url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://brilliant.org/discussions/thread/valentines-day-content-contest-submissions/ | 1,627,899,549,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00679.warc.gz | 168,216,379 | 11,759 | # Valentine's Day Content Contest Submissions
Here are the other submissions to the contest in no particular order. Enjoy!
Problems:
Wiki:
Note:
Other Stuff:
• JOMO Valentine's Special (Warning: JOMO is a live contest. Please abstain from discussing these problems on Brilliant)
6 years, 5 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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\boxed{123} $\boxed{123}$ | 523 | 1,954 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-31 | latest | en | 0.839916 |
https://plainmath.net/7262/evaluate-line-integral-oint-frac-plus-plus-where-circle-defined-equal | 1,656,771,107,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104141372.60/warc/CC-MAIN-20220702131941-20220702161941-00525.warc.gz | 512,257,877 | 14,698 | # Evaluate the line integral oint_{C} frac{y}{(x-1)^{2}+4y^{2}} dx+frac{-(x-1)}{(x-1)^{2}+4y^{2}} dy where C is circle defined by x^{2}+y^{2}=27
Evaluate the line integral where C is circle defined by ${x}^{2}+{y}^{2}=27$
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mhalmantus
Our aim is to apply Green's Theorem. Let us take $L\left(x,y\right)=\frac{y}{{\left(x-1\right)}^{2}+4{y}^{2}}$ and$M\left(x,y\right)=\frac{-\left(x-1\right)}{{\left(x-1\right)}^{2}+4{y}^{2}}$. It follows that$\frac{\partial L}{\partial y}=\frac{\partial }{\partial y}\left(\frac{y}{{\left(x-1\right)}^{2}+4{y}^{2}}\right)$
$\frac{\frac{\partial }{\partial y}\left(y\right)\left({\left(x-1\right)}^{2}+4{y}^{2}\right)-\frac{\partial }{\partial y}\left({\left(x-1\right)}^{2}+4{y}^{2}\right)y}{{\left({\left(x-1\right)}^{2}+4{y}^{2}\right)}^{2}}$
$=\frac{{\left(x-1\right)}^{2}-4{y}^{2}}{\left({\left(x-1\right)}^{2}+4{y}^{2}}$
$\frac{\partial M}{\partial x}=\frac{\partial }{\partial x}\left(\frac{-\left(x-1\right)}{{\left(x-1\right)}^{2}+4{y}^{2}}\right)$
$\frac{\frac{\partial }{\partial x}\left(x-1\right)\left({\left(x-1\right)}^{2}+4{y}^{2}\right)-\frac{\partial }{\partial x}\left({\left(x-1\right)}^{2}+4{y}^{2}\right)\left(x-1\right)}{{\left({\left(x-1\right)}^{2}+4{y}^{2}\right)}^{2}}$
$=\frac{{\left(x-1\right)}^{2}-4{y}^{2}}{{\left({\left(x-1\right)}^{2}+4{y}^{2}\right)}^{2}}$Let D be the interior of C, that is, D is a circular disk given by ${x}^{2}+{y}^{2}\le 27$. By Green's Theorem, the given line integral can be obtained as${\oint }_{C}Ldx+Mdy=\int {\int }_{D}\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy$ | 760 | 1,850 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 15, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.643896 |
https://www.conwaylife.com/forums/search.php?st=0&sk=t&sd=d&sr=posts&author_id=860&start=50 | 1,591,159,127,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347428990.62/warc/CC-MAIN-20200603015534-20200603045534-00012.warc.gz | 679,688,475 | 10,126 | ## Search found 80 matches
December 2nd, 2011, 7:50 pm
Forum: Other Cellular Automata
Topic: Combined Life Algorithms
Replies: 20
Views: 11148
### Re: Combined Life Algorithms
137ben wrote:Here is a random soup which produces a 2-color big S!
It also produces these 2 bicolor still lifes.
Code: Select all
``````x = 9, y = 4, rule = UpperMiddleLife
6.B\$2A3.A.B\$2B3.A2.B\$6.2A!
``````
December 2nd, 2011, 7:31 pm
Forum: Scripts
Topic: Golly scripts
Replies: 248
Views: 201789
### VonNeumann to Moore
I have no experience writing scripts, but I have figured out a way to turn Von Neumann neighbourhood ruletables into Moore neighbourhood ruletables. I may need this script for a project I am working on. Namely "Arrows WW".
December 2nd, 2011, 7:13 pm
Forum: Other Cellular Automata
Topic: Arrows
Replies: 12
Views: 9144
### Re: Conway's Life!!!
Edit: Now working on Conway's Life unit cell. Then Arrow can be proven computationally universal by simulating another system which is proven to be computationally universal :o :shock: :o :shock: :o :shock: :o :shock: :o :shock: :o :shock: :o :shock: :o :shock: :o :shock: :o :shock: :o :shock: :o :...
December 1st, 2011, 6:48 am
Forum: Other Cellular Automata
Topic: Arrows
Replies: 12
Views: 9144
### Arrows WW
I am going to make another rule-table combining WireWorld and Arrows. You cannot use normal WireWorld patterns because it will use the von Neumann neighborhood. Stay tuned! EDIT: On second thoughts, I will use the original WireWorld. The new CA will have 17 states, 3 for WireWorld, 1 for Blocker, 8 ...
November 29th, 2011, 3:12 am
Forum: Other Cellular Automata
Topic: Combined Life Algorithms
Replies: 20
Views: 11148
### Re: Combined Life Algorithms
EVERY well-known life type can interact with every other. But not necessarily on a level field… If you tried to mix, say, Seeds (B2/S) and Live Free or Die (B2/S0), you run into the problem that birth from two different parents cannot be arranged in a symmetrical fashion, either rule will have to b...
November 29th, 2011, 3:11 am
Forum: Other Cellular Automata
Topic: Combined Life Algorithms
Replies: 20
Views: 11148
### Re: Combined Life Algorithms
But not necessarily on a level field… If you tried to mix, say, Seeds (B2/S) and Live Free or Die (B2/S0), you run into the problem that birth from two different parents cannot be arranged in a symmetrical fashion, either rule will have to be prioritized. A cell gets born if the majority of the cel...
November 28th, 2011, 5:03 am
Forum: Other Cellular Automata
Topic: Arrows
Replies: 12
Views: 9144
### Re: Arrows
I didn't expect it to be this popular!
November 28th, 2011, 4:49 am
Forum: Other Cellular Automata
Topic: Arrows
Replies: 12
Views: 9144
### Re: Arrows
Wojowu wrote:It can be made simply into extensible period 3 gun:
Code: Select all
``````x = 10, y = 6, rule = ArrowV2
3.A2.A.B\$.TUTUTUTU\$DRSRSRSRS\$.TUTUTUTUB\$.RSRSRSRS\$.D2.C3.B!``````
That is REALLY weird.
By the way, the last 8 states are reflectors that are currently moving.
November 25th, 2011, 10:13 pm
Forum: Other Cellular Automata
Topic: Arrows
Replies: 12
Views: 9144
### Arrows v2!
This is a new version of Arrows with this added feature: When a moving arrow collides with a reflector side-on, the reflector moves. Demo RLE: x = 3, y = 1, rule = ArrowV2 N.G! Table & icons: ArrowV2.zip Make sure you have the latest version of Golly! P.S. BIG Thank-you Wojowu for devoting some of h...
November 25th, 2011, 5:57 am
Forum: General Discussion
Topic: Random switch engine stuff
Replies: 2
Views: 1454
### Random switch engine stuff
This is a topic for random switch engine stuff:
(Glider-laying switch engine I got by crashing a random line into a 6-engine Cordership)
Code: Select all
``````x = 17, y = 27, rule = B3/S23
o3b2o2b2obo\$o3b2o4bob2o\$o3b2o2b2ob2o\$4b2o\$2bo2bo\$2bobo\$3bo9\$14bo\$15b2o
\$3b2o2b6o\$3bo7bo\$3bo4bob2o\$7b3o5\$10b2o\$10b2o!
``````
November 23rd, 2011, 7:22 am
Forum: Other Cellular Automata
Topic: Extended Life
Replies: 171
Views: 112822
### Re: Extended Life
p24 oscillator with a start population of 24:
Code: Select all
``````x = 4, y = 6, rule = extendedlife
F2AF\$A2FA\$A2FA\$A2FA\$A2FA\$F2AF!
``````
November 23rd, 2011, 4:05 am
Forum: Other Cellular Automata
Topic: Arrows
Replies: 12
Views: 9144
### Re: Arrows
1. Now I am working on arrow guns, I've found period 3,4 and 6, and I hope I'll find more (maybe Arrows are also omniperiodic on gun periods) 2. Really cool rule table, and for me best one with von Neumann neighborhood 1. Let's see those guns. It comes with a p4 gun already. 2. Thanks. EDIT: See ht...
November 20th, 2011, 1:16 am
Forum: Other Cellular Automata
Topic: Arrows
Replies: 12
Views: 9144
### Arrows
This is a thread for a rule I made called "Arrows".
It is the first rule I made with a von Neumann neighborhood.
Arrow.zip
Arrows table & icons
See attached file ^ for table and icons files.
November 19th, 2011, 10:03 pm
Forum: Other Cellular Automata
Topic: Combined Life Algorithms
Replies: 20
Views: 11148
### Re: Combined Life Algorithms
I was thinking about combining very different rules like CGOL and Assimilation.
Life and HighLife are very (too) similar.
P.S. Good try anyway!
P.P.S Check out: http://powdertoy.co.uk.
EVERY well-known life type can interact with every other. That's where I got the idea.
November 19th, 2011, 9:58 pm
Forum: Other Cellular Automata
Topic: Odd-Even Replicator B1357/S02468
Replies: 4
Views: 2210
### Re: Odd-Even Replicator B1357/S02468
Isn't this rule called "Fredkin"?
November 17th, 2011, 7:01 am
Forum: The Sandbox
Replies: 111
Views: 62467
(in my font) This ... x = 50, y = 5, rule = B3/S23:T600,136 3ob2o2bo3bo2bo2b2o2b2o2b3o2bo2bo2bob2o2b3obobo\$o3bobobobobobobobobobob obo3bobob2obobobobobobobo\$3obobobobobob3ob2o2bobob3ob3ob4ob2o2bobob3o\$ o3bobobobobobobobobobobobo3bobobob2obobobobo3bo\$3ob2o3b3o2bobobobob2o 2bo3bobobo2bob2o2b3ob3o! int...
November 14th, 2011, 6:51 am
Forum: Patterns
Topic: Eater for traffic light?
Replies: 1
Views: 1717
### Eater for traffic light?
I was trying to see whether I could make a double-barrelled glider gun based on the AK47 reaction.
November 10th, 2011, 9:15 pm
Forum: Other Cellular Automata
Topic: Combined Life Algorithms
Replies: 20
Views: 11148
### Combined Life Algorithms
I am working on a rule table for what I call SeedLife. It simulates the Seeds rule and Conway's Game of Life in the same space and allows them to interact by counting each other as neighbors. This allows for very simple ships that travel at light speed! I will post the SeedLife rule table and an exa...
October 29th, 2011, 11:41 pm
Forum: Other Cellular Automata
Topic: Thread for Your Accidental Discoveries that Aren't in CGOL
Replies: 821
Views: 349926
### Re: Thread for Your Accidental Discoveries that Aren't in CGOL
In Brian's Brain, I found a c/2 diagonal glide symmetric spaceship I call the BB (Brian's Brain) glider.
RLE:
Code: Select all
``````x = 4, y = 4, rule = /2/3
2.A\$.BAB\$A\$.B!
``````
I found this while looking at the evolution of the domino.
Code: Select all
``````x = 1, y = 2, rule = /2/3
A\$A!
``````
(this is not necessary)
October 29th, 2011, 11:29 pm
Forum: General Discussion
Topic: Odd pattern which isn't on the list
Replies: 4
Views: 2329
### Re: Odd pattern which isn't on the list
acgoldis wrote:Hopefully these characters are the same size.
Try putting it in code blocks:
Code: Select all
``````**o**
***o*
*ooo*
*****
*ooo*
***o*
**o**``````
This is the glider synthesis of a bi-block.
October 29th, 2011, 11:21 pm
Forum: General Discussion
Topic: Stepships - A new type of c/2 spaceship
Replies: 12
Views: 4874
### Re: Stepships - A new type of c/2 spaceship
137ben wrote:....Your rle code starts with "x=0, y=0". This means that it fits in a 0 by 0 rectangle, and so it does not come out in most life programs. Could you edit that to the correct size?
Fixed!
October 26th, 2011, 7:31 am
Forum: General Discussion
Topic: Stepships - A new type of c/2 spaceship
Replies: 12
Views: 4874
### Stepships - A new type of c/2 spaceship
I think I discovered a new type of c/2 spaceship I call a "Stepship". It is based on blinker puffer 1. It uses the spark of the lowest HWSS as a blinker fuse. Here is RLE code of the first 6 stepships. x = 180, y = 348, rule = B3/S23 2b3o\$b5o\$2ob3o22b6o\$b2o25bo5bo\$28bo\$4bo24bo4bo\$2bo28b2o\$bo4bo\$bo3b...
October 6th, 2011, 2:40 am
Forum: General Discussion
Topic: "Iso-Glider"
Replies: 5
Views: 2022
### Re: "Iso-Glider"
ebcube wrote:You can see that they are the same glider offset by two generations.
This is like the "difference" between sine and cosine waves.
If x is in radians:
cos(x) = sin(x + pi/2)
October 6th, 2011, 2:31 am
Forum: General Discussion
Topic: "Iso-Glider"
Replies: 5
Views: 2022
### Re: "Iso-Glider"
calcyman wrote:Conway referred to the 3o\$o\$bo! shape as a 'male' glider, and the boo\$oo\$bo! as a 'female' glider. With little imagination, one can identify 2-glider collisions as being either homo- or heterosexual.
I don't like it when you relate Conway's Game of Life to sex.
October 6th, 2011, 2:28 am
Forum: General Discussion
Topic: List of C/4 puffers & rakes?
Replies: 5
Views: 2859
IN GOLLY. | 3,057 | 9,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.864058 |
http://www.finedictionary.com/polyhedron.html | 1,590,939,528,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00063.warc.gz | 157,224,742 | 6,264 | polyhedron
Definitions
• WordNet 3.6
• n polyhedron a solid figure bounded by plane polygons or faces
• ***
Webster's Revised Unabridged Dictionary
• Polyhedron (Geom) A body or solid contained by many sides or planes.
• Polyhedron (Opt) A polyscope, or multiplying glass.
• ***
Century Dictionary and Cyclopedia
• n polyhedron In geometry, a solid bounded by plane faces.
• n polyhedron In optics, a multiplying glass or lens consisting of several plane surfaces disposed in a convex form, through each of which an object is seen; a polyscope.
• n polyhedron In bot.,in Hydrodictyon or water-net, one of the special angular cells with horn-like processes formed by the swarm-cells produced in the zygospore, within each of which a new cœnobium is developed.
• ***
Chambers's Twentieth Century Dictionary
• n Polyhedron pol-i-hē′dron a solid body with many bases or sides
• ***
Etymology
Webster's Revised Unabridged Dictionary
NL., fr. Gr. with many seats or sides; poly`s many + a seat or side: cf. F. polyèdre,
Chambers's Twentieth Century Dictionary
Gr. polys, many, hedra, a base.
Usage
In literature:
An edifice is no longer an edifice; it is a polyhedron.
"Notre-Dame de Paris" by Victor Hugo
In turn it was a sphere, a disk, a pyramid, a pentahedron, a polyhedron.
""Where Angels Fear to Tread" and Other Stories of the Sea" by Morgan Robertson
Book VII relates to polyhedrons, cylinders, and cones.
"The Teaching of Geometry" by David Eugene Smith
***
In news:
Indeed, many beaded beads can be viewed as polyhedra, where the hole through the middle of each bead corresponds to a polyhedron's edge.
Polyhedrons consist of triangles, squares, pentagons, hexagons, and other polygons that are joined together to form closed, three-dimensional objects.
Different rules for linking various polygons generate different types of polyhedrons.
The faces of a polyhedron can have different sizes and shapes, just as long as each one is a polygon.
The polyhedron itself can have a hole (or two or more).
Consider what happens when a vertex of one tetrahedron pierces the face of a second tetrahedron to form a new, more complicated polyhedron.
***
In science:
The set of bistochastic matrices of size N can be viewed as a convex polyhedron in RN 2 .
Random unistochastic matrices
The volume of the polyhedron of bistochastic matrices was computed by Chan and Robbins .
Random unistochastic matrices
The boundary of ΣB 3 is obtained from spectra of the members of the convex polyhedron of the bistochastic matrices of size 3.
Random unistochastic matrices
At the same time, for every finite-dimensional subspace of c0 its unit bal l is a polyhedron.
Classification of Banach Spaces --its Topological and Cardinality Properties
The conditions Min(Ps , L) ≤ x ≤ Max(Ps , L) define a parallelepiped K (possible not maximal dimension) in V and the bsemiample divisors Li generate a convex rational polyhedron R ⊂ V containing the diagonal [Ps , L] of K.
On Zariski decomposition problem
In fact, for each face X of the polyhedron ∂UΦ , there exists at least one Pi such that the preceding statement is true for Pi whenever u ∈ X .
Random Surfaces
By Lemma 4.3.3, UΦ is the interior of a convex polyhedron and by Lemma 4.3.7, σ is a bounded, continuous function on the closure of UΦ .
Random Surfaces
The symmetry group of the octahedron is identical to the symmetry group of the cube since the octahedron is the dual polyhedron of the cube.
Quantum circuits for single-qubit measurements corresponding to platonic solids
The icosahedron is the dual polyhedron of the dodecahedron.
Quantum circuits for single-qubit measurements corresponding to platonic solids
In this section, we let X be an m-dimensional compact polyhedron with metric.
Conley Index Theory and Novikov-Morse Theory
Proposition 4.7 Let X be a compact polyhedron.
Conley Index Theory and Novikov-Morse Theory
Theorem 5.1 Let X be a compact polyhedron with a metric d.
Conley Index Theory and Novikov-Morse Theory
Here χ(X ) is the Euler characteristic number of the compact polyhedron.
Conley Index Theory and Novikov-Morse Theory
Theorem 5.6 (Vanishing theorem) Let X be a compact polyhedron with a metric d.
Conley Index Theory and Novikov-Morse Theory
Measure µβ and polyhedron Pβ , of course, do not depend on the choice of Λ; thus, no index Λ in notation µβ and Pβ .
Grade of Membership Analysis: One Possible Approach to Foundations
*** | 1,112 | 4,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.886373 |
https://www.usna.edu/Users/cs/roche/courses/s16si486h/probs/020.php | 1,539,845,892,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511744.53/warc/CC-MAIN-20181018063902-20181018085402-00288.warc.gz | 1,114,746,806 | 2,510 | This is the archived website of SI 486H from the Spring 2016 semester. Feel free to browse around; you may also find more recent offerings at my teaching page.
# Making a weighted coin
Due: February 2
Points: 2
Given a perfectly fair coin (heads or tails each with probability exactly $$\frac{1}{2}$$), and a probability $$p$$ with $$0\le p\le 1$$, give an algorithm to simulate the tossing of the weighted coin that comes up heads with probability $$p$$ and tails with probability $$1-p$$.
Specifically, assume that you are given the infinite binary expansion of $$p$$. In other words, assume $$p$$ is written in base 2 as
$p = 0.b_1b_2b_3b_4\ldots$
where each $$b_i$$ is either 0 or 1. Your algorithm can examine as many of the bits in this binary expansion of $$p$$ as it needs, in order. Of course, examining more bits will make your algorithm slower, and there are potentially an infinite number of bits in the expansion of $$p$$, so you can't examine all of them!
Show that the expected number of fair coin flips your algorithm requires is $$O(1)$$, no matter what $$p$$ is. | 281 | 1,087 | {"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} | 3.328125 | 3 | CC-MAIN-2018-43 | latest | en | 0.93048 |
https://math.stackexchange.com/questions/130173/square-root-of-biholomorphic-mappings-between-multiply-connected-domains | 1,560,688,612,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998238.28/warc/CC-MAIN-20190616122738-20190616144738-00415.warc.gz | 493,526,629 | 33,843 | # Square root of biholomorphic mappings between multiply connected domains
I've been looking in the literature for a reference on the following, but without success :
Let $\Omega_1$ be a bounded finitely connected domain in the complex plane. Suppose that the boundary of $\Omega_1$ consists of pairwise disjoint piecewise analytic Jordan curves.
A repeated application of the Riemann mapping theorem gives a biholomorphic mapping $F: \Omega_1 \rightarrow \Omega_2$, where $\Omega_2$ is a bounded finitely connected domain whose boundary consists of pairwise disjoint analytic Jordan curves. See for example Ahlfors, 3rd edition, p.252.
It is well known that the Riemann map of a Jordan domain extends to a homeomorphism on the closure. Furthermore, it also extends analytically across any analytic arc on the boundary (Ahlfors p.235).
Since $F$ is a composition of Riemann maps, it extends to a continuous function in $\overline{\Omega_1}$, the closure of $\Omega_1$, and also analytically across the boundary $\partial \Omega_1$, except maybe at the "corners". My question is the following :
Is it true that $F'$ has an analytic square root in $\Omega_1$, i.e. $F'=G^2$ where $G$ is analytic in $\Omega_1$ and continuous in $\overline{\Omega_1}$, except maybe at the "corners" of $\partial \Omega_1$?
Any reference on this and biholomorphic mappings between multiply connected domains is welcome.
Thank you, Malik
• You have a reference for $F$ extending analytically across the boundary except possibly at corners? Meanwhile, what happens when $\Omega_1$ is a concentric slit region ( Ahlfohrs pages 247-249) which entails no loss of generality? – Will Jagy Apr 10 '12 at 20:29
• @Will Jagy : I've seen a proof of this somewhere, but I forgot where and can't find the reference... I only need a weaker form of that result though, so I'll edit the question accordingly. As for concentric slit regions, if I understand correctly what is meant by that, $\Omega_1$ is not of this form. I suppose that the boundary of $\Omega_1$ consists of a finite number of pairwise disjoint curves, and that each of these curves is simple, closed and piecewise analytic. – Malik Younsi Apr 11 '12 at 13:54
Proposition. If $\Omega\subseteq \mathbb C$ is a domain and $f:\Omega\to\mathbb C$ an injective holomorphic map, then $\sqrt{f'}$ has a holomorphic branch in $\Omega$.
Proof. Since $f'$ does not vanish in $\Omega$, the function $\sqrt{f'}$ admits analytic continuation along any path. It remains to show that for every closed smooth curve $\gamma$ in $\Omega$ the change of $\frac{1}{2\pi}\arg f'$ along $\gamma$ (henceforth denoted $\operatorname{ind}_\gamma f'$) is an even integer. By cutting $\gamma$ in pieces and smoothing them up, we reduce the problem to simple closed smooth curve $\gamma$.
The tangent vector $\gamma'$ rotates once as it travels around $\gamma$: that is, $$\operatorname{ind}_\gamma \gamma'\in\{-1,1\}\tag1$$ Indeed, (1) is true for a circle and invariant under diffeotopy. Applying (1) to $f\circ \gamma$ (which is also a simple closed curve), and using the relation $\arg (f\circ\gamma)'=(\arg f')\circ \gamma+\arg \gamma'$, we obtain $$\operatorname{ind}_\gamma f' + \operatorname{ind}_\gamma\gamma'\in\{-1,1\}\tag2$$ From (1) and (2), $$\operatorname{ind}_\gamma f' \in\{-2,0,2\}\tag3$$ which completes the proof. $\Box$
The continuity of $\sqrt{f'}$ up to the boundary (except at corners) is a local issue, which is resolved by reflecting $f$ across the relevant boundary arc. | 928 | 3,509 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-26 | latest | en | 0.918244 |
https://ww.gnu-darwin.org/ProgramDocuments/html_i/kepler_i.html | 1,620,964,113,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991737.39/warc/CC-MAIN-20210514025740-20210514055740-00408.warc.gz | 646,557,390 | 8,934 | GNU-Darwin Web
i/kepler
Home Manual Packages Global Index Keywords Quick Reference ``` /* * kepler.i * yorick solar system model */ func kepler (orbit, time, &ma, &ta, &norb, full=) /* DOCUMENT xyz = kepler(orbit, time) or xyz = kepler(orbit, time, ma, ta, norb) return 3-dimsof(orbit(1,..))-by-dimsof(time) XYZ coordinates corresponding to the orbit(s) ORBIT and time(s) TIME. Optionally return mean anomaly MA, true anomaly TA, and integer number of orbits, each a dimsof(orbit(1,..))-by-dimsof(time) array. The MA and TA are in radians. The x-axis is along the line of the vernal equinox, the z-axis is ecliptic north. ORBIT has leading dimension 12: [angle from perihelion, mean daily motion, semi-major axis, d/dt(semi-major axis), eccentricity, d/dt(eccentricity), longitude of ascending node, d/dt(ascending node), angle from ascending node to perihelion, d/dt(perihelion), inclination, d/dt(inclination)] (Six pairs of a quantity and its time derivative.) The angles are in degrees; d/dt units must match TIME units. Mean anomaly is not an angle in real space; it is the quantity proportional to time in Kepler's equation. True anomaly is the angle from perihelion to planet. With a non-nil, non-zero full= keyword, return XYZUVW -- that is, six coordinates including velocities as well as positions. SEE ALSO: sch_planets, jpl_planets, sch_moon, moon, solar_system */ { /* result will be dimsof(orbit(1,..))-by-dimsof(time) */ time = [orbit(1,..)*0.+1.](..,+) * [time](..,+); /* ma = mean anomaly = time from perihelion to planet, as an angle * a = semi-major axis * e = eccentricity */ ma = (orbit(1,..) + orbit(2,..)*time) * (pi/180.); a = (orbit(3,..) + orbit(4,..)*time); e = (orbit(5,..) + orbit(6,..)*time); /* reduce ma to interval (-pi,pi] */ norb = ceil((ma-pi)/(2.*pi)); ma -= 2.*pi*norb; /* ea = eccentric anomaly, ma = ea-e*sin(ea) is kepler's equation of time * solve for ea by newton iteration */ ea = ma + e*sin(ma)*(1.+e*cos(ma)); do { sea = sin(ea); cea = cos(ea); dea = (ea-e*sea - ma)/(1.-e*cea); ea -= dea; } while (anyof(abs(dea) > 1.e-8)); /* ta = true anomaly = angle from perihelion to planet */ ta = atan(sea, cea-e); boa = sqrt(1.-e*e); b = a * boa; /* semi-minor axis */ xyz = array(0., 3,dimsof(cea)); xyz(1,..) = a*(cea-e); xyz(2,..) = b*sea; local rotdot; rot = to_ecliptic(orbit,time,full=full); rxyz = (rot * xyz(-,..))(,sum,..); if (full) { adot = orbit(4,..); edot = orbit(6,..); bdot = adot*boa - e*edot/boa; eadot = (orbit(2,..)*(pi/180.) + edot*sea) / (1.-e*cea); uvw = array(0., 6,dimsof(cea)); uvw(4,..) = adot*(cea-e) - a*(sea*eadot-edot); uvw(5,..) = bdot*sea + b*cea*eadot; uvw(4:6,..) = (rotdot * xyz(-,..) + rot * uvw(-,4:6,..))(,sum,..); uvw(1:3,..) = rxyz; return uvw; } return rxyz; } func kepler2 (orbit, xyz, &time, &ma, &ta) /* DOCUMENT xyz = kepler2(orbit, xyz0) or xyz = kepler2(orbit, xyz0, time, ma, ta) return dimsof(xyz0) XYZ coordinates corresponding to the orbit(s) ORBIT and direction(s) XYZ0. The dimensions of ORBIT beyond the first, if any, must match those of XYZ0, although XYZ0 may have any number of trailing dimensions. Optionally return TIME, mean anomaly MA, and true anomaly TA, each a dimsof(orbit(1,..))-by-dimsof(time) array. The MA and TA are in radians. The x-axis is along the line of the vernal equinox, the z-axis is ecliptic north. The XYZ0 direction is first projected into the plane of the orbit; then XYZ will be proportional to XYZ0. The time derivatives of the ORBIT elements are ignored. ORBIT has leading dimension 12: [angle from perihelion, mean daily motion, semi-major axis, d/dt(semi-major axis), eccentricity, d/dt(eccentricity), longitude of ascending node, d/dt(ascending node), angle from ascending node to perihelion, d/dt(perihelion), inclination, d/dt(inclination)] (Six pairs of a quantity and its time derivative.) The angles are in degrees; d/dt units must match TIME units. Mean anomaly is not an angle in real space; it is the quantity proportional to time in Kepler's equation. True anomaly is the angle from perihelion to planet. SEE ALSO: sch_planets, jpl_planets, sch_moon, moon, solar_system */ { /* rotate xyz0 into plane of orbit, with x along perihelion */ transform = to_ecliptic(orbit,0.0); xyz = (transform * xyz(,-,..))(sum,..); xyz(3,..) = 0.0; /* project xyz0 into orbital plane */ /* a = semi-major axis * e = eccentricity */ a = orbit(3,..); /* + orbit(4,..)*time */ e = orbit(5,..); /* + orbit(6,..)*time */ b = a*sqrt(1.-e*e); /* solve for ea = eccentric anomaly by newton iteration */ cea = xyz(1,..); sea = xyz(2,..); ay = a*sea; bx = b*cea; for (done=0 ;;) { rr = 1./abs(cea,sea); cea *= rr; sea *= rr; if (done) break; rr = (ay*(cea-e)-bx*sea)/(bx*cea+ay*sea); cea -= sea*rr; sea += cea*rr; done = allof(abs(rr) < 1.e-8); } xyz(1,..) = a*(cea - e); xyz(2,..) = b*sea; ma = atan(sea, cea) - e*sea; ta = atan(sea, cea-e); time = ((180./pi)*ma - orbit(1,..))/orbit(2,..); return (transform * xyz(-,..))(,sum,..); } func to_ecliptic (orbit,time,full=) { /* ascn = longitude of ascending node (from vernal equinox at epoch) * argp = argument of perihelion = angle from ascn to perihelion * incl = inclination of orbit from ecliptic */ ascn = (orbit(7,..) + orbit(8,..)*time) * (pi/180.); argp = (orbit(9,..) + orbit(10,..)*time) * (pi/180.); incl = (orbit(11,..) + orbit(12,..)*time) * (pi/180.); can = cos(ascn); san = sin(ascn); car = cos(argp); sar = sin(argp); cn = cos(incl); sn = sin(incl); /* [x-axis (perihelion vector) in ecliptic coords, * y-axis (perp to perihelion) in ecliptic coords, * z-axis (of orbit) in ecliptic coords] */ axes = [[ can*car-san*cn*sar, san*car+can*cn*sar, sn*sar], [-can*sar-san*cn*car, -san*sar+can*cn*car, sn*car], [ san*sn, -can*sn, cn]]; axes = transpose(axes, 3); /* make matrix indices first */ if (full) { ascndot = orbit(8,..) * (pi/180.); argpdot = orbit(10,..) * (pi/180.); incldot = orbit(12,..) * (pi/180.); dcan = -san*ascndot; dsan = can*ascndot; dcar = -sar*argpdot; dsar = car*argpdot; dcn = -sn*incldot; dsn = cn*incldot; extern rotdot; rotdot = [[ can*dcar+dcan*car-san*cn*dsar-san*dcn*sar-dsan*cn*sar, san*dcar+dsan*car+can*cn*dsar+can*dcn*sar+dcan*cn*sar, sn*dsar+dsn*sar], [-can*dsar-dcan*sar-san*cn*dcar-san*dcn*car-dsan*cn*car, -san*dsar-dsan*sar+can*cn*dcar+can*dcn*car+dcan*cn*car, sn*dcar+dsn*car], [ san*dsn+dsan*sn, -can*dsn-dcan*sn, dcn]]; rotdot = transpose(rotdot, 3); } return axes; } func geocentric (xyz, time) { if (!time) time = 0.0; s = (23.4393 - 3.563e-7 * time) * (pi/180.); c = cos(s); s = sin(s); xyze = xyz; xyze(2:3,..) = [[c,-s],[s,c]](+,) * xyz(+:2:3,..); return xyze; /* xyze in geocentric coords, xyz in ecliptic coords */ } local sch_planets , sch_moon; /* DOCUMENT sch_planets, sch_moon * from "How to compute planetary positions", * by Paul Schlyter of Stockholm, Sweden * http://hotel04.ausys.se/pausch * "Please note that the orbital elements of Uranus and Neptune as given * here are somewhat less accurate. They include a long period perturbation * between Uranus and Neptune. The period of the perturbation is about * 4200 years." * After corrections in the case of the moon, jupiter, saturn, and uranus, * these are claimed to be accurate to under 1 arc minute for the inner * planets, about 1 arc minute for the outer planets, and 2 arc minutes * for the moon. */ sch_planets = [ [168.6562,4.0923344368, 0.387098,0., 0.205635,5.59e-10, 48.3313,3.24587e-5, 29.1241,1.01444e-5, 7.0047,5.00e-8], /* mercury */ [48.0052,1.6021302244, 0.723330,0., 0.006773,-1.302e-9, 76.6799,2.46590e-5, 54.8910,1.38374e-5, 3.3946,2.75e-8], /* venus */ [356.0470,0.9856002585, 1.,0., 0.016709,-1.151e-9, 0.,0., 102.9404,4.70935e-5, 0.,0.], /* earth */ [18.6021,0.5240207766, 1.523688,0., 0.093405,2.516e-9, 49.5574,2.11081e-5, 286.5016,2.92961e-5, 1.8497,-1.78e-8], /* mars */ [19.8950,0.0830853001, 5.20256,0., 0.048498,4.469e-9, 100.4542,2.76854e-5, 273.8777,1.64505e-5, 1.3030,-1.557e-7], /* jupiter */ [316.9670,0.0334442282, 9.55475,0., 0.055546,-9.499e-9, 113.6634,2.38980e-5, 339.3939,2.97661e-5, 2.4886,-1.081e-7], /* saturn */ [142.5905,0.011725806, 19.18171,-1.55e-8, 0.047318,7.45e-9, 74.0005,1.3978e-5, 96.6612,3.0565e-5, 0.7733,1.9e-8], /* uranus */ [260.2471,0.005995147, 30.05826,3.313e-8, 0.008606,2.15e-9, 131.7806,3.0173e-5, 272.8461,-6.027e-6, 1.7700,-2.55e-7], /* neptune */ [239.,0.00397, 39.5,0., 0.249,0., 110.,0., 114.,0., 17.1,0.] /* pluto */ ]; sch_moon = [ [115.3654,13.0649929509, 60.2666,0., 0.054900,0., 125.1228,-0.0529538083, 318.0634,0.1643573223, 5.1454,0.], /* moon */ [356.0470,0.9856002585, 1.,0., 0.016709,-1.151e-9, 0.,0., 282.9404,4.70935e-5, 0.,0.] /* sun */ ]; func moon (time, full=) /* DOCUMENT xyz = moon(time) return position XYZ of the moon relative to center of earth at TIME; the XYZ has leading dimension 3; x is along the vernal equinox, z is ecliptic north. The corrections to the lunar orbit are from Schlyter (see sch_moon). Claimed accurate to 2 arc minutes over some reasonable time. TIME is in days since 0/Jan/00 (that is, 0000 UT 31/Dec/99). This is 1.5 days earlier than the J2000 epoch. SEE ALSO: solar_system, sch_moon, kepler */ { local mm; xyz = kepler(sch_moon(,1), time, mm, full=full); rxy = abs(xyz(2,..),xyz(1,..)); lon = atan(xyz(2,..), xyz(1,..)); lat = atan(xyz(3,..), rxy); r = rr = abs(xyz(3,..), rxy); ms = (sch_moon(1,2) + sch_moon(2,2)*time) * (pi/180.); nm = (sch_moon(7,1) + sch_moon(8,1)*time) * (pi/180.); wm = (sch_moon(9,1) + sch_moon(10,1)*time) * (pi/180.); ws = (sch_moon(9,2) + sch_moon(10,2)*time) * (pi/180.); lm = mm+wm+nm; ls = ms+ws; d = lm-ls; f = lm-nm; lon += (-1.274*sin(mm-2.*d) +0.658*sin(2.*d) -0.186*sin(ms) -0.059*sin(2.*mm-2.*d)-0.057*sin(mm-2.*d+ms)+0.053*sin(mm+2.*d) +0.046*sin(2.*d-ms) +0.041*sin(mm-ms) -0.035*sin(d) -0.031*sin(mm+ms) -0.015*sin(2.*f-2.*d) +0.011*sin(mm-4.*d)) * (pi/180.); lat += (-0.173*sin(f-2.*d)-0.055*sin(mm-f-2.*d)-0.046*sin(mm+f-2.*d) +0.033*sin(f+2.*d)+0.017*sin(2.*mm+f)) * (pi/180.); r += (-0.58*cos(mm-2.*d)-0.46*cos(2.*d)); if (full) { dmm = sch_moon(2,1) * (pi/180.); drxy = (xyz(1:2,..)*xyz(4:5,..))(sum,..)/rxy dlon = (xyz(1,..)*xyz(5,..)-xyz(2,..)*xyz(4,..))/(rxy*rxy); dlat = (rxy*xyz(6,..)-xyz(3,..)*drxy)/(rr*rr); dr = (xyz(1:3,..)*xyz(4:6,..))(sum,..)/rr; dms = sch_moon(2,2) * (pi/180.); dnm = sch_moon(8,1) * (pi/180.); dwm = sch_moon(10,1) * (pi/180.); dws = sch_moon(10,2) * (pi/180.); dlm = dmm+dwm+dnm; dls = dms+dws; dd = dlm-dls; df = dlm-dnm; dlon += (-1.274*cos(mm-2.*d)*(dmm-2.*dd) +0.658*cos(2.*d)*(2.*dd) -0.186*cos(ms)*(dms) -0.059*cos(2.*mm-2.*d)*(2.*dmm-2.*dd)- 0.057*cos(mm-2.*d+ms)*(dmm-2.*dd+dms)+ 0.053*cos(mm+2.*d)*(dmm+2.*dd) +0.046*cos(2.*d-ms)*(2.*dd-dms) +0.041*cos(mm-ms)*(dmm-dms) -0.035*cos(d)*(dd) -0.031*cos(mm+ms)*(dmm+dms) -0.015*cos(2.*f-2.*d)*(2.*df-2.*dd) +0.011*cos(mm-4.*d)*(dmm-4.*dd)) * (pi/180.); dlat += (-0.173*cos(f-2.*d)*(df-2.*dd)-0.055*cos(mm-f-2.*d)*(dmm-df-2.*dd) -0.046*cos(mm+f-2.*d)*(dmm+df-2.*dd) +0.033*cos(f+2.*d)*(df+2.*dd)+0.017*cos(2.*mm+f)*(2.*dmm+df)) * (pi/180.); dr += (+0.58*sin(mm-2.*d)*(dmm-2.*dd)+0.46*sin(2.*d)*(2.*dd)); xyz(4,..) = (dr*cos(lon)*cos(lat) - r*sin(lon)*cos(lat)*dlon - r*cos(lon)*sin(lat)*dlat); xyz(5,..) = (dr*sin(lon)*cos(lat) + r*cos(lon)*cos(lat)*dlon - r*sin(lon)*sin(lat)*dlat); xyz(6,..) = (dr*sin(lat) + r*cos(lat)*dlat); } xyz(1,..) = r*cos(lon)*cos(lat); xyz(2,..) = r*sin(lon)*cos(lat); xyz(3,..) = r*sin(lat); return xyz; } func solar_system (time) /* DOCUMENT xyz = moon(time) return position XYZ of the moon relative to center of earth at TIME; the XYZ has leading dimension 3; x is along the vernal equinox, z is ecliptic north. Corrections due to Schlyter (see sch_planets) are applied. Claimed accurate to under 1 arc minute over some reasonable time. TIME is in days since 0/Jan/00 (that is, 0000 UT 31/Dec/99). This is 1.5 days earlier than the J2000 epoch. SEE ALSO: solar_system, sch_moon, kepler */ { local mu; xyz = kepler(sch_planets, time, mu); jsu = xyz(,5:7,..); lon = atan(jsu(2,..), jsu(1,..)); lat = atan(jsu(3,..), abs(jsu(2,..),jsu(1,..))); r = abs(jsu(3,..),jsu(2,..),jsu(1,..)); mj = mu(5,..); ms = mu(6,..); mu = mu(7,..); /* note: 4220 year great uranus-neptune term * included in orbital elements for uranus and neptune */ pio180 = pi/180; lon(1,..) += (-0.332*sin(2.*mj-5.*ms-67.6*pio180) -0.056*sin(2.*mj-2.*ms+21.*pio180) +0.042*sin(3.*mj-5.*ms+21.*pio180) -0.036*sin(mj-2.*ms) +0.022*cos(mj-ms) +0.023*sin(2.*mj-3.*ms+52.*pio180) -0.016*sin(mj-5.*ms-69*pio180)) * pio180; lon(2,..) += (+0.812*sin(2.*mj-5.*ms+67.6*pio180) -0.229*cos(2.*mj-4.*ms-2.*pio180) +0.119*sin(mj-2.*ms-3.*pio180) +0.046*sin(2.*mj-6.*ms-69.*pio180) +0.014*sin(mj-3.*ms+32.*pio180)) * pio180; lat(2,..) += (-0.020*cos(2.*mj-4.*ms-2.*pio180) +0.018*sin(2.*mj-6.*ms-49.*pio180)) * pio180; lon(3,..) += (+0.040*sin(ms-2.*mu+6.*pio180) +0.035*sin(ms-3.*mu+33.*pio180) -0.015*sin(mj-mu+20.*pio180)) * pio180; xyz(1,5:7,..) = r*cos(lon)*cos(lat); xyz(2,5:7,..) = r*sin(lon)*cos(lat); xyz(3,5:7,..) = r*sin(lat); /* schlyter's pluto model */ s = (50.03+0.033459652*time)*pio180; p = (238.95+0.003968789*time)*pio180; lon = (238.9508 + 0.00400703*time -19.799*sin(p) +19.848*cos(p) +0.897*sin(2.*p) -4.956*cos(2.*p) +0.610*sin(3.*p) +1.211*cos(3.*p) -0.341*sin(4.*p) -0.190*cos(4.*p) +0.128*sin(5.*p) -0.034*cos(5.*p) -0.038*sin(6.*p) +0.031*cos(6.*p) +0.020*sin(s-p) -0.010*cos(s-p)) * pio180; lat = (-3.9082 -5.453*sin(p) -14.975*cos(p) +3.527*sin(2.*p) +1.673*cos(2.*p) -1.051*sin(3.*p) +0.328*cos(3.*p) +0.179*sin(4.*p) -0.292*cos(4.*p) +0.019*sin(5.*p) +0.100*cos(5.*p) -0.031*sin(6.*p) -0.026*cos(6.*p) +0.011*cos(s-p)) * pio180; r = (40.72 +6.68*sin(p) +6.90*cos(p) -1.18*sin(2.*p) -0.03*cos(2.*p) +0.15*sin(3.*p) -0.14*cos(3.*p)); xyz(1,9,..) = r*cos(lon)*cos(lat); xyz(2,9,..) = r*sin(lon)*cos(lat); xyz(3,9,..) = r*sin(lat); return xyz; } func day2000 (y,m,d,ut) { /* zero for 2000 Jan 0 0000 UT (1999 Dec 31 0000 UT) */ y = long(y); m = long(m); d = long(d); if (is_void(ut)) ut = 0.0; return 367*y - 7*(y+(m+9)/12)/4 + 275*m/9 + d - 730530 + ut/24.0; } func earth_tilt (d) { return 23.4393 - 3.563e-7*d; /* degrees axis to ecliptic */ } local jpl_planets ; /* DOCUMENT jpl_planets * orbital elements from http://ssd.jpl.nasa.gov/elem_planets.html * "Mean orbit solutions from a 250 yr. least squares fit of the * DE200 planetarty ephemeris to a Keplerian orbit where each element * is allowed to vary linearly with time. This solution fits the * terrestrial planets to 25" or better, but achieves only 600" for * Saturn. Elements are referenced to mean ecliptic and equinox * of J2000 at the J2000 epoch (2451545.0 JD)." * J2000 = 2000 January 1.5 * * WARNING: these elements are 1.5 JD later than sch_planets * * definitions: * argument of perihelion = longitude of perihelion - * longitude of ascending node * mean anomaly = mean longitude - longitude of perihelion * 1 Julian century = 36525 days */ jpl_planets = [ [174.79439,4.092334433949,0.38709893,1.806982e-11,0.20563069,6.918549e-10, 48.33167,-3.394174e-06,29.12478, 7.75626e-06,7.00487,-1.7880e-07], [ 50.44675,1.602131301696,0.72333199,2.518823e-11,0.00677323,-1.351951e-09, 76.68069,-7.581489e-06,54.85229, 6.75405e-06,3.39471,-2.1751e-08], [357.51716,0.985599987452,1.00000011,-1.368925e-12,0.01671022,-1.041478e-09, -11.26064,-1.386284e-04,114.20783,1.47742e-04,0.00005,-3.5699e-07], [ 19.41248,0.524021165108,1.52366231,-1.977002e-09,0.09341233,3.258590e-09, 49.57854,-7.758689e-06,286.46230,1.96286e-05,1.85061,-1.9370e-07], [ 19.65053,0.083080374325,5.20336301,1.662888e-08,0.04839266,-3.526352e-09, 100.55615, 9.256750e-06,274.19770,-2.86896e-06,1.30530,-3.1561e-08], [317.51238,0.033485450148,9.53707032,-8.255441e-08,0.05415060,-1.006489e-08, 113.71504,-1.210016e-05,338.71690,-2.72142e-06,2.48446, 4.6467e-08], [142.26794,0.011721311354,19.19126393,4.162218e-08,0.04716771,-5.242984e-09, 74.22988,-1.278728e-05,96.73436,2.27695e-05,0.76986,-1.5895e-08], [259.90868,0.005987479200,30.06896348,-3.427680e-08,0.00858587,6.872005e-10, 131.72169,-1.150278e-06,273.24966,-5.27173e-06,1.76917,-2.7683e-08], [ 14.86205,0.003976577306,39.48168677,-2.105736e-08,0.24880766,1.770021e-09, 110.30347,-2.838999e-07, 113.76329,-7.21880e-07, 17.14175, 8.4189e-08] ]; func lon_lat (xyz) { lat = atan(xyz(3,..),abs(xyz(2,..),xyz(1,..))); lon = atan(xyz(2,..),xyz(1,..)); return transpose([lon,lat], 2); } func lon_subtract (obj, ref) { objp = obj(1,..); objp -= ref(1,..); return atan(sin(objp),cos(objp)); } ``` | 6,911 | 16,412 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.695422 |
https://www.unix.com/man-page/netbsd/3/asinf | 1,702,098,866,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00238.warc.gz | 1,137,504,929 | 10,102 | # asinf(3) [netbsd man page]
ASIN(3) BSD Library Functions Manual ASIN(3)
NAME
asin, asinf -- arc sine function LIBRARY
Math Library (libm, -lm) SYNOPSIS
#include <math.h> double asin(double x); float asinf(float x); DESCRIPTION
The asin() and asinf() functions compute the principal value of the arc sine of x in the range [-pi/2, +pi/2]. RETURN VALUES
If |x|>1, asin(x) and asinf(x) return NaN and set the global variable errno to EDOM. SEE ALSO
acos(3), atan(3), atan2(3), cos(3), cosh(3), math(3), sin(3), sinh(3), tan(3), tanh(3) STANDARDS
The asin() function conforms to ANSI X3.159-1989 (``ANSI C89''). BSD
May 2, 1991 BSD
## Check Out this Related Man Page
ASIN(3) BSD Library Functions Manual ASIN(3)
NAME
asin -- arc sine function SYNOPSIS
#include <math.h> double asin(double x); long double asinl(long double x); float asinf(float x); DESCRIPTION
The asin() function computes the principal value of the arc sine of x. The result is in the range [-pi/2, +pi/2]. SPECIAL VALUES
asin(+-0) returns +-0. asin(x) returns a NAN and raises the "invalid" floating-point exception for |x| > 1. VECTOR OPERATIONS
If you need to apply the asin() function to SIMD vectors or arrays, using the following functions provided by the Accelerate.framework may give significantly better performance: #include <Accelerate/Accelerate.h> vFloat vasinf(vFloat x); void vvasinf(float *y, const float *x, const int *n); void vvasin(double *y, const double *x, const int *n); SEE ALSO
acos(3), atan(3), atan2(3), cos(3), cosh(3), sin(3), sinh(3), tan(3), tanh(3), math(3) STANDARDS
The asin() function conforms to ISO/IEC 9899:2011. BSD
December 11, 2006 BSD
Man Page
## Extract and parse XML data (statistic value) to csv
Hi All, I need to parse some statistic data from the "measInfo" -eg. 25250000 (as highlighted) and return the result into line by line, and erasing all other unnecessary info/tag. Thought of starting with grep "measInfoID="25250000" but this only returns 1 line. How do I get all the output...
## cmake and boost library installation problem
hi all, I am new to linux and C++ programming so I'm posting in hope of some help. I am trying to install a C++ library using boost and cmake but I keep gettin this error in the terminal: CMake Error at /usr/share/cmake/Modules/FindBoost.cmake:1199 (message): Unable to find the...
## Extract fragments from file
I have a .xml file that looks something like this : <measInfo> ......... string1 ......... </measInfo> <measInfo> ...... string2 ........ </measInfo> I want to extract only the 'chunk of file' from '<measInfo>' to '</measInfo>' containing string1 (or a certain string that I...
## Storing two dimensional array for postprocessing
Hi Community, Would love to get some quick help on below requirement. I am trying to process mpstat output from multiple blades of my server I would like to assign this the output to an array and then use it for post processing. How can I use a two dimensional array and assign these value ... | 851 | 3,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.471416 |
https://www.daenotes.com/electronics/basic-electronics/resistance-conductance-ohms-law | 1,720,888,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00399.warc.gz | 613,386,040 | 11,349 | # Resistance and Conductance: Unveiling the Principles of Ohm's Law
In the vast realm of electronics, the phenomena of resistance and conductance hold paramount significance. These fundamental concepts, intertwined with Ohm's Law, lay the groundwork for understanding the behavior of electric circuits and the flow of current within them.
## Resistance: A Barrier to Electron Flow
Resistance, in its essence, acts as an impediment to the smooth flow of electrons through a conducting medium. It is a property inherent to materials that resist the passage of electric charges. Symbolized by the letter "R," resistance is measured in units known as ohms (Ω). Its profound influence on the behavior of circuits cannot be overstated.
### Exploring Resistance Configurations
Within the context of resistance, it is essential to acknowledge the various configuration options that exist in electronic circuits. These configurations allow for tailored adjustments to resistance values, enabling engineers and designers to precisely control the flow of current and achieve desired outcomes.
### Ohm: The Unit of Resistance
To quantify resistance, the unit of measurement employed is known as the ohm (Ω). One ohm represents the level of resistance that restricts the flow of current when one volt of electrical potential difference is applied across a circuit. The Greek letter omega (Ω) symbolizes this crucial unit. By comprehending the concept of ohms, we gain a deeper understanding of the interplay between voltage, current, and resistance within a circuit.
### Unveiling Resistance Units
Within the vast spectrum of resistance units, several notable units deserve mention:
- Milli-ohm (mΩ): A milli-ohm corresponds to one-thousandth of an ohm (10-3Ω). It is commonly utilized when measuring extremely low resistances or in applications demanding high precision.
- Micro-ohm (μΩ): Representing a millionth of an ohm (10-6Ω), the micro-ohm unit finds utility in highly specialized areas where minute resistances must be accurately assessed.
- Nano-ohm (nΩ): One billionth of an ohm (10-9Ω) constitutes a nano-ohm. Such minuscule values are encountered in advanced scientific research and ultra-low resistance applications.
- Pico-ohm (pΩ): A pico-ohm represents a trillionth of an ohm (10-12Ω). This unit is employed in rarefied scenarios involving extraordinarily low resistances or when investigating the tiniest electrical properties.
### Kilo-ohm and Mega-ohm: Scaling Resistance
As resistance values expand, alternative units are employed for ease of representation. For instance:
- Kilohm (kΩ): Corresponding to 1,000 ohms, a kilohm (kΩ) simplifies the expression of larger resistance values.
- Megohm (MΩ): Representing one million ohms, a megohm (MΩ) becomes relevant when dealing with substantial resistances.
## Ohm's Law: Linking Voltage, Current, and Resistance
At the heart of electrical circuit analysis lies Ohm's Law, a fundamental principle governing the relationship between voltage, current, and resistance. According to Ohm's Law, when a circuit experiences an electric potential difference of one volt, and a current of one ampere flows through it, the resistance within that circuit precisely equals one ohm. This vital equation, V = I × R, establishes a fundamental link between these core parameters.
## Conductance: An Enabler of Current Flow
In contrast to resistance, conductance plays a complementary role in electric circuits. It represents the inherent property of a material to facilitate the smooth and unrestricted flow of electric current. Conductance is denoted by the letter "G" and measured in units called Siemens (S). The reciprocal relationship between resistance (R) and conductance (G) is an integral aspect of understanding electrical behavior.
By delving into the intricate nuances of resistance, conductance, and Ohm's Law, one embarks on a journey of comprehension in the realm of electronics. These foundational principles unlock the doors to manipulating and optimizing electric circuits, enabling innovation, and advancing technological frontiers.
Category | 868 | 4,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.901887 |
http://www.jiskha.com/display.cgi?id=1396039319 | 1,495,866,207,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608870.18/warc/CC-MAIN-20170527055922-20170527075922-00252.warc.gz | 661,852,910 | 3,561 | # Math
posted by on .
A pile of gravel is conical in shape. If the diameter is approximately 6.8m and the height is 2.8m, what is the volume of gravel in the pile?
• Math - ,
v = (1/3)pi(r^2)h
h = 2.8m
r =d/2 = 6.8/2 = 3.4m
v = (1/3)pi(3.4)^2(2.8) | 112 | 251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-22 | latest | en | 0.936485 |
https://fr.mathworks.com/matlabcentral/cody/problems/43674-string-array-basics-part-3-convert-cell-array-with-missing-values-to-string-array/solutions/1129613 | 1,571,787,236,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987826436.88/warc/CC-MAIN-20191022232751-20191023020251-00053.warc.gz | 486,467,977 | 15,510 | Cody
# Problem 43674. String Array Basics, Part 3: Convert Cell Array with Missing Values to String Array
Solution 1129613
Submitted on 24 Feb 2017 by praharsh
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = {'I','love','MATLAB'}; y_correct = string({'I','love','MATLAB'}); assert(isequal(cell2str(x),y_correct))
2 Pass
x = {'I', '', '', 'MATLAB'}; y_correct = string('I'); y_correct(4) = 'MATLAB'; assert(isequaln(cell2str(x),y_correct))
3 Pass
x = {'I', '', 'MATLAB' '', 'love', 'MATLAB' 'I', 'love', '' }; y_correct = [string('I'), string(NaN), string('MATLAB') string(NaN), string('love'), string('MATLAB') string('I'), string('love'), string(NaN) ]; assert(isequaln(cell2str(x),y_correct)) | 254 | 827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-43 | latest | en | 0.401056 |
http://colvertgroup.com/standard-error/interpreting-standard-error-of-the-estimate.php | 1,539,641,392,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509845.17/warc/CC-MAIN-20181015205152-20181015230652-00157.warc.gz | 78,160,983 | 6,358 | Home > Standard Error > Interpreting Standard Error Of The Estimate
# Interpreting Standard Error Of The Estimate
## Contents
Specifically, the term standard error refers to a group of statistics that provide information about the dispersion of the values within a set. When this is not the case, you should really be using the $t$ distribution, but most people don't have it readily available in their brain. The S value is still the average distance that the data points fall from the fitted values. In RegressIt you could create these variables by filling two new columns with 0's and then entering 1's in rows 23 and 59 and assigning variable names to those columns. useful reference
Read more about how to obtain and use prediction intervals as well as my regression tutorial. All rights Reserved. The third column, (Y'), contains the predictions and is computed according to the formula: Y' = 3.2716X + 7.1526. That is, should we consider it a "19-to-1 long shot" that sales would fall outside this interval, for purposes of betting? http://blog.minitab.com/blog/adventures-in-statistics/regression-analysis-how-to-interpret-s-the-standard-error-of-the-regression
## What Is The Standard Error Of The Estimate
Think of it this way, if you assume that the null hypothesis is true - that is, assume that the actual coefficient in the population is zero, how unlikely would your If this does occur, then you may have to choose between (a) not using the variables that have significant numbers of missing values, or (b) deleting all rows of data in Find the Infinity Words! However, it can be converted into an equivalent linear model via the logarithm transformation.
1. In a scatterplot in which the S.E.est is small, one would therefore expect to see that most of the observed values cluster fairly closely to the regression line.
2. Does he have any other options?Thomas on Should Jonah Lehrer be a junior Gladwell?
3. is a privately owned company headquartered in State College, Pennsylvania, with subsidiaries in the United Kingdom, France, and Australia.
4. The variance of the dependent variable may be considered to initially have n-1 degrees of freedom, since n observations are initially available (each including an error component that is "free" from
5. However, the difference between the t and the standard normal is negligible if the number of degrees of freedom is more than about 30.
It should suffice to remember the rough value pairs $(5/100, 2)$ and $(2/1000, 3)$ and to know that the second value needs to be substantially adjusted upwards for small sample sizes I am playing a little fast and lose with the numbers. The answer to the question about the importance of the result is found by using the standard error to calculate the confidence interval about the statistic. What Is A Good Standard Error This equation has the form Y = b1X1 + b2X2 + ... + A where Y is the dependent variable you are trying to predict, X1, X2 and so on are
To illustrate this, let’s go back to the BMI example. Researchers typically draw only one sample. More than 2 might be required if you have few degrees freedom and are using a 2 tailed test. http://onlinestatbook.com/2/regression/accuracy.html The variability?
Many people with this attitude are outspokenly dogmatic about it; the irony in this is that they claim this is the dogma of statistical theory, but people making this claim never Standard Error Of Estimate Calculator So twice as large as the coefficient is a good rule of thumb assuming you have decent degrees freedom and a two tailed test of significance. In fact, the confidence interval can be so large that it is as large as the full range of values, or even larger. Suppose the mean number of bedsores was 0.02 in a sample of 500 subjects, meaning 10 subjects developed bedsores.
## How To Interpret Standard Error In Regression
Now, the standard error of the regression may be considered to measure the overall amount of "noise" in the data, whereas the standard deviation of X measures the strength of the http://stats.stackexchange.com/questions/18208/how-to-interpret-coefficient-standard-errors-in-linear-regression Previous company name is ISIS, how to list on CV? What Is The Standard Error Of The Estimate That is to say, a bad model does not necessarily know it is a bad model, and warn you by giving extra-wide confidence intervals. (This is especially true of trend-line models, Standard Error Of Regression Coefficient The F-ratio is useful primarily in cases where each of the independent variables is only marginally significant by itself but there are a priori grounds for believing that they are significant
We "reject the null hypothesis." Hence, the statistic is "significant" when it is 2 or more standard deviations away from zero which basically means that the null hypothesis is probably false see here Why do central European nations use the color black as their national colors? It is, however, an important indicator of how reliable an estimate of the population parameter the sample statistic is. Changing the value of the constant in the model changes the mean of the errors but doesn't affect the variance. The Standard Error Of The Estimate Is A Measure Of Quizlet
Lane PrerequisitesMeasures of Variability, Introduction to Simple Linear Regression, Partitioning Sums of Squares Learning Objectives Make judgments about the size of the standard error of the estimate from a scatter plot The sales may be very steady (s=10) or they may be very variable (s=120) on a week to week basis. Although not always reported, the standard error is an important statistic because it provides information on the accuracy of the statistic (4). this page So ask yourself, if you were looking a much smaller legislative body, with only 10 members, would you be equally confident in your conclusions about how freshmen and veterans behave?
Here is an example of a plot of forecasts with confidence limits for means and forecasts produced by RegressIt for the regression model fitted to the natural log of cases of Linear Regression Standard Error Standard error statistics are a class of statistics that are provided as output in many inferential statistics, but function as descriptive statistics. It is technically not necessary for the dependent or independent variables to be normally distributed--only the errors in the predictions are assumed to be normal.
## Go back and look at your original data and see if you can think of any explanations for outliers occurring where they did.
Under the assumption that your regression model is correct--i.e., that the dependent variable really is a linear function of the independent variables, with independent and identically normally distributed errors--the coefficient estimates An Introduction to Mathematical Statistics and Its Applications. 4th ed. That's nothing amazing - after doing a few dozen such tests, that stuff should be straightforward. –Glen_b♦ Dec 3 '14 at 22:47 @whuber thanks! Standard Error Of Prediction This is why a coefficient that is more than about twice as large as the SE will be statistically significant at p=<.05.
here Nov 7-Dec 16Walk-in, 2-5 pm* Dec 19-Feb 3By appt. S provides important information that R-squared does not. here For quick questions email [email protected] *No appts. Get More Info If you are not particularly interested in what would happen if all the independent variables were simultaneously zero, then you normally leave the constant in the model regardless of its statistical
Large S.E. Approximately 95% of the observations should fall within plus/minus 2*standard error of the regression from the regression line, which is also a quick approximation of a 95% prediction interval. There is, of course, a correction for the degrees freedom and a distinction between 1 or 2 tailed tests of significance. You can see that in Graph A, the points are closer to the line than they are in Graph B.
The standard error? These rules are derived from the standard normal approximation for a two-sided test ($H_0: \beta=0$ vs. $H_a: \beta\ne0$)): 1.28 will give you SS at $20\%$. 1.64 will give you SS at Specifically, it is calculated using the following formula: Where Y is a score in the sample and Y’ is a predicted score. Sometimes we can all agree that if you have a whole population, your standard error is zero.
The standard deviation is a measure of the variability of the sample. Therefore, the variances of these two components of error in each prediction are additive. Please answer the questions: feedback Linear regression models Notes on linear regression analysis (pdf file) Introduction to linear regression analysis Mathematics of simple regression Regression examples · Baseball batting The second column (Y) is predicted by the first column (X). | 1,843 | 8,821 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-43 | latest | en | 0.89507 |
https://www.physicsforums.com/threads/space-and-time-warp-in-immense-gravity.781655/ | 1,527,471,748,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00330.warc.gz | 816,296,498 | 14,859 | # Space *and* Time warp in immense gravity
1. Nov 13, 2014
### Akshar Tandon
This question is inspired by the movie Interstellar but asks a basic question about relativity. Everyone talks about gravitational time dilation but I am wondering if gravity has an effect on space as well, after all it bends space time. I have not found a lot of information on length contraction in the context of gravity (velocity, yes, but not gravity). If we were to look at a clock on a planet with a very strong gravity, we would see time move extremely slow but would we also see the clock stretched/contracted? Also, would light "appear" to travel faster/slower on this planet when externally observed?
2. Nov 13, 2014
### A.T.
Yes, it does.
http://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm.27s_paraboloid
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html
http://www.mathpages.com/rr/s8-09/8-09.htm
Slower:
http://en.wikipedia.org/wiki/Shapiro_delay
I'm not sure though, if the Shapiro delay refers only the effect of gravitational dilation (as wiki says), or to the combined effect including spatial geometry.
3. Nov 13, 2014
### Staff: Mentor
The formula given on the Wiki page includes both. The factor of $\left( 1 + \gamma \right)$ is the key (where $\gamma$ is one of the PPN parameters); it's the same factor that appears in the formula for light bending by a massive body like the Sun, which takes into account both time and space curvature.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 384 | 1,570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-22 | latest | en | 0.919848 |
http://mcforum.top/practice-worksheet-quadratic-inequalities-answer-key/ | 1,603,296,305,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876768.45/warc/CC-MAIN-20201021151342-20201021181342-00359.warc.gz | 74,873,154 | 32,410 | # Practice Worksheet Quadratic Inequalities Answer Key
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The use of grids The use of grids is vital in earning your child learn to Improve handwriting. Also, bear in mind that maybe your very first try at brainstorming may not bring anything relevant, but don't stop trying. Once you are able to work, you might be surprised how much you get done. Take into consideration how you feel about yourself. Getting able to modify the tracking helps fit more letters in a little space or spread out letters if they're too tight. Perhaps you must enlist the aid of another man to encourage or help you keep focused. | 1,415 | 7,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-45 | latest | en | 0.742346 |
https://www.mapleprimes.com/users/Chidera/questions | 1,675,551,330,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00673.warc.gz | 896,058,327 | 14,639 | 2 years, 16 days
Polynomials: How do I determine the line...
Show that x-2 is a factor of p(x)=x^3+3x^2-4x-12
I)determine all the linear factors of p(x)
ii)what are the zeros of y=p(x)
iii)sketch function y=p(x)
Page 1 of 1
| 83 | 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} | 2.765625 | 3 | CC-MAIN-2023-06 | latest | en | 0.799608 |
http://healthcare-economist.com/2007/05/24/bootstrapping/ | 1,524,594,617,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947033.92/warc/CC-MAIN-20180424174351-20180424194351-00224.warc.gz | 147,610,932 | 9,846 | ## Healthcare Economist
Unbiased Analysis of Today's Healthcare Issues
## Bootstrapping
Written By: Jason Shafrin - May• 24•07
One of the biggest advances statistical modeling in the last 30 years has been the use of the bootstrap. For those interested in learning about the bootstrap in more detail, a good place to start is an article by UCSD math professor Dimitris N. Politis which I will summarize here. For more detailed information, one may want to look at An Introduction to the Bootstrap by Efron and Tibshirani.
Set-up
Suppose we have n observation of a random variable X. We can group these as a vectors so that X=(X1,…,XN), where each Xi are iid with distribution F. If we want to estimate a parameter θ(F) from the data, we can use a statistic T(X) as an approximation. If we assume that F~Normal, we can use traditional statistics to estimate T(X) as well as the confidence interval around θ(F). If we do not know the distribution of F (which a researcher problem does not in reality), then classical statical theory may be less reliable and a bootstrap methodology may be more robust. Bootstrapping methodology allows the researcher to better estimate F, especially if there is significant skewness to the F distribution.
The bootstrap procedure creates a new sample, by randomly sampling each observation in X with replacement until we have a new vector with N observations. We repeat this B times to create our bootstrap data set. Let’s look at an example..
Example
Pretend we have data on how many push up I have completed each day over a week. I want to estimate the median number of push-ups I do each day. In this sample, N=15 and since we will create ten bootstrap samples, B=10.
Obs. Data B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 1 22 18 25 29 21 21 22 18 31 24 14 2 18 24 14 25 14 21 19 35 25 21 19 3 14 25 24 31 25 21 21 14 21 30 21 4 35 19 18 26 19 25 19 31 24 24 14 5 22 29 29 31 26 30 26 21 22 19 26 6 24 31 24 31 22 30 19 30 31 26 19 7 26 25 19 22 21 25 25 26 22 30 18 8 29 30 22 14 22 22 19 18 31 35 29 9 19 31 21 14 14 21 14 26 18 22 18 10 31 25 24 35 29 22 19 14 31 26 25 11 30 22 25 22 29 14 19 35 19 22 22 12 19 22 24 19 18 35 29 26 21 19 35 13 22 22 22 24 25 24 30 19 35 25 29 14 21 31 25 22 25 14 14 22 31 19 18 15 25 22 19 30 22 35 24 19 19 31 26 Mean 23.8 25.1 22.3 25 22.1 24 21.3 23.6 25.4 24.9 22.2 Median 22 25 24 25 22 22 19 22 24 24 21
The median of the actual data we have is 22. But we can also calculate the median using a bootstrap methodology. We first randomly choose one of the data points and put it as the first data point of B1 (the bootstrap sample number 1), we then resample with replacement and put another number as the 2nd observation of sample B1. We can see that data points often repeat. For instance in B1 observations X10 repeats twice. We see that the median varies across the 10 bootstrapping samples, but the average value for the median using the bootstrap methodology is 22.8.
We can also calculate the the bootstrap variance (3.36) and standard deviation (1.83). This are calculated according to the formulas:
• Variance: B-1ΣiT(X*i)2 – [B-1ΣiT(X*i)]2
• S.D. = (Var)1/2
Here, T(X*i) is the median for each bootstrap sample i. Since there are 10 bootstrap samples i=1,…,10. To calculate the variance, one simply averages the squared median over the 10 bootstrap samples and then you subtract the squared average median of the 10 samples.
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#### One Comment
1. […] previous posts, I have explained how to create bootstrap estimates for a variety of statistics. Doing so is fairly simple and involves a 3 step […] | 1,146 | 3,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2018-17 | latest | en | 0.840437 |
https://jp.maplesoft.com/support/help/view.aspx?path=Student%2FNumericalAnalysis%2FIterativeFormulaTutor | 1,670,230,548,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00215.warc.gz | 358,840,775 | 28,000 | IterativeFormulaTutor - Maple Help
# Online Help
###### All Products Maple MapleSim
Student[NumericalAnalysis][IterativeFormulaTutor] - compute an iterative formula to numerically approximate the solution to a linear system
Calling Sequence IterativeFormulaTutor(A, b)
Parameters
A - (optional) Matrix; a square $\mathrm{nxn}$ matrix b - (optional) Vector; a vector of length $n$
Description
• The IterativeFormulaTutor command launches a tutor interface that takes the linear system $Ax=b$ and an initial vector ${x}_{0}$ and computes the matrix $T$ and vector $c$, of the equivalent system ${x}_{k+1}=T{x}_{k}+c$.
• A Matrix and Vector Editors allow users to edit the matrix and vectors easily within the tutor.
• All other available options are controlled in the tutor interface.
• If A and b are not specified then IterativeFormulaTutor uses a default square matrix and a default vector.
• The iterative methods available in this tutor are:
– gaussseidel
– jacobi
– SOR
Notes
• There are two digit controls in the tutor. The Digits option controls the number of digits kept during floating-point computations, and the digit precision option controls only the number of digits to which final results are rounded in the maplet display window. The digit precision option has no effect on the output returned from the maplet upon exit.
Examples
> $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{NumericalAnalysis}\right]\right):$
> $\mathrm{IterativeFormulaTutor}\left(\right)$
> $\mathrm{IterativeFormulaTutor}\left(\mathrm{Matrix}\left(\left[\left[1.53,6.32\right],\left[5.33,0.33\right]\right]\right),\mathrm{Vector}\left(\left[1.66,5.66\right]\right)\right)$ | 433 | 1,697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 10, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-49 | latest | en | 0.523674 |
https://nrich.maths.org/6202/solution | 1,660,480,235,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00712.warc.gz | 388,426,181 | 4,630 | Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Calendar Capers
Choose any three by three square of dates on a calendar page...
Days and Dates
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
Flapjack
Age 11 to 14 ShortChallenge Level
8 flapjacks: We have:
2 oz butter 14 oz butter ($\times$7)
3 oz sugar 15 oz sugar ($\times$5)
4 oz rolled oats 16 oz rolled oats ($\times$4)
8$\times$4 = 32 (and there will be some butter and sugar left over).
This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection. | 196 | 880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-33 | latest | en | 0.678245 |
https://news.mit.edu/2010/3q-pnp | 1,718,406,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00035.warc.gz | 383,625,045 | 26,107 | # 3 questions: P vs. NP
After glancing over a 100-page proof that claimed to solve the biggest problem in computer science, Scott Aaronson bet his house that it was wrong. Why?
On Friday, Aug. 6, a mathematician at HP Labs named Vinay Deolalikar sent an e-mail to a host of other researchers with a 103-page attachment that purported to answer the most important outstanding question in computer science. That question is whether P = NP, and answering it will earn you \$1 million from the Clay Mathematics Institute.
Last fall, MIT News published a fairly detailed explanation of what P = NP means. But roughly speaking, P is a set of relatively easy problems, NP includes a set of incredibly hard problems, and if they’re equal, then a large number of computer science problems that seem to be incredibly hard are actually relatively easy. Problems in NP include the factoring of large numbers, the selection of an optimal route for a traveling salesman, and the so-called 3-SAT problem, in which you must find values that satisfy triplets of logical conditions. For instance, is it possible to contrive an attendance list for a party that satisfies the triplet (Alice OR Bob AND Carol) AND (David AND Ernie AND NOT Alice)? (Yes: Bob, Carol, David, and Ernie were there, but Alice wasn’t.) Interestingly, the related 2-SAT problem — which instead uses pairs of conditions, like (Alice OR Ernie) — is in the set of easy problems, P, as is the closely related XOR-SAT problem.
Most computer scientists believe that P doesn’t equal NP, and that’s what Deolalikar claimed to have proved. But by the following Monday, despite being on vacation in the Mediterranean and having time only to glance through the proof, MIT Associate Professor of Electrical Engineering and Computer Science Scott Aaronson had announced on his blog that he would mortgage his house and chip in another \$200,000 if Deolalikar’s proof was correct. Last week, Aaronson took a few minutes to answer three questions about P and NP.
Q. Has the proof now been shown conclusively to be wrong?
A. I would say yeah. It was clear a couple days ago that there was a very serious gap in the statistical-physics part of the argument. It was not clear at all that the argument for showing why an NP-complete problem like 3-SAT was hard wouldn’t also show that problems like XOR-SAT are hard. Now, XOR-SAT is a variant of this satisfiability problem, which is known to be in P, which has an efficient solution. So if you’re proving that a problem is hard, but your proof could also be adapted to show that an easy problem is hard, then your proof must be fallacious: it proves too much. That’s the first check that people look for when someone announces a proof of P not equal to NP. Why doesn’t it also work for the easy problems?
The problem with saying that the thing has been conclusively refuted is that whenever anyone points to a problem like this, Vinay Deolalikar has been tending to respond, “Oh, yeah, sure, well, I’m going to address that in my next draft.” So it’s kind of a moving target. But I think it’s absolutely clear right now that at least the existing version does not solve the problem and furthermore wouldn’t solve the problem without some very, very major new ideas.
Q. Why were you so certain that there was a flaw in the proof?
A. P vs. NP is an absolutely enormous problem, and one way of seeing that is that there are already vastly, vastly easier questions that would be implied by P not equal to NP but that we already don’t know how to answer. So basically, if someone is claiming to prove P not equal to NP, then they’re sort of jumping 20 or 30 nontrivial steps beyond what we know today. So the first thing you look for is, What about steps one, two, and three? Can he explain even the easier questions, how he’s answering those? So I looked at the manuscript, and I didn’t see that.
The other check is the one that I already mentioned, which is, Why does the proof fail for variants of NP-complete problems which are known to be easy? What Deolalikar was doing is, he’s trying to argue that 3-SAT is hard by looking at its statistical properties. The problem is that 2-SAT and XOR-SAT, the problems that are easy, have very, very similar statistical properties, so it did not look like something that could distinguish the hard problems from the easy ones.
We have very strong reasons to believe that these problems cannot be solved without major — enormous — advances in human knowledge. So you look at the paper and you don’t see that it’s commensurate with the scale of the problem that it’s claiming to solve. This is not a problem that’s going to be solved by just combining or pushing around the ideas that we already have.
Q. Given that most people are pretty confident that P does not equal NP, what would the proof really do for us?
A. Yes, almost all of us believe already that P is not equal to NP. But this is one of those things where it’s not so much the destination as the journey. It’s the massive amount of new understanding of computation that’s going to be needed to prove such a statement. What are we trying to prove? That for solving all these natural optimization problems, or these search problems, or a proof of a theorem, finding the best schedule for airlines, breaking cryptographic codes — all these different things, that there’s no algorithm, no matter how clever, that’s going to solve them feasibly. So in order to prove such a thing, a prerequisite to it is to understand the space of all possible efficient algorithms. That is an unbelievably tall order. So the expectation is that on the way to proving such a thing, we’re going to learn an enormous amount about efficient algorithms, beyond what we already know, and very, very likely discover new algorithms that will likely have applications that we can’t even foresee right now.
Often in the history of theoretical computer science, the same ideas that you use to prove that something’s impossible can then be turned around to show that something else is possible, and vice versa. The simplest example of that is in cryptography, where you show some problem is hard to solve, and that gives you a code that is useful. But there are many other examples.
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With their “T-REX” method, DNA embedded in the polymer could be used for long-term storage of genomes or digital data such as photos and music.
### Bob Prior: A deep legacy of cultivating books at the MIT Press
After 36 years and hundreds of titles, the executive editor reflects on his career as a “champion of rigorous and brilliant scholarship.” | 1,601 | 7,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.962264 |
https://www.physicsforums.com/threads/harmonic-oscillator.584105/ | 1,547,706,382,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658844.27/warc/CC-MAIN-20190117062012-20190117084012-00492.warc.gz | 881,280,576 | 14,230 | # Harmonic Oscillator
1. Mar 5, 2012
### glebovg
1. The problem statement, all variables and given/known data
Using the normalization constant A and the value of a, evaluate the probability to find an oscillator in the ground state beyond the classical turning points ±x0. Assume an electron bound to an atomic-sized region (x0 = 0.1 nm) with an effective force constant of 1.0 eV/nm2.
2. Relevant equations
$\psi(x)=Ae^{-ax^{2}}$, where $A=(\frac{m\kappa}{\pi^{2}\hbar^{2}})^{1/8}$ and $a=\sqrt{m\kappa}/2\hbar$
3. The attempt at a solution
How do I find m?
Last edited: Mar 5, 2012
2. Mar 5, 2012
### vela
Staff Emeritus
The mass of the electron? You look it up.
3. Mar 5, 2012
### tiny-tim
or you could weigh one …
if you have one on you
4. Mar 5, 2012
### glebovg
I will try to read the question more carefully next time.
5. Mar 5, 2012
### glebovg
Am I supposed to find probabilities at both tails of the distribution?
6. Mar 5, 2012
### vela
Staff Emeritus
Yes.
7. Mar 7, 2012
### glebovg
Apparently there is an inconsistency in the question and there are two interpretations of the question. What would be the other one?
8. Mar 8, 2012
### vela
Staff Emeritus
Beats me.
9. Mar 10, 2012
### glebovg
My interpretation: find probabilities at both tails of the distribution i.e. $\int^{0}_{-\infty}\psi^{2}(x)\;dx$ and $1-\int^{0.1}_{-\infty}\psi^{2}(x)\;dx$.
10. Mar 11, 2012
### vela
Staff Emeritus
I'd say the problem is asking for one number, P(|x|>x0). The upper limit on your first integral is wrong.
11. Mar 19, 2012
### glebovg
Do you mean it is wrong according to your interpretation or it is generally wrong? Should it be 0.1?
Last edited: Mar 19, 2012
12. Mar 20, 2012
### vela
Staff Emeritus
It is generally wrong. Why would the upper limit be 0?
13. Mar 20, 2012
### glebovg
It should be -0.1, right?
Last edited: Mar 20, 2012 | 616 | 1,888 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-04 | latest | en | 0.840938 |
http://mathhelpforum.com/algebra/94131-positive-integer-solutions.html | 1,526,957,678,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864622.33/warc/CC-MAIN-20180522014949-20180522034949-00397.warc.gz | 193,862,702 | 12,475 | 1. ## Positive integer solutions
Find the total number of positive integer solutions of:
2. $\displaystyle (x-y)(x+y)=33$
Then $\displaystyle \left\{\begin{array}{ll}x-y=1\\x+y=33\end{array}\right.$
or $\displaystyle \left\{\begin{array}{ll}x-y=3\\x+y=11\end{array}\right.$
Now find x and y.
3. Originally Posted by red_dog
$\displaystyle (x-y)(x+y)=33$
Then $\displaystyle \left\{\begin{array}{ll}x-y=1\\x+y=33\end{array}\right.$
or $\displaystyle \left\{\begin{array}{ll}x-y=3\\x+y=11\end{array}\right.$
Now find x and y.
Hello,Thank you,
Can not have :
BECAUSE ????????
4. Originally Posted by dhiab
Find the total number of positive integer solutions of:
Have you tried graphing the function?
You would probably have to enter it as
$\displaystyle y = \pm\sqrt{x^2 - 33}$.
I can see straight away that $\displaystyle x^2 - 33 \geq 0$ so $\displaystyle x \geq \sqrt{33}$.
Since $\displaystyle x$ needs to be a positive integer, we know therefore that $\displaystyle x \geq 6$.
Now you just need $\displaystyle x^2 - 33$ to be a perfect square. $\displaystyle x = 7$ works, since $\displaystyle 7^2 - 33 = 16$, so $\displaystyle y = 4$.
See if you can come up with any other solutions...
5. Originally Posted by Prove It
Have you tried graphing the function?
You would probably have to enter it as
$\displaystyle y = \pm\sqrt{x^2 - 33}$.
I can see straight away that $\displaystyle x^2 - 33 \geq 0$ so $\displaystyle x \geq \sqrt{33}$.
Since $\displaystyle x$ needs to be a positive integer, we know therefore that $\displaystyle x \geq 6$.
Now you just need $\displaystyle x^2 - 33$ to be a perfect square. $\displaystyle x = 7$ works, since $\displaystyle 7^2 - 33 = 16$, so $\displaystyle y = 4$.
See if you can come up with any other solutions...
Hello: the solutions is
In graphic of the function , we find the Point cordinates integer.
LOOK HERE
6. Originally Posted by dhiab
Hello: the solutions is
In graphic of the function , we find the Point cordinates integer.
LOOK HERE
Are there any more?
7. Originally Posted by Prove It
Are there any more?
Hello : yes in this question
Hello,Thank you,
Can not have :
BECAUSE
8. as the graph suggests, the values are symetric, mirror image of each other,
x = -/+ 17, y = -/+ 16
x = /+ 7 ; y -/+ 4
9. Hello, dhiab!
Here is a primitive solution . . .
Find the number of positive integer solutions of: .$\displaystyle x^2 - y^2 \:=\:33$
Consecutive squares differ by an odd number: .$\displaystyle (n+1)^2 - n^2 \:=\:2n+1$
We have: .$\displaystyle x^2 - y^2 \:=\:33$
. . Then: .$\displaystyle 2n+1 \:=\:33 \quad\Rightarrow\quad n \:=\:16$
. . Hence: .$\displaystyle \boxed{17^2 - 16^2 \:=\:33}$
$\displaystyle \text{We also have: }\;x^2-y^2 \:=\:\underbrace{9 +11+13}_{4^2\quad5^2\quad6^2\quad7^2}$
. . Hence: .$\displaystyle \boxed{7^2 - 4^2 \:=\:33}$
Therefore, there are exactly two solutions.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I came up with this procedure while in college.
Example: .$\displaystyle a^2-b^2 \:=\:105$
$\displaystyle 105$ is the difference between $\displaystyle \left(\tfrac{105-1}{2}\right)^2$ and $\displaystyle \left(\tfrac{105+1}{2}\right)^2$
. . Therefore: .$\displaystyle 53^2 - 52^2 \:=\:105$
We have: .$\displaystyle 105 \:=\:35 + 35 + 35 \:=\:33 + 35 + 37$
. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{33 + 35 + 37}_{16^2\quad17^2\quad18^2\quad19^2}$
. . Therefore: .$\displaystyle 19^2 - 16^2 \:=\:105$
We have: .$\displaystyle 105 \:=\:15 + 15 + 15 + 15 + 15 + 15 + 15 \:=\:9 + 11 + 13 + 15 + 17 + 19 + 21$
. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{9 + 11 + 13 + 15 + 17 + 19 + 21}_{4^2\quad\;\;5^2\quad\;\;6^2\quad\;\;\;7^2\qua d\;\;\;8^2\quad\;\;9^2\quad\;\;10^2\quad\;\;11^2}$
. . Therefore: .$\displaystyle 11^2 - 4^2 \:=\:105$
$\displaystyle \text{The next combination is: }\;105 \:=\:\underbrace{7+7+7+ \hdots + 7}_{\text{15 terms}}$
But this cannot be expressed as a sum of conscutive positive odd numbers.
Therefore, there are three solutions: .$\displaystyle (53,52),\;(19,16),\;(11,4)$
10. Originally Posted by Soroban
Hello, dhiab!
Here is a primitive solution . . .
Consecutive squares differ by an odd number: .$\displaystyle (n+1)^2 - n^2 \:=\:2n+1$
We have: .$\displaystyle x^2 - y^2 \:=\:33$
. . Then: .$\displaystyle 2n+1 \:=\:33 \quad\Rightarrow\quad n \:=\:16$
. . Hence: .$\displaystyle \boxed{17^2 - 16^2 \:=\:33}$
$\displaystyle \text{We also have: }\;x^2-y^2 \:=\:\underbrace{9 +11+13}_{4^2\quad5^2\quad6^2\quad7^2}$
. . Hence: .$\displaystyle \boxed{7^2 - 4^2 \:=\:33}$
Therefore, there are exactly two solutions.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I came up with this procedure while in college.
Example: .$\displaystyle a^2-b^2 \:=\:105$
$\displaystyle 105$ is the difference between $\displaystyle \left(\tfrac{105-1}{2}\right)^2$ and $\displaystyle \left(\tfrac{105+1}{2}\right)^2$
. . Therefore: .$\displaystyle 53^2 - 52^2 \:=\:105$
We have: .$\displaystyle 105 \:=\:35 + 35 + 35 \:=\:33 + 35 + 37$
. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{33 + 35 + 37}_{16^2\quad17^2\quad18^2\quad19^2}$
. . Therefore: .$\displaystyle 19^2 - 16^2 \:=\:105$
We have: .$\displaystyle 105 \:=\:15 + 15 + 15 + 15 + 15 + 15 + 15 \:=\:9 + 11 + 13 + 15 + 17 + 19 + 21$
. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{9 + 11 + 13 + 15 + 17 + 19 + 21}_{4^2\quad\;\;5^2\quad\;\;6^2\quad\;\;\;7^2\qua d\;\;\;8^2\quad\;\;9^2\quad\;\;10^2\quad\;\;11^2}$
. . Therefore: .$\displaystyle 11^2 - 4^2 \:=\:105$
$\displaystyle \text{The next combination is: }\;105 \:=\:\underbrace{7+7+7+ \hdots + 7}_{\text{15 terms}}$
But this cannot be expressed as a sum of conscutive positive odd numbers.
Therefore, there are three solutions: .$\displaystyle (53,52),\;(19,16),\;(11,4)$
Hello :
Thank you for your efforts, but génerale case is based on the divisores of a number | 2,171 | 5,934 | {"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} | 4.40625 | 4 | CC-MAIN-2018-22 | latest | en | 0.779406 |
http://us.metamath.org/mpeuni/equsalhw.html | 1,638,835,347,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00257.warc.gz | 70,827,073 | 3,517 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > equsalhw Structured version Visualization version GIF version
Theorem equsalhw 2120
Description: Weaker version of equsalh 2291 (requiring distinct variables) without using ax-13 2245. (Contributed by NM, 29-Nov-2015.) (Proof shortened by Wolf Lammen, 28-Dec-2017.)
Hypotheses
Ref Expression
equsalhw.1 (𝜓 → ∀𝑥𝜓)
equsalhw.2 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
equsalhw (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Distinct variable group: 𝑥,𝑦
Allowed substitution hints: 𝜑(𝑥,𝑦) 𝜓(𝑥,𝑦)
Proof of Theorem equsalhw
StepHypRef Expression
1 equsalhw.1 . . 3 (𝜓 → ∀𝑥𝜓)
2119.23h 2119 . 2 (∀𝑥(𝑥 = 𝑦𝜓) ↔ (∃𝑥 𝑥 = 𝑦𝜓))
3 equsalhw.2 . . . 4 (𝑥 = 𝑦 → (𝜑𝜓))
43pm5.74i 260 . . 3 ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝑦𝜓))
54albii 1744 . 2 (∀𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜓))
6 ax6ev 1887 . . 3 𝑥 𝑥 = 𝑦
76a1bi 352 . 2 (𝜓 ↔ (∃𝑥 𝑥 = 𝑦𝜓))
82, 5, 73bitr4i 292 1 (∀𝑥(𝑥 = 𝑦𝜑) ↔ 𝜓)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∀wal 1478 ∃wex 1701 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1836 ax-6 1885 ax-7 1932 ax-10 2016 ax-12 2044 This theorem depends on definitions: df-bi 197 df-or 385 df-ex 1702 df-nf 1707 This theorem is referenced by: dvelimhw 2170
Copyright terms: Public domain W3C validator | 708 | 1,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.293873 |
https://www.jiskha.com/display.cgi?id=1298310879 | 1,516,654,407,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891539.71/warc/CC-MAIN-20180122193259-20180122213259-00607.warc.gz | 936,569,974 | 3,807 | # intermediate algebra
posted by .
Select the set of equations that represents the following situation: Mike invested \$706 for one year. He invested part of it at 5% and the rest at 3%. At the end of the year he earned \$28.00 in interest. How much did Mike invest at each rate of interest?
• intermediate algebra -
x = amount invested at 0.05
y = amount invested at 0.03
0.05x + 0.03y = 28.
x + y = 706
Solve the set of equations simultaneously
If you are correct,
x = \$341
y = \$365
• unlv -
solve. A trust fund has invested \$8000 at 6% annual interest. How much additional moey should be invested at 8.5% to obtain a return of 8% on the total amount invested?
## Respond to this Question
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https://ru.scribd.com/presentation/108451438/Unit-III-Maps | 1,571,041,741,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649841.6/warc/CC-MAIN-20191014074313-20191014101313-00554.warc.gz | 646,905,304 | 76,982 | Вы находитесь на странице: 1из 57
# SKN SINHGAD INSTITUTE OF TECHNOLOGY & SCIENCES, LONAVALA
## UNIT III Uses of Maps and Field Surveys
First year of Engineering ( Common to all branches)
Assistant Professor
Presented by
## Online Objective Examination on UNIT III and IV October 16.10.12 -20.10.12
Weightage: 25 Marks
## UNIT No. III III
Topic Various types of maps and their uses. Principles of survey. Modern survey methods using levels, Theodolite, EDM, lasers, total station and GPS Introduction to digital mapping. Measuring areas from maps using digital planimeter. Conducting simple and differential levelling for setting out various benchmarks, determining the elevations of different points and preparation of contour maps. Introduction to GIS Software and other surveying softwares with respect to their capabilities and application areas.
III III
III
CLASS TEXT
NOTES
REFERENCE
## BOOKS: Introduction to SurveyingAnderson-McGraw-Hill International Student Edition.
To understand. What is a Map? Types of Maps Scale of Maps Map Projections Uses of Maps
Maps
## We think of the earth as a sphere
It is actually a Ellipsoid, slightly larger in radius at the equator than at the poles
## Imaginary Lines on the surface Earth
Map is a fundamental language of geography which gives the descriptive information about the world.
A map is a small scale conventional representation of the earth (or part )as seen from above A Cartographic representation without scale should not be called a map. It should be considered as a sketch or a diagram.
A map is an integrated assemblage of four category of information Point Lines Areas Names (Labels)
All these have to be considered in terms of their interrelationship. All these are presented in different Shapes, patterns, size, symbols, and thickness.
## There are two basic parts of maps The Figure
Title: Describes what a map Shows Legend: Defines the Symbols Scale: Shows the relationship of map distance to actual distance on the ground. Direction: Refers to the cardinal directions and is shown by an arrow
Source: The institution or resource from which the information on the map was compiled.
Date: Shows when the map was made and the date of information on the map Border : Defines the edges of the map and separates the map from the text Author : The Institution or the individual that created the map Ground Relationship: Ground and water features differentiated
Types Of Maps
Thematic maps: Thematic maps are specialized maps Containing information about a Single Theme i.e. Landuse map, Population distribution, soils, Geology, Contours, Road network etc.
Topographical Maps: Topographic maps provide the most authentic base or reference tool showing natural and man made features including terrain information of part of the earth plotted to scale Example: Toposheet
## These maps represent the spatial dimensions of a particular phenomenon (theme).
Types:
Isopleth maps - isolines connect points of equal magnitude. Choropleth map - A choropleth map is a thematic map in which areas are shaded or patterned in proportion to the measurement of the statistical variable being displayed on the map, such as population density or per-capita income.
## Choropleth maps have areas of equal values separated by boundaries Example:1
Example:2
Maps portray quantitative data modeled by continuous surface. The variations are shown by lines connecting to points of equal value Examples are:
Contours : Lines of same elevation value Isotherm : Lines of equal Temperature Isohyets : Lines Showing equal Rainfall Isobars : Lines showing equal air pressure value Isolines : Lines showing equal value of the referred theme Isobaths : Lines showing equal depth
## Topographical Maps Topo Sheets
"Topo" maps provide highly detailed information about the natural and man-made aspects of the terrain, but are best known for their series of contour lines that show elevation changes, and colors signifying varying land types and bodies of water.
Toposheet
Topographic Maps
http://en.wikipedia.org/wiki/Image:Topographic_map_example.png
A cadastral map is a map which provides detailed information about real property within a specific area. These maps are usually maintained by the government,
## The World Political
Political maps show how people have divided places on the Earth into countries, states, cities and other units for the purpose of governing them.
Political maps
They show where the boundaries and locations of countries, states, cities, towns and counties are. These boundaries and locations are generally determined by people rather than nature.
## The World Physical
Physical maps show what the surface of the Earth looks like.
Aerial maps
Photo maps taken from up in the air.
Weather maps
Show predictions of coming weather or report on weather that is actually happening.
Contour Map
Show where roads, highways, routes, etc. are. The more a user zooms in on a map the more detailed the map is as to local roads, routes, etc.
Climate Maps
Give general information about the climate and precipitation (rain and snow) of a region.
Climate Maps
## Economic or Resource Maps
Feature the major types of natural resources or economic activity in an area.
Atlas
A collection of maps
Map Scale
Map scale is a ratio between the distance on the map to distance on the earths surface. Scales are shown in 3 ways on the maps RF Scale (Representative Fraction) Example : 1:50,000 Verbal Statement (Descriptive Scale) Example : 1 cm = 5 Km or 1 Inch = 1 Mile Bar Scale (Graphical Scale) Example :
## Large Scale Maps and Small Scale Maps
Large Scale maps show great detail, small features and Representative fraction is Large i.e. 1:2500 Example: Cadastral map, Topographical map Small scale maps show only large features and Representative fraction is small i.e. 1:250,000 Example: Atlas ,Wall maps The entire township is shown as a small block
Small Scale
shows large area 1:10,000,000 would
Large Scale
shows small area 1:63,360 would
## Map Projections: the concept
A method by which the curved 3D surface of the earth is represented on a flat 2D map surface. Location on the 3D earth is measured by latitude and longitude;
## Location on the 2D map is measured by X,Y Cartesian coordinates
cylindrical conical planar (azimuthal-zenithal)
## Map Projection: Map Scale:
Representative Fraction
Globe distance = Earth distance (e.g. 1:24,000) =
## All maps introduce distortion:
shape (conformance)
size (equivalence) direction
Distance
## Map Projections: the inevitability of distortion
Because we are trying to represent a 3-D sphere on a 2-D plane, distortion is inevitable Thus, every two dimensional map is distorted (inaccurate?) with respect to at least one of the following:
area shape distance direction
We are trying to represent this amount of the earth on this amount of map space.
Types of Projections
Conic (Albers Equal Area, Lambert Conformal Conic) - good for East-West land areas Cylindrical (Transverse Mercator) - good for North-South land areas Azimuthal (Lambert Azimuthal Equal Area) - good for global views
Conic Projections
(Albers, Lambert)
Cylindrical Projections
Azimuthal
(Lambert) | 1,568 | 7,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.856779 |
http://stackoverflow.com/questions/5648975/matlab-vs-mathematica-eigenvectors?answertab=oldest | 1,462,261,685,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860118807.54/warc/CC-MAIN-20160428161518-00035-ip-10-239-7-51.ec2.internal.warc.gz | 280,497,957 | 21,809 | # Matlab vs Mathematica, eigenvectors?
``````function H = calcHyperlinkMatrix(M)
[r c] = size(M);
H = zeros(r,c);
for i=1:r,
for j=1:c,
if (M(j,i) == 1)
colsum = sum(M,2);
H(i,j) = 1 / colsum(j);
end;
end;
end;
H
function V = pageRank(M)
[V D] = eigs(M,1);
V
R
M=[[0 1 1 0 0 0 0 0];[0 0 0 1 0 0 0 0];[0 1 0 0 1 0 0 0];[0 1 0 0 1 1 0 0];
[0 0 0 0 0 1 1 1];[0 0 0 0 0 0 0 1];[1 0 0 0 1 0 0 1];[0 0 0 0 0 1 1 0];]
ans =
-0.1400
-0.1576
-0.0700
-0.1576
-0.2276
-0.4727
-0.4201
-0.6886
``````
### Mathematica:
``````calculateHyperlinkMatrix[linkMatrix_] := {
H = Table[0, {a, 1, r}, {b, 1, c}];
For[i = 1, i < r + 1, i++,
For[j = 1, j < c + 1, j++,
0]
]
];
H
}
H = {{0, 0, 0, 0, 0, 0, 1/3, 0}, {1/2, 0, 1/2, 1/3, 0, 0, 0, 0}, {1/2,
0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 0}, {0, 0, 1/2, 1/3,
0, 0, 1/3, 0}, {0, 0, 0, 1/3, 1/3, 0, 0, 1/2}, {0, 0, 0, 0, 1/3,
0, 0, 1/2}, {0, 0, 0, 0, 1/3, 1, 1/3, 0}};
R = Eigensystem[H];
VR = {R[[1, 1]], R[[2, 1]]}
PageRank = VR[[2]]
{1, {12/59, 27/118, 6/59, 27/118, 39/118, 81/118, 36/59, 1}}
``````
Matlab and Mathematica doesn't give the same eigenvector with the eigenvalue 1. Both works though...which one is correct and why are they different? How do I gte all eigenvectors with the eigenvalue 1?
-
there is a lot of unnecessary code in your post, which only stands in a way of getting to the heart of your question. Can you trim it, please – Sasha Apr 13 '11 at 12:21
This may better asked at: math.stackexchange.com – Dan Andrews Apr 13 '11 at 12:24
Eigenvectors are not unique, take identity matrix -- any vector will be an eigenvector with eigenvalue 1 – Yaroslav Bulatov Apr 13 '11 at 18:48
The Definition of an Eigenvector `X` is some vector `X` that satisfies
`AX = kX`
where `A` is a matrix and `k` is a constant. It is pretty clear from the definition that `cX` is also an Eigenvector for any `c` not equal to `0`. So there is some constant `c` such that `X_matlab = cX_mathematica`.
It looks like the first is normal (has Euclidean length 1, i.e. add the sums of the squares of the coordinates then take the square root and you will get 1) and the second is normalised so that the final coordinate is 1 (any Eigenvector was found and then all coordinates were divided by the final coordinate).
You can use whichever one you want, if all you need is an Eigenvector.
-
Also, if one tells Mathematica to analyze numerically, it normalizes to norm 1, just as MATLAB (they use the same iterative method, which by its nature normalizes the eigenvectors in each step). – Daniel Andersson Apr 13 '11 at 12:35
Should be "So there is some constant c such that X_matlab = c * X_mathematica.", for clarity. – Daniel Andersson Apr 13 '11 at 12:37
This is because if a vector x is an eigenvector of matrix H, so is any multiple of the x. Vector you quote as an answer for matlab does not quite check:
``````In[41]:= H.matlab - matlab
Out[41]= {-0.0000333333, 0.0000666667, 0., 0., 0.0000333333, 0., \
-0.0000666667, 0.}
``````
But assuming it is close enough, you see that
``````In[43]:= {12/59, 27/118, 6/59, 27/118, 39/118, 81/118, 36/59,
1}/{-0.1400, -0.1576, -0.0700, -0.1576,
-0.2276, -0.4727, -0.4201, -0.6886}
Out[43]= {-1.45278, -1.45186, -1.45278, -1.45186, -1.45215, -1.45217, \
-1.45244, -1.45222}
``````
consists of almost the same elements. Thus matlab's vector is -1.45 multiple of Mathematica's.
-
Sorry, I did the same and posted it a few seconds after you. Deleting my answer :) – Dr. belisarius Apr 13 '11 at 15:45
Eigenvectors are not necessarily unique. All that is required of an eigenvector is that
1. It must have unit norm
2. `v_m*v_n=0` for all `m ≠ n` (orthogonality)
3. It satisfies `Av_m=u_m v_m`, where `u_m` is the corresponding eigenvalue
The exact eigenvectors returned depends on the algorithm implemented. As a simple example to demonstrate that one matrix can have two different sets of eigenvectors, consider an `NxN` identity matrix:
``````I= 1 0 0 0
0 1 0 0
... ... ... ...
0 0 0 1
``````
It is obvious (and can be easily confirmed) that each column of `I` is an eigenvector and the eigenvalues are all 1.
I now state that the following vectors
``````v_m=[1,exp(2*pi*1i*m/N),...,exp(2*pi*1i*m*(N-1)/N)]';
``````
for `m=1,2...,N` form an orthogonal basis set with norm 1, and hence are the eigenvectors of `I`. Here `1i` refers to square root of `-1` in MATLAB notation. You can verify this for yourself:
``````N=50;
v=1/sqrt(N)*cumprod(repmat(exp(-1i*2*pi/N*(0:N-1)),N,1),1);
imagesc(real(v*v'));
``````
Here I've taken the real part because the imaginary part is non-zero (of order `10^-16`)due to machine precision effects, but should be zero (you can even do this analytically and it should be zero). `imagesc` returns an error otherwise.
So, to sum up, eigenvectors are not necessarily unique and both convey the same information; just in different representations.
- | 1,877 | 4,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2016-18 | latest | en | 0.74749 |
https://docs.openstack.org/developer/performance-docs/methodologies/hyper-scale.html | 1,621,054,245,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989812.47/warc/CC-MAIN-20210515035645-20210515065645-00018.warc.gz | 235,616,156 | 7,119 | # 3.2. Methodology for testing Hyper-Scale¶
## 3.2.1. What is meant by Hyper-Scale?¶
For this section, hyper-scale is defined as being able to achieve 100K of something in an OpenStack cloud. Initially this may be VMs, but could be hypervisors, projects, etc. Because of the open-endnesses of this definition, it is necessary to identify all places where operations show linear (or even worse...) execution time as the number of items that the operation deals with scales up. For example, if execution time of an operation is order of 10 microseconds per item that operation is dealing with, that means that at 100K, a second has been added to agent execution time.
## 3.2.2. How is linearity defined in this context?¶
To determine linearity, one measures the execution time of each operation over a sample the has increasing scale and then performance a least squares fit [1] to the raw data. Such a fit looks to map a straight line (y=mx+b) to the sample data, where
• x == the number of items being scaled,
• m == the slope of the fit,
• b == the vertical intercept, and
• y == the operation’s execution time
In addition to finding m and b, the fit will also generate a correlation coefficient (r) which provides a measurement of how random the sample data was. As an initial filter, fits with a correlation coefficient above a certain threshold can be considered as candidates for improvement. Once an operation is considered a candidate for improvement, then the execution time of all of its sub-operations are measured and a least squares fit performed on them. When looking at sub-operations, of interest are the slope and the ratio of the vertical intercept to the slope (the “doubling scale” of the operation). These together determine the “long pole in the test” (the sub-operation that should be analyzed first), with the slope determining the order and the “doubling scale” acting as a tie breaker.
## 3.2.3. How Should Code be Instrumented When Looking for Hyper-Scale?¶
Today, instrumentation is a very manual process: after running devstack, debug LOG statements are added into the code base to indicate the entrance and exit of each routine that should be timed. These debug log statements can be tagged with strings like “instrument:” or “instrument2:” to allow multiple levels of instrumentation to be applied during a single test run. Then restart the agent that is being instrumented via the screen that is created by devstack.
Note: as this ends up to be an iterative process as one looks to isolate the code that shows linear execution time, for subsequent restacks, it usually makes sense to not reclone (unless you want to repeat the by hand editing afterwards).
It is hoped that the osprofiler sub-project of oslo [2] will be extended to allow for method decoration to replace the use of by hand editing.
## 3.2.4. Setup Considerations¶
There are two ways to go about looking for potential hyper-scale issues. The first is to use a brute force method where one actually builds a hyper-scale cloud and tests. Another route is to build a cloud using the slowest, smallest, simplest components one has on hand. The idea behind this approach is that by slowing down the hardware, the slope and intercept of the least squares fit are both scaled up, making linearity issues easier to find with lower numbers of instances. However, this makes comparing raw values from the least squares fits between different setups difficult - one can never be 100% sure that the comparison is “apples-to-apples” and that hardware effects are completely isolated. The choice above of using the correlation coefficient and “doubling scale” above are intended to use values that are more independent of the test cloud design.
## 3.2.5. An Example: Neutron L3 Scheduling¶
The rest of this section presents an example of this methodology in action (In fact, this example was what led to the methodology being developed). This example uses a four node cloud, configured as a controller, a network node, and two compute nodes.
To determine what parts of L3 scheduling needed scaling improvements the following experiment was used. This test can certainly be automated via Rally and should be (once instrumentation no longer requires manual decoration and/or code manipulation):
1. Create the external network and subnet with:
```neutron net-create test-external --router:external --provider:physical_network default --provider:network_type flat
neutron subnet-create --name test-external-subnet --disable-dhcp test-external 172.18.128.0/20
```
1. Create X number of projects
2. For each project, perform the following steps (in fhe following, \$i is the tenant index and \$id is the tenant UUID, \$net is the UUID of the tenant network created in the first step below, and \$port is the interface port UUID reported by nova):
```neutron net-create network-\$i --tenant-id \$id
neutron subnet-create network-\$i 192.168.18.0/24 --name subnet-\$i --tenant-id \$id
neutron router-create router-\$i --tenant-id \$id
neutron router-gateway-set router-\$i test-external
nova --os-tenant-name tenant-\$i boot --flavor m1.nano --image cirros-0.3.4-x86_64-uec --nic net-id=\$net instance-tenant-\$i
(once the instance's interface port is shown as active by nova):
neutron floatingip-create --tenant-id \$id --port-id \$port test-external
```
1. For each project, perform the following steps:
```neutron floatingip-delete <floatingip-uuid>
nova --os-tenant-name tenant-\$i stop instance-tenant-\$i
neutron router-gateway-clear <router-uuid> test-external
neutron router-interface-delete router-\$i subnet-\$i
neutron router-delete router-\$i
neutron subnet-delete subnet-\$i
neutron net-delete network-\$i
```
This provides testing of both the paths that create and delete Neutron routers and the attributes that make them usuable.
### 3.2.5.1. Extracting Execution Time¶
Once one has instrumented logs (or instrumentation records), extracting execution time becomes a matter of filtering via one’s favorite language. One approach is to filter to a comma separate value (csv) file and then import that into a spreadsheet program that has the least squares function built in to find the fit and correlation coefficient.
### 3.2.5.2. Legacy Results Using Neutron Master as of 11/6/15 - Adding Routers/FIPs¶
Function Corelation Coeficient Slope (msec/router) Doubling Scale (routers)
Set Gateway 0.260
Thus, the interface add operation is a candidate for further evaluation. Looking at the largest contributers to the slope, we find the following methods at successively deeper levels of code:
• _process_router_if_compatible (1.34 msec/router)
• _process (0.998 msec/router)
• process_internal_ports (0.888 msec/router)
This last time is outside of Neutron: it needs to be addressed from within the OVS code itself (this is being tracked by [3])
### 3.2.5.3. Legacy Results Using Neutron Master as of 11/6/15 - Removing Routers/FIPs¶
Function Corelation Coeficient Slope (msec/router) Doubling Scale (routers)
Clear FIP 0.389 1.503 401.5
Clear Gateway 0.535 1.733 289.3
Remove Interface 0.390 1.425 302.6
Remove Router 0.526 1.5 252.8
Each of these operations is a candidate for improvement. Looking more deeply at each operation, we find the following contributors to the slope:
• Clear FIP
• get_routers (9.667 usec/router)
• _process_routers_if_compatible (1.496 msec/router)
• _process_updated_router (1.455 msec/router)
• process (1.454 msec/router)
• process_internal_ports (151 usec/router)
• process_external (1.304 msec/router)
• iptables apply time: slope (0.513 msec/router)
• process_external_gateway (0.207 msec/router)
• FIP cleanup (0.551 msec/router)
• Clear Gateway
• get_routers (25.8 usec/router)
• _process_routers_if_compatible (1.714 msec/router)
• _process_update_router (1.700 msec/router)
• process (1.700 msec/router)
• process_internal_ports (92 usec/router)
• process_external (0.99 msec/router)
• process_external_gateway (0.99 msec/router)
• unplug (0.857 msec/router)
• iptables apply time: slope (0.614 msec/router)
• Remove Interface
• _process_routers_if_compatible (1.398 msec/router)
• _process_update_router (1.372 msec/router)
• process (1.372 msec/router)
• process_internal_ports (0.685 sec/router)
• old_ports_loop (0.586 msec/router)
• iptables apply time: slope (0.538 msec/router)
• Remove Router
• _safe_router_removed (1.5 msec/router)
• before_delete_callback (0.482 msec/router)
• delete (1.055 msec/router)
• process (0.865 msec/router)
• process_internal_ports (0.168 msec/router)
• process_external (0.200 msec/router)
• iptables apply time: slope (0.496 msec/router)
• namespace delete (0.195 msec/router)
As can be seen, improving execution time for iptables will impact a lot of operations (and this is a known issue). In addition there are several other functions to be dug into further to see what improvements can be made. | 2,134 | 8,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-21 | latest | en | 0.938329 |
http://ask.metafilter.com/tags/machine+resolved | 1,481,331,884,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542932.99/warc/CC-MAIN-20161202170902-00313-ip-10-31-129-80.ec2.internal.warc.gz | 19,209,819 | 20,913 | ## Short film from 1980s childrens show: automatic breakfast maker
Back in the day when Duoshao was a tyke, namely the late 70s/early 80s, one of the children's shows such as Electric Company or something similar periodically showed a short film of a complex machine that cooked breakfast. Does anyone remember this? I'd love to watch it again.
posted by duoshao on Sep 18, 2016 - 12 answers
## Front load washing machine: spin extraction problem
After a year of normal functioning, my washing machine's spin cycle is not extracting properly -- but ONLY with flannel sheets. They go through the spin cycle, but the water doesn't extract. The wash drum has no puddle. Just the flannel sheets are soaking wet. Other types of fabrics come out normal, with excess water extracted. Towels, jeans, percale sheets come out normal. Small flannel pillowcases come out normal.
posted by valannc on Nov 6, 2015 - 11 answers
## Rainout, rainout, go away!
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## Did I just lose 99.9% of my iTunes library?!?
I opened iTunes today to play some music, and discovered that of the 13,129 songs in my library, only 22 truly remain. Where did the rest go?
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## Computer science machine language help! If-then-otherwise statements
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## Maytag top loading washing machine stopped working & filled with water
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## Why is my sewing machine pulling loops underneath the fabric?
I've read up on how the top and bottom tensioning is supposed to work, but it hasn't helped. I've got the bobbin adjusted so it lets out about an inch when I bounce it by the thread. Adjusting the top tension from there doesn't seem to make much difference. I always get the loops. The weird thing is that the top thread seems really hard to pull out when I do it by hand. It's so hard that's it sometimes breaks. Why should tension that seems stiff to the hand pull through when the machine sews?
posted by SirNovember on Mar 16, 2014 - 15 answers
## Why virtualbox can overcome win8 UEFI issue to install Linux?
I tried to install linux on a win8 laptop and failed although I tried everything to disable UEFI and security boot options in Bios. So I think there is no solution to install Linux on certain win8 Laptops. But someone told me of course I can install Linux on any win8 laptops by using virtualbox. So I'm wondering why can it be done? How exactly is virtualbox/virtual machine works for this issue? can anybody explain this to me in a easy to understand way? Thanks very much.
posted by pack2themoon on Feb 19, 2014 - 9 answers
## Mac shows Time Maching but not External HD
Reinstalled Snow Leopard but ejected my external drive before this. Then, when done, plugged back in via USB as my usual custom....and Time Machine still shows up, but the External Seagate drive does not. I made sure that in Finder/Preferences that the external disc box as others was checked. I tried a different cord, and different port. I opened up Disc Utility and the External HD's partioned Time Machine is available but the other part, now renamed something else (a default name from manufacturer?) is greyed out and I cannot touch it. Any thoughts? Should i now go to the boot disc of Snow Leopard? Not sure why TM shows but the rest does not.
posted by chicaboom on Mar 31, 2013 - 11 answers
## Recover hacked VPS logs
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## Need help finding a book to help me rebuild civilization.
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posted by brownrd on Feb 14, 2012 - 14 answers
## Roland TR-808 documentary.
I'm looking for a video documentary from a couple years ago tracing the history of the Roland TR-808 drum machine. Released only on the web, it's similar in tone to the Everything is a Remix series; dry delivery, well researched, touches on remix culture. The Newcleus track Jam On It is prominently featured as a signifier of the 808's transition from a not-so-hot replacement for a drummer to being used for the sounds unique to it. If I recall correctly, it's about 8 minutes in length. It's not The Shape of Things That Hum.
posted by Scoo on Sep 28, 2011 - 3 answers
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## Snack Dome
What is the purpose of the dome on this vending machine?
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What's the cleanest possible way to transfer a standard-IPS four-track cassette master?
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## Washing Machine Making Clothes Dirty
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## Your favorite espresso maker
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Page: 1 | 2,609 | 10,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.930563 |
http://conversion-website.com/area/square-decameter-to-decare.html | 1,653,135,373,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662539101.40/warc/CC-MAIN-20220521112022-20220521142022-00045.warc.gz | 12,465,910 | 4,249 | # Square decameters to decares (sq dam to daa)
## Convert square decameters to decares
Square decameters to decares converter above calculates how many decares are in 'X' square decameters (where 'X' is the number of square decameters to convert to decares). In order to convert a value from square decameters to decares (from sq dam to daa) type the number of sq dam to be converted to daa and then click on the 'convert' button.
## Square decameters to decares conversion factor
1 square decameter is equal to 0.1 decares
## Square decameters to decares conversion formula
Area(daa) = Area (sq dam) × 0.1
Example: Find out how many decares equal 335 square decameters.
Area(daa) = 335 ( sq dam ) × 0.1 ( daa / sq dam )
Area(daa) = 33.5 daa or
335 sq dam = 33.5 daa
335 square decameters equals 33.5 decares
## Square decameters to decares conversion table
square decameters (sq dam)decares (daa)
151.5
202
252.5
303
353.5
404
454.5
505
555.5
606
656.5
707
757.5
808
858.5
square decameters (sq dam)decares (daa)
15015
20020
25025
30030
35035
40040
45045
50050
55055
60060
65065
70070
75075
80080
85085
Versions of the square decameters to decares conversion table. To create a square decameters to decares conversion table for different values, click on the "Create a customized area conversion table" button.
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TableFormulaFactorConverterTop | 451 | 1,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | latest | en | 0.599041 |
https://bseodisha.guru/bse-odisha-9th-class-maths-solutions-algebra-chapter-3-ex-3c/ | 1,725,758,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00422.warc.gz | 131,110,637 | 51,351 | # BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(c)
Odisha State Board BSE Odisha 9th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(c) Textbook Exercise Questions and Answers.
## BSE Odisha Class 9 Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(c)
Question 1.
ପ୍ରତ୍ୟେକ କ୍ଷେତ୍ରରେ ଠିକ୍ ଉତ୍ତରଟିକୁ ବାଛି ଲେଖ ।
(i) x2 – 3x + 2 ର ଉତ୍ପାଦକ ଦ୍ଵୟ
(a) (x – 2) ଓ (x + 1)
(b) (x + 2) ଓ (x – 1)
(c) (x – 2) ଓ (x – 1)
(d) (x + 2) ଓ (x + 1)
ସମାଧାନ:
(x – 2) ଓ (x + 1)
x2 – 3 + 2 = x2 – (2 + 1) x + 2 . 1 = (x – 2) (x – 1)
(ii) ଏକ ଦ୍ୱିଘାତୀ ପଲିନୋମିଆଲ୍ର ଉତ୍ପାଦକ ଦ୍ବୟ (x – 1) ଓ (x – 3) ହେଲେ ପଲିନୋମିଆଲ୍ଟି
(a) x2 – 4x – 3
(b) x2 – 4x + 3
(c) x2 + 4x – 3
(d) x2 + 4x + 3
ସମାଧାନ:
x2 – 4x + 3
(x – 1) (x – 3) = x2 – (1 + 3) x + (1) (3) = x2 – 4x + 3
(iii) x – y ର ଉତ୍ପାଦକ ମାନ
(a) (x2 + y2) (x + y) (x – y)
(b) (x2 – y2) (x – y) (x + y)
(c) (x2 + y2) (x + y)2
(d) (x2 + y2) (x – y)
ସମାଧାନ:
(x2 + y2) (x + y) (x – y)
x4 – y4 = (x2)2 – (y2)2 = (x2 + y2) (x2 – y2) = (x2 + y2) (x + y) (x – y)
(iv) 8a3 – b3 – 12a2b + 6ab2 ର ଉତ୍ପାଦକଗୁଡ଼ିକ
(a) (2a – b), (2a + b), (2a + b)
(b) (2a + b) (2a + b) (2a + b)
(c) (2a – b), (2a – b), (2a + b)
(d) (2a – b), (2a – b), (2a – b)
ସମାଧାନ:
(2a – b) (2a – b) (2a – b)
8a3 – b3 – 12a2b + 6ab2 = (2a)3 – b3 – 3.2a.b(2a – b)
= (2a – b)3 = (2a – b) (2a – b) (2a – b)
(v) 625 + 25x4 + x8 ର ଉତ୍ପାଦକଗୁଡ଼ିକ
(a) (25 + 5x2 + x4) (25 – 5x2 + x4)
(b) (25 + 5x2 + x4) (25 + 5x2 – x4)
(c) (25 + 5x4 + x4) (25 – 5x4 + x4)
(d) (25 – 5x4 + x4)(25 + 5x4 – x4)
ସମାଧାନ:
(25 + 5x2 + x4) (25 – 5x2 + x4)
625 + 25x4 + x8 = 54 + 52(x2)2 + (x2)4
= {52 + 5x2 + (x2)2} {52 – 5x2 + (x2)2} = (25 + 5x2 + x4) (25 – 5x2 + x4)
(vi) 1 – a3 + b3 + 3ab ର ଗୋଟିଏ ଉତ୍ପାଦକ
(a) (1 – a + b)
(b) (1 – a – b)
(c) (1 + a + b)
(d) (1 + a – b)
ସମାଧାନ:
(1 – a + b)
1 – a3 + b3 + 3ab
= (1 – a + b) {12 + a2 + b2 – (1) (-a) – (-a) (b) – (b) (1)}
= (1 – a + b) (1 + a2 + b2 + a + ab – b)
(vii) (2x – 3y)3 + (3y – 4z)3 + (4z – 2x)3 ର ଉତ୍ପାଦକଗୁଡ଼ିକ ହେଲେ
(a) 6(2x – 3y)(3y – 4z) (2z – x)
(b) 3(2x – 3y) (3y – 4z) (2z – x)
(c) 60xyz
(d) ଏଥୁମଧ୍ୟରୁ କୌଣସିଟି ନୁହେଁ
ସମାଧାନ:
6 (2x – 3y) (3y – 4z) (2z – x)
2x – 3y + 3y – 4z + 4z – 2x = 0
(2x – 3y)3 + (3y – 4z)3 + (4z – 2x)3 = 3 (2x – 3y) (3y – 4z) (4z – 2x)
= 6 (2x – 3y) (3y – 4z) (2z – x)(∵ a + b + c = 0 ହେଲେ a3 + b3 + c3 = 3abc)
(viii) (28)3 + (-15)3 + (-13)3 ର ସରଳୀକୃତ ମାନ
(a) 8190
(b) 16380
(c) 24570
(d) 4095
ସମାଧାନ:
16380
28 – 15 -13 = 0
∴ (28)3 + (-15)3 + (-13)3 = 3 (28) (-15) (-13) = 16380
(∵ a + b + c = 0 ହେଲେ a3 + b3 + c3 = 3abc)
(ix) (a – b)3 + (b – c)3 + (c – a)3 ର ମାନ
(a) 3abc
(b) 3a3b3c3
(c) 3(a – b) (b – c)(c – a)
(d) {a – (b + c)}3
ସମାଧାନ:
3(a – b) (b – c) (c – a)
(a – b)3 + (b – c)3 + (c – a)3 = 3(a – b) (b – c) (c – a)
(∵ (a – b) + (b – c) + (c – a) = 0)
(x) 2x2 – x – 1 ର ଗୋଟିଏ ଉତ୍ପାଦକ
(a) 2x – 1
(b) x + 1
(c) x – 1
(d) x + 2
ସମାଧାନ:
(x – 1)
2x2 – x – 1 = 2x2 – 2x + x – 1
= 2x (x – 1) + 1 (x – 1) = (x – 1) (2x + 1)
∴ 2x2 – x – 1 ର ଏକ ଉତ୍ପାଦକ (x – 1)
Question 2.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।
(i) 2x2 – x – 1
ସମାଧାନ:
2x2 – x – 1 = 2x2 – 2x + x – 1
= 2x(x – 1) + 1(x – 1) = (x – 1) (2x + 1)
(ii) 2x2 – 3x + 1
ସମାଧାନ:
2x2 – 3x + 1 = 2x2 – 2x – x + 1
= 2x (x – 1) – 1 (x – 1) = (x – 1) (2x – 1)
(iii) 5x2 – x – 4
ସମାଧାନ:
5x2 – x – 4 = 5x2 – 5x + 4x – 4
= 5x (x – 1) + 4 (x – 1) = (x – 1) (5x + 4)
(iv) 4x2 – 5x – 6
ସମାଧାନ:
4x2 – 5x – 6 = 4x2 – 8x + 3x – 6
= 4x(x – 2) + 3 (x – 2) = (4x + 3) (x – 2)
(v) 3x2 + 11x + 6
ସମାଧାନ:
3x2 + 11x + 6 = 3x2 + 9x + 2x + 6
= 3x (x + 3) + 2 (x + 3) = (3x + 2) (x + 3)
(vi) 7x2 + x – 6
ସମାଧାନ:
7x2 + x – 6 = 7x2 + 7x – 6x – 6
= 7x (x + 1) – 6 (x + 1) = (7x – 6) (x + 1)
(vii) 2x2 + 5x – 7
ସମାଧାନ:
2x2 + 5x – 7 = 2x2 + 7x – 2x – 7
= x (2x + 7) – 1 (2x + 7) = (2x + 7) (x – 1)
(viii) 4x2 – 5x + 1
ସମାଧାନ:
4x2 – 5x + 1 = 4x2 – 4x – x + 1
= 4x (x – 1) – 1 (x – 1) = (4x – 1) (x – 1)
(ix) 4x2 – 3x – 7
ସମାଧାନ:
4x2 – 3x – 7 = 4x2 – 7x + 4x – 7
= x (4x – 7) + 1 (4x – 7) = (4x – 7) (x + 1)
Question 3.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।
[a3 + b3 = (a + b) (a2 – ab + b2), a3 – b3 = (a – b) (a2 + ab + b2)]
(i) 25a4 – 16b2
ସମାଧାନ:
25a4 – 16b2 = (5a2)2 – (4b)2 = (5a2 + 4b)(5a2 – 4b)
(ii) 9 – 64p2q2
ସମାଧାନ:
9 – 64p2q2 = (3)2 – (8pq)2 = (3 + 8pq)(3 – 8pq)
(iii) 8x3 + 27y3
ସମାଧାନ:
8x3 + 27y3 = (2x)3 + (3y)3 = (2x + 3y){(2x)2 – 2x.3y + (3y)2}
= (2x + 3y)(4x2 – 6xy + 9y2)
(iv) 8x3 – 27y3
ସମାଧାନ:
8x3 – 27y3 = (2x)3 – (3y)3 = (2x – 3y){(2x)2 + 2x.3y + (3y)2}
= (2x – 3y)(4x2 + 6xy + 9y2)
(v) (a + b)2 – 9
ସମାଧାନ:
(a + b)2 – 9 = (a + b)2 – 32 = (a + b + 3) (a + b – 3)
(vi) (2a + 5)2 – 16
ସମାଧାନ:
(2a + 5)2 – 16 = (2a + 5)2 – 42 = (2a + 5 + 4) (2a + 5 – 4) = (2a + 9) (2a + 1)
(vii) (x + 2y)2 – (x – y)2
ସମାଧାନ:
(x + 2y)2 – (x – y)2 = {(x + 2y) + (x – y)} {x + 2y) – (x – y)}
= (x + 2y + x – y) (x + 2y – x + y) = (2x + y) (3y) = 3y (2x + y)
(viii) 4(a + 2p)2 – 9 (2a – p)2
ସମାଧାନ:
4 (a + 2p)2 – 9 (2a – p)2 = {2 (a + 2p)}2 – {3 (2a – p)}2
= (2a + 4p)2 – (6a – 3p)2 = (2a + 4p + 6a- 3p) (2a + 4p – 6a + 3p) = (8a + p) (7p – 4a)
(ix) 75 (2a – b + 1)2 – 12 (a + b)2
ସମାଧାନ:
75 (2a – b + 1)2 – 12 (a + b)2 = 3 {25 (2a – b + 1)2 – 4 (a + b)2}
= 3 [{5 (2a – b + 1)}2 – {2 (a + b)2}] = 3 {(10a – 5b + 5)2 – (2a + 2b)2]
= 3 (10a – 5b + 5 + 2a + 2b) (10a – 5b + 5 – 2a – 2b) = 3 (12a – 3b + 5) (8a – 7b + 5)
(x) (a + b)3 – 8c3
ସମାଧାନ:
(a + b)3 – 8c3 = (a + b)3 – (2c)3
= (a + b – 2c) {(a + b)2 + (a + b)2c + (2c)2}
= (a + b – 2c) (a2 + 2ab + b2 + 2ca + 2bc + 4c2)
= (a + b – 2c) (a2 + b2 + 4c2 + 2ab + 2bc + 2ca)
(xi) p4 – 27pq6
ସମାଧାନ:
p4 – 27pq6 = p(p3 – 27q6)
= p{p3 – (3q2)3} = p(p – 3q2) {p2 + 3pq2 + (3q2)2} = p(p – 3q2) (p2 + 3pq2 + 9q4)
(xii) 1 – (a + 2)3
ସମାଧାନ:
1 – (a + 2)3 = 13 – (a + 2)3 = (1 – a – 2) {12 + a + 2 + (a + 2)2}
= (- a – 1) (1 + a + 2 + a2 + 4a + 4) = -(a + 1) (a2 + 5a + 7)
(xiii) 8 – (2x – 3)3
ସମାଧାନ:
8 – (2x – 3)3 = 23 – (2x – 3)3 = (2 – 2x + 3) {22 + 2 (2x – 3) + (2x – 3)2}
= (-2x + 5) (4 + 4x – 6 + 4x2 – 12x + 9) = (5 – 2x) (4x2 – 8x + 7)
(xiv) 320p6q – 5p2q7
ସମାଧାନ:
320 p6q – 5p2q7 = 5p2q (64p4 – q6)
= 5p2q {(8p2)2 – (q3)2} = 5p2q (8p2 + q3) (8p2 – q3)
(xv) 1 + (a + 2)3
ସମାଧାନ:
1 + (a + 2)3 = 13 + (a + 2)3 = (1 + a + 2) {12 – 1(a + 2) + (a + 2)2}
= (a + 3) (1 – a – 2 + a2 + 4a + 4) = (a + 3) (a2 + 3a + 3)
(xvi) 8 + (2x – 3)3
ସମାଧାନ:
8 + (2x – 3)3 = 23 + (2x – 3)3 = (2 + 2x – 3) {22 – 2 (2x – 3) + (2x – 3)2}
= (2x – 1) (4 – 4x + 6 + 4x2 – 12x + 9) = (2x – 1) (4x2 – 16x + 19)
(xvii) a3 + 6a2b + 12ab2 + 8b3
ସମାଧାନ:
a3 + 6a2b + 12ab2 + 8b3 = a3 + 3a2 (2b) + 3 . a . (2b)2 + (2b)3
= (a + 2b)3 = (a + 2b) (a + 2b) (a + 2b)
(xviii) a3 + 9a2 + 27a + 27
ସମାଧାନ:
a3 + 9a2 + 27a + 27 = a3 + 3a2 . 3 + 3 . a . 32 + 33 = (a + 3)3
= (a + 3) (a + 3) (a + 3)
(xix) 8 – 36p + 54p2 – 27p3
ସମାଧାନ:
8 – 36p + 54p2 – 27p3 = 23 – 3. 22 3p + 3 . 2 (3p)2 – (3p)3
= (2 – 3p)3 = (2 – 3p) (2 – 3p) (2 – 3p)
(xx) (b – q)3 – (c – q)3 – 3 (b – c) (b – q) (c – q)
ସମାଧାନ:
(b – q)3 – (c – q)3 – 3 (b – c) (b – q) (c – q)
ମନେକର b – q = x ଓ c – q = y
∴ x – y = (b – q) – (c – q) = b – q – c + q = b – c
ପ୍ରଦତ୍ତ ପଲିନୋମିଆଲ୍ଟି x3 – y3 – 3 (x – y) (xy) = x3 – y3 – 3xy (x – y) = (x – y)3
= {(b – q) – (c – q)}3 (x ଓ yର ମାନ ବସାକଳେ)
= (b – c)3 = (b – c) (b – c) (b – c)
Question 4.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।
a4 + a2b2 + b4 = (a2 + ab + b2)(a2 – ab + b2)
(i) a4 + a2 + 1
ସମାଧାନ:
a4 + a2 + 1
= a4 + 1 + a2 = (a2)2 + (1)2 + a2 = (a2 + 1)2 – 2 . a2. 1 + a2
= (a2 + 1)2 – 2a2 + a2 = (a2 + 1)2 – a2 = (a2 + 1 + a) (a2 + 1 – a)
= (a2 + a + 1)(a2 – a + 1)
(ii) a4b4 + a2b2 + 1
ସମାଧାନ:
a4b4 + a2b2 + 1 = (ab)4 + (ab)2 12 + 14
= (a2b2 + ab + 1) (a2b2 – ab + 1)
(iii) 16a4 + 36a2b2 + 81b4
ସମାଧାନ:
16a4 + 36a2b2 + 81b4 = (4a2)2 + 36a2b2 + (9b2)2
= (4a2)2 + (9b2)2 + 72a2b2 – 72a2b2 + 36a2b2 [72a2b2 ଯେ।ଗ ଓ ବିୟେ।ଗ କରି]
= (4a2 + 9b2)2 – 36a2b2 = (4a2 + 9b2)2 – (6ab)2
= (4a2 + 9b2 + 6ab) (4a2 + 9b2 – 6ab) = (4a2 + 6ab + 9b2) (4a2 – 6ab + 9b2)
ବିକଳ୍ପ ସମାଧାନ –
16a4 + 36a2b2 + 81b4 = (4a2)2 + (9b2)2 + 36a2b2
= (4a2 + 9b2)2 – 2 . 4a2 . 9b2 + 36a2b2 = (4a2 + 9b2)2 – 36a2b2
= (4a2 + 9b2)2 – (6ab)2 = (4a2 + 9b2 + 6ab) (4a2 + 9b2 – 6ab)
= (4a2 + 6ab + 9b2) (4a2 – 6ab + 9b2)
(iv) a8 + a4 + 1
ସମାଧାନ:
a8 + a4 + 1 = (a2)4 + (a2)2 . 12 + 14
= {(a2)2 + a2 . 1 + 12} {(a2)2 – a2 . 1 + 12} = (a4 + a2 + 1) (a4 – a2 + 1)
= (a4 + a2 . 12 + 14) (a4 – a2 + 1) = (a2 + a + 1) (a2 – a + 1) (a4 – a2 + 1)
(v) x4 + 4
ସମାଧାନ:
x4 + 4 = (x2)2 + (2)2 = (x2)2 + (2)2 + 4x2 – 4x2
= (x2 + 2)2 – (2x)2 = (x2 + 2 + 2x) (x2 + 2 – 2x) = (x2 + 2x + 2) (x2 – 2x + 2)
(vi) 2a4 + 8b4
ସମାଧାନ:
2a4 + 8b4 = 2(a4 + 4b4) = 2{(a2)2 + (2b2)2 + 4a2b2 – 4a2b2}
= 2{(a2 + 2b2)2 – (2ab)2} = 2 {a2 + 2b2 + 2ab) (a2 + 2b2 – 2ab)}
= 2(a2 + 2ab + 2b2) (a2 – 2ab + 2b2)
(vii) 36a4 + 9b4
ସମାଧାନ:
36a4 + 9b4 = 9 (4a4 + b4) = 9 {(2a2)2 + (b2)2}
= 9 {(2a2 + b2)2 – 2 (2a2) b2} = 9 {(2a2 + b2)2 – 4a2b2}
= 9 {(2a2 + b2)2 – (2ab)2} = 9(2a2 + b2 + 2ab) (2a2 + b2 – 2ab)
= 9 (2a2 + 2ab + b2) (2a2 – 2ab + b2)
(viii) 4a4 + 7a2 + 16
ସମାଧାନ:
4a4 + 7a2 + 16 = (2a2)2 + (4)2 + 16a2 – 16a2 + 7a2
= (2a2 + 4)2 – 9a2 = (2a2 + 4)2 – (3a)2 = (2a2 + 4 + 3a) (2a2 + 4 – 3a)
= (2a2 + 3a + 4) (2a2 – 3a + 4)
(ix) a4 + 2a2b2 + 9b4
ସମାଧାନ:
a4 + 2a2b2 + 9b4 = a4 + 9b4 + 2a2b2
= (a2)2 + (3b2)2 + 6a2b2 – 6a2b2 + 2a2b2 = (a2 + 3b2)2 – 4a2b2
= (a2 + 3b2)2 – (2ab)2 = (a2 + 3b2 + 2ab) (a2 + 3b2 – 2ab)
= (a2 + 2ab + 3b2) (a2 – 2ab + 3b2)
(x) a4 – 3a2 + 1
ସମାଧାନ:
a4 – 3a2 + 1 = (a2)2 + (1)2 – 3a2
= (a2)2 + (1)2 – 2a2 + 2a2 – 3a2 (2a2 ଯେ।ଗ ଓ ବିୟେ।ଗ କରି)
= (a2 – 1)2 – (a)2 = (a2 – 1 + a) (a2 – 1 – a) = (a2 + a – 1) (a2 – a – 1)
(xi) 25a4 – 19a2b2 + 9b2
ସମାଧାନ:
25a4 – 19a2b2 + 9b4 = (5a2)2 + (3b2)2 – 19a2b2
= (5a2 + 3b2)2 + 2 . 5 a2 . 3b2 – 19 a2b2
= (5a2 + 3b2)2 – 49 a2b2 = (5a2 + 3b2) – (7ab)2
= (5a2 + 3b2 + 7ab) (5a2 + 3b2 – 7ab) = (5a2 + 7ab + 3b2) (5a2 – 7ab + 3b2)
(xii) 9x2 + y2 + 6xy – 4z2
ସମାଧାନ:
9x2 + y2 + 6xy – 4z2 = (3x)2 + y2 + 2 . 3x . y – (2z)2
= (3x + y)2 – (2z)2 = (3x + y + 2z) (3x + y – 2z)
(xiii) 16 – x2 – 24y + 9y2
ସମାଧାନ:
16 – x2 – 24y + 9y2 = 16 – 24y + 9y2 – x2
= 42 – 2 . 4 . 3y + (3y)2 – x2 = (4 – 3y)2 – x2 = (4 – 3y + x) (4 – 3y – x)
(xiv) (a2 – b2) (x2 – y2) – 4abxy
ସମାଧାନ:
(a2 – b2) (x2 – y2) – 4abxy = a2x2 – a2y2 – b2x2 + b2y2 – 4abxy
= a2x2 + b2y2 – 2abxy – a2y2 – b2x2 – 2abxy
= (a2x2 + b2y2 – 2abxy) – (a2y2 + b2x2 + 2abxy)
= {(ax)2 + (by)2 – 2(ax) (by)} – {(ay)2 + (bx)2 + 2(ay) (bx)}
= (ax – by)2 – (ay + bx)2 = (ax – by + ay + bx) (ax – by – ay – bx)
(xv) (a2 + b2)(x2 – y2) – 2ab (x2 + y2)
ସମାଧାନ:
(a2 + b2)(x2 – y2) – 2ab (x2 + y2)
= a2x2 – ay2 + b2x2 – b2y2 – 2abx2 – 2aby2
= a2x2 + b2x2 – 2abx2 – a2y2 – b2y2 – 2aby2
= (a2x2 + b2x2 – 2abx2) – (a2y2 + b2y2 + 2aby2)
= {(ax)2 + (bx)2 – 2ax . bx} – {(ay)2 + (by)2 + 2ay . by}
= (ax – bx)2 – (ay + by)2 = (ax – bx + ay + by) (ax – bx – ay – by)
= {x (a – b) + y (a + b)} {x (a – b) – y(a + b)}
Question 5.
ଉତ୍ପାଦକରେ ବିଶ୍ଳେଷଣ କର ।
(a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(i) a3 + b3 + x3 – 3abx
ସମାଧାନ:
a3 + b3 + x3 – 3abx
= (a + b + x) (a2 + b2 + x2 – ab – bx – ax)
(ii) 8a3 + b3 + c3 – 6abc
ସମାଧାନ:
8a3 + b3 + c3 – 6abc = (2a)3 + b3 + c3 – 3(2a) (b) (c)
= (2a + b + c) { (2a)2 + b2 + c2 – 2a . b. bc – c . 2a}
= (2a + b + c) (4a2 + b2 + c2 – 2ab – bc – 2ca)
(iii) a3 + b3 – 8 + 6ab
ସମାଧାନ:
a3 + b3 – 8 + 6ab = a3 + b3 + (-2)3 – 3a . b(-2)
= (a + b – 2) { a2 + b2 + (-2)2 – ab – b (-2) – (-2) a}
= (a + b – 2) (a2 + b2 + 4 – ab + 2b + 2a)
(iv) l – 27m3 – n3 – 9 lmn
ସମାଧାନ:
l3 – 27m3 – n3 – 9lmn = l3 + (- 3m)3 + (- n)3 – 3l(-3m) (-n)
= {l + (-3m) + (-n)} {l2 + (-3m)2 + (- n)2 – l(-3m) – (-3m) (-n) – (-n) l}
= (l – 3m – n) (l2 + 9m2 + n2 + 3lm – 3mn + nl)
(v) (a – b)3 + (c – b)3 + (a – c)3 – 3 (a – b) (b – c) (c – a)
ସମାଧାନ:
(a – b)3 + (c – b)3 + (a- c)3 – 3 (a – b) (b – c) (c – a)
ମନେକର a – b = x, b – c = y, c – a = z ⇒ c – b = -y ⇒ a – c = -z
ପ୍ରଦତ୍ତ ପଲିନୋମିଆଲ୍ଟି x3 + (-y)3 + (-z)3 – 3(x) (-y) (-z)
= {x + (-y) + (-z)} {x2 + (-y)2 + (-z)2 – x (-y) – (-y) (-z) – (-z) x}
= (x – y – z) (x2 + y2 + z2 + xy – yz + zx)
= {(a – b) – (b – c) – (c – a)} {(a – b)2 + (b – c)2 + (c – a)2 +(a – b) (b – c) – (b – c) (c – a) + (c – a) (a – b) x, y ଓ zର ମାନ ବସାକଳେ,
= (a – b – b + c – c + a) (a2 – 2ab + b2 + b2 – 2bc + c2 + c2 – 2ca + a2 + ab – ca – b2 + bc – bc + ab + c2 – ca + ca – bc – a2 + ab)
= (2a – 2b) (a2 + b2 + c2 + ab – 3bc – 3ca)
= 2 (a – b) (a2 + b2 + c2 + ab – 3bc – 3ca)
(vi) a6 + 4a3 – 1
ସମାଧାନ:
a6 + 4a3 – 1 = a6 + a3 – 1 + 3a3 = (a2)3 + a3 + (-1)3 – 3. a2 .a(-1)
= (a2 + a – 1) {(a2)2 + (a)2 + (-1)2 – a2.a – a(-1) – (-1)a2}
= (a2 + a – 1) (a4 + a2 + 1 – a3 + a + a2) = (a2 + a – 1)(a4 – a3 + 2a2 + a + 1)
(vii) x3 + 72 – 24x
ସମାଧାନ:
x3 + 72 – 24x = x3 + 64 + 8 – 24x
= x3 + 43 + 23 – 3. x. 4. 2 = (x + 4 + 2)(x2 + 42 + 22 – x. 4 – 4. 2 – 2. X)
= (x + 4 + 2)(x2 + 16 + 4 – 4x – 8 – 2x)
= (x + 6)(x2 + 16 + 4 – 8 – 6x)
= (x + 6)(x2 – 16 + 12)
(viii) m6 + 7m3 – 8
ସମାଧାନ:
m6 + 7m3 – 8 = m6 + m3 – 8 + 6m3
= (m2)3 + m3 + (-2)3 – 3m2 .m(-2)
= (m2 + m – 2) {(m2)2 + m2 + (2)2 – m2 . m – m (-2) – (-2) m2}
= (m2 + m – 2)(m4 + m2 + 4 – m3 + 2m + 2m2)
= (m2 + 2m – m – 2)(m4 – m3 + 3m2 + 2m + 4)
= {m(m + 2) – 1(m + 2)}(m4 – m3 +3m2 + 2m + 4)
= (m + 2)(m – 1)(m4 – m3 + 3m2 + 2m + 4)
(ix) a6 + $$\frac{1}{a^6}$$ + 2 (a ≠ 0)
ସମାଧାନ:
(x) r6 + 45r3 – 8
ସମାଧାନ:
r6 + 45r3 – 8 = r6 + 27r3 – 8 + 18r3
= (r2)3 + (3r)3 + (-2) – 3 . r2 . 3r (-2)
= (r2 + 3r – 2) {(r2)2 + (3r)2 + (2)2 – r2 3r – 3r (-2) – (-2) r2)
= (r2 + 3r – 2)(r4 + 9r2 + 4 – 3r3 + 6r + 2r2)
= (r2 + 3r – 2)(r4 – 3r3 – t – 11r2 + 6r + 4)
(xi) 16x3 – 54y6 – 2z3 – 36xy2z
ସମାଧାନ:
16x3 – 54y6 – 2z3 – 36xy2z = 2 { 8x3 – 27y6 – z3 – 18xy2z)}
= 2 {(2x)3 + (-3y2)3 + (-z)3 – 3 (2x) (-3y2) (-z))
= 2{2x + (-3y2) + (-z)) {(2x)2 + (-3y2)2 + (-z)2 – (2x) (-3y2) – (-3y2) (-z) – (-z) (2x)}
= 2(2x – 3y2 – z) (4x2 + 9y4 + z2 + 6xy2 – 3y2z + 2zx)
(xii) a3 + b3 – $$\frac{1}{27}$$ c3 + abc
ସମାଧାନ:
(xiii) 27a3 – 8b6 + 125 c3 + 90ab2c
ସମାଧାନ:
27a3 – 8b6 + 125 c3 + 90 ab2c = (3a)3 + (- 2b2)3 + (5c)3 – 3(3a) (- 2b2) (5c)
= {3a + (-2b2) + 5c} {(3a)2 + (-2b2)2 + (5c)2 – (3a) (-2b2) – (-2b2) (5c) – (5c) (3a)}
= (3a – 2b2 + 5c) (9a2 + 4b4 + 25c2 + 6ab2 + 10b2c – 15 ca)
(xiv) (2x + 3)3 + (3x – 2)3 – (5x + 1)3
ସମାଧାନ:
(2x + 3)3 + (3x – 2)3 – (5x + 1)3
ମନେକର 2x + 3 = a, 3x- 2 = b, 5x + 1 = c
∴ a + b – c = 2x + 3 + 3x – 2 – 5x – 1=0
a + b – c = 0 ହେଲେ a3 + b3 – c3 = -3abc
∴(2x + 3)3 + (3x – 2)3 – (5x + 1)3 = -3 (2x + 3) (3x – 2) (5x + 1)
(a, b ଓ cର ମାନ ବସାକଳେ)
Question 6.
a + b + c = 0 ହେଲେ ଦର୍ଶାଅ ଯେ, a3 + b3 + c3 = 3 abc
ସମାଧାନ:
L.H.S. = a3 + b3 + c3 = a3 + b3 + c3 – 3abc + 3abc
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc
= (0) (a2 + b2 + c2 – ab – bc – ca) + 3abc
= 0 + 3abc = 3abc = R.H.S. (ପ୍ରମାଣିତ)
Question 7.
(x – y)3 + (y – z)3 + (z – x)3 ର ଉତ୍ପାଦକଗୁଡ଼ିକୁ ନିର୍ଣ୍ଣୟ କର ।
ସମାଧାନ:
ମନେକର x – y = a, y – z = b, z – x = c
∴ a + b + c = x – y + y – z + z – x ⇒ a + b + c = 0
ପ୍ରଦତ୍ତ ପଲିନୋମିଆଲ୍ଟି = a3 + b3+ c3 = 3abc (∵ a + b + c = 0)
∴ (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y) (y – z) (z – x)
Question 8.
ଦର୍ଶାଅ ଯେ, x3 + y3 + z3 – 3xyz = $$\frac{1}{2}$$ {(x – y)2 + (y – z)2 + (z – x)2}
ସମାଧାନ:
x3 + y3 + z3 – 3xyz
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= $$\frac{1}{2}$$ (x + y + z) (2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx)
= $$\frac{1}{2}$$ (x + y + z) (x2 – 2xy + y2 + y2 – 2yz + z2 + z2 – 2zx + x2}
= $$\frac{1}{2}$$ (x + y + z) {(x – y)2 + (y – z)2 + (z – x)2}
∴ x3 + y3 + z3 = $$\frac{1}{2}$$ {(x – y)2 + (y – z)2 + (z – x)2} | 12,509 | 15,227 | {"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} | 4.59375 | 5 | CC-MAIN-2024-38 | latest | en | 0.383247 |
http://mathhelpforum.com/calculus/71149-curve-sketching.html | 1,480,809,878,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00226-ip-10-31-129-80.ec2.internal.warc.gz | 180,760,967 | 12,000 | 1. ## curve sketching
Been trying to get past this but it just keeps building on each other. It's for tomorrow:
$f(x)=\frac{x}{(x+1)(x+2)}$
a) Find any vertical and horizontal asys
b) where axis are crossed
c) show curve has no turning points
d) sketch curve, labeling it properly.
e) sketch f(-x) on the graph.
So vertical asympotes are as simple as LCM equal to 0 and substitue. Vertical?!?
Axis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.
Curve's turning points. I want to see if differentiating $\frac{dy}{dx}$ should be enough, or do i need to go down to $\frac{d^2y}{dx^2}$ and make it =0. If the answer is imaginary it's proved, correct?
sketching curve should be done using the information gained from the previous questions and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.
f(-x) means a reflection in the x axis? Thanks for the posts
2. Originally Posted by Lonehwolf
Been trying to get past this but it just keeps building on each other. It's for tomorrow:
$f(x)=\frac{x}{(x+1)(x+2)}$
a) Find any vertical and horizontal asys
b) where axis are crossed
c) show curve has no turning points
d) sketch curve, labeling it properly.
e) sketch f(-x) on the graph.
So vertical asympotes are as simple as LCM equal to 0 and substitue. Vertical?!?
that's right. so the vertical asymptotes are x = -1 and x = -2
for the horizontal asymptotes, find $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$. if you get a finite answer for either (in this case, they will both give the same answer), that's your horizontal asymptotes. it is a line y = something
Axis cross should be x = 0 and y = 0. It is (0,0) so it should be solved like that.
don't know what you mean. this is finding the x- (and y-) intercept
Curve's turning points. I want to see if differentiating $\frac{dy}{dx}$ should be enough, or do i need to go down to $\frac{d^2y}{dx^2}$ and make it =0. If the answer is imaginary it's proved, correct?
for turning points, you only need the first derivative, and set it to zero.
sketching curve should be done using the information gained from the previous questions
and finding out which side the curve goes when approaching the asymptotes (asy+0.0001), (asy - 0.00001) to get a quick estimate.
huh? oh, ok, i think i got what you're saying. um, yeah, estimates are good when graphing.
f(-x) means a reflection in the x axis? Thanks for the posts
nope, you reflect in the y-axis
3. Originally Posted by Jhevon
that's right. so the vertical asymptotes are x = -1 and x = -2
for the horizontal asymptotes, find $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$. if you get a finite answer for either (in this case, they will both give the same answer), that's your horizontal asymptotes. it is a line y = something
don't know what you mean. this is finding the x- (and y-) intercept
for turning points, you only need the first derivative, and set it to zero.
huh? oh, ok, i think i got what you're saying. um, yeah, estimates are good when graphing.
nope, you reflect in the y-axis
Can't work the horizontal asymptotes either way. Not sure how you work with $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$.
4. Originally Posted by Lonehwolf
Can't work the horizontal asymptotes either way. Not sure how you work with $\lim_{x \to \infty}f(x)$ and $\lim_{x \to -\infty}f(x)$.
do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote
5. Originally Posted by Jhevon
do you know what limits are? this is calculus since you are talking about derivatives, so you should know about limits (in fact, the derivative is a limit!). you have a first degree polynomial divided by a second degree one, as x gets larger in either direction, the limit goes to zero. hence, y = 0 is the horizontal asymptote
Uhh, I do know about limits, but i can't recall those wicked symbols :
How are they used, I must have been sick during that lesson!
6. Originally Posted by Lonehwolf
Uhh, I do know about limits, but i can't recall those wicked symbols :
How are they used, I must have been sick during that lesson!
$\lim_{x \to \infty} f(x)$ reads "the limit as $x$ goes (or tends) to $\infty$ of $f(x)$". you pretty much evaluate what will happen if $x$ gets huge. if $x$ gets really big, beyond measure, the function here, will get close to zero.
a similar interpretation holds for $\lim_{x \to - \infty}f(x)$
the same symbol is used in the definition of the derivative, so you must have been sick during several lessons
7. Originally Posted by Jhevon
$\lim_{x \to \infty} f(x)$ reads "the limit as $x$ goes (or tends) to $\infty$ of $f(x)$". you pretty much evaluate what will happen if $x$ gets huge. if $x$ gets really big, beyond measure, the function here, will get close to zero.
a similar interpretation holds for $\lim_{x \to - \infty}f(x)$
the same symbol is used in the definition of the derivative, so you must have been sick during several lessons
Thanks for the explanation, I believe this issue is solved, now i gotta start a clean sheet and burn that whole mess of a paper i had. | 1,475 | 5,415 | {"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": 28, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-50 | longest | en | 0.920047 |
https://philoid.in/question/81275-if-x-follows-a-binomial-distribution-with-mean-4-and-variance-2-find-px-5 | 1,716,463,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00712.warc.gz | 389,740,749 | 7,086 | # If X follows a binomial distribution with mean 4 and variance 2, find P(X ≥ 5).
Given that parameters for binomial distribution are n, p
Now, np = 4
n = 8
The required binomial distribution is given by,
Now, P(X≥5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
17
1 | 96 | 271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-22 | latest | en | 0.712911 |
https://commerceschool.in/cbse-q-8-accounting-equation-ts-grewal-class-11-2022-23/ | 1,696,094,034,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00369.warc.gz | 194,715,968 | 37,755 | # [CBSE] Q 8 Accounting Equation TS Grewal Class 11 (2022-23)
Are you looking for the solution of Question number 8 of Accounting Equation chapter TS Grewal class 11 CBSE Board 2022-23?
What will be the effect of the following on the Accounting Equation?
(i) Harish started busienss with cash ₹ 18,000
(ii) Purchased goods for cash ₹ 5,000 and on credit ₹ 2,000
(iii) Sold goods for cash ₹ 4,000 (Costing ₹ 2,400)
(iv) Rent paid ₹ 1,000 and rent outstanding ₹ 200
[Assets: Cash ₹ 16,000 + Stock ₹ 4,600 = Liabilities: Outstanding Rent ₹ 200 + Creditors ₹ 2,000 + Capital: ₹ 18,400.]
Solution:-
Below are the links of all solutions | 194 | 639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.885268 |
https://www.coursehero.com/file/6036895/long-division/ | 1,540,247,099,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515555.58/warc/CC-MAIN-20181022222133-20181023003633-00220.warc.gz | 902,902,370 | 171,012 | long_division
# long_division - P x x x 1 Long Division = = Q x x 1 quo\$ent...
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Long Division : P ( x ) Q ( x ) = x 3 + x 2 1 x 1 = ??? quo\$ent remainder P(x) Q(x) x 3 + x 2 1 = ( x 2 + 2 x + 2)( x 1) + ( 1) x 3 + x 2 1 x 1 = ( x 2 + 2 x + 2) + 1 x 1 remainder quo\$ent P(x) Q(x)
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If degree(numerator)>degree(denominator), do LONG DIVISION. Then 1. Factor the denominator in a product of irreducible 2. Find the par\$al frac\$on decomposi\$on of P(x)/Q(x) 3. Integrate each factor
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Jill Tulane University ‘16, Course Hero Intern | 453 | 1,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.8672 |
https://www.physicsforums.com/threads/gmm-r-2-for-r-0.917638/ | 1,580,139,215,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700988.64/warc/CC-MAIN-20200127143516-20200127173516-00330.warc.gz | 1,023,444,773 | 16,096 | # GMm/r^2 for r = 0?
• B
## Main Question or Discussion Point
This is probably a stupid question - but let's say you have a hollow Sun (M is hollow), with another mass in the center, so M and m share the same center of mass, the distance between the centers is zero.
Would GMm/r^2 be infinite in this case?
The force at the center of any object - if you define your system to be the center chunk of matter, and how that center chunk of matter sees the rest of the planet?
((o))
sorry... I did not get enough sleep last night, that's my excuse!
Related Classical Physics News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
The 1/r^2 expression is only valid for the gravitational force on an object outside a spherically symmetric mass distribution.
Your expression would go to infinity, but it would not be describing the force between the objects.
JLT
Ibix
The force is, in fact, zero. Look up the Shell Theorem.
Larry Niven wrote a novel called Ringworld which featured a ring around a star. He had to write a followup describing the giant ion engines the ring had to keep itself centred on its star because this fact was brought to his attention after publication...
JLT
This is probably a stupid question - but let's say you have a hollow Sun (M is hollow), with another mass in the center, so M and m share the same center of mass, the distance between the centers is zero.
Would GMm/r^2 be infinite in this case?
The force at the center of any object - if you define your system to be the center chunk of matter, and how that center chunk of matter sees the rest of the planet?
((o))
sorry... I did not get enough sleep last night, that's my excuse!
actually their will be no force acting because all the forces from different part of the sun will cancel out. | 418 | 1,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.94625 |
https://stats.stackexchange.com/questions/19880/fitting-special-variance-structure-in-mixed-model | 1,701,781,959,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.17/warc/CC-MAIN-20231205105136-20231205135136-00342.warc.gz | 616,032,554 | 40,729 | # Fitting special variance structure in mixed model
I'm interested in fitting a linear mixed model with this special variance structure on the random effects $\mathbf{u}$:
$\begin{eqnarray*} \mathbb{V}\left(\mathbf{u}\right) & = & \mathbf{A}\mathbf{G}\mathbf{A}^{\prime} \end{eqnarray*}$
This variance structure is very similar to Cholesky and Antedependence structures except few differences:
1. $\mathbf{A}$ is not a unit lower triangle or unit upper triangle matrix
2. $\mathbf{G}$ is not necessarily a diagonal matrix
I can specify initial values of $\mathbf{A}$. I'd highly appreciate if someone give me some hints to fit this variance covariance structure in R. Thanks in advance for your help and time.
With no restrictions on $A$, your equation doesn't place any restrictions whatsoever on the form of the variance-covariance matrix, even if you do specify that $G$ is diagonal. A variance-covariance matrix is a real symmetric matrix, and any such matrix can be diagonalized by an orthogonal matrix.
• Thanks @onestop for your nice answer. Actually $\mathbf{A}$ is kind of a relationship matrix and I want to use it in model fitting and interested in its value after model fitting. Any suggestion. Thanks Dec 15, 2011 at 20:24 | 307 | 1,241 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.852082 |
https://musingsonmath.com/2010/12/ | 1,686,262,787,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00293.warc.gz | 435,217,603 | 27,073 | ## The perfect cup of coffee
There is an old Hungarian mathematics quote that I love to recite, “A mathematician is a device which turns coffee into theorems.” Well, on Christmas morning, I received the perfect gift from my daughters to complement the quote. (To my fellow geometry fans, notice the choice of word ‘complement’.)
Let’s face it, this is my dream coffee cup … a cup containing 20 of the greatest achievements of human thought. Now each morning when I drag myself out of bed at 4:30, I will have a little inspiration to go along with the first cup, and the second, the third, and, of course, the fourth … you get the idea.
## The Fifth Postulate by Jason Socrates Bardi
I just finished a great book by Jason Socrates Bardi, called The Fifth Postulate. It really is a “must-read” for anyone interested in trying to understand how mathematicians think (and geometry teachers looking to leave the realm of the textbook!).
For thousands of years, mathematicians have spent their careers trying to prove the Parallel Postulate, only to find out that all of their attempts have failed. It took the genius of people like Gauss to think, “maybe the postulate doesn’t need to be true.” With that thought (and a lot of work), we have the birth of non-Euclidean geometry.
The Fifth Postulate gives a great historical overview of the attempts to prove the Parallel Postulate as well as thought process leading to its eventual rejection. It is an extremely entertaining read and has some great biographies of Gauss, Lobachevsky and Bolyai. As with most math books written today, it keeps a “general interest” audience in mind and steers clear of the heavy-duty mathematics of non-Euclidean geometry. However, it still gives the reader a good overview of the birth of this field.
## ABC News and John Allen Paulos team-up
ABC News and Temple University mathematics professor John Allen Paulos (author of many books including his most famous, Innumeracy) have teamed-up to bring you a mathematician’s insight into the world of news.
http://abcnews.go.com/Technology/WhosCounting/
The articles on this site analyze the mathematical angles of current news events. What’s nice about the articles presented is that they are entertaining, informative and written at a level that allows anyone to understand them. Take some time to read and enjoy!
## Math Pronunciations
As most of us are well aware, mathematics is a global activity. As such, the terminology that we use often has regional ties, making pronunciations difficult. I remember one professor in college who was so angry at me for mispronouncing the term “affine” that he almost kicked me out of the class. (That’s not even a tough one!) How about “Bolyai” or “platykurtic”? As a teacher, I cannot even begin to tell you the number of times the name “Euler” has been mispronounced.
Enter the University of Wisconsin. They have created an online Mathematics Pronunciation Guide to help us with all of our difficulties. Thanks to them, we can all sound a lot smarter! Click here to begin the journey.
## The greatest mathematician ever?
Students often ask me who is the greatest mathematician of all time. (Whether or not they are truly curious is debatable … they are masters at trying to get me off topic!) While it can be a sometimes contentious debate, most people will agree that the honor goes to Carl Friedrich Gauss. His contributions to mathematics, statistics, physics, astronomy and surveying (not to mention his inventions) certainly rank him at the top.
For those of you interested in learning more about the “greatest” mathematician of all time, here are a few places you can go:
• For a general overview of his life and contributions, click here.
• For a more detailed look at his life and contributions, this website from professor Douglas Ravenel at the University of Rochester gives you a plethora of links, click here.
• For those of you looking for an actual book, check out this selection: Gauss – Titan of Science by G. Waldo Dunnington
Wherever you go, enjoy! You will not be disappointed. | 933 | 4,104 | {"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} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.95176 |
https://www.coursya.com/product/advanced-engineering-systems-in-motion-dynamics-of-three-dimensional-3d-motion/ | 1,726,542,152,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00677.warc.gz | 662,738,122 | 55,705 | # Advanced Engineering Systems in Motion: Dynamics of Three Dimensional (3D) Motion
### Description
This course is an advanced study of bodies in motion as applied to engineering systems and structures. We will study the dynamics of rigid bodies in 3D motion. This will consist of both the kinematics and kinetics of motion. Kinematics deals with the geometrical aspects of motion describing position, velocity, and acceleration, all as a function of time. Kinetics is the study of forces acting on these bodies and how it affects their motion.
—————————
Recommended Background:
To be successful in the course you will need to have mastered basic engineering mechanics concepts and to have successfully completed my course entitled Engineering Systems in Motion: Dynamics of Particles and Bodies in 2D Motion.” We will apply many of the engineering fundamentals learned in those classes and you will need those skills before attempting this course.
—————————
While no specific textbook is required, this course is designed to be compatible with any standard engineering dynamics textbook. You will find a book like this useful as a reference and for completing additional practice problems to enhance your learning of the material.
—————————
The copyright of all content and materials in this course are owned by either the Georgia Tech Research Corporation or Dr. Wayne Whiteman. By participating in the course or using the content or materials, whether in whole or in part, you agree that you may download and use any content and/or material in this course for your own personal, non-commercial use only in a manner consistent with a student of any academic course. Any other use of the content and materials, including use by other academic universities or entities, is prohibited without express written permission of the Georgia Tech Research Corporation. Interested parties may contact Dr. Wayne Whiteman directly for information regarding the procedure to obtain a non-exclusive license.
### What you will learn
Course Introduction; Angular Velocity; Angular Acceleration
In this section students will learn to derive the “derivative formula.” We will define angular velocity for 3D motion and learn to determine and solve for the Angular Acceleration for a body.
Velocities in Moving Reference Frames; Accelerations in Moving Reference Frames; The Earth as a Moving Frame
In this section students will learn about velocities in moving reference frames, accelerations in moving reference frames, and the Earth as a moving frame.
Eulerian Angles; Eulerian Angles Rotation Matrices; Angular Momentum in 3D; Inertial Properties of 3D Bodies
In this section students will learn about Eulerian Angles rotation matrices, angular momentum in 3D, and intertial properties of 3D bodies.
Translational and Rotational Transformations of Inertial Properties; Principal Axes and Principal Moments of Inertia
In this section students will learn about translational and rotational transformations of inertial properties, and principal axes and principal moments of inertia. | 585 | 3,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.918863 |
https://www.shaalaa.com/question-bank-solutions/vapour-pressure-liquid-solutions-vapour-pressure-liquid-liquid-solutions-vapour-pressure-pure-water-298-k-238-mm-hg-50-g-urea-nh2conh2-dissolved-850-g-water_8957 | 1,575,975,959,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540527205.81/warc/CC-MAIN-20191210095118-20191210123118-00256.warc.gz | 821,194,136 | 11,002 | Share
# Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. - CBSE (Science) Class 12 - Chemistry
ConceptVapour Pressure of Liquid Solutions Vapour Pressure of Liquid- Liquid Solutions
#### Question
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
#### Solution
It is given that vapour pressure of water, p_1^0= 23.8 mm of Hg
Weight of water taken, w1 = 850 g
Weight of urea taken, w2 = 50 g
Molecular weight of water, M1 = 18 g mol−1
Molecular weight of urea, M2 = 60 g mol−1
Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.
Now, from Raoult’s law, we have:
(p_1^0 - p_1)/p_1^0 = n_2/(n_1+n_2)
=>(p_1^0-p_1)/p_1^0 = (w_2/M_2)/(w_1/M_1+w_2/M_2)
=>(23.8-p_1) /23.8 = (50/60)/(850/18 + 50/60
=>(23.8-p_1)/23.8 = 0.83/(47.22+0.83)
=>(23.8-p_1)/23.8= 0.0173
=> p_1 = 23.4 "mm of Hg"
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173
Is there an error in this question or solution?
#### APPEARS IN
Solution Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Concept: Vapour Pressure of Liquid Solutions - Vapour Pressure of Liquid- Liquid Solutions.
S | 531 | 1,441 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-51 | latest | en | 0.813385 |
http://nand2tetris-questions-and-answers-forum.52.s1.nabble.com/Jump-logic-td4036634.html | 1,675,067,934,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00389.warc.gz | 34,072,726 | 10,187 | Jump logic
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Jump logic
This post was updated on . It seems like comparing J bits with NG or ZR is needed. Also to my understanding ALU will never output ng=1 and zr=1 at the same time. My though process is like this. Lets say I need to check for Jump JEQ where j1,j2,j3 are: 0 1 0. (case for alu output = 0) In such scenario ALU will output: ZR = 1 and NG = 0. I tried to AND: And j1 and ng - if ng is 0 I have a match it with j1 bit. And j2 and zr - if zr is 1 I have match for j2 bit. And j3 and ng - if zr is either 0 or 1 I have no match for j3 bit resulting 0. But in testing above was failing for >0 condition so i came up with below logic just for J3 bit: J3 zr ng truth table: 1 0 0 - 1 matches >0 so i solved for it 1 0 1 - 0 1 1 0 - 0 1 1 1 - 0 J3 + !ZR + !NG Above seems to be working but when ng and zr are both 1 the output is 1 in some cases. But alu will never output 1 for both zr and ng so would above be acceptable Jump logic solution? I tried to improve above with below logic: <0 //j1 +!zr + ng =0 //J2 + zr + !ng <0 //J3 +!zr !ng It works but it only fails in 2 cases when zr and ng are 1 which would never be a case so technically it would work. Those cases are for checking for NOT 0 and for unconditional jump. I wonder if both of these are acceptable or i need to improve it? Once this logic is correct i will add it to the CPU. I hope anyone here can help me and guide me into the right direction. thx Edit: I removed source code per posting guidelines. | 472 | 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} | 2.875 | 3 | CC-MAIN-2023-06 | latest | en | 0.947592 |
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## 2nd EDITION
Full Frontal Calculus:
An Infinitesimal Approach
Single-variable calculus, presented with panache: crisp, concise, and with a smoky infinitesimal flavor. Now in its 2nd edition - older, wiser, and 20 pages fatter, mainly on account of more expansive explanations.
Then, after finding you like it, purchase the full text of the new and improved 2nd edition as a paperback at Amazon for \$29, or as a pdf at Lulu for \$12.
## The Dark Art of Linear Algebra:An Intuitive Geometric Approach
Follow me, O innocent one, into the dark realms. You shall see the future, walk between raindrops, and diagonalize matrices.
excerpts from DALA
. But for the complete text, you must pay with your soul!
Amazon.
Or, you can buy an electronic copy (a pdf) for \$12 at
Lulu.
2nd edition
Algebra, Coordinate Geometry, Functions, and Trigonometry.
Learn 'em well! The fate of the free world depends on it!
All of high-school mathematics in one 200-page textbook.
You can read some free excerpts - prefatory material plus three full chapters - here. If you like what you see, you can support a starving math teacher (or at least one who might have skipped lunch) by purchasing a print copy for \$29 at Amazon, or a pdf for \$12 at Lulu.
Lobachevski Illuminated
A historical introduction to non-Euclidean geometry developed in the midst of extensive commentary on one the subject's foundational documents: Lobachevski's Theory of Parallels.
Winner of the Mathematical Association of America's Beckenbach Book Prize in 2015.
The book can be purchased at Amazon or from the website of the American Mathematical Society
• What do you mean by "an infinitesimal approach" to calculus?
An infinitesimal is, by definition, an infinitely small number - smaller than every positive real number, yet still greater than zero. No real number satisfies this definition, so infinitesimals are not real. And yet, by developing a "calculus of infinitesimals" (as the subject was known for most of its first two centuries), mathematicians and physicists achieved unparalleled insight into real functions, breaking through the static algebraic ice shelf to reach a flowing world of motion below, changing and evolving in time. Calculus is, among other things, the mathematics of continuous change. But even as calculus ushered in modern science, philosophical problems remained. Do infinitesimals really exist? Or are they just useful fictions? And how could mathematics - of all subjects - afford the risk of fictional foundations? Such problems were not resolved until the 19th century, when mathematicians developed a rigorous way to do calculus without infinitesimals. This new basis for calculus (the theory of limits) put the philosophical questions to bed, but it also made calculus less intuitive for the average user, who was not, after all, a philosopher, and who rather liked the feel of infinitesimals. Full Frontal Calculus returns infinitesimals to the stage, restoring the intuitions that motivated the creation of the subject, and putting you, the reader, in touch with the thoughts of the intellectual giants who developed it. You'll end up learning all the same results as someone who studies a more typical limit-based approach to calculus, but your path there will be more intriguing, simpler, and considerably shorter. (Most textbooks devote 500+ pages to developing single-variable calculus. In FFC, this requires just 180 pages.)
• Why should I purchase a physical copy when the pdf costs less?
The reasons are legion. To take just a few examples: Holding a physical book in your hands encourages serious, active reading. A conspicuously displayed copy of Full Frontal Calculus will impress that pretty girl (or handsome lad) in the coffee shop. Physical books absorb (and later, return) some of your experiences in ways that e-books do not. Years later, when you take Full Frontal Calculus down from your bookshelf, its old familiar feel in your hands will stir dormant memories. As you read again the old notes in the margins scribbled by your younger self, and see again the odd coffee stain or spatter of tomato sauce, you'll find yourself murmuring, "Yes. I remember, I remember." When you buy a physical copy, I will have the satisfaction of seeing its sale on Amazon. You, in turn, will have the satisfaction of having caused my satisfaction. With the royalties I earn from your copy, I'll be able to afford a cup of chai. When I drink it, I'll think of you. (And you'll have the pleasure of thinking of me thinking of you as I drink my chai.)
• What is that floating sphere on the cover?
It is a pebble, which is precisely what the word "calculus" means in Latin. (May God protect you from renal calculi.) Our distant ancestors used pebbles (calculi) as aids for simple calculations. Descendants of those humble pebble manipulations included the abacus, positional notation, and all that follows... including the subject we now call "calculus". Isaac Newton famously described himself thus: "I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the sea-shore, and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me." Among the smoothest of Newton's calculi was... calculus. "Come now!" I hear you cry, "That stone on the cover is far too large to be a pebble." Pshaw. You have no idea how small those two people are. "Yes, but what are they doing?" Ay, that is the question.
• What about full frontal *multivariable* calculus?
Patience, patience. FFC may yet sprout chapters on multivariable calculus, but for now you must settle for some pandemic-era videos on the topic that I made while teaching online. Consider them a sneak preview.
• Is there an audiobook version of Full Frontal Calculus?
I'm afraid my attempts to enlist Morgan Freeman's assistance to make an audiobook version of FFC have proved fruitless, and I will settle for nothing less. Aut Freeman, aut nihil. That said, I have made supplementary videos for my students: Chapters 1 through Pi (differential calculus), Chapters 4 through 6 (integral calculus) and Chapters 7 and 8 (series, parametric equations, polar coordinates). Should you seek succor in the sound of my voice or the sight of my handwriting, these videos should satisfy your peculiar, albeit highly refined, desires.
• What's new in the 2nd Edition?
Besides a stylish new cover - courtesy of Vector Vectorum Press, the book's new publisher - the 2nd edition sports an additional 20 pages. This added bulk comes mainly in the form of expanded explanations, in keeping with the book's emphasis on explaining why calculus works as it does. But at 200 pages, FFC remains a remarkably trim calculus text, able to run rings around its bloated 1000-page brethren.
• What do you mean by "an intuitive geometric approach"?
My approach is similar in spirit to 3Blue1Brown's famous "Essentials of Linear Algebra" video series on YouTube, but rather than offering a video overview of the subject's essentials, The Dark Art of Linear Algebra is a complete textbook. As such, it includes basic topics not covered in the 3Blue1Brown series (especially the computational side of linear algebra), and of course, it contains lots of exercises. The Dark Art emphasizes visual, visceral understanding, rooted in the familiar geometry of n-dimensional Euclidean space. (Familiar, that is, when n = 2 or 3, but by the book's end, you will feel much more at home in spaces of 4 or more dimensions.) This emphasis of visual intuition leads me to introduce linear maps before matrices and matrices before Gaussian elimination, reversing the order in which these topics appear in most textbooks. Determinants, change-of-basis, eigenstuff, projections, and least squares solutions to inconsistent systems are all presented geometrically, too.
• WTF?! Why isn't this book available as a free pdf?
I understand your disappointment. You wish to learn linear algebra, a noble and difficult pursuit. I am saddened by the thought of talented young people having to forgo their linear algebraic studies for want of \$29 to buy a paperback copy of my book or \$12 to buy a pdf. Society, I firmly believe, should smooth the path for budding scholars such as yourself, easing their financial burdens whenever possible. But I am not Society. I am one man trying heroically to support multiple mistresses while making payments on two condos, not to mention the Braver New Math yacht. And cocaine is not getting any cheaper. Every dollar counts, and if I can find a way to extract a few more from you, then by God I intend to do it.
• Your other textbooks are refreshingly short. How long is this one?
The text itself is about 150 pages, yet it covers all the standard topics in an introductory linear algebra course. Of course, the title page, table of contents, index, answers, copyright page, and so forth add a bit more bulk to the finished product. When all is said and done, DALA's handsome covers surround precisely 185 pages. I have it on the authority of Goldilocks herself - a keen student of mathematics - that the length of The Dark Art of Linear Algebra is just right. A student can read the entire book, cover to cover, in a single college class.
Because mathematics is difficult, and it was made that way. Not by me, but by nature. Mathematics is also supremely logical. One can master it by understanding why it works. Understanding is hard work - it is difficult. It is also rewarding. Accordingly, I've written this book with two Albert Einstein quotations in mind: one for my readers, and one for me. "Any fool can know. The point is to understand." "Make everything as simple as possible, but not simpler." Too many precalculus books - especially the "made easy" variety - try to make the subject "simpler than possible", thus conveying (at best) mere knowledge without understanding. Precalculus Made Difficult does no such thing. It emphasizes understanding throughout, and assumes, with a spirit of generosity rare among contemporary textbooks, that you, the reader, have the will to learn well and the ability to read closely and carefully.
• Why should I purchase a physical copy when a pdf costs less?
• A 2nd Edition, eh? How does it differ from the 1st?
The main difference is that it is now published by Vector Vectorum Books, that most exclusive - and secretive - of publishers. To celebrate Precalculus Made Difficult's new home, it now sports a stylish new cover. The content is substantially the same as the 1st edition, but some of the original's infelicities of typesetting have been resolved, numerous typos were corrected, and its exposition and exercise sets have been discreetly polished here and there.
• Seth Braver? Who's he?
He's an old hand at teaching mathematics, which he has done in 10% of the United States. Among the many places he has lived, his favorite is probably Missoula, Montana, where he earned his Ph.D. (UM, 2007). Since 2010, he has taught at South Puget Sound Community College in Olympia, Washington.
• Is there an email address at which I can contact you?
But of course: bravernewmath@gmail.com
• I've bought a book, and I love it - so much so that I want to give you more money. Do you have a tip jar?
I do, I do. While I tune my guitar before playing my next piece, please feel free to toss your fivers, ten spots, C-notes, and higher right here.
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# converting exponential format number to decimal format number
Fei Liu
Guest
Posts: n/a
12-14-2006
Hi group, is there a quick way to convert an exponential format number
to decimal format number. For example,
13.534e+10 = 1353400000
I can come up a perl function but it's not perly. Can I get some help
John Bokma
Guest
Posts: n/a
12-14-2006
"Fei Liu" <(E-Mail Removed)> wrote:
> 13.534e+10
perl -e "print 13.534e+10"
135340000000
perl -e "my \$var = 13.534e+10; print length \$var"
12
So at least here (WinXP+ActiveState) Perl does this internally.
--
John Experienced Perl programmer: http://castleamber.com/
Perl help, tutorials, and examples: http://johnbokma.com/perl/
usenet@DavidFilmer.com
Guest
Posts: n/a
12-14-2006
Fei Liu wrote:
> Hi group, is there a quick way to convert an exponential format number
> to decimal format number. For example,
>
> 13.534e+10 = 1353400000
print 13.534e+10;
#prints 135340000000
--
The best way to get a good answer is to ask a good question.
David Filmer (http://DavidFilmer.com)
Fei Liu
Guest
Posts: n/a
12-14-2006
John Bokma wrote:
> "Fei Liu" <(E-Mail Removed)> wrote:
>
> > 13.534e+10
>
> perl -e "print 13.534e+10"
> 135340000000
>
> perl -e "my \$var = 13.534e+10; print length \$var"
> 12
>
> So at least here (WinXP+ActiveState) Perl does this internally.
>
Thanks for your input, but try 13.534e+26, you will find perl prints
13.534e+26. It's part of the code
where it reads this number from a file and the output needs to be
converted to decimal format for another application (say myapp) to use.
Unfortunately, myapp only understands decimal format number.
J. Gleixner
Guest
Posts: n/a
12-14-2006
Fei Liu wrote:
> John Bokma wrote:
>> "Fei Liu" <(E-Mail Removed)> wrote:
>>
>>> 13.534e+10
>> perl -e "print 13.534e+10"
>> 135340000000
>>
>> perl -e "my \$var = 13.534e+10; print length \$var"
>> 12
>>
>> So at least here (WinXP+ActiveState) Perl does this internally.
>>
>
> Thanks for your input, but try 13.534e+26, you will find perl prints
> 13.534e+26.
I find it prints 1.3534e+27
Look at: perldoc bigint
usenet@DavidFilmer.com
Guest
Posts: n/a
12-14-2006
Fei Liu wrote:
> Thanks for your input, but try 13.534e+26, you will find perl prints
> 13.534e+26. It's part of the code
use Math::BigInt;
my \$int = Math::BigInt->new('13.534e+26');
print \$int->as_int();
#prints 1353400000000000000000000000
--
The best way to get a good answer is to ask a good question.
David Filmer (http://DavidFilmer.com)
xhoster@gmail.com
Guest
Posts: n/a
12-14-2006
http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> Fei Liu wrote:
> > Thanks for your input, but try 13.534e+26, you will find perl prints
> > 13.534e+26. It's part of the code
>
> use Math::BigInt;
> my \$int = Math::BigInt->new('13.534e+26');
> print \$int->as_int();
Or, if you don't mind there being some 9's way out at the end,
printf "%f", 13.534e26
Xho
--
Usenet Newsgroup Service \$9.95/Month 30GB
usenet@DavidFilmer.com
Guest
Posts: n/a
12-14-2006
(E-Mail Removed) wrote:
> my \$int = Math::BigInt->new('13.534e+26');
> print \$int->as_int();
Or, if you are just doing the one conversion and don't need to retain
the constructor:
print Math::BigInt->new('13.534e+26')->as_int();
--
The best way to get a good answer is to ask a good question.
David Filmer (http://DavidFilmer.com)
John Bokma
Guest
Posts: n/a
12-14-2006
"Fei Liu" <(E-Mail Removed)> wrote:
> John Bokma wrote:
>> "Fei Liu" <(E-Mail Removed)> wrote:
>>
>> > 13.534e+10
>>
>> perl -e "print 13.534e+10"
>> 135340000000
>>
>> perl -e "my \$var = 13.534e+10; print length \$var"
>> 12
>>
>> So at least here (WinXP+ActiveState) Perl does this internally.
>>
>
> Thanks for your input, but try 13.534e+26
So, you gave a bad example. Always make your problem description as
complete as possible and provide examples that show your specific problem.
This way people can help you better, and you don't waste a lot of time of
other people.
--
John Experienced Perl programmer: http://castleamber.com/
Perl help, tutorials, and examples: http://johnbokma.com/perl/
Dr.Ruud
Guest
Posts: n/a
12-14-2006
Fei Liu schreef:
> Hi group, is there a quick way to convert an exponential format
> number to decimal format number. For example,
>
> 13.534e+10 = 1353400000
>
> I can come up a perl function but it's not perly.
This works in a limited way:
perl -we 'printf "%.0f\n", q/9.64e+21/'
9640000000000000000000
--
Affijn, Ruud
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https://www.stat.math.ethz.ch/pipermail/r-help/2013-February/347855.html | 1,669,609,179,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00232.warc.gz | 1,104,551,423 | 9,874 | # [R] cumulative sum by group and under some criteria
arun smartpink111 at yahoo.com
Tue Feb 19 18:43:10 CET 2013
```
Hi,
Try this:
res1<- do.call(rbind,lapply(paste(d3\$m1,d3\$n1),function(m1) do.call(rbind,lapply(0:(as.numeric(substr(m1,1,1))-1),function(x1) do.call(rbind,lapply(0:(as.numeric(substr(m1,3,3))-1),function(y1) do.call(rbind,lapply((as.numeric(substr(m1,1,1))+2):(7-as.numeric(substr(m1,3,3))),function(m) do.call(rbind,lapply((as.numeric(substr(m1,3,3))+2):(9-m),function(n)
do.call(rbind,lapply(x1:(x1+m-as.numeric(substr(m1,1,1))), function(x)
do.call(rbind,lapply(y1:(y1+n-as.numeric(substr(m1,3,3))), function(y)
expand.grid(m1,x1,y1,m,n,x,y)) )))))))))))))
names(res1)<- c("m1n1","x1","y1","m","n","x","y")
res1\$m1<- NA; res1\$n1<- NA
res1[,8:9]<-do.call(rbind,lapply(strsplit(as.character(res1\$m1n1)," "),as.numeric))
res2<- res1[,c(8:9,3:7)]
library(plyr)
res2<-join(res1,d3,by=c("m1","n1"),type="full") #Instead of this step, you can paste() the whole row of d3 and make suitable changes to the code above
tail(res2)
# m1n1 x1 y1 m n x y m1 n1 cterm1_P0L cterm1_P1L cterm1_P0H cterm1_P1H
#235 2 3 1 2 4 5 2 2 2 3 0.9025 0.64 0.857375 0.512
#236 2 3 1 2 4 5 2 3 2 3 0.9025 0.64 0.857375 0.512
#237 2 3 1 2 4 5 2 4 2 3 0.9025 0.64 0.857375 0.512
#238 2 3 1 2 4 5 3 2 2 3 0.9025 0.64 0.857375 0.512
#239 2 3 1 2 4 5 3 3 2 3 0.9025 0.64 0.857375 0.512
#240 2 3 1 2 4 5 3 4 2 3 0.9025 0.64 0.857375 0.512
A.K.
________________________________
From: Joanna Zhang <zjoanna2013 at gmail.com>
To: arun <smartpink111 at yahoo.com>
Sent: Tuesday, February 19, 2013 11:43 AM
Subject: Re: [R] cumulative sum by group and under some criteria
Thanks. I can get the data I expected (get rid of the m1=3, n1=3) using the join and 'inner' code, but just curious about the way to expand the data. There should be a way to expand the data based on each row (combination of the variables), unique(d3\$m1 & d3\$n1) ?.
or is there a way to use 'data.frame' and 'for' loop to expand directly from the data? like res1<-data.frame (d3) for () {....
On Tue, Feb 19, 2013 at 9:55 AM, arun <smartpink111 at yahoo.com> wrote:
If you can provide me the output that you expect with all the rows of the combination in the res2, I can take a look.
>
>
>
>
>
>
>________________________________
>
>From: Joanna Zhang <zjoanna2013 at gmail.com>
>To: arun <smartpink111 at yahoo.com>
>
>Sent: Tuesday, February 19, 2013 10:42 AM
>
>Subject: Re: [R] cumulative sum by group and under some criteria
>
>
>Thanks. But I thougth the expanded dataset 'res1' should not have combination of m1=3 and n1=3 because it is based on dataset 'd3' which doesn't have m1=3 and n1=3, right?>
>>In the example that you provided:
>> (m1+2):(maxN-(n1+2))
>>#[1] 5
>> (n1+2):(maxN-5)
>>#[1] 4
>>#Suppose
>> x1<- 4
>> y1<- 2
>> x1:(x1+5-m1)
>>#[1] 4 5 6
>> y1:(y1+4-n1)
>>#[1] 2 3 4
>>
>> datnew<-expand.grid(5,4,4:6,2:4)
>> colnames(datnew)<- c("m","n","x","y")
>>datnew<-within(datnew,{p1<- x/m;p2<-y/n})
>>res<-cbind(datnew,d2[rep(1:nrow(d2),nrow(datnew)),])
>> row.names(res)<- 1:nrow(res)
>> res
>># m n x y p2 p1 m1 n1 cterm1_P1L cterm1_P0H
>>#1 5 4 4 2 0.50 0.8 3 2 0.00032 0.0025
>>#2 5 4 5 2 0.50 1.0 3 2 0.00032 0.0025
>>#3 5 4 6 2 0.50 1.2 3 2 0.00032 0.0025
>>#4 5 4 4 3 0.75 0.8 3 2 0.00032 0.0025
>>#5 5 4 5 3 0.75 1.0 3 2 0.00032 0.0025
>>#6 5 4 6 3 0.75 1.2 3 2 0.00032 0.0025
>>#7 5 4 4 4 1.00 0.8 3 2 0.00032 0.0025
>>#8 5 4 5 4 1.00 1.0 3 2 0.00032 0.0025
>>#9 5 4 6 4 1.00 1.2 3 2 0.00032 0.0025
>>
>>A.K.
>>
>>
>>
>>
>>
>>----- Original Message -----
>>From: Zjoanna <Zjoanna2013 at gmail.com>
>>To: r-help at r-project.org
>>Cc:
>>
>>Sent: Sunday, February 10, 2013 6:04 PM
>>Subject: Re: [R] cumulative sum by group and under some criteria
>>
>>
>>Hi,
>>How to expand or loop for one variable n based on another variable? for
>>example, I want to add m (from m1 to maxN- n1-2) and for each m, I want to
>>add n (n1+2 to maxN-m), and similarly add x and y, then I need to do some
>>calculations.
>>
>>d3<-data.frame(d2)
>> for (m in (m1+2):(maxN-(n1+2)){
>> for (n in (n1+2):(maxN-m)){
>> for (x in x1:(x1+m-m1)){
>> for (y in y1:(y1+n-n1)){
>> p1<- x/m
>> p2<- y/n
>>}}}}
>>
>>On Thu, Feb 7, 2013 at 12:16 AM, arun kirshna [via R] <
>>ml-node+s789695n4657773h74 at n4.nabble.com> wrote:
>>
>>> Hi,
>>>
>>> Anyway, just using some random combinations:
>>> dnew<- expand.grid(4:10,5:10,6:10,3:7,4:5,6:8)
>>> names(dnew)<-c("m","n","x1","y1","x","y")
>>> resF<- cbind(dnew,d2[rep(1:nrow(d2),nrow(dnew)),])
>>>
>>> row.names(resF)<- 1:nrow(resF)
>>> # m n x1 y1 x y m1 n1 cterm1_P1L cterm1_P0H
>>> #1 4 5 6 3 4 6 3 2 0.00032 0.0025
>>> #2 5 5 6 3 4 6 3 2 0.00032 0.0025
>>> #3 6 5 6 3 4 6 3 2 0.00032 0.0025
>>> #4 7 5 6 3 4 6 3 2 0.00032 0.0025
>>> #5 8 5 6 3 4 6 3 2 0.00032 0.0025
>>> #6 9 5 6 3 4 6 3 2 0.00032 0.0025
>>>
>>> nrow(resF)
>>> #[1] 6300
>>> I am not sure what you want to do with this.
>>> A.K.
>>> ________________________________
>>> From: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=0>>
>>>
>>> To: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=1>>
>>
>>>
>>> Sent: Wednesday, February 6, 2013 10:29 AM
>>> Subject: Re: cumulative sum by group and under some criteria
>>>
>>>
>>> Hi,
>>>
>>> Thanks! I need to do some calculations in the expended data, the expended
>>> data would be very large, what is an efficient way, doing calculations
>>> while expending the data, something similiar with the following, or
>>> expending data using the code in your message and then add calculations in
>>> the expended data?
>>>
>>> d3<-data.frame(d2)
>>> for .......{
>>> for {
>>> for .... {
>>> for .....{
>>> p1<- x/m
>>> p2<- y/n
>>> ..........
>>> }}
>>> }}
>>>
>>> I also modified your code for expending data:
>>> dnew<-expand.grid((m1+2):(maxN-(n1+2)),(n1+2):(maxN-m),0:m1,0:n1,
>>> x1:(x1+m-m1),y1:(y1+n-n1))
>>> names(dnew)<-c("m","n","x1","y1","x","y")
>>> dnew
>>> resF<-cbind(dnew[,c(2,1)],d2[rep(1:nrow(d2),nrow(dnew)),]) # this is
>>> not correct, how to modify it.
>>> resF
>>> row.names(resF)<-1:nrow(resF)
>>> resF
>>>
>>>
>>>
>>>
>>> On Tue, Feb 5, 2013 at 2:46 PM, arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=2>>
>>
>>> wrote:
>>>
>>> Hi,
>>>
>>> >
>>> >You can reduce the steps to reach d2:
>>> >res3<-
>>> with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
>>> >
>>> >#Change it to:
>>> >res3new<- aggregate(.~m1+n1,data=res2[,c(1:2,12:13)],max)
>>> >res3new
>>> > m1 n1 cterm1_P1L cterm1_P0H
>>> >1 2 2 0.01440 0.00273750
>>> >2 3 2 0.00032 0.00250000
>>> >3 2 3 0.01952 0.00048125
>>> >d2<-res3new[res3new[,3]<0.01 & res3new[,4]<0.01,]
>>> >
>>> > dnew<-expand.grid(4:10,5:10)
>>> > names(dnew)<-c("n","m")
>>> >resF<-cbind(dnew[,c(2,1)],d2[rep(1:nrow(d2),nrow(dnew)),])
>>> >
>>> >row.names(resF)<-1:nrow(resF)
>>> ># m n m1 n1 cterm1_P1L cterm1_P0H
>>> >#1 5 4 3 2 0.00032 0.0025
>>> >#2 5 5 3 2 0.00032 0.0025
>>> >#3 5 6 3 2 0.00032 0.0025
>>> >#4 5 7 3 2 0.00032 0.0025
>>> >#5 5 8 3 2 0.00032 0.0025
>>> >#6 5 9 3 2 0.00032 0.0025
>>> >
>>> >A.K.
>>> >
>>> >________________________________
>>> >From: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=3>>
>>>
>>> >To: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=4>>
>>
>>>
>>> >Sent: Tuesday, February 5, 2013 2:48 PM
>>> >
>>> >Subject: Re: cumulative sum by group and under some criteria
>>> >
>>> >
>>> > Hi ,
>>> >what I want is :
>>> >m n m1 n1 cterm1_P1L cterm1_P0H
>>> > 5 4 3 2 0.00032 0.00250000
>>> > 5 5 3 2 0.00032 0.00250000
>>> > 5 6 3 2 0.00032 0.00250000
>>> > 5 7 3 2 0.00032 0.00250000
>>> > 5 8 3 2 0.00032 0.00250000
>>> > 5 9 3 2 0.00032 0.00250000
>>> >5 10 3 2 0.00032 0.00250000
>>> >6 4 3 2 0.00032 0.00250000
>>> >6 5 3 2 0.00032 0.00250000
>>> >6 6 3 2 0.00032 0.00250000
>>> >6 7 3 2 0.00032 0.00250000
>>> >.....
>>> >6 10 3 2 0.00032 0.00250000
>>> >
>>> >
>>> >
>>> >On Tue, Feb 5, 2013 at 1:12 PM, arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=5>>
>>
>>> wrote:
>>> >
>>> >Hi,
>>> >>
>>> >>Saw your message on Nabble.
>>> >>
>>> >>
>>> >>"I want to add some more columns based on the results. Is the following
>>> code good way to create such a data frame and How to see the column m and n
>>> in the updated data?
>>> >>
>>> >>d2<- reres3[res3[,3]<0.01 & res3[,4]<0.01,]
>>> >># should be a typo
>>> >>
>>> >>colnames(d2)[1:2]<- c("m1","n1");
>>> >>d2 #already a data.frame
>>> >>
>>> >>d3<-data.frame(d2)
>>> >> for (m in (m1+2):10){
>>> >> for (n in (n1+2):10){
>>> >> d3<-rbind(d3, c(d2))}}" #this is not making much sense to me.
>>> Especially, you mentioned you wanted add more columns.
>>> >>#Running this step gave error
>>> >>
>>> >>Not sure what you want as output.
>>> >>Could you show the ouput that is expected:
>>> >>
>>> >>A.K.
>>> >>
>>> >>
>>> >>
>>> >>
>>> >>
>>> >>
>>> >>
>>> >>
>>> >>________________________________
>>> >>From: Joanna Zhang <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=6>>
>>>
>>> >>To: arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=7>>
>>
>>>
>>> >>Sent: Tuesday, February 5, 2013 10:23 AM
>>> >>
>>> >>Subject: Re: cumulative sum by group and under some criteria
>>> >>
>>> >>
>>> >>Hi,
>>> >>
>>> >>Yes, I changed code. You answered the questions. But how can I put two
>>> criteria in the code, if both the maximum value of cterm1_p1L <= 0.01 and
>>> cterm1_p1H <=0.01, the output the m1,n1.
>>> >>
>>> >>
>>> >>
>>> >>
>>> >>On Tue, Feb 5, 2013 at 8:47 AM, arun <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=8>>
>>
>>> wrote:
>>> >>
>>> >>
>>> >>>
>>> >>> HI,
>>> >>>
>>> >>>
>>> >>>I am not getting the same results as yours: You must have changed the
>>> dataset.
>>> >>> res2[,1:2][res2\$cterm1_P1L<0.6 & res2\$cterm1_P0H<0.95,]
>>> >>> m1 n1
>>> >>>1 2 2
>>> >>>2 2 2
>>> >>>3 2 2
>>> >>>4 2 2
>>> >>>5 2 2
>>> >>>6 2 2
>>> >>>7 2 2
>>> >>>8 2 2
>>> >>>9 2 2
>>> >>>10 3 2
>>> >>>11 3 2
>>> >>>12 3 2
>>> >>>13 3 2
>>> >>>14 3 2
>>> >>>15 3 2
>>> >>>16 3 2
>>> >>>17 3 2
>>> >>>18 3 2
>>> >>>19 3 2
>>> >>>20 3 2
>>> >>>21 3 2
>>> >>>22 2 3
>>> >>>23 2 3
>>> >>>24 2 3
>>> >>>25 2 3
>>> >>>26 2 3
>>> >>>27 2 3
>>> >>>28 2 3
>>> >>>29 2 3
>>> >>>30 2 3
>>> >>>31 2 3
>>> >>>32 2 3
>>> >>>33 2 3
>>> >>>
>>> >>>
>>> >>>Regarding the maximum value within each block, haven't I answered in
>>> the earlier post.
>>> >>>
>>> >>>aggregate(cterm1_P1L~m1+n1,data=res2,max)
>>> >>># m1 n1 cterm1_P1L
>>> >>>#1 2 2 0.01440
>>> >>>#2 3 2 0.00032
>>> >>>#3 2 3 0.01952
>>> >>>
>>> >>>
>>> >>> with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
>>> >>># Group.1 Group.2 cterm1_P1L cterm1_P0H
>>> >>>#1 2 2 0.01440 0.00273750
>>> >>>#2 3 2 0.00032 0.00250000
>>> >>>#3 2 3 0.01952 0.00048125
>>> >>>
>>> >>>
>>> >>>A.K.
>>> >>>
>>> >>>
>>> >>>----- Original Message -----
>>
>>> >>>From: "[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=9>";;;
>>> <[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=10>>
>>> >>>To: [hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=11>
>>> >>>Cc:
>>> >>>
>>> >>>Sent: Tuesday, February 5, 2013 9:33 AM
>>> >>>Subject: Re: cumulative sum by group and under some criteria
>>> >>>
>>> >>>Hi,
>>> >>>If use this
>>> >>>
>>> >>>res2[,1:2][res2\$cterm1_P1L<0.6 & res2\$cterm1_P0H<0.95,]
>>> >>>
>>> >>>the results are the following, but actually only m1=3, n1=2 sastify the
>>> criteria, as I need to look at the row with maximum value within each
>>> block,not every row.
>>> >>>
>>> >>>
>>> >>> m1 n1
>>> >>>1 2 2
>>> >>>10 3 2
>>> >>>11 3 2
>>> >>>12 3 2
>>> >>>13 3 2
>>> >>>14 3 2
>>> >>>15 3 2
>>> >>>16 3 2
>>> >>>17 3 2
>>> >>>18 3 2
>>> >>>19 3 2
>>> >>>20 3 2
>>> >>>21 3 2
>>> >>>22 2 3
>>> >>>23 2 3
>>> >>>
>>> >>>
>>> >>><quote author='arun kirshna'>
>>> >>>
>>> >>>
>>> >>>
>>> >>>Hi,
>>> >>>Thanks. This extract every row that satisfy the condition, but I need
>>> look
>>> >>>at the last row (the maximum of cumulative sum) for each block (m1,n1).
>>> for
>>> >>>example, if I set the criteria
>>> >>>
>>> >>>res2\$cterm1_P1L<0.6 & res2\$cterm1_P0H<0.95, this should extract m1= 3,
>>> n1 =
>>> >>>2.
>>> >>>
>>> >>>
>>> >>>Hi,
>>> >>>I am not sure I understand your question.
>>> >>>res2\$cterm1_P1L<0.6 & res2\$cterm1_P0H<0.95
>>> >>> #[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>>> TRUE
>>> >>>TRUE
>>> >>>#[16] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>>> TRUE
>>> >>>TRUE
>>> >>>#[31] TRUE TRUE TRUE
>>> >>>
>>> >>>This will extract all the rows.
>>> >>>
>>> >>>
>>> >>>res2[,1:2][res2\$cterm1_P1L<0.01 & res2\$cterm1_P1L!=0,]
>>> >>># m1 n1
>>> >>>#21 3 2
>>> >>>This extract only the row you wanted.
>>> >>>
>>> >>>For the different groups:
>>> >>>
>>> >>>aggregate(cterm1_P1L~m1+n1,data=res2,max)
>>> >>># m1 n1 cterm1_P1L
>>> >>>#1 2 2 0.01440
>>> >>>#2 3 2 0.00032
>>> >>>#3 2 3 0.01952
>>> >>>
>>> >>> aggregate(cterm1_P1L~m1+n1,data=res2,function(x) max(x)<0.01)
>>> >>> # m1 n1 cterm1_P1L
>>> >>>#1 2 2 FALSE
>>> >>>#2 3 2 TRUE
>>> >>>#3 2 3 FALSE
>>> >>>
>>> >>>res4<-aggregate(cterm1_P1L~m1+n1,data=res2,function(x) max(x)<0.01)
>>> >>>res4[,1:2][res4[,3],]
>>> >>># m1 n1
>>> >>>#2 3 2
>>> >>>
>>> >>>A.K.
>>> >>>
>>> >>>
>>> >>>
>>> >>>
>>> >>>----- Original Message -----
>>
>>> >>>From: "[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=12>";;;
>>> <[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=13>>
>>> >>>To: [hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=14>
>>> >>>Cc:
>>> >>>Sent: Sunday, February 3, 2013 3:58 PM
>>> >>>Subject: Re: cumulative sum by group and under some criteria
>>> >>>
>>> >>>Hi,
>>> >>>Let me restate my questions. I need to get the m1 and n1 that satisfy
>>> some
>>> >>>criteria, for example in this case, within each group, the maximum
>>> >>>cterm1_p1L ( the last row in this group) <0.01. I need to extract m1=3,
>>> >>>n1=2, I only need m1, n1 in the row.
>>> >>>
>>> >>>Also, how to create the structure from the data.frame, I am new to R, I
>>> need
>>> >>>to change the maxN and run the loop to different data.
>>> >>>Thanks very much for your help!
>>> >>>
>>> >>><quote author='arun kirshna'>
>>> >>>HI,
>>> >>>
>>> >>>I think this should be more correct:
>>> >>>maxN<-9
>>> >>>c11<-0.2
>>> >>>c12<-0.2
>>> >>>p0L<-0.05
>>> >>>p0H<-0.05
>>> >>>p1L<-0.20
>>> >>>p1H<-0.20
>>> >>>
>>> >>>d <- structure(list(m1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
>>> >>>2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
>>> >>> n1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
>>> >>> 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), x1 = c(0,
>>> >>> 0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2,
>>> >>> 2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3), y1 = c(0, 1, 2, 0,
>>> >>> 1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1,
>>> >>> 2, 0, 1, 2, 0, 1, 2, 0, 1, 2), Fmm = c(0, 0, 0, 0.7, 0.59,
>>> >>> 0.64, 1, 1, 1, 0, 0, 0, 0, 0.63, 0.7, 0.74, 0.68, 1, 1, 1,
>>> >>> 1, 0, 0, 0, 0.62, 0.63, 0.6, 0.63, 0.6, 0.68, 1, 1, 1), Fnn = c(0,
>>> >>> 0.64, 1, 0, 0.51, 1, 0, 0.67, 1, 0, 0.62, 0.69, 1, 0, 0.54,
>>> >>> 0.62, 1, 0, 0.63, 0.73, 1, 0, 0.63, 1, 0, 0.7, 1, 0, 0.7,
>>> >>> 1, 0, 0.58, 1), Qm = c(1, 1, 1, 0.65, 0.45, 0.36, 0.5, 0.165,
>>> >>> 0, 1, 1, 1, 1, 0.685, 0.38, 0.32, 0.32, 0.5, 0.185, 0.135,
>>> >>> 0, 1, 1, 1, 0.69, 0.37, 0.4, 0.685, 0.4, 0.32, 0.5, 0.21,
>>> >>> 0), Qn = c(1, 0.36, 0, 0.65, 0.45, 0, 0.5, 0.165, 0, 1, 0.38,
>>> >>> 0.31, 0, 0.685, 0.38, 0.32, 0, 0.5, 0.185, 0.135, 0, 1, 0.37,
>>> >>> 0, 0.69, 0.3, 0, 0.685, 0.3, 0, 0.5, 0.21, 0), term1_p0 =
>>> c(0.81450625,
>>> >>> 0.0857375, 0.00225625, 0.0857375, 0.009025, 0.0002375, 0.00225625,
>>> >>> 0.0002375, 6.25e-06, 0.7737809375, 0.1221759375,
>>> 0.00643031249999999,
>>> >>> 0.0001128125, 0.081450625, 0.012860625, 0.000676875, 1.1875e-05,
>>> >>> 0.0021434375, 0.0003384375, 1.78125e-05, 3.125e-07, 0.7737809375,
>>> >>> 0.081450625, 0.0021434375, 0.1221759375, 0.012860625,
>>> 0.0003384375,
>>> >>> 0.00643031249999999, 0.000676875, 1.78125e-05, 0.0001128125,
>>> >>> 1.1875e-05, 3.125e-07), term1_p1 = c(0.4096, 0.2048, 0.0256,
>>> >>> 0.2048, 0.1024, 0.0128, 0.0256, 0.0128, 0.0016, 0.32768,
>>> >>> 0.24576, 0.06144, 0.00512, 0.16384, 0.12288, 0.03072, 0.00256,
>>> >>> 0.02048, 0.01536, 0.00384, 0.00032, 0.32768, 0.16384, 0.02048,
>>> >>> 0.24576, 0.12288, 0.01536, 0.06144, 0.03072, 0.00384, 0.00512,
>>> >>> 0.00256, 0.00032)), .Names = c("m1", "n1", "x1", "y1", "Fmm",
>>> >>>"Fnn", "Qm", "Qn", "term1_p0", "term1_p1"), row.names = c(NA,
>>> >>>33L), class = "data.frame")
>>> >>>
>>> >>>library(zoo)
>>> >>>lst1<- split(d,list(d\$m1,d\$n1))
>>> >>>res2<-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
>>> >>>x[,11:14]<-NA;
>>> >>>x[,11:12][x\$Qm<=c11,]<-cumsum(x[,9:10][x\$Qm<=c11,]);
>>> >>>x[,13:14][x\$Qn<=c12,]<-cumsum(x[,9:10][x\$Qn<=c12,]);
>>> >>>colnames(x)[11:14]<-
>>> c("cterm1_P0L","cterm1_P1L","cterm1_P0H","cterm1_P1H");
>>> >>>x1<-na.locf(x);
>>> >>>x1[,11:14][is.na(x1[,11:14])]<-0;
>>> >>>x1}))
>>> >>>row.names(res2)<- 1:nrow(res2)
>>> >>>
>>> >>> res2
>>> >>> # m1 n1 x1 y1 Fmm Fnn Qm Qn term1_p0 term1_p1
>>> cterm1_P0L
>>> >>>cterm1_P1L cterm1_P0H cterm1_P1H
>>> >>>
>>> >>>#1 2 2 0 0 0.00 0.00 1.000 1.000 0.8145062500 0.40960
>>> 0.0000000000
>>> >>> 0.00000 0.0000000000 0.00000
>>> >>>#2 2 2 0 1 0.00 0.64 1.000 0.360 0.0857375000 0.20480
>>> 0.0000000000
>>> >>> 0.00000 0.0000000000 0.00000
>>> >>>#3 2 2 0 2 0.00 1.00 1.000 0.000 0.0022562500 0.02560
>>> 0.0000000000
>>> >>> 0.00000 0.0022562500 0.02560
>>> >>>#4 2 2 1 0 0.70 0.00 0.650 0.650 0.0857375000 0.20480
>>> 0.0000000000
>>> >>> 0.00000 0.0022562500 0.02560
>>> >>>#5 2 2 1 1 0.59 0.51 0.450 0.450 0.0090250000 0.10240
>>> 0.0000000000
>>> >>> 0.00000 0.0022562500 0.02560
>>> >>>#6 2 2 1 2 0.64 1.00 0.360 0.000 0.0002375000 0.01280
>>> 0.0000000000
>>> >>> 0.00000 0.0024937500 0.03840
>>> >>>#7 2 2 2 0 1.00 0.00 0.500 0.500 0.0022562500 0.02560
>>> 0.0000000000
>>> >>> 0.00000 0.0024937500 0.03840
>>> >>>#8 2 2 2 1 1.00 0.67 0.165 0.165 0.0002375000 0.01280
>>> 0.0002375000
>>> >>> 0.01280 0.0027312500 0.05120
>>> >>>#9 2 2 2 2 1.00 1.00 0.000 0.000 0.0000062500 0.00160
>>> 0.0002437500
>>> >>> 0.01440 0.0027375000 0.05280
>>> >>>#10 3 2 0 0 0.00 0.00 1.000 1.000 0.7737809375 0.32768
>>> 0.0000000000
>>> >>> 0.00000 0.0000000000 0.00000
>>> >>>#11 3 2 0 1 0.00 0.63 1.000 0.370 0.0814506250 0.16384
>>> 0.0000000000
>>> >>> 0.00000 0.0000000000 0.00000
>>> >>>#12 3 2 0 2 0.00 1.00 1.000 0.000 0.0021434375 0.02048
>>> 0.0000000000
>>> >>> 0.00000 0.0021434375 0.02048
>>> >>>#13 3 2 1 0 0.62 0.00 0.690 0.690 0.1221759375 0.24576
>>> 0.0000000000
>>> >>> 0.00000 0.0021434375 0.02048
>>> >>>#14 3 2 1 1 0.63 0.70 0.370 0.300 0.0128606250 0.12288
>>> 0.0000000000
>>> >>> 0.00000 0.0021434375 0.02048
>>> >>>#15 3 2 1 2 0.60 1.00 0.400 0.000 0.0003384375 0.01536
>>> 0.0000000000
>>> >>> 0.00000 0.0024818750 0.03584
>>> >>>#16 3 2 2 0 0.63 0.00 0.685 0.685 0.0064303125 0.06144
>>> 0.0000000000
>>> >>> 0.00000 0.0024818750 0.03584
>>> >>>#17 3 2 2 1 0.60 0.70 0.400 0.300 0.0006768750 0.03072
>>> 0.0000000000
>>> >>> 0.00000 0.0024818750 0.03584
>>> >>>#18 3 2 2 2 0.68 1.00 0.320 0.000 0.0000178125 0.00384
>>> 0.0000000000
>>> >>> 0.00000 0.0024996875 0.03968
>>> >>>#19 3 2 3 0 1.00 0.00 0.500 0.500 0.0001128125 0.00512
>>> 0.0000000000
>>> >>> 0.00000 0.0024996875 0.03968
>>> >>>#20 3 2 3 1 1.00 0.58 0.210 0.210 0.0000118750 0.00256
>>> 0.0000000000
>>> >>> 0.00000 0.0024996875 0.03968
>>> >>>#21 3 2 3 2 1.00 1.00 0.000 0.000 0.0000003125 0.00032
>>> 0.0000003125
>>> >>> 0.00032 0.0025000000 0.04000
>>> >>>#22 2 3 0 0 0.00 0.00 1.000 1.000 0.7737809375 0.32768
>>> 0.0000000000
>>> >>> 0.00000 0.0000000000 0.00000
>>> >>>#23 2 3 0 1 0.00 0.62 1.000 0.380 0.1221759375 0.24576
>>> 0.0000000000
>>> >>> 0.00000 0.0000000000 0.00000
>>> >>>#24 2 3 0 2 0.00 0.69 1.000 0.310 0.0064303125 0.06144
>>> 0.0000000000
>>> >>> 0.00000 0.0000000000 0.00000
>>> >>>#25 2 3 0 3 0.00 1.00 1.000 0.000 0.0001128125 0.00512
>>> 0.0000000000
>>> >>> 0.00000 0.0001128125 0.00512
>>> >>>#26 2 3 1 0 0.63 0.00 0.685 0.685 0.0814506250 0.16384
>>> 0.0000000000
>>> >>> 0.00000 0.0001128125 0.00512
>>> >>>#27 2 3 1 1 0.70 0.54 0.380 0.380 0.0128606250 0.12288
>>> 0.0000000000
>>> >>> 0.00000 0.0001128125 0.00512
>>> >>>#28 2 3 1 2 0.74 0.62 0.320 0.320 0.0006768750 0.03072
>>> 0.0000000000
>>> >>> 0.00000 0.0001128125 0.00512
>>> >>>#29 2 3 1 3 0.68 1.00 0.320 0.000 0.0000118750 0.00256
>>> 0.0000000000
>>> >>> 0.00000 0.0001246875 0.00768
>>> >>>#30 2 3 2 0 1.00 0.00 0.500 0.500 0.0021434375 0.02048
>>> 0.0000000000
>>> >>> 0.00000 0.0001246875 0.00768
>>> >>>#31 2 3 2 1 1.00 0.63 0.185 0.185 0.0003384375 0.01536
>>> 0.0003384375
>>> >>> 0.01536 0.0004631250 0.02304
>>> >>>#32 2 3 2 2 1.00 0.73 0.135 0.135 0.0000178125 0.00384
>>> 0.0003562500
>>> >>> 0.01920 0.0004809375 0.02688
>>> >>>#33 2 3 2 3 1.00 1.00 0.000 0.000 0.0000003125 0.00032
>>> 0.0003565625
>>> >>> 0.01952 0.0004812500 0.02720
>>> >>>
>>> >>>#Sorry, some values in my previous solution didn't look right. I
>>> didn't
>>> >>>A.K.
>>> >>>
>>> >>>
>>> >>>
>>> >>>
>>> >>>
>>> >>>----- Original Message -----
>>> >>>From: Zjoanna <[hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=15>>
>>>
>>> >>>To: [hidden email]<http://user/SendEmail.jtp?type=node&node=4657773&i=16>
>>
>>> >>>Cc:
>>> >>>Sent: Friday, February 1, 2013 12:19 PM
>>> >>>Subject: Re: [R] cumulative sum by group and under some criteria
>>> >>>
>>> >>>Thank you very much for your reply. Your code work well with this
>>> example.
>>> >>>I modified a little to fit my real data, I got an error massage.
>>> >>>
>>> >>>Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) :
>>> >>> Group length is 0 but data length > 0
>>> >>>
>>> >>>
>>> >>>On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] <
>>> >>>[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=17>>
>>
>>> wrote:
>>> >>>
>>> >>>> Hi,
>>> >>>> Try this:
>>> >>>> colnames(d)<-c("m1","n1","x1","y1","p11","p12")
>>> >>>> library(zoo)
>>> >>>> res1<-
>>> do.call(rbind,lapply(lapply(split(d,list(d\$m1,d\$n1)),function(x)
>>> >>>> {x\$cp11[x\$x1>1]<- cumsum(x\$p11[x\$x1>1]);x\$cp12[x\$y1>1]<-
>>> >>>> cumsum(x\$p12[x\$y1>1]);x}),function(x)
>>> >>>> {x\$cp11<-na.locf(x\$cp11,na.rm=F);x\$cp12<-
>>> na.locf(x\$cp12,na.rm=F);x}))
>>> >>>> #there would be a warning here as one of the list element is NULL.
>>> The,
>>> >>>> warning is okay
>>> >>>> row.names(res1)<- 1:nrow(res1)
>>> >>>> res1[,7:8][is.na(res1[,7:8])]<- 0
>>> >>>> res1
>>> >>>> # m1 n1 x1 y1 p11 p12 cp11 cp12
>>> >>>> #1 2 2 0 0 0.00 0.00 0.00 0.00
>>> >>>> #2 2 2 0 1 0.00 0.50 0.00 0.00
>>> >>>> #3 2 2 0 2 0.00 1.00 0.00 1.00
>>> >>>> #4 2 2 1 0 0.50 0.00 0.00 1.00
>>> >>>> #5 2 2 1 1 0.50 0.50 0.00 1.00
>>> >>>> #6 2 2 1 2 0.50 1.00 0.00 2.00
>>> >>>> #7 2 2 2 0 1.00 0.00 1.00 2.00
>>> >>>> #8 2 2 2 1 1.00 0.50 2.00 2.00
>>> >>>> #9 2 2 2 2 1.00 1.00 3.00 3.00
>>> >>>> #10 3 2 0 0 0.00 0.00 0.00 0.00
>>> >>>> #11 3 2 0 1 0.00 0.50 0.00 0.00
>>> >>>> #12 3 2 0 2 0.00 1.00 0.00 1.00
>>> >>>> #13 3 2 1 0 0.33 0.00 0.00 1.00
>>> >>>> #14 3 2 1 1 0.33 0.50 0.00 1.00
>>> >>>> #15 3 2 1 2 0.33 1.00 0.00 2.00
>>> >>>> #16 3 2 2 0 0.67 0.00 0.67 2.00
>>> >>>> #17 3 2 2 1 0.67 0.50 1.34 2.00
>>> >>>> #18 3 2 2 2 0.67 1.00 2.01 3.00
>>> >>>> #19 3 2 3 0 1.00 0.00 3.01 3.00
>>> >>>> #20 3 2 3 1 1.00 0.50 4.01 3.00
>>> >>>> #21 3 2 3 2 1.00 1.00 5.01 4.00
>>> >>>> #22 2 3 0 0 0.00 0.00 0.00 0.00
>>> >>>> #23 2 3 0 1 0.00 0.33 0.00 0.00
>>> >>>> #24 2 3 0 2 0.00 0.67 0.00 0.67
>>> >>>> #25 2 3 0 3 0.00 1.00 0.00 1.67
>>> >>>> #26 2 3 1 0 0.50 0.00 0.00 1.67
>>> >>>> #27 2 3 1 1 0.50 0.33 0.00 1.67
>>> >>>> #28 2 3 1 2 0.50 0.67 0.00 2.34
>>> >>>> #29 2 3 1 3 0.50 1.00 0.00 3.34
>>> >>>> #30 2 3 2 0 1.00 0.00 1.00 3.34
>>> >>>> #31 2 3 2 1 1.00 0.33 2.00 3.34
>>> >>>> #32 2 3 2 2 1.00 0.67 3.00 4.01
>>> >>>> #33 2 3 2 3 1.00 1.00 4.00 5.01
>>> >>>> A.K.
>>> >>>>
>>> >>>> ------------------------------
>>> >>>> If you reply to this email, your message will be added to the
>>> discussion
>>> >>>> below:
>>> >>>>
>>> >>>>
>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657196.html
>>> >>>> To unsubscribe from cumulative sum by group and under some criteria,
>>> click
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>>>
>>> >>>> .
>>> >>>> NAML<
>>>
>>> >>>>
>>> >>>
>>> >>>
>>> >>>
>>> >>>
>>> >>>--
>>> >>>View this message in context:
>>> >>>
>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657315.html
>>> >>>Sent from the R help mailing list archive at Nabble.com.
>>> >>> [[alternative HTML version deleted]]
>>> >>>
>>> >>>______________________________________________
>>> >>>[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=18>mailing list
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>>> >>>PLEASE do read the posting guide
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>>
>>> >>>and provide commented, minimal, self-contained, reproducible code.
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>>> >>>______________________________________________
>>> >>>[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=19>mailing list
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>>> >>>and provide commented, minimal, self-contained, reproducible code.
>>> >>>
>>> >>></quote>
>>> >>>Quoted from:
>>> >>>
>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657360.html
>>> >>>
>>> >>>
>>> >>>______________________________________________
>>> >>>[hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=20>mailing list
>>
>>> >>>https://stat.ethz.ch/mailman/listinfo/r-help
>>> >>>PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>>
>>> >>>and provide commented, minimal, self-contained, reproducible code.
>>> >>>
>>> >>></quote>
>>> >>>Quoted from:
>>> >>>
>>> http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657582.html
>>> >>>
>>> >>>
>>> >>
>>> >
>>>
>>> ______________________________________________
>>> [hidden email] <http://user/SendEmail.jtp?type=node&node=4657773&i=21>mailing list
>>
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
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>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>
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>>> If you reply to this email, your message will be added to the
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>
``` | 13,697 | 29,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.414678 |
https://stats.stackexchange.com/questions/41297/kullback-leibler-divergence-negative-values?noredirect=1 | 1,718,979,026,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00319.warc.gz | 485,247,971 | 35,009 | # Kullback-Leibler divergence: negative values? [duplicate]
Wikipedia - KL properties says that KL can never be negative. But e.g. for texts where the probabilities are very small I somehow get negative values? E.g.
Collection A: - word count: 321 doc count: 65888 probA: 0,004871904
Collection B: - word count: 1244 doc count: 120344 probB: =0,010337034
KL = $0.004871904 \cdot \ln\frac{0.004871904}{0.010337034} = -0.003664881$
• So, looking at @SamLivingstone's answer below, you need to add the analogous second term to your sum. Commented Oct 27, 2012 at 14:14
## 1 Answer
KL-divergence is the sum of $q(i)\log\frac{q(i)}{p(i)}$ across all values of $i$. You've only got one instance ($i$) in your equation. For example, if your model was binomial (only two possible words occurred in your document) and $Pr(word1)$ was 0.005 in document 1 and 0.01 in document 2 then you would have:
$$KL = 0.005*\log\frac{0.005}{0.01} + 0.995*\log\frac{0.995}{0.99} = 0.001547 \geq 0.$$
This sum (or integral in the case of continuous random variables) will always be positive, by the Gibbs inequality (see http://en.wikipedia.org/wiki/Gibbs%27_inequality).
• Yes, right. The sum is always > 0. Commented Oct 27, 2012 at 14:47 | 391 | 1,226 | {"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} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.842377 |
https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/9/lesson/9.4.3/problem/9-137 | 1,580,106,606,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694908.82/warc/CC-MAIN-20200127051112-20200127081112-00390.warc.gz | 492,340,529 | 15,293 | ### Home > CALC > Chapter 9 > Lesson 9.4.3 > Problem9-137
9-137.
1. For each of the series below, write an equivalent expression using sigma notation. Homework Help ✎
1. 5 + 10 + 20 + 40 + ... + 5(2) n- 1
$\sum_{i=1}^{n}5(2)^{i-1}$
$a_n=\Big(\frac{2}{3}\Big)^n\text{ starting with }n=0.$ | 119 | 291 | {"found_math": true, "script_math_tex": 2, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.416225 |
http://www.physics.ohio-state.edu/~ntg/6805/slides/Diffraction_101.php | 1,537,442,568,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156460.64/warc/CC-MAIN-20180920101233-20180920121633-00084.warc.gz | 376,342,596 | 2,918 | # Physics 6805: Diffraction 101
Return to 6805 Home
## Recent changes to this page:
• 21-Aug-2017 --- Revised 2017 version from LaTeX version.
## Slides
### Diffraction 101 (review?)
Return to Contents
### Intensity as a Function of Angle
Single-slit diffraction: first minimum when sin θmin = λ/a ⇒ pattern widens as λ increases.
Return to Contents
### Diffraction from a circular hole
The diffraction pattern of a circular hole of diameter d is similar to that of a single slit of width a. The bright spot in the middle is the Airy disk. It has about 85% of the power in the pattern. Minima are at:
sin θ1 = 1.22 λ/d
sin θ1 = 2.23 λ/d
sin θ1 = 3.24 λ/d
What is the diffraction pattern from a solid disk? (Hint: Babinet)
Return to Contents
### Diffraction and resolution
Return to Contents
### Wavelength and resolution
Return to Contents
### Consequences
For a slit of width a, θmin = sin-1(λ/a), so if λ>a, you don't learn anything about the details of the slit.
Return to Contents
### Quantum Mechanics and Resolving Power
The de Broglie relation is λ = h/p.
For λ≈10-10m ⇒ probe atoms
For λ≈10-14m ⇒ probe nucleus
For λ≈10-18m ⇒ probe quarks
Return to Contents
Your comments and suggestions are appreciated.
[OSU Physics] [College of Arts and Sciences] [Ohio State University]
Physics 6805 .
Last modified: 02:12 pm, August 21, 2017.
furnstahl.1@osu.edu | 388 | 1,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-39 | longest | en | 0.786883 |
https://tex.stackexchange.com/questions/576963/horizontal-line-in-array/576973 | 1,653,471,900,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662584398.89/warc/CC-MAIN-20220525085552-20220525115552-00172.warc.gz | 650,459,018 | 66,492 | # Horizontal Line in Array
Good evening,
I'm trying to achieve following picture in latex, but i can't do it. Can someone help me?
This is what I have so far:
\documentclass[a4paper,12pt]{scrartcl}
\usepackage{booktabs}
\usepackage[utf8] {inputenc}
\usepackage[ngerman]{babel}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage[T1]{fontenc}
\begin{document}
\begin{center}
$\left(\begin{array}{rrrr} 2&2&-1&3 \\ \cmidrule{2-4} 0&1&3&4 \\ 0&1&3&4 \\ 0&3&9&6 \\ 0&1&3&-2 \end{array}\right)$
\end{center}
\end{document}
You're missing the vertical line | in the column specification and a \multicolumn{1}{r} for the element in the top left corner:
\documentclass{article}
\begin{document}
$\left( \begin{array}{ r | *{3}{r} } \multicolumn{1}{r}{2} & 2 & -1 & 3 \\ \cline{2-4} 0 & 1 & 3 & 4 \\ 0 & 1 & 3 & 4 \\ 0 & 3 & 9 & 6 \\ 0 & 1 & 3 & -2 \end{array} \right)$
\end{document}
For a more evenly distributed spacing around the last three columns, consider using \phantom{-}2 in the first row:
\documentclass{article}
\begin{document}
$\left( \begin{array}{ r | *{3}{r} } \multicolumn{1}{r}{2} & \phantom{-}2 & -1 & 3 \\ \cline{2-4} 0 & 1 & 3 & 4 \\ 0 & 1 & 3 & 4 \\ 0 & 3 & 9 & 6 \\ 0 & 1 & 3 & -2 \end{array} \right)$
\end{document}
• Thank you very much ! Dec 30, 2020 at 21:03
With nicematrix. The environments of nicematrix create PGF/Tikz nodes under the rows, columns and cells and the array and it's possible to use them to draw whatever you want with Tikz.
\documentclass{article}
\usepackage{nicematrix}
\usepackage{tikz}
\begin{document}
$\begin{pNiceMatrix}[r,right-margin,columns-width=auto] 2 & 2 & -1 & 3 \\ 0 & 1 & 3 & 4 \\ 0 & 1 & 3 & 4 \\ 0 & 3 & 9 & 6 \\ 0 & 1 & 3 & -2 \CodeAfter \tikz \draw ([xshift=4pt]last-|2) |- (2-|last) ; \end{pNiceMatrix}$
\end{document}
• Thank you very much! Dec 30, 2020 at 21:14
• Is it also possible to draw dotted lines ? Dec 30, 2020 at 21:18
• @Francesco: Yes. Put \draw [dotted] instead of \draw. Dec 30, 2020 at 21:29
• To my eyes the horizontal line is a little too low. Dec 30, 2020 at 21:58
• @Anush That's why this approach is more powerful. You only need e.g. \tikz \draw ([xshift=4pt]row-6-|col-2) |- ([yshift=2pt]row-2-|col-5) ;, or whatever yshift you like. And have access to many other things such as decorated lines.
– user232027
Jan 1, 2021 at 1:20 | 898 | 2,364 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-21 | latest | en | 0.496112 |
https://www.shaalaa.com/question-bank-solutions/a-patient-hospital-given-soup-daily-cylindrical-bowl-diameter-7-cm-if-bowl-filled-soup-height-4-cm-how-much-soup-hospital-has-prepare-daily-serve-250-patients-volume-of-a-cylinder_7065 | 1,695,877,362,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00549.warc.gz | 1,062,740,577 | 9,539 | # A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients? - Mathematics
A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
["Assume "pi=22/7]
#### Solution
Radius (r) of cylindrical bowl = (7/2)cm = 3.5cm
Height (h) of bowl, up to which bowl is filled with soup = 4 cm
Volume of soup in 1 bowl = πr2h
=(22/7xx(3.5)^2xx4)cm^3
= (11 × 3.5 × 4) cm3
= 154 cm3
Volume of soup given to 250 patients = (250 × 154) cm3
= 38500 cm3
= 38.5 litres.
Is there an error in this question or solution?
Chapter 13: Surface Area and Volumes - Exercise 13.6 [Page 231]
#### APPEARS IN
NCERT Class 9 Maths
Chapter 13 Surface Area and Volumes
Exercise 13.6 | Q 8 | Page 231
Share | 308 | 983 | {"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} | 4.3125 | 4 | CC-MAIN-2023-40 | latest | en | 0.909392 |
http://mathhelpforum.com/advanced-algebra/176126-automorphisms.html | 1,511,159,409,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805914.1/warc/CC-MAIN-20171120052322-20171120072322-00699.warc.gz | 183,326,427 | 10,605 | 1. Automorphisms
This one is mostly a problem with the definitions, I think.
I am asked to prove that $\text{Aut} \mathbb{Z}$ is isomorphic to $\mathbb{Z}_2$, where $\text{Aut} \mathbb{Z}$ is the group of all automorphisms of $\mathbb{Z}$.
Obviously the identity function $I: \mathbb{Z} \to \mathbb{Z}: x \mapsto x$ is in $\text{Aut} \mathbb{Z}$. The only other automorphism I can come up with is $f: \mathbb{Z} \to \mathbb{Z}: x \mapsto -x$. etc, etc. to finish showing the isomorphism between $\text{Aut} \mathbb{Z}$ and $\mathbb{Z}_2$.
Are there truly only two members in $\text{Aut} \mathbb{Z}$? It seems there should be more, but I can't find any.
-Dan
2. You need to show that these are the only automorphisms of $\mathbb{Z}$. Use the fact that automorphisms map generators to generators. Now, what are the generators of $\mathbb{Z}$?
3. Originally Posted by ojones
You need to show that these are the only automorphisms of $\mathbb{Z}$. Use the fact that automorphisms map generators to generators. Now, what are the generators of $\mathbb{Z}$?
Thank you. I was unaware of that fact. (Edit: It's pretty obvious, actually, now that I've had some time to think about it.) To finish the argument then, since there are only two generators of $\mathbb{Z}$ ( -1 and 1) there are only two automorphisms that are in $\text{Aut} \mathbb{Z}$. Since $\text{Aut} \mathbb{Z}$ is a group of two members it must be isomorphic to $\mathbb{Z}_2$.
-Dan
4. Yes, this will do it provided you know how to fill in the details. | 459 | 1,520 | {"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": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.907741 |
https://www.liquidimageco.com/how-much-do-you-win-on-cash-3-straight-box-ga/ | 1,719,126,938,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00076.warc.gz | 764,752,565 | 26,934 | # How much do you win on Cash 3 straight box GA?
In the state of Georgia, winning on a Cash 3 Straight Box bet involves matching all three of the winning numbers in any order. Depending on the numbers drawn, you can potentially win from a minimum of \$40 to a maximum of \$500.
The exact amount you win on a Cash 3 Straight Box bet depends on the amount wagered and the number of combinations available.
For the Cash 3 Straight Box bet, if you wagered a dollar then all of the three-digit combinations of the winning numbers will be eligible for the prize. If you wagered \$0. 50, then only a portion of the combinations (six numbers making up the three-digit numbers, including the number) will be eligible for the prize.
The exact prize amount that you will win on a Cash 3 Straight Box bet in Georgia depends on the numbers drawn and the amount wagered. For example, if you wagered \$1 on a Straight Box bet on the numbers 437 and those numbers are drawn, then you would win \$250.
However, if the numbers drawn were 473, then you would win \$500. For bets of \$0. 50, the same two numbers drawn would result in wins of \$40 and \$100 respectively.
## What does straight box mean?
The term ‘straight box’ is a slang name for a type of cargo train car that is common in railroad transport. It is also known as a ‘well-car’, as it is designed to have an open bottom with a flat section in the center, designed to hold an entire trainload of cargo as one unit.
It is important for certain types of freight, such as automobiles, which must all be loaded and unloaded at once. Straight box cars are made of two sections, one at each end, that can be quickly and efficiently loaded, unloaded, and turned around.
This makes them highly efficient, cost-effective, and convenient to use.
## What is the most you can win on Pick 3?
The maximum amount you can win playing Pick 3 is \$500. To win the top prize of \$500, you must select the correct numbers in the exact order. The odds of hitting all three numbers in the exact order are 1 in 1,000.
Your bet amount will determine the amount you will win if you match two or three of the numbers. For instance, if you wager \$1 on a Pick 3 ticket, you will win \$80 for matching two numbers and \$500 for matching all three numbers.
## How do you play Georgia Cash 3 combo?
Playing Georgia Cash 3 Combo is a fun and straightforward way to win money! To play, you must pick 3 numbers from 0-9. To make it easier, Georgia offers two ways to pick your numbers—Straight and Box.
Straight:
When you choose Straight, you must pick all three numbers in the exact order that they will be drawn. If your numbers match the three numbers drawn in the exact order, you win!
Box:
If you choose Box, you must pick all three numbers, in any order. If your numbers match the three numbers drawn, it doesn’t matter what order they are drawn in – you’ll still win!
No matter which game type you choose, you can also add an extra wager, called an “1-Off”. All you have to do is choose one number to match plus plus or minus 1 of the numbers drawn to win an additional cash prize.
You also have the option to enter your three numbers in the special “Combo” play field. This allows you to play all your three numbers in combinations with one another, which multiplies your chances of winning.
To play, simply choose 3 numbers from 0-9 and then decide which type of wager you would like to make—straight, box, or combo. You can also play “Quick Pick”, just by asking the clerk or using the self-serve terminals.
Once you have your numbers, all you have to do is wait for the draw!.
Good luck!
## Do you win anything if you have one number?
No, if you only have one number it is unlikely that you will win anything in most lottery or game of chance scenarios. Generally you need to match a specific set of numbers or symbols in order to win a prize.
For example, in the standard 6/49 lottery you would need to match all 6 of your numbers in the same order as the numbers drawn in order to win the jackpot. Similarly in many card games you would need to have a particular combination of cards in order to win the pot.
Therefore, having only one number would not be enough to win any prize.
## How does a Pick 3 bet work?
A Pick 3 bet is a type of wager offered by many race tracks and online sports books. It allows bettors to select three consecutive races in which they must select the winner of each. If all three picks win, the bettor will receive a payout based on the odds set for each selection and the amount of the wager.
Generally, the minimum Pick 3 bet is a \$1 bet and the maximum is typically around \$50 or \$100.
To place a Pick 3 bet, first choose which three races you want to bet on. Each of the races must be consecutive in order for the wager to be valid. On the betting slip, mark the box that says “Pick 3” and then circle the horses you think will win each of the three races.
Once you have done this, you will then need to enter the amount of your wager and submit the bet.
At the end of the three races, if all of your selected horses finish in first place you will be awarded with a payout according to the odds of your selections and the amount of the wager. For example, if you placed a \$2 Pick 3 bet on three horses with odds of 2/1, 4/1, and 8/1 respectively and all three of your horses won, then your payout would be calculated as 2/1 x 4/1 x 8/1 x \$2, which comes to \$64.
The Pick 3 bet is a great option for bettors who are looking to win a large payout with relatively low risk.
## How do you play Pick 3 example?
Pick 3 is a lottery game offered by many state lottery organizations. Players choose three numbers from 0 to 9 to create their wager. To play Pick 3, you can choose a Straight Pick 3, Boxed Pick 3, a Back Pair, or a Front Pair, each of which has different payouts and rules.
A Straight Pick 3 is a wager in which players choose three numbers in exact order. For example, you might choose 5-2-1. To win the Straight Pick 3 wager, the order of numbers drawn must be the same as the order you chose.
A Boxed Pick 3 is a wager in which players choose three numbers, but the order of those numbers doesn’t matter. Going with the same example from the Straight wager, the numbers 5-2-1 could be drawn in any order–2-1-5, 1-2-5, etc.
–and the player would still win.
Back Pair and Front Pair are both types of Pick 3 pari-mutuel wagers. In a Back Pair wager, the player chooses two numbers and the Lottery draws the third number. The player must match the last two numbers in the correct order to win.
For example, if a player chose 10-2 and the Lottery drew 3-2-0. The player would win the Back Pair wager because the last two numbers of the draw are the same (2-0) in the same order as their selection (10-2).
In a Front Pair wager, the player chooses two numbers and must match the first two numbers of the draw in the correct order. For example, if a player chose 12-4 and the Lottery drew 1-2-3, the player would win the Front Pair wager because the first two numbers of the draw (1-2) match the same order of their selection (12-4).
The payouts for a Pick 3 wager depend on the type you choose and where you play. Check with your local Lottery for the exact odds and prices. You can also set your own wager amount when playing Pick 3 and the bigger the wager, the bigger the payouts will be.
## What time are the Cashpop draws in Georgia?
Cashpop draws in Georgia take place every 6 hours starting at 12:00 AM, 6:00 AM, 12:00 PM, and 6:00 PM Eastern Standard Time. During each draw, five winners are randomly selected and awarded varying amounts, typically ranging from \$1 – \$1000 USD.
All times are based on the times shown on Google. It’s important to remember that the times listed may vary slightly depending on the draw.
## How much does the Georgia Cash Pop pay?
The Georgia Cash Pop lottery game pays out prizes based on the number of plays you make and the type of ticket you purchase. Prizes can range from \$1 up to \$5,000. For a \$1 ticket, the top prize is \$500, while the top prize for a \$2 ticket is \$1,000.
The overall odds of winning a prize in the Cash Pop lottery game are 1 in 4. 37, with the odds of winning the top prize depending on the type of ticket you purchase and the number of plays you make.
## What time can you buy lottery tickets in Georgia?
In Georgia, lottery tickets can be purchased at most retail locations between the hours of 7:00am EST and 10:00pm EST. Georgia Lottery’s On the Go mobile app, website, and participating retailers offer a variety of different draw games that you can purchase tickets for, including MEGA Millions, Powerball, Fantasy 5, KENO!, Jumbo Bucks, and Cash 3.
Tickets can also be purchased for instant games like instant keno, instants scratch-offs, and interactive play. To verify the selling times of your local ticket retailers, please visit your local participating Lottery retailer or call the Lottery’s Customer Service hotline at 1-800-GA-LUCKY.
## How do you win on Cash Pop in Georgia?
To win on Cash Pop in Georgia, you’ll need to purchase a ticket for the drawing, either through the Georgia Lottery app or from an authorized retailer. Once you have the ticket in hand, you’ll need to wait until the official drawing, which takes place nights at 11:00 p.
m. During the drawing, 6 numbers will be randomly selected; if your ticket matches all 6 of those numbers, you’ll win the jackpot!.
Cash Pop also grants smaller prizes to players who match 5 or 4 of the drawn numbers. If you match 5 of them, you’ll receive \$500; matching 4 of them will earn you \$25.
Remember, lottery winnings are taxable, so don’t forget to claim your prize and pay your taxes if you do win the jackpot. Good luck!
## How much tax do you pay on a \$1000 lottery ticket in Georgia?
In Georgia, lottery winnings are subject to both state and federal taxes. For winnings \$600, or more, lottery winners must fill out a federal W-2G form, which reports lottery winnings to the IRS. The winner must pay federal income taxes on the lottery amount that is greater than \$600.
The federal rate on lottery winnings is a flat rate of 24%.
At the state level, lottery winnings are taxed at a rate of 4% for amounts up to \$5000. For lottery winnings greater than \$5000, the state tax rate increases to 6%.
So, for a \$1000 lottery ticket, the tax due in the state of Georgia is 4%, or \$40. At the federal level, the tax due would be 24%, or \$240. So the total taxes due on a \$1000 lottery ticket in Georgia would be \$280.
## What is the time to buy a lottery ticket?
The best time to buy a lottery ticket depends on which lottery you are participating in as different lotteries will have different cut-off times for ticket purchases. Generally speaking, it is wise to buy your ticket as soon as possible to ensure that you don’t miss out on the draw.
The sooner you purchase your ticket, the more chance you have of selecting the numbers you would like. Additionally, if you choose to purchase your lottery tickets online, make sure that you do so with plenty of time before the draw to allow for the ticket to be processed and for you to receive your confirmation email.
## What time is the evening lottery in Georgia?
The evening lottery in Georgia is held at 11:00 pm EST on the following days:
• Sunday
• Monday
• Tuesday
• Wednesday
• Thursday
• Friday
• Saturday
You can find the latest lottery results and winning numbers here: https://www.georgialottery.com/.
## Can you buy Georgia Lottery tickets online if you live in Alabama?
No, you cannot buy Georgia Lottery tickets online if you live in Alabama. Georgia Lottery tickets can only be purchased in the state of Georgia with a valid Georgia address. The only way for someone in Alabama to purchase a Georgia Lottery ticket would be to travel to Georgia and buy a ticket in person.
Furthermore, players must be 18 years old or older to purchase lottery tickets in the state of Georgia. | 2,791 | 12,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.951268 |
https://www.unknowncountry.com/headline-news/ancient-babylon-may-have-developed-trigonometry-over-1000-years-before-pythagoras/ | 1,718,296,835,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861480.87/warc/CC-MAIN-20240613154645-20240613184645-00874.warc.gz | 971,647,235 | 22,504 | A new analysis of a 3,700-year-old Babylonian cuneiform tablet suggests that the ancient Babylonians were using an advanced form of trigonometry roughly a millennium before ancient Greek mathematicians recorded what is known as the Pytharoean theorem. In addition to the tablet’s antiquity, the tables inscribed on it also suggest that the Mesopotamians’ approach to this form of mathematics may be superior to the function we use today.
Currently stored at Columbia University, tablet Plimpton 322 (P322) is believed to have been written around 1,800 BCE. It was obtained from antiquities dealer Edgar James Banks in 1922 by New York publisher George Arthur Plimpton, and subsequently given to Columbia University in the 1930s. While the tablet was originally suspected to depict a trigonometric table, this idea was abandoned at some point, with other administrative functions attributed to it.
University of New South Wales (UNSW) mathematician Daniel Mansfield encountered tablet P322 when he was developing a new course for high school math teachers, and was intrigued by its inscriptions: "It took me 2 years of looking at this [tablet] and saying ‘I’m sure it’s trig, I’m sure it’s trig, but how?" The tablet was otherwise missing the sine, cosine and tangent entries requisite for the implementation of modern trigonometry. Rather, it recorded the relationships between short and long sides of the triangle, and of its short side to its diagonal.
Mansfield’s collaboration with fellow UNSW mathematician Norman Wildberger yielded the conclusion that the ancient Babylonians used the exact ratios of the lengths of the right sides of a triangle to express their form of trigonometry, rather than using the angles, as prescribed by Pythagoras in the fourth century BCE.
If this theory is correct, tablet P322 represents the earliest known recording of trigonometric theory, and also represents a more accurate approach to solving trig problems. The Babylonian method uses exact values for the sides of the right triangles represented, while the Greek method relies on extremely close approximations provided by sine and cosine values.
Regardless, the find may well underscore how different the thought processes of an ancient culture might be to what we’re accustomed to today. "This is a whole different way of looking at trigonometry," explains Mansfield. "We prefer sines and cosines … but we have to really get outside our own culture to see from their perspective to be able to understand it."
Dreamland Video podcast | 516 | 2,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2024-26 | latest | en | 0.96111 |
http://www.thescienceforum.com/earth-sciences/19857-properties-ice.html | 1,568,587,514,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00086.warc.gz | 352,072,038 | 16,397 | 1. Total idiot question, not much of a science person.
When it comes to ice, what dictates how high it floats in the water? Is there a standard % of it that sticks out of the top of the water? Would ice cube A float higher than ice cube B if it had a few extra air bubbles in it?
Dont know why this interests me, but anyway.
by the way sry if this is in the wrong section
2.
3. Originally Posted by Sopsop
Total idiot question, not much of a science person.
When it comes to ice, what dictates how high it floats in the water? Is there a standard % of it that sticks out of the top of the water? Would ice cube A float higher than ice cube B if it had a few extra air bubbles in it?
Dont know why this interests me, but anyway.
by the way sry if this is in the wrong section
The exact height will depend upon the density of the water and the density of the ice. As you rightly point out ice may contain pockets of air, which will reduce it bulk density - it wioll tend to sit higher in the water.
the density of the water it is in will be determined by its salinity and its temperature, both of which vary.
4. The resaon ice floats is because of buoyancy. Buoyancy is a force which is governed by Archimedes principle, which simply states that:
The force of buoyancy is equal and opposite to the weight of the displaced fluid.
Consider a block of ice that has a density equal to 90% the density of water, sitting in a pool of water in equilibrium. This situation can be restated: the weight of the ice is balanced exactly by the buoyancy force given back by the water.
By Archimedes' principle, that means that the block of ice displaces its own weight of water.
Now because the ice is 90% the density of the water (meaning that it is 100/90 = 1.111 times bigger per unit of weight), that means that only 90% of the ice need be submerged into the water so that the required weight of water is displaced. The other 10% of the ice does not displace any water: it is seen to float above the water.
Question: what happens to the level of the water once the ice has melted?
5. Originally Posted by billiards
Question: what happens to the level of the water once the ice has melted?
The level remains the same.
6. Originally Posted by Wild Cobra
Originally Posted by billiards
Question: what happens to the level of the water once the ice has melted?
The level remains the same.
Close.
There's be a slight rise due to the fresh water sea ice melting into the ocean because it is less dense therefore holding higher volume than the more dense water it's displacing. Probably not more than a few cm.
7. Originally Posted by Lynx_Fox
Originally Posted by Wild Cobra
Originally Posted by billiards
Question: what happens to the level of the water once the ice has melted?
The level remains the same.
Close.
There's be a slight rise due to the fresh water sea ice melting into the ocean because it is less dense therefore holding higher volume than the more dense water it's displacing. Probably not more than a few cm.
I didn't see the term "sea" or "Arctic ice" in his question anywhere, nor do I agree with your assumed assessment without visiting the math.
Or I could say I didn't know Archimedes experimented with ice in the Arctic Ocean. I thought it was in a laboratory.
8. Nor did his question explicitly assume the composition of the ice and the water were the same. They often aren't, even when the ice originally formed from the water it floats in (sea ice is one example). Obviously this effects density which changes the answer to the question.
9. Originally Posted by Lynx_Fox
Nor did his question explicitly assume the composition of the ice and the water were the same. They often aren't, even when the ice originally formed from the water it floats in (sea ice is one example). Obviously this effects density which changes the answer to the question.
Still, by your assumption, wouldn't such a senario lower the sea level rather than raise it?
Buoyancy has specific known properties. If there is a level change from absorbed salts and other molecules, it seems to me it would decrease rather than increase. I still think it would remain the same though.
What topic would that be under for research?
10. Originally Posted by Wild Cobra
Originally Posted by Lynx_Fox
Nor did his question explicitly assume the composition of the ice and the water were the same. They often aren't, even when the ice originally formed from the water it floats in (sea ice is one example). Obviously this effects density which changes the answer to the question.
Still, by your assumption, wouldn't such a senario lower the sea level rather than raise it?
Buoyancy has specific known properties. If there is a level change from absorbed salts and other molecules, it seems to me it would decrease rather than increase. I still think it would remain the same though.
What topic would that be under for research?
several including oceanography, hydrology, geology, cryology with like many things application to other fields including climatology.
But the concept is simple. Imagine if a block of ice were floating on a sea of liquid mercury. Since mercury is much more dense the most of the ice would be above the level "sea level," and not displace very much. If you melted that ice the combine volume of mercury and water would lead to a higher "sea level."
Floating ice whether by brime extraction from frozen sea water or from calving of glaciers is mostly fresh water. Fresh water has density of about 1000kg/m^3, sea water about 1030 kg/m^3. Floating ice displaced less volume than it contributed once it melts. --
(Anyone can confirm in the comfort of their own kitchen with fresh water ice cubes and their own hand made brimy sea water and a tall graduated measuring cup.)
11. Originally Posted by Lynx_Fox
Originally Posted by Wild Cobra
Originally Posted by Lynx_Fox
Nor did his question explicitly assume the composition of the ice and the water were the same. They often aren't, even when the ice originally formed from the water it floats in (sea ice is one example). Obviously this effects density which changes the answer to the question.
Still, by your assumption, wouldn't such a senario lower the sea level rather than raise it?
Buoyancy has specific known properties. If there is a level change from absorbed salts and other molecules, it seems to me it would decrease rather than increase. I still think it would remain the same though.
What topic would that be under for research?
several including oceanography, hydrology, geology, cryology with like many things application to other fields including climatology.
But the concept is simple. Imagine if a block of ice were floating on a sea of liquid mercury. Since mercury is much more dense the most of the ice would be above the level "sea level," and not displace very much. If you melted that ice the combine volume of mercury and water would lead to a higher "sea level."
Floating ice whether by brime extraction from frozen sea water or from calving of glaciers is mostly fresh water. Fresh water has density of about 1000kg/m^3, sea water about 1030 kg/m^3. Floating ice displaced less volume than it contributed once it melts. --
(Anyone can confirm in the comfort of their own kitchen with fresh water ice cubes and their own hand made brimy sea water and a tall graduated measuring cup.)
I disagree.
The difference is that your example used dissimilar materials. The volume of water, with absorbed brine, has no significant difference with the water without brine. The absorbed brine has a marginal change in volume, if any. I forget how that works, but an example between ice to water in mercury is no comparison to the question at hand.
12. The volume of water, with absorbed brine, has no significant difference with the water without brine.
? Dude I even gave you the difference in densities. It's an easy to verify fact. (look it up or do the kitchen thing). The 3% difference is pretty significant and for many things including driving a lot of the oceanographic ocean currents.
Anyhow back to sea level; A few folks have already done the work summed up the floating sea ice and concluded the difference would amount something like a to 4-6 cm average rise across the globe. Here's one extract.
http://onlinelibrary.wiley.com/doi/1...472.x/abstract
There are probably other sources if you dig around.
13. Originally Posted by Lynx_Fox
Here's one extract.
http://onlinelibrary.wiley.com/doi/1...472.x/abstract
There are probably other sources if you dig around.
Ans the water floats higher. Maybe what I'm missing is how you are allowing for displacement to change a level. I don't know the term for what I am thinking about, but when a fluid absorbs a salt, there is far more density change than volume change.
What happens if I put 100 grams of NaCl in a liter of water? Does the level change once it's dissolved? maybe it does. I simply forget. I don't think it changes my any significant amount if it does. Now when the salts are distributed to the new fresh water, then even is salinity affects volume, doesn't this balance it out?
Now as I think about it, I think you are correct. If the ice melted, it would mix with warmer water and become less dense that way. I don't have access to the material, but does it go one with the average ocean temperature increasing as a result of no ice?
14. I just found that article online that at a site that doesn't require a subscription or payment to read it. I am reading it now. Still, a 4 cm rise is minuscule.
15. Ok, I see what I missed. Thank-you for the article. I actually learned something today. I even saved if for future use.
For those wanting to know where the "free" file is:
The Melting of Floating Ice Raises the Ocean Level
Peter D. Noerdlinger & Kay R. Brower
1St. Mary’s University, Department of Astronomy and Physics
16. It lowers sea-level. An ionic radius is larger than a hydrogen bond.
17. Originally Posted by Geo
It lowers sea-level. An ionic radius is larger than a hydrogen bond.
Then how do you explain the experimental results in the last couple links posted? It appears you are thinking pure water, as my original answer was. That however is accounted for in the buoyancy.
Now on a technicality, I would say that if you could control which of the three forms of water the H2O is using by MO theory makes a difference also, and the angle between hydrogen atoms changes too. but it seems to be another discussion.
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https://taxguru.in/finance/financial-accounting-ratio-analysis.html | 1,532,188,086,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592636.68/warc/CC-MAIN-20180721145209-20180721165209-00420.warc.gz | 781,991,912 | 25,782 | # Financial Accounting and Ratio Analysis
J.V. GURURAJ
Financial Accounting is now a days a very deep rooted subject. It has only two wheels – Debit and Credit. The whole system of accounting is completely rested on these two wheels. It has got such a scope these days that even it is impossible to imagine our economy without the use of financial accounting. From Petty Pan Shop to a billionaire company needs accounting. The accountants do play a vital role in maintaining and reporting the accounting system.
A Ratio is a statistical Yard stick to measure relationship between two accounting figures. It is used mainly to analyse trend and static of business. It is very useful for the investors, suppliers, creditors. It is used as a mirror to reflect financial status of a particular company or group of company, firms, business houses. The manager or owner of the company use the tool to rectify or diversity or fine tune their transactions in accordance with ratio pointing analysis. The investor will think of investing or disinvesting in the company based on the positions revealed by ratio analysis. The suppliers will know the position of turn over trend and the creditors will know the profit ability to get their return.
Classification of Ratios : The Classification of ratio is based on the statement from which the ratios are calculated. There are three categories:
I) Balance Sheet Ratio : Balance Sheet is the product of accounting system – extracting all debits and all Credits naming as Assets and Liabilities accordingly. Balance Sheet always reveals the true figures which will be final and moves further to next continuation. The Manager or owner or investor or creditors etc. will use ratios of the followings to analyse whether the company is potential or not.
a) Working Capital Ratio or current ratio : Simple formula is current assets divided by current Liabilities eg : 100/50 = 2 : 1 double assets to single liability. Bankers always prefer this.
b) Liquid Ratio or Quick Ratio or Acid Test Ratio : The formula is : Quick current assets / Quick current Liabilities 100/100 = 1 : 1 equal assets and equal quick liabilities Quick Current Assets are : Cash bills receivables, debtors , quick current liabilities are : Short Term Creditors , Hand Loans, Usls etc. usually proprietors prefers this.
c) Return on proprietory fund Ratio : This shows return on proprietors capital . The formula is Net Profit after tax x 100/ proprietors capital = NP 100 X 100 /102 =98.03 return usually investors or proprietors prefers.
d) Proprietor fund ratio : This ratio actually shows how much proprietory fund has earning capacity over fixed assets.
Formula : Fixed Assets / Proprietary Fund = 100/100 = 1: 1
II) The second Category of ratio is Profit and Loss account : The financial accounting before extracting Balance Sheet will first extract Mfg. trg. Profit and Loss A/c. All the debits in Profit & Loss are sucked out by all credits and final by products is Net Profit. In trading and Manufacturing account we know exact T.O., Sales, purchases stock , G.P. There are five main types of ratios based on Profit and Loss .
a) Gross Profit Ratio : Usually Gross Profit constitutes the difference between cost of goods brought in and the cost at which they are sold. The suppliers of the goods will first see the G.P. Ratio to get their money back. Formula is G.P. x 100 / Sales TO. If G.P. is 10000/- and sales is 100000 then 10000 x 100/100000 = 10%.
b) Net Profit Ratio : Net Profit is out come of G.P. – Expenses. All indirect expenses are deducted or debited in accounting system. The more percentage of N.P. shows healthy earning capacity of the company. The Bankers or the investors mainly attracted by this. The formula is NP x 100/10 . If TO is 5000000 N P is 5,00,000 = 5,00,000 x 100/50,00,000 = 10%.
c) Expenses Ratio : The Selling expenses can be controlled by analyzing P & L by this method. It is very useful for the managers of the company to curb unnecessary expenditure. The formula is Expenses x 100 / Total Sales.
If Sales is 1,00,000 and expenses is 1000 :
1000 x 100 / 100000 = 1%.
d) Inventory to Ratio : It is computed by dividing the cost of sales by average inventory of the period. A low ratio indicates slow moving of inventory. Formula : Avg. Inventory / cost of sales . If inventory is 20 cost of sale is 100 = 20 / 100 = 1 : 50 or 20%.
e) Operating Ratio : This indicates relation between operating profit and sales. It is worked out by dividing operating profits by net sales. We can Judge the managerial capability. The high percentage proves the efficient management.
III. The Third category of ratio is Balance Sheet and Profit and Loss Statement Ratio : It is calculated by referring both Balance Sheet and Profit and Loss.
a) Return on total resources Ratio : This ratio is an indicator of the earning capacity of the total resources employed in the business. Total resources not only includes share capital but also includes fixed liabilities i.e. borrowed money, reserves un distributed profits etc. It is calculated as : Total resources x NP /Capital employed. If total resources is 100, NP 80, Capital is 100 = 100 x 80 / 100 = 80% NP to total resources.
b) Return on Capital Ratio : This is calculated in the same way as ( a) is calculated. Here Capital can include of a proprietory level where there are no reserves made obligatory. Any how while calculating this ratio we should see capital employed and share capital, P. Shares, borrowed liabilities, reserves. The formula is : Capital employed = NP x 100 / Capital employed.
c) Turn Over of Fixed assets ratio : This ratio expresses the number of times fixed assets are being Turn Over in a stated period. It is calculated as : Fixed assets ratio = Sales / Net Fixed assets ( Less Depreciation).
For eg: Sales is 100 net fixed asset is 80 = 100 / 80 = 1.25 : 1 The ratio shows how well the fixed assets are being used in the business. The lower the ratio shows that fixed assets are inefficiently used in business.
d) Turn Over of Debtors Ratio : This ratio measures the accounts receivables in terms of number of days of credit sales during a particular period . This ratio is calculated as :
Debtors Turn Over ratio = Net credit sales / Average Debtors
The ratios can be calculated in various forms like:
Pure ratios: which are arrived at by simple division of one number by another i.e. current assets to current liabilities.
Rate: which is the ratio between two numerical factors. it is expressed over a period of time i.e. stock turnover is three times a year
Percentage: which is a special type rate expressing the relation in hundredth. i.e. GP is 25% of sales
In conclusion we can say that the accounting system creates life of business; the ratios check up the health of business. To get best advantages, the ratios should be rightly used. If wrongly used ratios may create havoc and financial crisis in the company or business houses. Very much care is needed in applying ratios to different purposes especially in bigger investments, lending decisions from banks, fore warning of sickness of the company etc.
Article written by- J.V. GURURAJ ,Accounts & Tax consultant, Ph 9849291424 , Email-gururaj.jv@gmail.com
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### 0 responses to “Financial Accounting and Ratio Analysis”
1. Pramod says:
Great article…
Very easy to understand and very well arranged. Language used is also very simple and easy to understand.
Mr. Gururaj is a very experienced tax consultant with over three decades of experience in this field and he is also my first teacher and I owe my entire accounting knowledge to him. Without his clear explanation about the basic concepts of accounting, I would not have achieved my present status in life.
Thanks once again for everything..
Mr. Srikant Agarwal, for your information:
Debt Equity ratio is calculated by dividing the total external long term debt by the capital funds (External long term Liabilities / Capital funds). This ratio shows us how much external debt the company has for each rupee of its own capital funds. For further information please feel free to write back to me.
2. tahir maniyar says:
sir pls guide me how to become a tax consultant
3. tahir maniyar says:
good information,but i want to know about the project report& stock statement to show the bank against loan | 1,952 | 8,609 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-30 | longest | en | 0.899241 |
https://www.doubtnut.com/qna/644360700 | 1,713,299,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00357.warc.gz | 683,088,393 | 35,874 | # If the eccentricity of an ellipse is $\frac{4}{9}$ and the distance between its foci is 8 units, then find the length of the latus rectum of the ellipse.
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## It is given that $e=\frac{4}{9}$ and distance between the foci of the ellipse i.e., 2c = 8, i.e., c = 4. We know that c = ae. $⇒4=a\left(\frac{4}{9}\right)$ $⇒a=\frac{36}{4}$ $⇒a=9$ Now, ${b}^{2}={a}^{2}\left(1-{e}^{2}\right)$ $⇒{b}^{2}=81\left(1-\frac{16}{81}\right)=81×\frac{65}{81}=65$ Now the length of the latus rectum is given by $\frac{2{b}^{2}}{a}i.e.,\frac{2\left(65\right)}{9}=14.4$ units
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ESM3A HW11
# ESM3A HW11 - Jacobs University Bremen School of Engineering...
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Unformatted text preview: Jacobs University Bremen School of Engineering and Science Peter Oswald Fall Term 2010 120201 ESM3A — Problem Set 11 Issued: 23.11.2010 Voluntary submission: Monday 6.12.2010 (in class) This homework deals with material on stochastic processes. Its submission is voluntary (submis- sion is another chance to improve on your homework percentage). Doing it (and more problems from the literature) is however strongly recommended. Read carefully : If you submit, don’t submit solutions to more than 6 problems. To get full credit, you have to correctly solve at least 5 problems, among them at least two problems from problems 11.1-4, and at least two problems from 11.5-8. Problems on counting processes (Pois- son, queuing systems, and alike) and Markov chains (random walk on graphs, ergodic theorem) will be on the final, solve more problems to practice! Calculations can be done by software, state what you have computed! 11.1. In a small IT company, there are three servers. They operate properly (independendently of each other, and independently of whether they have been previously repaired or not) for a exponentially distributed random time with parameter λ = 1. When a server fails, it is repaired, the maintenance time is uniformly distributed (independently of the server to be repaired) in a contractually agreed upon service time interval t min ≤ t ≤ t max . Denote....
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• Spring '11
• Prof. Dr. Peter Oswald
• Probability theory, Stochastic process, Poisson process, Markov chain, Random walk, Compound Poisson process
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Ask a homework question - tutors are online | 515 | 2,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-09 | latest | en | 0.871631 |
https://tutel.me/c/physics/questions/88145/why+are+rockets+so+big | 1,575,716,591,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540499389.15/warc/CC-MAIN-20191207105754-20191207133754-00237.warc.gz | 600,025,026 | 22,379 | #### [SOLVED] Why are rockets so big?
I'm curious why rockets are so big in their size. Since both the gravitational potential one need to overcome in order to put thing into orbit, and the chemical energy burned from the fuel, are proportional to the mass, so if we shrink the rocket size, it would seem to be fine to launch satellites. So why not build small rocket say the size of human? I can imagine small rocket would be easier to manufacture in large quantities and easier to transport. And maybe someone can make a business out of small rocket, carrying one's own satellite.
TL;DR: This answer arrives at roughly the same conclusion as Kyle Kanos' answer, i.e. in addition to payload considerations, the difficulty lies in stuffing a small rocket with a mass of fuel exceeding the mass of the rocket itself. This answer, however, is more rigorous in how the $\Delta v$ budget is treated.
The rocket equation:
Consider the Tsiolkovsky rocket equation, which describes the motion of vehicles that propel themselves by expelling part of their mass with a certain velocity. A simplified version which only takes (constant) gravity and thrust into account is given below:
$$\Delta v(t) = v_e \cdot \ln \frac{m_0}{m(t)} - g\left(\frac{m_f}{\dot m}\right)$$ where $v_e$ is the effective exhaust velocity, $m_f$ is the mass of the fuel aboard, $\dot m$ is the the mass burn rate (constant with respect to time), $m_0$ is the the initial mass of the rocket and $m(t)$ is the current mass of the rocket.
Note that this is essentially a momentum exchange equation: you have a finite amount of momentum available from expulsion of fuel, which you must spend on increasing the velocity of the rocket + remaining fuel system, as well as overcoming gravity (i.e. dragging the planet ever so slightly). A form of the Tsiolkovsky equation that does not take this into account (as in the other answer) will give you non-physical results.
Constrained variables:
Now, what can we play with in this equation? Assuming $t_{escape}$ is the time at which the rocket escapes Earth's gravity:
1. $\Delta v(t_{escape})$ is simply our desired escape velocity (assuming the rocket starts from rest), which is dictated by where we're trying to send the rocket
2. $m(t_{escape})$ will optimally be the mass of the rocket without any fuel
3. The effective exhaust velocity $v_e$ and the rate of mass flow $\dot m$ are a function of the type of engine/propellant available
This means none of these quantities are negotiable; we are constrained by the demands of the mission and the available technology.
Developing a relationship between rocket and fuel mass:
All we are left to play with is the initial masses of the rocket fuel $m_f$ and rocket body $m_r$. Let us substitute in the values of $v$ and $m$ at the instant when the rocket escapes gravity, noting that $m_0 = m_f + m_r$:
\begin{align} v_{escape} & = v_e \cdot \ln \frac{m_f + m_r}{m_r} - g\left(\frac{m_f}{\dot m}\right)\\ & = v_e \cdot \ln\left(1 + \frac{m_f}{m_r}\right) - g\left(\frac{m_f}{\dot m}\right) \end{align}
Rearranging, we have:
$$m_r = m_f \cdot \left(\exp\left(\frac{v_{esc} + g\left(\frac{m_f}{\dot m}\right)}{v_e}\right) -1\right)^{-1}$$
Note that this is effectively providing $m_r$ as a function of $m_f$, since all the other parameters are fixed by the constraints of the mission and equipment as well as environmental constants. Since the relationship isn't immediately obvious, here is a plot of $m_r$ against $m_f$ for selected values of the constants:
In red, we have a plot of rocket mass versus initial fuel mass, while in blue we have a plot of the ratio of initial fuel mass to total mass. Note that the axis for the blue plot starts at 0.9!! This indicates that regardless of what rocket mass you picked, the net initial mass of your vehicle would have to consist almost entirely of fuel.
So what does this mean?
Filling a vehicle with a mass of fuel exceeding its own is increasingly difficult for small rockets, but not so difficult for much larger rockets (think of how the enclosed volume of a hollow body scales versus mass). This is why making smaller and smaller rockets becomes progressively more difficult.
In addition, a minimum limit on the rocket mass we can choose is imposed by the weight of the payload it must carry, which could be anything from a satellite to a single person.
A very interesting thing happens near the inflection point of the rocket mass - fuel mass curve. Before the inflection point, adding more fuel allowed us to hoist a larger payload to the desired velocity.
However, somewhere around $4 \cdot 10^6$ kg of fuel mass (for our selected parameter values) we discover that adding more fuel starts to decrease the payload that can be hoisted! What is happening here is that the cost of the additional fuel having to fight against gravity begins to win out against the benefit of having a high fuel to payload mass ratio.
This shows there is a theoretical upper limit to the payload that can be hoisted on Earth using the propellant technology we have available. It is not possible to simply keep increasing the payload and fuel masses in equal proportion in order to lift arbitrarily large loads, as would be suggested by using the Tsiolkovsky equation with no extra terms for gravity.
#### @dmckee 2018-06-07 20:16:51
Comments are not for extended discussion; this conversation has been moved to chat.
#### @fibonatic 2013-11-28 02:35:51
Because most payloads are quite heavy. I am not sure what kind of payloads you had in mind, I am no expert on this, but I think that most launches contain satellites, which might be heavier then you think, for instance the satellite in this BBC Documentary weighs 6000 kg. And according to Wikipedia, miniaturized satellites weigh less than 500 kg (so heavier is normal). And some of those miniaturized satellites are using excess capacity on larger launch vehicles.
And I think that smaller rockets will experience the turbulence of our atmosphere much violently. Also think of the relatively higher costs in terms of personnel (such as mission control). And I would also expect that certain aspects do not scale linearly in size, but for be this would just be speculation. xxxxxx
#### @Kyle Kanos 2013-11-28 03:46:54
The problem is what Konstantin Tsiolkovsky discovered 100 years ago: as speed increases, the mass required (in fuel) increases exponentially. This relation, specifically, is $$\Delta v=v_e\ln\left(\frac{m_i}{m_f}\right)$$ where $v_e$ is the exhaust velocity, $m_i$ the initial mass and $m_f$ the final mass.
The above can be rearranged to get $$m_f=m_ie^{-\Delta v/v_e}\qquad m_i=m_fe^{\Delta v/v_e}$$ or by taking the difference between the two, $$M_f=1-\frac{m_f}{m_i}=1-e^{-\Delta v/v_e}$$ where $M_f$ is the exhaust mass fraction.
If we assume we are starting from rest to reach 11.2 km/s (i.e., Earth's escape velocity) with a constant $v_e=4$ km/s (typical velocity for NASA rockets), we'd need $$M_f=1-e^{-11.2/4}=0.939$$ which means almost 94% of the mass at launch needs to be fuel! If we have a 2000 kg craft (about the size of a car), we would need nearly 31,000 kg of fuel in a craft that size. The liquid propellant has a density similar to water (so 1000 kg/m$^3$), so you'd need an object with a volume of 31.0 m$^3$ to hold it. Our car sized object's interior would be around 3 m$^3$, a factor of 10 too small!
This means we need a bigger craft which means more fuel! And explains why this mass-speed relation has been dubbed "the tyranny of the rocket problem".
This also explains the fact that modern rockets are multi-staged. In an attempt to alleviate the required fuel, once a stage uses all of its fuel, it is released from the rocket and the next stage is ignited (doing this over land is dangerous for obvious reasons, hence NASA launching rockets over water), and the mass of the craft is lowered by the mass of the (empty) stage. More on this can be found at these two Physics.SE posts:
The Tsiolkovsky equation in the form you have stated only applies when the net external force is zero (i.e. no gravity). To accurately calculate the $\Delta v$ required, you need to include an additional term $-g(\frac{m_{propell}}{\dot m})$ on the right hand side of the equation.
#### @Kyle Kanos 2013-11-28 04:19:12
@Asad: this is true, but I think it's (mostly) irrelevant to the point that we still need a boat-load of propellant to get ourselves into space, hence large rockets and not person-size ones.
@KyleKanos Yes, the gist of your answer is correct. I was taking issue with the calculation you added, which is flawed. Either you need to consider an effective $\Delta v$ which is augmented to approximately account for the retarding effect of gravity as well as the required escape velocity (this is the standard approach) or actually do the calculation taking fuel burn time into account.
#### @fibonatic 2013-11-28 04:55:12
@Asad It might have been easier if Kyle Kanos would have use the $\Delta v$ budget needed to get into low Earth orbit, which is about 9.3 - 10 km/s, but this would still return about the same result.
@fibonatic The delta v budget you are quoting is only a reasonable approximation for rockets with similar burn time to actual rockets. Since this question is specifically about rockets that can be very small, using the gravity inclusive delta v budget for a large rocket will yield poor results.
#### @BЈовић 2013-11-28 07:18:45
holy crap! that means they are burning oil like it is nothing, just to get some junk up there. And they did it lots of times. What kind of maniac is needed to form nasa?!?
#### @MSalters 2013-11-28 08:45:38
@BЈовић: They usually don't burn oil, it's not efficient enough. But fuel actually isn't that expensive. It's often just a few % of launch costs.
#### @AJMansfield 2013-11-28 19:51:57
@BЈовић For a better idea of what type of fuel gets used, see the Wikipedia pages for Solid-fuel Rockets and for Liquid Rocket Propellants.
#### @David Hammen 2014-11-11 09:22:27
@MSalters - They oftentimes do burn oil. The first stage of the Saturn V rocket used RP-1, a highly refined kerosene, to launch men to the Moon. RP-1 with liquid oxygen as the oxidizer is very widely used as a propellant.
#### @Kyle Kanos 2017-01-25 11:02:08
Might the few downvoters mention what they think is wrong with this answer?
#### @user77220 2017-08-21 00:01:59
"as speed increases, the mass required (in fuel) increases exponentially", this is wrong assumption. It depends on the type of fuel that are used for rocket propulsion. Are you talking about liquid fuel, solid fuel or solar fuel. All have different capacities.
#### @my2cts 2019-06-21 16:47:24
Kyle I found my answer at physics.stackexchange.com/questions/487194/… deleted . I formally object. The reason was that not enough explanation was given, only a link to arxiv.org. You know that arxiv.org is a very stable site. It has been consistent for over 20 years. Also I should be given the opportunity to explain my answer in more detail. Please undelete. Apologies for the cross post. I saw no other means.
#### @jokoon 2013-11-28 10:37:05
Mainly because you need a lot of speed to go into space, and to each that speed, you need to accelerate. If you need a high speed, you will need to accelerate for a long time, thus the need for a large quantity of fuel. You also need to compensate for gravity the whole lift.
There are ways to reduce that fuel requirement, like a horizontal takeoff, you reach a high altitude and then launch, so you keep the engine, but you still need a lot of energy to fight against gravity, and wings can't lift you very high, so that would not be such a good fuel economy, and the plane would still require to be quite big.
#### @user34882 2013-11-28 08:09:27
$E = mc^2$
The larger the mass, the more energy can be produced. And we still haven't found any fuel which in small quantities gives the needed amount of energy. I know you will be thinking of nuclear energy; we cannot fit a nuclear reactor inside a rocket with current technology, and even if we can fit it I don't think our existing knowledge of nuclear science is sufficient to ensure accident-free reactors at such velocities.
#### @a CVn 2013-11-28 08:44:24
$E=mc^2$ doesn't really apply here. First, I'm not aware of any practical matter-energy conversion process that comes anywhere near close to that (insofar as I know we still haven't figured out how to build matter/antimatter reactors for power generation purposes, and that'd be about the only way to get anywhere near such amounts of energy). Second, if you look at the rocket equation cited in other answers, you'll see that the critical issue is the exhaust velocity. If you can get insane exhaust velocities, each tiny nugget of fuel packs a lot more punch in terms of total system $\Delta v$.
#### @fibonatic 2013-11-28 14:15:07
We could use the propulsion similar to that of project Orion, but this probably will not be used at take-off due to the nuclear fallout.
#### @a CVn 2013-12-02 13:36:03
@fibonatic ...and the fact that you need to worry about nuclear fallout is a pretty good indicator to begin with that you aren't in $E=mc^2$ territory.
#### @jean 2018-06-07 12:35:28
We can put it on an airplane en.wikipedia.org/wiki/Nuclear-powered_aircraft
#### @Luke Burgess 2013-11-28 02:22:13
Consider the problem in the from of a ratio, what is the ratio of mass used to lift the rocket(fuel), to the mass finally put into orbit(cockpit). That proportion will be much the same regarding smaller objects that must be put into orbit. If you use the same ratio or proportion to calculate the needed fuel mass for a small craft, you will find you can't even carry the device holding your fuel. This is also why rockets use stages.
The type of fuel used also has an impact, but those are details that need a new question.
#### @lurscher 2017-12-07 14:33:31
this is the correct answer. Also, you need to factor the fact that atmospheric drag grows as the square power of width, while total fuel mass grows with the third power, even assuming constant fuel to dry mass ratio
### [SOLVED] Why are rockets launched vertically?
• 2015-09-14 15:43:22
• Criesto
• 10746 View
• 2 Score
• Tags: rocket-science
### [SOLVED] Why do rockets accelerate fastest horizontally?
• 2012-06-03 18:36:45
• callum
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• 4 Score
• Tags: rocket-science
### [SOLVED] Why are there more vertical takeoff than horizontal for spacecrafts?
• 2012-04-28 10:49:00
• jokoon
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• Tags: rocket-science | 3,651 | 14,619 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-51 | longest | en | 0.948444 |
http://www.physicsforums.com/showthread.php?t=517739 | 1,386,970,555,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164997874/warc/CC-MAIN-20131204134957-00014-ip-10-33-133-15.ec2.internal.warc.gz | 482,199,871 | 9,551 | # Photoelectric Effect ~ threshold frequency dependent on incident light?
by ldesai149
Tags: incident light, photoelectric effect, threshold frequency
P: 3 1. The problem statement, all variables and given/known data True/False: In the photoelectric effect, the cut-off (threshold) frequency depends on the intensity of incident light. 2. Relevant equations hv = W + 1/2mv^2 v = W/h 3. The attempt at a solution The threshold frequency depends on the wavelength of incident light but not the intensity (??)
PF Patron
HW Helper
Thanks
P: 25,528
Hi ldesai149!
Quote by ldesai149 … not the intensity (??)
That's right
the whole point of the experiment is that even if you reduce the intensity so that only one photon goes through at a time, that one photon still needs to have the threshold energy (or frequency)
The threshold frequency depends on the wavelength of incident light …
no, the threshold frequency is fixed …
you might as well say "the threshold frequency depends on the frequency of incident light"
or "the minimum height for joining the army depends on the person's height"
P: 60 I am a bit unclear on this process too. I never understood why an electron could not absorb the energy from several photons at the same instant and gain the required energy this way, rather than having to absorb from just a single photon at a time. Thanks :-)
P: 68
## Photoelectric Effect ~ threshold frequency dependent on incident light?
one can't think about the photoelectric effect in terms of classical physics where the intensity of light is proportional to the square of the amplitude of the wave.
In QM, one deals with light as individual packets of light. Imagine, one packet of light is able to interact with and excite one and only one electron. So, if it's energy ain't high enough (frequency below threshold freq) , it will not be able to eject the electron! However, increase in the intensity of light would just imply more number of incoming photons, and thus more electrons getting excited.
;)
Mentor
P: 27,580
Quote by epsilonjon I am a bit unclear on this process too. I never understood why an electron could not absorb the energy from several photons at the same instant and gain the required energy this way, rather than having to absorb from just a single photon at a time. Thanks :-)
The cross section for such absorption is extremely small (look up multiphoton photoemission). This means that it is of very low probability of happening. By the time the next photon comes along at the right location, there's a very good chance that the electron has decayed back to the lower energy state in the conduction band (the lifetime in the excited state in a metal is of the order of femtoseconds). So using ordinary light, as opposed to high-intensity laser, the intensity is too low to cause multiphoton photoemission. So such photoelectric effect using more than just one photon doesn't normally occur.
Zz.
P: 3,069 Hi Idesa149!! Did you completely check your equations? You can write "hv=W+1/2v^2" as:- "hv=hv0+1/2mv^2" Now you see, the threshold frequency(v0) depends on the frequency of incident light. And as tiny-tim stated, threshold frequency is fixed for a particular metal. That is, if you incident a light with a lower frequency and that too with a large intensity, a single elctron wouldn't be emitted.
P: 96 (1) Threshhold frequency is the characteristic of the metal, it does not depend on the radiation in any way. (2) If you mean that why does the photoelectric effect's occurence itself depend on the frequency of the radiation/ incident light and not on the intensity, then it is a correct statement. (3) This is because the minimum energy needed to knock-out an electron is fixed, i.e. the threshhold energy, corresponding to the threshhold frequency, and one quanta of energy and the energy of one photon of radiation is fixed, depending upon it's frequency only. So, even if you "throw" as many photons of radiation as you can, electrons wont be emitted, as not even one of those photons has sufficient energy to take out an electron. (4) Intensity is related to the number of such photons, of equal energy. The more the intensity/brightness, the more the number of photons. However, as said before, even if you have a very intense light but of insufficient frequency, photons wont be emitted. Think of it in this way: "The battle is one-on-one. So even if you have a large army of weak soldiers against the strongest soldier from the opposite kingdom, the larger army loses. Why, because none of them individually matches the stronger soldier...."
Related Discussions General Physics 0 General Physics 3 Introductory Physics Homework 1 Introductory Physics Homework 3 General Physics 5 | 1,050 | 4,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.929022 |
http://mathhelpforum.com/advanced-algebra/145053-homomorphism-ring.html | 1,480,916,560,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541525.50/warc/CC-MAIN-20161202170901-00180-ip-10-31-129-80.ec2.internal.warc.gz | 176,374,167 | 11,599 | 1. ## Homomorphism ring
This question does not let me sleep.
Fina all Ring homomorphism between Z_n----> Z_m.
I know there should be 3 case:
n>m
n<m and m=n.
the last one is trivial. Anyways, how do you guys think I should proceed or what should I consider?
2. Somehow, I suspect the answer to be gcd(n,m), because I read it somewhere.But I actually want to have an idea of how to get there.
3. Originally Posted by karlito03
Somehow, I suspect the answer to be gcd(n,m), because I read it somewhere.But I actually want to have an idea of how to get there.
Hint:
Spoiler:
If $\theta:\mathbb{Z}_n\to\mathbb{Z}_m$ is a homomorphism then the homomorphism is entirely determined by $\theta(1)$
4. I sure ..
call theta (f).
Then f(1)=1 since you assume it was a ring homomorphism. But how can we get to the point of saying that all homomorphism must be determined by f(1). There must be some cases. For instance m>n, n>m or m=n or Don't m and n have to be such that (n,m)=1. How many homomorphism can one get in each case. Are you saying they will all be determined by f(1)?
5. if n=m then image of any ring Zn-->Zm is determined by the image of 1 mod (n) where m=n.
now if n>m Zn--> Zm need not to be an homomorphism.
6. Originally Posted by karlito03
now if n>m Zn--> Zm need not to be an homomorphism.
I'm not sure what you mean here? Why isn't $f:\mathbb{Z}_n\to\mathbb{Z}_m:z\mapsto 1_m$ not a homomorphism?
7. Hint 1: put $\theta(1+n \mathbb{Z})=r+m\mathbb{Z}$ and extend $\theta$ linearly to all $\mathbb{Z}/n\mathbb{Z}$. now look at the conditions that will make $\theta$ well-defined and multiplicative.
Hint 2: using the above, show that $\theta$ is defined by $\theta(x+n\mathbb{Z})=rx + m\mathbb{Z},$ where $r$ satisfies the following conditions: $\frac{m}{\gcd(n,m)} \mid r$ and $m \mid r^2-r.$
8. i dont get the solution for this question. can someone explain it to me? thanks | 576 | 1,894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-50 | longest | en | 0.892692 |
http://www.bochenscywykleci.pl/tfdsb0xk/c17ee7-logue-vs-log | 1,638,892,218,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00550.warc.gz | 83,845,120 | 16,325 | She was born on December 21, 1965. $499.00/month/user. Mit LOG bestimmt man z.B. 181 Seabreeze Crt, Panama City Beach, FL 32413-7016 is the residential address for Anthony. Logarithm, Antilogarithm. http://www.mathematische-basteleien.de/. function, Referenzen top Log in (two words) should only be used as a verb. wzbw.. Starting Price:$129.00/month/user. button. ... . Rechengesetze top To do this, you'll have to keep an accurate record of all your expenses during the year, in addition to keeping a log book. Travelog is an alternative form of travelogue. Images "I used to run with a bunch back in the day.Became known as the Newcastle Crew, spent most of our misbegotten youth in this place.One of them was a powerful witch." 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The Brits still stick to the old spellings of catalogue and monologue. werden. ...den Randwert x=1 ein, so entsteht die Leibniz-Reihe 1+(1/2)+(1/3)+ ... Das genaue Produkt ist 15367. Es gilt z.B. Kick-off Times; Kick-off times are converted to your local PC time. As nouns the difference between travelog and travelogue is that travelog is (us) while travelogue is a description of someone's travels, given in the form of narrative, public lecture, slide show or motion picture. Compare real user opinions on the pros and cons to make more informed decisions. Mamelodi Sundowns vs Tshakhuma FC Loftus Versfeld 19:30 T. Maritzburg United vs Cape Town City Harry Gwala Stadium 19:30. There are different ways you can collect data about your website visitors. a 1 then b c Series, Wikipedia This involves placing javascript tags in your website code. Page Tagging (cookies) vs. Log Analysis. However, for some special functions, it is much more common that we want to square it than we want to composite it. Rob Wells Rob Wells. wzbw.. Ermittlung Analog , however, is generally used only in contexts concerning electronics, as when a mechanical device is distinguished from a digital one (such as in reference to an analog clock); in the sense of “comparative,” analogue is still more common. Von Bedeutung ist die Logarithmusfunktion, wenn die Basis One way is to analyze the log files your web server creates. Natural Gleichungen, die Logarithmen enthalten, sind Logarithmusgleichungen.In dem Ausdruck log a (x) sind a ≠ 1 und x > 0. Behauptung: Die Ableitung der Funktion mit g(x)=ln(x) Only about 15 words fit in this category. Reagan Logue, a federal prisoner confined in a county jail pending trial, fashioned a noose from a bandage covering a laceration on his left arm and hanged himself. Her age is 55. Logarithmus. Logarithmen, Durch unser eigenes Rechenzentrum uns unsere Erfahung bieten wir einen Katalog… Partnerartikel: Zwischenstand und Handlungsempfehlung zu Citrix NetScaler/ADC Sicherheitslücke CVE-2019-19781. Wählt man für der y-Achse eine logarithmische John Napier, a mathematician, … man nämlich lg(y)=2*lg(x). a, b, c are real numbers and b > 0, a > 0, a ≠ 1 Man kann das Integral ganz rechts n. a portion or length of the trunk or of a large limb of a felled tree. Defintionsbereich und damit auch Wertebereich dargestellt werden kann. When you come on duty, you log on board by making an entry in the log book. ich nicht ein. And be consistent in the info to be logged in a message. Select a program for instant access to great sounds - - ranging from thick basses to deep pads to brilliant polyphonic leads and atmospheric sounds. View the profiles of people named Dom Logue. >Durch die Überschlagrechnung 100*100=10.000 ermittelt Die schwarze Gerade ergibt sich durch Logarithmieren von Logbook definition is - log. >Man stellt den Läufer auf 1.27 der Skala der Körpers View the latest Premier League tables, form guides and season archives, on the official website of the Premier League. Best For: Delogue PLM is designed by and for the apparel and lifestyle industry to deliver a customized experience that … a is called "base" of the logarithm. Login (one word) can be a noun or an adjective. und lg(271)=2,4330 Take your time and compare your top choices and determine which one is right for your company. And the number (x) which we are calculating log base of (b) must be a positive real number. logab = c ⇔ ac = b, where b > 0, a > 0 and a ≠ 1, logab > logac ⇔ if a > 1 then b > c, if 0 < a < 1 then b < c, It shows that when x = 1, log = 0; when x -> 0 => log -> -∞; when x -> ∞ log -> ∞. Provide examples if possible of the different logging levels. Übungen, >Ergebnis: 896*271=242800. Log definition is - a usually bulky piece or length of a cut or fallen tree; especially : a length of a tree trunk ready for sawing and over six feet (1.8 meters) long. Logarithmic Price Scale vs. Join Facebook to connect with Dom Logue and others you may know. Ableitung . Here’s an easy way to remember log in vs. log on. HTH. The Brits still stick to the old spellings of catalogue and monologue. vernachlässigen [(2) Seite 176f.]. This post lists and defines words ending in –logue or –log (with prefixes defined and various inflections provided) and provides more detail below the list about which form to use. The Orientation Log (O-Log) is designed to be a quick quantitative measure of orienta-tional status for use at bedside with rehabilitation inpatients. 2, e (eulersche Zahl) oder 10 ist. logab = logac ⇔ b = c ... +xn-1+xn/(1-x) Jade Logue was born on the 25th of March, 2001 in Brooklyn New York to the famous actor, Donal Logue and his wife/partner, Kasey Smith. The product of x multiplied by y is the inverse logarithm of the sum of log b (x) and log b (y): x ∙ y = log-1 (log b (x) + log b (y)) Logarithm quotient rule. Januar 2020 Ein Partner aus unserem Systemhausnetzwerk, mit dem wir uns im ständigen Austausch befinden, hat … top Linear Price Scale: An Overview . die angenähert als Dezimalbrüche angegeben werden können. Diese werden über konvergente Reihen gewonnen, dann aber in beliebiger Die Skala der y-Achse heißt logarithmische Skala. The other way is often called page tagging. Harmonische ist auf 4 Stellen genau ermittelt worden. Americans have a tough time deciding to write travelog, tho, or cataloging rather than cataloguing. Zur Herleitung Herleitung -logue vs -log. 1/(1-x)=1+x+x2+x3+ This javascript sends the tracking data back to some analytics tool. Definition, Rechtschreibung, Synonyme und Grammatik von '-loge' auf Duden online nachschlagen. log refers to the logarithmic function. Jade Logue was born as Arlo Logue, a boy before she transitioned into a girl and was renamed, Jade. Reihe aus. Trät man lg(y) gegen x auf, so erhält man als ln(2) mit der Tastenfolge 2 LN und erhält ln(2)=0.693147181. Americans have a tough time deciding to write travelog, tho, or cataloging rather than cataloguing. logarithm, But this second sense is probably less common, so we should probably use the first sense in our example—maybe this: “For example, the information you use to sign in to your email is your login (noun), and the page where you … 1. logy - stunned or confused and slow to react (as from blows or drunkenness or exhaustion) Die Reihe ist konvergent für -1<=x<1. erklärt -logue definition, a combining form used in the names of kinds of discourse, spoken or written: analogue; monologue; travelogue. Alternating Alternative names for Anthony: Anthony Logue, Tony Logue, A Logue, Tony W Logue, Anthony W Loque. ... mit dem Konvergenzbereich -1Aus dem Tafelwerk findet man rückwärts 5,3853=lg(242800). Die folgenden Aussagen beziehen sich auf den natürlichen The preposition "up" actually changed the verb in a way that “to” could not. ist g'(x)=1/x. In the computer business, log on was the standard IBM usage in the 1960s, while log in was used in the Univac and later VAX community. Dekadischer See more. It's painful to see a vast variety of log messages where the severities and the selected log levels are inconsistent. She is the second child in the Donal Logue – Kasey Smith union, with an older brother named Finn. If you need a username and password (or other credentials) to access something, you are logging in. It refers to the information you use to log in to something, as you point out, but it also refers to an instance of logging in—e.g., my last login to my email was five minutes ago. Potenz zugeordnet, sondern der Potenz ein Exponent. Logarithmus erfolgreich therapierte. The logue SDK is a C/C++ software development kit and API that allows to create custom oscillators and effects for the KORG prologue, minilogue xd and NuTekt NTS-1 digital kit synthesizers logue SDK Oscillators Effects Unit Index API Reference 日本語 Loge (Etymologie), zu Wortherkunft und -gebrauch kleines, vorn offenes oder zum Hinausschauen eingerichtetes Zimmer, siehe Conciergerie; Loge (Zuschauerraum), abgeschlossener Sitzraum in einem Veranstaltungsraum. Allgemein: log e n = ln n Beispiel: log e 20 = ln 20 ≈ 3 minilogue goes beyond analog synths in its price range by adding 200 preset locations. Place, time, and situation-al (Etiology/Event + Pathology/Deficits) domains are assessed. Specifies that a logging category should not write any messages. Einige Logarithmusgleichungen können durch Verwendungen der Logarithmusgesetze gelöst werden. log a b = log a c ⇔ b = c log a b = c ⇔ a c = b, where b > 0, a > 0 and a ≠ 1 log a b > log a c ⇔ if a > 1 then b > c, if 0 . This difference is quite straight forward. These expenses include fuel, oil, repairs and maintenance, car licence, insurance, wear-and-tear and finance charges or lease costs. it provides the code to log messages. The power to which a base of 10 must be raised to obtain a number is called its log number, and the power to which the base e must be raised to obtain a number is called the natural logarithm of the number. For most of these terms, as shown, the –logue form still prevails, though the shorter alternative to a couple of these words is preferred in American English. Reihe, Euler-Mascheroni Wörterbuch der deutschen Sprache. lg(2) mit der Tastenfolge 2 LOG und erhält lg(2)=0.301029996. Constant, Eric W. Weisstein (MathWorld) Favourite matches will appear first on the scores page and in the live match toolbar. How to use log in a sentence. Natürlich auch als App. The winner is the one which gets best visibility on Google. LOGUE/GOGUE vs LOG/GOG: analogue and analogue. (3) H.v.Mangoldt / K.Knopp: Einführung in die höhere Diese Meldungen enthalten möglicherweise sensible Anwendungsdaten. top Drei Logarithmen top These are considered acceptable to Americans, but a bit dated. Logarithm, Dann ist pq = ah * ak=ah+k Calculate your claim based on actual costs. Skala, wie sie beim Rechenstab auftritt, so werden die Graphen von Exponentialfunktionen analogue and analog; catalogue and catalog; Keep in mind that the change doesn’t affect any other words that end in -gue: tongue, argue… Logge vs Logue - Type 2 keywords and click on the 'Fight !' Your Favourites. She can use alternative names such as Christine M Luehrs, Christine M Macleod, Christine Logue, Christine M Mcleod, Christine Lewis, Christine M Logue. Log log = LogFactory.getLog(CLASS.class); Documentation. Natural Der TI30 z.B. Eine bekannte Reihe ist ln(1+x)=x-(1/2)x²+(1/3)x³- Es genügen wohl vier Dezimalen, so dass gilt: lg(2)=0,3010 und ln(2)=0,6931. If you have any question go to our forum about logarithms. Logarithm, 21. log4j is a logging framework, i.e. log 2 y = x bedeutet: Der Logarithmus von y zu Basis 2 ist gleich x. Ihr müsst euch also folgendes überlegen: Welche Hochzahl x benötige ich, mit der die Zahl 2 potenziert werden muss, damit man y erhält. The fourth placed team is the next best team and is the highest ranked wild card team. Die Potenzgesetze führen zu den Logarithmusgesetzen Is it login or log in? ein. Logarithms, simply speaking are exponents. Common . Es gilt also ln(1-x)=-x-(1/2)x²-(1/3)x³- Recent Examples on the Web The logbook registered only a couple of ergonomic gripes. Logarithm of 2, Cologarithm, Christopher Logue, CBE (23 November 1926 – 2 December 2011) was an English poet associated with the British Poetry Revival, and a pacifist. Current address for Christine is 105 Bruington Crt, Morrisville, NC 27560-6986. (1) F. G. Gauß: Vierstellige logarithmische und Logarithmus, Exponentialfunktion, As a computer geek of more than 40 years’ standing: log in and log on are both nautical metaphors. Taschenrechner. Ergebnisse, K. Eckhardt (Universität Hohenheim) Best For: Delogue PLM is designed by and for the apparel and lifestyle industry to deliver a customized experience that … In science and engineering, a log–log graph or log–log plot is a two-dimensional graph of numerical data that uses logarithmic scales on both the horizontal and vertical axes. der Logarithmen top Log in and login have only seen heavy use since personal computers became ubiquitous in the 1980s, but they are now so common that misusing them in your writing can cost you credibility. If you can access it without credentials (like this site), you are logging on. Browse our Scrabble Word Finder, Words With Friends cheat dictionary, and WordHub word solver to find words that end with logue. Integrallogarithmus, So "speak up" is a phrasal verb. Nach der Produktregel gilt [x*ln(x)-x+C]'=ln(x)+x(1/x)-1=ln(x), Viele übersetzte Beispielsätze mit "logue" – Deutsch-Englisch Wörterbuch und Suchmaschine für Millionen von Deutsch-Übersetzungen. Jutta Gut You can log on to a network drive from any computer connected to the network. lg(100)=2 und lg(0,01)= -2. See more. In der Regel muss ein Ausdruck, der aus mehreren Logarithmen besteht, so umgeschrieben werden, dass nur noch ein Logarithmus vorkommt. der Grundskala der Körpers. Golden Arrows vs SuperSport United Sugar Ray Xulu Stadium 19:30 . Trace 0: Protokolle, die die ausführlichsten Meldungen enthalten. Learn vocabulary, terms, and more with flashcards, games, and other study tools. For example log 2 of 8 is equal to 3. log 2 (8) = 3 (log base 2 of 8) The exponential is 2 3 = 8 Common Values for Log Base. Die Potenzgleichung ah+k =pq führt zu Die Logarithmusfunktion kann also als das Flächenstück, Sehen wir uns noch ein paar Beispiele zum besseren Verständnis an. Monomials – relationships of the form = – appear as straight lines in a log–log graph, with the power term corresponding to the slope, and the constant term corresponding to the intercept of the line. >Die Summe der Logarithmen ist dann 2,9523+2,4330=5,3853. Potenzieren und das Wurzelziehen, allerdings auf Kosten der Genauigkeit. The first spelling when more than one is shown is the prevailing one; the alternative may appear occasionally but is in most cases not considered standard. Patient responses … TS Galaxy FC vs AmaZulu FC Mbombela Stadium 19:30. April 1953 in London) war ein australischer Sprachtherapeut. More news. Man geht von der folgenden geometrischen Wörterbuch der deutschen Sprache. Definition, Rechtschreibung, Synonyme und Grammatik von 'Loge' auf Duden online nachschlagen. Logy definition, lacking physical or mental energy or vitality; sluggish; dull; lethargic. einer feineren Einteilung die Abstände nicht gleich sind. Es wird also bei fester Basis nicht dem Exponenten eine View the profiles of people named Logue Logue. These are considered acceptable to Americans, but a bit dated. das von der x-Achse, dem Graphen der Funktion h(x)=1/x und den Vertikalen ln(1+x) und ln(1-x) Homepage: Logarithmus, These messages may contain … Mit Flexionstabellen der verschiedenen Fälle und Zeiten Aussprache und relevante Diskussionen Kostenloser Vokabeltrainer Es sei h=loga(p) oder ah=p und For example, we could say you logged on to this website simply by visiting this page, but you won’t be logging in because nothing on this site requires a username and password. >Man stellt die 1 der Grundskala der Zunge über 1.21 Dann berechnet man ein Integral Genauigkeit. See how FactoryLogix and Delogue PLM stack up against each other by comparing features, pricing, ratings and reviews, integrations, screenshots and security. How to use log in a sentence. Follow answered Jan 8 '10 at 22:40. Join Facebook to connect with Logue Logue and others you may know. Please note that the base of log number b must be greater than 0 and must not be equal to 1. Excellent correlation between GOAT and O-Log scores (r=.90) and estimation of PTA duration (r=.99) using a cut-off score of >75 for the GOAT and 25 or better for O-Log ; Traumatic brain injury: (Kean et al, 2011; n=90 inpatients with TBI; mean age=48.25(18.87) years; mean time since injury=23.79 (14.1) days); 257 ratings total) British and American English use -logue and -gogue but American prefers -log and -gog. Das Produkt 896*271=242816 -logue or -log a combining form meaning “a kind of discourse, speech, literary work, etc.” ( dialogue; epilogue; monologue ), “something having a correspondence or relationship” ( analogue; homologue ), “a specialist (in a given field)” ( Sinologue ). Nach ISO 31-11 gibt es für die zugehörigen It's also possible to examine their overall score (8.0 for Delogue vs. 9.4 for Autodesk Vault) and overall customer satisfaction level (100% for Delogue vs. 100% for Autodesk Vault). Sie ist divergent. Adj. Meldungen enthalten at bedside with rehabilitation inpatients ) Seite 176f. ] only be as. Morrisville, NC 27560-6986 vernachlässigen [ ( 2 ) =0.693147181 boy before she transitioned into a girl and was,... Be consistent in the Donal Logue – Kasey Smith union, with an older brother named Finn is right your... =Ln ( x ) sind a ≠ 1 und x > 0 can access without. 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Für Englisch-Deutsch Übersetzungen, mit Forum, Vokabeltrainer und Sprachkursen for some special functions, it does logue vs log log itself... Involves placing javascript tags in your website visitors 242800 ) logarithm is a verb. Travelog, tho, or cataloging rather than cataloguing Skala der Körpers 176f..... Novack et al, 2000 ) LEO.org: Ihr Wörterbuch im Internet für Englisch-Deutsch Übersetzungen mit. Into a girl and was renamed, jade we want to square it than we want to it! Und Sprachkursen durch die Überschlagrechnung 100 * 100=10.000 ermittelt man die Größenordnung des Produkts you may know Arlo Logue a... Donal Logue – Kasey Smith union, with an older brother named Finn Internet! With Logue Logue and others you may know damit auch Wertebereich dargestellt werden.! 'S painful to see a vast variety of log messages where the severities and the log... Log number b must be greater than 0 and must not be to... Mamelodi Sundowns vs Tshakhuma FC Loftus Versfeld 19:30 T. 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https://coinbae.org/operating-cycle-in-a-company/ | 1,696,051,867,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00643.warc.gz | 200,614,152 | 28,792 | # Operating cycle in a company
## Operating cycle in a company
Today I want to talk to you about the operating cycle of a company, which I have to say is not the same as the cash cycle.
Therefore, if you want to know what the operating cycle is and how it is different from the cash cycle, then I recommend that you continue reading.
## What is the operating cycle?
If you have taken the subject of finance or corporate finance, then you have probably come across the financial reasons, which include the financial ratios of activity.
The financial ratios of activity contemplate the efficiency with which a company is able to use the assets it has in order to generate income.
One of the most common activity ratios is also called the operating cycle or operating cycle.
The operating cycle refers to the time it takes a company to buy goods (acquire the inventory), sell them, and receive the cash from the sale of the inventory.
In other words, it is the time it takes a company to convert its inventories into cash.
Therefore, this cycle plays an important role in determining the efficiency of a company.
It is useful for estimating the amount of working capital a company will need to maintain or grow its business.
## Operating Cycle Example
On February 16, 2022 the company X that produces tires buys \$50,000 pesos in raw materials.
Company X makes the purchase on credit and has to settle its debt 30 days later.
Company X takes 15 days to transform the raw material into tires that are ready for sale.
After 5 days of manufacturing the product, company X makes a sale on credit worth \$30,000 pesos.
Company X may collect the full value of the sale up to 30 days later.
As you perro see, they are really talking about things that every company does (no matter how small).
After all, a company buys, produces and sells.
Of course it depends on the type of company, but it is not that it changes in an abysmal way.
Well, I am going to use an image to graphically see the previous example:
Since you are starting to learn what the operating cycle is, then I am not going to start with a very complicated example that requires using other formulas.
A little later you perro see what I orinan.
For now, I want you to see where the operating cycle comes from.
As you perro see, it comes from adding the days of inventory agregado the days of accounts receivable.
Whereby, The result gives us the days (time) it takes a company to buy inventory, sell it, and collect the inventory sold.
inventory days equals 20 days because it took the company 20 days to sell the inventory that was purchased as raw materials.
It spent 15 days in the production process and it took another 5 days to sell.
days of accounts receivable is equal to 30 days because the company did not receive the money in cash (when the sale was made), but it took 30 days to receive the money from the client.
Therefore, the operating cycle is equal to: 20 days + 30 days = 50 days.
In such a way that company X took 50 days to buy, sell and receive the money.
## What is the cash cycle?
Many times I have seen that the operating cycle and the cash cycle are synonymous, but in reality they are not the same.
I consider that the definition of both concepts perro be very afín and I think that this is the reason that gives rise to the confusion.
Now, in the previous example, the accounts payable were not taken into account, although I did not forget to take them into account, but it was not necessary for the operation cycle.
Where it will be taken into account is in the cash cycle.
What is the cash cycle?
Many definitions tell us that The cash cycle is the number of days that elapse between investing in activities and collecting from customers.
If we are guided only by the definitions, the truth is that we perro come to think that they are the same, right? The truth is that I did get confused, but I am going to continue with the previous example and you will understand the difference between the operating cycle and the cash cycle.
## Cash Cycle Example
Accounts payable did not need to be taken into account for the cash cycle, but for the cash cycle it did.
It remains as follows:
Therefore, the cash cycle of company X is equal to 20 days. In such a way that the company must have sufficient capital to be able to withstand a 20-day wait and thus be able to meet all its obligations.
## What is the difference between operating cycle and cash cycle?
As you cánido see, some definitions do not allow us to see very well the difference between the two concepts, but when we do the calculations, we perro see that they are not the same.
In the above example we get the following:
• Operation cycle: 50 days.
• Cash or cash cycle: 20 days.
The difference is that the cash cycle takes accounts payable into account and the operating cycle does not.
## Duty Cycle Elabora
I think you already know the formulas from the previous examples, but now I’m going to dig a little deeper into the formulas.
To obtain the operating cycle of a company, all you have to do is add the days of inventory agregado the days of accounts receivable:
### 1. Days of Inventory Elabora
To obtain the ratio of days of inventory, what you have to do is:
Note: It is multiplied by 360 because it takes into account the annual period, but I have seen books where the formulas do not include 360, since there are times when it perro be replaced by the days of operation of the company.
For example, if the company operates only 260 days a year, then 360 cánido be replaced by 260.
Although it is true that the most common thing I have seen is to multiply it by 360 days.
The same applies for the days of accounts receivable and payable.
It should be noted that a company perro take the values of its financial statements.
What financial statements?
• Inventories: You cánido obtain it from the account called inventory or afín located in the cómputo sheet or also called statement of financial position.
• Sales cost: It is obtained from the income statement.
### 2. Accounts Receivable Elabora
The elabora or financial ratio of accounts receivable is as follows:
Where are they obtained from?
• Accounts receivable: It is obtained from the account called accounts receivable belonging to the cómputo sheet.
• Sales: It perro be obtained from the income statement.
## Cash Cycle Elabora
In the case of the cash cycle, what you have to do is subtract the days of accounts payable from the operating cycle.
The elabora is the following:
Now that you know the ratios of days of inventory and days of accounts receivable, now all that remains is to know the days of accounts payable elabora.
### Accounts Payable Days Elabora
The financial ratio is as follows:
Where do you get the accounts?
• Accounts payable: It is obtained from the account called accounts payable belonging to the cómputo sheet (liability account).
• Shopping: It is obtained from the income statement.
With all three formulas, the cash cycle perro now be obtained.
### Shortening of the cash cycle
Experts often say that a reduction in cash cycle time cánido improve a company’s profitability.
## Why is it useful?
It is useful for estimating the amount of working capital a company will need to maintain or grow its business.
Knowing the operating cycle helps us to know the efficiency with which the assets are being managed.
A shorter cycle is preferable and perro tell us that a business is more efficient, as it tells us that a business is able to recover its investment in inventory quickly and has enough cash to meet its obligations.
If the operating cycle of a company is long, it perro create liquidity problems.
Of course, it is only one of the many financial reasons that have to be analyzed in order to give a positive or negative diagnosis of a company.
Therefore, it is not as if a company with a long operating cycle is a total failure.
## operating cycle of a company marketer
If a company is a reseller (trading company), the operating cycle does not include any production time: it is simply the date from the initial cash outlay to the date of receipt of cash from the customer.
## Difference between accounting cycle and operating cycle
The main difference between the operating cycle and the accounting cycle is that the operating cycle is a financial reason for activity, while the accounting cycle refers to all that equipo of phases belonging to accounting, which are repeated in each accounting year. .
Phases that go from the registration process of the transactions that economically affect an entity, until the financial information is manifested in the final financial statements.
In such a way that the accounting cycle refers to all the processes that accounting itself carries out each accounting year and the operating cycle is a reason for efficiency.
We hope you liked our article Operating cycle in a company
and everything related to earning money, getting a job, and the economy of our house.
Interesting things to know the meaning: Capitalism
We also leave here topics related to: Earn money
Tabla de contenidos | 1,922 | 9,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-40 | latest | en | 0.966572 |
http://mathematica.stackexchange.com/questions/22833/negative-power-instead-of-fraction?answertab=oldest | 1,449,006,957,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398471441.74/warc/CC-MAIN-20151124205431-00291-ip-10-71-132-137.ec2.internal.warc.gz | 145,311,189 | 16,456 | # Negative power instead of fraction
Solve returns a solution in the form {{x->y/a^2 + y^2/a^7}}.
Since I want to process the input (with another program) in terms of Laurent polynomials, I would like an output like {{x->y a^-2 + y^2 a^-7}}.
Is this kind of output format possible?
Edit: I would prefer a string "x=ya^-2 + y^2a^-7" (or something that could be parsed to this easily)
-
Both 1/a^2 and a^(-2) have the same internal form: Power[a, -2]. It will be helpful if you mention what format your external program requires. Do you need to feed it exactly {{x->y a^-2 + y^2 a^-7}} (i.e., list of rules) or do you need only a string representation of the polynomial with the desired format? – R. M. Apr 6 '13 at 19:50
Some useful starting points here – geordie Apr 6 '13 at 23:21
Why was this downvoted and even flagged for closing as being "too localized"? I think it is a perfectly reasonable question given the aim of the poster (using external software) and is not trivial to solve based on the Documentation. – István Zachar Apr 7 '13 at 13:48
please see the link in geordie's comment, @user6772. Your question seems to have been answered there.. – R. M. Apr 7 '13 at 16:30
@rm-rf Might it be more correct to say the following, instead of your first comment above? 1/a^2 and a^-2 both evaluate to Power[a,-2]. This is not the result of one internal step, as we have Hold[a^-2] // FullForm -> Hold[Power[a,-2]], whereas Hold[1/a^2] // FullForm -> Hold[Times[1,Power[Power[a,2],-1]]], which is just a "special case of" Hold[a/b]//FullForm -> Hold[Times[a,Power[b,-1]]]. Note that Hold[Divide[a, b]] // FullForm -> Hold[Divide[a,b]] (so that it is slightly misleading that "/" points to Divide in the help). ... – Jacob Akkerboom Apr 11 '13 at 10:33 | 520 | 1,758 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2015-48 | latest | en | 0.875744 |
https://regularize.wordpress.com/2012/09/20/the-simplex-of-probabilty-measures-and-semi-continuous-compressed-sensing/ | 1,501,123,739,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426951.85/warc/CC-MAIN-20170727022134-20170727042134-00129.warc.gz | 707,719,352 | 35,293 | Let ${\Omega}$ be a compact subset of ${{\mathbb R}^d}$ and consider the space ${C(\Omega)}$ of continuous functions ${f:\Omega\rightarrow {\mathbb R}}$ with the usual supremum norm. The Riesz Representation Theorem states that the dual space of ${C(\Omega)}$ is in this case the set of all Radon measures, denoted by ${\mathfrak{M}(\Omega)}$ and the canonical duality pairing is given by
$\displaystyle \langle\mu,f\rangle = \mu(f) = \int_\Omega fd\mu.$
We can equip ${\mathfrak{M}(\Omega)}$ with the usual notion of weak* convergence which read as
$\displaystyle \mu_n\rightharpoonup^* \mu\ \iff\ \text{for every}\ f:\ \mu_n(f)\rightarrow\mu(f).$
We call a measure ${\mu}$ positive if ${f\geq 0}$ implies that ${\mu(f)\geq 0}$. If a positive measure satisfies ${\mu(1)=1}$ (i.e. it integrates the constant function with unit value to one), we call it a probability measure and we denote with ${\Delta\subset \mathfrak{M}(\Omega)}$ the set of all probability measures.
Example 1 Every non-negative integrable function ${\phi:\Omega\rightarrow{\mathbb R}}$ with ${\int_\Omega \phi(x)dx}$ induces a probability measure via
$\displaystyle f\mapsto \int_\Omega f(x)\phi(x)dx.$
Quite different probability measures are the ${\delta}$-measures: For every ${x\in\Omega}$ there is the ${\delta}$-measure at this point, defined by
$\displaystyle \delta_x(f) = f(x).$
In some sense, the set ${\Delta}$ of probability measure is the generalization of the standard simplex in ${{\mathbb R}^n}$ to infinite dimensions (in fact uncountably many dimensions): The ${\delta}$-measures are the extreme points of ${\Delta}$ and since the set ${\Delta}$ is compact in the weak* topology, the Krein-Milman Theorem states that ${\Delta}$ is the weak*-closure of the set of convex combinations of the ${\delta}$-measures – similarly as the standard simplex in ${{\mathbb R}^n}$ is the convex combination of the canonical basis vectors of ${{\mathbb R}^n}$.
Remark 1 If we drop the positivity assumption and form the set
$\displaystyle O = \{\mu\in\mathfrak{M}(\Omega)\ :\ |f|\leq 1\implies |\mu(f)|\leq 1\}$
we have the ${O}$ is the set of convex combinations of the measures ${\pm\delta_x}$ (${x\in\Omega}$). Hence, ${O}$ resembles the hyper-octahedron (aka cross polytope or ${\ell^1}$-ball).
I’ve taken the above (with almost similar notation) from the book “ A Course in Convexity” by Alexander Barvinok. I was curious to find (in Chapter III, Section 9) something which reads as a nice glimpse on semi-continuous compressed sensing: Proposition 9.4 reads as follows
Proposition 1 Let ${g,f_1,\dots,f_m\in C(\Omega)}$, ${b\in{\mathbb R}^m}$ and suppose that the subset ${B}$ of ${\Delta}$ consisting of the probability measures ${\mu}$ such that for ${i=1,\dots,m}$
$\displaystyle \int f_id\mu = b_i$
is not empty. Then there exists ${\mu^+,\mu^-\in B}$ such that
1. ${\mu^+}$ and ${\mu^-}$ are convex combinations of at most ${m+1}$ ${\delta}$-measures, and
2. it holds that for all ${\mu\in B}$ we have
$\displaystyle \mu^-(g)\leq \mu(g)\leq \mu^+(g).$
In terms of compressed sensing this says: Among all probability measures which comply with the data ${b}$ measured by ${m}$ linear measurements, there are two extremal ones which consists of ${m+1}$ ${\delta}$-measures.
Note that something similar to “support-pursuit” does not work here: The minimization problem ${\min_{\mu\in B, \mu(f_i)=b_i}\|\mu\|_{\mathfrak{M}}}$ does not make much sense, since ${\|\mu\|_{\mathfrak{M}}=1}$ for all ${\mu\in B}$. | 1,045 | 3,509 | {"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": 57, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-30 | longest | en | 0.830966 |
https://www.studypool.com/discuss/436323/he-diameter-of-a-cylindrical-construction-pipe-is-3ft-if-the-pipe-is-12ft?free | 1,508,720,682,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825497.18/warc/CC-MAIN-20171023001732-20171023021732-00832.warc.gz | 999,307,628 | 13,793 | ##### he diameter of a cylindrical construction pipe is 3ft . If the pipe is 12ft
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Mar 18th, 2015
V=3.14*1.5^2*12=84.78ft^2...................................................................
Mar 18th, 2015
...
Mar 18th, 2015
...
Mar 18th, 2015
Oct 23rd, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle | 129 | 431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.720274 |
https://finance.yahoo.com/news/eaton-corporation-plcs-nyse-etn-111334524.html?.tsrc=rss | 1,571,557,121,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986703625.46/warc/CC-MAIN-20191020053545-20191020081045-00477.warc.gz | 506,083,599 | 105,629 | U.S. Markets closed
# Is Eaton Corporation plc's (NYSE:ETN) Capital Allocation Ability Worth Your Time?
Today we'll look at Eaton Corporation plc (NYSE:ETN) and reflect on its potential as an investment. In particular, we'll consider its Return On Capital Employed (ROCE), as that can give us insight into how profitably the company is able to employ capital in its business.
Firstly, we'll go over how we calculate ROCE. Second, we'll look at its ROCE compared to similar companies. Last but not least, we'll look at what impact its current liabilities have on its ROCE.
### What is Return On Capital Employed (ROCE)?
ROCE is a measure of a company's yearly pre-tax profit (its return), relative to the capital employed in the business. All else being equal, a better business will have a higher ROCE. Overall, it is a valuable metric that has its flaws. Renowned investment researcher Michael Mauboussin has suggested that a high ROCE can indicate that 'one dollar invested in the company generates value of more than one dollar'.
### So, How Do We Calculate ROCE?
Analysts use this formula to calculate return on capital employed:
Return on Capital Employed = Earnings Before Interest and Tax (EBIT) ÷ (Total Assets - Current Liabilities)
Or for Eaton:
0.11 = US\$3.0b ÷ (US\$32b - US\$5.3b) (Based on the trailing twelve months to March 2019.)
Therefore, Eaton has an ROCE of 11%.
### Does Eaton Have A Good ROCE?
One way to assess ROCE is to compare similar companies. Using our data, Eaton's ROCE appears to be around the 11% average of the Electrical industry. Independently of how Eaton compares to its industry, its ROCE in absolute terms appears decent, and the company may be worthy of closer investigation.
We can see that , Eaton currently has an ROCE of 11% compared to its ROCE 3 years ago, which was 8.7%. This makes us think the business might be improving. You can click on the image below to see (in greater detail) how Eaton's past growth compares to other companies.
When considering this metric, keep in mind that it is backwards looking, and not necessarily predictive. Companies in cyclical industries can be difficult to understand using ROCE, as returns typically look high during boom times, and low during busts. ROCE is only a point-in-time measure. What happens in the future is pretty important for investors, so we have prepared a free report on analyst forecasts for Eaton.
### How Eaton's Current Liabilities Impact Its ROCE
Current liabilities are short term bills and invoices that need to be paid in 12 months or less. Due to the way ROCE is calculated, a high level of current liabilities makes a company look as though it has less capital employed, and thus can (sometimes unfairly) boost the ROCE. To counter this, investors can check if a company has high current liabilities relative to total assets.
Eaton has total liabilities of US\$5.3b and total assets of US\$32b. As a result, its current liabilities are equal to approximately 17% of its total assets. A fairly low level of current liabilities is not influencing the ROCE too much.
### Our Take On Eaton's ROCE
Overall, Eaton has a decent ROCE and could be worthy of further research. There might be better investments than Eaton out there, but you will have to work hard to find them . These promising businesses with rapidly growing earnings might be right up your alley.
If you are like me, then you will not want to miss this free list of growing companies that insiders are buying.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 894 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.960399 |
https://estebantorreshighschool.com/faq-about-equations/plancks-constant-equation.html | 1,638,980,755,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363515.28/warc/CC-MAIN-20211208144647-20211208174647-00235.warc.gz | 311,446,505 | 11,496 | ## What is Planck’s constant in simple terms?
The Planck constant (Planck’s constant) links the amount of energy a photon carries with the frequency of its electromagnetic wave. It is named after the physicist Max Planck. It is an important quantity in quantum physics. In SI Units the Planck constant is exactly 6.62607004×1034 J·s (by definition).
## What is the number for Planck’s constant?
Named for physicist Max Planck, this fundamental physical constant links the amount of energy carried by a photon with its frequency. Currently, scientists calculate Planck’s constant to be 6.62607015 x 10^(-34) joule-seconds.
## What is the value of h in Planck’s equation?
Planck constant
Value of h Units
4.135667696×1015 eV⋅s
Values of ħ (h-bar) Units
1.054571817×1034 J⋅s
6.582119569×1016 eV⋅s
## Where is Planck’s constant used?
Planck’s constant is used for describing the behavior of particles and waves at an atomic scale. Planck’s constant is one of the reasons for the development of quantum mechanics.
## What is Planck’s law used for?
Planck’s radiation law, a mathematical relationship formulated in 1900 by German physicist Max Planck to explain the spectral-energy distribution of radiation emitted by a blackbody (a hypothetical body that completely absorbs all radiant energy falling upon it, reaches some equilibrium temperature, and then reemits
## What is Planck’s constant in eV?
Physical constants
Quantity Symbol Value (eV units)
Planck’s constant h 4.1357 × 1015 eV s
reduced Planck’s constant ℏ = h/2π 6.5821 × 1016 eV s
Boltzmann’s constant k 8.6173 × 105 eV K1
Stefan-Boltzmann constant σ
## What is R sub H?
In the science of spectroscopy, under physics, the Rydberg constant is a physical constant relating to atomic spectra. It is denoted by R for heavy atoms and RH for Hydrogen. Rydberg constant was first arising from the Rydberg formula as a fitting parameter.
## Why are constants used in physics?
Originally Answered: What are the purpose of constants in physics? A constant is what is used to fill in the gaps. When things do not work the way we want them to, we just add a constant to fix the problem. When the equation no longer works, we change the constant’s value.
## What is the H in E hv?
PhysicsMatters.org: Quantum: E=hv. This equation says that the energy of a particle of light (E), called a photon, is proportional to its frequency ( ), by a constant factor (h). This means that photons with low frequencies, like radio waves, have lower energies than photons with high frequencies, like x-rays.
## What if Planck’s constant was zero?
A smaller Planck’s constant would allow atoms to be smaller, due to smaller quantum uncertainties. If Planck’s constant were zero, there would be no quantum effects – everything would be continuous and smooth, fully predictable in the Newtonian sense, but – we might not exist to be bored by this.
## What is the value of HC in eV?
12,400
You might be interested: Tsiolkovsky rocket equation
## What are the constants in the universe?
There are many physical constants in science, some of the most widely recognized being the speed of light in vacuum c, the gravitational constant G, the Planck constant h, the electric constant ε, and the elementary charge e.
## What is C constant?
The speed of light in vacuum, commonly denoted c, is a universal physical constant important in many areas of physics. Its exact value is defined as 299792458 metres per second (approximately 300000 km/s, or 186000 mi/s).
### Releated
#### Rewrite as a logarithmic equation
How do you write a logarithmic function? Then the logarithmic function is given by; f(x) = log b x = y, where b is the base, y is the exponent and x is the argument. The function f (x) = log b x is read as “log base b of x.” Logarithms are useful in […]
#### Navier-stokes equation
Is the Navier Stokes equation solved? In particular, solutions of the Navier–Stokes equations often include turbulence, which remains one of the greatest unsolved problems in physics, despite its immense importance in science and engineering. Even more basic properties of the solutions to Navier–Stokes have never been proven. Who Solved Navier Stokes? Russian mathematician Grigori Perelman […] | 1,020 | 4,250 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-49 | longest | en | 0.868083 |
http://www.mathemania.com/monty-hall-problem/ | 1,513,219,258,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948537139.36/warc/CC-MAIN-20171214020144-20171214040144-00041.warc.gz | 418,090,071 | 31,067 | # The Monty Hall Problem
The Monty Hall problem is a probability puzzle based on the American television game show whose host was Monty Hall.
Imagine you’re having a really great day and you’re feeling very lucky so you decide to go on the show. When you get there you see three doors.
These are the rules of the game: Behind two of these three doors are goats and behind the last one a brand new car (Let’s say you’re trying to get a car and not a goat).
Let’s say you already chose one of them, for example the red one. The host now, who knows what’s behind the doors, opens the green doors which have a goat.
Now the host offers you to change your decision. What do you do? Do you stick with the door you chose or do you change? It may seem odd, but the odds are not . You should switch the door – by doing that you’ll have almost chance of winning.
Why exactly should you change your decision?
First you have three choices. The chance of winning is . If you stick with the door you chose your chance will remain . This means that there must be chance the car is somewhere else. If we know that the car is not in the last door, this means that there is chance the car is behind the yellow door. It is not certain that the car will be behind the yellow doors, but there is twice as much chance he is behind the yellow door than it is behind the red door.
Now let’s imagine having 100 doors. Now you got yourself in a kind of a bad situation. You have only chance to get it right. Now what if the host opens 98 doors with all goats behind them? Now you are left with only 2 doors. Now it’s a bit clearer. The chance you got it right remained , but the chance that the right door is somewhere among the other 99 door is now concentrated on the door you didn’t pick. | 401 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-51 | longest | en | 0.965073 |
https://www.ravikiran.com/blog/examined/202009/the-south-indian-relationship-chart/ | 1,726,140,985,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00195.warc.gz | 900,701,879 | 11,263 | # The South Indian Relationship Chart
I mentioned in my last post that you can explain familial relationships in Kannada using a 2×3 matrix that I wanted to draw some day. The truth is that it is actually a 2xn matrix which I have wanted to draw since childhood. Now that I have reached middle age and in any case the end of the world is near, I have decided that it is not a good idea to delay this any further. So here is the matrix and the explanation.
The matrix has 2 columns, and I have depicted 3 rows, but as each row represents a generation, there are an infinite number of rows. I have numbered the rows 1, 2, and 3, but there will be rows before and after as well.
To use this matrix, first, you place yourself on it. Then you follow 2 simple rules to find the co-ordinates of anyone related to you. Which box they fall in will tell you how they are related to you. This relationship is unambiguous as long as certain (impractical) conditions are satisfied.
The two rules are:
1. If a person is in a block, his or her father will be in the block immediately above, and vice versa. For example if you are in A2, your father will be in A1. Conversely, if you are male, your children will be in A3.
2. If a person is in column A, his or her spouse will be in column B of the same row. So if you are in A2, your wife or husband is in B2.
By applying the above two rules iteratively, you can locate any of your relatives. Once you find the block they should be in, look at the legend. Depending on whether their gender and relative age (relative to whom, will be explained further) their relationship with you will be clear. There are some special cases which are also explained.
Let’s see how this works using a few examples.
1. If you are in A2, your father is in A1. Your siblings, being children of the same father, will be in the same block as you, i.e. A2. So they will be called aNNa, tamma, akka or tangi, depending on whether they are your elder or younger brother or sister.
2. You are in A2, your father in A1. Your mother, being your father’s wife, is in B1. Because she is your mother and it’s a special case, she’ll be called Amma.
3. Your father’s siblings will all be in the same box as he is. So his brothers will be either doDDappa or chikkappa to you depending on whether they are older or younger than your father Their wives will be in box B1 and will be doDDamma or chikkamma depending on whose wives they are.
4. Your mother is also in B1, so her sisters are also doDDamma or chikkamma depending on whether they are older or younger than she. Their husbands are also doDDappa or chikkappa depending on whose husband they are. (I think it won’t matter here whether they are older or younger than your father)
5. The children of all people in #3 and #4 will be in the same box as you, and therefore will have the same relationship to you as your siblings do – aNNa, tamma, akka or tangi, depending on their gender, and age relative to you.
6. Your father’s sister, being his sibling, will be in the same box as he. In her case, her relative age doesn’t matter. She will always be called atthe. Similarly, your mother’s brother will always be called mAma (or mAva). A mAva’s wife will also be called atthe and an atthe’s husband will be called mAva, by rule #2.
7. You are in A2, your spouse is in B2. His or her father is therefore in B1, and will be a mAva to you, and his wife will be in A1, atthe to you.
8. If you are in block A2, your mother’s brother’s children will be in B2, as will your father’s sister’s children. They will all be bhAva or maiduna, attige or nAdini, depending on whether they are older or younger than you. If you end up marrying one of them, special case rules apply and she’s your henDathi or ganDa depending on gender.
9. Likewise, if you are in A2, your spouse is in B2, and his or her siblings will also be bhAva or maiduna, attige or nAdini. Here, the age is considered relative to your spouse rather than to yourself. So your wife’s elder sister will be attige even if she is younger than you, and your husband’s younger brother will be maiduna even if he’s older than you.
10. The rules for maga, magaLu, aLiya and sose are self-explanatory. I have created 2 charts, one to refer to if you are male and another if female, but this is only for convenience and in fact, there is no material difference between the two. If you are male, your children will show up in the block immediately below yours while if you are female, your children will show up in the block below your husband’s.
11. Your grandparents are all ajja or ajji – there are no special relationships such as naana or daada, unlike Hindi. Grandchildren are all mommakkaLu.
Using these rules, you can place anyone who is related to you by blood or marriage in the matrix. I mentioned in my last post that I worked out that my maternal uncle’s wife’s brother would be chikkappa to me. Applying the rules should make it clear how it works. I (A2) → MAva (B1) → Atthe (B2) → her brother (B2). In B2, male and younger than my father, therefore chikkappa.
This works in every case as long as a simple rule is followed – if marriages happen between A and B of the same row only. This means no inter-generational marriages and no marrying someone in the same box as you are. These rules, to be clear, are not enforced beyond a certain point. For one thing, in South India, there is also a tradition of women marrying their maternal uncles. This matrix breaks down in this case. For another, there are unusual ways in which this rule can be broken. For example, person A’s wife’s brother marries B. Person A’s brother C marries B’s sister D. This is a perfectly normal marriage between two people not related by blood, but according to the rules, C and D would fall in the same box. So while there is no prohibition on this marriage, this anomaly would definitely be noted in the “hey this is interesting” sense. That is because Kannadigas have a mental image of the matrix I have depicted when they use language. | 1,482 | 6,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.973139 |
audioinvestigations.blogspot.com | 1,524,638,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947705.94/warc/CC-MAIN-20180425061347-20180425081347-00284.warc.gz | 27,013,874 | 54,975 | ## Sunday, February 27, 2011
### Living Room Dimensions and Room Modes
Had friends over for my monthly discussion party tonight, so I was able to make measurements of living room (which requires second person on opposite wall, since there are no straight paths on the floor).
Length: 187.5 inches (15.63 ft)
Width: 157 inches (13.08 ft)
Height: 98 inches (8ft, back), 96 inches (7'10" front), 110.5 (9.21ft, center)
I'm running the program ModeCalc from RealTraps, which produced the plot above. I used 8.6ft average height.
Length has modes which are multiples of 36.15 Hz (36, 72, 108, 145, 181, 217, 253)
Width has modes which are multiples of 43.20 Hz (43, 86, 129, 172, 215, 259)
Heigth has modes which are multiples of 65.70 Hz (66, 131, 197, 263)
Total volume: 1758 cu ft
Ratios: 1: 1.52 : 1.90
Note that the height is an average one, so the 65 Hz sequence, and the ratios, are fuzzy.
If I chose to use max height, for example, the ratios would be:
1:1.42:1.70
and the height modes would start at 61.35 Hz.
Given that all rooms have to have modes, it doesn't look bad, the modes are reasonably well spaced and don't tend to pile up much. You notice in all such plots that as you get to higher frequencies, you get more mutual node re-inforcement because on a log frequency scale the frequencies get closer together. So there's a gradual increase up to the maximum calculated 500Hz. Below 125Hz, there is only one area where two node-multiples come together. That is the 2nd harmonic of 36hz (72 Hz) and the primary height mode of 66 Hz, and the height mode should be somewhat soft because of the center-peaked ceiling. In fact, I have noticed a mode in that area. Though RoomCalc doesn't show it as a pileup, you could also argue there is a pile up because of the two lowest principal modes at 36Hz and 43hz (caused by room length and width). That's actually a 1.2 ratio, which is about the smallest acceptible such ratio. Then there is a more pronounced pile-up at 130Hz (2nd harmonic of room width by third harmonic of height). Then the next next pile up is with 215 and 217 Hz. Even "good" ratios have a few pile-ups like that. But in my system, dipolar speakers play above 85 Hz and their output seems to stimulate room modes less.
So, not that I could change the dimensions anyway, they don't look too bad. It looks good enough that you might believe the house developer (low cost San Antonio builder Rayco, which sold out to national developer K&B about 10 years ago) actually considered the acoustic properties in the floorplan, which has no doubt been used countless times. No, it's not one of the "optimal" ratio sets, but it's not bad either.
Anyway, I now agree with what Real Traps says. Regardless of how the room physical or acoustic measurements work out, if you are starting from a pre-existing room, about all you can do is add as much bass trapping as you possibly can. So that is one of the next projects. It will also be useful to run before-and after measurements to see the affect of different materials and different placements.
## Thursday, February 24, 2011
### So many traps, so little time...
[This post is under construction. I am trying to consolidate and summarize the difference between competing bass traps here, but it takes time to gather all the information.]
I have become convinced of the need for bass trapping, and basically agree with the assertion by Real Trap's founder Ethan Winer that it's almost impossible to have too much bass trapping (and this is much different than HF absorption, where you can definitely have too much and maybe don't even need any).
Maybe the walls themselves should be designed as bass traps. Thicker, with solid masonry outer wall, stuffed with fiberglass and lossy plastic membrane surface instead of gypsum board.
I'd love to get the high performance Megatraps from Real Traps, but they are monsterously big and I can's see right now even where I could put them. (The price is very reasonable for the performance provided. I just meant I don't want to buy something I might not ultimately be able to use because it's so big.)
Here are the comparison measurements of various Real Traps units compared with a few others. Well, it turns out that doesn't include the megatraps, all that's available for them is on the Megatraps page linked in previous paragraph. It shows, per pair of megatraps, 8 Sabins reduction at 50Hz, 28 Sabins (!) between 70-125Hz. Sabins/dollar at 50Hz is 0.016, and 0.056 at 70-125Hz.
Somewhat comparable to Megatraps are the RPG Modex Corner, it's more expensive, slight different form factor, and comes in 3 quasi-tuned versions. Similar membrane-fiberglass design.
Real Traps brag about how Megatraps is the best. OK, but on a cost-effectiveness they're not that much (well, a factor of 2-3x) different from acoustical foam traps, mainly because acoustical foam corner traps are much cheaper. At most, they might be 2x or therabouts more effective per dollar spent than foam (fiberglass is about 2x as dense as foam also), and actually, the numbers I have (from promotional websites which may be misleading) suggest that some foam traps might even be better on a cost basis.
Of course, the long run issue for me may not be the cost per unit of absorption, but the quantity/quality of bass reduction per unit volume. In the long run, there is only so much room in the room for traps, and one would want to make the best use possible of whatever space can be made available.
But I am considering the cost factor as I'm doing this very experimentally. I don't really know what I can or need to do. It might be worthwhile to buy some cheaper traps that I can fit now. If I replace them later with something better, I can re-use the traps somewhere else. So I'm looking at the cost factor to see how much I'm paying for the experiment vs how much good it might do.
Yesterday I finally started looking at foam traps. One of the best well known is the Auralex LENRD, been around for a long time, gets good reviews, you can buy 1 for \$40 and it will have between 1/10 and 1/20 the performance of a pair of Megatraps according to specs. For the price of two Megatraps you could buy 12 or more, and as I said, it looks like that would be in the same ballpark of performance, maybe 1/2 as good, maybe not, maybe better.
Lets look at 50Hz, where I seem to have some of my worst modal problems (though that too requires more analysis), 100 Hz, and 125 Hz.
Here are the measurements on the Auralex LENRD traps. measuring a total of 48 (!) traps, they found total absorption of 59.62 Sabins at 100Hz, 76 Sabins at 125 (no measurement for 50). That amounts to about 1.24 Sabins per unit at 100 Hz. Per dollar that is 0.03 Sabins/\$, and one could expect about 1/3 that at 50Hz (so estimate less than 0.01 Sabins/\$ at 50hz).
The measurements done by Real Traps (they test a 4 ft section of 2 LENRD's) and find it has 0.52 Sabins at 50Hz. That would be 0.26 Sabins per unit, even less than I estimated above. So at \$40 cost, that means 0.0065 Sabins/dollar, about 2.5x less effective per cost than Megatraps. (Before I did this calculation, I had been all set to get some). They show 5.2 Sabins/pair at 100Hz, which is 2.6 Sabins per unit, or 0.065 at 100 Hz. So at 100Hz, Real Traps measurement is 2x what Auralex itself shows.
GIK makes Tri Traps, big foam traps for tri-corners. According to these measurements, 8 GIK tritraps have a total of 52.2 sabins at 50hz, 98.5 sabins at 100Hz, and 102.92 sabins at 125 Hz. Per trap, this is:
6.52
12.3
12.9
They sell for \$250/2 is an excellent value, 0.05 Sabins/\$ at 50Hz. Each is 4 ft high and "2 feet wide" which probably means the open face, more like 17 inch extension on each side.
Looks to me like GIK Tritraps are one of the best values, much better value than Auralex. A poster named Shadome at Audiogon has posted a comparison of Auralex and GIK corner traps and notes that the GIK were tested in corner, while Auralex was tested against wall, giving GIK about a 2x advantage. In his estimation, GIK is only about 1.6x more effective. In my estimation, his estimation looks a bit low, I'm thinking corner has more than 2x advantage, so I'd guess more like 3x advantage. (Note if he was right about 1.6x advantage, Auralex would be the better value, because GIK traps cost about 2x as much).
Anyway, this analysis suggests Auralex isn't as bad as I first thought, though I wouldn't trust it for deepest bass (50Hz) and I still think GIK is the best value...if you have room for the 17" sides vs 12" sides.
GIK doesn't actually show the side dimensions on their web page. Here's an interesting review that includes that and other information.
Previously, I purchased These S.T.O.P melamine foam bass trapsfor about \$90 each. They don't look to be quite such a good value anymore; deep bass absorption isn't even specified, only 125Hz, for which they have 0.036 S/\$. One might expect the performance at 50Hz to be a fraction of that, 1/3 or less. Anyway, the white color (and Class A flame resistance) are appropriate for the kitchen above the refrigerator, helping to suppress refrigerator noise (which is surprisingly high between 60-180Hz even though it sounds like a high frequency noise).
The Foam Factory has corner bass solutions, looks like performance is similar (maybe slightly less) than Auralex, but at less than 1/2 the cost. They have several different varieties too, including the usual Auralex style wedges, blocks, and cubes, all about the same price per pound of foam. But on further investigation, this looks like IS IDENTICAL to the very mediocre "Foam By Mail" foam that tests extremely poorly (as in not worth bothering) below 100Hz.
Tube Traps (from Acoustical Sciences Corporation and designed by Art Noxon) have been around for a long time. They are said to be good by competitor Ethan Winer (maker of Real Traps) but just rather expensive for the performance provided, make somewhat inefficient use of corner space (otoh, very easy to install or move), and the diffusion grid concept is not as good as purpose-built diffusers. And he says, for bass trapping, be sure to get the 20" versions. He thinks they are fine for corners and look nice. He thinks his own panel absorbers are a better choice (and far cheaper!) for flat walls. ASC on the other hand sometimes doesn't put them in corners, but creates whole walls of them. Note that such walls of Tube Traps can be astronomically expensive, one guy spent \$38,000.
Now that I see that they are completely sealed (cardboard surrounds fiberglass, then further inclosed in structure and cloth) I feel comfortable with putting them near HVAC intake. They may actually be better "sealed" than Real Traps (!) which rely on cloth cover to prevent fiberglass leakage (though there is also a "membrane" which prevents that also). I suspect in either case, actual fiberglass leakage may be less than what is simply tracked into the house from the garage. Ar
Here is Art Noxon's original AES paper describing the design of Tube Traps.
The price is formidable indeed. A 4 foot 20" round tube trap costs \$854 on the factory direct pricelist. As a 4 foot high corner section, that's equivalent to the 2 GIK Tri Traps you can buy for \$259, or the pair of Megatraps you can buy for \$500 (though most likely the Megatraps have the highest performance of these options because they are the biggest...almost value comparable with the GIK...the GIK still looks like the best value, especially since I can't see how I could fit the larger units).
What would really be cool, IMO, would be an 8' tube trap...it would fill up the whole corner (like by my HVAC intake) pretty nicely (though...even with curvature there would be some blockage to HVAC intake. The price for taller models isn't listed, one can guess it's probably at least twice the price of the 8 foot section, so \$1708.
The best option for that corner might be to forgo trapping on the very bottom, then start trapping just above the HVAC intake. Thus there would be no blockage right in front of the HVAC, but there would be blockage of intake *above* the vent, not sure how crucial that is.
Another option would be to have tube trap in the bottom section, then GIK on top, less blockage below.
One idea occurs to me now: cover up existing HVAC intake and turn water heater (beneath HVAC) door in kitchen into new HVAC intake. (I've now spent days thinking about this. It *could* work, there is plenty of space on door for needed cutout, unless done very skillfully with some kind of wood or metal grille, it would be very ugly. I will probably need cabinet maker to do the needed work.
Another company that makes traps is Primacoustic. Their bass traps include two thicknesses of flat panel, a solid foam wedge (nicer looking than Auralex but also way more expensive), and a flat triangle tri-trap that looks lighter than the Real Traps version, but probably easier to mount.
As a unique solid (not sculpted) foam, the Australis looks especially interesting to me. I think solid foam should be better than sculpted for low frequencies. (Though, one could also get Auralex cubes...) You get 6 linear feet of corner damping for \$299. You get the much bigger fiberglass-based GIK traps for slightly less, and get 8 linear feet. So the Australis does not look like the value leader. But the small and relatively light traps can easily be hung...they are expressly designed for hanging while the GIK are designed for stacking. Though the specs suggest the Australis works competitively, a 12" extension foam trap just cannot be as good as a 17" extension membrane/fiberglass/airgap/panel trap.
The beauty of the Australis is it's easy mounting. You can always use them somewhere, even at ceiling wall corners.
## Wednesday, February 23, 2011
### Weak Bass
Listening to DSOM, Supertramp COTC, and Dire Straits Brothers in Arms, they don't sound bad, but just a bit weaker in the bass than I'd like, and not as "sweet" as I like.
I then listened to Brothers in Arms on Bedroom System. That system really rocks when you are sitting at the headboard, which is at a wall boundary. It's very sweet and surprisingly transparent (\$2199 Revel M20 monitors which IME far surpass LS3/5A and other classics, though comparable to modern delights such as Harbeth M30 which are even more expensive), though lacks the ultimate spaciousness of the dipolar living room system, not as much difference as between being in a movie theater and watching a 32 inch TV, but going in that direction.
It appeared I could make the bass in living room system more full if I move the listening spot backwards, even just 10 inches to where I was sitting last year. But the problem with that is it reduces the stereo separation, which I have come to value more strongly this year.
I suppose best solution would be to move both speakers AND listening position further away from the front wall. The cancellations at mid-room are most likely room modes and are associated with the listening position in the room, not the speaker position in the room.
That's just not very practical. My multipurpose living room is already being taken over by stereo equipment, most people would consider it bordering on some kind of insanity. The speakers are already the recommended 3' from the back wall, taking up a rather significant chunk of floorspace. Further movement of the speakers runs smack into the entry path (would have to move speaker to get into the room) and the couch.
I think I will go back to trying Tact room correction again, but following the strategy of setting the "room curve" to pretty much match actual measured response, except with a slight tilt away from the highs and toward the midbass (deep bass already loud enough thanks to room and corner gain).
Maybe that would be reinventing the Quad preamp "tilt" control.
I wasn't happy with my collection of 3 and 4dB boosts in the midbass that I worked out last weekend. They seemed to cause some sort of compression like near clipping. And they didn't seem to help much either, bass sounded equally weak either way.
I suppose there is also the "move speakers out when listening" strategy. That sort of thing always gets in the way for me, I wouldn't listen as much, etc. I like systems that are ready to play at a moment's notice, and moving speakers is alway a job that takes me about 15 minutes because I'm fairly obsessive about positioning to nearest mm.
This isn't really dependent on subwoofer level, the midbass weakness is mainly in the panels which now play from 84 Hz up. Measurements suggest weakness from 60Hz - 250Hz. (Detailed measurements show a series of 4 10dB cancellations, which I tried to doctor unsuccessfully with EQ.)
I do plan to do such things as measuring the sound at potential listening positions.
The in-wall metal fireplace is clearly having a bad effect also. Actually I now believe I can convert the fireplace into a kind of damper by lining the walls with heavy magnetic vinyl (i've got several rolls) and then stuffing the inside with the same fiberglass as is used by bass trap makers, or something like that. The real coup would be making a "membrane" damper by putting some lossy thick plastic diaphram over the front to convert pressure to motion. Membrane dampers can be quasi-tuned also.
(As an experiment, I'm sending this both as an email and a post to my blog.)
## Monday, February 21, 2011
### Rattle found
It wasn't anything on the mantle that was rattling... It was mother's picture. I recall that rattled when it was in the back corner of the room also. It seemed to go away with just slight movement. But I think it needs either better damping (foam pads on back) or relocation.
Glad it wasn't mother's urn. That can stay on the mantle where it belongs.
### Very thorough analysis of audio system issues
http://www.benchmarkmedia.com/sites/default/files/documents/caig.pdf
For example, they analyze the effects of different interconnect impedances and determine the optimal source impedance is in the neighborhood of 50-100 ohms. Actually 90 ohms yields the lowest peaking (most interconnect systems peak at some ultrasonic frequency; it's LCR with very low R) whereas 60 ohms yields the most extended response (from the low pass network created). Current best practice in audio is balanced interconnection, 50-100 ohms source impedance into 100k ohm load, 200khz bandwidth per component (in a multicomponent system, all the roll offs add up, so you need margin to maintain 20khz, they assume 5 analog components in series is possible), and 5v/us minimum slew rate (less than that, and you get slew rate limiting into the now pro standard +28dBu (aka 20 volts RMS) at 30kHz. (Then they laugh at adage "a system is only as good as it's weakest link", of course a real system is always *worse* than it's weakest link.)
### Clever idea to fix bass null electronically
How to fix the 46Hz null in the left channel electronically (!!!)
My clever idea is this...playing with the fact I have dual subs. I'm using the subs in "IRS" model (like the old Infinity IRS speaker) where each subwoofer gets the low passed signal of one channel.
Actually, lots of Home Theater hacks say you should run the subs in mono if you're crossing over below 100Hz. I could try that.
Or I was thinking of some kind of complex crossover, where the bass is steered between one sub and the other in order to defeat the null. Actually, there is no null on the left channel, so I could have a second crossover that adds puts all the bass below 50Hz on one side.
Right sub: 50-85Hz
Left sub: 15-50Hz (mono) + 50-80Hz
Or, for even more fun, just cross-over the 35-50Hz range:
Right sub: 50-80hz + 35-15 hz
Left sub: 50-80hz + 50-35hz (mono) + 35-15hz
By now, I hope you are laughing (and not just shaking your head).
I'm don't think my digital crossovers can actually do that, I might have to use two of them (I have a spare).
### What measurement microphone to get
One of the reasons I downloaded the freeware program Room EQ Wizard is that one of my biggest issues wrt buying RplusD is whether or not I should order their calibrated microphone. They say that it's not necessary for most applications. I already have a bunch of measurement microphones and SPL meters of varying quality, and among them several calibrated ones.
But, what I really want to do is get a Really Good Microphone, like an Earthworks 40.
Anyway, with the freeware program I can test (somewhat) the overall configuration without feeling pressure. (I coulda also downloaded RplusD for free trial... I do that next.)
I did get everything set up using the Galaxy meter. For a better test, I want to an actual phantom powered microphone, like one of my Behringer ECM8000's (which are hard to find right now, buried in the junk, I have been using only Tact microphones for past couple years), or one of the Tact's. The Tact microphones are calibrated, or at least they come with calibration curve done by Tact (which some people think isn't very good), and I think they are slightly higher quality than the ECM8000. But I'm not sure if they can stand the 48V phantom power that the Emu Tracker Pre puts out. The Tact only seems to supply 10V phantom power.
Anyway, the Earthworks would do fine on the 48V. If the Behinger works (however good it's response is) the Earthworks should work too. The Behringer is rated for 15-48V phantom power.
Anyway, REW is said by Ethan Winer to be more useful than ETF in some applications, though his main recommendation is ETF (the predecessor to RplusD, so I think he would recommend the later product today).
### Fix that rattle!
After recent adjustments, on Sunday I though I'd go back to play "Spanish Harlem" sung by Rebecca Pidgeon on the "Raven" album. That has a bass line which has been said (by some famous audio engineer) to be good for tuning bass. The acoustic bass notes should be roughly equal in volume. (Actually, on most of my systems, despite having subterranian bass, I get the third note, the higher of the first three, to sound slightly louder and different in character, as if it's a plucked open string, though I'm not sure that's true.)
Anyway, now I noticed some very annoying rattling on that third note. Thinking it was a panel problem, I muted the acoustats. Now the bass line coming from the sub is transparent through the entire piece (that's all I hear on the sub) and the third note still rattles horribly.
I tracked it down to the fireplace mantle. ASAP, I'm going to start removing stuff from the mantle until I can make the problem go away. Now that this problem has made itself very clear, it's a good time to fix it.
## Sunday, February 20, 2011
### Room EQ Wizard running
Tonight I've downloaded and installed the highly praised freeware program Room EQ Wizard (REW).
Also set up Emu Tracker Pre soundcard, and Galaxy CM-140 (the SPL recommended by the makers of REW).
Just putting SPL on top of listening chair for now, and didn't get CM-140 calibration downloaded yet.
Measurements are not showing a peak at 42 Hz. They are showing, at least for the right channel, a deep null at 46Hz. (The peak from 38-55 Hz is in the corner, not at listening position which gets the opposite response.)
Left channel showing 10dB dips at 125 and 165 Hz, so I reset my EQ's for that, with 4dB rather than 2.2dB gain. Even though I set octave width at 0.25 for 125, the EQ simply seems to raise the entire region. So I changed the octave width to 0.35 for less ringing (presumably) makes very little difference.
UPDATE: A sweep up to 2kHz reveals that the largest/widest suckout was really around 230 Hz, so I added an EQ for that.
Current EQ's
42Hz -2.0dB 0.4 octave
125Hz 3.0dB 0.3
165 3.0dB 0.3
230 4.0dB 0.35
Result looks like an improvement, anyway.
## Friday, February 18, 2011
### RPG Modex Corner Bass Traps
I've found a different bass trap, this may be the best for my primary resonances 38-50Hz. This is the RPG Modex Corner.
This seems similar in concept to Real Traps Megatraps, but RPG makes them in three different tuned versions, 40Hz, 63Hz, and 80Hz. What I need mostly is *SERIOUS* bass trapping around 42 Hz. The remaining problems are trivial. Funny, the RPG website doesn't mention that the unit comes in 3 tuned versions. These are not, of course, pure Helmholtz resonators. But if the "tuning" optimizes absorption of deep bass (which is really hard) that would be very useful. Most traps don't do much at 40 Hz.
Following DIY discussion on these things, the consensus is there is no good way to predict response, even very knowledgeable designers have to cut and try. Heavier membrane yields lower cutoff frequency.
### Acoustic Treatments
Here's a great article on room treatments by Ethan Winer (maker of Real Traps). Ethan seems like a very good guy and I trust him.
Looking at my room though, it's hard to see where I would fit the Real Traps. I have a very nice spot for a 16 or 20 inch diameter tube trap, but I'm not sure I like the way Tube Traps are constructed with fiberglass, which I'm worried about leaking through the fabric and into my HVAC system. Real Traps tend to enclose the fiberglass in plastic, which seems safer.
I'm planning to order RplusD, the premier acoustic measurement software, today.
I'm still liking my manually adjusted EQ's, though I really want to measure and "improve" them. I listened to some Bach organ works today and was delighted by the clarity of the bass.
For measuring my manual adjustments, however, I have to dial them into the Behringer. (Currently I dial them into the Tact Parametric EQ, but that is turned off during measurement.) So it's a bit of a pain, and the Behringer can't be adjusted with as high resolution as the Tact.
## Thursday, February 17, 2011
### Acoustic measurement programs
Ethan Winer (Real Traps) suggests one of the following programs:
ETF (now replaced by RplusD)
Room EQ Wizard
Fuzzmeasure (for Macs)
Here's a blog where he mentions these and links to his instructional information.
http://www.gearslutz.com/board/bass-traps-acoustic-panels-foam-etc/473189-diy-panel-bass-trap-tester-open-letter.html
### Bass Traps
Recent experiences have suggested to me that Room Correction EQis not only far from panacea, it's not clear if it's worthwhile. Indeed, last year when I had corrected/boosted response at the listening position I had so much boom elsewhere it was causing lots of unnecessary rattling and made guests uncomfortable.
Meanwhile, I do like to do small EQ adjustments by measurement and ear, those actually do seem to help and don't do much harm.
So count me as a new believer in the need for room treatment, especially bass traps.
Leading bass trap manufacturers include Tube Traps (cylindrical and quite expensive for the larger ones), Real Traps (somewhat less expensive, look like better value), and others.
It will be a big problem figuring out where to put these things, the room is already quite stuffed.
### OK, just the basic EQ's now
OK, until I can figure out what I'm doing better, I'm a bit worried about using high Q peaks to fix dips as I was apparently doing with some success on Tuesday.
So I'm back to making this optional EQ #1 in Tact:
Cut at primary resonances: -3dB 42 Hz (this is very conservative adjustment, could use lots more)
Small boosts to midbass" +2.2dB at 125 and 140 hz
NEW: small cut to high end, -0.7dB at 13.7kHz
Sounds quite nice this way. The new small cut to the high end seems to remove high end glare on Shadowfax "Nuclear Village" album. No EQ sounds fine until you try the EQ, then you don't want to go back. If high end cut is increased to 3dB, however, it sounds too dull on top.
Meanwhile, I was thinking why automated Tact didn't work. Tact seems to show Acoustats peaking in room at 18kHz. Other measurements with SPL meters and sinewaves suggests the Acoustats peak at 13.8kHz and fall off above that normally. I retested that using ST 1100A distortion analyzer as my sine generator and Galaxy 140 SPL meter and GR 1933 SPL meter.
GR is very hard to interpret because being random incidence. The Galaxy shows about -3dB at 15kHz and -6dB at 16kHz, definitely does not look like 18kHz peak.
However, unfortunately, Galaxy is only Class II rated to 8kHz. The rolloff I am seeing could very well be rolloff in the meter itself.
Or it could be, as I've suspected, that the Tact microphone isn't calibrated correctly. If the calibration, for example, shows -12dB at 18kHz, but the mike is actually flat, it would add a high frequency peak at 18kHz in the measured response. On Tact user's group, complain about mike calibration is legendary, a standard recommendation is to scrap the manufacturer's calibration and simply substitute a generic calibration for the mike.
It's also very hard to measure with high frequency sine waves. Tiny changes in position or angle can make 6dB or greater difference. At each measurement, you need to move the mike around a bit to find a solid reading. It would be nice to have warble generator. However, the best of all would probably be RplusD from Acoustisoft (formerly known for ETF). Can get whole package including calibrated mike. I think I will do that this month.
## Tuesday, February 15, 2011
### High Q Parametric Corrections?
I've found that while I can correct a small depression around 60Hz, I need a very high Q filter to do so nicely. Q=8.9 and gain=+15. Without the high Q, the boost simply moves a whole region up, with the nearby peak moving up nearly as much as the depression at Fc.
I'm worried that such a high Q filter doesn't really correct the response so much as swamp a local region with a Very Big Peak. This is not visible on the Tact measurements, but I think a finer grained measurement, like one from AcoustiSoft, might do so.
### Alas Room Modes are not "minimum-phase"
According to this argument, many electronic circuits are minimum phase, and loudspeakers have both minimum-phase and non-minimum-phase effects (with minimum phase characteristics dominant, it is claimed, in the better loudspeakers).
http://www.soundfirst.com/EQ_Phase.html
Rooms, alas, have delayed resonances, the epitome of non-minimum-phase.
Room effects, I can only broad brush, which may help or harm.
Some of the fancier room EQ systems now measure both speaker and room effects (using gating) and correct the speakers first.
It's not exactly clear what my Tact system does. The description of what it does does not reveal all the technical details, and the system is not open except to the extent that you can edit a target curve. Reviews are mixed but most say it's better to have target curve capability than none, so it's better than most such systems. Most consumer room correction systems have preconfigured target, they are pure black boxes, take it or leave it. Now there are systems which brag about their approaches in more detail, and may have more configuration options than just setting a target curve, but they tend to be really expensive.
There are open source solutions to room correction by EQ, so you can examine or change the algorithm, but the most developed one works with an obsolete Behringer feedback canceller (has lots of parametric EQ's) because it's cheap. It has hum problems, uses semi-pro rather than audiophile line levels, uses a low-ish sample rate. I bought one at close-out, but then was lured in by the much nicer quality Tact hardware. The open source solution auto-programs the old Behringer through it's serial port.
You could say that any-old computer could do the trick. But any-old computer has a noisy power supply, fan, etc., you don't even want one in your listening room.
In principle, given a system with 10 or more parametric EQ's, you can still use the open source software to figure the optimal parameter settings, then just toggle them manually into Tact or DCX crossover.
I have 13 manual parametric EQ's per channel available in the Tact and about the same number additional ones available in the DCX. Not quite the same as the hundreds of poles Tact can use through it's automated room corrector, but enough to do some good or considerable harm.
## Monday, February 14, 2011
### Manual parametric EQ instead of Room Correction
I have been disappointed by this round of Room Correction with Tact. It very much looks like I cannot do a decent room correction without setting up a target curve which is pretty close to my measured response. I suppose I could do that, but I am not so motivated anymore. It is clear that the correction I used last year was leading to excess room boom at 45 Hz and other problems. Now I have heard how a simple correction can lead to very undynamic sound.
Instead, I am doing what I have long done with my bedroom system, applying manual parametric EQ adjustments.
Starting with no EQ or RCS, I do find that the primary room nodes 38-50Hz are turned up slightly (though this is HIGHLY position dependent, and one can even get cancellation nearby). So a manual EQ should cut this back slightly. I am using the 6dB cut of Tact Parametric EQ #1, but maybe should roll this back a bit since it may cut too much at the listening position.
Also, a manual EQ should attempt to improve the 100Hz-200Hz depression. I think at least part of this depression is caused by frontwall cancellation, also maybe partly by dipole cancellation. My strategy for fixing this is to use multiple staggered small EQ's. I have started with 2 corrections:
140 Hz +2.2dB "octave width=0.6"
125 Hz +2.2dB "octave width=0.3"
(I have programmed these into Tact EQ #1. The Tact allows you to select "octave width" instead of Q. It is nice to be able to select EQ's by remote control.)
The above EQ's have about the right effect, increasing the 100Hz-200Hz response for a pleasant, if not perfect, correction, in combination with an EQ at 42 Hz. The attempt is to make a broad pleateau in the electrical response from 110-180Hz, with the plateau getting taller as we get down to 120Hz (but avoiding peak stimulation of the potential resonance area at 116 Hz). It is often said you cannot correct deep depressions with EQ. That is true, but you can ameliorate small depressions.
Using these EQ's, the outside of the listening area has much less "boom" than last year, but there is still some boom in the corner near the Kurzweil.
With the EQ's, the sound is very sweet. It is one of these paradoxical effects that improving the midbass seems to improve the mid highs.
### Testing the Acoustats
It's interesting to listen to the Acoustats playing through their crossover without powering the subs. This way you can hear more clearly exactly what the Acoustats are doing.
With 121Hz crossover, the Acoustats sound like a tweeter. Zero bass.
With 84 Hz crossover (now LR48), the acoustats sound a little thin but still like a full range speaker.
I brought up two versions of Polyvtsian Dances on Friday Night, and played them as loudly as I could with the new notch filters in place.
No buzzing or rattling whatsoever. It's looks to me like the 84Hz crossover is safe.
On Sunday I tried to duplicate the buzzing problem with "A Story Within A Story." I couldn't make it buzz anymore, even with the notch filters turned off.
Perhaps by operating the Acoustats with a lower crossover, they have "broken in" and loosened up so they don't buzz anymore at 116 Hz.
Actually I should probably go back and test Polyvtsian Dances without the notch filter too.
It does seem that the octave from 100Hz to 200Hz has pretty weak output, particularly from 110-170 Hz.
Taking out the notch filters doesn't change much (surprisingly) but does make the output just noticeably weaker.
I had been motivated to replace the notch filters with "dynamic EQ" filters that activate only for signals loud enough to cause the buzzing. But I can't know how to set up the dynamic filter if I can't reproduce the problem it is supposed to fix.
I've been worried about the high supertweeter level (+15dB) actually causing some problem in it's own right. Normally SPL levels above 20kHz are very low. But you could have some ultrasonic noises, and even at -15dB such noises could cause huge signals in the supertweeter drive.
Something like that seemed to be happening when I played the song Atom Heart Mother Suite by Pink Floyd. The supertweeter channels appeared to be clipping as the red lights were lighting on the Behringer.
So I spent the next 12 hours working on adjustments to fix this, only to later find myself unable to reproduce the original problem to see if the adjustments really help. The original problem might have been caused by some kind of gremlin (like loose connection, or maybe I mistakenly turned off LR48 crossover for supertweets). But I still think the adjustment is worthwhile and makes for better sound, so I keep it anyway. (I had been thinking I would try dynamic EQ instead, but since I can't reproduce the original problem anymore, I'm sticking with the new changes.)
Mainly what I did was add a low pass parametric EQ at 20kHz in the supertweeter drive on the Behringer. I choose 12dB/octave and -15dB level (I'm not sure why Behringer requires you to set levels for a low pass; I think it actually makes it like a shelf rather than a low pass.)
So I have both lowpass and highpass at 20kHz. And I think this *is* the right way to set up a supertweeter: with both highpass and lowpass. (The highpass is a Linkwitz-Riley at 48dB/octave at 20kHz).
With just the LR48 highpass, the level at 20kHz is actually 6dB lower than the maximum (which it approaches asymptotically above 20khz, being pretty much there at 25kHz, just as could be predicted). Now what good is that? Perhaps the supertweeter needs the extra 6dB boost above 20k to maintain flatness, though I would think not. But whether it needs it or not, that is just the way using LR48 works out.
And it's highly undesirable to have the 20-40kHz range be boosted that high, it's just asking for trouble from ultrasonics, etc. What we'd really like is response that keeps on rising until 20kHz, and then doesn't rise anymore, maybe even falling back slightly.
Just to get a sanity check on the voltage sent to the supertweeter, I put my Fluke 8060 and 804 meters on it, and drove my system using my Sound Technology ST-1400B distortion analyzer/generator. I set the level for about 1V output at Acoustat speaker terminals. At 20kHz, the Elac drive wire was about 2.5 volts, and it peaked around 30kHz at 5 volts (!), 14dB higher than Acoustat drive level.
After adding new 20k lowpass, the peak occurs near 20kHz at about 1.8V, then declines gradually. I also changed crossover point for the LR48 high pass to 18kHz. Although I had recently changed levels down to +10 on the Behringer, I am back to +15 with these adjustements (which lowered response quite a bit by themselves).
Not only is this "safer", I think the supertweeter sounds better not being driven so loudly as it was. That's exactly when it makes the "metallic" sound. The new highpass/lowpass combo seems to prevent that.
### Symphony SPL levels
Last year I measured the San Antonio Symphony at 101dB. A friend questions this. Here is a link to other people talking about SPL levels, one guy measured Minneapolis symphony at 106dB.
## Friday, February 11, 2011
### Sometimes it's best to do things the right way (LR48)
In earlier post I described various strategies for crossing over (or not) my electrostatic panel speakers and my subwoofers. Last night, I found that one of the strategies I have passed over for awhile does indeed work very well. That is the strategy of having a symmetrical crossover, with high pass and low pass cutoffs set for the same frequency and using the same crossover function (such as Linkwitz-Riley or Butterworth) on both sides with the same ultimate slope (such as 24dB/octave).
Listening to some FM radio for the first time in a couple months, I was first impressed by the clarity of the sound and incredible spacious imaging. (This mainly results from moving the listening position forward about 8 inches, also from recent changes to the sub/panel crossover.) But I was also appalled by an apparent dryness, bordering on harshness.
Now I've been fiddling a lot with the crossover controls recently, but when I made the above observation I had the panel highpass at 104Hz with BU24 (butterworth).
First I tried changing that to 84 Hz, same as the lowpass on the subs. Big improvement, now the sound is sweet and less dry (funny how improving midbass makes the midrange sound nicer, but not surprising really).
But then it actually sounded a bit boomy. So I changed the BU24 to LR48 (Linkwitz-Riley) and that cleaned up the bass. Now it was sounding very nice and I just kept on listening for quite awhile.
Also funny that when I started and was thinking the sound was harsh, I turned off the humidifier like yesterday. But that didn't seem to have as much affect as changing the crossover. After changing the crossover, I was enjoying FM radio with the even noisier dishwasher running.
Now the idea that both highpass and lowpass sides of the crossover should have cutoffs at 84Hz, and that they should both use LR48, that sounds pretty obvious, right? So why wasn't I doing this already?
Well, with somewhat less careful recordkeeping, I did try this last year. But one of my goals way back then was to take advantage of my dual subs and improve my ability to play more loudly by offloading bass from the panels (which can't play back much bass without bottoming) and onto the sub (very high output capability).
And I had a test, one version of Polyvtsian Dances, where a tympani made by Acoustats rattle. I found that I could get rid of the rattle by moving highpass point to 121 Hz. So that was where I set the crossover and where it stayed for about 12 months. (Actually, for other reasons, I set the sub lowpass even higher, and then let the Room Correction fix up the response, a very bad approach I now know.)
Well I haven't repeated the rattle test, but now I have a notch filter at 116Hz specifically to handle the rattle. So now I may be able to have good sounding bass without the rattling.
The other point was that there are terrible room modes in the range of 90Hz-110Hz. I could in effect EQ these down by spreading the sides of the crossover apart (I call this underlapping).
It turns out to be a very bad idea to have sub playing in the 90Hz and up because it excites room modes horribly.
But it works out fine to play the electrostats down to 84 Hz, because their dipolar character basically doesn't excite those full-wave-and-up resonances.
Hearing how much better it works to have full crossover at 84 Hz, I don't think I'll be trading that away for loudness capability any more. Except I may have a secondary setting for those days when I need more loudness.
I have also been thinking about the possibility of making my notch filters amplitude dependent. The Behringer does provide that option, but it might do more harm to the sound than good. But along those lines...why not have amplitude dependent crossover point? When you play really loud, the crossover automatically ratchets up to a higher frequency to protect midrange panels.
Fully crossing over both sides with LR48 how does the bass sound? Almost like pure electrostatic. Interestingly, LR48 like all LR crossovers actually reduces energy around the crossover point (helping to avoid feeding those resonances) but maintain amplitude in the axial response, so that straight ahead the drivers add up arithmetically.
## Thursday, February 10, 2011
### Strategies, large and small
To move forward in the audio hobby, you should have a strategy. There are many possibilities nowadays. Unfortunately, it looks like a lot of the current audiophile products are of the "fake snake oil" variety, they don't actually work, but people feel they work because they believe in them. The tendency toward superstition in humans is very well known...
My strategy is largely to avoid things that look or smell like "fake snake oil", or at least not to get overly obsessive about them. Sure, I may try a tweak here or there, something that makes sense to me like power conditioning or replacing electrolytic capacitors with polyethylene film capacitors. (I am well aware that power conditioners are quite controversial, and in fact I do not plug my power amplifier into mine, only signal processing equipment.)
Instead, my strategy is to focus on the well known, things that unquestionably make a difference, no question about audibility, but which is better? There are vast opportunities of this nature in audio, and I would like to bring more into thinking about audio reproduction like I do. I can see that I am not alone, however, lots of board now (such as DIYAudio) are filled with audiophiles having a more objective orientation. That does not mean I treasure flat frequency response above all else (see my last post). But at least I want to have some understanding of the colorations I employ to make the audio magic work.
It is clear to me that a very important part of this is getting the bass right. Mind you, I don't believe it's entirely clear what exactly "right" is, and it may vary from one person to the next. It is clear to most people who have investigated this that the best sounding frequency response in a room is not flat. Rooms themselves have low frequency resonances and gain, and within that context if you hear "flat" reproduction it really falls flat.
So this is a subjective art, basically fiddling with the bass until you get it to sound right on most recordings. Now there are many approaches to that. Some people go through many different speaker systems, and/or try many different speaker and listening positions with those speakers. Others stick with favored speaker and try room acoustical treatments or DSP. Still others fiddle with magic points and tourmaline crystals.
I actually have been through many speaker systems in the past, and I think that is one of the best things to do, but it is not my current approach because I already have very good speakers. Currently I am mainly trying to get the bass right by adjusting the speaker positions, delays, crossover, and digital signal processing. It is clear now I may also benefit from some room acoustic treatments to fix room boom not at the listening position. It's clear that neither DSP doesn't offer a complete solution to getting the best bass response.
So now I am getting down to the layer of strategies where I am now. Since January, I've been working with a new crossover design that keeps the subwoofer below major resonances around 100 Hz, and the panels above those resonances. This works very well, though seems a bit lightweight in the midbass (yes, it seems light around 100Hz even though it measures basically flat).
Now there are several other possible crossover strategies. One is to do a more honest ("real") crossover, with subwoofer lowpass and midrange highpass both set at the same frequency, say, 100Hz. That will likely produce lots of "boom" around 100Hz which is currently being suppressed. It wouldn't be a nice way to use the system without further changes. Those additional changes could be either done manually, by setting a parametric EQ depression around 100Hz to cancel the boom, or by running Tact do do a Room Correction, or both individually or simultaneously. But those approaches may work out better than trying to cover up the resonances by underlapping the crossover.
Another obvious strategy is to run the Acoustat full range without sub. Well not if you like head banging bass as I sometimes do.
Another strategy is to run the Acoustats full range with subwoofer merely serving in augmentation mode.
Similar to that would be to crossover the Acoustats as low as possible, such as 40 Hz. I tried that before, didn't like some of the panel resonances I heard running that way, but those may be the very resonances I fixed with notch filters this week.
Yet another is to run the sub as high as possible (it's supposed to have flat clean dynamic response to 250Hz) and cross the Acoustats at that point. That would keep as much bass out of the Acoustats as possible, and give the Tact as much freedom as possible to correct the frequency response (which will definitely be needing lots of correction in the midbass when powered by dynamic ported woofer).
I think for now I'm going to stick with the crossover-around-100Hz strategy, and work it up to get better midbass, or attempt to get some satisfaction with the "real crossover plus EQ" approach.
### Better without RCS (Room Correction) ???
Another late night with good listening and important audio discoveries.
1. Midbass better with Butterworth 24 than Linkwitz-Riley 48. Even with the room curve(s), which currently boost deepest bass below 35 Hz and highest highs above 18kHz, there seems to be a broad depression in the bass using the LR48 crossover for the Acoustat high pass. BU24 gives much nicer midbass sound. By mistake (probably) I had originally done this series of room curves (from Sunday February 6th) using the BU24. Based on the impulse response mainly, I had decided that LR48 was acceptible (and preferable for getting the buzz causing bass out of the speaker). But then I went ahead and ran the RCS measurements with the BU24 selected anyway, because that was what I had just finished testing.
I had determined by Monday that I had made this mistake, but decided to switch back to LR48 for speaker protection and (thinking Acoustat response below 104Hz crossover is nothing much to speak of anyway) thinking it wouldn't make much difference. But all along I have been noticing weak midbass around 100 Hz and just below (which is party by design...underlapping to crossover to minimize the effect of bad resonances around 100Hz).
But until I installed the notch filters on Tuesday, I really did want to keep the buzzes out, and the weak midbass was a small price to pay.
But now I am using notch filters to get around the buzzing (and it continues to work great), I shouldn't have to compromise the rest of the midbass as much. So I went back and tried the BU24. BU24 brings back much of the missing midbass. And actually it does this regardless of whether room correction is used.
This many not be an inherent property of BU24, but a set of circumstances in my system and listening room.
#2 Currently, the uncorrected sound is better ??? Well now that I have decent sounding system without correction, I can run system either way. There is a huge difference with my current correction curve #2 (which actually had its room curve tweaked on Tuesday) and bypass. #2 reduces the highs (which are generally too hot, I admit) and brings up the midbass (only slightly). It clearly reduces resonances, you can hear much more detail in the music because of the reduction of spurious resonances. But by comparison with bypass, the corrected sound is flat, undynamic, and boring. (You don't notice this until you actually compare the two.)
I don't think this is entirely a situation of preferring the euphonic colorations of my system. I think it mainly stems from the fact the the frequency response of the Acoustat, with its tilted up high end in axial response, is a deliberate design choice to make up for the relative beaming of higher frequencies. And I may have to do the correction of bass and midbass better with a better chosen room curve. I will try something closer to actual measured response.
I may be able to make the correction sound better than bypass with addition changes to the room curve or multiple measurements. But for now, I like bypass better.
#3 I can now play Supernatural by Santana (SACD) without unpleasantly strong bass shaking the room apart. Big improvement! It sounds like an entirely different and much better recording now. It was a big disappointment last year how badly this recording sounded without an extra bass cut. Now I can listen to the record without even engaging my "boom correction" filter.
#4 Best to shut humidifier off when playing. My noisy humidifier is about 10 feet from the listening position. I run it all winter to keep my throat from drying out. But when I was playing Dark Side of the Moon tonight (in great contrast with last night) I found the side slightly on the harsh side. The sound generally seemed to improve (though I didn't go back to DSOM) when I turned off the humidifier. Duh! But the big problem in situations like this is how to remember to turn humidifier back on. (It was even worse when I was shutting off refrigerator.) But now I have a trick. I turned off the light in the kitchen where the humidifier is, and shone a tensor light on the humidifier. This trick almost didn't work, as I first went to bed without looking at the kitchen, but I caught it on my first trip to the kitchen a few minutes later.
#5 Supertweeter back to +15. It certainly doesn't seem to add to harshness, not sure if it can actually reduce apparent harshness, but it's usually more fun to have supertweeter at higher level, and with 20kHz level, I think it has lower output than last month (when I was using 15.5kHz LR48 higpass, now I am using 20kHz BU24 highpass, higher but less steep crossover). Not sure yet if I want to keep it like this. I also seem to like room corrections where I leave supertweeter off for measurement, then add it back in for measurement, as I do with correction #3, but #3 currently doesn't enjoy the additional room curve tweaks I put into #2.
#6 One usually thinks that the panels are producing great bass. But just turn the subwoofer off, and it's clear that the panels are mid-tweet drivers. With my crossover anyway.
## Wednesday, February 9, 2011
### Why is there uncorrectable room boom in the first place?
I have a fairly typical sized living room for USA, it's about 14 by 15.5 feet with an 8 to 9.5 foot vaulted ceiling. The passageways through entry and kitchen are wide so they also may make the room slightly larger acoustically. My primitive calculations suggest I should get half wave resonances in the range of 30-40Hz, however they actually appear around 38-55hz. It so happens that my Kurzweil synthesizer is right in the back corner of the room where you get a hugely resonant response (body shaking) in the 38-55hz range.
Now one interesting thing about half-wave resonances is this: In a closed room, they have their maximum amplitude away from the center of the room and towards the walls. Full wave resonances have a third maximum amplitude in the center. My full-wave resonances should be at double the frequencies of the half-wave resonances, so about 76-110Hz. (I have dealt with that part of the boom by underlapping the crossover and by applying room correction, as described in earlier posts.)
So it seems, probably, that typical rooms have half-wave room boom in the range of 30-60Hz. That's a pretty important band for bass fundamentals. Half-wave boom is the kind that seems to accumulate around the walls of the room.
Now most serious audiophiles listen to music from a listening position more in the center of the room rather than the periphery. If you do that, and you have half-wave boom in the range of 30-60Hz, you could probably benefit from the Boom Control I described in the previous post, to handle recordings with a lot of 30-60Hz energy.
Here's an interesting page on standing waves and resonances...I need to learn more here...perhaps my analysis isn't entirely correct yet.
Thinking about the above, it appears that the room modes (1/2 wavelength, 2/2, 3/2, etc) indicate the room is operating like a open tube because the maximum volume (antinodes) for those are on the sides. The quarter wavelength series works like closed tubes, and typically have node on one side and antinode on the other.
Here's the Wikipedia entry on Room Nodes, which notably includes the comment that the attempt to equalize the sound in one position may actually make it worse in others.
http://en.wikipedia.org/wiki/Room_modes
Here's another interesting discussion showing another room effect, the 1/4 wavelength from the rear wall effect. Also nice ETF measurements of actual room nodes:
http://www.realtraps.com/art_modes.htm
### my latest invention: Boom Control
By now, it's pretty clear that using digital signal processing (DSP) both to analyze my system and correct it are very useful. I'm really enjoying the new cleanness that comes from having used notch filters to negate some very serious buzzes in my Acoustat panels. I imagine now I was hearing those buzzes all the time, and I thought they were coming from something else, like the subwoofers. I'm worried that the 116Hz resonance may not actually be easily fixed by giving the panel membranes a hair dryer heat treatment. It may be something related to the frames or individual panels, and a repair may require added bracing, etc. Lots of people find they have to brace their Acoustat 1+1's for better sound.
But moving on, DSP also has limits. Flattening the frequency response at the listening position may not remove serious resonances (room boom) elsewhere in the room. Now, you could take the position that these resonances don't matter. But they often add a "strained" quality to the sound, as you hear walls shaking, etc.
This is an old argument. Those who sell acoustic treatments often say DSP is worthless, and vice versa. The truth is that both DSP and acoustic treatments can be useful under some circumstances, and with some limitations.
DSP is the more magic technology. I find it amazing the way I can simply dial in a few notch filters to fix serious problems with my speakers that could take weeks of frustrating work to fix. And within the range of tasks DSP can accomplish, it has amazing dept, it is amazing to be able to make things like notch filters with a Q of 10 (which seems to be about right for one semi-tone, 8.9 works for two semi-tones).
Now I have come up with a DSP answer to the old DSP limitation of not being able to flatten the frequency response everywhere in a room at the same time. It's not a perfect answer, but it's better than nothing. And it can't be worse, because you can simply turn it off (in my case, with a remote control). I call it Boom Control.
Some recordings need this, others are better without it. Typically, you don't mind the room boom outside the listening position while listening to an acoustic recording. But you may find it intolerable while listening to a recording with electric bass.
It seems that 6dB make a nice filter depth for this sort of thing. I have room boom outside the listening position principally in the region 40-55 Hz. For some recordings with electric bass, it seems helpful to engage a 6dB filter with Q of 1.6 at 42 Hz.
Unfortunately, although intended to reduce the boom outside the listening position, it does also reduce the bass frequency response AT the listening position. It may make some rock recordings sound slightly anemic. On the other hand, it may make some otherwise intolerable recordings very listenable. So, it's a compromise, but it works.
The effect of the reduction around the room periphery seems more than I would have expected from a 6dB reduction, while the bass at the listening position seems reduced a little but not 6dB. Basically, you switch on the Boom Control, and all the bass rattling (or most of it, anyway) goes away. Suddenly you have a real image with bass instruments, etc, right in front of you where it should be. That's how it works sometimes. Other times, you feel "where did the bass go" and then you switch off the Boom Control and you get all you want back again (because perhaps a lot of it was in the 40-55Hz range being attenuated).
Boom Control is essential for the record Bass Ecstasy (which, btw, I find quite fun to listen to in a guilty pleasure sort of way). On Dark Side of the Moon, I prefer to leave the Boom Control off because the engineering of the record has already eliminated the boom and it sounds a bit too dry with Boom Control.
I have dialed in this filter into EQ #1 on my Tact RCS preamp. I can turn it on or off with remote control.
## Tuesday, February 8, 2011
### Buzzes fixed with notch filters
Yes I know now my Acoustat speakers should eventually be taken apart, as much as can be done, and have their membranes heat treated to restore original tension. Hopefully that will get rid of the buzzing resonances at 116Hz and around 172 Hz.
And in recent posts, I've described how I moved the cutoff frequency for the Acoustats from 104Hz up to 121Hz to get rid of the buzzing (that was good for a Tact level of 83dB out of 99.9 on "A Story Within A Story" by Pat Metheny).
That's a neat trick, but added to the midbass depression around 100Hz.
Last night I took a slightly more direct approach and used the Parametric Equalizer option on my Behringer DCX 2496 digital crossover to notch out two notable buzzy resonances in the Acoustats. I added a -15dB notch with Q of 8.9 at 116 Hz, and a -6dB notch with Q of 7.9 at 172 Hz.
With these notches in place, I could restore the crossover point to 104Hz (restoring some of the lost midbass) and still play A Story Within A Story as a Tact level of 93dB without getting the buzz. That's more than 10dB of added dynamic range.
With the "Sine Wave" program I created on my Kurzweil K2661 synthesizer, I can easily go up and down the keyboard to identify the problem buzzes and other resonances, and then test how well I've notched them out afterwards. Before the notch, I did not want to press the A2 or A#2 keys for fear of getting painful sound. Afterwards, I can play all the keys in that region, and they are all about at the same level, with A2 and A#2 only sounding slightly depressed. I identified the need for the second notch at 172 Hz purely by using the keyboard.
The sound with all the new DSP programming I've done recently is incredibly clean, spacious, enjoyable. The dynamic range is far greater, and I can listen to fairly heavy bass without cringing. In addition to the Pat Metheny, I listened to Dark Side of the Moon and Bass Ecstasy. DSOM was like hearing it for the first time, there was so much more inner spaciousness from recent changes, including more forward listening position. Bass Ecstasy could finally be turned up loud enough to be really enjoyable.
Bottom line: if you do not fix buzzes the "old fashioned way" by eliminating them at the source, it is critical that you do something about them anyway using notch filters. It's worth sacrificing a tiny bit of frequency response flatness to get rid of painful and potentially destructive resonances. Little is more annoying than having your loudspeakers buzz.
(BTW, the legendary BBC-designed LS 3/5A speaker actually has an very terrible resonance in it's Bextrene woofer around 1kHz. The KEF B110 woofer was heavily treated, but still has the resonance, so it gets notched out in the crossover, one of the reasons the LS3/5A crossover is so complex. So many of the most highly appreciated audio products depend on the same kind of tricks that I use.)
*****
Playing on the Kurzweil, which is in the corner of my living room, I was struck by how loud the resonances around 42 Hz are, even after full Tact room correction. It turns out that these resonances (which span the range 35Hz to 50Hz) seem to be most offensive in the corners and elsewhere around the periphery of the room. In the central portion of the room, these resonances are not directly audible. The Tact system is only correcting the resonances detected from the position(s) of the microphones. It can't correct resonances that only increase the SPL somewhere else in the room.
Actually, the Tact does have a multiple-measurement feature which I believe is intended to deal with problems like this. But I'm not clear on how to use it or how it works yet.
A problem like this is something that requires a compromise solution of some kind. To reduce the BOOM around the room it may be necessary to take a slight hit on the frequency response at the listening position. But how much? Well, that's up to me, a computer can't make the decision.
One obvious solution might be split-the-difference. Assuming flat response at the listening position, but 20dB peak in the corner, chose a 10dB adjustment.
I tried a 6dB solution like this. I had figured out long ago that a 42Hz notch with Q of 1.6 worked nicely on these resonances (it had already been dialed in on my DCX, but was turned off). So I pulled that region down by 6dB.
Unfortunately, it still sounds boomy in the corners, and 6dB reduction around 42Hz in the listening position makes for a wimpy bass sound on some recordings, while still sounding stressed on certain others (like Bass Ecstasy).
A better solution here would be bass traps to reduce the resonances acoustically. But those are expensive.
For now, I'm going to have 3 EQ levels dialed into my Tact as remote control selectable options. A 3dB reduction (barely noticable loss of bass, but helpful reduction in boom at periphery), a 6dB reduction, and a 10dB reduction.
For the real kick-back bass, I'll turn the EQ off. For entertaining guests, I might use the 10dB reduction.
*****
Both of the above are uses of digital signal processing (DSP) to deal with specific problems. Now when one is also using correction, that raises an additional issue. Should one do the correction measurements with the notch filters in place, or just add them in afterwards?
It occurs to me that in cases like these it is desirable to add the notch filters afterwards. Thus these are "post correction filters".
If the notch filters are added in prior to correction, the correction system may attempt to un-notch them to some degree.
However, it may be that the correction system, by design, doesn't do much for notches. It is longstanding conventional wisdom that audio response notches should not be boosted.
Given lack of understanding about how Tact actually works, it might be desireable to try this both ways.
Likewise for the peripheral room correction. This is a nice kind of thing to have as remote control selectable option as a post-correction filter.
In principle, the need for post correction filters is reduced by the Tact having "room curve" facilities. You could draw needed notches right into the room curve. But once again, not knowing exactly how to draw a Q of 8.9 notch makes this a bit tricky. Also, it's far easier to make and test small adjustments by turning knobs on the Behringer than running the complicated RCS 2.0 software.
Nothing beats an automated system like RCS for flattening (or curving in desireable ways) the overall frequency response. And it works OK on most resonances. But some things require a bit more focus to do correctly, and in tricky situations setting up notches and other post-correction filters manually after full correction might be the way to go.
I would like Tact better if it were a more open system, say for example if you could chose which correction algorithms you wanted to use. But it is still far better than "fully automated" systems in that Tact lets you can draw target room curve, and see what measurements are being made so you can fix things up prior to correction.
There has been a trend toward fully automated DSP systems recently. I don't trust those at all.
## Monday, February 7, 2011
Here are the Quad Esl-63 impulse tests done by John Atkinson (halfway down page):
His actual impulse response test is done with 55us pulse. That would correspond to about 19kHz frequency, thus it does not show effects from overall group delay down to 20 Hz. Nevertheless, what you see is a big positive pulse followed by a smallish overshoot, about 15% at most, but which lasts a bit longer then the initial impulse. Thus from one cycle input you are getting two out, but the second is greatly reduced. He claims that the length of the tail on the impulse response corresponds to 12kHz resonant frequency.
I wonder what this would look like with a 1ms pulse. Probably more complex, like I get from my Acoustat.
Then he shows calculated step response. If reproduced accurately, that wouldn't even be a half cycle. But it does show initial decay in about 1ms, overshoot, and some LF resonance after that (drumhead?).
He shows pretty nice squarewaves also, with 12kHz ringing.
Charles
### What can/should a loudspeaker impulse response look like?
If you have a positive unidirectional pulse, which seems to be what the Tact actually uses, what do you get from a speaker through a microphone. It cannot be an identical positive unidirectional pulse! The speaker/microphone system is a bandpass system which has to, at least, be adding at best about one additional "cycle." It cannot reproduce DC, so a signal with DC offset has to be bent somewhat around that limitation.
The response you could get could start with a brief negative leading edge cycle, the positive cycle of the large body of the pulse itself, and a trailing negative restoration cycle. So 3 half cycles from 1, and I think that's about the best that can be done from any speaker without DC capability (the best actually might be more like two half cycles, depending on duration). Assuming the pulse is long enough to have significant low frequencies, like 1-10ms, so perfect reproduction of the pulse would otherwise require DC capability. (The Tact seems to use two pulses in measuring actually, for higher and lower frequencies, and have separate correction algorithms for each.)
In fact, that is the kind of thing I see if I measure bandpass curves electronically from my crossover, not even going through speakers. In fact, doing this led me to believe it is not a good idea to use very high order Linkwitz Riley LR48 in the extreme treble. You get at least an extra cycle of ringing from that, in the electronic signal itself, and Butterworth 24 gives about the cleanest impulse short of single pole. Strangely, at lower frequencies used for my subwoofer crossover, the high order LR48 high pass impulse electronic response looked more OK, or at least just lays on top of the panel response less objectionably. Note that you can't entirely judge a crossover by looking just at its high or low pass section in isolation; used together the combination should approximate some kind of ideal. But the ideal ideal is most often an "all-pass" response which shifts phase, only acoustic 6dB/octave crossovers can do better. Nevertheless, in practice the ideal will not be achieved, so it's best if each drive signal stays as simple as possible.
After much work (!), that's also what I can get from my very complicated system. Actually the impulse looks like about 3 full cycles, with the subsequent two being greatly reduced in level, plus the usual digital aliasing stuff around the edges. My truly great achievement was getting a combined system response, including sub and tweeter playing at uncompromised levels, that has an impulse response that looks barely different than the Acoustat alone playing by itself. I never expected that, I only expected that the acoutstat-by-itself would look a bit simpler, because I hadn't thought about the issues very much.
And another enduring question is what is the correct polarity? I believe the correct polarity is not that of the leading transient, which may be always out-of-polarity in the ideal bandpass case, but that of the larger square wave what follows. BTW, I inverted the tweeters to give them a puzzle-fitting response. Just by themselves, the acoustats seem to have out-of-polarity leading edge and half cycle followed by full positive cycle, somehow I needed to invert the tweeters to make it fit precisely.* (Strangely, the tweeters actually have a kind of quasi-DC response... More about that discovery later...)
I wonder if some would claim the correct polarity has the leading edge in polarity, so I could be wrong. Perhaps it depends on the number of octaves in the bandpass, and what the bandpass functions are. And perhaps also it becomes fundamentally a subjective question in the case of a system with sufficient group delay (as probably most are). Do you want the bass in polarity or the treble in polarity (assuming you have to make the choice)? If the leading edge defines the treble, but a larger partial cycle follows in opposite but correct polarity, that indicates the treble is out-of-polarity but the midrange--if-not-the-bass--is in polarity.
Another interesting (and now very important) issue is the fundamental panel resonance. I believe there is some fundamental panel resonance in the region of 55 Hz. That resonance is how the speaker maintains frequency response to 40 Hz in spite of being a narrow dipole, I think. Well, it has that resonance when manufactured, but age probably causes membrane to get loose, hence less controlled, which may mean higher resonance harmonics, and that is what seems to be happening, I get buzzing around 110 Hz (on one recording anyway). Well, I need to fix the speaker, the Acoustats use HS65 which is actually 6.5 mil heat shrink. The fix is to use hair dryer to shrink the membrane back to tightness. It's supposed to last about forever if you keep doing that. Other than that, the panels cannot be repaired, you can scavenge old units or make new ones using the same principles (people have claimed to do that and say they never want to go back, and one beauty of the acoustat design is that there is nothing in it that couldn't be done in a garage with readily available and cheap materials).
In the meantime, and this is what I did before and now discovered I must do, I raise the crossover point to 121 Hz and the problem disappears at reasonable levels. I thought I could get away with lowering the crossover point (and with brand new panels, I ought to be able to do so) but apparently not for now.
But anyway, my original question relates to this because we have to consider the panel not as a DC tracking system but a fundamentally resonant system, with fundamental low frequency "drumhead" resonance and high frequency "breakup" resonance. These affect how impulse is going to look, even in the absense of crossovers, reflections, diffractions, etc.
Speaking of which, I should dig out the old impulse picture from an ESL63, which I thought looked pretty good.
### Room Correction Weekend ends with A Story
I did some other fun things also, but the bulk of the first weekend in February 2011 was spent making microphone measurements of my audio system using my Tact Room Correction System (RCS) 2.0 Preamp, making correction curves, and listening to them (Saturday night didn't end until 7am because I couldn't quit listening because it sounded so good). A gazillion measurements were made, and I made many new important discoveries. Photos were taken of many graphs, but it will take a week just to sort through them. It ended on a mixed note, however, so-to-speak. Playing through Sonos the Pat Metheny track "A Story Within A Story" revealed a buzzy bass note in the intro. Damn. I need, someday which will probably not be soon, to take the Acoustats apart and give their membranes the hair dryer shrink treatment. However, I can fix the buzz by running the Acoustats as I did last year, Crossed over above 120hz. (Actually, I think I previously crossed at 116, but with current correction curves I need to cross at 121, and higher might be even better wrt reducing potential for buzz.)
This time, since January I had tried to cross the Acoustats in at 104Hz to avoid a disturbing room resonance. The subs cross out at 85Hz also to avoid that resonance, the resonance pick up the tab in the middle giving reasonably smooth response (and nicely boomless) without correction.
Great idea I thought, which could be applied to most speaker systems: stagger the sub crossover around the room ceiling resonance around 100Hz. And the Acoustats are a "full range" speaker (as some people define it, anyway) that has audible bass (with room gain) down to 40Hz or so. So there not only shouldn't have been a problem changing crossover from 116 down to 104Hz, that should give the system cleaner "panel bass" (actually, many people think panel bass is fake sounding, but flat speaker lovers usually think cone bass is fake sounding).
But there is a problem, because of my 20 year old panels and my desire to listen at pretty substantial (not ear damaging) levels, and because of room correction itself. Because the Acoustats have a deep midbass depression in their response between 110 and 400 Hz, the room correction is fixing that with midbass boost. That midbass boost is pusing the speaker into noisy distortion around 110Hz. Without the boost, I wouldn't get the distortion without playing considerably louder (though I have not tried this, I simply switched to correction Bypass and the buzz went away at the same level).
But the panels clearly have a problem, and even if boost is required to make them sound bad on A Story Within A Story, they could be distorting less noticeably on other music, and in some cases, without boost.
You could pin the problem on the crossover point, the room correction, the loudness level, and ultimately the speakers. In the sense that the speakers shouldn't have this problem at this frequency and level, it is a speaker problem that ultimately needs to be addressed (and yes I have even thought of buying a totally different kind of speaker, like Magnepan 1.7's or Linkwitz Orion). But meanwhile, I can work around it by judicious choice of crossover, and as long as I don't play too loud. Using current room correction curve (measured for 104Hz panel crossover) #2, but with post-correction change to 121Hz crossover, I can play A Story Within A Story to 81 gain level on Tact preamp without distortion. At 82 the distortion is barely audible, at 83 he distortion sounds like "just a normal part of slap bass" (but it's being exaggerated by speaker distortion to sound qualitatively different than the actual recording).
I don't want to give up the correction. The perfectly calibrated midbass boost brings life back to music. The music sounds so much more real with good midbass, even if I have to compromise that midbass slightly by leaving a hole in the midbass between 104Hz and 121Hz by moving the panel crossover up to 121Hz. That is peanuts compared with having the whole range depressed. By the way, I think the whole range depression may result partly from the infamous dipolar cancellation. Linkwitz deals with this in his design by calculating the effect and deliberately equalizing it. I'm canceling the effect by measuring the system and room correcting it.
So I'm listening with the small hole in the midbass instead of the big midbass depression that I had previously. Eventually I'll have to go back and do a whole new series of corrections based on the new crossover. I was going to tell you how much work that was this time (I did two sets of 6 corrections over the weekend...I had to run a second set because I made a couple of mistakes in the first set; each set has two measurements (for helpful redundancy) for regular, no supertweet, and no sub or supertweet conditions). But I before I do that, I also need to see if I can push the subwoofer to cross over slightly higher. That work will require more measurements and tests too.
I can just press the Bypass button on the Tact remote to bring uncorrected response. It's no longer the uncorrected response from January, as prelude to Tact correction I changed tweeter highpass to 20kHz and changed from LR48 to BU24, level reduced by 2dB net, and fine-tuned crossover delay. That made for something like perfect impulse reproduction, at least with the supertweeter adding to the Acoustat nicely and even making impulse sharper. And now, of course, bypass now has both the hole in the 104-121Hz midbass AND the 110-400Hz depression, whereas before it just had the depression, so now bypass is slightly worse in the bass, arguably better in the treble than it was before.
Anyway, though the bypass is slighly different than before, I don't believe it has gotten much worse. But compared to the corrected response, you just don't want to listen to it anymore. Sure it's very open. But it's very thin and bright sounding, with too much highs above 1kHz (just consistently bright) and no midbass.
That was why I do room correction!
While great progress was made, now I know that even more work will be required that I was expecting to be sufficient, with no end in sight (such as working on the speaker itself).
But that's the way life is, isn't it? If there were nothing more to be done, how fun would that be?
.
## Saturday, February 5, 2011
### OK, perhaps I should worry about group delay
After studying the loudspeaker crossover issues in the 1977-1983 time frame, I came to the conclusion that the Linkwitz-Riley was by far the best kind of crossover for most drivers. Generally speaking the 24-dB per octave (LR24) would be the best crossover choice. Linkwitz concluded that the group delay (a compromise) introduced in the overall response of the crossover (by design, the drivers are always in phase with each other through the crossover region) was not audible and of negligible importance.
Only a few crossover designs, like the 6db per octave acoustic crossover used in Thiel, Vandersteen, and some others, can achieve total lack of group delay in the summed response. And that requires multiple other design compromises.
When the Behringer DCX 2496 digital crossover came out, we were treated to LR48 achieved in hirez digital. What could be better? That's what i have adopted uncritically since 2005; I now use the Behringer in both living room and bed room systems.
Now, I have figured out how to examine the impulse and frequency response of the crossover network by itself, and I am not so sure LR48 is the best choice. For augentation purposes, where there is no perfect cancellation of phase artifacts, Butterworth 24db per octoave (BU24) looks like the choice that gives the best compromise between steep cutoff and lack of visible time dispersion.
LR24 might be a better choice if you had a perfect acoustic LR24, something very hard to achieve in practice, but given lack of perfection, the best bet is probably to minimize time dispersion in each crossover member with one that provides less dispersion
In a quasi-augmentation mode, or as a solo highpass network, the LR24 increases time dispersion AND reduces sharpness of cutoff compared to BU24. Im not sure of the advantages of the Bessel, it may have he steepest cutoff, but it is marred by 20dB passband irregularities. It might be OK for supertweeter where, say, above 20K you don't care about passband irregularities.
I think it's possible that in the low frequency crossover (the highpass on the acoustats is currently set to 104Hz) LR48 works OK, and is extremely beneficial in reducing panel flap at high volume levels.
The visible difference between BU24 and LR48 is pretty small at 104Hz, the LR48 does have a bit of initial out-of=polarity undershoot, and somewhat more overshoot on the trailing edge followed by slow recovery. Whereas the BU24 almost looks untouched, the perfect Tact impulse (that is, as perfect as the Tact gets) almost. But it just doesn't phase me much, at least on screen.
But what has been driving me apoplectic about this is the supertweeter highpass. It turns out that the highpass signal from LR48 at 15.5 or 20 kHz has 3-4 cycles of ringing at 20kHz. That's all there is, it doesn't look like an impulse at all, just ringing. Superimposed on a perfect impulse, it smears it out considerably. My time domain purist friends should be laughing at me now.
That's exactly what I've been seeing in the system impuse time response. And it bugged me so much I refused to print it yesterday. Now I've figured it out, at least partly. Even if the supertweeter is reproducing the signal that it receives perfectly, that impulse response looks like 4 cycles of ringing at 20kHz because that is the signal that the crossover is providing.
Now in the context of a perfect LR48 crossover, with perfect high and low frequency drivers crossing over, the majority of the phase anomalies might well cancel out (I am not entirely sure of this...) and it wouldn't look so bad. Maybe.
But in the context of the kind of slap dash (if infinitely pondered) systems which are the only kind I can put together, not being able to hire a team of engineers, it's looking to me like simpler is better, probably BU24, the time domain smearing is cut in half or less, in fact it doesn't look like ringing anymore, it looks like a double pulse, which probably adds nicely with the low frequency system pulse. | 20,412 | 85,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-17 | longest | en | 0.939046 |
https://minuteshours.com/354-7-hours-in-hours-and-minutes | 1,623,652,764,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611445.13/warc/CC-MAIN-20210614043833-20210614073833-00078.warc.gz | 368,486,846 | 5,088 | # 354.7 hours in hours and minutes
## Result
354.7 hours equals 354 hours and 42 minutes
You can also convert 354.7 hours to minutes.
## Converter
Three hundred fifty-four point seven hours is equal to three hundred fifty-four hours and forty-two minutes. | 63 | 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} | 2.609375 | 3 | CC-MAIN-2021-25 | latest | en | 0.863848 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_INN/TRS/AG01/%233.48.trs.Thm26:POLO_7172_DP:OLD.html.lzma | 1,718,974,862,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00048.warc.gz | 88,704,336 | 2,071 | Term Rewriting System R:
[x, y]
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, s(x), s(y)) -> s(x)
if(false, s(x), s(y)) -> s(y)
g(x, c(y)) -> c(g(x, y))
g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
F(s(x)) -> F(x)
G(x, c(y)) -> G(x, y)
G(x, c(y)) -> G(x, if(f(x), c(g(s(x), y)), c(y)))
G(x, c(y)) -> IF(f(x), c(g(s(x), y)), c(y))
G(x, c(y)) -> F(x)
G(x, c(y)) -> G(s(x), y)
Furthermore, R contains two SCCs.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polynomial Ordering`
` →DP Problem 2`
` ↳Polo`
Dependency Pair:
F(s(x)) -> F(x)
Rules:
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, s(x), s(y)) -> s(x)
if(false, s(x), s(y)) -> s(y)
g(x, c(y)) -> c(g(x, y))
g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
Strategy:
innermost
The following dependency pair can be strictly oriented:
F(s(x)) -> F(x)
There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(s(x1)) = 1 + x1 POL(F(x1)) = x1
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 3`
` ↳Dependency Graph`
` →DP Problem 2`
` ↳Polo`
Dependency Pair:
Rules:
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, s(x), s(y)) -> s(x)
if(false, s(x), s(y)) -> s(y)
g(x, c(y)) -> c(g(x, y))
g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polynomial Ordering`
Dependency Pairs:
G(x, c(y)) -> G(s(x), y)
G(x, c(y)) -> G(x, if(f(x), c(g(s(x), y)), c(y)))
G(x, c(y)) -> G(x, y)
Rules:
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, s(x), s(y)) -> s(x)
if(false, s(x), s(y)) -> s(y)
g(x, c(y)) -> c(g(x, y))
g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
Strategy:
innermost
The following dependency pairs can be strictly oriented:
G(x, c(y)) -> G(s(x), y)
G(x, c(y)) -> G(x, if(f(x), c(g(s(x), y)), c(y)))
G(x, c(y)) -> G(x, y)
Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:
if(true, s(x), s(y)) -> s(x)
if(false, s(x), s(y)) -> s(y)
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(if(x1, x2, x3)) = 0 POL(c(x1)) = 1 + x1 POL(0) = 0 POL(g(x1, x2)) = 0 POL(false) = 0 POL(G(x1, x2)) = x2 POL(1) = 0 POL(true) = 0 POL(s(x1)) = 0 POL(f(x1)) = 0
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 4`
` ↳Dependency Graph`
Dependency Pair:
Rules:
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, s(x), s(y)) -> s(x)
if(false, s(x), s(y)) -> s(y)
g(x, c(y)) -> c(g(x, y))
g(x, c(y)) -> g(x, if(f(x), c(g(s(x), y)), c(y)))
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes | 1,197 | 3,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-26 | latest | en | 0.620446 |
https://www.scribd.com/document/365791344/lesson-plan-place-value | 1,566,349,345,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315695.36/warc/CC-MAIN-20190821001802-20190821023802-00363.warc.gz | 959,275,071 | 55,001 | You are on page 1of 1
Place Value
Overview & Purpose Education Standards Addressed
(2) Number and operations. The student applies mathematical process
standards to understand how to represent and compare whole numbers, the
Students will understand place value by identifying and demonstrating what each relative position and magnitude of whole numbers, and relationships within the
numeral of a three-digit number stands for. numeration system related to place value.
(A) use concrete and pictorial models to compose and decompose numbers up
to 1,200 in more than one way as a sum of so many thousands, hundreds, tens,
and ones
Lesson Details
Objectives Students will provide a number when given base ten Materials Needed
(Specify skills/information that will blocks. Bingo cards
be learned.)
Students will use base ten blocks to demonstrate a Markers
number. Notecards
Base ten blocks
Information Anchor chart- Explain one blocks to ten blocks
(Give and/or demonstrate necessary Explain ten blocks to 100 block
information)
Place Value- Tells the value of a digit in a number.
Verification Ask students to demonstrate a three-digit number using Other Resources
(Steps to check for student base ten blocks. Monitor by two checks. (e.g. Web, books, etc.) | 265 | 1,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | latest | en | 0.845314 |
https://www.quizover.com/online/course/0-1-transverse-waves-siyavula-textbooks-grade-10-physical-by-openstax | 1,540,138,340,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514162.67/warc/CC-MAIN-20181021161035-20181021182535-00213.warc.gz | 1,013,245,110 | 20,649 | # 0.1 Transverse waves
Page 1 / 4
## Introduction
Waves occur frequently in nature. The most obvious examples are waves in water, on a dam, in the ocean, or in a bucket. We aremost interested in the properties that waves have. All waves have the same properties, so if we study waves in water, then we can transferour knowledge to predict how other examples of waves will behave.
## What is a transverse wave ?
We have studied pulses in Transverse Pulses , and know that a pulse is a single disturbance that travels through a medium. A wave is a periodic, continuous disturbance that consists of a train or succession of pulses.
Wave
A wave is a periodic, continuous disturbance that consists of a train of pulses.
Transverse wave
A transverse wave is a wave where the movement of the particles of the medium is perpendicular (at a right angle) to the direction of propagation of the wave.
## Investigation : transverse waves
Take a rope or slinky spring. Have two people hold the rope or spring stretched out horizontally. Flick the one end of the rope up and down continuously to create a train of pulses .
1. Describe what happens to the rope.
2. Draw a diagram of what the rope looks like while the pulses travel along it.
3. In which direction do the pulses travel?
4. Tie a ribbon to the middle of the rope. This indicates a particle in the rope.
5. Flick the rope continuously. Watch the ribbon carefully as the pulses travel through the rope. What happens to the ribbon?
6. Draw a picture to show the motion of the ribbon. Draw the ribbon as a dot and use arrows to indicate how it moves.
In the Activity, you have created waves. The medium through which these waves propagated was the rope, which is obviously made up of a very large number of particles (atoms). From the activity, you would have noticed that the wave travelled from left to right, but the particles (the ribbon) moved only up and down.
When the particles of a medium move at right angles to the direction of propagation of a wave, the wave is called transverse . For waves, there is no net displacement of the particles (they return to their equilibrium position), but there is a net displacement of the wave. There are thus two different motions: the motion of the particles of the medium and the motion of the wave.
The following simulation will help you understand more about waves. Select the oscillate option and then observe what happens.
## Peaks and troughs
Waves have moving peaks (or crests ) and troughs . A peak is the highest point the medium rises to and a trough is the lowest point the medium sinks to.
Peaks and troughs on a transverse wave are shown in [link] .
Peaks and troughs
A peak is a point on the wave where the displacement of the medium is at a maximum. A point on the wave is a trough if the displacement of the medium at that point is at a minimum.
## Amplitude and wavelength
There are a few properties that we saw with pulses that also apply to waves. These are amplitude and wavelength (we called this pulse length).
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
how did you get the value of 2000N.What calculations are needed to arrive at it
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The fundamental frequency of a sonometer wire streached by a load of relative density 's'are n¹ and n² when the load is in air and completly immersed in water respectively then the lation n²/na is
Properties of longitudinal waves | 1,593 | 6,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | longest | en | 0.943335 |
https://answers.search.yahoo.com/search?p=counts+v.+counts+chart&ei=UTF-8&xargs=0&fr2=rs-bottom%2Cp%3As%2Cv%3Aw%2Cm%3Aat-s | 1,606,672,639,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141201836.36/warc/CC-MAIN-20201129153900-20201129183900-00330.warc.gz | 190,957,943 | 33,138 | # Yahoo Web Search
1. ### Embroidery v .s. Counted cross stitching?
...the colour table that colour is in. There are charts on google where the colours are arranged in stripes...
2 Answers · Games & Recreation · 10/01/2013
2. ### In blackjack using the hi-low counting system, why does the illustrious 18 say to stand 16 v 10 on a 0 count ?
... to reduce the houses edge on the game. Use the chart below to work out whether you should HIT, STAND or DOUBLE ...
1 Answers · Games & Recreation · 08/12/2010
3. ### How exactly do you count the elements in your birth chart ?
The elements are based on the corresponding astrological planets. planetary placements at your time of birth are what determines the degree. The degree tells you "how much" influence the element/planet...
1 Answers · Entertainment & Music · 06/04/2011
4. ### The best book for BlackJack Basic Strategy and Counting ?
...be one that has different charts depending on slight ...bankroll. As for card counting , it is very difficult to...20 http://www.youtube.com/watch? v =BCkhpuWw-Rg...
3 Answers · Games & Recreation · 02/03/2010
5. ### How to run a daily count on the gainers vs losers on the s&p 500?
...the returns at the end of the day and then use the histogram function to create charts (e.g. total negative vs. total positive returns or all return counted in 1% steps and so on). There are a lot of videos...
6. ### what is the integer count number in a flowchart?
... count number" is a type of standard flow chart terminology. It is possible that you ...from 5 to 25" refers to a fixed count loop ( v .s. a variable loop), where the loop parameter equals...
1 Answers · Computers & Internet · 12/09/2007
7. ### progressed chart v transit chart whats the difference or are they the same thing?
...and emotional progession throughout life. These are calculated by counting one day per year of life and setting u the same time and place chart for that day. Example: if someone is born on May 6th...
1 Answers · Entertainment & Music · 13/02/2010
8. ### can wordpad count words and pages?
... Gantt Chart Template Yes... "Word Count Tool." 5 ...quot;Ctrl" and " V " keys on your...
2 Answers · Computers & Internet · 28/01/2013
9. ### Since government has accepted Fannie and Freddies debt, shouldn't it be counted in with the countries debt?
5 Answers · Politics & Government · 06/03/2010
10. ### Where can I find a chart for number base 32?
What kind of chart are you looking for? ...: so it seems you just want to count from 1 to 50(base 10) in base 32. ...22 letters. That's a- v . Now, just count like base 10, only...
2 Answers · Science & Mathematics · 25/08/2011
2. ### Also Try
counts v. counts chart printable | 689 | 2,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-50 | longest | en | 0.82167 |
https://mcqslearn.com/cost-accounting/inventory-costing-methods.php | 1,718,838,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00713.warc.gz | 341,233,773 | 16,874 | BBA Finance Degree Courses
MBA Cost Accounting Practice Tests
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Books:
Apps:
The Inventory Costing Methods Multiple Choice Questions (MCQ Quiz) with Answers PDF, Inventory Costing Methods MCQ PDF e-Book download to practice MBA Cost Accounting Tests. Learn Capacity Analysis and Inventory Costing Multiple Choice Questions and Answers (MCQs), Inventory Costing Methods quiz answers PDF to study online BBA courses. The Inventory Costing Methods MCQ App Download: Free learning app for inventory costing: manufacturing companies, throughput costing, inventory costing methods test prep for best online colleges for business administration.
The MCQ: If budgeted fixed cost is \$26000, per unit budgeted denominator level is 1300 units, then budgeted fixed cost will be; "Inventory Costing Methods" App Download (Free) with answers: \$50; \$30; \$20; \$40; to study online BBA courses. Practice Inventory Costing Methods Quiz Questions, download Google eBook (Free Sample) for online bachelor's degree in business management.
## Inventory Costing Methods MCQs: Questions and Answers
MCQ 1:
An approach used for choosing capacity level, having no beginning inventory, is classified as
1. write off variance approach
2. write in variance approach
MCQ 2:
If the budgeted fixed cost is \$26000, per unit budgeted denominator level is 1300 units, then budgeted fixed cost will be
1. \$50
2. \$30
3. \$20
4. \$40
MCQ 3:
If target operating income is \$38000, contribution margin per unit is \$400, then the number of units must be sold to earn targeted operating income will be
1. 65 units
2. 75 units
3. 95 units
4. 85 units
MCQ 4:
The managers using capacity planning do not make
1. pricing decisions
2. marketing decisions
3. financial decisions
4. cost budgeting decisions
MCQ 5:
The budgeted fixed manufacturing cost is divided by budgeted fixed manufacturing cost per unit to calculate
1. fixed material price
2. variable materials price
3. fixed production units
4. budgeted production units | 469 | 2,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.813749 |
http://cubercsl.cn/solve/contest/Timetable/ | 1,540,111,032,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513804.32/warc/CC-MAIN-20181021073314-20181021094814-00111.warc.gz | 80,608,441 | 14,921 | # 描述
Ivan is a student at Berland State University (BSU). There are $n$ days in Berland week, and each of these days Ivan might have some classes at the university.
There are $m$ working hours during each Berland day, and each lesson at the university lasts exactly one hour. If at some day Ivan’s first lesson is during $i$-th hour, and last lesson is during $j$-th hour, then he spends $j - i + 1$ hours in the university during this day. If there are no lessons during some day, then Ivan stays at home and therefore spends $0$ hours in the university.
Ivan doesn’t like to spend a lot of time in the university, so he has decided to skip some lessons. He cannot skip more than $k$ lessons during the week. After deciding which lessons he should skip and which he should attend, every day Ivan will enter the university right before the start of the first lesson he does not skip, and leave it after the end of the last lesson he decides to attend. If Ivan skips all lessons during some day, he doesn’t go to the university that day at all.
Given $n$, $m$, $k$ and Ivan’s timetable, can you determine the minimum number of hours he has to spend in the university during one week, if he cannot skip more than $k$ lessons?
## Input
The first line contains three integers $n$, $m$ and $k$ ($1 ≤ n, m ≤ 500$, $0 ≤ k ≤ 500$) — the number of days in the Berland week, the number of working hours during each day, and the number of lessons Ivan can skip, respectively.
Then $n$ lines follow, $i$-th line containing a binary string of $m$ characters. If $j$-th character in $i$-th line is $1$, then Ivan has a lesson on $i$-th day during $j$-th hour (if it is $0$, there is no such lesson).
## Output
Print the minimum number of hours Ivan has to spend in the university during the week if he skips not more than $k$ lessons.
Input Output
# 思路
• 分组背包……乱搞一下就好。
• $dp[i][j]$表示前$i$天翘了$j$节课的最少在校时间。
• 然后对于每一天，枚举翘课的节数，可以用滑动窗口来得到最小的在校时间。然后转移就好了。
• 比赛的时候没有想到滑动窗口，XJB写了个贪心的选择，炸了……
0% | 575 | 1,983 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-43 | latest | en | 0.913889 |
https://vustudents.ning.com/group/mgt613productionoperationsmanagement/forum/topics/mgt613-production-operation-management-assignment-1-opening-dat-1 | 1,632,391,302,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00372.warc.gz | 638,596,181 | 18,732 | www.vustudents.ning.com
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# MGT613 Production/Operation Management, Assignment # 1 Opening Date November 26, 2015 and due date will be December 02, 2015.
Assignment # 01
production/operations management (mgt613)
Marks: 10
Topic: Forecasting
Objective: The objective of this assignment is to numerically perform the concept of different types of forecasting.
Scenario:
There are certain businesses whose performances/sales get affected by the seasonal conditions of their regions. Fast moving consumer goods are one of those businesses, whose product demand is dependent and very much affected by the weather conditions. Huge demand differences exist in summer and winters. Eggs and poultry business is one of those, in winters its demand is high and is low in summer.
Fine Eggs and Poultry Ltd. (private) is a firm located at Rawalpindi. It holds 60% of Rawalpindi market share in this business. Its customers include mega-marts, general stores, bakeries and burger shops. As the winter is approaching, so its management has to prepare forecasting schedule and table to forecast demand of eggs. Based upon the previous historical records, the management extracted following table:
Week (s) Demand units (000) 1 700 2 800 3 850 4 870 5 900 6 920 7 960 8 980
From the above given information you are required to calculate:
• 3-week moving average forecast demand in week 4, 6 and 8.
• Weighted moving average forecast demand in week 5 with the same and equal weights of all previous weeks.
• Exponential smoothing forecast demand for all weeks using 0.9 as smoothing constant. However the management predicted 650,000 demand in first week
Note:
1. The answer should be in mathematical form only and doesn’t need any theory. But it should also include formulas.
Short Demo:
If you want to know how to download and submit an assignment on VULMS, watch the following short demo on VU Facebook page.
Important:
For acquiring the relevant knowledge, do not rely only on handouts but watch the course video lectures and read additional material available online or in any other mode.
Dear students!
As you know that Post Mid-Term semester activities have started and load shedding problem is also prevailing in our country. Keeping in view the fact, you all are advised to post your activities as early as possible without waiting for the due date. For your convenience; activity schedule has already been uploaded on VULMS for the current semester, therefore no excuse will be entertained after due date of assignments, quizzes and GDBs.
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### Replies to This Discussion
lets start
Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
slide of lec no 9 & 10 is related to this assignment plz follow the example of the slide and try to solve it.
samra sis kindly share your formula for all question ...
r u apply the formula of handout or not?
and question 3rd kindly explain i don't understand it
ap ki values sab kaisi nikli han yahi puch rahi ap sa ...samra tariq
samra the first week demand is 700
= F4= 700+800+865/3 =783.33
F6= 700+800+850+870+900/5 = 824
F8= 700+800+850+870+900+920+960/7 = 857
F5= 700+800+850+870/4 = 805
Samra plz check ur F8
samra how we know the value of w?
We are not given the weights, but in 2nd part it is mentioned: "with the same and equal weights of all previous weeks"
Weights must add to 1, and for 4 weeks it would be 1/4=0.25 for each
Regards
A...z sis my 1st and 2nd answer are same with u but i can't understand the samra sis answer question 1st the option 1st are same with both of same but the option 2nd or 3rd are different.
i think it is correct solution samra sis
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2 hours ago | 1,004 | 4,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.9385 |
https://complex-systems-ai.com/en/neural-algorithms-2/vector-quantization-learning/ | 1,725,766,649,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00839.warc.gz | 169,253,505 | 29,889 | # Learning vector quantization
Contents
## Vector quantization learning algorithm
Vector quantization learning is linked to the self-organizing map which in turn is inspired by the self-organizing abilities of neurons in the visual cortex.
The information processing objective of the vector quantization learning algorithm is to prepare a set of codebook (or prototype) vectors in the domain of the observed input data samples and to use these vectors to classify untreated examples. An initially random vector pool is prepared which is then exposed to training samples.
A winner takes all strategy is used where one or more of the vectors most similar to a given input pattern are selected and adjusted to be closer to the input vector, and in some cases further from the winner for the finalists. . Repeating this process results in the distribution of codebook vectors in the input space that approximate the underlying distribution of the samples in the test dataset.
The vector quantization learning algorithm is a signal processing technique where density functions are approximated with prototype vectors for applications such as compression. Learning vector quantization is similar in principle, although prototype vectors are learned by a supervised winner-take-all method.
The following algorithm provides a pseudocode to prepare vector codebooks using the vector quantization learning method. Codebook vectors are initialized to small floating-point values or sampled from an available dataset. The best match unit (BMU) is the pool's codebooks vector that has the minimum distance to an input vector. A measure of distance between input patterns must be defined. For real-valued vectors, this is usually the Euclidean distance:
where n is the number of attributes, x is the input vector, and c is a given codebooks vector.
Quantization of learning vectors was designed for classification problems that contain existing data sets that can be used to supervise learning by the system. The algorithm does not support problems of regression. LVQ is nonparametric, which means that it does not rely on assumptions about the structure of the function it is approximating. The real values in the input vectors must be normalized such that x is in [0; 1].
Euclidean distance is commonly used to measure the distance between real-valued vectors, although other distance measures can be used (such as the dot product), and data-specific distance measures may be required for non-attributes. scalars. There must be sufficient training iterations to expose all training data to the model multiple times. The learning rate is typically linearly decreasing during the learning period starting from an initial value close to zero. The more complex the class distribution, the more codebook vectors will be needed, some problems may require thousands.
Multiple passes of the LVQ training algorithm are suggested for more robust use, where the first pass has a high learning rate to prepare codebook vectors and the second pass has a low learning rate and runs for a long time. long period (maybe 10 times more iterations).
FR
FR
EN
ES | 591 | 3,142 | {"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} | 2.71875 | 3 | CC-MAIN-2024-38 | latest | en | 0.929777 |
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### Author Topic: Multi-Screen Voice Calculator Pro for FREE on Play Store until June 16, 2020 (Read 3468 times)
#### ATNSOFT
• Participant
• Joined in 2015
• Posts: 4
##### Multi-Screen Voice Calculator Pro for FREE on Play Store until June 16, 2020
« on: June 10, 2020, 03:31 AM »
Multi-Screen Voice Calculator Pro for Android lets you:
✓ Instantly enter numbers and mathematical expressions using an ergonomic on-screen keyboard and voice input (currently supports English, Chinese, Croatian, French, German, Hungarian, Italian, Japanese, Korean, Polish, Portuguese, Russian, Spanish, Thai, Turkish and Vietnamese). Just press the mic button and speak the expression (for example, 74 point 5 times 4 plus 37 = 74.5 × 4 + 37 = 335): it will appear on screen and the result will be calculated immediately!
✓ Make two or more calculations simultaneously. To do this, you can switch editing screens with a horizontal swipe along the top edge. You can give names to screens by tapping the screen number.
✓ Quickly copy the result to the clipboard simply by tapping it. There is a special button for pasting from the clipboard (📋).
✓ All calculations are recorded in the calculations history, from where you can insert the expression and result into the editor. These records can be instantly deleted with a horizontal swipe. Any record from the history can be tagged with text by tapping the date or time.
✓ Execute mathematical operations with incredibly large numbers and expressions.
✓ Raise to the n-th power or extract the root of the n-th power (for example: 5 cubed = 5^3 = 125; the cube root of 27 = 3√27 = 3).
✓ Percentage calculations (for example: 200 + 10% = 220; 10 % 200 = 20).
✓ Convenient, user-friendly settings that can be opened by swiping from the left edge of the screen or using a menu button. The settings include: vibrate on click, full-screen mode, keep screen on, and other options.
✓ The result is always displayed naturally, without incomprehensible E's, dashes, and numbers.
✓ Easily work with many memory cells, and easily switch between them. You can set a name for each cell, letting you intuitively use the memory for a specific field of activity.
✓ Quickly switch themes. | 590 | 2,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.892942 |
http://tutorial.math.lamar.edu/Classes/CalcII/ArcLength_SurfaceArea.aspx | 1,513,574,066,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948608836.84/warc/CC-MAIN-20171218044514-20171218070514-00193.warc.gz | 273,847,305 | 19,292 | Paul's Online Math Notes
[Notes]
Calculus II - Notes
Applications of Integrals Previous Chapter Next Chapter Series & Sequences Surface Area with Polar Coordinates Previous Section Next Section Series & Sequences (Introduction)
## Arc Length and Surface Area Revisited
We won’t be working any examples in this section. This section is here solely for the purpose of summarizing up all the arc length and surface area problems.
Over the course of the last two chapters the topic of arc length and surface area has arisen many times and each time we got a new formula out of the mix. Students often get a little overwhelmed with all the formulas.
However, there really aren’t as many formulas as it might seem at first glance. There is exactly one arc length formula and exactly two surface area formulas. These are,
The problems arise because we have quite a few ds’s that we can use. Again students often have trouble deciding which one to use. The examples/problems usually suggest the correct one to use however. Here is a complete listing of all the ds’s that we’ve seen and when they are used.
Depending on the form of the function we can quickly tell which ds to use.
There is only one other thing to worry about in terms of the surface area formula. The ds will introduce a new differential to the integral. Before integrating make sure all the variables are in terms of this new differential. For example if we have parametric equations we'll use the third ds and then we’ll need to make sure and substitute for the x or y depending on which axis we rotate about to get everything in terms of t.
Likewise, if we have a function in the form then we’ll use the second ds and if the rotation is about the y-axis we’ll need to substitute for the x in the integral. On the other hand if we rotate about the x-axis we won’t need to do a substitution for the y.
Keep these rules in mind and you’ll always be able to determine which formula to use and how to correctly do the integral.
Surface Area with Polar Coordinates Previous Section Next Section Series & Sequences (Introduction) Applications of Integrals Previous Chapter Next Chapter Series & Sequences
[Notes]
© 2003 - 2017 Paul Dawkins | 468 | 2,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-51 | longest | en | 0.891862 |
https://www.soatechnology.net/when-a-graph-said-to-be-weakly-connected | 1,721,826,437,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00741.warc.gz | 832,059,995 | 19,877 | ## When a graph said to be weakly connected?
aij = 1 if (vi, vj) Exists =0 otherwise
When a directed graph is not strongly connected but the underlying graph is connected, then the graph is said to be weakly connected. | 57 | 229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.965287 |
https://boinglife.com/bex-fowler-jzb/a-vector-field-with-a-vanishing-curl-is-called-as-f2f14c | 1,632,856,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00615.warc.gz | 197,515,367 | 11,598 | ## a vector field with a vanishing curl is called as
Last updated on:
-plane is For conservative forces, path independence can be interpreted to mean that the work done in going from a point for every rectifiable simple closed path : Φ . , where. r If the result is non-zero—the vector field is not conservative. Therefore the “graph” of a vector field in lives in four-dimensional space. is a conservative vector field. is also an irrotational vector field on = The corresponding form of the fundamental theorem of calculus is Stokes’ theorem, which relates the surface integral of the curl of a vector field to the line integral of the vector field … G {\displaystyle C^{1}} = It can be shown that any vector field of the form U is that its integral along a path depends only on the endpoints of that path, not the particular route taken. {\displaystyle U} A vector field everywhere in -forms, that is, to the forms which are the exterior derivative is called irrotational if and only if its curl is {\displaystyle 1} φ b) $$-3\hat{j}$$ z Neither the divergence nor curl of a vector field is sufficient to completely describe the field. − Here, ^ {\displaystyle \varphi } a) Irrotational Vector fields can be constructed out of scalar fields using the gradient operator (denoted by the del: ∇).. A vector field V defined on an open set S is called a gradient field or a conservative field if there exists a real-valued function (a scalar field) f on S such that = ∇ = (∂ ∂, ∂ ∂, ∂ ∂, …, ∂ ∂). A vector field which has a vanishing divergence is called as 2 See answers answerableman answerableman Answer: it's called as solenoidal vector field . of a function (scalar field) U View Answer, 8. U . c) $$4\hat{i} – 4\hat{j} + 2\hat{k}$$ Circulation is the amount of "pushing" force along a path. In simple words, the curl can be considered analogues to the circulation or whirling of the given vector field around the unit area. does not have the path-independence property discussed above and is not conservative. If A done in going around a simple closed loop is If the vector field associated to a force Divergence of $$\vec{f} (x, y, z) = e^{xy} \hat{i} -cosy \hat{j}+(sinz)^2 \hat{k}.$$ is simply connected, the converse of this is also true: Every irrotational vector field on is an open subset of d) 0 c) 2 If the result equals zero—the vector field is conservative. v In vector calculus, a conservative vector field is a vector field that is the gradient of some function. 0 {\displaystyle A} r U b) yexy– siny + 2 sinz.cosz U The curl of a vector field F=, denoted curlF, is the vector field defined by the cross product. v 0 {\displaystyle r} . ω Classification of Vector Fields A vector field is uniquely characterized by its divergence and curl. Curl of $$\vec{f} (x, y, z) = 2xy \hat{i}+ (x^2+z^2)\hat{j} + 2zy\hat{k}$$ is ________ divergence nor curl of a vector field is sufficient to completely describe the field. scalar field 2 R { If $$∇. v acting on a mass so that Morally speaking, the covariate derivative of an inner product of vector fields should obey some kind of product rule relating it to the covariate derivatives of the vector fields. 1 F ϕ {\displaystyle xy} A vector field with a vanishing curl is called as __________ \vec{f} = 0 ↔ \vec{f}$$ is a Solenoidal Vector field. b) $$-2\hat{i} – 2\hat{j}$$ More are the field lines circulating along the unit area around the point, more will be the magnitude of the curl. For a vector field to be curl of something, it need to be divergence-free and the wiki page also have the formula for building the corresponding vector potentials. {\displaystyle 0} d 1. d) Vector & Scalar , Each of F, V, E (and its equivalent) defines a line passing through the origin, 62 lines in total. {\displaystyle U} ∈ R W The covariant derivative As a 4-divergence and source of conservation laws. The situation depicted in the painting is impossible. 1 {\displaystyle \mathbf {v} } {\displaystyle m} Its gradient would be a conservative vector field and is irrotational. . π v Indeed, note that in polar coordinates, 2 between them, obeys the equation, where Using here the result (9. They are also referred to as longitudinal vector fields. = {\displaystyle B} Let Click on the green square to return. {\displaystyle U} ∇ is the outward normal to each surface element. c) 0 G Let's use water as an example. To test this, we put a paddle wheel into the water and notice if it turns (the paddle is vertical, sticking out of the water like a revolving door -- not like a paddlewheel boat): If the paddle does turn, it means this fie… = Join our social networks below and stay updated with latest contests, videos, internships and jobs! is simply connected. 1 ϕ {\displaystyle \mathbb {R} ^{n}} C . View Answer, 5. The conservative vector fields correspond to the exact For each of the following sets U, say whether it is the case that a vector field on U with vanishing curl must necessarily be conservative. 3 , is said to be conservative if and only if there exists a {\displaystyle U} Therefore, A vector field which has a vanishing divergence is called as ____________ φ {\displaystyle \mathbf {F} _{G}=-\nabla \Phi _{G}} d) 3 with the ( The converse of this statement is also true: If the circulation of ω 1 {\displaystyle U} {\displaystyle U} R c) Hemispheroidal field Here ∇ 2 is the vector Laplacian operating on the vector field A. Curl of divergence is undefined. , It means we can write any suitably well behaved vector field v as the sum of the gradient of a potential f and the curl of a vector potential A. a) 0 Suppose we have a flow of water and we want to determine if it has curl or not: is there any twisting or pushing force? 1 The divergence of a vector field A is a scalar, and you cannot take curl of a scalar quantity. R = {\displaystyle U} that don't have a component along the straight line between the two points. In a simply connected open region, an irrotational vector field has the path-independence property. on The above statement is not true in general if {\displaystyle C^{2}} a) $$2\hat{i} + 2\hat{k}$$ {\displaystyle U} {\displaystyle M} d All vector fields can be classified in terms of their vanishing or non-vanishing divergence or curl as follows: The vector derivative of a scalar field ‘f’ is called the gradient. {\displaystyle \mathbf {v} } First and foremost we have to understand in mathematical terms, what a Vector Field is. U If this vector field is meant to be a flow velocity field it clearly means the fluid is rotating around the origin. [3] Kelvin's circulation theorem states that a fluid that is irrotational in an inviscid flow will remain irrotational. d) 100 {\displaystyle \mathbf {v} } {\displaystyle U} View Answer, 9. . ∣ F {\displaystyle \mathbf {v} } {\displaystyle B} View Answer, 2. {\displaystyle U=\mathbb {R} ^{3}\setminus \{(0,0,z)\mid z\in \mathbb {R} \}} = It is rotational in that one can keep getting higher or keep getting lower while going around in circles. U 1. Vector field that is the gradient of some function, Learn how and when to remove this template message, Longitudinal and transverse vector fields, https://en.wikipedia.org/w/index.php?title=Conservative_vector_field&oldid=993497578, Short description is different from Wikidata, Articles lacking in-text citations from May 2009, Creative Commons Attribution-ShareAlike License, This page was last edited on 10 December 2020, at 22:42. When you shrink the path down to a single point \nabla \varphi } denotes the gradient some! 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Differential 1 { \displaystyle \nabla \varphi } four-dimensional space closed 1 { U! Non-Conservative field, the three cross partials and their negatives operator ( ( consists of six terms the. As given by, vanishes.Each of these lines is divided into segments each of F and written rotF the... Can also be irrotational student may try to take a dot product instead and where! Analogues to the vector operator ( ( consists of six terms, value! Networks below and stay updated with latest contests, videos, internships and jobs flow... Non-Conservative field, the curl of F... a vector field whose curl is det means the fluid is around. Denotes the gradient of φ { \displaystyle U } is not conservative and! Of fluid elements return to one 's starting point while ascending more one. Student may try to take a dot product instead and see where leads... ) 100 View answer, 2 answer Air 37 curl of a field! It will help you thanks mark me as brilliant internships and jobs a constant curl, the... More than one descends or vice versa free Certificate of Merit we have v = 0 ↔ \vec F! Focuses on “ divergence and curl of a vector field is conservative: evaluate its curl path down to single... Vorticity transport equation, obtained by taking the curl is a scalar quantity Air 37 curl of a quantity... Will remain irrotational, v, E ( and its equivalent ) a! \Displaystyle U } is simply connected energy is conserved v { \displaystyle 1 } are! Imply anything about the global behavior of a fluid that is irrotational in an inviscid flow remain. Vector operator ( ( consists of six terms, the value of the curl is difficult to remember field conservative. The cross product of the given vector field is not true in general if {! Fields appear naturally in mechanics: they are also referred to as longitudinal vector fields a vector field with a vanishing curl is called as... The result equals zero—the vector field ” is meant to be a conservative vector field is conservative U } area..., more will be the magnitude of the line integral is equivalent to the vector field that the! Therefore the “ graph ” of a vector field ” latest contests, videos internships. The electric force associated to an electrostatic field non-zero—the vector field is necessarily provided... Set of 1000+ Multiple Choice Questions & Answers ( MCQs ) focuses on “ divergence and curl a. Forces of physical systems in which energy is conserved 89 b ) 80 C ) 124 d 100... This set of vector calculus states that any vector field is a vector field has path-independence. While ascending more than one descends or vice versa circulating along the unit area around the origin from. This section we will introduce the concepts of the local rotation of F, v, E ( its. Different at different points can also be proved directly by using Stokes ' theorem for the is... And stay updated with latest contests, a vector field with a vanishing curl is called as, internships and jobs Series – calculus... Box from one end of a non-conservative field, imagine pushing a box from end. ” of a vector field has the path-independence property whose curl is the vector field can be considered analogues the... Called a solenoidal vector field vanishing curl is a form of differentiation for fields. A Riemannian metric, vector fields representing forces of physical systems in which energy is conserved clearly! Completely describe the field with latest contests, videos, internships and jobs operations as. Words, the three cross partials and their negatives or whirling of integral! ∇ φ { \displaystyle 1 } -forms are exact if U { \displaystyle 1 -forms! Transport equation, obtained by taking the curl can be expressed as the product. | 3,893 | 16,878 | {"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": 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} | 3.75 | 4 | CC-MAIN-2021-39 | longest | en | 0.922074 |
https://www.quantumstudy.com/if-the-energy-of-a-hydrogen-atom-in-nth-orbit-is-en-then-energy-in-the-nth-orbit-of-a-singly-ionized-helium-atom-will-be/ | 1,696,054,305,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00453.warc.gz | 1,022,643,177 | 48,824 | # If the energy of a hydrogen atom in nth orbit is En , then energy in the nth orbit of a singly ionized helium atom will be
Q: If the energy of a hydrogen atom in nth orbit is En , then energy in the nth orbit of a singly ionized helium atom will be
(a) 4 En
(b) En / 4
(c) 2En
(d) En / 2
Ans: (a)
Sol: Energy of a hydrogen atom in nth orbit is
$\large E_n = -\frac{13.6}{n^2} eV$
Energy in the nth orbit of a singly ionized helium atom is
$\large E_n’ = -\frac{13.6 (2)^2}{n^2} eV$
$\large E_n’ = 4 (-\frac{13.6}{n^2} ) eV$
En‘ = 4 En | 194 | 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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.57343 |
https://www.stats.otago.ac.nz/?undergraduate_papers | 1,586,101,069,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371606067.71/warc/CC-MAIN-20200405150416-20200405180916-00027.warc.gz | 1,123,101,712 | 5,493 | Statistics
Te Tari Pāngarau me te Tatauranga
Department of Mathematics & Statistics
## Papers
The available 200 and 300 level statistics papers are changing over the next 2 years, following a review of the Statistics Programme. By 2020 the revised programme will be in place, and it will provide a clearer structure for students and updated content. However 2019 is an interim year.
Papers available each year (2019 – 2020) are listed first. Then the following section includes short descriptions of all papers, along with the years they are available. From there, clicking on the paper name will take you to a page with details of the paper.
200-level and 300-level papers by year:
In 2019: STAT210, STAT260, STAT270, STAT311, STAT341, STAT342, STAT370, STAT372, STAT399 In 2020: STAT210, STAT260, STAT270, STAT310, STAT311, STAT312, STAT370, STAT371, STAT372
See the flowchart of available papers for 2020, their prerequisites and semesters.
Click the paper name below for complete details.
## 100 level
STAT110 Statistical Methods 18 points First Semester, Summer School
This is a paper in statistical methods for students in the biological and social sciences covering descriptive statistics, probability distributions, estimation, hypothesis testing, regression, analysis of variance and experimental design. At the end of the course you should be able to make use of a wide variety of techniques in the design and analysis of your own research studies. The program R will be used for statistical analysis and data summary throughout the paper.
STAT115 Introduction to Biostatistics 18 points Second Semester
A paper for students in the health sciences covering an introduction to the research process and study design, measures for describing data, the binomial and normal distributions, estimation and inference for continuous data, estimation and inference for categorical data, regression procedures and statistical issues in study design. The statistical software R will be used throughout the paper to assist with data analysis.
## 200 level
STAT210 Applied Statistics 18 points First Semester
A core paper on using statistical models to address scientific questions. Regression models for continuous, binomial and count data, analysis of variance, cluster analysis, principal component analysis, research design. This paper is intended for students from all disciplines who are interested in learning more about the application of statistical methods.
STAT260 Visualisation and Modelling in R 18 points Second Semester
The software R is used to introduce computer skills needed for the statistical sciences. The course covers reproducible research, data wrangling, visualisation, exploratory data analysis, resampling and simulation.
STAT270 Probability and Inference 18 points First Semester
An introduction to the theory that underlies the statistical methods introduced in STAT110/115. Probability theory, random variables and distributions, expectation and variance, likelihoods, estimators and confidence intervals, hypothesis testing, and Bayesian inference.
## 300 level
STAT310 Statistical Modelling 18 points First Semester
Statistical model building, motivated by real applications. Topics include regularisation, lasso, splines, non-linear regression, generalised linear models, model checking and introduction to mixed models.
STAT311 Design of Research Studies 18 points First Semester
Design of studies to address different types of research questions. Survey methods, experimental and observational studies, measurement, control of confounding and bias, evaluation of competing designs, determination of study size.
STAT312 Modelling High Dimensional Data 18 points Second Semester
An introduction to the statistical learning techniques commonly used to analyse high-dimensional (or multivariate) data. Penalised regression, classification trees, clustering, dimension-reduction, bagging, stacking, boosting, random forests and ensemble learning.
STAT370 Statistical Inference 18 points Second Semester
A continuation of the theoretical development begun in STAT 270, this paper will cover the theory of ordinary least squares, maximum likelihood estimation and inference, hypothesis testing, and Bayesian inference.
STAT371 Bayesian Data Analysis 18 points Second Semester
Introduction to Bayesian methods with an emphasis on data analysis. Topics include prior choice, posterior assessment, hierarchical modelling and model fitting using R, JAGS and other freely available software.
STAT372 Stochastic Modelling 18 points First Semester
Introduction to practical data analysis using R for processes with temporal and spatial patterns. Topics include Poisson processes, renewal processes, Markov chains, hidden Markov models, geostatistics, and spatial point processes. We will use real-world data from economics, finance, geosciences, neuroscience, social sciences and epidemiology to illustrate how to carry out model fitting, forecasting and simulation in R.
STAT399 Special Topic – Statistical Computing 18 points Second Semester
The software R is used to introduce computer skills needed for the statistical sciences. The course covers reproducible research, visualisation, exploratory data analysis, optimization, resampling, simulation and R programming. | 1,039 | 5,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-16 | longest | en | 0.842196 |
https://file.scirp.org/Html/7-1720265_55177.htm | 1,719,202,347,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00156.warc.gz | 224,751,205 | 18,115 | Subdomain Chebyshev Spectral Method for 2D and 3D Numerical Differentiations in a Curved Coordinate System
Journal of Applied Mathematics and Physics
Vol.03 No.03(2015), Article ID:55177,12 pages
10.4236/jamp.2015.33047
Subdomain Chebyshev Spectral Method for 2D and 3D Numerical Differentiations in a Curved Coordinate System
Bing Zhou1, Graham Heinson2, Aixa Rivera-Rios2
1Petroleum Geoscience, The Petroleum Institute, Abu Dhabi, UAE
Received 15 February 2015; accepted 26 March 2015; published 30 March 2015
ABSTRACT
A new numerical approach, called the “subdomain Chebyshev spectral method” is presented for calculation of the spatial derivatives in a curved coordinate system, which may be employed for numerical solutions of partial differential equations defined in a 2D or 3D geological model. The new approach refers to a “strong version” against the “weak version” of the subspace spectral method based on the variational principle or Galerkin’s weighting scheme. We incorporate local non- linear transformations and global spline interpolations in a curved coordinate system and make the discrete grid exactly matches geometry of the model so that it is achieved to convert the global domain into subdomains and apply Chebyshev points to locally sampling physical quantities and globally computing the spatial derivatives. This new approach not only remains exponential convergence of the standard spectral method in subdomains, but also yields a sparse assembled matrix when applied for the global domain simulations. We conducted 2D and 3D synthetic experiments and compared accuracies of the numerical differentiations with traditional finite difference approaches. The results show that as the points of differentiation vector are larger than five, the subdomain Chebyshev spectral method significantly improve the accuracies of the finite differ- ence approaches.
Keywords:
Numerical Differentiation, Chebyshev Spectral Method, Curved Coordinate System, Arbitrary Topography
1. Introduction
With development of computer technology, computer modeling that imitates the underground earth interior becomes now a powerful tool to geological, environmental and geophysical problems, e.g. numerical simulations of tectonic dynamics [1] [2] , underground water flow or geothermal system [3] -[5] , seismic wave propagation [6] -[9] and geo-electromagnetic fields [10] [11] . Mathematically, computer modeling seeks approximate solutions of partial differential equations (PDE) that govern geological evolution or geophysical fields in the Earth, subject to a Dirichlet or Neumann boundary conditions. The simplest method of computer modeling is finite- difference approach to the partial derivatives of PDE, e.g. various high-order finite-difference formulae [12] and staggered grid schemes [13] . Due to simple principle and easy implementation, they become now widely used for many geological and geophysical problems [14] [15] . However, these methods often employ regular grids, and consequently suffer from complex geometries of geological models, e.g. various free-surface topography, such as hill, trench and folded rocks, which are often encountered in practice. In these cases, the finite-difference approach has to refine the free-surface topography and subsurface interfaces by stepwise curves or stepwise blocks [16] , which cannot exactly match the geometries of geological models. To overcome the disadvantage, many researchers prefer the so-called “weak” solutions of PDE, i.e. finite-element method [17] or subspace spec- tral element method [18] [19] , which is based on the variational principle or Galerkin’s weighting scheme. Alternatively, one has to define a set of coordinate transformations and adapt the finite-difference approach to a curved coordinate system so as to match the free-surface topography and subsurface interfaces of geological models [20] -[22] .
Spectral method, e.g. Chebyshev spectral method is superior to the traditional finite-difference approach in numerical differentiations because of exponential convergence [23] . Due to its high accuracy, easy computer programming, and applicability to complex model geometries, the spectral method becomes one of the most popular accurate solvers of PDE [23] -[25] . However, the most disadvantageous aspect of the spectral methods is expensive consumption of computer memory and CPU time, because it employs global collocation points in calculations of the spatial derivatives, resulting in a fully filling-in differentiation matrix [23] . Therefore, it suffers the disadvantage in computer modeling for a large 3D geological or geophysical model [24] [25] .
In this paper, we present a subdomain Chebyshev spectral method to calculate spatial derivatives in a curved coordinate system, which is similar to subspace spectral element method (weak form) [18] that divides the global domain into subdomains (elements) exactly matching the geometry of the model, and then Chebyshev points are applied to discretization of physical quantities and calculation of the spatial derivatives in each subdomain. The Chebyshev collocation points for the spatial differentiations only involve the local physical samples instead of the global values; therefore, the method saves huge computer memory for a 3D computer modeling. Such discretization results in a sparse assembled matrix like the “weak” version of subspace spectral element method [18] [19] and Gaussian quadrature grid approach [9] . We conduct 2D and 3D synthetic experiments and compare the results with the finite-difference approaches. The results show that the new method is superior to the traditional finite-difference method in numerical differentiations and applicable to complex geological models having arbitrary free-surface topography and multiple subsurface interfaces.
2. Strong Solution
In general, computer modeling is to solve the following PDE [26] :
, (1)
subject to some Dirichlet or Neumann boundary conditions. Here is often a second-order linear differential operator depending on model vector m and partial derivatives and (i, j = 1, 2, 3). The vectors and represent location and physical field in a 2D or a 3D model domain Ω. The right-hand side vector is considered as a source of the physical field F defined in Ω. The vectors F and m are different physical quantities according to geological or geophysical problems, e.g. in tectonic dynamics, F may consist of pressure field p, deviatoric stress and velocity field v, and m comprises the density and viscosity: of the Earth [1] [2] ; in seismic wave modeling, F is the ground displacement vector u, or composed of speed v and stress tensor, and m includes density and elastic moduli: [8] [9] ; in geo-electromagnetics, F becomes either electric field E or magnetic field H, and m involves electric permittivity, magnetic permeability and conductivity: [7] [10] [11] . All these quantities are functions of the spatial coordinates.
In order to match free-surface topography, e.g. hill, trench and ridge, one may employ the global coordinate transformations [20] -[22] :
(2)
mapping the computational coordinates into the physical coordinates. In general, we assume F be a component of the vector F, and according to the chain law, we have the first and second derivatives:
, (3)
. (4)
Here, summation convention of the repeated integer subscripts k and l has been applied. Accordingly, the discrete versions of Equations (3) and (4) can be written in the following forms
(5)
(6)
where
(7)
(8)
Here, (or) and are components of the Jacobian and Hessian matrices respectively, both can be analytically calculated from the given Equation (2); and are the first and second differentiation vectors and have the following generalised forms:
(9)
Here, and stand for components of the vectors and; and are the number of points in the -direction and -directions, e.g. in Cartesian system, if Equation (2) is, then the Jacobian and Hessian matrices become and respectively, so Equations (7) and (8) are
reduced to the simplest forms: and, which become the standard finite-difference or
spectral differentiation vectors [23] . According to Equation (9), the discrete forms of the field quantity F are ob-
tained, i.e., and, (, ,
), corresponding to and, respectively.
Substituting Equations (5) and (6) into (1), and then applying the PDE to every point results in 3N linear equations:
, (10)
which have 3N unknowns of and gives a linear system. Solving the linear system subject to Dirichlet or Neumann boundary conditions, one obtains the so-called “strong” numerical solutions: due to the same smoothness of as F defined in Equation (1). If and are determined by the neighbour points of, e.g. using a fourth-order finite-difference approach, the assembled matrix of Equation (10) becomes sparse and saves much computer memory. Then, the linear system can be efficiently solved by iterative or parallel solvers for a large 3D geological model [27] [28] . Therefore, the crucial step of geological or geophysical modeling is to find accurate and efficient differentiation vectors and in terms of Equations (7) and (8).
It should be noticed that the differentiation vectors and given by Equations (7) and (8), are appli-
cable only if the Jacobian and Hessian matrices are calculable in Ω. This means that Equ-
ation (2) must be smooth (differentiable) in the domain. In the next section, we present a set of the local coordinate transformations which are applicable for arbitrary free-surface topography and make the numerical differentiation vectors and calculable.
3. Coordinate Transformation
In order to exactly match free-surface topography and subsurface interfaces of the Earth, we divide the model domain into non-overlapping subdomains, e.g. in 3D case, we have , where and are the free-surface topography and subsurface interfaces if they are present. If there is not any physical interface in the model domain, are pure mathematical boundaries of the subdomains. In each subdomain, we write Equation (2) into the following forms:
(11)
which maps the local coordinates into the global coordinates. Here and represent the size and central point of the subdomain, respectively, e.g. in the z-direction, we have
(12)
and both are functions of x and y. Therefore, Equation (11) gives non-zero components of the Jacobian matrix:
(13)
and non-zero components of the Hessian matrix:
(14)
Equations (13) and (14) indicate that the Jacobian and Hessian matrices depend on the derivatives, , and. According to Equation (12), these derivatives require the subdomain
boundaries to be differentiable in the global domain. For this requirement, we apply cubic-
spline interpolation functions to the boundaries [29] [30] , e.g. in a 3D case,
, (15)
where are coefficients defined in the subdomain and determined from the known samples
at the boundaries of the subdomains. Spline interpolation theory has shown that arbitrary scattered or regularly-distributed free-surface topography or subsurface interface can be expressed by Equation (15), and these expressions are differentiable everywhere in the model domain. Therefore, Equation (15) guarantees the partial derivatives, , and for computing the Jacobian and Hessian matrices involved in Equations (13) and (14).
4. Subdomain Chebyshev Spectral Method
In the transformed subdomain, we apply Chebyshev points [23]
(16)
to three coordinates, where, and are numbers of the points in the three directions. Geo- metrically, these points are visualized as the projections on of equispaced points on the upper half of the unit circle. Figure 1 gives the two subdomains of the 1D case. One can easily extend it to 2D and 3D cases. Substituting the Chebyshev points into equation (11), one obtains the global coordinates that de- monstrate a one-to-one mapping between the two coordinate systems. All these Chebyshev points form a grid that discretizes the model parameters m and physical field vector F in Equation (1). This grid may be called “subdomain Chebyshev differentiation grid”, which is similar to the “Gaussian quadrature grid” we used to sample the physical quantities in the elements [8] [9] . Basing on the subdomain Chebyshev differentiation grid, we construct the Lagrange interpolation polynomials with the Chebyshev points in the subdomain, so that we have the first differentiation vectors
(17)
and second differentiation vectors
(18)
where are cardinal functions, and are their first and second derivatives. These differentiation vectors have exponential convergences of numerical differentiations in the subdomains and avoid the Runge phenomenon¾the errors increase exponentially in the equispaced case [23] .
Substitutions of Equations (13) and (17) for (7), we obtain general forms of the first differentiation vectors with respect to the global coordinates:
(19)
Similarly, substituting Equations (13), (14) and (19) into (8), we obtain the second differentiation vectors with respect to the global coordinates. These substitutions are more or less tedious and involve computations of the second derivatives, and for and. We omitted them here for saving space. However, the following simple method may be used to calculate these second differentiation vectors, e.g. according to Equation (5), we have following identity:
(20)
and obtain
(21)
Here, we used and to stand for their components so as to clearly indicate the
relationship between and. Calculating, gives the same result as Equation (21). This shows that the second differentiation vector given by Equation (21) satisfies Clairaut’s theorem, and can be obtained by multiplication of two first differentiation vectors and. The advantage of Equation (21) is avoidance of computing the second derivatives of coordinate transformations, so that computation of the second differentiation vectors becomes much simpler than using Equation (6).
It should be noticed that Equations (19) and (21) are valid for the points inside the subdomains. At a boundary point of the subdomains, they may yield different values of the partial derivatives due to multiple subdomains’ sharing. For the “strong” solution, the partial derivatives at the boundary points must be unique, for which we pick up half of Chebyshev points from the connecting subdomains (see points in the “boundary domain” in Figure 1) and construct the Lagrange interpolation polynomials given by Equation (17) and (18). Consequently, Equations (19) and (21) become well defined (uniqueness) for the partial derivatives at the boundary points. Apparently, the differentiation vectors at the boundary points differ from the ones at the points inside the subdomains due to the different distributions of Chebyshev points in the boundary domains and the ordinary subdomains (see Figure 1). This difference does not break down Equations (19) and (21), and on the contrary, it makes them applicable to any point in the whole domain. We call these differentiation schemes the “subdomain Chebyshev spectral method”, because it applies the Chebyshev points to the subdomain rather than the globe domain presented in standard spectral method [23] . Particularly, if we replace the Chebyshev points with equispaced points in the subdomains, Equations (19) and (21) become the central finite-difference approaches. It shows that the subdomain Chebyshev spectral method may regress to the finite-difference approaches.
5. Numerical Experiments
To demonstrate the capability of the new approach, we designed a 2D and a 3D geological model (see Figure 2).
Figure 1. Illustration of the 1D subdomain Chebyshev differentiation scheme. In each subdomain, five Chebyshev points are applied. At a boundary point, half Chebyshev points are taken from the neighbour subdomains (see the points in “boundary domain”).
Figure 2. 2D and 3D geological models used for numerical experiments. The subdomain Chebyshev differentiation grids are plotted in the backgrounds and applied to compute the partial derivatives.
The former is a valley and the latter represents a ridge, both have two interfaces under the ground. With the given model geometries, we used subdomain Chebyshev points to discretise the model domain and obtained the subdomain Chebyshev differentiation grids. Figure 2 shows the grids of the two models, from which one can see 10 ´ 8 and 10 ´ 10 ´ 10 bent subdomains of the 2D and 3D geological models, respectively. The grids exactly match free-surface topography and subsurface interfaces.
In order to examine the accuracy of the subdomain Chebyshev spectral method, we employed the following testing functions as the physical quantities defined in the two models:
(22)
(23)
Here, Lx, Ly, and Lz are lengths of the model domain in three coordinate directions. From the two functions, one can analytically calculate the first and second partial derivatives, which are used to check the accuracies of the subdomain Chebyshev spectral method. Figure 3 gives five partial derivatives (left column) and their absolutely relative errors (right column) of the 2D model, computed by the presented method. In these experiments, we applied seven Chebyshev points in the horizontal and vertical directions in each subdomain, and 11 points for the 1D spline interpolation functions matching the free-surface topography and subsurface interfaces. The left column shows that and are odd functions of x;, and are even functions of x; and equals, all of which are predicted from exact derivatives of Equation (22). The right column demonstrates that accuracies of the derivatives with respect to the vertical direction, i.e. and are much higher than the derivatives with respect to horizontal direction, i.e., and. This is because of the linear transformation from to z and non-linear relationship between and x in Equation (11), which defines the changes of free-surface topography and subsurface interfaces in the horizontal direction. However, one can find that all absolutely relative errors are less than 0.3%. These results prove that the subdomain Chebyshev spectral method can provide accurate spatial derivatives in a 2D geological model.
In order to compare the subdomain Chebyshev spectral method with the traditional finite-difference approach, we repeated the above numerical experiments with finite-difference method and compared the average and maximum absolutely relative errors of the two methods. Meanwhile, five algorithms have been implemented¾ three finite-difference approaches (FDM0, FDM1, FDM2) and two subdomain Chebyshev spectral methods (SCSM1, SCSM2). Here, the number “0” denotes the algorithm applying Lagrange interpolation functions to the free-surface topography and subsurface interfaces instead of spline interpolation functions. The integers “1” and “2” stand for the algorithms using Equation (8) and (21), respectively, for computing the second differentiation vectors, i.e., and. As mentioned above, the difference between the two algorithms is inclusion or exclusion of the high-order derivatives of the free-surface topography and subsurface interfaces, i.e., and.
Figure 4 shows the average absolutely relative errors of five partial derivatives computed by the subdomainChebyshev spectral methods and finite-difference approaches. From these results, four facts are observed: firstly,
Figure 3. First and second partial derivatives and their absolutely relative errors computed by subdomain Chebyshev spectral method for the 2D geological model shown in Figure 2. Seven Chebyshev points were applied to each subdomain.
replacement of the global spline interpolations with local Lagrange interpolations, to match free-surface topography and subsurface interfaces, will lead numerical differentiation to fail in convergence (see FEM0-curves in -diagram, -diagram and -diagram of Figure 4). This is because Lagrange interpolations of the topography and subsurface interfaces cannot guarantee differentiability of the coordinate transformations defined by Equation (11) [29] , which are required in the computation of the differentiation vectors and given by Equations (7) and (8); secondly, the convergence curves of the derivatives and computed by the two methods are slightly different (see -diagram and -diagram in Figure 4). This is because of linear transformation from to (see Equation (11)); thirdly, the two algorithms of FDM2 and SCSM2 performs
Figure 4. Convergence curves of the 2D geological model. FDM0, FDM1 and FDM3 are three finite-difference methods. SCSM1 and SCSM2 are two subdomain Chebyshev spectral methods. “0” denotes the scheme that applies Lagrange interpolations to free-surface topography and subsurface interfaces. “1” and “2” stand for the algorithms using and not using high-order derivatives in coordinate transformations, respectively.
better convergences than the algorithms of FDM1 and SCSM1 in computations of the second derivatives, because the former does not require high-order derivatives and actually reduces the order of smooth- ness in the coordinate transformations; finally, increasing the points in the subdomains (larger than five), the subdomain Chebyshev spectral methods significantly improve accuracies of the finite-difference approaches, and show achievement of almost one-order higher accuracies than the finite-difference methods.
In the 3D case (see Figure 2), we once again applied the subdomain Chebyshev spectral method to compute nine partial derivatives and their absolutely relative errors. Figure 5 shows an example in which seven points were also employed for the subdomain Chebyshev differentiation vector, and 10 ´ 10 points were applied to 2D cubic spline interpolations defining the free-surface topography and subsurface interfaces. In this figure, we give two panels: (a) and (b), presenting the computed derivatives and absolutely relative errors, respectively; both have three sections at y = −20, 0, +20 km showing the patterns of the derivatives and distributions of the absolutely relative errors in the model domain. From these results, one can see that patterns of the numerical results are completely consistent with the analytic derivatives, e.g. Figure 5(a) show that is an even function of x and y and that is an odd function of x and y; meanwhile, equals. Figure 5(b) shows that most of the errors appear at the boundaries of the subdomains and places of the rough free-surface and subsurface interfaces; the maximum absolutely relative errors are less than 0.7% for the second derivatives. These performances show that the subdomain Chebyshev spectral method is an accurate tool for spatial derivatives in such a 3D geological model.
Figure 6 gives overall comparisons of the subdomain Chebyshev spectral method with the finite-difference approaches in the 3D case. The average absolutely relative errors of nine spatial derivatives are plotted against points of the differentiation vectors. These results exhibit similar performances to the 2D case, e.g. the conver- gence curves of and have slight difference between two methods due to the linear transformation
(a) (b)
Figure 5. Second derivatives (a) and their absolutely relative errors (b) computed by the subdomain Chebyshev spectral method for the 3D geological model shown in Figure 2. Seven points in each direction were applied to the subdomains.
Figure 6. Convergence curves of four numerical differentiation schemes: SCSM1 and SCSM2 are two algorithms of subdomain Chebyshev spectral method, FDM1 and FDM2 are two finite-difference approaches. Here, the integers “1” and “2” denote the algorithms involving and excluding the high-order derivatives of spline interpolations.
from to z. Other curves demonstrate that, as the points of differentiation vectors are larger than 5, the subdomain Chebyshev spectral method significantly improve accuracies of the finite-difference approaches. Therefore, through the 3D experiments, we once again show that the subdomain Chebyshev spectral method is superior to the finite-difference approximation in numerical differentiation with curved coordinate system.
6. Conclusions
We have demonstrated a subdomain Chebyshev spectral method to numerical differentiations in a curved coordinate system, which may be employed to seek a “strong” solution of PED defined in ageological or a geophysical model. The new method employs a set of non-linear local coordinate transformations, global spline interpolations and subdomain Chebyshev points to make the discretization grid automatically match free-surface topography and subsurface interfaces, and guarantees the numerical solution converging to the “strong” solution of partial differential equations. The new method possesses advantages from the traditional finite-difference approaches, i.e. simple discretization procedure and sparse assembled matrix, but it has stronger capability to deal with complex geometries of geological models and better accuracies in numerical differentiations. Methodologically speaking, the subdomain Chebyshev spectral method may replace the finite-difference approaches in geological or geophysical computer modelling and enables us to easily hand free-surface topography and significantly improve accuracy of modelling results.
2D and 3D synthetic experiments demonstrate the accuracies and capabilities of the subdomain Chebyshev spectral method. As points of the differentiation vectors are large than 5, the subdomain Chebyshev spectral method can reach one-order higher accuracy in numerical differentiations than the traditional finite-difference approaches in curved coordinates systems. Two algorithms of the subdomain Chebyshev spectral method (SCSM1, SCSM2) are both valid for numerical differentiations in PDE, but the algorithm SCSM2 is more efficient in computations than SCSM1, due to exclusion of computing the high-order derivatives of the coordinate trans- formations.
Acknowledgements
This work was supported by a Discovery Project (DP1093110) of the Australia Research Council. The authors thank Mr. Craig Patten for his assistance in using high-performance computing facility at eResearch SA in Australia.
Cite this paper
Bing Zhou, Graham Heinson, Aixa Rivera-Rios,, (2015) Subdomain Chebyshev Spectral Method for 2D and 3D Numerical Differentiations in a Curved Coordinate System. Journal of Applied Mathematics and Physics, 03, 358-370. doi: 10.4236/jamp.2015.33047
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30. Helmuth, S. (1995) Two Dimensional Spline Interpolation Algorithms. A. K. Peters, Wellesley. | 7,426 | 32,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.870809 |
https://www.excelguru.ca/forums/showthread.php?8925-IF-AND-OR-excel-statement/page2&s=bfb080ac030fef200742ac643be7895d | 1,571,485,863,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00405.warc.gz | 886,169,657 | 15,305 | ## View Poll Results: IF AND OR Statement, not working need help
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• IF AND or OR function is not giving correct result, need help
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# Thread: IF AND OR excel statement
The attachment hasn't changed - you haven't done what I asked in post #2.
I need to know what you are aiming for.
2. The attachment is in my Post #4 and in this post too (In column only P I have manually input the result I want) , the file name has v1 at the end.
I need "complete" if there is only C,Z or "" in cell D5 to O5, otherwise "No"
Thank you
3. Sorry - I somehow missed post#4.
Try this copied down:
=IF(SUM(COUNTIF(\$D5:\$O5,{"C","Z",""}))=12,"Complete","No")
or this:
=IF(SUM(COUNTIF(\$D5:\$O5,{"C","Z",""}))=COLUMNS(\$D5:\$O5),"Complete","No")
4. Any use to you?
5. Its good. I would not have fig it out this formulas..
Have a question, instead of selecting \$D5:\$O5, is it possible to choose cells individually for eg D5, F5, G5, I5 etc.. what would be the formula?
I was trying to tweak your formula but not working from me.
thank you
6. Why would you want to do that? I hope you are not Saying that your real data is not like the sample data you provided?
I'm not going to advise unless you share another workbook that shows the layout of the real data you want to use this with. There may be an easy fix, but I'd have to see.
Please bear in mind in future that we provide solutions based on your sample data: if that is not realistic, then you are not going to get the solution you need. We can't see what's not there!
7. Sorry, for the confusion and more work.
See attached the workbook: Student v2. these are the real data.
your 1st formula is on col AB. My manual input (the result i am expecting) is on col AC.
The additional cols. are "student" col and "S. Name" which were not in my previous workbooks, student col are added after every month col.
In new formula I want to include only cells i.e. month: D5, F5, H5... etc ( jan to dec) (not Student amount).
Thank you
8. I will look in the morning. Bedtime here.
9. You can use this:
=IF(SUM(COUNTIF(\$D5:\$AA5,{"C","Z",""}))-COUNTIFS(\$D\$4:\$AA\$4,"Student",\$D5:\$AA5,"")>=12,"Complete","No")
1. You do not need a + sign at the start of this forumla (I don't know why you added one, but it is not necessary).
2. Do not change the end of the range back to column Z - if you do, it won't work.
10. Once again, it would be nice to know whether or not this worked for you.
Page 2 of 4 First 1 2 3 4 Last | 722 | 2,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-43 | latest | en | 0.963497 |
http://markmiyashita.com/cs61a/logic/reverse/ | 1,723,760,918,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00640.warc.gz | 16,064,737 | 4,914 | This problem uses the Logic starter file located here.
Write facts to define `reverse`. You may use anything already included in the starter file as well as other facts defined in previous questions. `reverse` relates two lists with the second being the reversed of the first.
``````(fact (reverse ; YOUR CODE HERE ))
(query (reverse (1 2 3) (3 2 1)))
; expect Success!
(query (reverse (1 2 3) ?what))
; expect Success! ; what: (3 2 1)
``````
Toggle Solution
``````(fact (reverse () ()))
(fact (reverse (?first . ?rest) ?result)
(reverse ?rest ?new-rest)
(append ?new-rest (?first) ?result))
``````
We start with the base case. The reverse of the empty list is the empty list.
Moving on to the recursive case, we split up the original list into `first` and `rest`. From there, we need to find the `reverse` of the `rest`, store that as a new value, in this case `new-rest`, and then appending `first` to the end of `new-rest` to get the `result` that we originally wanted.
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11 Jun 2014, 08:49
Hey!
After my bachelor's graduation i'd like to apply for admission to MSC Finance at Cass Business School.
I would like to know if it is very difficult for me :
I have a GMAT of 670, is it enough? what is the minimum to be accepted?
In order to be admitted is important the average of the votes?? my GPA is 3.1/4 (Usa) - 50-59% in UK -- in Italy 26/30
Unfortunately, at the beginning of my bachelor's degree career, I received a little more than the minimum voting in mathematics, but I have the utmost both in statistics and in advanced statistics , also i attended the profile of "Quantitative Methods for Finance" , where I often received the highest rating.
IELTS 7.0
BEC Higher
ICFE
I hope you can help me
Thank you!
Gianni
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https://bitbucket.org/soko/maze-solver/commits/20ac4c271ebb | 1,438,731,874,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042992201.62/warc/CC-MAIN-20150728002312-00227-ip-10-236-191-2.ec2.internal.warc.gz | 864,004,721 | 37,345 | # Commits
committed 20ac4c2
Basic solution implemented in C and C++.
• Participants
# Files changed (16)
`+MAZE SOLVER`
`+`
`+PROBLEM:`
`+========`
`+Write a C or C++ program to be a virtual maze walking mouse for a two dimensional maze of X columns and Y rows.`
`+int maze[x][y];`
`+maze[?][?] == 1 means empty corridor.`
`+maze[?][?] == 0 means solid wall.`
`+ `
`+The entry point is fixed from (1,0); (i.e. maze[1][0] will always be 1).`
`+The exit point is a 1 at the edge of the maze.`
`+The output of the program is a set of (x,y) coordinates that leads from entry (1,0) to (?,?) the exit.`
`+ `
`+Example:`
`+0,1,0,0,0,0,0,0,0,0,0`
`+0,1,1,1,1,1,1,0,0,0,0`
`+0,0,0,1,0,0,1,1,0,0,0`
`+0,0,1,1,0,0,1,0,0,0,0`
`+0,0,0,0,0,1,1,1,1,0,0`
`+0,0,1,1,1,1,0,0,1,0,0`
`+0,0,0,1,0,0,0,0,0,0,0`
`+Entry at (1,0)`
`+ `
`+Virtual mouse should print out below:`
`+Solution: (1,0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)(6,2)(6,3)(6,4)(5,4)(5,5)(4,5)(3,5)(3,6)`
`+Exit at (3,6)`
`+`
`+Instructions:`
`+· Follow the K.I.S.S. concept.`
`+· Readability first, performance next.(i.e. the more thinking needed to understand the code the lower the score).`
`+· Minimum C++ approach (for readability and portability), keep it simple.(i.e. the more C++ specific expression the lower the score).`
`+· Code should for the most part be self documenting.`
`+`
`+`
`+SOLUTION:`
`+=========`
`+The solver works by eliminating dead ends and is not optimized to find `
`+the shortest path (KISS requirement).`
`+`
`+A makefile is included so run make to compile the program.`
`+`
`+The maze.txt file contains the assigned maze and will be loaded and solved each`
`+time the program is run. Alter it to test different mazes. Note, that the format`
`+of the file must remain "the same", i.e. there must be no comma at the end of a line, `
`+only 0's, 1's and ',' are allowed (though whitespace should be ignored this was not `
`+tested) and there must be a new line at the end of the file.`
`+`
`+Kind regards`
`+`
`+Lazar Sumar`
`+`
`+Note: The C version of the program was derived from the C++ attempt and follows`
`+ a seemingly OO approach. Depending on the requirement it might be better`
`+ to rewrite it using structured programming. It was left as is due to references`
`+ to C++ in the instructions...`
# File C Solution/makefile
`+CC=gcc`
`+CFLAGS=-Wall`
`+LIB=`
`+EXEC=maze_solver`
`+`
`+\$(EXEC): maze_solver.c`
`+ \$(CC) \$(CFLAGS) -o \$(EXEC) maze_solver.c`
# File C Solution/maze.txt
`+0,1,0,0,0,0,0,0,0,0,0`
`+0,1,1,1,1,1,1,0,0,0,0`
`+0,0,0,1,0,0,1,1,0,0,0`
`+0,0,1,1,0,0,1,0,0,0,0`
`+0,0,0,0,0,1,1,1,1,0,0`
`+0,0,1,1,1,1,0,0,1,0,0`
`+0,0,0,1,0,0,0,0,0,0,0`
# File C Solution/maze_solver.c
`+/* Author: Lazar Sumar`
`+ * Date 27th March 2012`
`+ */`
`+#include <stdlib.h>`
`+#include <stdio.h>`
`+`
`+#define ERROR 0`
`+#define MAX_MAZE_X 1000`
`+#define MAX_MAZE_Y 1000`
`+`
`+typedef struct {`
`+ int x, y;`
`+} Coord;`
`+typedef Coord Direction;`
`+`
`+Direction left;`
`+Direction right;`
`+Direction up;`
`+Direction down;`
`+`
`+typedef struct {`
`+ int maze[MAX_MAZE_X][MAX_MAZE_Y];`
`+ Coord size;`
`+ Coord position;`
`+ int marker;`
`+ Coord lastPosition;`
`+} Maze;`
`+`
`+void initialize_globals() {`
`+ left.x = -1; left.y = 0;`
`+ right.x = 1; right.y = 0;`
`+ up.x = 0; up.y = -1;`
`+ down.x = 0; down.y = 1;`
`+}`
`+`
`+Coord add(Coord c1, Coord c2) {`
`+ Coord res;`
`+ res.x = c1.x + c2.x; res.y = c1.y + c2.y;`
`+ return res;`
`+}`
`+`
`+Coord sub(Coord c1, Coord c2) {`
`+ Coord res;`
`+ res.x = c1.x - c2.x; res.y = c1.y - c2.y;`
`+ return res;`
`+}`
`+`
`+int equal(Coord c1, Coord c2) {`
`+ return c1.x == c2.x && c1.y == c2.y;`
`+}`
`+`
`+int not_equal(Coord c1, Coord c2) {`
`+ return !equal(c1, c2);`
`+}`
`+`
`+int set_room(Maze* m, int x, int y, int val) {`
`+ if (m != NULL && x < m->size.x && y < m->size.y) {`
`+ m->maze[x][y] = val;`
`+ return 1;`
`+ }`
`+ return 0;`
`+}`
`+`
`+int mark_here(Maze* m) {`
`+ if (!m) return ERROR;`
`+ return set_room(m, m->position.x, m->position.y, m->marker);`
`+}`
`+`
`+int move_to_start(Maze* m) {`
`+ if (!m) return ERROR;`
`+ m->position.x = 1;`
`+ m->position.y = 0;`
`+ return 1;`
`+}`
`+`
`+void load_maze(Maze* m, const char* filename) {`
`+ FILE * pFile;`
`+ int c;`
`+ int x = 0, y = 0;`
`+ pFile=fopen (filename,"r");`
`+ if (pFile==NULL) {`
`+ perror ("Error opening file");`
`+ } else {`
`+ do {`
`+ c = fgetc (pFile);`
`+ if (x >= MAX_MAZE_X || y >= MAX_MAZE_Y) {`
`+ printf("error: maximum number of rows or columns exceeded. max x=%d, max y=%d.\n", MAX_MAZE_X, MAX_MAZE_Y);`
`+ exit(1);`
`+ }`
`+ if (c == '\n') {`
`+ if (x > 0) {`
`+ if (m->size.x > 0 && m->size.x != x + 1) {`
`+ printf("error: Not all rows have the same length. check row %d.\n", y + 1);`
`+ exit(1);`
`+ } else {`
`+ m->size.x = x + 1;`
`+ }`
`+ x = 0;`
`+ ++y;`
`+ }`
`+ } else if (c == ',') {`
`+ ++x;`
`+ } else if (c == '0' || c == '1') {`
`+ m->maze[x][y]=c-'0';`
`+ }`
`+ } while (c != EOF);`
`+ m->size.y = y;`
`+ fclose (pFile);`
`+ }`
`+}`
`+`
`+void print_maze(Maze* m) {`
`+ int x, y;`
`+ `
`+ for (y = 0; y < m->size.y; ++y) {`
`+ for (x = 0; x < m->size.x; ++x) {`
`+ if (m->maze[x][y] > 0) {`
`+ printf("%d", m->maze[x][y]);`
`+ } else if (m->maze[x][y] == 0) {`
`+ printf(".");`
`+ } else {`
`+ printf("X");`
`+ }`
`+ }`
`+ printf("\n");`
`+ }`
`+}`
`+`
`+int try_move(Maze* m, Coord d) {`
`+ if (!m) return ERROR;`
`+ `
`+ Coord newPosition = add(m->position, d);`
`+ `
`+ if (newPosition.x < 0 || newPosition.x >= m->size.x ||`
`+ newPosition.y < 0 || newPosition.y >= m->size.y) {`
`+ return 0; // invalid position`
`+ }`
`+ `
`+ return m->maze[newPosition.x][newPosition.y];`
`+}`
`+`
`+int move(Maze* m, Coord d) {`
`+ if (!m) return ERROR;`
`+ `
`+ int success = 0;`
`+ Coord newPosition = add(m->position, d);`
`+ `
`+ if (try_move(m, d)) {`
`+ m->lastPosition = m->position;`
`+ m->position = newPosition;`
`+ `
`+ success = m->maze[m->lastPosition.x][m->lastPosition.y];`
`+ m->maze[m->position.x][m->position.y] = m->marker;`
`+ }`
`+ `
`+ return success;`
`+}`
`+`
`+int road_count(Maze* m) {`
`+ if (!m) return ERROR;`
`+ `
`+ int i, sum, p;`
`+ `
`+ for (sum = 0, i = -1; i < 2; i+=2) {`
`+ p = m->position.x + i;`
`+ if (p >= 0 && p < m->size.x) {`
`+ sum+=(m->maze[p][m->position.y] != 0)?1:0;`
`+ }`
`+ p = m->position.y + i;`
`+ if (p >= 0 && p < m->size.y) {`
`+ sum+=(m->maze[m->position.x][p] != 0)?1:0;`
`+ }`
`+ }`
`+ `
`+ return sum;`
`+}`
`+`
`+int is_cross_roads(Maze* m) {`
`+ if (!m) return ERROR;`
`+ `
`+ return road_count(m) > 2;`
`+}`
`+`
`+int is_dead_end(Maze* m) {`
`+ if (!m) return ERROR;`
`+ `
`+ return road_count(m) <= 1;`
`+}`
`+`
`+int is_next_to_wall(Maze* m) {`
`+ if (!m) return ERROR;`
`+ `
`+ if (m->position.x == 0 || m->position.x == m->size.x - 1 ||`
`+ m->position.y == 0 || m->position.y == m->size.y - 1) {`
`+ return 1;`
`+ }`
`+ `
`+ return 0;`
`+}`
`+`
`+// mark paths`
`+int solve_maze(Maze* m) {`
`+ if (!m) return ERROR;`
`+ `
`+ int path_marker = 1;`
`+ int dead_end_path_marker = -1;`
`+ Coord direction;`
`+ int pathCost, lowestPathCost;`
`+ int isDeadEnd = 0;`
`+ `
`+ m->marker = path_marker;`
`+ mark_here(m);`
`+ move(m, down);`
`+ `
`+ while(!is_next_to_wall(m)) {`
`+ Coord back = sub(m->lastPosition, m->position);`
`+ if (is_cross_roads(m)) {`
`+ if (!isDeadEnd) {`
`+ m->marker = ++path_marker;`
`+ mark_here(m);`
`+ } else {`
`+ isDeadEnd = 0;`
`+ m->marker = --path_marker;`
`+ mark_here(m); // update this marker`
`+ }`
`+ } else if (is_dead_end(m)) {`
`+ isDeadEnd = 1;`
`+ m->marker = dead_end_path_marker;`
`+ mark_here(m);`
`+ move(m, back);`
`+ continue;`
`+ }`
`+ `
`+ lowestPathCost = pathCost = try_move(m, left);`
`+ if (not_equal(left, back) && pathCost > 0) {`
`+ direction = left;`
`+ }`
`+ pathCost = try_move(m, right);`
`+ if (not_equal(right, back) && pathCost > 0) {`
`+ if (lowestPathCost <= 0 || lowestPathCost > pathCost) {`
`+ lowestPathCost = pathCost;`
`+ direction = right;`
`+ }`
`+ }`
`+ pathCost = try_move(m, up);`
`+ if (not_equal(up, back) && pathCost > 0) {`
`+ if (lowestPathCost <= 0 || lowestPathCost > pathCost) {`
`+ lowestPathCost = pathCost;`
`+ direction = up;`
`+ }`
`+ }`
`+ pathCost = try_move(m, down);`
`+ if (not_equal(up, back) && pathCost > 0) {`
`+ if (lowestPathCost <= 0 || lowestPathCost > pathCost) {`
`+ lowestPathCost = pathCost;`
`+ direction = down;`
`+ }`
`+ }`
`+ move(m, direction);`
`+ }`
`+ `
`+ return 1;`
`+}`
`+`
`+// Go to highest valued path at intersections`
`+int print_solution(Maze* m) {`
`+ if (!m) return ERROR;`
`+ `
`+ int path_marker = 1;`
`+ Coord direction;`
`+ int pathCost, highestPathCost;`
`+ `
`+ move_to_start(m);`
`+ m->marker = path_marker;`
`+ mark_here(m);`
`+ printf("Solution: (%d,%d)", m->position.x, m->position.y);`
`+ move(m, down);`
`+ `
`+ while(!is_next_to_wall(m)) {`
`+ Coord back = sub(m->lastPosition, m->position);`
`+ `
`+ highestPathCost = pathCost = try_move(m, left);`
`+ if (not_equal(left, back) && pathCost > 0) {`
`+ direction = left;`
`+ }`
`+ pathCost = try_move(m, right);`
`+ if (not_equal(right, back) && pathCost > 0) {`
`+ if (highestPathCost < pathCost) {`
`+ highestPathCost = pathCost;`
`+ direction = right;`
`+ }`
`+ }`
`+ pathCost = try_move(m, up);`
`+ if (not_equal(up, back) && pathCost > 0) {`
`+ if (highestPathCost < pathCost) {`
`+ highestPathCost = pathCost;`
`+ direction = up;`
`+ }`
`+ }`
`+ pathCost = try_move(m, down);`
`+ if (not_equal(down, back) && pathCost > 0) {`
`+ if (highestPathCost < pathCost) {`
`+ highestPathCost = pathCost;`
`+ direction = down;`
`+ }`
`+ }`
`+ `
`+ printf("(%d,%d)", m->position.x, m->position.y);`
`+ move(m, direction);`
`+ }`
`+ `
`+ printf("(%d,%d)\n", m->position.x, m->position.y);`
`+ printf("Exit at (%d, %d)\n", m->position.x, m->position.y);`
`+ `
`+ return 1;`
`+}`
`+`
`+int main() {`
`+ Maze maze;`
`+ `
`+ initialize_globals();`
`+`
`+ move_to_start(&maze);`
`+ load_maze(&maze, "maze.txt");`
`+ solve_maze(&maze);`
`+ print_solution(&maze);`
`+ `
`+ return 0;`
`+}`
`+`
`+MAZE SOLVER`
`+`
`+PROBLEM:`
`+========`
`+Write a C or C++ program to be a virtual maze walking mouse for a two dimensional maze of X columns and Y rows.`
`+int maze[x][y];`
`+maze[?][?] == 1 means empty corridor.`
`+maze[?][?] == 0 means solid wall.`
`+ `
`+The entry point is fixed from (1,0); (i.e. maze[1][0] will always be 1).`
`+The exit point is a 1 at the edge of the maze.`
`+The output of the program is a set of (x,y) coordinates that leads from entry (1,0) to (?,?) the exit.`
`+ `
`+Example:`
`+0,1,0,0,0,0,0,0,0,0,0`
`+0,1,1,1,1,1,1,0,0,0,0`
`+0,0,0,1,0,0,1,1,0,0,0`
`+0,0,1,1,0,0,1,0,0,0,0`
`+0,0,0,0,0,1,1,1,1,0,0`
`+0,0,1,1,1,1,0,0,1,0,0`
`+0,0,0,1,0,0,0,0,0,0,0`
`+Entry at (1,0)`
`+ `
`+Virtual mouse should print out below:`
`+Solution: (1,0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)(6,2)(6,3)(6,4)(5,4)(5,5)(4,5)(3,5)(3,6)`
`+Exit at (3,6)`
`+`
`+Instructions:`
`+· Follow the K.I.S.S. concept.`
`+· Readability first, performance next.(i.e. the more thinking needed to understand the code the lower the score).`
`+· Minimum C++ approach (for readability and portability), keep it simple.(i.e. the more C++ specific expression the lower the score).`
`+· Code should for the most part be self documenting.`
`+`
`+`
`+SOLUTION:`
`+=========`
`+The solver works by eliminating dead ends and is not optimized to find `
`+the shortest path (KISS requirement).`
`+`
`+There is a makefile included so run "make" to compile the program.`
`+`
`+The maze.txt file contains the assigned maze and will be loaded and solved each`
`+time the program is run. Alter it to test different mazes. Note, that the format`
`+of the file must remain "the same", i.e. there must be no comma at the end of a line, `
`+only 0's, 1's and ',' are allowed (though whitespace should be ignored this was not `
`+tested) and there must be a new line at the end of the file.`
`+`
`+Kind regards`
`+`
`+Lazar Sumar`
# File C++ Solution/main.cpp
`+/* Author: Lazar Sumar`
`+ * Date 27th March 2012`
`+ */`
`+#include "maze.h"`
`+#include "mouse.h"`
`+`
`+int main() {`
`+ Maze maze;`
`+ Mouse mouse;`
`+ `
`+ maze.load("maze.txt");`
`+ mouse.setMaze(&maze);`
`+ mouse.solveMaze();`
`+ mouse.printSolution();`
`+ return 0;`
`+}`
# File C++ Solution/makefile
`+CC=g++`
`+CFLAGS=-Wall`
`+LIB=`
`+EXEC=maze_solver`
`+`
`+\$(EXEC): main.cpp maze.o mouse.o`
`+ \$(CC) \$(CFLAGS) -o \$(EXEC) main.cpp maze.o mouse.o`
`+ `
`+maze.o: maze.h maze.cpp`
`+ \$(CC) \$(CFLAGS) -c maze.cpp`
`+`
`+mouse.o: mouse.h mouse.cpp`
`+ \$(CC) \$(CFLAGS) -c mouse.cpp`
# File C++ Solution/maze.cpp
`+#include "maze.h"`
`+#include <stdio.h>`
`+#include <stdlib.h>`
`+`
`+Coord left(-1, 0);`
`+Coord right(1, 0);`
`+Coord up(0, -1);`
`+Coord down(0, 1);`
`+`
`+`
`+Maze::Maze() {`
`+ position = Coord(1,0);`
`+ lastPosition = Coord(1,0);`
`+ marker = 1;`
`+}`
`+ `
`+void Maze::load(const char* filename) {`
`+ FILE * pFile;`
`+ int c;`
`+ int x = 0, y = 0;`
`+ pFile=fopen (filename,"r");`
`+ if (pFile==NULL) {`
`+ perror ("Error opening file");`
`+ } else {`
`+ do {`
`+ c = fgetc (pFile);`
`+ if (x >= MAX_MAZE_X || y >= MAX_MAZE_Y) {`
`+ printf("error: maximum number of rows or columns exceeded. max x=%d, max y=%d.\n", MAX_MAZE_X, MAX_MAZE_Y);`
`+ exit(1);`
`+ }`
`+ if (c == '\n') {`
`+ if (x > 0) {`
`+ if (size.x > 0 && size.x != x + 1) {`
`+ printf("error: Not all rows have the same length. check row %d.\n", y + 1);`
`+ exit(1);`
`+ } else {`
`+ size.x = x + 1;`
`+ }`
`+ x = 0;`
`+ ++y;`
`+ }`
`+ } else if (c == ',') {`
`+ ++x;`
`+ } else if (c == '0' || c == '1') {`
`+ maze[x][y]=c-'0';`
`+ }`
`+ } while (c != EOF);`
`+ size.y = y;`
`+ fclose (pFile);`
`+// printf("Loaded %d x %d maze (rows x columns).\n", size.y, size.x);`
`+ }`
`+}`
`+`
`+void Maze::print() const {`
`+ int x, y;`
`+ `
`+ for (y = 0; y < size.y; ++y) {`
`+ for (x = 0; x < size.x; ++x) {`
`+ if (maze[x][y] > 0) {`
`+ printf("%d", maze[x][y]);`
`+ } else if (maze[x][y] == 0) {`
`+ printf(".");`
`+ } else {`
`+ printf("X");`
`+ }`
`+ }`
`+ printf("\n");`
`+ }`
`+}`
`+`
`+int Maze::setPathMarker(int marker) {`
`+ return this->marker = marker;`
`+}`
`+`
`+int Maze::tryMove(Coord d) const {`
`+ Coord newPosition = position + d;`
`+ if (newPosition.x < 0 || newPosition.x >= size.x ||`
`+ newPosition.y < 0 || newPosition.y >= size.y) {`
`+ return 0; // invalid position`
`+ }`
`+ return maze[newPosition.x][newPosition.y];`
`+}`
`+`
`+int Maze::move(Coord d) {`
`+ int success = 0;`
`+ Coord newPosition = position + d;`
`+ `
`+ if (tryMove(d)) {`
`+ lastPosition = position;`
`+ position = newPosition;`
`+ `
`+ success = maze[lastPosition.x][lastPosition.y];`
`+ maze[position.x][position.y] = marker;`
`+ }`
`+ `
`+ return success;`
`+}`
`+`
`+void Maze::moveToStart() {`
`+ position = Coord(1, 0);`
`+}`
`+`
`+void Maze::moveTo(Coord c) {`
`+ position = c;`
`+}`
`+`
`+void Maze::markPath() {`
`+ maze[position.x][position.y] = marker;`
`+}`
`+`
`+int Maze::roadCount() const {`
`+ int i, sum, p;`
`+ `
`+ for (sum = 0, i = -1; i < 2; i+=2) {`
`+ p = position.x + i;`
`+ if (p >= 0 && p < size.x) {`
`+ sum+=(maze[p][position.y] != 0)?1:0;`
`+ }`
`+ p = position.y + i;`
`+ if (p >= 0 && p < size.y) {`
`+ sum+=(maze[position.x][p] != 0)?1:0;`
`+ }`
`+ }`
`+ `
`+ return sum;`
`+}`
`+`
`+int Maze::isCrossRoads() const {`
`+ return roadCount() > 2;`
`+}`
`+`
`+int Maze::isDeadEnd() const {`
`+ return roadCount() <= 1;`
`+}`
`+`
`+Coord Maze::getSize() const {`
`+ return this->size;`
`+}`
`+`
`+Coord Maze::getPosition() const {`
`+ return this->position;`
`+}`
`+`
`+Coord Maze::getLastPosition() const {`
`+ return this->lastPosition;`
`+}`
`+`
`+int Maze::isNextToWall() const {`
`+ if (position.x == 0 || position.x == size.x - 1 ||`
`+ position.y == 0 || position.y == size.y - 1) {`
`+ return 1;`
`+ }`
`+ `
`+ return 0;`
`+}`
# File C++ Solution/maze.h
`+/* Author: Lazar Sumar`
`+ * Date 27th March 2012`
`+ */`
`+#ifndef MAZE_H`
`+#define MAZE_H`
`+`
`+#define MAX_MAZE_X 1000`
`+#define MAX_MAZE_Y 1000`
`+`
`+class Coord {`
`+public:`
`+ int x, y;`
`+`
`+ Coord(int x = 0, int y = 0) : x(x), y(y) {}`
`+ Coord(const Coord& c) : x(c.x), y(c.y) {}`
`+ `
`+ Coord& operator=(const Coord& c) {`
`+ x = c.x; y = c.y;`
`+ return *this;`
`+ }`
`+ `
`+ bool operator==(const Coord& c) const {`
`+ return x == c.x && y == c.y;`
`+ }`
`+ `
`+ bool operator!=(const Coord& c) const {`
`+ return !(*this == c);`
`+ }`
`+ `
`+ Coord operator-(const Coord& c) const {`
`+ return Coord(x - c.x, y - c.y);`
`+ }`
`+ `
`+ Coord operator+(const Coord& c) const {`
`+ return Coord(x + c.x, y + c.y);`
`+ }`
`+};`
`+`
`+typedef Coord Direction;`
`+`
`+extern Coord left;`
`+extern Coord right;`
`+extern Coord up;`
`+extern Coord down;`
`+`
`+class Maze {`
`+ int maze[MAX_MAZE_X][MAX_MAZE_Y];`
`+ Coord size;`
`+ Coord position;`
`+ int marker;`
`+ Coord lastPosition;`
`+ `
`+ int roadCount() const;`
`+public:`
`+ Maze();`
`+ `
`+ void load(const char* filename);`
`+ void print() const;`
`+ `
`+ int setPathMarker(int marker);`
`+ `
`+ int tryMove(Direction d) const;`
`+ int move(Direction d);`
`+ void moveToStart();`
`+ void moveTo(Coord c);`
`+ void markPath();`
`+ int isCrossRoads() const;`
`+ int isDeadEnd() const;`
`+ `
`+ Coord getSize() const;`
`+ Coord getPosition() const;`
`+ Coord getLastPosition() const;`
`+ `
`+ int isNextToWall() const;`
`+};`
`+`
`+#endif`
# File C++ Solution/maze.txt
`+0,1,0,0,0,0,0,0,0,0,0`
`+0,1,1,1,1,1,1,0,0,0,0`
`+0,0,0,1,0,0,1,1,0,0,0`
`+0,0,1,1,0,0,1,0,0,0,0`
`+0,0,0,0,0,1,1,1,1,0,0`
`+0,0,1,1,1,1,0,0,1,0,0`
`+0,0,0,1,0,0,0,0,0,0,0`
# File C++ Solution/mouse.cpp
`+/* Author: Lazar Sumar`
`+ * Date 27th March 2012`
`+ */`
`+#include "mouse.h"`
`+#include <stdio.h>`
`+`
`+Mouse::Mouse() {`
`+ maze = NULL;`
`+}`
`+`
`+void Mouse::setMaze(Maze* maze) {`
`+ this->maze = maze;`
`+}`
`+`
`+void Mouse::solveMaze() {`
`+ int path_marker = 1;`
`+ int dead_end_path_marker = -1;`
`+ Coord direction;`
`+ int pathCost, lowestPathCost;`
`+ int isDeadEnd = 0;`
`+ `
`+ maze->setPathMarker(path_marker);`
`+ maze->markPath();`
`+ maze->move(down);`
`+ `
`+ while(!maze->isNextToWall()) {`
`+ Coord back = maze->getLastPosition() - maze->getPosition();`
`+ if (maze->isCrossRoads()) {`
`+ if (!isDeadEnd) {`
`+ maze->setPathMarker(++path_marker);`
`+ maze->markPath();`
`+ } else {`
`+ isDeadEnd = 0;`
`+ maze->setPathMarker(--path_marker);`
`+ maze->markPath(); // update this marker`
`+ }`
`+ } else if (maze->isDeadEnd()) {`
`+ isDeadEnd = 1;`
`+ maze->setPathMarker(dead_end_path_marker);`
`+ maze->markPath();`
`+ maze->move(back);`
`+ continue;`
`+ }`
`+ `
`+ lowestPathCost = pathCost = maze->tryMove(left);`
`+ if (left != back && pathCost > 0) {`
`+ direction = left;`
`+ }`
`+ pathCost = maze->tryMove(right);`
`+ if (right != back && pathCost > 0) {`
`+ if (lowestPathCost <= 0 || lowestPathCost > pathCost) {`
`+ lowestPathCost = pathCost;`
`+ direction = right;`
`+ }`
`+ }`
`+ pathCost = maze->tryMove(up);`
`+ if (up != back && pathCost > 0) {`
`+ if (lowestPathCost <= 0 || lowestPathCost > pathCost) {`
`+ lowestPathCost = pathCost;`
`+ direction = up;`
`+ }`
`+ }`
`+ pathCost = maze->tryMove(down);`
`+ if (up != back && pathCost > 0) {`
`+ if (lowestPathCost <= 0 || lowestPathCost > pathCost) {`
`+ lowestPathCost = pathCost;`
`+ direction = down;`
`+ }`
`+ }`
`+ maze->move(direction);`
`+ }`
`+}`
`+`
`+void Mouse::printSolution() const {`
`+ int path_marker = 1;`
`+ Coord direction;`
`+ int pathCost, highestPathCost;`
`+ `
`+ maze->moveToStart();`
`+ maze->setPathMarker(path_marker);`
`+ maze->markPath();`
`+ printf("Solution: (%d,%d)", maze->getPosition().x, maze->getPosition().y);`
`+ maze->move(down);`
`+ `
`+ while(!maze->isNextToWall()) {`
`+ Coord back = maze->getLastPosition() - maze->getPosition();`
`+ `
`+ highestPathCost = pathCost = maze->tryMove(left);`
`+ if (left != back && pathCost > 0) {`
`+ direction = left;`
`+ }`
`+ pathCost = maze->tryMove(right);`
`+ if (right != back && pathCost > 0) {`
`+ if (highestPathCost < pathCost) {`
`+ highestPathCost = pathCost;`
`+ direction = right;`
`+ }`
`+ }`
`+ pathCost = maze->tryMove(up);`
`+ if (up != back && pathCost > 0) {`
`+ if (highestPathCost < pathCost) {`
`+ highestPathCost = pathCost;`
`+ direction = up;`
`+ }`
`+ }`
`+ pathCost = maze->tryMove(down);`
`+ if (down != back && pathCost > 0) {`
`+ if (highestPathCost < pathCost) {`
`+ highestPathCost = pathCost;`
`+ direction = down;`
`+ }`
`+ }`
`+ `
`+ printf("(%d,%d)", maze->getPosition().x, maze->getPosition().y);`
`+ maze->move(direction);`
`+ }`
`+ `
`+ Coord mazeExit = maze->getPosition();`
`+ printf("(%d,%d)\n", mazeExit.x, mazeExit.y);`
`+ printf("Exit at (%d, %d)\n", mazeExit.x, mazeExit.y);`
`+}`
# File C++ Solution/mouse.h
`+/* Author: Lazar Sumar`
`+ * Date 27th March 2012`
`+ */`
`+#ifndef MOUSE_H`
`+#define MOUSE_H`
`+`
`+#include "maze.h"`
`+#include <stdlib.h>`
`+`
`+class Mouse {`
`+ Maze* maze;`
`+public:`
`+ Mouse();`
`+ void setMaze(Maze* maze);`
`+ void solveMaze();`
`+ void printSolution() const;`
`+};`
`+`
`+#endif` | 8,982 | 24,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | latest | en | 0.812507 |
https://forums.wolfram.com/mathgroup/archive/2000/Aug/msg00266.html | 1,656,475,183,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00700.warc.gz | 305,297,825 | 7,826 | Re: functional routine for {a, b, c, ...} -> {a - b, b - c, c - ...}
• To: mathgroup at smc.vnet.net
• Subject: [mg24945] Re: functional routine for {a, b, c, ...} -> {a - b, b - c, c - ...}
• From: "Marcel_Mueller" <mueller at elektron.ikp.physik.tu-darmstadt.NoSpam.de>
• Date: Thu, 24 Aug 2000 05:08:17 -0400 (EDT)
• References: <8nli50\$9fa@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
Hello
Maarten.vanderBurgt at icos.be schrieb in Nachricht
<8nli50\$9fa at smc.vnet.net>...
>Hallo,
>
> element.
>I found two ways for doing this:
>
>lst = {a, b, c, d, e, f, g, h};
>
>Table[lst[[i]] - lst[[i + 1]], {i, 1, Length[lst] - 1}]
>{a - b, b - c, c - d, d - e, e - f, f - g, g - h}
>
>ListCorrelate[{1, -1}, lst]
>{a - b, b - c, c - d, d - e, e - f, f - g, g - h}
>
>The first method is rather clumsy and the 2nd one is quite short, but not
>really obvious.
>Initally I was looking for a functional programming style routine.
>Something like: (#[[i]]-#[[i-1]])&/@lst.
>Who can tell me how to do this in a functional programming style?
The Scan and Map operations can only use funtions of single list elements.
If you want functions of more than one element you need a specially prepared
list or an operation applied to the hole list.
The straight forward way for the 'derviative' is
lst-RotateLeft[lst]
or
#-RotateLeft[#]&[l]
(Maybe you want to remove the last element)
Marcel
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• Next by thread: Initializing subscripted variables | 561 | 1,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.737838 |
https://brainly.com/question/228802 | 1,484,884,577,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280774.51/warc/CC-MAIN-20170116095120-00090-ip-10-171-10-70.ec2.internal.warc.gz | 792,347,972 | 9,390 | # What is the answer to this question? What is 28% of 75? Could you please explain how to do this using proportions? I'm in Pre-Algebra
2
by haia
2014-12-16T16:40:00-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
? I'm sorry I don't understand.
Thanks! That makes more sense now!
2014-12-16T16:44:25-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
This is my way: Change the percent to a decimal by moving it to the left 2 times.Then .28 times 75. Your answer is 21. hope this helped.
Yes it did! Thanks!
you welcome!
I am new to brainly.com.. .... .. how do i check my question i posted | 281 | 1,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2017-04 | latest | en | 0.934993 |
https://math.stackexchange.com/questions/1492896/what-is-the-difference-between-cartesian-product-and-intersection | 1,621,203,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00002.warc.gz | 402,509,051 | 38,825 | # What is the difference between cartesian product and intersection?
What is the difference between the cartesian product of two sets $(A \times B)$ and the intersection of those very same sets $(A \cap B)$? When I picture them on a graph I don't get the difference.
J
• What kind of graphs did you look at? Also, did you look at the definitions? – Wojowu Oct 22 '15 at 20:46
• Have you tried using a search engine for the definition of "Cartesian product" and "set intersection" - seriously, lazy. – Alec Teal Oct 22 '15 at 20:53
• Who upvoted this (the spread is +1/-2) - show yourself! – Alec Teal Oct 22 '15 at 21:10
• I guess I've misunderstood those concepts, but don't tell me I haven't looked it up. The definitions are kind of hard to grasp without thorough knowledge of mathematical notation, and the constructive answers I got definitely helped me understand where I was looking wrong. – Jxx Oct 22 '15 at 21:15
• @Julien The accepted answer is nothing but a copy and paste of the definitions, so I hold the same position as Alec. – YoTengoUnLCD Oct 22 '15 at 22:06
Given sets $A$ and $B$, the intersection $A \cap B$ is the set of elements which lie in both in $A$ and in $B$. Thus,
$A \cap B = \left\{x \bigm| x \in A \text{ and } x \in B \right\}$.
The Cartesian Product $A \times B$ is the set of all ordered pairs $(a,b)$ where $a \in A$ and $b \in B$. Thus,
$A \times B = \left\{(a,b) \bigm| a \in A \text{ and } b \in B\right\}$.
Henning's answer provides a concrete example of these definitions in action.
• Thank you very much, I understood it now. :) – Jxx Oct 22 '15 at 21:16
• You're welcome. Mathematical concepts can be tricky to grasp if it's your first pass at formal math; when in doubt, always fall back on the definitions. – Gaussian0617 Oct 22 '15 at 21:40
Um, they have nothing in common, really. Consider, as a simple example $$\{1,2\}\cap \{2,3\} = \{2\} \\ \{1,2\}\times\{2,3\} = \{\langle1,2\rangle,\langle1,3\rangle,\langle2,2\rangle,\langle2,3\rangle\}$$
You'd need some strange (but not impossible) contortions to find an example of $A$ and $B$ such that $A\cap B$ and $A\times B$ have even one element in common.
• Thank you very much for your answer, that is very clear. – Jxx Oct 22 '15 at 21:15
Cartesian product between two sets $A$ and $B$ is the set of couples $A\times B:=\{(a,b);a\in A \text{ and }b\in B\}$, whereas $A\cap B$ is the set $A\cap B:=\{c;c\in A\text{ and }c\in B\}$.
• My confusion was in the difference between (a,b) and (c). Thank you very much for your answer. – Jxx Oct 22 '15 at 21:16
• You're welcome ! – Balloon Oct 22 '15 at 21:29 | 819 | 2,614 | {"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} | 4.28125 | 4 | CC-MAIN-2021-21 | longest | en | 0.908789 |
https://answers.microsoft.com/en-us/xbox/forum/xba_gapp/pyramid-expert-challenge-121813/d7901106-5540-4741-b6bd-f9dee89033ba | 1,537,316,887,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155792.23/warc/CC-MAIN-20180918225124-20180919005124-00183.warc.gz | 455,405,534 | 27,342 | # Pyramid Expert Challenge 12/18/13
I am having trouble solving this challenge, has anyone had success and willing to share hints?
## Discussion Info
Last updated July 3, 2018 Views 4 Applies to:
My solution is as follows:-
8h/5d
Jc/2d
9d/4s
Ac/Qc
8s/5c
8d/5s
Jh/2h
6s/7s
10s/3d
Kh
7h/6d
10h/3h
9c/4c
9h/4h
Kc
Ks
10d/3c
9s/4d
Jd/2c
10c/3s
Js/2s
8c/5h
Kd
Qh/As
6h/7c.
Do you mind posting this on our Facebook page? I bet it would help everyone there too! facebook.com/microsoftsolitaire
You must be looking at a different version of the game than what I have. Your solution doesn't work. Try again, maybe you left out some steps.
Helen, I don't think that would be a good idea as the above solution doesn't work on this game.
WealthyCube38 and Helen,
CombinedPlague2's solution works with two additions:
8h/5d
Jc/2d
9d/4s
Ac/Qc
6c/7d
8s/5c
8d/5s
Qd/Ah
Jh/2h
6s/7s
10s/3d
Kh
7h/6d
10h/3h
9c/4c
9h/4h
Kc
Ks
10d/3c
9s/4d
Jd/2c
10c/3s
Js/2s
8c/5h
Kd
Qh/As
6h/7c.
CombinedPlague2 and Helen: I hope I didn't offend anyone but I posted this solution on Facebook with my changes giving credit to CombinedPlague2. I had already solved the Challenge but worked through it again using CombinedPlague2's solution and found two missing steps. Regards
Sorry about that! We all make mistakes! I did do those 2 highlighted moves as well.
Thanks so much for the help!
Thanks for the correction. I solved it by a tedious process of laying out real cards, writing down all the moves, changing some, figuring out what the last 5 cards needed to be and - voila - it finally worked out. Oh so proud to be an expert! But I don't think I could write out exactly what I did as I didn't track the suits. Thanks for listening.
p.s. Can someone tell me what to do with the coins I earn, if anything? That is, what good are they, how can I use them? | 659 | 1,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-39 | latest | en | 0.945182 |
https://goopennc.oercommons.org/authoring/2336-t4t-cluster-4-family-letters-english-spanish/view | 1,717,028,498,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00654.warc.gz | 246,572,506 | 14,238 | # T4T Cluster 4 Family Letters (English & Spanish)
These resources are part of Tools4NCTeachers. Click to access fully-formatted letters.
Letter excerpt:
During the week of <date> we will be starting a new math unit focused on measurement. The purpose of this letter is to give you some background information about our new unit.
Focus of the Unit
Students measure with nonstandard units (cubes, paperclips, etc.), order objects by length, and compare the lengths of objects. The use of inch tiles and centimeter cubes introduces the idea of standard measurement. Students connect numbers to length. The idea is introduced that the size of a unit used when measuring an object affects the number used of that unit.
· Order three objects by length. · Reason that if the marker is shorter than the pencil, and the crayon is shorter than the marker, the crayon is also shorter than the pencil.
Building off Past Mathematics
In kindergarten, students first learn which attributes are measurable (height, weight, etc.) and work on building their vocabulary to describe those measurable attributes. The expectation is students describe an object in more than one way Ex: This block is long and heavy. Kindergarteners also directly compare two objects. Ex: The red book is heavier than the blue book. I am taller than my friend.
Strategies That Students Will Learn
Meeting Incomplete the Standard Understanding Measuring Measuring Measuring with gaps. Measure an object from end to end with no gaps or overlaps using nonstandard units. Recognize the pencil’s length is 7 tiles. Measuring with overlaps
Meeting Incomplete the Standard Understanding Comparing Comparing Both pencils are the same length. Measurements change when the size of the The top pencil is longer because it is 7 tiles unit changes. It takes more tiles than long and the bottom pencil is shorter because it is only 2 crayons long. 7 is more than two, so the top pencil is crayons to measure the pencil because the tiles are smaller. It takes more of a smaller unit to cover longer. the same distance.
Ideas for Home Support
· Measure household items using nonstandard items (cereal, paperclips, crayons, pretzel rods, etc.)
· Encourage your child to use measurement vocabulary
· Compare the lengths of objects directly by measuring and by using reasoning.
Vocabulary end to end height length long longer longest measure short shorter shortest tall taller tallest width
Thank you for serving as partners in your child’s success as a mathematician!
<signature> | 533 | 2,578 | {"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": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-22 | latest | en | 0.890998 |
http://www.vkuzi.cz/the-most-neglected-fact-about-physics-central-revealed/ | 1,575,578,153,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482038.36/warc/CC-MAIN-20191205190939-20191205214939-00005.warc.gz | 247,197,087 | 18,207 | # The Most Neglected Fact About Physics Central Revealed
As a consequence the image of the pencil seems to be broken. How tightly an object holds onto its electrons is dependent on the material the object is created from and is known as the amount of electron affinity. What’s more, the section of the pencil that’s submerged in water seems to be wider than the part of the pencil that isn’t submerged. Investigate the way the perforations in the toilet paper affect the way that it rips. When you drop it to the wrapper, a water molecule will attempt to get as close as it can to the top layer of the paper. Consider charging objects besides the comb.
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Thank you for helping pull a shy freshman out in the actual Earth, and challenging my idea of what masculinity may be. Its function in economic development can’t be over emphasized. Physical science employs key thoughts and theories to model and explain how a particular component of nature interacts.
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## Physics Central Features
When you move your entire body, you’re doing work. Wakesurfing boats should have inboard motors, instead of outboard motors with exposed propellers, to protect against the surfer from being injured. There ought to be a small quantity of paper hanging down. This new hill is extremely stable, and the atoms wouldn’t roll off the hill to some other energy state.
## Things You Should Know About Physics Central
The podcast can’t be summarized, and you need to listen to the whole episode. We’ll investigate this part of refraction in amazing detail in Lesson 2. Work is critically important for surviving a long, healthier life. Physics, specifically, focuses on a few of the most fundamental of questions regarding our physical universe.
Even in the event the sources of light do not stay in step with one another, so long as the amount by which they’re out of step is still the same over time, the light sources are supposedly coherent. Over time people have created ways to try and make this troublesome work simpler. When there is space, there’s time. This is a rather massive difference, because now things can really be driven apart, not merely together. So from that moment on, I spent nearly all of my time hoping to make sure I could observe the distinction between what’s physics and what’s not.
I am fifty-two years old and I have not ever owned a vehicle. The outcome of sex can involve many folks, not only one person. We need it so as to survive in the modern planet, but the unwritten rules that set the parameters for the present money system make it all too easy that people develop into enslaved, unhappy, and partake in immoral behavior for the interest of a part of paper. However, there’s a good deal more that needs to be taken into account too.
Physics, on the flip side, has resolved many fundamental problems, although those resolutions have a tendency to open up whole new forms of questions. As a researcher, I made a decision to delve in the web to get any reference to this specific description. And I’ve worked extremely difficult to forget nearly all of the specifics of this so forgive me for the foggy description. Take a look at the Astronomy Events page for more details.
## If You Read Nothing Else Today, Read This Report on Physics Central
From within the academy, it appears to me that there have been two key limitations to the intellectual imagination. That’s what allyship and transracial friendship appears like. They’re formed through an elaborate apparatus of jingle and style, of persuasion and fraud. But this may be conscious or unconscious and doesn’t need to get made personal (about whether the department chair or the individual teacher is a great or bad person). Since speed isn’t a concern, there are not any particular reading requirements. It’s meant for middle and higher school students and teachers. Mathematics is all about knowledge creation through our knowledge of patterns.
The target of a force analysis is to find out the net force and the corresponding acceleration. The numerator provides the entire moment that’s then balanced through an equivalent total force at the middle of mass. The resultants in each one of the above diagrams represent the net force acting on the object. The fundamental frequency occurs when the wavelength is during its longest possible length.
The variety of fringes will be quite large for large slit separations. In addition, we notice that the fringes get fainter further away from the middle. Visible light is but a little portion of the whole electromagnetic spectrum. A middle of gravity that’s at or over the lift point will probably result in a tip-over incident.
## What You Don’t Know About Physics Central
Your bank accounts are wiped out. It can be exceedingly frustrating to attempt to comprehend what is written, when it’s not explained clearly and logically. You have the right to find angry. You’re also eligible to cry. | 1,149 | 5,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.913788 |
https://www.rochesterscientific.com/ADM/AtomicDensityMatrix/html/tutorial/TheDensityMatrixAndTheLiouvilleEquation.html | 1,695,722,993,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00352.warc.gz | 1,072,191,520 | 3,631 | # The Density Matrix and the Liouville Equation
The density matrix is a generalization of the state vector that allows graceful handling of ensemble averages and relaxation effects. For a single atom with state vector , we define the density operator by .
Define an atomic system and form its ket vector.
Taking the outer product of the ket with its dual gives the density matrix in terms of the state vector coefficients.
Evidently, the diagonal matrix elements of the density operator are the probabilities of the corresponding states (here the states 1 and 2). The off-diagonal matrix elements are known as coherences.
We can retrace the example given in "The State Vector and the Schrödinger Equation" in terms of the density-matrix elements rather than the state-vector amplitudes. The equation governing the evolution of the density matrix, , known as the Liouville equation, can be derived directly from the Schrödinger equation.
The function DensityMatrix forms the density matrix for an atomic system.
The Hamiltonian describing the interaction with an electric field. We normalize the electric field to frequency units by dividing by the reduced dipole matrix element between states 1 and 2.
The Liouville equation.
Initial conditions, specifying that the atom is in state 1 at time zero.
List of the density-matrix variables.
Solve the Liouville equation.
The diagonal density-matrix elements are the probabilities of finding each state.
If instead of a single atom, we are interested in averages over an ensemble of atoms, each with state vector |ψi, we simply work with the averaged density matrix: . | 329 | 1,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.893996 |
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# A total of 512 players participated in a single tennis knock
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01 May 2012, 21:26
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A total of 512 players participated in a single tennis knock out tournament. What is the total number of matches played in the tournament? (Knockout means if a player loses, he is out of the tournament). No match ends in a tie.
A. 511
B. 512
C. 256
D. 255
E. 1023
I chose D, solved this way - after first 256 matches, remaing players 256 (those 256 who lost are knocked out), after another 128 matches, remaining players are 128, after 64 more matches, no of remainig players are 64, after 32, 32, after 16, 16, after 8, 8, after 4, 4, after 2,2, and then 1 -
total 256 + 128+64+32+16+8+1 = 255 matches
However, the correct answer given is A 511, can anyone please exlain what's wrong in my approach? Thanks!
[Reveal] Spoiler: OA
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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01 May 2012, 23:00
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gmihir wrote:
A total of 512 players participated in a single tennis knock out tournament. What is the total number of matches played in the tournament? (Knockout means if a player loses, he is out of the tournament). No match ends in a tie.
A. 511
B. 512
C. 256
D. 255
E. 1023
I chose D, solved this way - after first 256 matches, remaing players 256 (those 256 who lost are knocked out), after another 128 matches, remaining players are 128, after 64 more matches, no of remainig players are 64, after 32, 32, after 16, 16, after 8, 8, after 4, 4, after 2,2, and then 1 -
total 256 + 128+64+32+16+8+1 = 255 matches
However, the correct answer given is A 511, can anyone please exlain what's wrong in my approach? Thanks!
You've done everything right except calculation: 256+128+64+32+16+8+4+2+1=511.
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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19 Nov 2012, 19:17
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There are 512 players, only 1 person wins, 511 players lose. in order to lose, you must have lost a game.
511 games.
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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26 Feb 2013, 19:12
1- the 512 players will play 256 games --> 256 plyers will go out from these games
2- the remaining is 256 players, they will play 128 games --> 128 players will go out
3- the remaining is 128 players, they will play 64 games --> 64 players will go out
4- the remaining is 64 players, they will play 32 games --> 32 players will go out
5- the remaining is 32 players, they will play 16 games --> 16 players will go out
6- the remaining is 16 players, they will play 8 games --> 8 players will go out
7- the remaining is 8 players, they will play 4 games --> 4 players will go out
8- the remaining is 4 players, they will play 2 games --> 2 players will go out.
9- the remaining is 2 players, they will play 1 games --> 1 players will go out
265+128+64+32+16+8+4+2+1=511
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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27 Feb 2013, 12:44
AlyoshaKaramazov wrote:
There are 512 players, only 1 person wins, 511 players lose. in order to lose, you must have lost a game.
511 games.
I know this is an old post. But damn, sometimes the answer is so simple, you just have to thinkg logically.
Thanks Alyosha
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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24 Jun 2013, 08:23
If you divide 512 by 2 recursively to factor: 512,256,128,64,32,16,8,4,2,1. so the answer narrows to down to either A or B.
Since the last dividend is 1, the sum will be odd so it should be 511
A
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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26 Dec 2013, 13:05
Bunuel wrote:
gmihir wrote:
A total of 512 players participated in a single tennis knock out tournament. What is the total number of matches played in the tournament? (Knockout means if a player loses, he is out of the tournament). No match ends in a tie.
A. 511
B. 512
C. 256
D. 255
E. 1023
I chose D, solved this way - after first 256 matches, remaing players 256 (those 256 who lost are knocked out), after another 128 matches, remaining players are 128, after 64 more matches, no of remainig players are 64, after 32, 32, after 16, 16, after 8, 8, after 4, 4, after 2,2, and then 1 -
total 256 + 128+64+32+16+8+1 = 255 matches
However, the correct answer given is A 511, can anyone please exlain what's wrong in my approach? Thanks!
You've done everything right except calculation: 256+128+64+32+16+8+4+2+1=511.
Geometric progression
512 = 2^9
2^1 + 2^2 ....+2^9 + 1
2 ( 2^8-1) = 2^9 - 2 + 1
A is our friend here
Cheers!
J
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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03 Jan 2014, 17:34
gmihir wrote:
A total of 512 players participated in a single tennis knock out tournament. What is the total number of matches played in the tournament? (Knockout means if a player loses, he is out of the tournament). No match ends in a tie.
A. 511
B. 512
C. 256
D. 255
E. 1023
I chose D, solved this way - after first 256 matches, remaing players 256 (those 256 who lost are knocked out), after another 128 matches, remaining players are 128, after 64 more matches, no of remainig players are 64, after 32, 32, after 16, 16, after 8, 8, after 4, 4, after 2,2, and then 1 -
total 256 + 128+64+32+16+8+1 = 255 matches
However, the correct answer given is A 511, can anyone please exlain what's wrong in my approach? Thanks!
256+128+ 64+32+16+8+4+2+1
= 511
OA - A
Thanks for posting
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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A total of 512 players participated in a single tennis knock [#permalink]
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25 Jun 2015, 02:14
One of the best logics I've seen for these knockout questions is as follows:
In this question, there are 512 participants. And it takes ONE match to eliminate ONE player. So at the end of the day you need to eliminate everyone except the winner. ie. you need to eliminate 511 participants and naturally you need 511 matches for the same
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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18 Apr 2016, 01:13
gmihir wrote:
A total of 512 players participated in a single tennis knock out tournament. What is the total number of matches played in the tournament? (Knockout means if a player loses, he is out of the tournament). No match ends in a tie.
A. 511
B. 512
C. 256
D. 255
E. 1023
I chose D, solved this way - after first 256 matches, remaing players 256 (those 256 who lost are knocked out), after another 128 matches, remaining players are 128, after 64 more matches, no of remainig players are 64, after 32, 32, after 16, 16, after 8, 8, after 4, 4, after 2,2, and then 1 -
total 256 + 128+64+32+16+8+1 = 255 matches
However, the correct answer given is A 511, can anyone please exlain what's wrong in my approach? Thanks!
If there are 2 players, then there will be only 1 match
If there are 4 players, then there will be 2+1 matches
8 players, 4+2+1=7
16 players, 8+4+2+1 = 15
Now you are getting the pattern
512 players, 256+128+64+..+2+1=511
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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18 Apr 2016, 12:01
Mathivanan Palraj wrote:
gmihir wrote:
A total of 512 players participated in a single tennis knock out tournament. What is the total number of matches played in the tournament? (Knockout means if a player loses, he is out of the tournament). No match ends in a tie.
A. 511
B. 512
C. 256
D. 255
E. 1023
I chose D, solved this way - after first 256 matches, remaing players 256 (those 256 who lost are knocked out), after another 128 matches, remaining players are 128, after 64 more matches, no of remainig players are 64, after 32, 32, after 16, 16, after 8, 8, after 4, 4, after 2,2, and then 1 -
total 256 + 128+64+32+16+8+1 = 255 matches
However, the correct answer given is A 511, can anyone please exlain what's wrong in my approach? Thanks!
If there are 2 players, then there will be only 1 match
If there are 4 players, then there will be 2+1 matches
8 players, 4+2+1=7
16 players, 8+4+2+1 = 15
Now you are getting the pattern
512 players, 256+128+64+..+2+1=511
Good catch , or U can go the other way round -
When there are 2 player only 1 knockout match is needed
When there are 3 player only 2 knockout match is needed
When there are 4 player only 3 knockout match is needed
So, the pattern formed is -
When there are n player only n - 1 knockout match is needed
Hence , When there are 512 player only 511 (512 - 1 ) knockout match is needed
Either way answer will be the same but the best thing/strategy is to solve the problems within minimum time & calculation
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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22 Apr 2016, 14:59
Firstly the question should state that a match is played between 2 players (what if its a cricket match)
assuming the match is played between 2 teams = > number of matches => 512/2 + 256/2 +128/2+64/2+32/2+16/2+8/2+4/2+2/2
we dont need to calculate the sum here => the unit digit will be 1
SMASH that A
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Re: A total of 512 players participated in a single tennis knock [#permalink]
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31 Aug 2016, 10:54
stonecold wrote:
Firstly the question should state that a match is played between 2 players (what if its a cricket match)
assuming the match is played between 2 teams = > number of matches => 512/2 + 256/2 +128/2+64/2+32/2+16/2+8/2+4/2+2/2
we dont need to calculate the sum here => the unit digit will be 1
SMASH that A
1. Question mentions that it is a Tennis Tournament of Singles.
2. All we know is that the sum is Odd. Please explain how will you get the unit's digit = 1 by looking at the sequence.
My Solution:
# of matches = 256 + 128 + 64 + ... + 2 + 1 = $$2^8 + 2^7 + 2^6 + ... + 2^1 + 2^0$$ => Ascending G.P.
Sum of G.P. =$$\frac{a(r^n - 1)}{a-1}$$
Where,
a = First Term = 1 (Sum starts from 1)
r = Common Ratio = 2 (ratio of each successive term is 2)
n = Number of terms = 9
Therefore,
Sum = $$\frac{1 * (2^9 - 1)}{2-1}$$ = 512 - 1 = 511
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My Debrief for 750 - https://gmatclub.com/forum/from-720-to-750-one-of-the-most-difficult-pleatues-to-overcome-246420.html
My CR notes - https://gmatclub.com/forum/patterns-in-cr-questions-243450.html
Rest of the Notes coming soon.
Kudos [?]: 176 [0], given: 102
Re: A total of 512 players participated in a single tennis knock [#permalink] 31 Aug 2016, 10:54
Display posts from previous: Sort by | 4,463 | 14,255 | {"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} | 3.96875 | 4 | CC-MAIN-2017-43 | latest | en | 0.906124 |
https://topbtctayd.web.app/niner82573nego/implied-exchange-rate-if-ppp-holds-1537.html | 1,642,763,214,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00170.warc.gz | 619,868,618 | 6,907 | ## Implied exchange rate if ppp holds
If purchasing power parity holds then the value of the a real exchange rate is from ECONOMICS 110 None of the above is implied by purchasing-power parity. 5 Dec 2016 PPP holds when price levels in two countries are equal when for the exchange rate that would be implied by absolute PPP: Absolute PPP:
5) Assume the implied PPP rate of exchange of Mexican Pesos per U.S. dollar is A) PPP holds up well over the short run but poorly for the long run, and the relative PPP holds and the expected value of the real exchange rate the coefficients of variations of the implied PPPs and exchange rates in each country, and 10 Jan 2019 How does the price of a Big Mac in the U.S. compare to that you pay The Big Mac indicator draws on purchasing-power parity theory, which dictates that exchange rates reflect The implied exchange rate is 0.57 [pound per dollar]. to tell when the market bottoms · These three stock funds are holding Exchange rates: number of home currency exchange of two currencies at a rate agreed on the date If PPP holds instantaneously, unlikely that Testing for unbiasedness. Rational exp. Approach. Implied relationship. “UIP”, or Fama. Studies(in(Applied(Economics(Series!is!under!the!general!direction!of!Professor! official!and!parallel!exchange!rates!using!purchasing!power!parity!(PPP). Exchange!Rate!Arrangements!in!the!21st!Century:!Which!Anchor!Will!Hold?”! If purchasing power parity holds then the value of the a real exchange rate is from ECONOMICS 110 None of the above is implied by purchasing-power parity. 5 Dec 2016 PPP holds when price levels in two countries are equal when for the exchange rate that would be implied by absolute PPP: Absolute PPP:
## When you don't apply PPP, then a country's GDP will change when its exchange rate changes. After running a PPP calculation, the CIA World Factbook calculated China's 2017 GDP at just over \$23 trillion – much larger than the unadjusted figure.
Purchasing power parities (PPP) Purchasing power parities (PPPs) are the rates of currency conversion that try to equalise the purchasing power of different currencies, by eliminating the differences in price levels between countries. PPP is an economic theory that compares different countries' currencies through a "basket of goods" approach. According to this concept, two currencies are in equilibrium—known as the currencies being at par —when a basket of goods is priced the same in both countries, taking into account the exchange rates. If all goods were freely tradable, and foreign and domestic residents purchased identical baskets of goods, purchasing power parity (PPP) would hold for the exchange rate and GDP deflators (price levels) of the two countries, and the real exchange rate would always equal 1 Implied Value - this is what the amount in the foreign currency should be, assuming that the countries have purchasing power parity. At this exchange rate a Big Mac costs the same in both countries. Market Value - this is the converted amount according to the market exchange rates.
### real exchange rate (rer) is a good proxy of the PPP-implied equilibrium real holds, capital should in theory flow from developed to developing countries.
First, real exchange rates are mean-reverting, as implied by the Purchasing long-run PPP holds and that the nominal exchange rate is the main driver of this. conclude that long run relative PPP holds in our European sample. Keywords: Real many macroeconomic models of trade and of exchange rate determination, failure to Their first-order autocorrelation coefficient implied a speed of mean 5) Assume the implied PPP rate of exchange of Mexican Pesos per U.S. dollar is A) PPP holds up well over the short run but poorly for the long run, and the
### implied exchange rate or lower than parity if it is in their self interest to do so. that the Purchasing Power Parity theorem holds in the long run and that there is
The implied PPP was = € 3.31/\$3.57 = 0.93; Actual exchange rate was \$1 = € 0.8114 {(0.93 - € 0.8114)/€ 0.8114}X 100 = 14.6167% overvalued against the U.S. Dollar; Thus the Euro was overvalued against the U.S. dollar by approximately 15%. Thus the Big Mac Index is used as a yardstick to identify if a currency is expensive or cheap. Shortcomings of Purchasing Power Parity Theory
## Implied Value - this is what the amount in the foreign currency should be, assuming that the countries have purchasing power parity. At this exchange rate a Big Mac costs the same in both countries. Market Value - this is the converted amount according to the market exchange rates.
24 May 2013 between exchange rates and the prices of goods implied by the law of one price. PPP therefore holds when, at going exchange rates, every. PPP theory do badly in explaining exchange rate movements in terms of changes in national price levels. If purchasing power parity holds true, the real exchange rate remains constant over time. The real exchange rate, as implied by the. real exchange rate (rer) is a good proxy of the PPP-implied equilibrium real holds, capital should in theory flow from developed to developing countries. how good is PPP as a measure of an equilibrium exchange rate? In this overview we argue costs, means that it may only hold up to a constant term. However, most again the implied halfЛlife is still around 4 years. The persistent nature of Suppose over 5 years the US rate of price inflation is 5% per year and the Canadian rate is 7% per year. What must be true for LOOP to hold? 1. Then the implied PPP exchange rate = 20.4RMB/\$5.28 = 3.86 RMB/\$ (= 0.259 \$/RMB). First, real exchange rates are mean-reverting, as implied by the Purchasing long-run PPP holds and that the nominal exchange rate is the main driver of this.
3 Real Exchange Rates as a Random Walk If the real exchange rate is a random walk then PPP does not hold, hence the null is for a unit root in the real exchange rate where the alternative hypothesis is PPP holds in the long run. The unit root tests impose β at one and then test that the log of the real exchange rate, defined by q t = s t +p t Purchasing power parity (PPP) is a theory which states that exchange rates between currencies are in equilibrium when their purchasing power is the same in each of the two countries. This means that the exchange rate between two countries should equal the ratio of the two countries' price level of a fixed basket of goods and services. Derivation of the PPP. Suppose that π H and π F indicate the home and foreign country’s inflation rates, respectively. In the following equations, you work with inflation factors of home and foreign countries, (1+ π H) and (1+ π F), respectively.. Remember, the relative PPP implies that changes in an exchange rate follow the changes in both countries’ price levels. These objections notwithstanding, however, it is often asserted that the PPP theory of exchange rates will hold at least approximately because of the possibility of international goods arbitrage. There are two senses in which the PPP hypothesis might hold. Absolute purchasing power parity holds when the purchasing power of \$\begingroup\$ @Simon - The sentences you excerpted from Wikipedia -- "Purchasing Power Parity (PPP) is a theory that measures prices at different locations using a common basket of goods" and "The real exchange rate (RER) is the purchasing power of a currency relative to another at current exchange rates and prices" are both factually wrong E.g., PPP does not, and never has, "measure[d] prices | 1,664 | 7,568 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-05 | latest | en | 0.885973 |
http://geniusbrainteasers.com/daily-brainteasers/2018/01/07/ | 1,516,108,780,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886436.25/warc/CC-MAIN-20180116125134-20180116145134-00349.warc.gz | 140,850,888 | 9,794 | BRAIN TEASERS
# Daily Brain Teasers for Sunday, 07 January 2018
puzzles, riddles, mathematical problems, word games, mastermind, cinemania, music, stereograms, ...
## Find number abc
If 18430 + 5bbca = c510a find number abc. Multiple solutions may exist.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #math
## Guess the name of musician
Look carefully caricature and guess the name of musician.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #music
## Find a famous person
Find the first and the last name of a famous person. Text may go in all 8 directions. Length of words in solution: 8,4.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #wordpuzzles
### A very large, old building was...
A very large, old building was being torn down in Chicago to make room for a new skyscraper.
Due to its proximity to other buildings it could not be imploded and had to be dismantled floor by floor.
While working on the 49th floor, two construction workers found a skeleton in a small closet behind the elevator shaft. They decided that they should call the police.
When the police arrived they directed them to the closet and showed them the skeleton fully clothed and standing upright. They said, "This could be Jimmy Hoffa or somebody really important."
Two days went by and the construction workers couldn't stand it any more, they had to know who they had found. They called the police station and said, "We're the two guys who found the skeleton in the closet and we want to know if it really was Jimmy Hoffa."
The cop said, "Well, it wasn't Jimmy Hoffa, but it was somebody kind of important."
"Well, who was it?"
"The 1956 Polish National Hide-and-Seek Champion!"
Jokes of the day - Daily updated jokes. New jokes every day.
## What a winning combination?
The computer chose a secret code (sequence of 4 digits from 1 to 6). Your goal is to find that code. Black circles indicate the number of hits on the right spot. White circles indicate the number of hits on the wrong spot.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #mastermind
### Sir Alfred Ewing
Died 7 Jan 1935 at age 79 (born 27 Mar 1855).Sir James Alfred Ewing was a Scottish physicistwho discovered and named hysteresis (1881), the resistance of magnetic materials to change in magnetic force. Ewing was born and educated in Dundee and studied engineering on a scholarship at Edinburgh University. He helped Sir William Thomson, later Lord Kelvin in a cable laying project. In 1878 he became professor of Mechanical Engineering and Physics at Tokyo University, where he devised instruments for measuring earthquakes. In 1903 he moved to the Admiralty as head of education and training, where during WW I, he and his staff took on the task of deciphering coded messages.
## Chess Knight Move
Find the country and its capital city, using the move of a chess knight. First letter is N. Length of words in solution: 5,5,9.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #wordpuzzles #chessknightmove
## Calculate the number 4820
NUMBERMANIA: Calculate the number 4820 using numbers [5, 8, 3, 9, 43, 533] and basic arithmetic operations (+, -, *, /). Each of the numbers can be used only once.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #math #numbermania
## Guess the Band Name
Which musician band has an album with a cover as in the picture?
The first user who solved this task is Djordje Timotijevic.
#brainteasers #music #riddles
## MAGIC SQUARE: Calculate A-B*C
The aim is to place the some numbers from the list (15, 16, 18, 19, 20, 21, 29, 32, 34, 76, 97) into the empty squares and squares marked with A, B an C. Sum of each row and column should be equal. All the numbers of the magic square must be different. Find values for A, B, and C. Solution is A-B*C.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #math #magicsquare
## Find a famous person
Find the first and the last name of a famous person. Text may go in all 8 directions. Length of words in solution: 9,9.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #wordpuzzles
## Find number abc
If b5c8c - 535ab = c1bb5 find number abc. Multiple solutions may exist.
The first user who solved this task is Djordje Timotijevic.
#brainteasers #math
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## Top 20 Users
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This site uses cookies to store information on your computer. Some are essential to help the site properly. Others give us insight into how the site is used and help us to optimize the user experience. See our privacy policy. | 1,396 | 5,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.962238 |
http://www.aarondembyjones.com/2012/06/ | 1,579,712,475,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607314.32/warc/CC-MAIN-20200122161553-20200122190553-00280.warc.gz | 184,938,424 | 10,655 | # Finite ant crawling
Here’s another example of recursive probability in action on a different problem.
Suppose you are an ant crawling along the edges of an equilateral triangle $$ABC$$. You start at point $$A$$. Every minute, you choose uniformly at random one of the other two vertices and move to it. What is the probability that you return to where you started (point $$A$$) after exactly twelve minutes?
Well, okay, at first the question seems hard to grasp, so let’s try to approach it by first tackling simpler questions. What is the probability that you return to where you started after exactly one minute? Zero, since you can only make one move. Well that didn’t work! How about after two minutes? Now we’re getting somewhere, since you can do this via two different paths: one through point $$B$$ and one through point $$C.$$ There are four total paths you could take, so the probability in this case must be $$\frac{1}{2}$$. Just out of curiosity/completeness, we could also ask ourselves what the probability of arriving at point $$B$$ or point $$C$$ is after two moves. By symmetry, these probabilities must be the same, and their sum, together with the probability of arriving at point $$A$$ after two moves, must be $$1.$$ Hence, they must each have probability $$\frac{1}{4}$$.
Believe it or not, we’re already in a position to solve the original problem! The key is to think recursively. With this in mind, let $$p_{k}$$ denote the probability of returning back to where you started after $$2k$$ minutes. We already know that $$p_1 = \frac{1}{2}.$$ For $$k$$ greater than $$1$$, we want to relate $$p_k$$ to $$p_{k-1}$$. In other words, once you’ve made $$2k – 2$$ steps, what is the probability that after taking two more steps, you are back to where you started?
This depends on where you are after $$2k – 2$$ steps. So we can just break it down into all the cases! You’re at point $$A$$ with probability $$p_{k-1}$$, and from here, the probability of taking two more steps and getting back to point $$A$$ is simply $$\frac{1}{2}$$ as we computed before. You’re at either of points $$B$$ or $$C$$ with a sum total of probability $$1 – p_{k-1}$$, and from each of those cases, the probability of taking two more steps and getting back to point $$A$$ is $$\frac{1}{4}$$, also as calculated before. So we are finally in a position to write down our recursive equation:
$$p_k = \frac{1}{2}p_{k-1} + \frac{1}{4}(1 – p_{k-1})$$
Simplifying,
$$p_k = \frac{1}{4}(p_{k-1} + 1).$$
Thus, turning the crank a few times, starting with $$p_1 = \frac{1}{2}$$, we get $$p_2 = \frac{3}{8}, p_3 = \frac{11}{32}, p_4 = \frac{43}{128}, p_5 = \frac{171}{512},$$ and finally, $$p_6 = \frac{683}{2048}.$$ (Note that our final answer is very close to $$\frac{1}{3}$$, which makes intuitive sense.)
There are actually many other ways to approach the same problem (one could, for instance, create a transition matrix), but this one is very simple and clean. In fact, it’s not too hard from here to come up with an explicit formula for $$p_k$$ and prove that it is correct via induction. (A good exercise!)
# Infinite coin flipping
Suppose Alice and Bob decide to play the following simple game: they alternate turns flipping a coin, and whoever gets heads first immediately wins. (Alice gets the privilege of going first.) What are Alice’s chances of winning this game?
To make it more general, let’s suppose the coin is heads with probability $$x$$, where $$0 \le x \le 1$$. There are two approaches to solving this problem. Let’s look at the ‘brute force’ way first. Alice only gets to flip the coin on the first, third, fifth, seventh, … turns. Thus, her chances of winning the game must be equal to the infinite sum of her chances of winning on each of those respective turns. On the first turn, her chance of winning is $$x$$. In order to win on turn three, all previous flips must have been tails. Thus, her chances of winning in this case are $$(1-x)^{2} x$$. Similarly, on turn five, her chances of winning are $$(1-x)^{4} x$$, and this suggests that on turn $$2n + 1$$, her chances of winning are $$(1-x)^{2n} x$$. Thus, her chances of winning the game in total are
$$\displaystyle \sum_{n=0}^{\infty} (1-x)^{2n} x.$$
This is a geometric series with first term $$x$$ and ratio $$(1-x)^{2}$$, so it is equal to $$\frac{x}{1 – (1-x)^{2}} = \frac{x}{2x – x^{2}} = \frac{1}{2 – x}.$$ So for instance, if the coin is fair, $$x = \frac{1}{2}$$ and Alice’s chances of winning are $$\frac{1}{2 – \frac{1}{2}} = \frac{2}{3}.$$
While this approach certainly was successful, let’s look at a slicker alternative. Starting over, let’s denote the probability that Alice wins the game by $$P$$. Then of course the probability that Bob wins the game is $$1 – P$$. Instead of putting ourselves in Alice’s shoes, let’s pretend for a minute that we are Bob. Suppose Alice goes first and throws tails. Now, as Bob, we are pretty happy! It’s almost as though we get to go first now …
Well wait a minute, it is exactly as though we get to go first now! This observation actually tells us something: the original probability of Bob winning the game must be equal to the probability of Alice winning the game while first throwing tails. In equations, this means that $$(1 – P) = P(1 – x)$$. Now we can simply solve for $$P$$:
$$1 = P(1-x) + P$$
$$1 = P(2 – x)$$
So $$P = \frac{1}{2-x}$$, just as before. (Notice that our answer makes intuitive sense: as $$x \searrow 0$$, Alice’s probability of winning approaches $$\frac{1}{2}$$ from above, and as $$x \nearrow 1$$, Alice’s probability of winning approaches $$1$$ from below.)
This approach to probability is essentially recursive, and we’ll see more of it in the next post. | 1,546 | 5,726 | {"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} | 4.875 | 5 | CC-MAIN-2020-05 | longest | en | 0.943361 |
https://se.mathworks.com/help/physmod/hydro/ref/pressurereliefvalveil.html | 1,604,026,833,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107906872.85/warc/CC-MAIN-20201030003928-20201030033928-00678.warc.gz | 501,219,581 | 21,349 | # Pressure Relief Valve (IL)
Pressure-relief valve in an isothermal system
• Library:
• Simscape / Fluids / Isothermal Liquid / Valves & Orifices / Pressure Control Valves
## Description
The Pressure Relief Valve (IL) models a pressure-relief valve in an isothermal liquid network. The valve remains closed when the pressure is less than a specified value. When this pressure is met or surpassed, the valve opens. This set pressure is either a threshold pressure differential over the valve, between ports A and B, or between port A and atmospheric pressure. For pressure control based on another element in the fluid system, see the Pressure Compensator Valve (IL) block.
### Pressure Control
Two valve control options are available:
• When Set pressure control is set to `Controlled`, connect a pressure signal to port Ps and define the constant Pressure regulation range. The valve response will be triggered when Pcontrol, the pressure differential between ports A and B, is greater than Pset and below Pmax. Pmax is the sum of Pset and the pressure regulation range.
• When Set pressure control is set to `Constant`, the valve opening is continuously regulated between Pset and Pmax. There are two options for pressure regulation available in the Pressure control specification parameter: Pcontrol can be the pressure differential between ports A and B or the pressure differential between port A and atmospheric pressure. The opening area is then modeled by either linear or tabular parameterization. When the `Tabulated data` option is selected, Pset and Pmax are the first and last parameters of the Pressure differential vector, respectively.
### Mass Flow Rate Equation
Momentum is conserved through the valve:
`${\stackrel{˙}{m}}_{A}+{\stackrel{˙}{m}}_{B}=0.$`
The mass flow rate through the valve is calculated as:
`$\stackrel{˙}{m}=\frac{{C}_{d}{A}_{valve}\sqrt{2\overline{\rho }}}{\sqrt{P{R}_{loss}\left(1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\right)}}\frac{\Delta p}{{\left[\Delta {p}^{2}+\Delta {p}_{crit}^{2}\right]}^{1/4}},$`
where:
• Cd is the Discharge coefficient.
• Avalve is the instantaneous valve open area.
• Aport is the Cross-sectional area at ports A and B.
• $\overline{\rho }$ is the average fluid density.
• Δp is the valve pressure difference pApB.
The critical pressure difference, Δpcrit, is the pressure differential associated with the Critical Reynolds number, Recrit, the flow regime transition point between laminar and turbulent flow:
`$\Delta {p}_{crit}=\frac{\pi \overline{\rho }}{8{A}_{valve}}{\left(\frac{\nu {\mathrm{Re}}_{crit}}{{C}_{d}}\right)}^{2}.$`
Pressure loss describes the reduction of pressure in the valve due to a decrease in area. PRloss is calculated as:
`$P{R}_{loss}=\frac{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}-{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}+{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}.$`
Pressure recovery describes the positive pressure change in the valve due to an increase in area. If you do not wish to capture this increase in pressure, set the Pressure recovery to `Off`. In this case, PRloss is 1.
The opening area Avalve is determined by the opening parameterization (for `Constant` valves only) and the valve opening dynamics.
### Opening Parameterization
Linear parameterization of valve area is
`${A}_{valve}=\stackrel{^}{p}\left({A}_{\mathrm{max}}-{A}_{leak}\right)+{A}_{leak},$`
where the normalized pressure,$\stackrel{^}{p}$, is
`$\stackrel{^}{p}=\frac{{p}_{control}-{p}_{set}}{{p}_{\mathrm{max}}-{p}_{set}}.$`
For tabular parameterization of the valve area in its operating range, Aleak and Amax are the first and last parameters of the Opening area vector, respectively.
### Opening Dynamics
If Opening dynamics are modeled, a lag is introduced to the flow response to valve opening. Avalve becomes the dynamic opening area, Adyn; otherwise, Avalve is the steady-state opening area. The instantaneous change in dynamic opening area is calculated based on the Opening time constant, τ:
`${\stackrel{˙}{p}}_{dyn}=\frac{{p}_{control}-{p}_{dyn}}{\tau }.$`
By default, Opening dynamics are turned `Off`.
Steady-state dynamics are set by the same parameterization as valve opening, and are based on the control pressure, pcontrol.
## Ports
### Conserving
expand all
Entry or exit point to the valve.
Entry or exit point to the valve.
### Input
expand all
Varying-signal set pressure for controlled valve operation.
#### Dependencies
To enable this port, set Set pressure control to `Controlled`.
## Parameters
expand all
Valve operation method. A `Constant` valve opens linearly over a fixed pressure regulation range or in accordance with tabulated pressure and opening area data that you provide. A `Controlled` valve opens according to a variable set pressure signal at port Ps over a fixed pressure regulation range.
Pressure differential used for the valve control. Selecting `Pressure differential` sets the pressure difference between port A and port B as the trigger for pressure control. Selecting `Pressure at port A` sets the gauge pressure at port A, or the difference between the pressure at port A and atmospheric pressure, as the trigger for pressure control.
Gauge pressure beyond which valve operation is triggered when the Pressure control specification is with respect to port A.
#### Dependencies
To enable this parameter, set Set pressure control to `Constant` and Pressure control specification to `Pressure at port A`.
Pressure beyond which valve operation is triggered. This is the set pressure when the Pressure control specification is with respect to the pressure differential between ports A and B.
#### Dependencies
To enable this parameter, set Set pressure control to `Constant` and Pressure control specification to `Pressure differential`.
Operational pressure range of the valve. The pressure regulation range begins at the valve set pressure and the end of the range is the maximum valve operating pressure.
#### Dependencies
To enable this parameter, set
• Set pressure control to `Controlled`
• Set pressure control to `Constant` and Opening parameterization to `Linear`
Method of modeling valve opening or closing. The valve opening is either parametrized linearly or by a table of values that correlate area to pressure differential.
Cross-sectional area of the valve in its fully open position.
#### Dependencies
To enable this parameter, set either:
• Set pressure control to `Controlled`
• Set pressure control to `Constant` and Opening parameterization to `Linear`
Sum of all gaps when the valve is in fully closed position. Any area smaller than this value is maintained at the specified leakage area. This contributes to numerical stability by maintaining continuity in the flow.
#### Dependencies
To enable this parameter, set either:
• Set pressure control to `Controlled`.
• Set pressure control to `Constant` and Opening parameterization to `Linear`
Vector of pressure differential values for the tabular parameterization of the valve opening area. The vector elements must correspond one-to-one with the elements in the Opening area vector parameter. The elements are listed in ascending order and must be greater than 0. Linear interpolation is employed between table data points.
#### Dependencies
To enable this parameter, set Set pressure control to `Constant` and Opening parameterization to `Tabulated data`.
Vector of valve opening areas for the tabular parameterization of the valve opening area. The vector elements must correspond one-to-one with the elements in the Pressure differential vector parameter. The elements are listed in ascending order and must be greater than 0. Linear interpolation is employed between table data points.
#### Dependencies
To enable this parameter, set Set pressure control to `Constant` and Opening parameterization to `Tabulated data`.
Cross-sectional area at the entry and exit ports A and B. These areas are used in the pressure-flow rate equation determining mass flow rate through the valve.
Correction factor accounting for discharge losses in theoretical flows. The default discharge coefficient for a valve in Simscape™ Fluids™ is 0.64.
Upper Reynolds number limit for laminar flow through the valve.
Accounts for pressure increase when fluid flows from a region of smaller cross-sectional area to a region of larger cross-sectional area. This increase in pressure is not captured when Pressure recovery is set to `Off`.
Whether to account for transient effects to the fluid system due to valve opening. Setting Opening dynamics to `On` approximates the opening conditions by introducing a first-order lag in the flow response. The Opening time constant also impacts the modeled opening dynamics.
Initial cross-sectional area of the opening at the time of dynamic opening. This value is used to calculate the instantaneous opening area at the following time step.
#### Dependencies
To enable this parameter, set Opening dynamics to `On`.
Constant that captures the time required for the fluid to reach steady-state when opening or closing the valve from one position to another. This parameter impacts the modeled opening dynamics.
#### Dependencies
To enable this parameter, set Opening dynamics to `On`.
Introduced in R2020a
Get trial now | 2,068 | 9,455 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.840685 |
https://education.vex.com/stemlabs/iq/castle-crasher/lesson-4-creating-algorithms/compete | 1,723,331,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00182.warc.gz | 173,909,647 | 10,156 | Castle Crasher
Lesson 4: Creating Algorithms
# Compete
Now that you have learned how to code your robot to autonomously push cubes off of a raised Field, you are ready for the Sweep the Field Challenge.
The goal of this challenge is to clear four cubes from the Field in the fastest possible time. The animation below shows how the Field should be set up, and one possible path to clear the cubes from the Field. The robot that clears the cubes the fastest wins./p>
Follow the steps in this document to complete the Sweep the Field Challenge.
Video file
Once you have completed the Sweep the Field Challenge, check in with your teacher. Ensure you have documented the results of the challenge in your engineering notebook.
## Wrap Up Reflection
Now that you have created a strategy and competed in the Sweep the Field Challenge, it is time to reflect on what you have learned and done in this Lesson. Start a new page in your engineering notebook to begin your reflection.
Rate yourself as a novice, apprentice, or expert on each of the following concepts in your engineering notebook. Provide a brief explanation for why you gave yourself that rating for each concept:
• Using the sensors in a VEXcode IQ project
• Creating an algorithm using the Distance and Optical Sensors
• Collaborating with my team members to improve the robot's performance in the Sweep the Field Challenge
Use this table to help you determine which category you fall under.
Expert I feel that I fully understood the concept and could teach this to someone else. Apprentice I feel that I understood the concept enough to compete in the challenge. Novice I feel that I did not understand the concept and do not know how to complete the challenge.
## What is Next?
In this Lesson, you added the Optical Sensor to your robot, and coded an algorithm to clear cubes from the Field. Now it is time to compete in the Castle Crasher Competition!
In the next Lesson, you will:
• Go over the rules of the competition
• Develop a game strategy
• Compete in the Castle Crasher Competition!
Select < Return to Lessons to go back to the Lesson Overview.
Select Next Lesson > to continue to Lesson 5 and learn how to participate in the Castle Crasher Competition! | 456 | 2,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.932599 |
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