.sub.v](1) (recall that [d.sup.v]x = [r.sup.v-1]dr du, with du embodying the angular part of the spherical Jacobian and the differentials of v - 1 angles); then we insert a factor of 1 = exp([r.sup.2]/2s)exp(-[r.sup.2]/2s) under the integral sign. Hence, [[alpha].sub.k] reads
[mathematical expression not reproducible]. (A.1)
The next step consists in expanding the inner exponential in Taylor series (in his original proof, Ruben considers a more general setup, with the center of the Euclidean ball shifted by a vector b [member of] [R.sup.v] from the center of the distribution. In that case, the Gaussian exponential looks different and must be expanded in series of Hermite polynomials. Here, we work in a simplified setup, where the Hermite expansion reduces to Taylor's); namely,
exp(-[Q (u) [r.sup.2]]/2) = [[infinity].summation over (m=0)] 1/m! [r.sup.2m]/[2.sup.m][(-Q).sup.m]. (A.2)
This series converges uniformly in u. We review the estimate just for the sake of completeness:
[mathematical expression not reproducible], (A.3)
where [q.sub.0] = [max.sub.i][absolute value of (1/s - 1/[[lambda].sub.i])]. It follows that we can integrate the series term by term. With the help of the uniform average operator introduced in (53), [[alpha].sub.k] is recast to
[mathematical expression not reproducible]. (A.4)
The presence of an additional factor of [u.sup.2.sub.k] in the angular average is harmless, since [absolute value of ([u.sup.2.sub.k])] < 1. We finally notice that the radial integral can be expressed in terms of a cumulative chi-square distribution function on replacing r [right arrow] [square root of (rs)]; namely,
[[integral].sup.[square root of ([rho])].sub.0] dr [r.sup.v+2m+1] exp(- [r.sup.2]/2s) = [2.sup.v/2+m][s.sup.v/2+m+1] [GAMMA] (v/2 + m + 1) [F.sub.v+2(m+1)] ([rho]/s). (A.5)
Inserting (A.5) into (A.4) results in Ruben's representation of [[alpha].sub.k]. This completes the first part of the proof.
As a next step, we wish to demonstrate that the function [[psi].sub.k] of (61) is the generating function of the coefficients [c.sub.k;m]. To this aim, we first recall the identities
[mathematical expression not reproducible], (A.6)
valid for a > 0. On setting [a.sub.i] = [1 - (1 - s/[[lambda].sub.i])z], we see that [[psi].sub.k] can be represented in the integral form
[mathematical expression not reproducible], (A.7)
provided [absolute value of (z)] < [min.sub.i][[absolute value of (1 - s/[[lambda].sub.i])].sup.-1]. As previously done, we introduce spherical coordinates x = ru and expand exp{-(1/2)zsQ(x)} = exp{-(1/2)zs[r.sup.2]Q(u)} in Taylor series.
By the same argument as above, the series converges uniformly in u (the factor of zs does not depend on u), thus allowing term-by-term integration. Accordingly, we have
[mathematical expression not reproducible]. (A.8)
We see that the right-hand side of (A.8) looks similar to (A.4), the only relevant differences being the presence of the factor of [z.sup.m] under the sum sign and the upper limit of the radial integral. With some algebra, we arrive at
[[psi].sub.k] (z) = [[infinity].summation over (m=0)] [z.sup.m] {[2/m!] [s/[[lambda].sub.k] [s.sup.v/2+m]/[[absolute value of ([LAMBDA])].sup.1/2] [[GAMMA] (v/2 + m + 1)]/[[GAMMA] (v/2)] x M [[(-Q).sup.m] [u.sup.2.sub.k]}. (A.9)
The series coefficients are recognized to be precisely those of (59).
In the last part of the proof, we derive the recursion fulfilled by the coefficients [c.sub.k;m]. To this aim, the mth derivative of [[psi].sub.k] has to be evaluated at z = 0 and then identified with m![c.sub.k;m]. The key observation is that differentiating [[psi].sub.k] reproduces [[psi].sub.k] itself; that is to say,
[[psi]'.sub.k] (z) = [[PSI].sub.k] (z) [[psi].sub.k] (z), (A.10)
with
[[PSI].sub.k] (z) = 1/2 [v.summation over (i=1)] [e.sub.k;i] (1 - s/[[lambda].sub.i]) [[1 - (1 - s/[[lambda].sub.i]) z].sup.-1], (A.11)
and the auxiliary coefficient [e.sub.k;i] being defined as in (67). Equation (A.10) lies at the origin of the recursion. Indeed, from (A.10), it follows that that [[psi]".sub.k] is a function of [[psi]'.sub.k] and [[psi].sub.k]; namely, [[psi]".sub.k] = [[PSI]'.sub.k][[psi].sub.k] + [[PSI].sub.k][[psi].sub.'k]. Proceeding analogously yields the general formula
[mathematical expression not reproducible]. (A.12)
At z = 0, this reads
m![c.sub.k;m] = [m-1.summation over (r=0)] (m-1)!/(m - r - 1)!r! [[PSI].sup.(m-r-1).sub.k] (0) r![c.sub.k;r]. (A.13)
The last step consists in proving that
[[PSI].sup.(m).sub.k] (0) = 1/2 m![g.sub.k;m+1], (A.14)
with [g.sub.k;m] defined as in (65). This can be done precisely as explained in [2].
Having reiterated Ruben's proof explicitly in a specific case, it is now easy to see how the theorem is extended to any other Gaussian integral. First of all, from (A.1), we infer that each additional subscript in [[alpha].sub.klm ...] enhances the power of the radial coordinate under the integral sign by 2 units. This entails a shift in the number of degrees of freedom of the chi-square distributions in Ruben's expansion, amounting to twice the number of subscripts. For instance, since [[alpha].sub.jk] has two subscripts, its Ruben's expansion starts by [F.sub.v+4], independently of whether j = k or j [not equal to] k. In second place, we observe that in order to correctly identify the generating functions of Ruben's coefficients for a higher-order integral [[alpha].sub.klm ...], we need to take into account the multiplicities of the indices k, l, m, .... As an example, consider the case of [[psi].sub.jk] (j [not equal to] k) and [[psi].sub.kk]. By going once more through the argument presented in (A.7), we see that (A.6) are sufficient to show that (63) is the generating function of [[alpha].sub.jk]. By contrast, in order to repeat the proof for the case of [[psi].sub.kk], we need an additional integral identity; namely,
[a.sup.-5/2] = [1/3] [(2[pi]).sup.-1/2] [[integral].sup.+[infinity].sub.- [infinity]] dx [x.sup.4] exp (-a/2 [x.sup.2]), (A.15)
valid once more for a > 0. Hence, we infer that [[psi].sub.kk] must depend upon [[lambda].sub.k] via a factor of [[1 - (1 - s/[[lambda].sub.k])z].sup.-5/2], whereas [[psi].sub.jk] (j [not equal to] k) must depend on [[lambda].sub.j] and [[lambda].sub.k] via factors of, respectively, [[1 - (1 - s/[[lambda].sub.j])z].sup.-3/2] and [[1 - (1 - s/[[lambda].sub.k])z].sup.-3/2]. The different exponents are ultimately responsible for the specific values taken by the auxiliary coefficients [e.sub.kk;i] and [e.sub.jk;i] of (68).
To conclude, we observe that the estimates of the residuals [R.sub.k;m] and [R.sub.jk;m], presented in Section 4 without an explicit proof, do not require any further technical insight than already provided by [2] plus our considerations. We leave them to the reader, since they can be obtained once more in the tracks of the original derivation of [R.sub.m].
Competing Interests
The authors declare that they have no competing interests.
Acknowledgments
The authors are grateful to A. Reale for encouraging them throughout all stages of this work and to G. Bianchi for technical support at ISTAT. They also thank R. Mukerjee and S. H. Ong for promptly informing them about their proof of (22) and (23). The computing resources used for their numerical study and the related technical support at ENEA have been provided by the CRESCO/ENEAGRID High Performance Computing infrastructure and its staff [28]. CRESCO (Computational Research Centre for Complex Systems) is funded by ENEA and by Italian and European research programmes.
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Filippo Palombi, (1,2) Simona Toti, (1) and Romina Filippini (1)
(1) Istituto Nazionale di Statistica (ISTAT), Via Cesare Balbo 16, 00184 Rome, Italy
(2) Italian Agency for New Technologies, Energy and Sustainable Economic Development (ENEA), Via Enrico Fermi 45, 00044 Frascati, Italy
Correspondence should be addressed to Filippo Palombi; filippo.palombi@enea.it
Received 7 July 2016; Revised 7 October 2016; Accepted 12 October 2016; Published 10 January 2017
Academic Editor: Ramon M. Rodriguez-Dagnino
Caption: Figure 1: A classical particle beam with elliptical transversal profile is cut off upon entering a circular coaxial pipeline.
Caption: Figure 2: (a) Numerical reconstruction of D([[tau].sup.-1.sub.[rho]]) in v = 2 dimensions. (b) Numerical reconstruction of D([[tau].sup.-1.sub.[rho]]) in v = 3 dimensions.
Caption: Figure 3: Monte Carlo simulation of the probability density function of the ordered eigenvalues [[lambda].sub.k] (even rows) and their truncated counterparts [[mu].sub.k] at [rho] = 1 (odd rows) for the Wishart ensemble [W.sub.10](20, [20.sup.-1] x [I.sub.10]). The last two plots (bottom right) display the distribution of the sum of eigenvalues.
Caption: Figure 4: Monte Carlo simulation of the probability mass function of the parameter kth for the Gaussian probability content [alpha]. The histograms refer to the eigenvalue ensemble [lambda] of [SIGMA] ~ [W.sub.10](20, [20.sup.-1] x [I.sub.10]), with [rho] chosen as in Table 1 and [epsilon] = 1.0 x [10.sub.-14].
Caption: Figure 6: Log-log plots of [[bar.n].sub.it] versus [rho] in the GJOR scheme at [[epsilon].sub.T] = 1.0 x [10.sub.-7]. The parameter [rho] has been chosen as in Table 1. The (red) dashed line in each plot represents our best jackknife linear fit to (76) of data points with [rho] [less than or equal tol] 1.
Caption: Figure 7: Scaling parameters a and b of the modified GJOR scheme as functions of v at [[epsilon].sub.T] = 1.0 x [10.sup.-7], with [beta] varying in the range 0/2 in steps of 1/5. Each trajectory (represented by a dashed curve) refers to a different value of [beta]. Those with darker markers correspond to smaller values of [beta] and the other way round.
Caption: Figure 8: The parameter [kappa] as a function of [beta]. Estimates of [kappa] are obtained from least-squares fits of data to a linear model a = [a.sub.0] + [kappa]x v.
Caption: Figure 9: (a) Numerical reconstruction of the failure probability of the iterative procedure as V = 4 and [SIGMA] = diag(0.1, 0.3, 0.8, 2.2), for several values of [rho] and for N = 200, 250, ..., 1000. (b) Failure probability of the perturbative regularization with the same parameters.
Caption: Figure 10: (a) Set difference D([T.sup.-1.sub.[rho]]) D([[tau].sup.-1.sub.[rho]) in v = 2 dimensions. (b) Set difference D([T.sup.-1.sub.[rho]]) \ D([[tau].sup.-1.sub.[rho]]) in v = 3 dimensions.
```Table 1: Numerical estimates of the mode of the ordered eigenvalues
{[[lambda].sub.1] ..., [[lambda].sub.v]} of [SIGMA] ~ [W.sub.v](2v,
[(2v).sup.-1] x [I.sub.V]) with v = 3, ..., 10. The estimates have
been obtained from Grenander's mode estimator [23].
v = 3 v = 4 v = 5 v = 6
[??]([[lambda].sub.1]) 0.1568 0.1487 0.1435 0.1383
[??]([[lambda].sub.2]) 0.6724 0.4921 0.4017 0.3424
[??]([[lambda].sub.3]) 1.6671 1.0112 0.7528 0.6071
[??]([[lambda].sub.4]) -- 1.8507 1.2401 0.9621
[??]([[lambda].sub.5]) -- -- 2.0150 1.4434
[??]([[lambda].sub.6]) -- -- -- 2.1356
[??]([[lambda].sub.7]) -- -- -- --
[??]([[lambda].sub.8]) -- -- -- --
[??]([[lambda].sub.9]) -- -- -- --
[??]([[lambda].sub.10]) -- -- -- --
v = 7 v = 8 v = 9 v = 10
[??]([[lambda].sub.1]) 0.1344 0.1310 0.1269 0.1258
[??]([[lambda].sub.2]) 0.3039 0.2745 0.2554 0.2399
[??]([[lambda].sub.3]) 0.5138 0.4543 0.4048 0.3693
[??]([[lambda].sub.4]) 0.7854 0.6684 0.5858 0.5288
[??]([[lambda].sub.5]) 1.1269 0.9263 0.7956 0.7032
[??]([[lambda].sub.6]) 1.5789 1.2559 1.0527 0.9111
[??]([[lambda].sub.7]) 2.2190 1.6764 1.3673 1.1603
[??]([[lambda].sub.8]) -- 2.2763 1.7687 1.4624
[??]([[lambda].sub.9]) -- -- 2.3210 1.8473
[??]([[lambda].sub.10]) -- -- -- 2.3775
Figure 5: (a) Box-plot of [n.sub.it] in the GJOR scheme at v = 10,
with [[epsilon].sub.T] = 1.0 x [10.sup.-7] and [rho] chosen as in
Table 1. The distributions have been reconstructed from a sample of
N [equivalent] [10.sup.3] eigenvalue spectra extracted from
[W.sub.10](20, [20.sup.-1] x [I.sub.10]). The whiskers extend to the
most extreme data point within (3/2)(75%-25%) data range. (b) Numerical
estimates of the scaling parameters a and b of the GJOR scheme, as
obtained from jackknife fits to (76) of data points with [rho] [less
than or equal tol] 1 and [[epsilon].sub.T] = 1.0 x [10.sup.-7]. We
quote in parentheses the jackknife error.
(b)
v [??](jk. err.) [??](jk. err.)
3 4.93(3) 0.940(1)
4 5.25(2) 0.940(1)
5 5.44(2) 0.948(1)
6 5.65(1) 0.953(1)
7 5.85(1) 0.948(1)
8 6.02(1) 0.951(1)
9 6.18(1) 0.946(1)
10 6.31(1) 0.948(1)
```
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The Benefits of Lenses Physics
Within this lesson, we’ll observe a similar technique for constructing ray diagrams for double concave lenses. There are good reasons to feel this paradigm should apply to economic systems generally and financial markets specifically. The previous position is known as neutral equilibrium. But because the lens is an incredibly important region of the eye, and difficulties with eyesight (vision) are often treated with the usage of spectacles or contact lenses, this introduction to two very simple forms of lenses is included as background info.
We’ll use these 3 rays throughout the remainder of this lesson, merely since they’re the easiest rays to draw. The polarized filters on such lenses preferentially block the horizontal part of light oscillation when transmitting the vertical component. As we start to talk about the refraction of light rays and the formation of images by both of these kinds of lenses, we should use an assortment of terms.
There are many sorts of chemical reactions. The imaging properties of a lenses aren’t elaborate and here’s a list which can help you distinguish a lens. Figure out the average of both distances.
It is essential that ALL eigenvalues have negative real components. This kind of equilibrium is known as unstable equilibrium. It may be stable or unstable.
A triangle is a great way to observe how light behaves in a prism. All you should know is that the sum of the 3 angles in a triangle is 180. This point is called the focus of the converging lens.
It needs to be noted that the procedure for constructing a ray diagram is the exact same regardless of where the object is situated. Eigenvalues are usually complex numbers. A microscope arrangement is shown below, as well as the ray diagram showing the way the very first lens makes an actual image. This is in agreement with the ray diagram.
Researchers now plan to check the ECE technique on a broad number of plasma discharges. Begin by doing what you understand how to do and attempt to work towards a solution. You are able to even use completely free software that could give the readers that have many functions to the reader than merely a very simple platform to read the desired eBooks.
The ray diagram above illustrates that if the object is situated at a situation past the 2F point, the image is going to be located at a post between the 2F point and the focal point on the other side of the lens. Hence the result is just confined to the exact top section of the curve. Since light does not really pass through this point, the image is called a digital image. After the image is smaller than the object, it’s believed to be diminished. An actual image is one where it is possible to project it on a bit of paper, and so, it forms on the other side of the lens to the object. Mark the image of the surface of the object.
Vital Pieces of Lenses Physics
So, it’s essential to present your eyes rest for a little while by taking breaks after particular time intervals. We won’t use over two lenses, and we can do two or three examples to observe how you analyze problems similar to this. Welcome to the true world.
This must be carried out very carefully. Speak to the department chair to talk about this alternative.
The important point is reached while the cg is no longer over the base of support. The effect is utilized in navigation to keep stability for all types of machines. Like the term lens, the term focus first begun to appear in scientific writing in the Seventeenth Century. Inside this chapter, both of these concepts will be explained in details using ray diagrams alongside their applications in actual life circumstances.
Here is an extra information. The sign convention is only a little different. Inside this chapter, you’ll also learn common examples of such reactions like rusting and corrosion.
|
This is the procedure by which lenses do the job. A concave eyepiece is utilized to correct this issue. We can analyze what a lens is going to do in a variety of means. They are able to bend light because of a phenomenon called refraction. So this lens may be used as a magnifying lens. They are used to correct farsighted vision problems. In the same way, concave lenses are utilised to correct nearsightedness.
The image is the point where the rays intersect. All these were created employing the exact same 3 rays I just described. A ray passing straight through the middle of the lens.
An alternative way of reducing the spherical aberration of a lens is just to lower the aperture working with a diaphragm (also known as an iris or a simply a stop). This kind of axis is known as the principal axis.
Then they say this would produce the atom unstable. It is called stable equilibrium. An equilibrium is supposed to be unstable if the least departures produce forces or torques that generally boost the displacement.
The point is to focus light to a single point irrespective of the wavelength. So that the focus is the area where rays that arrive in parallel to the principal axis go. We’ll discuss later how to find out if the focal point is negative or positive connected to the lens.
As soon as the conducting wire touches the sinker, it gives a path for those charges to flow. This optimum state isn’t only elusive, it’s also hyper-fragile to small changes in the health of the planet, and for that reason often irrelevant to understanding what’s going on. Overcompensation is due to large time steps. The greater pressure region underneath the wing attempts to move toward the very low pressure region, and because of this lifts the wing. Also, it might not be the issue.
This must be carried out very carefully. Speak to the department chair to talk about this alternative.
The important point is reached while the cg is no longer over the base of support. Contrary to other life processes like respiration and nutrition, reproduction isn’t necessary for maintaining the life span of a person organism. Since it’s a linear relationship, it’s also simpler to handle mathematically. Inside this chapter, both of these concepts will be explained in details using ray diagrams alongside their applications in actual life circumstances.
Find more details on the Altmetric Attention Score and the way the score is figured. Topics you’ll need to understand as a way to pass the quiz include the capacity of an object to keep its balance and discerning once an object has reached its tipping point. For instance, if you know that there is going to be just a 5 marks question from a specific section, you do not need to dedicate a week practicing those sums. Here’s a list of some vital topics which you need to study for scoring good marks in your boards.
Top Lenses Physics Secrets
It’s even feasible to discover the focus of a Jello lens. This will certainly help to create reading easier. Lots of people are conversant with the capacity of magnifying glasses to concentrate sunlight to the point at which it can burn paper. It is going to be useful to have a fantastic eBook reader to be in a position to truly have an excellent reading experience and top quality eBook display.
This is found employing a little bit of geometry. We can bring the middle of gravity down merely by adding more weight to the base of the cup. In the event the filter on the left is placed in addition to the filter on the right, no light will have the ability to pass through whatsoever. Think about a ball kept at the peak of a sphere. It follows that the quantity of spin required to be sure it stays stable becomes greater.
It’s real, inverted and diminished in dimension. A marble on a level horizontal surface is a good example. Tall glasses are much more inclined to tip over than short ones.
|
It’s even feasible to discover the focus of a Jello lens. Despite the fact that this may mean that you will have less text on every page and increased amount of page turning, you will be able to read your desirable eBook with amazing convenience and have a superb reading experience with better eBook screen. Lots of people are conversant with the capacity of magnifying glasses to concentrate sunlight to the point at which it can burn paper. Alright now observe this light ray isn’t going towards the principal axis so something different is happening.
Though it is in beta, it has all of the functionality of the prior version plus some extra characteristics that are still in development. Strategy First, we will need to graph the prospective energy for a function of x. Science Explained The Leyden jar acts as a capacitor a system that stores charges.
The size of the image is dependent upon the place of the object. Within this chapter, you can know about what constitutes electricity and the way that it flows in an electric circuit. It’s a method of illustrating the direction light is moving and is significantly simpler than drawing the intricate waves which make it up. It is merely a vertical line. An actual image is one where it is possible to project it on a bit of paper, and so, it forms on the other side of the lens to the object. Repeat the process for the base of the object.
Most Noticeable Lenses Physics
And, they’ve given sight to folks who may hardly see in any respect. Lenses have made a great effect on our everyday lives and are almost utilised in everyday life. Welcome to the true world.
Problems and questions are increased by 22% and lots of authentic applications of physics principles are added. They’ve developed an effective selection of study tools that will provide you with an excess advantage in your physics class. These living structures will likely collapse over time as a result of environmental consequences. The goal of physics is to realize the world we dwell in.
The majority of the times, it’s been believed that the readers, that are employing the eBooks for first time, happen to truly have a troublesome time before getting used to them. You will be able to handle videos in your Watchlist, keep tabs on your favourite shows, watch PBS in high definition, and a lot more! You need to make sure that you put in more time into subjects that require lots of practice. Get accustomed to the numerical part.
Here is an extra information. The sign convention is only a little different. Inside this chapter, you’ll also learn common examples of such reactions like rusting and corrosion.
What Is So Fascinating About Lenses Physics?
This is the procedure by which lenses do the job. A concave eyepiece is utilized to correct this issue. They work according to their shape. A lens is double sided, however, and the 2 sides may or might not have the exact same curvature. Camera lenses are made to minimize this effect by careful design and application of special coatings. Convex lenses may also form virtual images, the kind you’ve got to examine the lens to see. In the same way, concave lenses are utilised to correct nearsightedness.
The signs linked with magnification also work the exact same way for lenses and mirrors. By rotating a polarizing filter over the crystal, it’s possible to view 1 image at a moment.
Centrifugal force can be raised by increasing either the speed of rotation or the mass of the human body or by decreasing the radius, that is the distance of the human body from the middle of the curve. The dispersion brought on by the convex lens is simply cancelledby the dispersion resulting from the concave lens.
Following that, you can demonstrate the thin lens equation works. This very first step is precisely the same, whatever the sort of lens. Clearly, this is a significant component that explains a better part of the considerable difference in thickness between contacts and eyeglasses.
This imaginary line is called the principal axis. So that the focus is the area where rays that arrive in parallel to the principal axis go. We’ll discuss later how to find out if the focal point is negative or positive connected to the lens.
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Lo que se dice en Tuiter de este blog/libro: | 2,489 | 12,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.926 |
https://fixedpointtheoryandapplications.springeropen.com/articles/10.1186/1687-1812-2013-41 | 1,606,631,947,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197278.54/warc/CC-MAIN-20201129063812-20201129093812-00285.warc.gz | 291,400,089 | 44,494 | # A common fixed point theorem for two weakly compatible pairs in G-metric spaces using the property E.A
## Abstract
In view of the fact that the fixed point theory provides an efficient tool in many fields of pure and applied sciences, we use the notion of the property E.A to prove a common fixed point theorem for weakly compatible mappings. The presented results are applied to obtain the solution of an integral equation and the bounded solution of a functional equation arising in dynamic programming.
MSC:47H10, 54H25.
## 1 Introduction
Inspired by the fact that the metric fixed point theory provides an efficient tool in many fields of pure and applied sciences, many authors investigated the possibility to generalize the notion of a metric space. In this direction, Gahler [1, 2] introduced the notion of a 2-metric space, while Dhage [3] introduced the concept of a D-metric space. Later on, Mustafa and Sims [4] showed that most of the results concerning Dhage’s D-metric spaces are invalid. Therefore, they introduced a new notion of a generalized metric space, called G-metric space. After then, many authors studied fixed and common fixed points in generalized metric spaces; see [415].
Here, we give preliminaries and basic definitions which are helpful in the sequel. First, we introduce the concepts of a G-metric and a G-metric space.
Definition 1.1 [4]
Let X be a nonempty set and $G:X×X×X→[0,+∞)$ be a function satisfying the following properties:
1. (G1)
$G(x,y,z)=0$ if $x=y=z$;
2. (G2)
$0 for all $x,y∈X$ with $x≠y$;
3. (G3)
$G(x,x,y)≤G(x,y,z)$ for all $x,y,z∈X$ with $z≠y$;
4. (G4)
$G(x,y,z)=G(x,z,y)=G(y,z,x)=⋯$ (symmetry in all three variables);
5. (G5)
$G(x,y,z)≤G(x,a,a)+G(a,y,z)$ for all $x,y,z,a∈X$ (rectangle inequality).
Then the function G is called a generalized metric or, more specifically, a G-metric on X, and the pair $(X,G)$ is called a G-metric space.
Definition 1.2 A G-metric space $(X,G)$ is said to be symmetric if $G(x,y,y)=G(y,x,x)$ for all $x,y∈X$.
Example 1.3 Let $X={2,3}$ and $G:X×X×X→[0,+∞)$ be defined by $G(2,2,2)=G(3,3,3)=0$, $G(2,2,3)=G(2,3,2)=G(3,2,2)=1$, $G(2,3,3)=G(3,2,3)=G(3,3,2)=2$. It is easy to show that the function G satisfies all properties of Definition 1.1, but $G(x,x,y)≠G(x,y,y)$ for all $x,y∈X$ with $x≠y$. Therefore, G is not symmetric.
Definition 1.4 [4]
Let $(X,G)$ be a G-metric space, and let ${ x n }$ be a sequence of points of X. A point $x∈X$ is said to be the limit of the sequence ${ x n }$ if $lim n , m → + ∞ G(x, x n , x m )=0$, and we say that the sequence ${ x n }$ is G-convergent to x or ${ x n }$ G-converges to x.
Thus, $x n →x$ in a G-metric space $(X,G)$ if for any $ε>0$, there exists $k∈N$ such that $G(x, x n , x m )<ε$ for all $m,n≥k$.
Proposition 1.5 [4]
Let $(X,G)$ be a G-metric space. Then the following are equivalent:
1. (1)
${ x n }$ is G-convergent to x;
2. (2)
$G( x n , x n ,x)→0$ as $n→+∞$;
3. (3)
$G( x n ,x,x)→0$ as $n→+∞$.
Definition 1.6 [4]
Let $(X,G)$ be a G-metric space. A sequence ${ x n }$ is called G-Cauchy if for every $ε>0$, there is $k∈N$ such that $G( x n , x m , x l )<ε$ for all $n,m,l≥k$; that is, $G( x n , x m , x l )→0$ as $n,m,l→+∞$.
Proposition 1.7 [4]
Let $(X,G)$ be a G-metric space. Then the following are equivalent:
1. (1)
the sequence ${ x n }$ is G-Cauchy;
2. (2)
for every $ε>0$, there is $k∈N$ such that $G( x n , x m , x m )<ε$ for all $n,m≥k$.
Proposition 1.8 [4]
Let $(X,G)$ be a G-metric space. Then the function $G(x,y,z)$ is jointly continuous in all three of its variables.
Definition 1.9 [4]
A G-metric space $(X,G)$ is called G-complete if every G-Cauchy sequence in $(X,G)$ is G-convergent in $(X,G)$.
Proposition 1.10 [4]
Let $(X,G)$ be a G-metric space. Then, for any $x,y,z,a∈X$, it follows that
1. (i)
if $G(x,y,z)=0$, then $x=y=z$;
2. (ii)
$G(x,y,z)≤G(x,x,y)+G(x,x,z)$;
3. (iii)
$G(x,y,y)≤2G(y,x,x)$;
4. (iv)
$G(x,y,z)≤G(x,a,z)+G(a,y,z)$;
5. (v)
$G(x,y,z)≤ 2 3 [G(x,y,a)+G(x,a,z)+G(a,y,z)]$;
6. (vi)
$G(x,y,z)≤G(x,a,a)+G(y,a,a)+G(z,a,a)$.
An interesting observation is that any G-metric space $(X,G)$ induces a metric $d G$ on X given by
Moreover, $(X,G)$ is G-complete if and only if $(X, d G )$ is complete.
It was observed that in the symmetric case ($(X,G)$ is symmetric), many fixed point theorems on G-metric spaces are particular cases of the existing fixed point theorems in metric spaces. This allows us to readily transport many results from the metric spaces into the G-metric spaces.
On the other hand, by reasoning on the properties of the mappings, the practice of coining weaker forms of commutativity to ensure the existence of a common fixed point for self-mappings on metric spaces is still on. To read more in this direction, we refer to [16] and the references therein. Here, for our further use, we recall only the two fundamental notions of ‘weakly compatible mappings’ and ‘property E.A’; see also [17, 18].
In 1976, Jungck [19] introduced the notion of weakly compatible mappings as follows.
Definition 1.11 Let S and T be two self-mappings of a metric space $(X,d)$. Then the pair $(S,T)$ is said to be weakly compatible if they commute at their coincidence points, that is, if $Su=Tu$ for some $u∈X$, then $TSu=STu$.
In 2002, Amari and El Moutawakil [20] introduced a new concept of the property E.A in metric spaces to generalize the concept of noncompatible mappings. Then, they proved some common fixed point theorems.
Definition 1.12 Let S and T be two self-mappings of a metric space $(X,d)$. Then the pair $(S,T)$ is said to satisfy the property E.A if there exists a sequence ${ x n }$ in X such that $lim n → + ∞ S x n = lim n → + ∞ T x n =t$ for some $t∈X$.
Example 1.13 Let $X=[0,+∞)$. Define $S,T:X→X$ by $Sx= 3 4 x$ and $Tx= x 4$ for all $x∈X$. Consider the sequence ${ x n }={ 1 n }$ in X. Clearly, $lim n → + ∞ S x n = lim n → + ∞ T x n =0∈X$, and so S and T satisfy the property E.A.
Example 1.14 Let $X=[2,+∞)$. Define $S,T:X→X$ by $Sx=2x+1$ and $Tx=x+1$ for all $x∈X$. Suppose that the property E.A holds. Then there exists a sequence ${ x n }$ in X such that $lim n → + ∞ S x n = lim n → + ∞ T x n =t$ for some $t∈X$. It follows that $lim n → + ∞ x n = t − 1 2$ and $lim n → + ∞ x n =t−1$ and so, by Definition 1.12, $t=1$ but $t∉X$. Therefore, S and T do not satisfy the property E.A.
In conclusion of this preliminary section, we consider the following set:
Let Φ denote the set of all the functions $φ:[0,+∞)→[0,+∞)$ such that:
1. (1)
φ is non-decreasing;
2. (2)
$lim n → + ∞ φ n (r)=0$ for all $r∈[0,+∞)$.
If $φ∈Φ$, then it is an easy matter to show that $φ(0)=0$ and $φ(r) for all $r∈(0,+∞)$; see Matkowski [21].
In this paper, by merging the above notions, we prove a common fixed point theorem for two pairs of weakly compatible mappings in a G-metric space. The presented results are applied to obtain the solution of an integral equation and the bounded solution of a functional equation arising in dynamic programming.
## 2 Main results
The following is the main result of this section.
Theorem 2.1 Let $(X,G)$ be a G-metric space and $A,B,S,T:X→X$ be four self-mappings such that:
1. (i)
$A(X)⊆T(X)$ and $B(X)⊆S(X)$;
2. (ii)
one of the pairs $(A,S)$ and $(B,T)$ satisfies the property E.A;
3. (iii)
for all $x,y∈X$, $G(Ax,By,By)≤φ(max{G(Sx,Ty,Ty),G(Sx,By,By),G(Ty,By,By)})$, where $φ∈Φ$;
4. (iv)
one of $A(X)$, $B(X)$, $S(X)$ and $T(X)$ is a complete subset of X.
Then the pairs $(A,S)$ and $(B,T)$ have a coincidence point. Further, if $(A,S)$ and $(B,T)$ are weakly compatible, then A, B, S and T have a unique common fixed point in X.
Proof Suppose the pair $(B,T)$ satisfies the property E.A. Then there exists a sequence ${ x n }$ in X such that $lim n → + ∞ B x n = lim n → + ∞ T x n =t$ for some $t∈X$. Since $B(X)⊆S(X)$, there exists a sequence ${ y n }$ in X such that $B x n =S y n$. Hence $lim n → + ∞ S y n =t$. We will show that $lim n → + ∞ A y n =t$. From (iii), we have
$G(A y n ,B x n ,B x n )≤φ ( max { G ( S y n , T x n , T x n ) , G ( S y n , B x n , B x n ) , G ( T x n , B x n , B x n ) } ) .$
Taking the limit as $n→+∞$ (upper limit) and using the fact that $φ(r)$ is continuous at $r=0$, we get
$lim n → + ∞ G(A y n ,t,t)≤φ ( max { G ( t , t , t ) , G ( t , t , t ) , G ( t , t , t ) } ) =φ(0)=0,$
and so $lim n → + ∞ A y n =t$. Thus, we have $lim n → + ∞ A y n = lim n → + ∞ B x n = lim n → + ∞ S y n = lim n → + ∞ T x n =t$. Suppose that $S(X)$ is a complete subset of X. Then $t=Su$ for some $u∈X$. Now, we will show that $Au=Su=t$. From (iii), we have
$G(Au,B x n ,B x n )≤φ ( max { G ( S u , T x n , T x n ) , G ( S u , B x n , B x n ) , G ( T x n , B x n , B x n ) } ) .$
Taking the limit as $n→+∞$, by the property of φ, we get
$G(Au,Su,Su)≤φ ( max { G ( t , t , t ) , G ( t , t , t ) , G ( t , t , t ) } ) =φ(0)=0,$
which implies $Au=Su$. Therefore, u is a coincidence point of the pair $(A,S)$. The weak compatibility of A and S implies that $ASu=SAu$ and hence $AAu=ASu=SAu=SSu$. Since $A(X)⊆T(X)$, there exists $v∈X$ such that $Au=Tv$. We claim that $Tv=Bv$. Suppose not, from (iii) and using the fact that $φ(r), we have
$G ( A u , B v , B v ) ≤ φ ( max { G ( S u , T v , T v ) , G ( S u , B v , B v ) , G ( T v , B v , B v ) } ) = φ ( max { 0 , G ( A u , B v , B v ) , G ( A u , B v , B v ) } ) = φ ( G ( A u , B v , B v ) ) < G ( A u , B v , B v ) .$
This implies that $Au=Bv$ and hence $Tv=Bv$. It follows that also the pair $(B,T)$ has a coincidence point. Thus, we have $Au=Su=Tv=Bv$.
Now, if B and T are weakly compatible, then we obtain $BTv=TBv=TTv=BBv$ and show that Au is a common fixed point of A, B, S and T. For $x=Au$ and $y=v$, from (iii) and the property of φ, we get
$G ( A A u , A u , A u ) = G ( A A u , B v , B v ) ≤ φ ( max { G ( S A u , T v , T v ) , G ( S A u , B v , B v ) , G ( T v , B v , B v ) } ) = φ ( max { G ( A A u , B v , B v ) , G ( A A u , B v , B v ) , G ( B v , B v , B v ) } ) = φ ( max { G ( A A u , B v , B v ) , G ( A A u , B v , B v ) , 0 } ) = φ ( G ( A A u , B v , B v ) ) < G ( A A u , B v , B v ) ,$
which implies $Au=AAu=Bv$. Therefore, $Au=AAu=SAu$ is a common fixed point of A and S. Similarly, one can prove that Bv is a common fixed point of B and T. Since $Au=Bv$, we deduce that Au is a common fixed point of A, B, S and T. Now, we have only to show that the common fixed point is unique. Suppose to the contrary that w and z, with $w≠z$, are two common fixed points of A, B, S and T. Then, from (iii) and the property of φ, we have
$G ( w , z , z ) = G ( A w , B z , B z ) ≤ φ ( max { G ( S w , T z , T z ) , G ( S w , B z , B z ) , G ( T z , B z , B z ) } ) = φ ( max { G ( w , z , z ) , G ( w , z , z ) , G ( z , z , z ) } ) = φ ( G ( w , z , z ) ) < G ( w , z , z ) ,$
that is a contradiction and so must be $w=z$. Therefore, A, B, S and T have a unique common fixed point. Clearly, proceeding on the foregoing lines, one can obtain the same conclusion in case (instead of $S(X)$) one of $A(X)$, $B(X)$ and $T(X)$ is a complete subset of X, and in case (instead of $(B,T)$) $(A,S)$ satisfies the property E.A. □
If we assume $S=T$ in above Theorem 2.1, we deduce the following result involving three self-mappings.
Corollary 2.2 Let $(X,G)$ be a G-metric space and $A,B,S:X→X$ be three mappings such that:
1. (i)
$A(X)⊆S(X)$ and $B(X)⊆S(X)$;
2. (ii)
one of the pairs $(A,S)$ and $(B,S)$ satisfies the property E.A;
3. (iii)
for all $x,y∈X$, $G(Ax,By,By)≤φ(max{G(Sx,Sy,Sy),G(Sx,By,By),G(Sy,By,By)})$, where $φ∈Φ$;
4. (iv)
one of $A(X)$, $B(X)$ and $S(X)$ is a complete subset of X.
Then the pairs $(A,S)$ and $(B,S)$ have a coincidence point. Further, if $(A,S)$ and $(B,S)$ are weakly compatible, then A, B and S have a unique common fixed point in X.
Example 2.3 Let $X=[0,2]$ and $G:X×X×X→[0,+∞)$ be defined by $G(x,y,z)=max{|x−y|,|y−z|,|z−x|}$ for all $x,y,z∈X$. Define also $A,B,S:X→X$ by $Ax=1$, $Bx=2−x$ and $Sx=x$ for all $x∈X$ and $φ:[0,+∞)→[0,+∞)$ by $φ(t)=t/2$ for all $t≥0$. Clearly, the hypotheses (i) and (iv) of Corollary 2.2 hold trivially. Moreover, the pair $(A,S)$ satisfies the property E.A. Here we show only that the hypothesis (iii) holds. In fact, for all $x,y∈X$, we have $G(Ax,By,By)=G(1,2−y,2−y)=|1−y|$, $G(Sx,Sy,Sy)=G(x,y,y)=|x−y|$, $G(Sx,By,By)=G(x,2−y,2−y)=|2−x−y|$, $G(Sy,By,By)=G(y,2−y,2−y)=2|1−y|$, and consequently,
$G ( A x , B y , B y ) = | 1 − y | = 1 2 2 | 1 − y | = φ ( G ( S y , B y , B y ) ) ≤ φ ( max { G ( S x , S y , S y ) , G ( S x , B y , B y ) , G ( S y , B y , B y ) } ) .$
Then, by Corollary 2.2, the pairs $(A,S)$ and $(B,S)$ have a coincidence point, that is, $u=1$. Moreover, since $(A,S)$ and $(B,S)$ are weakly compatible, then $u=1$ is the unique common fixed point of A, B and S in X.
## 3 Applications to integral and functional equations
In this section we illustrate two useful applications of our presented results. Firstly, we show how it is possible to obtain the solution of an integral equation by applying Corollary 2.2. Let $Ω=[0,1]$ and $C(Ω)$ be the space of all the real continuous functions defined on Ω. Obviously, this space endowed with the G-metric given by
is a G-complete metric space.
Let $p:Ω×R→R$ and $q:Ω×Ω×R→R$ be two continuous functions. Consider an integral equation of the following type:
$p ( t , x ( t ) ) = ∫ Ω q ( t , s , x ( s ) ) ds.$
(3.1)
We will prove the following theorem.
Theorem 3.1 Suppose there exists $H:Ω×R→[0,+∞)$ such that:
1. (i)
$H(s,v(t))≤ ∫ Ω q(t,s,u(s))ds≤p(s,v(t))$ for all $s,t∈Ω$;
2. (ii)
$p(s,v(t))−H(s,v(t))≤k|p(s,v(t))−v(t)|$, where $k∈(0,1)$.
Then integral equation (3.1) has a solution in $C(Ω)$.
Proof Define $(Ax)(t)= ∫ Ω q(t,s,x(s))ds$ and $(Bx)(t)=p(t,x(t))$. Now
$G ( A x , B y , B y ) = 2 sup t ∈ Ω | ( A x ) ( t ) − ( B y ) ( t ) | = sup t ∈ Ω | p ( t , y ( t ) ) − ∫ Ω q ( t , s , x ( t ) ) d t | ≤ 2 sup t ∈ Ω | p ( t , y ( t ) ) − H ( t , y ( t ) ) | ≤ 2 k sup t ∈ Ω | p ( t , y ( t ) ) − y ( t ) | = k G ( y , B y , B y ) .$
Thus, all the hypotheses of Corollary 2.2 are satisfied with $S= I C ( Ω )$, the identity mapping on Ω, and $φ(r)=kr$ for all $r≥0$ and $k∈(0,1)$. Therefore, there is a unique solution of integral equation (3.1) in $C(Ω)$. □
Now, we study the existence and uniqueness of the bounded solution of a functional equation using again Corollary 2.2. Here we assume that U and V are Banach spaces, $W⊆U$ is a state space and $D⊆V$ is a decision space.
It is well known that the dynamic programming provides useful tools for mathematical optimization and computer programming as well; see [2224]. In particular, the problem of dynamic programming related to a multistage process reduces to the problem of solving the functional equation
$Q(x)= sup y ∈ D { f ( x , y ) + K ( x , y , Q ( τ ( x , y ) ) ) } ,x∈W,$
(3.2)
where $τ:W×D→W$, $f:W×D→R$, $K:W×D×R→R$.
Let $B(W)$ denote the space of all bounded real-valued functions on W. Clearly, this space endowed with the G-metric given by
is a G-complete metric space.
We will prove the following theorem.
Theorem 3.2 Let $K:W×D×R→R$ and $f:W×D→R$ be two bounded functions and let $A:B(W)→B(W)$ be defined by
$A(hx)= sup y ∈ D { f ( x , y ) + K ( x , y , h ( τ ( x , y ) ) ) }$
(3.3)
for all $h∈B(W)$ and $x∈W$. Assume that the following condition holds:
$|K(x,y,hx)−K(x,y,kx)|≤φ ( | h x − k x | ) ,$
where $x∈W$, $y∈D$ and $φ∈Φ$. Then functional equation (3.2) has a unique bounded solution.
Proof Note that $(B(W),G)$ is a complete G-metric space. Let ε be an arbitrary positive number, $x∈W$ and $h 1 , h 2 ∈B(W)$, then there exist $y 1 , y 2 ∈D$ such that
(3.4)
(3.5)
(3.6)
(3.7)
Then from (3.4) and (3.7), it follows easily that
$A ( h 1 x ) − A ( h 2 x ) < K ( x , y 1 , h 1 ( τ ( x , y 1 ) ) ) − K ( x , y 1 , h 2 ( τ ( x , y 1 ) ) ) + ε ≤ | K ( x , y 1 , h 1 ( τ ( x , y 1 ) ) ) − K ( x , y 1 , h 2 ( τ ( x , y 1 ) ) ) | + ε ≤ φ ( | h 1 x − h 2 x | ) + ε .$
Hence we get
$A( h 1 x)−A( h 2 )x<φ ( | h 1 x − h 2 x | ) +ε.$
(3.8)
Similarly, from (3.5) and (3.6), we obtain
$A( h 2 x)−A( h 1 x)<φ ( | h 1 x − h 2 x | ) +ε.$
(3.9)
Therefore, from (3.8) and (3.9), we have
$|A( h 1 x)−A( h 2 x)|<φ ( | h 1 x − h 2 x | ) +ε,$
(3.10)
which implies
$G ( A ( h 1 x ) , A ( h 2 x ) , A ( h 2 x ) ) <φ ( G ( h 1 x , h 2 x , h 2 x ) ) +ε.$
Since the above inequality is true for any $x∈W$ and $ε>0$ is taken arbitrary, then we conclude immediately that
$G ( A ( h 1 x ) , A ( h 2 x ) , A ( h 2 x ) ) ≤φ ( G ( h 1 x , h 2 x , h 2 x ) ) .$
Thus, all the hypotheses of Corollary 2.2 are satisfied with $A=B$ and $S= I B ( W )$, the identity mapping on $B(W)$. Therefore, functional equation (3.2) has a unique bounded solution. □
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Di Bari C, Vetro P: Common fixed points in generalized metric spaces. Appl. Math. Comput. 2012, 218: 7322–7325. 10.1016/j.amc.2012.01.010
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Mustafa, Z, Sims, B: Some remarks concerning D-metric spaces. In: Proc. Int. Conf. on Fixed Point Theory and Applications, Valencia, Spain, July 2003, pp. 189-198 (2003)
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## Acknowledgements
The authors thank the editor and the referees for their useful comments and suggestions.
## Author information
Authors
### Corresponding author
Correspondence to Saurabh Manro.
### Competing interests
The authors declare that they have no competing interests.
### Authors’ contributions
All authors contributed equally and significantly to writing this manuscript. All authors read and approved the manuscript.
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Manro, S., Bhatia, S.S., Kumar, S. et al. A common fixed point theorem for two weakly compatible pairs in G-metric spaces using the property E.A. Fixed Point Theory Appl 2013, 41 (2013). https://doi.org/10.1186/1687-1812-2013-41 | 7,720 | 21,098 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 275, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.840041 |
https://www.physicsforums.com/threads/linear-first-order-differential-equation-with-unknown-function.769192/ | 1,519,604,661,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817523.0/warc/CC-MAIN-20180225225657-20180226005657-00112.warc.gz | 934,716,804 | 16,082 | # Linear first order differential equation with unknown function
1. Sep 5, 2014
### DMT69
I was wondering if there is a way to get specific numerical values for the following differential equation:
$$f'(x)+ \frac{1}{x-20}\cdot f(x)=\frac{1}{x-20}\cdot g(x)$$
I have numerical values for g(x) for about 10 different x values. I need to find f(x) numerically for those same values.
Is this even possible from this information?
2. Sep 5, 2014
### HallsofIvy
The simplest thing to do is to find the "integrating factor".
That is, you want a function u(x) such that
$$\frac{d(uf)}{dx}= uf'+ u'f= uf'+ \frac{u}{x- 20}f$$
Clearly, that requires that $u'= \frac{u}{x- 20}$
$$\frac{du}{u}= \frac{dx}{x- 20}$$
$$ln(u)= ln(x- 20)+ C$$
$$u= c(x- 20)$$
Since we only need one such function, we can take c= 1. The integrating factor is u= x- 20.
Multiplying both sides of the equation by x- 20, we have
$$((x- 20)f)'= g(x)$$
So that $(x- 20)f= \int_a^x g(t)dt$ where the choice of a gives the "constant of integration".
The solution, then, is $f(x)= \frac{1}{x- 20}\int_a^x g(t) dt$
Use the values for g(x) that you are given to do a numerical integral.
3. Sep 6, 2014
### DMT69
Thanks, I'll give that a try
4. Sep 6, 2014
### DMT69
Okay, I've been reading up on numerical integration, and I have a fair handle on that part. And I am able to follow how you solved the differential equation. But there is one thing that I am still confused about, and this is maybe more a general misunderstanding on my part for differential equation solutions.
It has to do with the lower limit "a" of the integration. How is this chosen appropriately for a practical situation?
Say, for example, I have some values $g(x_0), g(x_1) ... g(x_n)$ and I would like good approximations for $f(x_0), f(x_1) ... f(x_n)$, how would I choose the lower limit of integration in these cases (I assume the upper limit is the value of f(x) I am trying to approximate).
If it helps, I have values for f(20), which is the same as g(20) in all cases. In all cases (I have a bunch of similar cases to calculate) the values I have are for x = {20,50,100,140,200,250,300,350,400,450,500}
5. Sep 6, 2014
### pasmith
The lower limit corresponds to the arbitrary constant of integration.
To obtain a unique solution for $f$, you need to specify the value of $f$ at some $x_0$. You then have $$f(x) = \frac{1}{x - 20}\left((x_0 - 20)f(x_0) + \int_{x_0}^x g(t)\,dt\right)$$
EDIT: If you take $x_0 = 20$ then you must take $f(20) = g(20)$ to avoid a singularity.
6. Sep 6, 2014
### DMT69
Okay, that was what I was thinking since I do have f(20)=g(20) in all cases. Now if I use, for example, Newton-Cotes to approximate the integral for the different points, it seems that I will have a bad approximation for x=50, with increasingly better ones up to x=500. This seems a little arbitrary to me in some sense (shouldn't it be reversible?). Also, why would it get more accurate the further I get away from my "starting point", even if I have more intermediate values at that point? Or could I oscillate somehow? For example, use all the values to get an approximation for 500, but then taken 500 as my starting point and change the sign in front of the integral to a minus to a good value for 50? Or does that not make work?
Thinking a bit more on this, shouldn't having many points increase the accuracy equally for all the points I have?
Last edited: Sep 6, 2014 | 1,023 | 3,440 | {"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} | 4.15625 | 4 | CC-MAIN-2018-09 | latest | en | 0.887759 |
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```(* Title: HOL/Library/Sorting_Algorithms.thy
Author: Florian Haftmann, TU Muenchen
*)
theory Sorting_Algorithms
imports Main Multiset Comparator
begin
section ‹Stably sorted lists›
abbreviation (input) stable_segment :: "'a comparator ⇒ 'a ⇒ 'a list ⇒ 'a list"
where "stable_segment cmp x ≡ filter (λy. compare cmp x y = Equiv)"
fun sorted :: "'a comparator ⇒ 'a list ⇒ bool"
where sorted_Nil: "sorted cmp [] ⟷ True"
| sorted_single: "sorted cmp [x] ⟷ True"
| sorted_rec: "sorted cmp (y # x # xs) ⟷ compare cmp y x ≠ Greater ∧ sorted cmp (x # xs)"
lemma sorted_ConsI:
"sorted cmp (x # xs)" if "sorted cmp xs"
and "⋀y ys. xs = y # ys ⟹ compare cmp x y ≠ Greater"
using that by (cases xs) simp_all
lemma sorted_Cons_imp_sorted:
"sorted cmp xs" if "sorted cmp (x # xs)"
using that by (cases xs) simp_all
lemma sorted_Cons_imp_not_less:
"compare cmp y x ≠ Greater" if "sorted cmp (y # xs)"
and "x ∈ set xs"
using that by (induction xs arbitrary: y) (auto dest: compare.trans_not_greater)
lemma sorted_induct [consumes 1, case_names Nil Cons, induct pred: sorted]:
"P xs" if "sorted cmp xs" and "P []"
and *: "⋀x xs. sorted cmp xs ⟹ P xs
⟹ (⋀y. y ∈ set xs ⟹ compare cmp x y ≠ Greater) ⟹ P (x # xs)"
using ‹sorted cmp xs› proof (induction xs)
case Nil
show ?case
by (rule ‹P []›)
next
case (Cons x xs)
from ‹sorted cmp (x # xs)› have "sorted cmp xs"
by (cases xs) simp_all
moreover have "P xs" using ‹sorted cmp xs›
by (rule Cons.IH)
moreover have "compare cmp x y ≠ Greater" if "y ∈ set xs" for y
using that ‹sorted cmp (x # xs)› proof (induction xs)
case Nil
then show ?case
by simp
next
case (Cons z zs)
then show ?case
proof (cases zs)
case Nil
with Cons.prems show ?thesis
by simp
next
case (Cons w ws)
with Cons.prems have "compare cmp z w ≠ Greater" "compare cmp x z ≠ Greater"
by auto
then have "compare cmp x w ≠ Greater"
by (auto dest: compare.trans_not_greater)
with Cons show ?thesis
using Cons.prems Cons.IH by auto
qed
qed
ultimately show ?case
by (rule *)
qed
lemma sorted_induct_remove1 [consumes 1, case_names Nil minimum]:
"P xs" if "sorted cmp xs" and "P []"
and *: "⋀x xs. sorted cmp xs ⟹ P (remove1 x xs)
⟹ x ∈ set xs ⟹ hd (stable_segment cmp x xs) = x ⟹ (⋀y. y ∈ set xs ⟹ compare cmp x y ≠ Greater)
⟹ P xs"
using ‹sorted cmp xs› proof (induction xs)
case Nil
show ?case
by (rule ‹P []›)
next
case (Cons x xs)
then have "sorted cmp (x # xs)"
moreover note Cons.IH
moreover have "⋀y. compare cmp x y = Greater ⟹ y ∈ set xs ⟹ False"
using Cons.hyps by simp
ultimately show ?case
by (auto intro!: * [of "x # xs" x]) blast
qed
lemma sorted_remove1:
"sorted cmp (remove1 x xs)" if "sorted cmp xs"
proof (cases "x ∈ set xs")
case False
with that show ?thesis
next
case True
with that show ?thesis proof (induction xs)
case Nil
then show ?case
by simp
next
case (Cons y ys)
show ?case proof (cases "x = y")
case True
with Cons.hyps show ?thesis
by simp
next
case False
then have "sorted cmp (remove1 x ys)"
using Cons.IH Cons.prems by auto
then have "sorted cmp (y # remove1 x ys)"
proof (rule sorted_ConsI)
fix z zs
assume "remove1 x ys = z # zs"
with ‹x ≠ y› have "z ∈ set ys"
using notin_set_remove1 [of z ys x] by auto
then show "compare cmp y z ≠ Greater"
by (rule Cons.hyps(2))
qed
with False show ?thesis
by simp
qed
qed
qed
lemma sorted_stable_segment:
"sorted cmp (stable_segment cmp x xs)"
proof (induction xs)
case Nil
show ?case
by simp
next
case (Cons y ys)
then show ?case
by (auto intro!: sorted_ConsI simp add: filter_eq_Cons_iff compare.sym)
(auto dest: compare.trans_equiv simp add: compare.sym compare.greater_iff_sym_less)
qed
primrec insort :: "'a comparator ⇒ 'a ⇒ 'a list ⇒ 'a list"
where "insort cmp y [] = [y]"
| "insort cmp y (x # xs) = (if compare cmp y x ≠ Greater
then y # x # xs
else x # insort cmp y xs)"
lemma mset_insort [simp]:
"mset (insort cmp x xs) = add_mset x (mset xs)"
by (induction xs) simp_all
lemma length_insort [simp]:
"length (insort cmp x xs) = Suc (length xs)"
by (induction xs) simp_all
lemma sorted_insort:
"sorted cmp (insort cmp x xs)" if "sorted cmp xs"
using that proof (induction xs)
case Nil
then show ?case
by simp
next
case (Cons y ys)
then show ?case by (cases ys)
qed
lemma stable_insort_equiv:
"stable_segment cmp y (insort cmp x xs) = x # stable_segment cmp y xs"
if "compare cmp y x = Equiv"
proof (induction xs)
case Nil
from that show ?case
by simp
next
case (Cons z xs)
moreover from that have "compare cmp y z = Equiv ⟹ compare cmp z x = Equiv"
by (auto intro: compare.trans_equiv simp add: compare.sym)
ultimately show ?case
using that by (auto simp add: compare.greater_iff_sym_less)
qed
lemma stable_insort_not_equiv:
"stable_segment cmp y (insort cmp x xs) = stable_segment cmp y xs"
if "compare cmp y x ≠ Equiv"
using that by (induction xs) simp_all
lemma remove1_insort_same_eq [simp]:
"remove1 x (insort cmp x xs) = xs"
by (induction xs) simp_all
lemma insort_eq_ConsI:
"insort cmp x xs = x # xs"
if "sorted cmp xs" "⋀y. y ∈ set xs ⟹ compare cmp x y ≠ Greater"
using that by (induction xs) (simp_all add: compare.greater_iff_sym_less)
lemma remove1_insort_not_same_eq [simp]:
"remove1 y (insort cmp x xs) = insort cmp x (remove1 y xs)"
if "sorted cmp xs" "x ≠ y"
using that proof (induction xs)
case Nil
then show ?case
by simp
next
case (Cons z zs)
show ?case
proof (cases "compare cmp x z = Greater")
case True
with Cons show ?thesis
by simp
next
case False
then have "compare cmp x y ≠ Greater" if "y ∈ set zs" for y
using that Cons.hyps
by (auto dest: compare.trans_not_greater)
with Cons show ?thesis
qed
qed
lemma insort_remove1_same_eq:
"insort cmp x (remove1 x xs) = xs"
if "sorted cmp xs" and "x ∈ set xs" and "hd (stable_segment cmp x xs) = x"
using that proof (induction xs)
case Nil
then show ?case
by simp
next
case (Cons y ys)
then have "compare cmp x y ≠ Less"
then consider "compare cmp x y = Greater" | "compare cmp x y = Equiv"
by (cases "compare cmp x y") auto
then show ?case proof cases
case 1
with Cons.prems Cons.IH show ?thesis
by auto
next
case 2
with Cons.prems have "x = y"
by simp
with Cons.hyps show ?thesis
qed
qed
lemma sorted_append_iff:
"sorted cmp (xs @ ys) ⟷ sorted cmp xs ∧ sorted cmp ys
∧ (∀x ∈ set xs. ∀y ∈ set ys. compare cmp x y ≠ Greater)" (is "?P ⟷ ?R ∧ ?S ∧ ?Q")
proof
assume ?P
have ?R
using ‹?P› by (induction xs)
auto simp add: sorted_Cons_imp_sorted intro: sorted_ConsI)
moreover have ?S
using ‹?P› by (induction xs) (auto dest: sorted_Cons_imp_sorted)
moreover have ?Q
using ‹?P› by (induction xs) (auto simp add: sorted_Cons_imp_not_less,
ultimately show "?R ∧ ?S ∧ ?Q"
by simp
next
assume "?R ∧ ?S ∧ ?Q"
then have ?R ?S ?Q
by simp_all
then show ?P
by (induction xs)
(auto simp add: append_eq_Cons_conv intro!: sorted_ConsI)
qed
definition sort :: "'a comparator ⇒ 'a list ⇒ 'a list"
where "sort cmp xs = foldr (insort cmp) xs []"
lemma sort_simps [simp]:
"sort cmp [] = []"
"sort cmp (x # xs) = insort cmp x (sort cmp xs)"
lemma mset_sort [simp]:
"mset (sort cmp xs) = mset xs"
by (induction xs) simp_all
lemma length_sort [simp]:
"length (sort cmp xs) = length xs"
by (induction xs) simp_all
lemma sorted_sort [simp]:
"sorted cmp (sort cmp xs)"
by (induction xs) (simp_all add: sorted_insort)
lemma stable_sort:
"stable_segment cmp x (sort cmp xs) = stable_segment cmp x xs"
by (induction xs) (simp_all add: stable_insort_equiv stable_insort_not_equiv)
lemma sort_remove1_eq [simp]:
"sort cmp (remove1 x xs) = remove1 x (sort cmp xs)"
by (induction xs) simp_all
lemma set_insort [simp]:
"set (insort cmp x xs) = insert x (set xs)"
by (induction xs) auto
lemma set_sort [simp]:
"set (sort cmp xs) = set xs"
by (induction xs) auto
lemma sort_eqI:
"sort cmp ys = xs"
if permutation: "mset ys = mset xs"
and sorted: "sorted cmp xs"
and stable: "⋀y. y ∈ set ys ⟹
stable_segment cmp y ys = stable_segment cmp y xs"
proof -
have stable': "stable_segment cmp y ys =
stable_segment cmp y xs" for y
proof (cases "∃x∈set ys. compare cmp y x = Equiv")
case True
then obtain z where "z ∈ set ys" and "compare cmp y z = Equiv"
by auto
then have "compare cmp y x = Equiv ⟷ compare cmp z x = Equiv" for x
by (meson compare.sym compare.trans_equiv)
moreover have "stable_segment cmp z ys =
stable_segment cmp z xs"
using ‹z ∈ set ys› by (rule stable)
ultimately show ?thesis
by simp
next
case False
moreover from permutation have "set ys = set xs"
by (rule mset_eq_setD)
ultimately show ?thesis
by simp
qed
show ?thesis
using sorted permutation stable' proof (induction xs arbitrary: ys rule: sorted_induct_remove1)
case Nil
then show ?case
by simp
next
case (minimum x xs)
from ‹mset ys = mset xs› have ys: "set ys = set xs"
by (rule mset_eq_setD)
then have "compare cmp x y ≠ Greater" if "y ∈ set ys" for y
using that minimum.hyps by simp
from minimum.prems have stable: "stable_segment cmp x ys = stable_segment cmp x xs"
by simp
have "sort cmp (remove1 x ys) = remove1 x xs"
by (rule minimum.IH) (simp_all add: minimum.prems filter_remove1)
then have "remove1 x (sort cmp ys) = remove1 x xs"
by simp
then have "insort cmp x (remove1 x (sort cmp ys)) =
insort cmp x (remove1 x xs)"
by simp
also from minimum.hyps ys stable have "insort cmp x (remove1 x (sort cmp ys)) = sort cmp ys"
also from minimum.hyps have "insort cmp x (remove1 x xs) = xs"
finally show ?case .
qed
qed
lemma filter_insort:
"filter P (insort cmp x xs) = insort cmp x (filter P xs)"
if "sorted cmp xs" and "P x"
using that by (induction xs)
lemma filter_insort_triv:
"filter P (insort cmp x xs) = filter P xs"
if "¬ P x"
using that by (induction xs) simp_all
lemma filter_sort:
"filter P (sort cmp xs) = sort cmp (filter P xs)"
by (induction xs) (auto simp add: filter_insort filter_insort_triv)
section ‹Alternative sorting algorithms›
subsection ‹Quicksort›
definition quicksort :: "'a comparator ⇒ 'a list ⇒ 'a list"
where quicksort_is_sort [simp]: "quicksort = sort"
lemma sort_by_quicksort:
"sort = quicksort"
by simp
lemma sort_by_quicksort_rec:
"sort cmp xs = sort cmp [x←xs. compare cmp x (xs ! (length xs div 2)) = Less]
@ stable_segment cmp (xs ! (length xs div 2)) xs
@ sort cmp [x←xs. compare cmp x (xs ! (length xs div 2)) = Greater]" (is "_ = ?rhs")
proof (rule sort_eqI)
show "mset xs = mset ?rhs"
by (rule multiset_eqI) (auto simp add: compare.sym intro: comp.exhaust)
next
show "sorted cmp ?rhs"
by (auto simp add: sorted_append_iff sorted_stable_segment compare.equiv_subst_right dest: compare.trans_greater)
next
let ?pivot = "xs ! (length xs div 2)"
fix l
have "compare cmp x ?pivot = comp ∧ compare cmp l x = Equiv
⟷ compare cmp l ?pivot = comp ∧ compare cmp l x = Equiv" for x comp
proof -
have "compare cmp x ?pivot = comp ⟷ compare cmp l ?pivot = comp"
if "compare cmp l x = Equiv"
using that by (simp add: compare.equiv_subst_left compare.sym)
then show ?thesis by blast
qed
then show "stable_segment cmp l xs = stable_segment cmp l ?rhs"
by (simp add: stable_sort compare.sym [of _ ?pivot])
(cases "compare cmp l ?pivot", simp_all)
qed
context
begin
qualified definition partition :: "'a comparator ⇒ 'a ⇒ 'a list ⇒ 'a list × 'a list × 'a list"
where "partition cmp pivot xs =
([x ← xs. compare cmp x pivot = Less], stable_segment cmp pivot xs, [x ← xs. compare cmp x pivot = Greater])"
qualified lemma partition_code [code]:
"partition cmp pivot [] = ([], [], [])"
"partition cmp pivot (x # xs) =
(let (lts, eqs, gts) = partition cmp pivot xs
in case compare cmp x pivot of
Less ⇒ (x # lts, eqs, gts)
| Equiv ⇒ (lts, x # eqs, gts)
| Greater ⇒ (lts, eqs, x # gts))"
using comp.exhaust by (auto simp add: partition_def Let_def compare.sym [of _ pivot])
lemma quicksort_code [code]:
"quicksort cmp xs =
(case xs of
[] ⇒ []
| [x] ⇒ xs
| [x, y] ⇒ (if compare cmp x y ≠ Greater then xs else [y, x])
| _ ⇒
let (lts, eqs, gts) = partition cmp (xs ! (length xs div 2)) xs
in quicksort cmp lts @ eqs @ quicksort cmp gts)"
proof (cases "length xs ≥ 3")
case False
then have "length xs ∈ {0, 1, 2}"
by (auto simp add: not_le le_less less_antisym)
then consider "xs = []" | x where "xs = [x]" | x y where "xs = [x, y]"
by (auto simp add: length_Suc_conv numeral_2_eq_2)
then show ?thesis
by cases simp_all
next
case True
then obtain x y z zs where "xs = x # y # z # zs"
by (metis le_0_eq length_0_conv length_Cons list.exhaust not_less_eq_eq numeral_3_eq_3)
moreover have "quicksort cmp xs =
(let (lts, eqs, gts) = partition cmp (xs ! (length xs div 2)) xs
in quicksort cmp lts @ eqs @ quicksort cmp gts)"
using sort_by_quicksort_rec [of cmp xs] by (simp add: partition_def)
ultimately show ?thesis
by simp
qed
end
subsection ‹Mergesort›
definition mergesort :: "'a comparator ⇒ 'a list ⇒ 'a list"
where mergesort_is_sort [simp]: "mergesort = sort"
lemma sort_by_mergesort:
"sort = mergesort"
by simp
context
fixes cmp :: "'a comparator"
begin
qualified function merge :: "'a list ⇒ 'a list ⇒ 'a list"
where "merge [] ys = ys"
| "merge xs [] = xs"
| "merge (x # xs) (y # ys) = (if compare cmp x y = Greater
then y # merge (x # xs) ys else x # merge xs (y # ys))"
by pat_completeness auto
qualified termination by lexicographic_order
lemma mset_merge:
"mset (merge xs ys) = mset xs + mset ys"
by (induction xs ys rule: merge.induct) simp_all
lemma merge_eq_Cons_imp:
"xs ≠ [] ∧ z = hd xs ∨ ys ≠ [] ∧ z = hd ys"
if "merge xs ys = z # zs"
using that by (induction xs ys rule: merge.induct) (auto split: if_splits)
lemma filter_merge:
"filter P (merge xs ys) = merge (filter P xs) (filter P ys)"
if "sorted cmp xs" and "sorted cmp ys"
using that proof (induction xs ys rule: merge.induct)
case (1 ys)
then show ?case
by simp
next
case (2 xs)
then show ?case
by simp
next
case (3 x xs y ys)
show ?case
proof (cases "compare cmp x y = Greater")
case True
with 3 have hyp: "filter P (merge (x # xs) ys) =
merge (filter P (x # xs)) (filter P ys)"
show ?thesis
proof (cases "¬ P x ∧ P y")
case False
with ‹compare cmp x y = Greater› show ?thesis
next
case True
from ‹compare cmp x y = Greater› "3.prems"
have *: "compare cmp z y = Greater" if "z ∈ set (filter P xs)" for z
using that by (auto dest: compare.trans_not_greater sorted_Cons_imp_not_less)
from ‹compare cmp x y = Greater› show ?thesis
by (cases "filter P xs") (simp_all add: hyp *)
qed
next
case False
with 3 have hyp: "filter P (merge xs (y # ys)) =
merge (filter P xs) (filter P (y # ys))"
show ?thesis
proof (cases "P x ∧ ¬ P y")
case False
with ‹compare cmp x y ≠ Greater› show ?thesis
next
case True
from ‹compare cmp x y ≠ Greater› "3.prems"
have *: "compare cmp x z ≠ Greater" if "z ∈ set (filter P ys)" for z
using that by (auto dest: compare.trans_not_greater sorted_Cons_imp_not_less)
from ‹compare cmp x y ≠ Greater› show ?thesis
by (cases "filter P ys") (simp_all add: hyp *)
qed
qed
qed
lemma sorted_merge:
"sorted cmp (merge xs ys)" if "sorted cmp xs" and "sorted cmp ys"
using that proof (induction xs ys rule: merge.induct)
case (1 ys)
then show ?case
by simp
next
case (2 xs)
then show ?case
by simp
next
case (3 x xs y ys)
show ?case
proof (cases "compare cmp x y = Greater")
case True
with 3 have "sorted cmp (merge (x # xs) ys)"
then have "sorted cmp (y # merge (x # xs) ys)"
proof (rule sorted_ConsI)
fix z zs
assume "merge (x # xs) ys = z # zs"
with 3(4) True show "compare cmp y z ≠ Greater"
by (clarsimp simp add: sorted_Cons_imp_sorted dest!: merge_eq_Cons_imp)
qed
with True show ?thesis
by simp
next
case False
with 3 have "sorted cmp (merge xs (y # ys))"
then have "sorted cmp (x # merge xs (y # ys))"
proof (rule sorted_ConsI)
fix z zs
assume "merge xs (y # ys) = z # zs"
with 3(3) False show "compare cmp x z ≠ Greater"
by (clarsimp simp add: sorted_Cons_imp_sorted dest!: merge_eq_Cons_imp)
qed
with False show ?thesis
by simp
qed
qed
lemma merge_eq_appendI:
"merge xs ys = xs @ ys"
if "⋀x y. x ∈ set xs ⟹ y ∈ set ys ⟹ compare cmp x y ≠ Greater"
using that by (induction xs ys rule: merge.induct) simp_all
lemma merge_stable_segments:
"merge (stable_segment cmp l xs) (stable_segment cmp l ys) =
stable_segment cmp l xs @ stable_segment cmp l ys"
by (rule merge_eq_appendI) (auto dest: compare.trans_equiv_greater)
lemma sort_by_mergesort_rec:
"sort cmp xs =
merge (sort cmp (take (length xs div 2) xs))
(sort cmp (drop (length xs div 2) xs))" (is "_ = ?rhs")
proof (rule sort_eqI)
have "mset (take (length xs div 2) xs) + mset (drop (length xs div 2) xs) =
mset (take (length xs div 2) xs @ drop (length xs div 2) xs)"
by (simp only: mset_append)
then show "mset xs = mset ?rhs"
next
show "sorted cmp ?rhs"
next
fix l
have "stable_segment cmp l (take (length xs div 2) xs) @ stable_segment cmp l (drop (length xs div 2) xs)
= stable_segment cmp l xs"
by (simp only: filter_append [symmetric] append_take_drop_id)
have "merge (stable_segment cmp l (take (length xs div 2) xs))
(stable_segment cmp l (drop (length xs div 2) xs)) =
stable_segment cmp l (take (length xs div 2) xs) @ stable_segment cmp l (drop (length xs div 2) xs)"
by (rule merge_eq_appendI) (auto simp add: compare.trans_equiv_greater)
also have "… = stable_segment cmp l xs"
by (simp only: filter_append [symmetric] append_take_drop_id)
finally show "stable_segment cmp l xs = stable_segment cmp l ?rhs"
qed
lemma mergesort_code [code]:
"mergesort cmp xs =
(case xs of
[] ⇒ []
| [x] ⇒ xs
| [x, y] ⇒ (if compare cmp x y ≠ Greater then xs else [y, x])
| _ ⇒
let
half = length xs div 2;
ys = take half xs;
zs = drop half xs
in merge (mergesort cmp ys) (mergesort cmp zs))"
proof (cases "length xs ≥ 3")
case False
then have "length xs ∈ {0, 1, 2}"
by (auto simp add: not_le le_less less_antisym)
then consider "xs = []" | x where "xs = [x]" | x y where "xs = [x, y]"
by (auto simp add: length_Suc_conv numeral_2_eq_2)
then show ?thesis
by cases simp_all
next
case True
then obtain x y z zs where "xs = x # y # z # zs"
by (metis le_0_eq length_0_conv length_Cons list.exhaust not_less_eq_eq numeral_3_eq_3)
moreover have "mergesort cmp xs =
(let
half = length xs div 2;
ys = take half xs;
zs = drop half xs
in merge (mergesort cmp ys) (mergesort cmp zs))"
using sort_by_mergesort_rec [of xs] by (simp add: Let_def)
ultimately show ?thesis
by simp
qed
end
end
``` | 5,724 | 18,019 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.77958 |
https://socratic.org/questions/how-do-you-find-an-equivalent-algebraic-expression-for-the-composition-cos-arcsi#196994 | 1,726,004,695,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00873.warc.gz | 498,772,595 | 5,879 | # How do you find an equivalent algebraic expression for the composition cos(arcsin(x))?
##### 1 Answer
Dec 6, 2015
$\sqrt{1 - {x}^{2}}$.
#### Explanation:
Since ${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$, you have that $\cos \left(x\right) = \setminus \pm \setminus \sqrt{1 - {\sin}^{2} \left(x\right)}$. So, your expression becomes
sqrt(1-sin^2(arcsin(x))
And since $\sin \left(\arcsin \left(x\right)\right) = x$, then ${\sin}^{2} \left(\arcsin \left(x\right)\right) = {x}^{2}$.
So, your expression becomes $\sqrt{1 - {x}^{2}}$. | 205 | 554 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.566225 |
http://www.talkstats.com/threads/same-flat-tire.28544/?p=151919 | 1,516,255,314,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887067.27/warc/CC-MAIN-20180118051833-20180118071833-00094.warc.gz | 562,810,773 | 11,246 | # Same flat tire
#### dina786
##### New Member
Hello, I am unsure about how to solve this one:
It's Friday afternoon and six fraternity brothers at UNC have just attended their last class of the semester. To unwind before the final exam Monday morning, they decide to spend the weekend at the beach. After a weekend of immature, irresponsible behavior, they oversleep Monday morning and miss the final exam. Later that day they go to their professor (who received her Ph.D. from NC State) and explain that they went to the beach and were on the way back for the final exam but their car (a BMW of course) had a flat tire. The professor says that they can come back the next morning and she will give them a make-up final exam consisting of one short-answer question! The students are ecstatic and high-five each other all the way back to the frat house. The next morning the professor puts each student in a different room and hands them their one-question test. The question: Which tire?
What is the probability that all 6 students will say that the same tire was flat?
Hint: the answer is NOT (1/4)^6.
Thanks.
##### New Member
(1/4)^5 since the first one can answer any tire, then other 5 must match
#### Dason
Nope! But now that I'm at ISU I'm obliged to say that clearly ISU is the better institution. Also you should join in on the chatbox at some point.
#### dina786
##### New Member
you are welcome, i made the same mistake the first time i got a similar question
#### statsnut
##### New Member
Can anyone help?
Four students failed to show up in Final Exam. They claimed that they shared a car on a field trip one week before the exam and had a flat tire, thus could not get back in time. They asked for a make-up exam. The professor agreed. He asks: Which tire? So, If all four of them claimed a same tire, how confident will the professor be to believe they are telling the truth?
My questions are:
1. Which one is the null hypothesis? Students told the truth or students lied?
2. What is the appropriate rejection region of the selected hypothesis test?
3. What is the Type I Error rate α of the rejection region you defined in the previous question?
4. Having collected the four students' answers, how do you define the p-value of such answers for this test?
@statsnut! | 517 | 2,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-05 | latest | en | 0.972647 |
https://www.physicsforums.com/threads/momentum-is-quantity-of-motion-m-v-does-it-leave-room-for-improvement.515823/ | 1,511,096,902,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805578.23/warc/CC-MAIN-20171119115102-20171119135102-00609.warc.gz | 830,375,739 | 20,358 | # Momentum is (quantity of motion), = ( m * v). Does it leave room for improvement?
1. Jul 21, 2011
### simpatico
I know definition is hard job, expecially
I) verbal definition,
which must find the essence (nature, substance,...) of an entity (being: [ object, thing..], [concept, idea, category...]),
and needs interdisciplinary help from ontology, linguistics (morpho-syntax, semantics), logic and epistemology.....
but I do hope a phisicist could explain why Quantity of Motion ?
just two questions:
1) why motion? it doesnt seem a technical (precise,defined...) term, I never found it elsewhere. Is it different from movement, speed ,velocity v ?
I know v is a vector, speed is not. But motion?
2) Why add quantity? is "quantity of motion" a different category from motion
is "quantity of energy", (books,pears...) different from energy (...)
II) numeric definition : p = m * v (mass-by-velocity)
seems simpler (just arithmetics) and more precise, effective, decisive, conclusive...
But please read the following with care, it is not a joke but a very profound issue
1) we have ......
......... 10 (sweets) and 5 (children), let's multiply :
........ 10 * 5 = 50 (fifty what? I say: children, and I am not joking)
someone says: NO WAY! it's sweets.He is using commonsense and (probably) he's right, but can he disprove me with scientific foolproof arguments?
2) Now we have 10 (crates) and 5 (boxes) , we get 50 what?
3)are we allowed to create a new category just by multiplication?...
so we'd have sweet-by-child, crate-by-box, jam-by-jar ... ad lib ?
2. Jul 21, 2011
### mikeph
You would multiply the units as well as the numerical values. So 10 sweets times 5 children equals 50 sweets.children. The units (what you call category) don't make any sense which reflects the nonsensical multiplication. But if you stick to this process, you will at least be self-consistent. Look at some physical constants... Stefan-Boltzmann constant, 5.67...*10^-8 J m^-2 s^-1 K^-4. To me that number doesn't make any sense. It's an energy per unit area per second per temperature^4. But if you discard the units, you are losing valuable physical information that could possible be very useful later on.
I've never heard it "quantity of motion", I guess you can call it whatever you want as long as your understanding of its meaning matches everyone else's. I don't like that definition for reasons that you posted. "Motion" is neither scalar nor vector, and "quantity" does not give you any idea that we are measuring mass.
3. Jul 21, 2011
### Staff: Mentor
No, he is wrong. It is 50 children sweets. Is he not aware of how to multiply units?
Of course you are allowed. You may even find something useful for your new units (e.g. predicting the number of screaming children), but you are allowed regardless.
4. Jul 21, 2011
### simpatico
1) unit is different from category, you might call it "a basic category" (space, time...)
or other(joule,...)
I purposedly avoided this word because
a) in physics it has a precise technical meaning, a 'connotation' I want to avoid in a verbal definition
b) It brings along the shadow of dimension
c) I do not wish that students who do not grasp the issue bring in elements, (such as inertia,unit,dimension,energy....) foreign to the discussion
It was started by Newton and Wallis, who,as it seems independently used the Latin 'quantitas motum'.It was then translated in most languages as
( 'cantitad de movimiento, quantità di moto, quantitè de mouvement, bewegungsmenge,quantity of motion..English prefers the short term,...)
3)I'm sorry about that too, because I asked for a better definition not for a worse!
Thank you anyway, your post is very useful for future posts
Last edited: Jul 21, 2011
5. Jul 21, 2011
### jambaugh
Here's the thing about formal definitions. Since you have to use terms to define terms then definitions are either cyclic (as in common language) or you start with certain undefined terms (as in formal mathematics) or a mixture of both (science).
Ultimately in physics observable quantities are defined by how they are observed. Other derivative quantities are defined in terms of observable quantities.
• time is what we measure with clocks,
• distance is what we measure with measuring rods,
• speed is distance/time or time rate of change of distance,
• velocity is displacement/time (keeping track of direction as well.)
momentum is usually a primary observable (what we measure in momentum observation experiments e.g with a http://en.wikipedia.org/wiki/Ballistic_pendulum" [Broken]) and we use the relationship between momentum and velocity to define (inertial) mass.
Of course the conventions of definitional primacy vary depending on the sub-discipline.
Last edited by a moderator: May 5, 2017
6. Jul 21, 2011
### simpatico
Hi, Sir, I'm honoured you're here... (:surprised)
...your arguments are so sound!, it's the first time I hear a scientist (mad or not) acknowledge his limits. Now..
..
1)(Common language), Linguists are lucky because most words mean objects, so they can use ostensive definitions, pictorial dictionaries, also use basic intuitive terms and so often are able to avoid circular (or cyclic) definitions.
2)Chemists are the luckiest,as , in a way,
they are Ontologists because they deal with essences
they can create new categories by addition.... what is water ? (H+H+O)
............wonderful! .
3) pardon my ignorance , I don't understand what is a derivative quantity.
I guess an observable quantity may be
g, (the effect, the phenomenon)
but from that you derive (deduce), I suppose,
(the cause) e.g. force, but not quantities.
or do you mean the formula ?
......
Now, coming to physics, the question is:
Can we ( I mean you, specialists, I am just alien,an autistic savant)
accept that physics (science by definition) give
circular definitions :
"energy is capacity to do work" and "work is....transferred energy" ?
(isn't it dangerous for the development of any sound theory?..If you acknowledged what really momentum is you wouldn't entangle in the photon-momentum senseless debate), or
intuitive definitions: Force is an influence, a push or a pull...
once,discussing with a friend, I questioned the force of gravity and its formula but could not conclude the debate, because he wouldn't produce a definition (and so a justification of its formula).
.....
Lastly, what do you think, we get 50 what ?
(please do not mention units: that's simple arithmetic, not definition.)
Last edited: Jul 21, 2011
7. Jul 21, 2011
### Staff: Mentor
You missed the most important part of his post.
That is what keeps physics definitions from being circular.
8. Jul 21, 2011
### simpatico
the reason for this is very simple : unit is an ambiguous word, in physics:
it has pole-asunder meanings
abstract idea : space
concrete thing: meter, rod
multiple (concrete): kilometer
......
it is also used as an inappropriate synonym for dimension
which is a meaningless word ,a non-existing category, when you use, as you should, ONE system of units.
It is a tool which is necessary only when you keep more than one, to convert from one to the other avoiding BABEL
It is dangerous in a formal (theoretical) discussion
9. Jul 22, 2011
### Staff: Mentor
In physics, when you plug numbers into equations those numbers have units. If you are not talking about units then your topic seems to be not relevant to physics. In particular, you seem concerned about whether or not it is valid to multiply a mass by a velocity to get a new quantity: it is.
Are you actually interested in discussing physics here or not?
10. Jul 22, 2011
### mikeph
Units are anything but ambiguous. 'space' is not a unit, nor is 'rod'. Without units, numbers have no connection to observations. If I weigh some apples and someone forbids me to use units, I'll end up saying "three ____ weigh seven ___ on this scale". The sentence is missing all its nouns. They're a crucial part of any coherent thought about reality.
11. Jul 22, 2011
### BruceW
I)
I went on wikipedia, and apparently "quantity of motion" was a term that Rene Descartes used to describe the product of size and speed. (Not the same meaning as momentum). And then when the concept of momentum was beginning to be invented, the term "quantity of motion" was initially used in several places. (I guess because the phrase seemed similar to what they were trying to describe).
So I guess the meaning of the phrase "quantity of motion" has changed since Descartes first invented it.
1) The term "motion" could mean several things. Einstein used it to mean the velocity, I think.
II) 3) Yeah, we can make whatever units we want, for example I could use foxes times rabbits as a unit useful for measuring the biodiversity. | 2,100 | 8,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-47 | longest | en | 0.891682 |
http://www.dreamincode.net/forums/topic/276183-priority-queues-arrays/page__p__1605502 | 1,462,066,483,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860113541.87/warc/CC-MAIN-20160428161513-00021-ip-10-239-7-51.ec2.internal.warc.gz | 494,148,523 | 22,339 | Page 1 of 1
## Priority Queues- Arrays Rate Topic: 1 Votes //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax§ion=topics&do=rateTopic&t=276183&s=1659fa1d2565da7ee7158480d9e72a83&md5check=' + ipb.vars['secure_hash'], cur_rating: 4, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>
### #1 karabasf
• D.I.C Regular
Reputation: 202
• Posts: 417
• Joined: 29-August 10
Posted 21 April 2012 - 04:09 PM
Level: Beginner - Intermediate
Assumed knowledge: before you can start this tutorial, you are assumed to have knowledge of:
• Arrays
• The difference between primitive data types and Objects (e.g. difference between int and Integer)
• Conditional statements
• Loops (for and while loop constructions)
• Basic Knowledge of Object Oriented Programming (mainly methods)
This tutorial will cover the implementation of a Sorted Array by means of (surprise!) an Array. Sorted Arrays can be used to to count sequence of numbers, retrieve the maximum values of an array and to sort data in either an ascending or a descending trend. Furthermore, the insertion method of a Sorted Array ensures that the array does not need to be sorted after an arbitrary element is inserted. Possible applications of a Sorted Array would be a Priority Queue or a ranked list (think of a high score list for example).
Although a linked list could be used for the implementation (which would then be a Sorted List instead), an array provides basic insight to implement the (insertion and removal) in a linked list. As this is not the scope of this tutorial, this assignment is left for the reader.
So before we start, we need to identify variables we need. First, we need a value to denote the capacity of the array. As we will be using an Object array (we're not using an array of primitive datatypes, this will become clear later in this tutorial), we will also need a variable denoting the size (amount of stored elements) in the array. Finally, we need an Object (for this example we'll be using the Integer object) to store the elements. So, to summarize, we need the following variables:
• size, denotes the amount of stored items in the array
• capacity, denotes the amount of items which can be stored in the array
• Integer[] sortedArray, to actually store the elements in the array
Now the variables are identified, the operations of a Sorted Array. We should be able to add and to remove an item. Obviously, we need a string representation and a constructor which takes the capacity of the Sorted Array. So we can already identify two methods and one constructor. Next to these values, I'll use some helper methods to determine whether the Sorted Array is empty or full, and a method to double the capacity of the Sorted Array when it is full. So, in short, we can identify the following methods:
• boolean isEmpty(), to determine if the Sorted Array is empty
• boolean isFull(), to determine if the Sorted Array is full
• void doubleCapacity(), to double the capacity of the array and to copy the original elements in the new array with twice the capacity
• void add(int i), method to add an integer (note that I use a primitive data type here!) to the Sorted Array
• Integer remove(int i), method to remove an integer at position i of the Sorted Array
• int size(), a method to retrieve the amount of elements in the Sorted Array
So now all the methods and variables are defined, lets start with the constructor. The constructor should take the capacity as an argument and instantiate our Integer[] sortedArray with that size.
However, we want to prevent that the user enters a negative number or a number equal to 0. If we would instantiate the Sorted Array with a negative number, we would get a NegativeArraySizeException. This can be easily prevented by checking the input of capacity and if the capacity is smaller or equal to 0, we instantiate the sorted array with a standard capacity of lets say 10 (any value is possible). This check is fully optional, so you can omit it if you want.
Using this knowledge, our constructor looks like:
```public class SortedArray{
private Integer[] elements; //The Integer array to store the elements
private int storedElements; //An integer denoting the amount of stored elements of the Sorted Array
private int capacity; //An integer denoting the capacity of the Sorted Array
//Default constructor, creates a sorted array with the capacity 10
public SortedArray(){
//Call the second constructor to create a Sorted Array with a capacity of 10
this(10);
}
//Overloaded constructor, creates a Sorted Array with an user specified capacity
public SortedArray(int capacity){
//Create the Sorted Array with the user capacity. If capacity is 0 or negative, construct a default Sorted Array
if(capacity > 0){
this.capacity = capacity;
this.elements= new Integer[this.capacity];
}
else{
this.capacity = 10;
this.elements= new Integer[this.capacity];
}
this.storedElements = 0;
}
```
Now, we can add the helper methods isEmpty(), isFull() and the method size(). Before we add these, we need to determine to determine when the Sorted Array is empty. This is only when the amount of storedElements is equal to 0. Same for when the list is full, this is when the amount of storedElements equals to the capacity of the Sorted Array. Thus, we can add:
```//Determine if the SortedArray is full
public boolean isFull(){
return (storedElements == capacity);
}
//Determine if the SortedArray is empty
public boolean isEmpty(){
return (storedElements == 0);
}
//Retrieve the amount of stored elements in the Sorted Array
public int size(){
return storedElements;
}
```
So now we defined our helper methods, lets define the doubleCapacity() method. We will be using the Arrays.copyof() method, which can be found in java.util, meaning that we have to import this package. Add import java.util.* in order to do so. So, as we want to double our capacity, this means that the variable capacity needs to become twice as big as it originally was. Thus:
```//Double the capacity of the array and retain the original elements
private void doubleCapacity(){
capacity = 2 * capacity; //Double the capacity
this.elements= Arrays.copyOf(elements, capacity);
}
```
Note that it is not necessary to use a doubleCapacity() method, the implementation of it is optional for the functionining of the Sorted Array. Furthermore, I made this method private as I do not want to have users modifying the capacity of the Sorted Array manually; the Sorted Array should increase the capacity only when it is needed.
Now we get to the more interesting part, the add and remove algorithms. Lets consider the add algorithm first for a descending Sorted Array:
```add(int newElement)
//Create an integer object in order to insert it in our Integer[] elements array
Integer newEntry = new Integer(newElement);
//Double the capacity of the array when it is full
if(sortedArray is full)
doubleCapacity(); //Note that this is optional. We could also decide to throw an exception or just return the method after applying some logic
//If the list is empty, we can add the element at first index
if(sortedArray is empty){
elements[0] = newEntry;
}
else{ //the element needs to be added elsewhere
//We will start the add from the tail and shift elements
int i= this.size();
//Start looping from the end of the Sorted Array
for(;(i>=1) && (elements[i-1].intValue() < newEntry.intValue()); i--)
elements[i] = elements[i-1]; //Shift the lower elements to the right
//Add the new entry at position i
elements[i] = newEntry;
}
storedElements++;
```
Note that when one wants to achieve an ascending sorted array, the sign of the element value comparison can be swicthed, e.g "<" becomes ">"
So how does this look like? Well, consider we have the following Sorted Array and we want to add the number 8
```[10][8][8][7][7][5][null] //Note that null elements can be present in our Integer array
```
So, if we want to add 8, we have to move the elements 7, 7, 5 one step towards the right. Visualizing this:
```#Iteration 1
i = 6
[10][8][8][7][7][5][5]
#Iteration 2
i = 5
[10][8][8][7][7][7][5]
#Iteration 3
i = 4
[10][8][8][X][7][7][5]
//Stop iterations
//Add the new entry at X (the 4th element)
[10][8][8][8][7][7][5] //Our final result
```
Implementing our add method in Java results in:
```//Method to add an element in the sorted array
//Create an Integer object to insert the entry in the Sorted Array
Integer newEntry = new Integer(newElement);
//Check if the list is full. If so, double the capacity
if(this.isFull())
doubleCapacity();
//If the array is empty, the new element needs to be added at the first index
if(this.isEmpty()){
element[0] = newEntry;
}
else{
//Get the amount of stored elements and loop from that element
int i = this.size();
//Loop till till the condition is met where the element should be added
for(;((i >= 1) && (elements[i-1].intValue() > newElement)); i--)
elements[i] = elements[i-1]; //Swap the elements
//Add the element at position i
elements[i] = newEntry;
}
//Increment the amount of stored elements
storedElements++;
}
```
Now, one of our final methods will be the remove method. Consider this algorithm:
```remove (int i)
if(this.isEmpty() || (i < 0) || (i >= size))
return null; //If the index i is outside the regions where the elements are stored or the array is empty we can return null, as no element was removed (this was one of the reasons of why to use an object Integer rather than a primitive datatype).
//Store the temp element to return
Integer deletedElement = elements[i];
//Start removing the element and shift the subsequent elements to the "empty spot"
for(; i < this.size() - 1 ; i++)
elements[i] = elements[i+1];
//Set the last element (located at size - 1) equal to null
elements[this.size() - 1] = null;
//Decrease the amount of stored elements
storedElements--;
//Return the removed Integer object
return deletedElement;
```
So, lets visualize this again. Suppose we want to remove the first element (i = 0) from the following Sorted Array
```[10][5][6][7][null]
```
Therefore, if we remove element [10] we need to shift all of our elements to the left. Furthermore, we need to make the direct element left of the null element equal to null. Visualizing this:
```#Iteration 1
[5][5][6][7][null]
#Iteration 2
[5][6][6][7][null]
#Iteration 3
[5][6][7][7][null]
//Stop iterations
//Make the element at the position size-1 (remember that the original size was 4. Also, note that arrays are 0 based, so we need to set at index i = 3) is equal to null
[5][6][7][null][null]
```
Implementing this in Java:
```public Integer remove(int i){
//If the index i is outside the regions where the elements are stored or the array is empty we can return null, as no element is apparent to be removed
if(this.isEmpty() || (i < 0) || (i >= size))
return null;
//Store the temp element to return
Integer deletedElement = elements[i];
//Start removing the element and shift the subsequent elements to the "empty spot"
for(; i < this.size() - 1 ; i++)
elements[i] = elements[i+1];
//Set the last element at the position of size-1 equal to null
elements[this.size()-1] = null;
//Decrease the amount of stored elements
storedElements--;
//Return the removed Integer object
return deletedElement;
}
```
Finally, we need a string representation of our SortedArray in Java. We know we need to loop from 0 till the amount of stored elements -1 (remember that arrays are zero based). A string representation of our SortedArray may look like:
```//The string representation of the Sorted Array. Note that our string representation only shows the stored elements and not the full array
//Customizable
public String toString(){
if(this.isEmpty())
return "[ ]";
else{
String representation = "["
//Append the string with entries
for(int i = 0; i < this.size() ; i++)
representation += elements[i] + ", ";
//Trim away the last comma and return the string
return representation.substring(0, representation.length() - 2) +"]";
}
}
```
The final code should look like:
Spoiler
To summarize everything, we designed a Sorted Array. Although some of its operations are still limited, could be expanded or can be differently specified, the reader should now be able to easily expand or modify the functionality of the Sorted Array. Some ideas would be:
• getIndex(int value), get the index of where an element is stored (Think smart, the array is sorted!
• remove(int value), instead of removing an integer by passing the position, remove the integer based on the value
• toArray(), make a method which transforms the Sorted Array from i = 0 till i = amount of stored elements to an array
• countNr(int value), count the occurence of a arbitrary number
Furthermore, with the logic of the proposed algorithms, the reader should be able to design its own Sorted (Double) Linked List. Adding and removing elements however differs from an ordinary array. As this is not the scope of this tutorial, the excercise is left to the reader.
Finally, it should be noted that a sorted array of objects can only be implemented when the objects can be compared against eachother, e.g. there should be a definition for when an Object1 has a "larger value/is bigger" than Object2.
I hope this tutorial was helpful for you
References:
Data Structures & Algorithms in Java by Michael T. Goodrich and Roberto Tamassia
Is This A Good Question/Topic? 2
Page 1 of 1
.related ul{list-style-type:circle;font-size:12px;font-weight:bold;}.related li{margin-bottom:5px;background-position:left 7px!important;margin-left:-35px;}.related h2{font-size:18px;font-weight:bold;}.related a{color:blue;} | 3,260 | 13,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-18 | latest | en | 0.809701 |
https://support.nag.com/numeric/nl/nagdoc_28.4/flhtml/g02/g02cdf.html | 1,702,073,248,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00445.warc.gz | 614,931,452 | 7,892 | # NAG FL Interfaceg02cdf (linregs_noconst_miss)
## ▸▿ Contents
Settings help
FL Name Style:
FL Specification Language:
## 1Purpose
g02cdf performs a simple linear regression with no constant, with dependent variable $y$ and independent variable $x$, omitting cases involving missing values.
## 2Specification
Fortran Interface
Subroutine g02cdf ( n, x, y,
Integer, Intent (In) :: n Integer, Intent (Inout) :: ifail Real (Kind=nag_wp), Intent (In) :: x(n), y(n), xmiss, ymiss Real (Kind=nag_wp), Intent (Out) :: result(21)
#include <nag.h>
void g02cdf_ (const Integer *n, const double x[], const double y[], const double *xmiss, const double *ymiss, double result[], Integer *ifail)
The routine may be called by the names g02cdf or nagf_correg_linregs_noconst_miss.
## 3Description
g02cdf fits a straight line of the form
$y=bx$
to those of the data points
$(x1,y1),(x2,y2),…,(xn,yn)$
that do not include missing values, such that
$yi=bxi+ei$
for those $\left({x}_{i},{y}_{i}\right)$, for $i=1,2,\dots ,n\text{ }\left(n\ge 2\right)$ which do not include missing values.
The routine eliminates all pairs of observations $\left({x}_{i},{y}_{i}\right)$ which contain a missing value for either $x$ or $y$, and then calculates the regression coefficient, $b$, and various other statistical quantities by minimizing the sum of the ${e}_{i}^{2}$ over those cases remaining in the calculations.
The input data consists of the $n$ pairs of observations $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right),\dots ,\left({x}_{n},{y}_{n}\right)$ on the independent variable $x$ and the dependent variable $y$.
In addition two values, $\mathit{xm}$ and $\mathit{ym}$, are given which are considered to represent missing observations for $x$ and $y$ respectively. (See Section 7).
Let ${w}_{\mathit{i}}=0$, if the $\mathit{i}$th observation of either $x$ or $y$ is missing, i.e., if ${x}_{\mathit{i}}=\mathit{xm}$ and/or ${y}_{\mathit{i}}=\mathit{ym}$; and ${w}_{\mathit{i}}=1$ otherwise, for $\mathit{i}=1,2,\dots ,n$.
The quantities calculated are:
1. (a)Means:
$x¯=∑i=1nwixi ∑i=1nwi ; y¯=∑i=1nwiyi ∑i=1nwi .$
2. (b)Standard deviations:
$sx=∑i= 1nwi (xi-x¯) 2 ∑i= 1nwi- 1 ; sy=∑i= 1nwi (yi-y¯) 2 ∑i= 1nwi- 1 .$
3. (c)Pearson product-moment correlation coefficient:
$r=∑i=1nwi(xi-x¯)(yi-y¯) ∑i=1nwi (xi-x¯) 2∑i=1nwi (yi-y¯) 2 .$
4. (d)The regression coefficient, $b$:
$b=∑i=1nwixiyi ∑i=1nwixi2 .$
5. (e)The sum of squares attributable to the regression, $SSR$, the sum of squares of deviations about the regression, $SSD$, and the total sum of squares, $SST$:
$SST=∑i=1nwiyi2; SSD=∑i=1nwi (yi-bxi) 2; SSR=SST-SSD.$
6. (f)The degrees of freedom attributable to the regression, $DFR$, the degrees of freedom of deviations about the regression, $DFD$, and the total degrees of freedom, $DFT$:
$DFT=∑i=1nwi; DFD=∑i=1nwi-1; DFR=1.$
7. (g)The mean square attributable to the regression, $MSR$, and the mean square of deviations about the regression, $MSD$:
$MSR=SSR/DFR; MSD=SSD/DFD.$
8. (h)The $F$ value for the analysis of variance:
$F=MSR/MSD.$
9. (i)The standard error of the regression coefficient:
$se(b)=MSD ∑i= 1nwixi2 .$
10. (j)The $t$ value for the regression coefficient:
$t(b)=bse(b) .$
11. (k)The number of observations used in the calculations:
$nc=∑i= 1nwi.$
## 4References
Draper N R and Smith H (1985) Applied Regression Analysis (2nd Edition) Wiley
## 5Arguments
1: $\mathbf{n}$Integer Input
On entry: $n$, the number of pairs of observations.
Constraint: ${\mathbf{n}}\ge 2$.
2: $\mathbf{x}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Input
On entry: ${\mathbf{x}}\left(\mathit{i}\right)$ must contain ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
3: $\mathbf{y}\left({\mathbf{n}}\right)$Real (Kind=nag_wp) array Input
On entry: ${\mathbf{y}}\left(\mathit{i}\right)$ must contain ${y}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
4: $\mathbf{xmiss}$Real (Kind=nag_wp) Input
On entry: the value $xm$, which is to be taken as the missing value for the variable $x$ (see Section 7).
5: $\mathbf{ymiss}$Real (Kind=nag_wp) Input
On entry: the value $ym$, which is to be taken as the missing value for the variable $y$ (see Section 7).
6: $\mathbf{result}\left(21\right)$Real (Kind=nag_wp) array Output
On exit: the following information:
${\mathbf{result}}\left(1\right)$ $\overline{x}$, the mean value of the independent variable, $x$; ${\mathbf{result}}\left(2\right)$ $\overline{y}$, the mean value of the dependent variable, $y$; ${\mathbf{result}}\left(3\right)$ ${s}_{x}$, the standard deviation of the independent variable, $x$; ${\mathbf{result}}\left(4\right)$ ${s}_{y}$, the standard deviation of the dependent variable, $y$; ${\mathbf{result}}\left(5\right)$ $r$, the Pearson product-moment correlation between the independent variable $x$ and the dependent variable, $y$; ${\mathbf{result}}\left(6\right)$ $b$, the regression coefficient; ${\mathbf{result}}\left(7\right)$ the value $0.0$; ${\mathbf{result}}\left(8\right)$ $se\left(b\right)$, the standard error of the regression coefficient; ${\mathbf{result}}\left(9\right)$ the value $0.0$; ${\mathbf{result}}\left(10\right)$ $t\left(b\right)$, the $t$ value for the regression coefficient; ${\mathbf{result}}\left(11\right)$ the value $0.0$; ${\mathbf{result}}\left(12\right)$ $SSR$, the sum of squares attributable to the regression; ${\mathbf{result}}\left(13\right)$ $DFR$, the degrees of freedom attributable to the regression; ${\mathbf{result}}\left(14\right)$ $MSR$, the mean square attributable to the regression; ${\mathbf{result}}\left(15\right)$ $F$, the $F$ value for the analysis of variance; ${\mathbf{result}}\left(16\right)$ $SSD$, the sum of squares of deviations about the regression; ${\mathbf{result}}\left(17\right)$ $DFD$, the degrees of freedom of deviations about the regression; ${\mathbf{result}}\left(18\right)$ $MSD$, the mean square of deviations about the regression; ${\mathbf{result}}\left(19\right)$ $SST$, the total sum of squares ${\mathbf{result}}\left(20\right)$ $DFT$, the total degrees of freedom; ${\mathbf{result}}\left(21\right)$ ${n}_{c}$, the number of observations used in the calculations.
7: $\mathbf{ifail}$Integer Input/Output
On entry: ifail must be set to $0$, $-1$ or $1$ to set behaviour on detection of an error; these values have no effect when no error is detected.
A value of $0$ causes the printing of an error message and program execution will be halted; otherwise program execution continues. A value of $-1$ means that an error message is printed while a value of $1$ means that it is not.
If halting is not appropriate, the value $-1$ or $1$ is recommended. If message printing is undesirable, then the value $1$ is recommended. Otherwise, the value $0$ is recommended. When the value $-\mathbf{1}$ or $\mathbf{1}$ is used it is essential to test the value of ifail on exit.
On exit: ${\mathbf{ifail}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6Error Indicators and Warnings
If on entry ${\mathbf{ifail}}=0$ or $-1$, explanatory error messages are output on the current error message unit (as defined by x04aaf).
Errors or warnings detected by the routine:
${\mathbf{ifail}}=1$
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 2$.
${\mathbf{ifail}}=2$
After observations with missing values were omitted, fewer than two cases remained.
${\mathbf{ifail}}=3$
After observations with missing values were omitted, all remaining values of at least one of x and y were identical.
${\mathbf{ifail}}=-99$
See Section 7 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library FL Interface for further information.
${\mathbf{ifail}}=-999$
Dynamic memory allocation failed.
See Section 9 in the Introduction to the NAG Library FL Interface for further information.
## 7Accuracy
g02cdf does not use additional precision arithmetic for the accumulation of scalar products, so there may be a loss of significant figures for large $n$.
You are warned of the need to exercise extreme care in your selection of missing values. g02cdf treats all values in the inclusive range $\left(1±{0.1}^{\left({\mathbf{x02bef}}-2\right)}\right)×{xm}_{j}$, where ${\mathit{xm}}_{j}$ is the missing value for variable $j$ specified in xmiss.
You must, therefore, ensure that the missing value chosen for each variable is sufficiently different from all valid values for that variable so that none of the valid values fall within the range indicated above.
If, in calculating $F$ or $t\left(b\right)$ (see Section 3), the numbers involved are such that the result would be outside the range of numbers which can be stored by the machine, then the answer is set to the largest quantity which can be stored as a real variable, by means of a call to x02alf.
## 8Parallelism and Performance
g02cdf is not threaded in any implementation.
The time taken by g02cdf depends on $n$ and the number of missing observations.
The routine uses a two-pass algorithm.
## 10Example
This example reads in eight observations on each of two variables, and then performs a simple linear regression with no constant, with the first variable as the independent variable, and the second variable as the dependent variable, omitting cases involving missing values ($0.0$ for the first variable, $99.0$ for the second). Finally the results are printed.
### 10.1Program Text
Program Text (g02cdfe.f90)
### 10.2Program Data
Program Data (g02cdfe.d)
### 10.3Program Results
Program Results (g02cdfe.r) | 2,974 | 9,693 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 151, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-50 | latest | en | 0.619067 |
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Fiscal sustainability exercise
# Fiscal sustainability exercise - EC 1500 Fiscal...
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EC 1500: Fiscal sustainability exercise From class notes, you know the following: Deficit = B(t) - B(t-1) = REAL government debt Where: B(t) = Public debt (bonds) outstanding at time t Deficit = B(t) - B(t-1) = rB(t-1) + G(t) - T(t) Where: r = real interest rate G(t) = government spending at time t T(t) = government revenues(from taxes)at time t rB(t-1) = REAL interest payments G(t) - T(t) = REAL primary deficit And we made these transformations: B(t) - B(t-1) = rB(t-1) + G(t) - T(t) B(t) = (1+r)B(t-1) + G(t) - T(t) B(t)/Y(t) = (1+r)B(t-1)/Y(t) + [G(t) - T(t)]/Y(t) Where Y(t)= output, or income, or GDP in year t now multiply first term on the RHS by Y(t-1)/Y(t-1) B(t)/Y(t) = (1+r)[Y(t-1)/Y(t)][B(t-1)/Y(t-1)] + [G(t) - T(t)]/Y(t) note that [Y(t)-Y(t-1)]/Y(t-1) = growth = g Y(t)/Y(t-1) - 1 = g, 1+g =Y(t)/Y(t-1) 1/[1+g] = Y(t-1)/Y(t), thus … (1) B(t)/Y(t) = [(1+r)/(1+g)][B(t-1)/Y(t-1)] + [G(t) - T(t)]/Y(t) Using the approximation (1+r)/1+g) = 1+ r - g B(t)/Y(t) = [1 + r - g][B(t-1)/Y(t-1)] + [G(t) - T(t)]/Y(t) B(t)/Y(t) - B(t-1)/Y(t-1) = [r - g][B(t-1)/Y(t-1)] + [G(t) - T(t)]/Y(t) (2) [B(t)/Y(t)] = [r-g][B(t-1)/Y(t-1)] + [G(t) - T(t)]/Y(t)
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# Answer to Question #233499 in Chemical Engineering for Lok
Question #233499
b) A first order reactionA→B, with the rate constant k is taking place in CSTR fed with A at concentration Car which remains unchanged. There are likely to be some deviations in the feed rate (F) of A. Derive linearized transfer function between. concentration of A in the outlet and feed rate of A assuming that volume V of reacting mixture remains unchanged.
1
2021-09-21T02:06:58-0400
Is the model linear? If we decide to solve the model numerically, we do not have to linearize; in fact, the non-linear model would be more accurate. However, in this problem we seek the insight obtained from the approximate, linear model.
All terms involve a constant times a variable (linear) except for the following term, which is linearized using the Taylor series..
"(1) C0.5 \u2248 (C0.5 ) + 0.5(C\u22120.5 ) (C \u2212 C )"
higher order terms AAs AsAAs
This approximation can be substituted into equation 2, and the initial steady-state model subtracted to obtain the following, with C’A = CA - CAS.
"(2) V dC'A = F(C' \u2212FC' ) \u2212 Vk(0.5C\u22120.5 )C' dt" A0
This linear, first order ordinary differential equation model can be arranged into the standard form, given in the following.
"\u03c4(sC'A (s) \u2212 C'A (t) |t=0 )+ C'A (s) = KC'A0 (s)"
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for any assignment or question with DETAILED EXPLANATIONS! | 424 | 1,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.913518 |
https://www.convert-measurement-units.com/convert+Square+micrometer+to+Skilodge.php | 1,725,960,963,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00581.warc.gz | 678,918,411 | 14,103 | Convert µm² to Skilodge (Square micrometer to Skilodge)
## Square micrometer into Skilodge
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Square+micrometer+to+Skilodge.php
# Convert µm² to Skilodge (Square micrometer to Skilodge)
1. Choose the right category from the selection list, in this case 'Area'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Square micrometer [µm²]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Skilodge'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
### Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '765 Square micrometer'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Square micrometer' or 'µm2'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Area'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '18 µm2 to Skilodge' or '11 µm2 into Skilodge' or '76 Square micrometer -> Skilodge' or '35 µm2 = Skilodge' or '93 Square micrometer to Skilodge' or '69 Square micrometer into Skilodge'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(62 * 21) µm2'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '45 Square micrometer + 4 Skilodge' or '79mm x 38cm x 96dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 5.985 222 167 756 7×1021. For this form of presentation, the number will be segmented into an exponent, here 21, and the actual number, here 5.985 222 167 756 7. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 5.985 222 167 756 7E+21. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 5 985 222 167 756 700 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 915 | 3,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.861115 |
https://www.katacoda.com/scylla/scenarios/high-availability-lab | 1,642,921,237,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00156.warc.gz | 866,994,427 | 13,266 | Difficulty: Beginner
Estimated Time: 13 minutes
This lab demonstrates, using a hands-on example, how Availability works in Scylla. You’ll try setting the Replication Factor and Consistency Levels in a three-node cluster and you’ll see how they affect read and write operations when all of the nodes in the cluster are up, and also when some of them are unavailable.
This lab is part of Scylla University.
To summarize, you saw what happens when some of the nodes are down in our three node cluster with RF=3, and with different Consistency Levels.
CL=ALL will provide higher consistency, while CL=1 will provide higher availability.
You can read more about the Replication Factor and Consistency Levels here.
In the next lesson you will take a closer look at the Scylla architecture.
Step 1 of 8
#### Create the Cluster
Scylla offers high availability by replicating data across multiple nodes. The Replication Factor (RF) is equivalent to the number of nodes where data (rows and partitions) are replicated. Data is replicated to multiple (RF=N) nodes.
An RF of ONE means there is only one copy of a row in a cluster, and there is no way to recover the data if the node is compromised or goes down. RF=2 means that there are two copies of a row in a cluster. An RF of at least three is used in most systems.
Data is always replicated automatically. Read or write operations can occur to data stored on any of the replicated nodes. The Consistency Level (CL) determines how many replicas in a cluster must acknowledge a read or write operation before it is considered successful.
In this part, you’ll bring up a three-node cluster.
## Create a Three-Node Cluster, CQLSH
First, you'll bring up a three-node Scylla cluster using Docker. Start with one node, called Node_X:
`docker run --name Node_X -d scylladb/scylla:4.3.0 --overprovisioned 1 --smp 1`
Create two more nodes, Node_Y and Node_Z, and add them to the cluster of Node_X.
The command “\$(docker inspect –format='{{ .NetworkSettings.IPAddress }}’ Node_X)” translates to the IP address of Node-X:
`docker run --name Node_Y -d scylladb/scylla:4.3.0 --overprovisioned 1 --smp 1 --seeds="\$(docker inspect --format='{{ .NetworkSettings.IPAddress }}' Node_X)"`
`docker run --name Node_Z -d scylladb/scylla:4.3.0 --overprovisioned 1 --smp 1 --seeds="\$(docker inspect --format='{{ .NetworkSettings.IPAddress }}' Node_X)"`
Wait a minute or so and check the node status:
`docker exec -it Node_X nodetool status`
You’ll see that eventually, all the nodes have UN for status. U means up, and N means normal. If you get a message "nodetool: Unable to connect to Scylla API server: java.net.ConnectException: Connection refused (Connection refused)", it means you have to wait a bit more for the node to be up and responding.
`docker exec -it Node_X cqlsh` | 692 | 2,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.907379 |
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